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[email protected] Visit www.mtg.in for buying books online. Other recommended books: JEE Main Mathematics (30 days) JEE Main Chemistry JEE Advanced JEE Main Physics (30 days) JEE Main Mathematics JEE Advanced Physics JEE Main Chemistry (30 days) JEE Main Explorer JEE Main Physics 12 + 12 JEE (Main + Advanced) JEE Advanced Chemistry JEE Advanced Mathematics PHYSICS 1. Physics and Measurement ........................................................................................................................................................ 1 2. Kinematics .................................................................................................................................................................................... 3 3. Laws of Motion ............................................................................................................................................................................ 9 4. Work, Energy and Power ........................................................................................................................................................ 14 5. Rotational Motion ...................................................................................................................................................................... 20 6. Gravitation .................................................................................................................................................................................. 26 7. Properties of Solids and Liquids ............................................................................................................................................ 30 8. Thermodynamics ....................................................................................................................................................................... 37 9. Kinetic Theory of Gases .......................................................................................................................................................... 42 10. Oscillations and Waves ........................................................................................................................................................... 44 11. Electrostatics .............................................................................................................................................................................. 54 12. Current Electricity...................................................................................................................................................................... 64 13. Magnetic Effects of Current and Magnetism ........................................................................................................................ 72 14. Electromagnetic Induction and Alternating Currents ........................................................................................................... 79 15. Electromagnetic W aves ........................................................................................................................................................... 86 16. Optics .......................................................................................................................................................................................... 88 17. Dual Nature of Matter and Radiation .................................................................................................................................... 94 18. Atoms and Nuclei ..................................................................................................................................................................... 99 19. Electronic Devices .................................................................................................................................................................. 106 20. Experimental Skills................................................................................................................................................................... 111 CHEMISTRY 1. Some Basic Concepts in Chemistry ........................................................................................................................................ 1 2. States of Matter ........................................................................................................................................................................... 3 3. Atomic Structure .......................................................................................................................................................................... 7 4. Chemical Bonding and Molecular Structure ..........................................................................................................................11 5. Chemical Thermodynamics ..................................................................................................................................................... 16 6. Solutions ..................................................................................................................................................................................... 22 7. Equilibrium .................................................................................................................................................................................. 28 8. Redox Reactions and Electrochemistry ................................................................................................................................ 35 9. Chemical Kinetics ..................................................................................................................................................................... 41 10. Surface Chemistry .................................................................................................................................................................... 45 11. Nuclear Chemistry .................................................................................................................................................................... 47 12. Classification of Elements and Periodicity in Properties ................................................................................................... 49 iv 13. General Principles and Processes of Isolation of Metals .................................................................................................. 53 14. Hydrogen .................................................................................................................................................................................... 55 15. sBlock Elements ...................................................................................................................................................................... 57 16. pBlock Elements ...................................................................................................................................................................... 59 17. d and fBlock Elements .......................................................................................................................................................... 65 18. Coordination Compounds ........................................................................................................................................................ 70 19. Environmental Chemistry ......................................................................................................................................................... 74 20. Purification and Characterisation of Organic Compounds ................................................................................................. 76 21. Some Basic Principles of Organic Chemistry ...................................................................................................................... 78 22. Hydrocarbons ............................................................................................................................................................................. 82 23. Organic Compounds Containing Halogens ........................................................................................................................... 87 24. Alcohols, Phenols and Ethers ................................................................................................................................................ 91 25. Aldehydes, Ketones and Carboxylic Acids ........................................................................................................................... 95 26. Organic Compounds Containing Nitrogen ............................................................................................................................ 99 27. Polymers ................................................................................................................................................................................... 101 28. Biomolecules ............................................................................................................................................................................ 103 29. Chemistry in Everyday Life ................................................................................................................................................... 107 30. Principles Related to Practical Chemistry .......................................................................................................................... 109 MATHEMATICS 1. Sets, Relations and Functions .................................................................................................................................................. 1 2. Complex Numbers ...................................................................................................................................................................... 7 3. Matrices and Determinants ..................................................................................................................................................... 12 4. Quadratic Equations ................................................................................................................................................................. 19 5. Permutations and Combinations ............................................................................................................................................. 24 6. Mathematical Induction and its Application .......................................................................................................................... 28 7. Binomial Theorem ..................................................................................................................................................................... 30 8. Sequences and Series ............................................................................................................................................................. 34 9. Differential Calculus .................................................................................................................................................................. 40 10. Integral Calculus ....................................................................................................................................................................... 52 11. Differential Equations ............................................................................................................................................................... 63 12. Two Dimensional Geometry .................................................................................................................................................... 67 13. Three Dimensional Geometry ................................................................................................................................................. 82 14. Vector Algebra ........................................................................................................................................................................... 90 15. Statistics ..................................................................................................................................................................................... 96 16. Probability ................................................................................................................................................................................. 100 17. Trigonometry ............................................................................................................................................................................ 105 18. Mathematical Logic .................................................................................................................................................................. 111 v « S Y L L A B U S PHYSICS SECTION A Unit 1: Physics and Measurement Physics, technology and society, S. I. units, fundamental and derived units, least count, accuracy and precision of measuring instruments, errors in measurement. Dimensions of physical quantities, dimensional analysis and its applications. Unit 2: Kinematics Frame of reference, motion in a straight line: positiontime graph, speed and velocity, uniform and nonuniform motion, average speed and instantaneous velocity. Uniformly accelerated motion, velocitytime, positiontime graphs, relations for uniformly accelerated motion. Scalars and vectors, vector addition and subtraction, zero vector, scalar and vector products, unit vector, resolution of a vector, relative velocity, motion in a plane, projectile motion, uniform circular motion. Unit 3: Laws of Motion Force and inertia, Newton’s first law of motion, momentum, Newton’s second law of motion, impulse, Newton’s third law of motion, law of conservation of linear momentum and its applications, equilibrium of concurrent forces. Static and kinetic friction, laws of friction, rolling friction. Dynamics of uniform circular motion: centripetal force and its applications. Unit 4: Work, Energy and Power Work done by a constant force and a variable force, kinetic and potential energies, workenergy theorem, power. Potential energy of a spring, conservation of mechanical energy, conservative and nonconservative forces, elastic and inelastic collisions in one and two dimensions. Unit 5: Rotational Motion Centre of mass of a twoparticle system, centre of mass of a rigid body, basic concepts of rotational motion, moment of a force, torque, angular momentum, conservation of angular momentum and its applications, moment of inertia, radius of gyration, values of moments of inertia for simple geometrical objects, parallel and perpendicular axes theorems and their applications. Rigid body rotation, equations of rotational motion. Unit 6: Gravitation The universal law of gravitation. Acceleration due to gravity and its variation with altitude and depth. Kepler’s laws of planetary motion. Gravitational potential energy, gravitational potential. Escape velocity, orbital velocity of a satellite, geostationary satellites. Unit 7: Properties of Solids and Liquids Elastic behaviour, stressstrain relationship, Hooke’s law, Young’s modulus, bulk modulus, modulus of rigidity. Pressure due to a fluid column, Pascal’s law and its applications. Viscosity, Stokes’s law, terminal velocity, streamline and turbulent flow, Reynolds number, Bernoulli’s principle and its applications. « For latest information refer prospectus 2014 vi Surface energy and surface tension, angle of contact, application of surface tension drops, bubbles and capillary rise. Heat, temperature, thermal expansion, specific heat capacity, calorimetry, change of state, latent heat. Heat transferconduction, convection and radiation, Newton’s law of cooling. Unit 8: Thermodynamics Thermal equilibrium, zeroth law of thermodynamics, concept of temperature, heat, work and internal energy, first law of thermodynamics. Second law of thermodynamics, reversible and irreversible processes, Carnot engine and its efficiency. Unit 9: Kinetic Theory of Gases Equation of state of a perfect gas, work done on compressing a gas. Kinetic theory of gases assumptions, concept of pressure, kinetic energy and temperature, rms speed of gas molecules, degrees of freedom, law of equipartition of energy, applications to specific heat capacities of gases, mean free path, Avogadro’s number. Unit 10: Oscillations and Waves Periodic motion period, frequency, displacement as a function of time, periodic functions, simple harmonic motion (S.H.M.) and its equation, phase, oscillations of a spring restoring force and force constant, energy in S.H.M. kinetic and potential energies, simple pendulum derivation of expression for its time period, free, forced and damped oscillations, resonance. Wave motion, longitudinal and transverse waves, speed of a wave, displacement relation for a progressive wave, principle of superposition of waves, reflection of waves, standing waves in strings and organ pipes, fundamental mode and harmonics, beats, Doppler effect in sound. Unit 11: Electrostatics Electric charges, conservation of charge, Coulomb’s lawforces between two point charges, forces between multiple charges, superposition principle and continuous charge distribution. Electric field, electric field due to a point charge, electric field lines, electric dipole, electric field due to a dipole, torque on a dipole in a uniform electric field. Electric flux, Gauss’s law and its applications to find field due to infinitely long, uniformly charged straight wire, uniformly charged infinite plane sheet and uniformly charged thin spherical shell. Electric potential and its calculation for a point charge, electric dipole and system of charges, equipotential surfaces, electrical potential energy of a system of two point charges in an electrostatic field. Conductors and insulators, dielectrics and electric polarization, capacitor, combination of capacitors in series and in parallel, capacitance of a parallel plate capacitor with and without dielectric medium between the plates, energy stored in a capacitor. Unit 12: Current Electricity Electric current, drift velocity, Ohm’s law, electrical resistance, resistances of different materials, VI characteristics of ohmic and nonohmic conductors, electrical energy and power, electrical resistivity, colour code for resistors, series and parallel combinations of resistors, temperature dependence of resistance. Electric cell and its internal resistance, potential difference and emf of a cell, combination of cells in series and in parallel. Kirchhoff’s laws and their applications, Wheatstone bridge, metre bridge. Potentiometer principle and its applications. Unit 13: Magnetic Effects of Current and Magnetism Biot Savart law and its application to current carrying circular loop. Ampere’s law and its applications to infinitely long current carrying straight wire and solenoid. Force on a moving charge in uniform magnetic and electric fields, cyclotron. Force on a currentcarrying conductor in a uniform magnetic field, force between two parallel currentcarrying conductorsdefinition of ampere, torque experienced by a current loop in uniform magnetic field, moving coil galvanometer, its current sensitivity and conversion to ammeter and voltmeter. vii Current loop as a magnetic dipole and its magnetic dipole moment, bar magnet as an equivalent solenoid, magnetic field lines, earth’s magnetic field and magnetic elements, para, dia and ferro magnetic substances . Magnetic susceptibility and permeability, hysteresis, electromagnets and permanent magnets. Unit 14: Electromagnetic Induction and Alternating Currents Electromagnetic induction, Faraday’s law, induced emf and current, Lenz’s law, Eddy currents, self and mutual inductance. Alternating currents, peak and rms value of alternating current/ voltage, reactance and impedance, LCR series circuit, resonance, quality factor, power in AC circuits, wattless current. AC generator and transformer. Unit 15: Electromagnetic Waves Electromagnetic waves and their characteristics, transverse nature of electromagnetic waves, electromagnetic spectrum (radio waves, microwaves, infrared, visible, ultraviolet, Xrays, gamma rays). Applications of electromagnetic waves. Unit 16: Optics Reflection and refraction of light at plane and spherical surfaces, mirror formula, total internal reflection and its applications, deviation and dispersion of light by a prism, lens formula, magnification, power of a lens, combination of thin lenses in contact, microscope and astronomical telescope (reflecting and refracting) and their magnifying powers. Wave optics wavefront and Huygens principle, laws of reflection and refraction using Huygens principle, interference, Young’s double slit experiment and expression for fringe width, coherent sources and sustained interference of light, diffraction due to a single slit, width of central maximum, resolving power of microscopes and astronomical telescopes, polarisation, plane polarized light, Brewster’s law, uses of plane polarized light and polaroids. Unit 17: Dual Nature of Matter and Radiation Dual nature of radiation, photoelectric effect, Hertz and Lenard’s observations, Einstein’s photoelectric equation, particle nature of light. Matter waveswave nature of particle, de Broglie relation, DavissonGermer experiment. Unit 18: Atoms and Nuclei Alphaparticle scattering experiment, Rutherford’s model of atom, Bohr model, energy levels, hydrogen spectrum. Composition and size of nucleus, atomic masses, isotopes, isobars, isotones, radioactivityalpha, beta and gamma particles/rays and their properties, radioactive decay law, massenergy relation, mass defect, binding energy per nucleon and its variation with mass number, nuclear fission and fusion. Unit 19: Electronic Devices Semiconductors, semiconductor diode IV characteristics in forward and reverse bias, diode as a rectifier, IV characteristics of LED, photodiode, solar cell. Zener diode, Zener diode as a voltage regulator, junction transistor, transistor action, characteristics of a transistor, transistor as an amplifier (common emitter configuration) and oscillator, logic gates (OR, AND, NOT, NAND and NOR), transistor as a switch. Unit 20: Communication Systems Propagation of electromagnetic waves in the atmosphere, sky and space wave propagation, need for modulation, amplitude and frequency modulation, bandwidth of signals, bandwidth of transmission medium, basic elements of a communication system (Block Diagram only). SECTION B Unit 21 : Experimental Skills Familiarity with the basic approach and observations of the experiments and activities: l Vernier callipersits use to measure internal and external diameter and depth of a vessel. l Screw gaugeits use to determine thickness/diameter of thin sheet/wire. l Simple Pendulumdissipation of energy by plotting a graph between square of amplitude and time. Coefficient of Viscosity of a given viscous liquid by measuring terminal velocity of a given spherical body. real gases. Surface tension of water by capillary rise and effect of detergents. molecule. Gas laws Boyle’s law. physical quantities and their measurements in chemistry. Potentiometer : (i) Comparison of emf of two primary cells. UNIT 2: STATES OF MATTER Classification of matter into solid. Capacitor from mixed collection of such items. element and compound. Focal length of: (i) Convex mirror (ii) Concave mirror and (iii) Convex lens using parallax method. units. Resistance of a given wire using Ohm’s law. chemical equations and stoichiometry. Characteristic curves of a pn junction diode in forward and reverse bias. van der Waals equation. Plotting a cooling curve for the relationship between the temperature of a hot body and time. Liquid State Properties of liquids vapour pressure. Solid State Classification of solids molecular. Using multimeter to: (i) Identify base of a transistor (ii) Distinguish between npn and pnp type transistor (iii) See the unidirectional flow of current in case of a diode and an LED. Resistance and figure of merit of a galvanometer by half deflection method. magnetic and dielectric properties. significant figures. Transistor. viscosity and surface tension and effect of temperature on them (qualitative treatment only). atomic and molecular masses. IC.viii l l l l l l l l l l l l l l l l l l l Metre Scale mass of a given object by principle of moments. concept of atom. Speed of sound in air at room temperature using a resonance tube. Ideal gas equation. S. deviation from Ideal behaviour. Gaseous State Measurable properties of gases. Bragg’s Law and its applications. concept of average. liquid and gaseous states. Avogadro’s law. unit cell and lattices. empirical and molecular formulae. Charle’s law. molar mass. imperfection in solids. Graham’s law of diffusion. dimensional analysis. Young’s modulus of elasticity of the material of a metallic wire. mole concept. Specific heat capacity of a given (i) solid and (ii) liquid by method of mixtures. transistor or IC). calculations involving unit cell parameters.I. Dalton’s atomic theory. packing in solids (fcc. compressibility factor. electrical. bcc and hcp lattices). voids. Resistor. . amorphous and crystalline solids (elementary idea). (iv) Check the correctness or otherwise of a given electronic component (diode. kinetic theory of gases (only postulates). precision and accuracy. (ii) Determination of internal resistance of a cell. Refractive index of a glass slab using a travelling microscope. Dalton’s law of partial pressure. ionic. Characteristic curves of a Zener diode and finding reverse break down voltage. CHEMISTRY SECTION A (Physical Chemistry) UNIT 1: SOME BASIC CONCEPTS IN CHEMISTRY Matter and its nature. LED. percentage composition. Identification of Diode. root mean square and most probable velocities. Resistivity of the material of a given wire using metre bridge. Laws of chemical combination. concept of absolute scale of temperature. covalent and metallic solids. Characteristic curves of a transistor and finding current gain and voltage gain. Plot of angle of deviation vs angle of incidence for a triangular prism. state functions. liquid gas and solid gas equilibria. ionization and solution. molarity. general characteristics of equilibrium involving physical processes. effect of catalyst. nature of electromagnetic radiation. quantum mechanical model of atom. molar heat capacity. Equilibria involving chemical processes Law of chemical equilibrium. molecular orbital electronic configurations of homonuclear diatomic molecules. dipole moment. UNIT 7: EQUILIBRIUM Meaning of equilibrium. Ionic equilibrium Weak and strong electrolytes. UNIT 4: CHEMICAL BONDING AND MOLECULAR STRUCTURE Kossel Lewis approach to chemical bond formation. spectrum of hydrogen atom. equilibrium constants (K p and K c) and their significance. hydration. types of processes. Bronsted Lowry and Lewis) and their ionization. concept of atomic orbitals as one electron wave functions. Resonance. vapour pressure of solutions and Raoult’s law Ideal and nonideal solutions. formation. Covalent Bonding concept of electronegativity. van’t Hoff factor and its significance. types of molecular orbitals (bonding. various concepts of acids and bases (Arrhenius. pressure. Hess’s law of constant heat summation. elementary ideas of quantum mechanics. Le Chatelier’s principle. bond length and bond energy. concept of bond order. hydrogen bonding and its applications. DG° (standard Gibbs energy change) and equilibrium constant. phase transition. extensive and intensive properties. atomization. shapes of s. Pauli’s exclusion principle and Hund’s rule. Ionic Bonding Formation of ionic bonds. dual nature of matter. angular momentum and magnetic quantum numbers) and their significance. calculation of lattice enthalpy. Fajan’s rule. concept of dynamic equilibrium. Henry’s law. extra stability of halffilled and completely filled orbitals. electron spin and spin quantum number. elevation of boiling point and osmotic pressure. antibonding). factors affecting the formation of ionic bonds. Quantum mechanical approach to covalent bonding valence bond theory its important features. electronic configuration of elements. depression of freezing point. heat internal energy and enthalpy. photoelectric effect. UNIT 6: SOLUTIONS Different methods for expressing concentration of solution molality. Heisenberg uncertainty principle. LCAOs. enthalpies of bond dissociation. Second law of thermodynamics Spontaneity of processes. deBroglie’s relationship. acid base equilibria (including multistage ionization) and ionization . derivation of the relations for energy of the electron and radii of the different orbits. variation of Y and Y 2 with r for 1s and 2s orbitals. factors affecting equilibrium concentration. rules for filling electrons in orbitals – Aufbau principle. concept of hybridization involving s. Molecular Orbital Theory its important features. First law of thermodynamics Concept of work. ionization of electrolytes.ix UNIT 3: ATOMIC STRUCTURE Thomson and Rutherford atomic models and their limitations. temperature. vapour pressure composition plots for ideal and nonideal solutions. concept of ionic and covalent bonds. sigma and pibonds. Bohr model of hydrogen atom its postulates. Valence Shell Electron Pair Repulsion (VSEPR) theory and shapes of simple molecules. mole fraction. DS of the universe and DG of the system as criteria for spontaneity. significance of DG and DG° in chemical equilibria. percentage (by volume and mass both). combustion. Equilibria involving physical processes Solid liquid. Elementary idea of metallic bonding. sublimation. various quantum numbers (principal. determination of molecular mass using colligative properties. heat capacity. abnormal value of molar mass. colligative properties of dilute solutions relative lowering of vapour pressure. limitations of Bohr’s model. p and d orbitals. UNIT 5: CHEMICAL THERMODYNAMICS Fundamentals of thermodynamics: system and surroundings. p and d orbitals. its important features. dialysis. ores. rules for assigning oxidation number. . periodic trends in properties of elementsatomic and ionic radii. Zn and Fe. Preparation and properties of some important compounds sodium carbonate. steps involved in the extraction of metals concentration. activation energy and its calculation. order and molecularity of reactions. oxidation states and chemical reactivity. p. reactions and uses of hydrogen peroxide. electron gain enthalpy. lyophobic. UNIT 14: s BLOCK ELEMENTS (ALKALI AND ALKALINE EARTH METALS) Group 1 and 2 Elements General introduction. electrode potentials including standard electrode potential. different types of electrodes. dry cell and lead accumulator. Cu. preparation and properties of colloids Tyndall effect. minerals. valence. activity and selectivity of solid catalysts. relationship between cell potential and Gibbs’ energy change. enzyme catalysis and its mechanism. Biological significance of Na. Brownian movement. structure. physical and chemical properties of water and heavy water. collision theory of bimolecular gaseous reactions (no derivation). conductance in electrolytic solutions. Mg and Ca. adsorption from solutions. buffer solutions. differential and integral forms of zero and first order reactions. balancing of redox reactions. UNIT 8 : REDOX REACTIONS AND ELECTROCHEMISTRY Electronic concepts of oxidation and reduction. anomalous properties of the first element of each group. their characteristics and half lives. s. ionization enthalpy. coagulation and flocculation. elementary and complex reactions. fuel cells. sodium hydroxide and sodium hydrogen carbonate. solubility of sparingly soluble salts and solubility products. isotopes. multi molecular. colloids and suspensions. common ion effect. oxidation number. UNIT 9 : CHEMICAL KINETICS Rate of a chemical reaction. properties and uses of hydrogen. pH scale. reduction (chemical and electrolytic methods) and refining with special reference to the extraction of Al. effect of temperature on rate of reactions Arrhenius theory. Catalysis Homogeneous and heterogeneous. Plaster of Paris and cement. emulsions and their characteristics. . Freundlich and Langmuir adsorption isotherms. Industrial uses of lime. Nernst equation and its applications. hydrogen as a fuel. temperature. UNIT 10 : SURFACE CHEMISTRY Adsorption Physisorption and chemisorption and their characteristics.x constants. factors affecting the rate of reactions concentration. diagonal relationships. emf of a galvanic cell and its measurement. electrophoresis. Electrolytic and metallic conduction. Electrochemical cells electrolytic and galvanic cells. Colloidal state distinction among true solutions. preparation. rate law. thermodynamic and electrochemical principles involved in the extraction of metals. rate constant and its units. d and f block elements. UNIT 12: GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF METALS Modes of occurrence of elements in nature. classification of hydrides ionic. preparation. pressure and catalyst. UNIT 15: p BLOCK ELEMENTS Group 13 to Group 18 Elements General Introduction Electronic configuration and general trends in physical and chemical properties of elements across the periods and down the groups. limestone. electronic configuration and general trends in physical and chemical properties of elements. redox reactions. K. ionization of water. Section B (Inorganic Chemistry) UNIT 11: CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES Modern periodic law and present form of the periodic table. UNIT 13: HYDROGEN Position of hydrogen in periodic table. half cell and cell reactions. unique behaviour of the first element in each group. macromolecular and associated colloids (micelles). hydrolysis of salts and pH of their solutions. molar conductivities and their variation with concentration: Kohlrausch’s law and its applications. covalent and interstitial. factors affecting adsorption of gases on solids. classification of colloids lyophilic. denticity. UNIT 17: COORDINATION COMPOUNDS Introduction to coordination compounds. their harmful effects and prevention. ligands. boron trifluoride. trends in the acidic nature of hydrogen halides. structures and uses of sulphuric acid (including its industrial preparation). fumes. phosphine and phosphorus halides. Werner’s theory. alloy formation. properties and uses of boron and aluminium. electronic configuration. Water Pollution Major pollutants such as. preparation. structures of fluorides and oxides of xenon. boric acid. ionization enthalpy. general trends in properties of the first row transition elements physical properties. dust. importance of coordination compounds (in qualitative analysis. hydrocarbons. magnetic properties. green house effect and global warming. properties. colour. IUPAC nomenclature of mononuclear coordination compounds. Stratospheric pollution Formation and breakdown of ozone. properties. their sources. Strategies to control environmental pollution. zeolites and silicones. complex formation. allotropic forms of sulphur. oxidation states and lanthanoid contraction. chelation.xi Group 13 Preparation. Particulate pollutants Smoke. Structures of oxoacids of sulphur. Soil Pollution Major pollutants like pesticides (insecticides. silicon tetrachloride. organic wastes and chemical pollutants. atomic radii. UNIT 16: d and f BLOCK ELEMENTS Transition Elements General introduction. structure. coordination number. bonding valence bond approach and basic ideas of crystal field theory. preparation. silicates. properties and uses of allotropes and oxides of carbon. depletion of ozone layer its mechanism and effects. PCl 5 ). water and soil. harmful effects and prevention. aluminium chloride and alums. structure. Group 14 Tendency for catenation. catalytic behaviour. nitrogen and sulphur. herbicides and fungicides). diborane. allotropic forms of phosphorus. structures and uses of ozone. their sources. Tropospheric pollutants Gaseous pollutants: oxides of carbon. properties and uses of hydrochloric acid. Group 16 Preparation. structure and uses of ammonia. (PCl 3 . Actinoids Electronic configuration and oxidation states. properties and uses of borax. acid rain. UNIT 18: ENVIRONMENTAL CHEMISTRY Environmental Pollution Atmospheric. properties and uses of K 2 Cr 2 O 7 and KMnO 4 . their harmful effects and prevention. properties. preparation. Inner Transition Elements Lanthanoids Electronic configuration. harmful effects and prevention. . Group 15 Properties and uses of nitrogen and phosphorus. mist. structures of interhalogen compounds and oxides and oxoacids of halogens. pathogens. smog. oxidation states. nitric acid. structures of oxides and oxoacids of nitrogen and phosphorus. isomerism. interstitial compounds. Atmospheric pollution tropospheric and stratospheric. Group 18 Occurrence and uses of noble gases. occurrence and characteristics. extraction of metals and in biological systems). Group 17 Preparation. colour and magnetic properties. UNIT 24: ORGANIC COMPOUNDS CONTAINING NITROGEN General methods of preparation. ozonolysis. Qualitative analysis Detection of nitrogen. Common types of organic reactions Substitution. classification. mechanisms of substitution reactions. Alkynes Acidic character. halogens. homologous series. Aromatic hydrocarbons Nomenclature. mechanism of dehydration. aldol condensation. electromeric effect. reactions and uses. Alkanes Conformations: Sawhorse and Newman projections (of ethane). freons and DDT. Aldehydes and Ketones Nature of carbonyl group. UNIT 20: SOME BASIC PRINCIPLES OF ORGANIC CHEMISTRY Tetravalency of carbon. halogens. numerical problems in organic quantitative analysis. reactions and uses. nitration. polymerization. haloform reaction. addition of hydrogen. Cannizzaro reaction. NH 3 and its derivatives). chemical tests to distinguish between aldehydes and ketones. elimination and rearrangement. electrophilic substitution reactions: halogenation. reduction (Wolff Kishner and Clemmensen). acidity of ahydrogen. properties. secondary and tertiary alcohols. Carboxylic acids Acidic strength and factors affecting it. Calculations of empirical formulae and molecular formulae. general methods of preparation. iodoform. nitrogen. relative reactivities of aldehydes and ketones. and polymerization. distillation. properties and reactions. Grignard reagent. Alkenes Geometrical isomerism. Alcohols Identification of primary. shapes of simple molecules hybridization (s and p). oxidation. . important reactions such as nucleophilic addition reactions (addition of HCN. water. sublimation. differential extraction and chromatography principles and their applications. hydrogen halides (Markownikoff’s and peroxide effect). oxidation. nature of CX bond. Electronic displacement in a covalent bond Inductive effect. sulphur. mechanism of halogenation of alkanes.xii SECTION C (Organic Chemistry) UNIT 19: PURIFICATION AND CHARACTERISATION OF ORGANIC COMPOUNDS Purification Crystallization. Reimer Tiemann reaction. addition. Diazonium Salts Importance in synthetic organic chemistry. Ethers Structure. stability of carbocations and free radicals. UNIT 22: ORGANIC COMPOUNDS CONTAINING HALOGENS General methods of preparation. phosphorus and halogens. UNIT 23: ORGANIC COMPOUNDS CONTAINING OXYGEN General methods of preparation. properties. UNIT 21: HYDROCARBONS Classification. Nomenclature (trivial and IUPAC) Covalent bond fission Homolytic and heterolytic: free radicals. secondary and tertiary amines and their basic character. sulphur. nitrogen and sulphur. properties and reactions. isomerism. resonance and hyperconjugation. Uses/environmental effects of chloroform. benzene structure and aromaticity. Isomerism structural and stereoisomerism. hydrogen. phosphorus. Amines Nomenclature. mechanism of electrophilic addition: addition of hydrogen. oxygen. carbocations and carbanions. IUPAC nomenclature. halogens. mechanism of electrophilic substitution: halogenation. nitration and sulphonation. directive influence of functional group in monosubstituted benzene. classification of organic compounds based on functional groups: and those containing halogens. Phenols Acidic nature. Quantitative analysis (Basic principles only) Estimation of carbon. water and hydrogen halides. nucleophilic addition to >C O group. electrophiles and nucleophiles. structure basic character and identification of primary. Friedel – Craft’s alkylation and acylation. Mohr’s salt vs KMnO 4 . pnitroacetanilide. oneone. nature of roots. peptide bond. some important polymers with emphasis on their monomers and uses polythene. potash alum. Chemistry involved in the titrimetric exercises Acids. enzymes. union. Carbohydrates Classification: aldoses and ketoses. AI 3+ . CI – . Enthalpy of solution of CuSO 4 2. polypeptides. disinfectants. antimicrobials. relations. Zn 2+ . proteins primary. antacids. functions. carbonyl (aldehyde and ketone). antiseptics. constituent monosaccharides of oligosaccharides (sucrose. halogens) in organic compounds. Chemistry involved in the preparation of the following: Inorganic compounds Mohr’s salt. square root of a complex number. Organic compounds Acetanilide. Cu 2+ . artificial sweetening agents common examples. natural and synthetic rubber and vulcanization. NH 4 + . Chemicals in food Preservatives. into and onto functions. SO 4 2– . Preparation of lyophilic and lyophobic sols. Ca 2+ . quadratic equations in real and complex number system and their solutions. intersection and complement of sets and their algebraic properties. Br – . denaturation of proteins. Vitamins Classification and functions. MATHEMATICS UNIT 1 : SETS. Fe 3+ . Chemical principles involved in the qualitative salt analysis: Cations Pb 2+ . Ni 2+ . lactose. maltose). types of relations. triangle inequality. I – (insoluble salts excluded). UNIT 26: BIOMOLECULES General introduction and importance of biomolecules. NO 3 – . Ba 2+ . cleansing action. iodoform. nylon. composition of functions. antihistamins their meaning and common examples. bases and the use of indicators. carboxyl and amino groups in organic compounds. Argand diagram. general methods of polymerization addition and condensation. S. NO 2 – . Cleansing agents Soaps and detergents. oxalic acid vs KMnO 4 . . representation of complex numbers in the form a + ib and their representation in a plane. Nucleic acids Chemical constitution of DNA and RNA. UNIT 27: CHEMISTRY IN EVERYDAY LIFE Chemicals in medicines Analgesics. UNIT 2 : COMPLEX NUMBERS AND QUADRATIC EQUATIONS Complex numbers as ordered pairs of reals. Enthalpy of neutralization of strong acid and strong base. copolymerization. power set. relation between roots and coefficients. monosaccharides (glucose and fructose). 4. antibiotics. UNIT 28: PRINCIPLES RELATED TO PRACTICAL CHEMISTRY Detection of extra elements (N. equivalence relations. polyester and bakelite. aniline yellow. modulus and argument (or amplitude) of a complex number. RELATIONS AND FUNCTIONS Sets and their representation. tertiary and quaternary structure (qualitative idea only). secondary. Anions – CO 3 2– . algebra of complex numbers. Kinetic study of reaction of iodide ion with hydrogen peroxide at room temperature. biological functions of nucleic acids.xiii UNIT 25: POLYMERS General introduction and classification of polymers. detection of the following functional groups: hydroxyl (alcoholic and phenolic). 3. antifertility drugs. S 2– . Mg 2+ . Chemical principles involved in the following experiments: 1. tranquilizers. formation of quadratic equations with given roots. Proteins Elementary Idea of a amino acids. . ò x2 ± a2 . Graphs of simple functions. å n . ò ax 2 + dx + c ò a 2 ± x 2 dx and ò x 2 . derivatives of order upto two.r) and C (n. simple applications. product and quotient of two functions. equation of family of lines passing through the point of intersection of two lines. insertion of arithmetic. Integration using trigonometric identities. permutation as an arrangement and combination as selection. ò a2 . area of triangles using determinants. Evaluation of definite integrals. determinants and matrices of order two and three.x 2 . algebra of matrices. locus and its equation. types of matrices. Rolle’s and Lagrange’s Mean value theorems. Properties of determinants. general term and middle term. Evaluation of simple integrals of the type ( px + q )dx ( px + q ) dx dx dx dx dx dx dx ò x 2 ± a2 . UNIT 5 : MATHEMATICAL INDUCTION Principle of Mathematical Induction and its simple applications. orthocentre and circumcentre of a triangle.r). solution of homogeneous and linear differential equations of the type: dy + p( x )y = q ( x ) dx UNIT 11: COORDINATE GEOMETRY Cartesian system of rectangular coordinates in a plane. UNIT 8 : LIMITS.M. . algebra of functions. Fundamental integrals involving algebraic. angles between two lines. Adjoint and evaluation of inverse of a square matrix using determinants and elementary transformations. Test of consistency and solution of simultaneous linear equations in two or three variables using determinants and matrices.x2 . Differentiation of trigonometric. ò ax2 + bx + c . intersection of lines. UNIT 4 : PERMUTATIONS AND COMBINATIONS Fundamental principle of counting. properties of Binomial coefficients and simple applications. and G. exponential and logarithmic functions. tangents and normals. rational. Sum upto n terms of special series: 2 3 å n.a 2 dx Integral as limit of a sum. equations of internal and external bisectors of angles between two lines. distance formula. Fundamental theorem of calculus. coordinates of centroid. å n . Properties of definite integrals. Limits. CONTINUITY AND DIFFERENTIABILITY Real valued functions. UNIT 10: DIFFERENTIAL EQUATIONS Ordinary differential equations. continuity and differentiability. determining areas of the regions bounded by simple curves in standard form. exponential. logarithmic and exponential functions. ò ax 2 + bx + c . Relation between A. translation of axes. UNIT 7 : SEQUENCES AND SERIES Arithmetic and Geometric progressions. composite and implicit functions. trigonometric. intercepts of a line on the coordinate axes. Meaning of P (n. ò a2 . logarithmic. their order and degree. evaluation of determinants. Solution of differential equations by the method of separation of variables. Straight lines Various forms of equations of a line. section formula. by parts and by partial fractions. Applications of derivatives: Rate of change of quantities. monotonic increasing and decreasing functions. Formation of differential equations. UNIT 6 : BINOMIAL THEOREM AND ITS SIMPLE APPLICATIONS Binomial theorem for a positive integral index. Arithmetic Geometric progression.M. Differentiation of the sum. ò ax2 + bx + c . inverse trigonometric. inverse functions.xiv UNIT 3 : MATRICES AND DETERMINANTS Matrices. Maxima and minima of functions of one variable. Integration by substitution. trigonometric. conditions for concurrence of three lines. slope of a line. polynomials. parallel and perpendicular lines. UNIT 9 : INTEGRAL CALCULUS Integral as an anti derivative. distance of a point from a line. difference. geometric means between two given numbers. UNIT 16: MATHEMATICAL REASONING Statements. ellipse and hyperbola) in standard forms. Baye’s theorem. Heights and Distances. equation of a circle when the end points of a diameter are given.xv Circles. UNIT 15: TRIGONOMETRY Trigonometrical identities and equations. intersection of a line and a plane. implied by. points of intersection of a line and a circle with the centre at the origin and condition for a line to be tangent to a circle. equation of the tangent. distance between two points. UNIT 14: STATISTICS AND PROBABILITY Measures of Dispersion Calculation of mean. Skew lines. equations of conic sections (parabola. Inverse trigonometrical functions and their properties. variance and mean deviation for grouped and ungrouped data. UNIT 12: THREE DIMENSIONAL GEOMETRY Coordinates of a point in space. logical operations and. converse and contrapositive. scalar and vector triple product. Understanding of tautology. components of a vector in two dimensions and three dimensional space. general form of the equation of a circle. probability distribution of a random variate. direction ratios and direction cosines. Sections of cones. section formula. median. condition for y = mx + c to be a tangent and point(s) of tangency. addition of vectors. Probability Probability of an event. its radius and centre. or. implies. Calculation of standard deviation. addition and multiplication theorems of probability. coplanar lines. angle between two intersecting lines. Trigonometrical functions. Bernoulli trials and Binomial distribution. if and only if. UNIT 13: VECTOR ALGEBRA Vectors and scalars. conic sections Standard form of equation of a circle. scalar and vector products. contradiction. mode of grouped and ungrouped data. Equations of a line and a plane in different forms. vvv . the shortest distance between them and its equation. xvi . PHYSICS . 7. (2004) 8. 0. 4. (2006) 1 .0003 and 2. (a) . which one does not have identical dimensions? (a) moment of inertia and moment of a force (b) work and torque (c) angular momentum and Planck’s constant (d) impulse and momentum (2005) 2. 1. (2010) The dimension of magnetic field in M. L. (d) (b) 3. The respective number of significant figures for the numbers 23. Let [Î 0] denote the dimensional formula of the permittivity of vacuum. 2 (c) 5. are (a) [L –1 T] (b) [L –2 T 2 ] (c) [L 2 T –2 ] (d) [LT –1 ]. (a) torque and work (b) stress and energy (c) force and stress (d) force and work. 2. (2002) Answer Key 1.1 × 10 –3 are (a) 4. The physical quantities not having same dimensions are (a) torque and work (b) momentum and Planck's constant (c) stress and Young's modulus (d) speed and (m 0e 0 ) –1/2 . Dimensions of 3. (2003) 10. 8.1 Physics and Measurement PHYSICS AND MEASUREMENT CHAPTER 1 1. T = time and A = electric current. 5. Identify the pair whose dimensions are equal. (a) 5. where symbols have their usual m 0e 0 meaning. (c) 6. Which one of the following represents the correct dimensions of the coefficient of viscosity? (a) ML –1 T –2 (b) MLT –1 –1 –1 (c) ML T (d) ML –2 T –2 . Resistance of a given wire is obtained by measuring the current flowing in it and the voltage difference applied across it. 4. Out of the following pairs. (c) (c) 2. T and C (coulomb) is given as (a) MT –2 C –1 (b) MLT –1 C –1 2 –2 (c) MT C (d) MT –1 C –1 . 1. If the percentage errors in the measurement of the current and the voltage difference are 3% each. If M = mass. L = length. (d) 10. where Q denotes the electric charge? (a) weber (Wb) (b) Wb/m 2 (c) henry (H) (d) H/m 2 . 9. then (a) [Î 0] = [M –1 L 2 T –1 A] (b) [Î 0 ] = [M –1 L –3 T 2 A] (c) [Î 0] = [M –1 L –3 T 4 A 2 ] (d) [Î 0] = [M –1 L 2 T –1 A –2 ] (2013) 6. (2008) Which of the following units denotes the dimensions ML 2 /Q 2 . then error in the value of resistance of the wire is (a) zero (b) 1% (c) 3% (d) 6% (2012) 7. 5 (d) 5. 5.023. 2 (b) 5. (b) (c) 4. (2003) 9. if at all. (a) : Moment of inertia (I) = mr 2 \ [I] = [ML 2 ] Moment of force (C) = r F \ [C] = [r][F] = [L][MLT –2 ] or [C] = [ML 2 T –2 ] Moment of inertia and moment of a force do not have identical dimensions. or [L][LT -1 ] 8. (c) : Viscous force F = 6phrv [ F ] F h= \ or [h] = [ r ][ v ] 6 p rv [MLT -2 ] [h] = or [h] = [ML –1 T –1 ].023 has 5 significant figures. 9. (b) : [Momentum] = [MLT –1 ] [Planck's constant] = [ML 2 T –1 ] Momentum and Planck's constant do not have same dimensions. 7. 2 [AT][AT] = [M -1L-3T 4 A 2 ] [MLT -2 ][L]2 2. (c) : According to Coulomb’s law F = é H ù = [MT -2 A -2 ] êë m 2 úû 1 q1q 2 1 q1q 2 \ Î0 = 4 p Î 0 r 2 4 p Fr 2 [Î0 ] = 2 Obviously henry (H) has dimensions ML .0003 has 1 significant figures. (c) : Velocity of light in vacuum = D R = D V + D I R V I The percentage error in R is DR ´ 100 = DV ´ 100 + DI ´ 100 = 3% + 3% = 6% R V I 3. (iv)The power of 10 is irrelevant to the determination of significant figures. 23.2 JEE MAIN CHAPTERWISE EXPLORER [henry] = [ML 2 T –2 A –2 ] 1. the zero(s) on the right of decimal point but to the left of the first nonzero digit are not significant. 2. (d) : R = V \ I Q 6. no matter where the decimal point is. 10. . (d) : Lorentz force =| F |=| qv ´ B | \ 5. (a) : Torque and work have the same dimensions.1 × 10 –3 has 2 significant figures. (b) : (i) All the nonzero digits are significant. r r r 4. According to the above rules. 0. (ii) All the zeros between two nonzero digits are significant. [ B ] = -2 -2 [ F ] = MLT = MLT = [MT –1 C –1 ] [ q ][v ] C ´ LT -1 C LT -1 (c) : [ML 2 Q –2 ] = [ML 2 A –2 [Wb] = [ML 2 T –2 A –1 ] é Wb ù –2 –1 ê 2 ú = [M T A ] ë m û T –2 ] or é 1 ù [LT -1 ] = ê ú ú ëê m 0 e0 û or é ù [L2 T -2 ] = ê 1 ú m e ë 0 0 û \ Dimensions of 1 = [L2 T -2 ] m 0 e 0 1 m 0 e 0 . (iii)If the number is less than 1. 2 2 ^ ^ (b) .2 m s –2 20 m P (x. Which of the following graphs correctly describes (x 1 – x 2 ) as a function of time t? (x1 – x 2 ) (x1 – x 2 ) x ^ (d) .5 v . Its speed after 10 s is (a) 10 units (b) 7 2 units (c) 7 units (d) 8. The position of the first body is given by x 1 (t) after time t and that of the second body by x 2 (t) after the same time interval. If the speed of water coming out of the fountain is v.3 $ j . 3. where v is the instantaneous dt 5. j and k are unit vectors along x. 2. where s is in metres and t is in seconds. to come to rest. A point P moves in counterclockwise direction on a circular path as shown in the figure. The general equation for its path is (a) y 2 = x 2 + constant (b) y = x 2 + constant 2 (c) y = x + constant (d) xy = constant (2010) (b) (a) O t O t (x1 – x 2 ) (x1 – x 2 ) (c) (d) O t O t (2008) . it starts moving in the positive xdirection with a constant acceleration.25 m s –1 . dv = - 2. At the same instant another body passes through x = 0 moving in the positive xdirection with a constant speed. would be (a) 1 s (b) 2 s (c) 4 s (d) 8 s (2011) 4. the total area around the fountain that gets wet is v 4 p v 4 v 2 v 2 (a) p g (b) p g 2 (c) 2 g 2 (d) p g 2 (2011) A small particle of mass m is projected at an angle q with the xaxis with an initial velocity v 0 in the xy plane as shown in v 0 sin q the figure. y and z‑axis respectively. (c) 12 m s –2 (d) 7. The maximum horizontal distance that the boy can throw the same stone up to will be (d) 20 2 m (a) 10 m (b) 10 2 m (c) 20 m (2012) 7. At a time t < g . the acceleration a at a point P(R. q) on the circle of radius R is (Here q is measured from the xaxis) ^ ^ r A particle is moving with velocity v = K ( y i + x j ). A water fountain on the ground sprinkles water all around it. 2 ^ 2 ^ (a) v i + v j R R 8. the angular momentum of y the particle is ^ (a) 1 mgv0 t 2 cos q i 2 (b) . where A projectile is given an initial velocity of ($ $ i is along the ground and $ j is along the vertical.5 units (2009) 10.v cos q i + v sin q j R R 2 2 2 2 ^ ^ ^ ^ (c) . is decelerated at a rate given by r For a particle in uniform circular motion. The time taken by the object. (2010) (c) mgv0 t cos q k^ 6. y ) x O (2010) A particle has an initial velocity 3$ i + 4 $ j and an acceleration of 0. If g = 10 m/s 2 . The radius of the path is 20 m. A body is at rest at x = 0. i + 2 $ j )m s.v sin q j R R R R (2010) An object moving with a speed of 6.1 mg v0 t 2 cos q k 2 ^ ^ ^ where i . the equation of its trajectory is (a) 4y = 2x – 25x 2 (b) y = x – 5x 2 2 (c) y = 2x – 5x (d) 4y = 2x – 5x 2 (2013) A boy can throw a stone up to a maximum height of 10 m. where K is a constant.mgv0 t 2 cos q ^ j v 0 q y B (b) 13 m s –2 9.v sin q i + v cos q j (d) . At t = 0. The movement of P is such that it sweeps out a length s = t 3 + 5. The acceleration of P when t = 2 s is nearly (a) 14 m s –2 speed.4$ i + 0.v cos q i .3 Kinematics CHAPTER KINEMATICS 2 1. 4 JEE MAIN CHAPTERWISE EXPLORER 11. i. A ball is thrown from a point with a speed v 0 at an angle of projection q. Which of the following statements is false for a particle moving in a circle with a constant angular speed? (a) The velocity vector is tangent to the circle. A projectile can have the same range R for two angles of projection. Will the person be able to catch the ball? If yes. A projectile can have the same range R for two angles of projection. then its displacement after unit time (t = 1) is (a) v 0 + g/2 + f (b) v 0 + 2g + 3f (c) v 0 + g/2 + f/3 (d) v 0 + g + f (2007) 12.66 m. (2004) 13. the stopping distance will be (a) 20 m (b) 40 m (c) 60 m (d) 80 m. The relation between time t and distance x is t = ax 2 + bx where a and b are constants. He reaches the ground with a speed of 3 m/s. (2004) 19.20 m (c) 2. The acceleration is (a) –2av 3 (b) 2av 2 (c) –2av 2 (d) 2bv 3 (2005) 16. A car. A car moving with a speed of 50 km/hr. What is the position of the ball in T/3 second? 25. (2004) (d) 182 m (2005) 21. 120 km/h. If the same car is moving at a speed of 100 km/hr. (b) The acceleration vector is tangent to the circle (c) the acceleration vector points to the centre of the circle (d) the velocity and acceleration vectors are perpendicular to each other. The coordinates of a moving particle at any time t are given by x = at 3 and y = bt 3 . The displacement of the particle varies with time as (a) t 3 (b) t 2 (c) t (d) t 1/2 . (2005) 18. (2006) (a) h/9 metre from the ground (b) 7h/9 metre from the ground (c) 8h/9 metre from the ground (2004) (d) 17h/18 metre from the ground. If t 1 and t 2 be the time of flights in the two cases. (2003) 26. (2003) . (2003) ms –2 towards northwest ms –2 towards northeast 1 ms –2 towards north 2 23. 60° (b) yes. did he bail out? (a) 293 m (b) 111 m (c) 91 m 15. If its position is x = 0 at t = 0. When parachute opens. then 1 2 1 2 (b) s = f t (a) s = f t 2 4 1 2 (2005) (c) s = ft (d) s = f t 6 22. The average acceleration in this time is (a) zero (b) (c) (d) 1 2 1 2 24. then the product of the two time of flights is proportional to (a) 1/R (b) R (c) R 2 (d) 1/R 2 . A particle is moving eastwards with a velocity of 5 m/s. It takes T second to reach the ground. then the product of the two time of flights is directly proportional to (a) 1/R 2 (b) 1/R (c) R (d) R 2 . (2004) (2005) 17. If the car is going twice as fast. 45°. sin30° = 1/2. How far from the throwing point will the ball be at the height of 10 m from the ground ? [g = 10 m/s 2 . A parachutist after bailing out falls 50 m without friction. what should be the angle of projection? (a) yes. If T 1 and T 2 be the time of flights in the two cases.33 m (d) 8. (2004) 14. cos30° = (a) 5. r r r r 20. accelerates at the rate f through a distance s. The speed of the particle at time t is given by (a) 3t a 2 + b 2 (b) 3t 2 a 2 + b 2 (c) t 2 a 2 + b 2 (d) a 2 + b 2 . then the angle between A and B is (a) p (b) p/3 (c) p/2 (d) p/4. If the total distance traversed is 15 s. A boy playing on the roof of a 10 m high building throws a ball with a speed of 10 m/s at an angle of 30° with the horizontal.e. the minimum stopping distance is (a) 12 m (b) 18 m (c) 24 m (d) 6 m.60 m 3 / 2 ] (b) 4. then continues at constant speed for time t and then decelerates at the rate f/2 to come to rest. In 10 s the velocity changes to 5 m/s northwards. The velocity of a particle is v = v 0 + gt + ft 2 . A ball is released from the top of a tower of height h metre. At what height. 30° (c) no (d) yes. From the same point and at the same instant a person starts running with a constant speed v 0 /2 to catch the ball. it decelerates at 2 m/s 2 . A particle located at x = 0 at time t = 0. starting from rest. can brake to stop within a distance of 20 m. starts moving along the positive xdirection with a velocity v that varies as v = a x . can be stopped by brakes after at least 6 m. If A ´ B = B ´ A . An automobile travelling with a speed of 60 km/h. 10. 22. 18. (c) (a) (*) (a) (b) 3. If a body looses half of its velocity on penetrating 3 cm in a wooden block. the ratio of the respective distances in which the two cars are stopped from that instant is (a) 1 : 1 (b) 1 : 4 (c) 1 : 8 29. 15. 11. (c) (d) (a) (c) (d) 2. From a building two balls A and B are thrown such that A is thrown upwards and B downwards (both vertically). 9. 20. then (a) v B > v A (b) v A = v B (c) v A > v B (d) their velocities depend on their masses. 26. (2002) 4. 23. 25. 29. then how much will it penetrate more before coming to rest? (a) 1 cm (b) 2 cm (c) 3 cm (d) 4 cm. 27. 5 N (c) 10 N. Two forces are such that the sum of their magnitudes is 18 N and their resultant is 12 N which is perpendicular to the smaller force. 8. 14. 21. (a) (b) (d) (c) (b) . Speeds of two identical cars are u and 4u at a specific instant. 7. 28. Then the magnitudes of the forces are (a) 12 N. 19. 12. 13. (d) (c) (b) (a) (a) 6. (b) (c) (b) (b) (d) 5. 16. (2002) 30. If the same deceleration is applied on both the cars. (b) (b) (a) (c) (b) (d) 1 : 16. (2002) Answer Key 1. 30. (2002) 28. 24. If v A and v B are their respective velocities on reaching the ground. 6 N (b) 13 N.5 Kinematics 27. 17. 2 N. 8 N (d) 16 N. we get dx = Ky; dy = Kx dt dt dy dy dt \ = ´ dx dt dx For a particle in uniform circular motion.6 JEE MAIN CHAPTERWISE EXPLORER 1. ^ ^ r ^ ^ v2 v 2 a = .sin q j R R 8.25 = - (2.a cos q i ... v = dt r d ^ ^ 1 v= (v cos qt i + (v0 sin qt . The maximum height a boy can throw a stone is u 2 H max = = 10 m . (d) : The position vector of the particle from the origin at any time t is r ^ ^ 1 r = v0 cos qt i + (v0 sin qt .mgv0 t cos q k ë 2 û 2 r ^ ^ 6. 2 (12) 2 144 a c = v = = = 7. a = (ac ) 2 + (at )2 = (7. q) q a asinq q x 4.a sin q j = .25 m s ò ydy = ò xdx y 2 = x 2 + constant y t 0 2 ´ [v1/ 2 ] 6.5 v 2 sin 90 ° v 2 = g g 4 p v 2 Area = p ( R max ) = 2 g 5. (b) : or dt v On integrating.2)2 + (12) 2 » 14 m/s 2 9.gt 2 ) j ) dt 0 2 ^ ^ = v0 cos q i + (v0 sin q - gt ) j The angular momentum of the particle about the origin is r r r L = r ´ mv r r r L = m (r ´ v ) 7.v0 gt cos q k ú = ..5 v dv = - 2. (d) : O acosq P( R . (a) : Here. (b) : v = u + at r v = (3$i + 4 $j ) + (0.5) t Þ t = -2 ´ (6..4$i + 0.(i) 2 g The maximum horizontal distance the boy can throw the same stone is u 2 R max = = 20 m (Using (i)) g 1 dv = - 2.3 $ j ) ´ 10 . within limit (v 1 = 6.v0 gt 2 cos q) k ^ 1 + (v02 sin q cos qt .v02 sin q cos qt k + v0 gt 2 cos q k ú û 2 ^ ù ^ é 1 1 2 2 = m ê ..v0 gt 2 cos q k ^ ^ ù 1 .2 m/s 2 R 20 20 Net acceleration. a = v towards the centre R From figure.(iii) Integrating both sides of the above equation. v = K ( y i + x j ) r ^ ^ v = Ky i + Kx j r dx ^ dy ^ Q v = i+ j dt dt Equating equations (i) and (ii).(i) .5 ò dt . (c) : Let u be the velocity of projection of the stone.gt 2 2 \ x = t 1 ´ 10 ´ t 2 = 2t - 5 t 2 2 Equation of trajectory is y = 2x – 5x 2 and y = 2t - 2.1 0 ^ ^ é = m ëv02 sin q cos qt k . (a) : s = t 3 + 3 \ v = ds = d (t 3 + 3) = 3 t 2 dt dt dv = d (3t 2 ) = 6 t Tangential acceleration. (b) : R max = ^ ^ 1 = m éê(v0 cos qt i + (v0 sin qt . v = 3(2) 2 = 12 m/s..25) 1/2 = 2 s - 2.5 dt 3..gt ) j ) ùû ^ é = m ë(v02 cos q sin qt .25 m s –1 to v 2 = 0) \ v 2 = 0 -1/ 2 òv dy Kx x = = dx Ky y v 1 = 6. a t = 6(2) = 12 m/s 2 Centripetal acceleration.. we get dv = -2.(ii) .gt 2 v0 cos q)( .k ) ùú û 2 .gt 2 ) j 2 r r dr \ Velocity vector. 2 Acceleration.cos q i .gt 2 ) j ) ë 2 ^ ^ ´ (v0 cos q i + (v0 sin q . at = dt dt At t = 2 s. (c) : Given : u = $ i + 2 $ j r As u = u x $i + u y $ j \ u x = 1 and u y = 2 Also x = u xt 1 and y = u y t . . (ii) r 5 ˆj . For the second it is proportional to t. v 2 = constant for the other particle.. Curve (c) satisfies this.. . u 2 = 2as 2 2 .t.. Therefore it is a parabola after crossing xaxis again.. 10.. t1t 2 = 18. 14. (i) .. (iv) s 1 + s 2 + s 3 = 15s or s + f t 1 t + 2s = 15s or f t 1 t = 12s From (i) and (v). (c) : Given : velocity v = v 0 + gt + ft 2 \ v = x dx dt 0 t or x = ò ( v0 + gt + ft 2 ) dt 0 gt 2 ft 3 + + C 2 3 where C is the constant of integration Given : x = 0.75 = 292..75 m \ Total height = 50 + 242.. v 1 = at. (a) : Initially. 2 is velocity of 1 – velocity of 2. the parachutist falls under gravity \ u 2 = 2ah = 2 × 9. Initially both are zero. (c) : As u = 0.. Relative velocity of particle 1 w. (b) : Velocity in eastward direction = 5iˆ velocity in northward direction = 5 ˆ j g f \ x = v0 + + ..75 = 293 m.7 Kinematics r v = (3 + 4)$ i + (4 + 3) $ j r Þ | v | = 49 + 49 = 98 = 7 2 units (This value is about 9... a = –2 ms –2 2 2 \ (3) = u – 2 × 2 × h 1 or 9 = 980 – 4 h 1 971 or h1 = 4 or h 1 = 242..q) 2 u sin q t1 = and t 2 = \ g g or \ \ 4 u 2 sin q cos q 2 æ u 2 sin 2 q ö 2 R = ´ç ÷ = g g è g g 2 ø t 1 t 2 is proportional to R. (*) : For first part of journey. (b) : v = a x dx dx = a dt or dt = a x or x dx = a ò dt or 2 x1 / 2 = a t or ò x j 5 ˆ a 45° W 2 æ a ö 2 or x = ç 2 ÷ t è ø or displacement is proportional to t 2 . At first the velocity of first particle is less than that of 2. - 5iˆ + 5 iˆ ... \ C = 0 x = v0 t + or x = v0 t + gt 2 ft 3 + 2 3 At t = 1 sec 2 ft 1 s = 12 s 2 ´ ft1 t t or t1 = 6 or s = 2 ft 2 1 2 1 æ t ö ft1 = f ç ÷ = 2 2 è 6 ø 72 ft 2 72 None of the given options provide this answer. t or ò dx = ò vdt 0 1 æ f ö 4 f t 1 2 (2t1 ) 2 or s3 = 1 ´ ç ÷ 2 è 2 ø 2 2 or s 3 = 2s 1 = 2s..8 × 50 = 980 m 2 s –2 He reaches the ground with speed = 3 m/s. If (a) is given that is also not wrong).. s3 = \ s2 æ 120 ö 2 s 2 æ u2 ö ç u ÷ = s Þ ç 60 ÷ = 20 è ø 1 è 1ø or s 2 = 80 m. (v) 15..... 11.9 units close to 10 units. 2 17... s 2 = vt or s 2 = f t 1 t For the third part of journey.÷ö 2 2 è 2 ø è 2 ø 2 r 1 | a |= ms -2 towards northwest........ s = s 1 . (d) : Let a be the retardation for both the vehicles For automobile. Then the distance travelled by particle 1 increases as x 1 = (1/2) at 1 2 .... v 2 = u 2 – 2 as \ u 1 2 – 2as 1 = 0 Þ u 1 2 = 2as 1 2 Similarly for car.. t = 0... ....... 1 s1 = f t1 2 = s 2 v = f t 1 For second part of journey. (b) : Range is same for angles of projection q and (90 – q) 2u sin (90 . (a) : t = ax 2 + bx Differentiate the equation with respect to t dx dx \ 1 = 2 ax + b dt dt dx or 1 = 2 axv + bv as = v dt 1 v = or 2 ax + b dv = -2a ( dx / dt ) = -2 av ´ v 2 or dt (2ax + b) 2 or Acceleration = –2av 3 ...r.... (iii) E \ 2 13.5 i ˆ Acceleration a = 10 or r r 1 1 1 1 a = ˆj . or s = 16.... 2 3 N 12..iˆ or | a | = çæ ÷ö + çæ . . (a) : The person will catch the ball if his speed and horizontal speed of the ball are same v 1 = v0 cos q = 0 Þ cos q = = cos60 ° \ q = 60°... 25.. The given statement is false. It completes its journey to ground under gravity.. u1 = 50 hour 60 ´ 60 9 sec 2 2 u \ Acceleration a = . (i) s = 0 + . (b) : Resultant R is perpendicular to smaller force Q and (P + Q) = 18 N \ P 2 = Q 2 + R 2 by right angled triangle Q 90° or or R (P 2 – Q 2 ) = R 2 (P + Q)(P – Q) = R 2 P or (18)(P – Q) = (12) 2 or (P – Q) = 8 Hence P = 13 N and Q = 5 N.. (i) 2 u 0 = æç ö÷ – 2 ax è 2 ø or u 2 = 8ax From (i) and (ii) 8ax = 8a Þ x = 1 cm. \ q = p... (d) : Height of building = 10 m The ball projected from the roof of building will be back to roof height of 10 m after covering the maximum horizontal range..1 = .. .. (ii) 21...(i) Ball B starts with downwards velocity u and reaches ground after travelling a vertical distance h \ v B 2 = u 2 + 2gh . p.. 2p. s 1 = 2 a (4 u ) 2 16 u 2 = For second car. (b) : The acceleration vector acts along the radius of the circle. 9 9 r r r r 20. (d) : Both are given the same deceleration simultaneously and both finally stop.... \ v A 2 = u 2 + 2gh .. (ii) 30... or 10 26..866 or R = 8.. 23. Formula relevant to motion : u 2 = 2 as 2 u For first car. (c) : For first case.. 2 æ u ö = u 2 – 2 a (3) ç ÷ è 2 ø or 3u 2 = 24a Þ u 2 = 8a For latter part of penetration.. \ 2 gT 2 1 æ T ö ´gç ÷ = 2 18 è 3 ø or 18 s = gT 2 From (i) and (ii)... (b) : Ball A projected upwards with velocity u falls back to building top with velocity u downwards.. (a) : For first part of penetration.. 9 h 8 h \ Height from ground = h .. 18 s = 2h h or s = m from top.æç 125 ö÷ ´ 1 = -16 m/sec 2 2 s1 è 9 ø 2 ´ 6 For second case. 2 Maximum horizontal range ( R ) = u sin 2 q g (10)2 ´ sin 60 ° R= = 10 ´ 0...(ii) From (i) and (ii) v A = v B . . [Q P + Q = 18] .... (c) : Equation of motion : s = ut + 1 2 gt 2 1 h = 0 + gT 2 2 or 2h = gT 2 After T/3 sec. (c) : Range is same for angles of projection q and (90º – q) 2u sin (90º -q) T1 = 2u sin q and T 2 = \ g g \ \ 4 u 2 sin q cos q 2 æ u 2 sin 2 q ö 2 R = ´ç ÷ = g g è g g 2 ø T 1 T 2 is proportional to R. by equation of motion.....u -1 æ 250 ö æ 1 ö s 2 = 2 = ´ = 24 m \ 2a 2 çè 9 ÷ø çè 16 ÷ø or s 2 = 24 m.... s 2 = 2a 2 a s 1 1 = \ s2 16 .8 JEE MAIN CHAPTERWISE EXPLORER 19...= m.. 28.. (b) : Q x = at 3 dx = 3at 2 Þ v x = 3 at 2 \ dt Again y = bt 3 dy Þ v y = 3 b t 2 \ v 2 = v x 2 + v y 2 \ dt or v 2 = (3at 2 ) 2 + (3bt 2 ) 2 = (3t 2 ) 2 (a 2 + b 2 ) or v = 3t 2 a2 + b 2 . (a) : A ´ B = B ´ A AB sin q nˆ = AB sin( -q ) nˆ or or sinq = – sinq or 2 sinq = 0 or q = 0.. \ 29. T1T 2 = 22. 2 2 km 50 ´ 1000 125 m = = 24. 27.......66 m.... u2 = 100 km = 100 ´1000 = 250 m hour 60 ´ 60 9 sec 2 2 . .... 4 kg.5] (a) 100 m (b) 400 m (c) 800 m (d) 1000 m (2005) 30° 4. the force of the blow exerted by the ball on the hand of the player is equal to (a) 300 N (b) 150 N (c) 3 N (d) 30 N. What is the relative vertical acceleration of A with respect to B? A A player caught a cricket ball of mass 150 g moving at a rate of 20 m/s.4 N s (c) 0. Find the force of the block of mass m. B 60° (d) 1. Then a constant force F starts acting on the block of mass M to pull it. Two blocks A and B are placed on the two planes. (2010) The figure shows the position time (xt) graph of one dimensional motion of a body of mass 0. The ratio of their centripetal acceleration is (a) m 1 : m 2 (b) r 1 : r 2 (c) 1 : 1 (d) m 1 r 1 : m 2 r 2 (2012) A particle of mass m is at rest at the origin at time t = 0.2 m which applying the force and the ball goes upto 2 m height further.9 ms –2 in vertical direction (a) (b) 4. The magnitude of its momentum is recorded as (a) 17. Then a is equal to (a) g (b) g tana (c) g/tana (d) g coseca mg sin a a (2005) . The blocks are kept on a smooth horizontal plane. (2006) 8. MF mF (a) ( m + M ) (b) M mF ( M + m ) F (c) (2007) (d) ( m + M ) m A ball of mass 0.2 N s (b) 0. Their speeds are such that they make complete circles in the same time t. Two cars of masses m 1 and m 2 are moving in circles of radii r 1 and r 2 . Initially the blocks are at rest and the spring is unstretched. If the hand moves 0. 2.9 Laws of Motion CHAPTER LAWS OF MOTION 3 1.8 ms –2 in vertical direction (d) zero 4. A block is kept on a frictionless inclined surface with angle of ma inclination a.565 kg ms (d) 17. It is subjected to a force F(t) = F o e –bt in the x direction.57 kg ms –1 (b) 17.2 kg is thrown vertically upwards by applying a force by hand. (2006) 9. Consider a car moving on a straight road with a speed of 100 m/s.9 ms –2 in horizontal direction (c) 9.00 ms –1 .8 N s ma cosa 10. A body of mass m = 3. The distance at which car can be stopped is [m k = 0. respectively.6 kg ms –1 –1 (c) 17.513 kg is moving along the xaxis with a speed of 5. 6. The magnitude of each impulse is 2 x (m) 0 2 4 8 6 t(s) 10 12 14 16 A block of mass m is connected to another block of mass M by a spring (massless) of spring constant k. find the magnitude of the force. The incline is given an acceleration a to keep the block stationary. Two fixed frictionless inclined planes making an angle 30° and 60° with the vertical are shown in the figure. v(t ) F o mb 7.56 kg ms –1 . Its speed v(t) is depicted by which of the following curves? (b) v( t ) t t (c) (d) v(t ) F o mb v(t ) t t (2012) 3.6 N s (2010) 5. If the catching process is completed in 0.1 s. Consider g = 10 m/s 2 (a) 22 N (b) 4 N (c) 16 N (d) 20 N. (2008) F o mb Fo b m (a) (a) 0. An annular ring with inner and outer radii R 1 and R 2 is rolling without slipping with a uniform angular speed. (2003) 22. (2003) 21.0 (c) 1. (2003) 23. If a force P is applied at the free end of the rope.06 (d) 0.0 cm (d) 2.0 cm (2005) 13. A block of mass M is pulled along a horizontal frictionless surface by a rope of mass m. The coefficient of friction is 1 1 (b) m s = 1 - 2 (a) m s = 1 - 2 n n (c) m k = 1 - 1 n 2 (d) m k = 1 - 1 n 2 (2005) 15. A body starting from rest at the top will again come to rest at the bottom if the coefficient of friction for the lower half is given by (a) 2tanf (b) tanf (c) 2sinf (d) 2cosf (2005) 14.8 kg tied to a string are hanging over a light frictionless pulley. Then the true statement about the scale reading is (a) both the scales read M kg each (b) the scale of the lower one reads M kg and of the upper one zero (c) the reading of the two scales can be anything but the sum of the reading will be M kg (d) both the scales read M/2 kg. A rocket with a liftoff mass 3.10 JEE MAIN CHAPTERWISE EXPLORER 11. A light spring balance hangs from the hook of the other light spring balance and a block of mass M kg hangs from the former one.2. the mass of the block (in kg) is (take g = 10 m/s 2 ) (a) 2.8 m/s 2 (c) 5 m/s 2 (d) 4. A bullet fired into a fixed target loses half its velocity after penetrating 3 cm.01. What will be its initial acceleration if it is released from a point 20 cm away from the origin? (a) 5 m/s 2 (b) 10 m/s 2 (c) 3 m/s 2 (d) 15 m/s 2 (2005) 12. Then the initial thrust of the blast is (a) 3.5 (2004) 17. How many bullets can he fire per second at the most? (a) one (b) four (c) two (d) three. A marble block of mass 2 kg lying on ice when given a velocity of 6 m/s is stopped by friction in 10 s. A horizontal force of 10 N is necessary to just hold a block stationary against a wall.0 (b) 4. (2004) 19. The coefficient of static friction between the block and the plane is 0. The time taken to slide is n times as much to slide on rough incline than on a smooth incline.75 × 10 5 N. 18. The ratio of the forces experienced by the two particles situated on the inner and outer parts of the ring. A particle of mass 0.02 (b) 0. F 1 /F 2 is 2 æ R 1 ö R 1 R 2 (a) 1 (b) (c) R (d) ç ÷ R 2 1 è R2 ø (2005) 16. The upper half of an inclined plane with inclination f is perfectly smooth while the lower half is rough. Then the coefficient of friction is (a) 0. A block rests on a rough inclined plane making an angle of 30° with the horizontal.5 × 10 5 N (b) 7. A spring balance is attached to the ceiling of a lift.3 kg is subjected to a force F = – kx with k = 15 N/m. (2003) m 1 m 2 (2004) 24.0 × 10 5 N 5 (c) 14. How much further it will penetrate before coming to rest assuming that it faces constant resistance to motion? (a) 1.8 m/s 2 .8.5 × 10 4 kg is blasted upwards with an initial acceleration of 10 m/s 2 . A machine gun fires a bullet of mass 40 g with a velocity 1200 ms –1 .6 (d) 2. (2003) 20.8 m/s 2 ) (a) 0. What is the acceleration of the masses when lift free to move? (g = 9. when the lift is stationary. The weight of the block is (a) 20 N (b) 50 N (c) 100 N (d) 2 N. A smooth block is released at rest on a 45 o incline and then slides a distance d. If the frictional force on the block is 10 N. the reading of the spring balance will be (a) 24 N (b) 74 N (c) 15 N (d) 49 N.0 × 10 N (d) 1. If the lift moves downward with an acceleration of 5 m/s 2 .0 cm (c) 3.03 (c) 0. the force exerted by the rope on the block is Pm Pm (a) M + m (b) M - m (c) P PM (d) M + m . The man holding it can exert a maximum force of 144 N on the gun.2 m/s 2 (b) 9. A man hangs his bag on the spring and the spring reads 49 N. Two masses m 1 = 5 kg and m 2 = 4. (2003) . The coefficient of friction between 10 N the block and the wall is 0.5 cm (b) 1. F 3 are acting on a particle of mass m such that F 2 and F 3 are mutually perpendicular. A light string passing over a smooth light pulley connects two blocks of masses m 1 and m 2 (vertically). then the ratio of the masses is (a) 8 : 1 (b) 9 : 7 (c) 4 : 3 (d) 5 : 3. 17. 11. These forces are represented in magnitude and direction by the three sides of a triangle ABC (as shown). 29. 7. (d) (b) (d) (a) (c) . 10. (2003) 26. The particle will now move with velocity C r (a) less than v r (b) greater than v r (c) | v | in the direction of the largest A B force BC r (d) v . If the acceleration of the system is g/8. The acceleration of the ball as observed by the man in the lift and a man standing stationary on the ground are respectively (a) g. g.6 ms –2 on a frictionless surface. (2002) 29. g – a (c) g – a. One end of a massless rope. remaining unchanged. 13. 21. 16. A man in the lift drops a ball inside the lift. 24. 14. 28. 25. 18. 27. 8. A lift is moving down with acceleration a. (c) (b) (a) (c) (b) 5. g (d) a. 30. 23. (b) (d) (a) (a) (d) (b) 2. The minimum velocity (in ms –1 ) with which a car driver must traverse a flat curve of radius 150 m and coefficient of friction 0. 22. 15.8 (2002) 27. which passes over a massless and frictionless pulley P is tied to a hook C while the other end is free.2 N with an acceleration of 0. then the particle remains stationary. (a) (b) (a) (d) (a) 6.8 B A (c) 4 F (d) 9. P C (a) 16 (b) 6 (c) 4 (d) 8.6 to avoid skidding is (a) 60 (b) 30 (c) 15 (d) 25. 31. g (b) g – a.2 (b) 7. (b) (d) (c) (a) (b) 3. (2002) Answer Key 1.11 Laws of Motion With what value of maximum safe acceleration (in ms –2 ) can a man of 60 kg climb on the rope? 25. When forces F 1. 26. (2002) 31. Maximum tension that the rope can bear is 960 N. (2002) 30. 20. F 2. (a) (d) (b) (d) (b) 4. Three forces start acting simultaneously on a particle moving r with velocity v . 12. Three identical blocks of masses m = 2 kg are drawn by a force F = 10. If the force F 1 is now removed then the acceleration of the particle is (a) F 1/m (b) F 2 F 3 /mF 1 (c) (F 2 – F 3)/m (d) F 2/m. 19. (2002) 28. then what is the tension (in N) in the string between the blocks B and C? C (a) 9. 9. 9 m s -2 in vertical direction. \ 8. mF . From graph.513 kg ; v = 5.. m + M (d) : Work done by hand = Potential energy of the ball .2 fx è 2 ø u 2 = 8fx From (i) and (ii) 3 × (8 fx) = 24 f x = 1 cm. (b) : The incline is given an acceleration a. (b): F(t) = F o e –bt (Given) ma = F o e –bt F a = o e . 17.4 N s Final momentum. component of g down the incline = gsinf . .2 (d) : Force × time = Impulse = Change of momentum \ 9.57. (a) : Momentum is mv.bt t t F m + M Force = Impulse 3 = = 30 N .. Pseudo acceleration a acts on block to the left... (b) : For first part of penetration. (b) : F = – kx F = -15 ´ æç 20 ö÷ = -3 N è 100 ø Initial acceleration is over come by retarding force. u = = 1 m s -1 (2 - 0) (0 .1 (d) : Retardation due to friction = mg v 2 = u 2 + 2as Q \ 0 = (100) 2 – 2(mg)s or 2 mgs = 100 × 100 or s= 100 ´ 100 = 1000 m ..bt m F dv F o . where q is the angle of inclination \ The vertical component of acceleration a is a (along vertical) = (gsinq)sinq = gsin 2q For block A a A(along vertical) = gsin 2 60° For block B a B(along vertical) = gsin 2 30° The relative vertical acceleration of A with respect to B is a AB(along vertical) = a A(along vertical) – a B(along vertical) = gsin 2 60° – gsin 2 30° æ 3 2 1 2 ö ÷ = g çè 2 2 ø g = = 4.565 = 17.1 m s -1 (4 - 2) Initial momentum.57 m s –1 Because the values will be accurate up to second decimel place only.0) Initial velocity. 2 ´ 0.. p i = mu = 0.8 N s FS = mgh Þ F = ( ) ( ) mgh 0.4) N s = – 0. (c) : Here. Because of impulse direction of velocity changes as can be seen from the slopes of the graph.3 or 12.2 f (3) ç 2 ÷ è ø or 3u 2 = 24 f For latter part of penetration.5 ´ 10 10. 2 æ u ö = ( u )2 . or m 0. Acceleration of the block is to the right. . m = 0.. (b): Centripetal acceleration.4 N s Impulse = Change in momentum = p f – p i = – 0. v = = .00 m/s \ mv = 17. mass of the body. (ii) 13. s 0.. 11.. time 0.4 × 1 = 0.4 × (–1) = – 0..bt ] m ë - b û 0 mb 0 0 3. or m × (acceleration a) = 3 3 3 a = = = 10 ms -2 . (i) 2 or or u 0 = æç ö÷ . (a) : For upper half smooth incline.12 JEE MAIN CHAPTERWISE EXPLORER 1.4 – (0. by equation of motion. we get v 5. \ macos a = mgsina or a = gtana.e . (d) : Acceleration of the system a = Force on block of mass m = ma = 7.bt F F ò dv = ò m e dt Þ v = o éê e ùú = o [1 . 6.bt dt dt m m Integrating both sides.2 ´ 10 ´ 2 = = 20 N. (2 . Equate resolved parts of a and g along incline.8 N s |Impulse| = 0. (a) : The acceleration of the body down the smooth inclined plane is a = gsinq It is along the inclined plane.2) Final velocity. p f = mv = 0. m = 3.bt = e or dv = o e .. a c = w 2 r 2 p Q w = T As T 1 = T 2 Þ w 1 = w 2 Q a c 1 r 1 = ac 2 r2 2.4 kg Since positiontime (xt) graph is a straight line. F o . so motion is uniform. 2 4. 2 × 10 = 2 N. 9. Frictional retardation up the plane = m k (gcosq) R 20. r Hence the particle velocity remains unchanged at v .2 9.m 2 ) (5 . Consequently the acceleration will be zero. v = mRg \ v = 0.8 9.8 = = = \ W1 mg 9. or g g 10 or d= P ( M + m ) Force on block a = 4. (c) : Component of g down the plane = gsinq \ For smooth plane. (a) : g = ( m + m ) = (5 + 4.2 17. 24. acceleration = g.m 2 ) 27.m k g cos q)( nt ) 2 2 \ Force applied Total mass MP = Mass of block × a = ( M + m ) . (d) : Acceleration of block ( a ) = l l + 2g(sinf – m k cosf) 2 2 or 19.8 ´ 0..5 4.6) or T BC = 7. frictional f retardation = m k gcosf \ Resultant acceleration = gsinf – m kgcosf l 2 \ 0 = v + 2 (gsinf – m k gcosf) 2 f g g cosf 0 = 2(gsinf) or or or 0 = sinf + sinf – m k cosf m k cosf = 2sinf m k = 2tanf.m ) k 2 2 or 22. (d) : Suppose he can fire n bullets per second \ Force = Change in momentum per second 144 = n ´ æç 40 ö÷ ´ (1200) è 1000 ø 144 ´ 1000 n= or 40 ´ 1200 or n = 3..2 or a = g´ = = 0.. 29. a ( m1 .5 × 10 5 N.06 . 16. (a) : When lift is standing.. 21. W 1 = mg When the lift descends with acceleration a.8 N. 1 ( g sin q . f = mgsinq \ 10 = m × 10 × sin30° or m = 2 kg. 14.2 – (2 × 2 × 0. W 2 = m (g – a) W 2 m ( g . (b) : g = ( m + m ) 1 2 m 1 9 1 ( m1 . m 30.8 25.8 49 ´ 4.mk g cos q ) n 2t 2 2 2 or sinq = n 2 (sinq – m k cosq) Putting q = 45° or sin45° = n 2 (sin45° – m kcos45°) or 1 .a ) 9.8) = 9. (i) 2 For rough plane. (c) : Frictional force provides the retarding force \ m mg = ma a u / t 6 /10 m= = = = 0.8 18. 23. (b) : Centripetal force on particle = mRw 2 m k = 1 - \ F1 mR1w2 R 1 = = F2 mR2 w 2 R 2 . Obviously the resultant force on the particle will be zero. (a) : F 2 and F 3 have a resultant equivalent to F 1 F \ Acceleration = 1 .8 = = 24 N.5 × 10 4 ) × (10) = 3. 31.8 1 2 0. (a) : Both the scales read M kg each.8 . or \ g sin q mk g cos q q q g g cosq \ 1 1 ( g sin q)t 2 = ( g sin q .6 ´ 150 ´ 10 = 30 m .13 Laws of Motion \ v 2 = 2(gsinf) m k g cos f R l 2 g sin f For lower half rough incline. (d) : By triangle of forces.. n 2 15.8) 0.8 W2 = W1 ´ 26. 9. (b) : Q Force = mass × acceleration \ F – T AB = ma and T AB – T BC = ma \ T BC = F – 2 ma or T BC = 10.8 9. s .4. (d) : Weight of the block is balanced by force of friction \ Weight of the block = mR = 0. (b) : For no skidding along curved track. (a) : Initial thrust = (Lift off mass) × acceleration = (3. (c) : For observer in the lift.8 9. (b) : T – 60g = 60a or 960 – (60 × 10) = 60a or 60a = 360 or a = 6 ms –2 . 1 = n 2 (1 .. R mgsin q q f q sq mg mg co a ( m1 .m 2 ) = = . (a) : For equilibrium of block. acceleration = (g – a) For observer standing outside. \ 8 ( m1 + m2 ) or m2 7 28.2 ms -2 . the particle will be in equilibrium under the three forces.. 1 d = ( g sin q ) t 2 . choose the one that best describes the two statements. Statement1 : A point particle of mass m moving with speed v collides with stationary point particle of mass M. The energy loss during the collision is (a) 0.000 N/m. StatementII is false. The spring compresses by (a) 8. Then the velocity as a function of time and the height as function of time will be This question has StatementI and StatementII. will be more than that on S 2 . StatementII is a correct explanation of StatementI. Statement 2 is true. StatementII is true. (a) Statement1 is true. WORK. Statement2 : Principle of conservation of momentum holds 7. A block of mass 0. (c) StatementI is true. Statement2 is true. Statement 2 is not the correct explanation of Statement 1. and compresses it till the block is motionless.000 J (b) 200 J 500 J (c) 2 × 10 5 J 3 × 10 5 J (d) 20. respectively.34 J (b) 0. (b) Statement1 is true.67 J. It strikes another mass of 1.00 J (d) 0. His kinetic energy can be estimated to be in the range (a) 2. (c) Statement 1 is true. Of the four choices given after the Statements. choose the one that best describes the two Statements. ENERGY AND POWER collision is negligible and the collision with the plate is totally elastic. (c) Statement1 is true. (2008) Statement1 : Two particles moving in the same direction do not lose all their energy in a completely inelastic collision. StatementII is true.000 J 5.5 cm (c) 2. 3. Statement 2 : k 1 < k 2 . (2012) y h (a) t t O v y +v 1 h t (b) O t –v 1 v y +v 1 (c) h t O t 1 2t 3t 1 4t 1 1 –v 1 t y h (d) t 1 2t 1 3t 1 4t 1 t (2009) t 5. Statement 2 is true.00 ms –1 on a smooth surface. StatementII is true. Statement 1 : If stretched by the same amount. Statement2 is true; Statement2 is 8. not the correct explanation of Statement1.50 kg is moving with a speed of 2. Statement 2 is the correct explanation of Statement 1. (d) Statement 1 is false.16 J (c) 1. Statement2 is true; Statement2 is the correct explanation of Statement1.5 cm (d) 11. (2013) This question has Statement 1 and Statement 2. work done on S 1 . 2. Statement 2 is true. Assume that the duration of v v 1 A particle is projected at 60° to the horizontal with a kinetic energy K. (2008) 6. Statement 2 is false. The kinetic friction force is 15 N and spring constant is 10. 4. Statement2 is false. (d) StatementI is true. are stretched by the same force. An athlete in the olympic games covers a distance of 100 m in 10 s.14 JEE MAIN CHAPTERWISE EXPLORER CHAPTER 4 1. Of the four choices given after the statements.0 cm (2007) . The kinetic energy at the highest point is (a) K/2 (b) K (c) zero (d) K/4 (2007) A 2 kg block slides on a horizontal floor with a speed of 4 m/s. it is found that more work is done on spring S 1 than on spring S 2. (a) StatementI is false. (2010) Consider a rubber ball freely falling from a height h = 4. (b) Statement 1 is true. StatementII is not a correct explanation of statementI. true for all kinds of collisions. (a) Statement 1 is true.000 J 50.5 cm (b) 5.00 kg and then they move together as a single body. If the maximum energy loss possible is given as æ 1 ö æ m ö h è mv 2 ø then h = è 2 M + m ø Statement2 : Maximum energy loss occurs when the particles get stuck together as a result of the collision. (d) Statement1 is false. It strikes a uncompressed spring. (b) StatementI is true. If two springs S 1 and S 2 of force constants k 1 and k 2 .9 m onto a horizontal elastic plate.000 J. A force F = (5iˆ + 3 ˆ j + 2kˆ ) N is applied over a particle which r displaces it from its origin to the point r = (2iˆ . the maximum speed (in m/s) is (a) 2 (b) 3/ 2 (c) (d) 1/ 2 . (2004) 21. The velocity of the 12 kg mass is 4 ms –1 . The kinetic energy of the other mass is (a) 96 J (b) 144 J (c) 288 J (d) 192 J. (2006) 12. The (c) work done on the particle in joule is (a) –7 (b) +7 (c) +10 (d) +13. (2003) .x ÷ J .5 J (c) –1. (2006) 11.25 Nm.00 Nm (d) 6. accelerates uniformly from rest to v 1 in time t 1 . (2004) 22. A particle of mass 100 g is thrown vertically upwards with a speed of 5 m/s. The maximum momentum of the block after collision is (a) zero (b) ML 2 K (c) MK L (d) KL 2 2 M (2005) 16. A particle moves in a straight line with retardation proportional to its displacement.15 Work. It follows that (a) its velocity is constant (b) its acceleration is constant (c) its kinetic energy is constant (d) it moves in a straight line.75 Nm (c) 25. A mass m moves with a velocity v and collides v / 3 inelastically with another after collision identical mass. The potential energy of a 1 kg particle free to move along the xaxis is given by 2 ö æ 4 V ( x ) = ç x . 2 ø è 4 The total mechanical energy of the particle 2 J. A bomb of mass 16 kg at rest explodes into two pieces of masses of 4 kg and 12 kg. The instantaneous power delivered to the body as a function of time t is (a) mv1t t 1 (b) mv1 2t t 1 2 mv1 t 2 mv1 2t . The velocity attained by the ball is (a) 10 m/s (b) 34 m/s (c) 40 m/s (d) 20 m/s (2005) 17. The block of mass M moving on the frictionless horizontal surface collides M with the spring of spring constant K and compresses it by length L. the motion of the particle takes place in a plane.50 Nm (b) 18. (2004) 20. A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle.5 J (b) –0. The instantaneous power delivered to the body as a function of time is given by 1 mv 2 1 mv 2 2 t t (b) (a) 2 T 2 2 T 2 mv 2 mv 2 2 × t × t (c) (d) (2005) 2 T T 2 14. then climbs up another hill of height 30 m and finally rolls down to a horizontal base at a height of 20 m above the ground. A spring of spring constant 5 × 10 3 N/m is stretched initially by 5 cm from the unstretched position. Energy and Power 9. (2004) 19. The work done by the force of gravity during the time the particle goes up is (a) 0. A body is moved along a straight line by a machine delivering a constant power. The distance moved by the body in time t is proportional to (a) t 3/4 (b) t 3/2 (c) t 1/4 (d) t 1/2 . (2004) (d) t 1 t 1 r 18. A spherical ball of mass 20 kg is stationary at the top of a hill of height 100 m. A mass of M kg is suspended by a weightless string. A body of mass m is accelerated uniformly from rest to a speed v in a time T. A body of mass m. (2006) 2 10. 2 (2006) 13.25 J (d) 1. (2003) 23. Find the speed of the 2 nd mass after collision 2 v v (b) (c) v (d) 3v (a) 3 3 (2005) 15. A uniform chain of length 2 m is kept on a table such that a length of 60 cm hangs freely from the edge of the table. Then the work required to stretch it further by another 5 cm is (a) 12.25 J. It rolls down a smooth surface to the ground. What is the work done in pulling the entire chain on the table? (a) 7.ˆ j ) m . Its loss of kinetic energy for any displacement x is proportional to (a) x 2 (b) e x (c) x (d) log e x. The horizontal force that is required to displace it until the string making an angle of 45º with the initial vertical direction is (a) Mg ( 2 - 1) (b) Mg ( 2 + 1) (c) Mg 2 (d) Mg .2 J (b) 3. After collision the first mass m moves with velocity in a m direction perpendicular to before collision the initial direction of motion.6 J (c) 120 J (d) 1200 J. The total mass of the chain is 4 kg. Then. Linear momentum of a system of particles is zero. (2002) 26. 26. 25. 14. A ball whose kinetic energy is E. Then (a) A does not imply B and B does not imply A (b) A implies B but B does not imply A (c) A does not imply B but B implies A (d) A implies B and B implies A. Kinetic energy of a system of particles is zero. 8. (2003) 25. (c) (c) (b) (b) 5. (d) (c) (b) (b) 6. when water is cooled to form ice. (a) (a) (b) (c) . 9.16 JEE MAIN CHAPTERWISE EXPLORER 24. If massenergy equivalence is taken into account. A. (a) (d) (c) (b) (b) 2. 10. 20. 7. 16. 11. B. 27. is projected at an angle of 45° to the horizontal. The work done in extending it from 5 cm to 15 cm is (a) 16 J (b) 8 J (c) 32 J (d) 24 J. the mass of water should (a) increase (b) remain unchanged (c) decrease (d) first increase then decrease. (d) (b) (a) (c) (c) 3. Consider the following two statements. 21. 23. 19. 17. (2002) Answer Key 1. The kinetic energy of the ball at the highest point of its flight will be (a) E (b) E / 2 (c) E/2 (d) zero. 24. 18. A spring of force constant 800 N/m has an extension of 5 cm. 13. (b) (b) (c) (a) (c) 4. 22. (2002) 27. 15. 12. = 1 ´ 1.00 (1. 2 Final K.17 Work. 2 . It is fixed. (b) : Total energy E T = 2 J. The change from +v to –v is almost instantaneous. decreases. Assuming an athelete has about 50 to 100 kg. v increases. Energy and Power 1.50 × (2. kinetic energy is maximum The potential energy should therefore be minimum. 2 av v v Hence. his kinetic Maximum energy loss = 1 mM u 2 2 ( M + m ) \ f = mM ( M + m) energy would have been 1 2 mv . 1. the value of v becomes +ve. .(i) t (Using (i)) For 100 kg. F = k 1 x 1 = k 2 x 2 Work done on spring S 1 is ( k x ) 2 F 2 1 W1 = k1x 1 2 = 1 1 = 2 2 k1 2 k1 Work done on spring S 2 is 2 (k x ) 2 W2 = 1 k 2 x 2 2 = 2 2 = F 2 2k 2 2 k2 W1 k 2 \ = W2 k1 As W 1 > W 2 \ k 2 > k 1 or k 1 < k 2 Statement 2 is true..(ii) 8.50) v. V ( x ) = .50) 6.00 2 = 1.1 J 4 (Kinetic energy) max = E T – V min or 1 9 (Kinetic energy) max = 2 – æç . Statement1 is false. comes to zero and increases.. v = gt when coming down. For the same extension. (c) : v = u + gt.E.5 cm. (1/2) × 100 × 100 = 5000 J.33 2 3. 2.50 × 2. V(x) = 0 or At \ x = ± 1.ö÷ = J è 4 ø 4 or 1 m v 2 = 9 m 2 4 or v m 2 = 9 ´ 2 = 9 ´ 2 = 9 m ´ 4 1´ 4 2 \ vm = 3 m/s. Therefore (c). Therefore kinetic energy of a particle at highest point is K K H = Kcos 2q = Kcos 2 60° = 4 .E. 9. For maximum speed. (b) : Let the spring be compressed by x Initial kinetic energy of the mass = potential energy of the spring + work done due to friction 1 1 ´ 2 ´ 4 2 = ´ 10000 ´ x 2 + 15 x 2 2 or 5000 x 2 + 15x – 16 = 0 or x = 0.50 = v Initial K.00 0. or x = 0.33 = 0.x = x ( x 2 - 1) dx 4 2 dV = 0 For V to be minimum.00 = (1 + 0. (a) : Loss of energy is maximum when collision is inelastic. 3. makes collision. (c) is correct. (d) : The kinetic energy of a particle is K At highest point velocity has its horizontal component. (b) 4.00) 2 = 1.00 J. dx \ x(x 2 – 1) = 0. v average = (v/2) = 10 m/s. = (1/2) × 0. Statement2 is true. (d) : For the same force. (Using (ii)) (Using (ii) As k 1 < k 2 \ W 1 < W 2 Statement 1 is false.00 – 0. x 1 = x 2 = x Work done on spring S 1 is W1 = 1 k1 x12 = 1 k1 x 2 2 2 v/2 .67 J. Using –ve signs when coming down. (Using (i)) 7. \ (v/2) = 10 m/s.. 1 2 Further h = gt is a parabola. ± 1 At x = 0. Q V ( x ) = x 4 x 2 4 2 dV 4 x 3 2 x 3 = = x .. As the ball is dropped.50 ´ 1. (d) : By the law of conservation of momentum mu = (M + m)v 1.00 = 0. \ Loss of energy = 1. Work done on spring S 2 is W2 = 1 k2 x22 = 1 k 2 x 2 2 2 W1 k 1 \ = W2 k2 time (1/2)mv a2 v = (1/2) × 50 × 100 = 2500 J.055 m = 5. t = 10 s. 2 5. It could be in the range 2000 to 5000 J. (a) : v = v / 2 is average velocity s = 100 m. 3 = 3. (b) : The centre of mass of the hanging part is at 0. 2 16.... (c) : Elastic energy stored in spring = 1 KL2 2 \ kinetic energy of block E = 1 KL2 2 mass of hanging part = 4 ´ 0.18 JEE MAIN CHAPTERWISE EXPLORER 10.6 = 1.1 ÷ 2 ø è Fl Mgl ( 2 .2 × 10 × 0. 2 12.. Hence kinetic energy of the particle remains constant. (a) : Let v 1 = speed of second mass after collision Momentum is conserved Along Xaxis. (c) : No work is done when a force of constant magnitude always acts at right angles to the velocity of a particle when the motion of the particle takes place in a plane. or Mg Work done = F ´ l sin 45 ° = Fl 2 Gain in potential energy = Mg(l – lcos45°) æ ö = Mgl ç 1 . 20. 19. Potential energy measures the workdone against the force of gravity during rise.3 m from table 1. mv 1sinq ..4 m 0. (b) : mgh = 1 æ 100 ö 2 5 ´ 5 ´ ( 5) = 2 ´ 10 = 1. (b) : Work done = F × r or work done = (5iˆ + 3 ˆj + 2kˆ ) × (2iˆ – ˆ j ) or work done = 10 – 3 = 7 J. 17...25 J 2 çè 1000 ÷ø Work done by force of gravity = – 1. (c) : Kinetic energy at projection point is converted into potential energy of the particle during rise.(i) mv = Along Yaxis..25 J 1 2 7 = 2 mv ∙ 5 = \ 11.. è t1 ø è t 1 ø t1 r r 18. 21.6 m 13. (ii) 3 From (i) and (ii) 2 2 æ mv ö 2 2 ( mv ) + \ (mv 1 cosq) + (mv 1 sinq) = ç 3 ÷ è ø 2 2 m 2 v1 2 = 4 m v or 3 2 v1 = v ...6 J.1) = \ 2 2 or F = Mg ( 2 – 1) . (b) : Acceleration a = \ l sin 45° 1 mv 2 æ1 + k 2 ö ç 2 ÷ 2 è R ø \ 5 5 = 1600 × 7 7 v 1 t1 v 1 velocity (v) = 0 + at = t t 1 Power P = Force × velocity = m a v 2 æ v ö æ v t ö mv t P = m ç 1 ÷ ´ ç 1 ÷ = 2 1 . 1 2 \ (– work done) = Kinetic energy = 2 mv or (– work done) Since p 2 = 2ME \ 2 p = 2 ME = 2 M ´ KL = MK L . or 3 15. (c) : Linear momentum is conserved \ 0 = m 1 v 1 + m 2 v 2 = (12 × 4) + (4 × v 2 ) or 4v 2 = – 48 Þ v 2 = – 12 m/s 1 2 \ Kinetic energy of mass m2 = m2 v2 2 1 = ´ 4 ´ (–12) 2 = 288 J. (a) : Work done in displacement is equal to gain in potential energy of mass l cos 45° 45° l 100 m 30 m 20 m \ F 1 2 æ 7 ö mv ç ÷ = mg ´ 80 2 è 5 ø or v 2 = 2 × 10 × 80 × or v = 34 m/s. (a) : Given : Retardation µ displacement dv = kx or dt æ dv öæ dx ö = kx ç dx ÷ç dt ÷ or è øè ø or dv (v) = kx dx v 2 or x ò vdv = k ò x dx v 1 0 .2 kg 2 \ W = mgh = 1. mv 1cosq = mv . (c) : Power = Force × velocity = (ma) (v) = (ma) (at) = ma 2 t 2 or 2 v mv Power = m æç ö÷ (t ) = 2 t è T ø T 14.. 05 k = ´ [0.05 0.05) 2 ù ë û W = 8 J. 1 2 27. By mass energy equivalent. but B is not true. mass should decrease.(0.0075) = 18. (b) : Power = Time = Time \ Force × velocity = constant (K) or (ma) (at) = K 1/ 2 K a = çæ ÷ö è mt ø Q s = 1 at 2 2 or 1/ 2 1/ 2 \ 1 K s = æç ö÷ 2 è mt ø 1 K t 2 = æç ö÷ t 3/ 2 2 è m ø or s is proportional to t 3/2 . u 1(a ) m m u 1(b ) 1(a ) explodes Total momentum = 0 ( ) 1 2 But total kinetic energy = 2 2 mu But if total kinetic energy = 0. (c) : When water is cooled to form ice.15 W = ò 800 xdx = x 2 ë û0. or 22.19 Work. 26. x 2 0.05 25.75 N m . Here A is true.0025] 2 (5 ´ 103 ) Workdone = ´ (0. (c) : Kinetic energy point of projection ( E ) = 2 mu At highest point velocity = u cosq \ Kinetic energy at highest point 1 = m (u cos q ) 2 2 1 2 E = mu cos2 45 ° = . A does not imply B. (c) : A system of particles implies that one is discussing total momentum and total energy. its thermal energy decreases. 2 Work Force× distance = Force× velocity 23. but B implies A.15)2 – (0.01 - 0. (b) : Force constant of spring (k) = F/x or F = kx \ dW = kxdx or or 0. 2 2 . 24.05) ùû 0.1 = 2 2 2 mv22 mv 1 2 mkx 2 = or 2 2 2 mk 2 ( K 2 – K1 ) = x or 2 or Loss of kinetic energy is proportional to x 2 .1 k 2 2 ò dW = ò kx dx = 2 éë (0.15 0.05 = 400 é (0.15 x 1 0.1) . Energy and Power v 22 v 2 k x 2 . velocities are zero. (b) : W = ò Fdx = ò kx dx \ or 800 é 2 ù0. A mass m hangs with the help of a string wrapped around a pulley on a frictionless bearing. is g 3 2 (b) g (c) g (d) (a) g 2 3 3 (2011) 8. the number of rotations made by the pulley before its direction of motion if reversed. Its maximum angular speed is w. An insect is at rest at a point near the rim of the disc. What will be the velocity of the centre of the hoop when it ceases to slip? rw 0 rw 0 rw 0 (a) rw 0 (b) (c) (d) 4 3 2 (2013) x as k(x/L) n where n can be zero or any positive number. During the journey of the insect. If the moment of inertia of the pulley about its axis of rotation is 10 kg m 2 . Then its acceleration is . 2. whose centre is O. the angular speed of the disc (a) remains unchanged (b) continuously decreases (c) continuously increases (d) first increases and then decreases (2011) 9. The insect now moves along a diameter of the disc to reach its other end. Assuming pulley to be a perfect uniform circular disc. which of the following graphs best approximates the dependence of x CM on n? A pulley of radius 2 m is rotated about its axis by a force F = (20t – 5t 2 ) newton (where t is measured in seconds) applied tangentially. is (a) less than 3 (b) more than 3 but less than 6 (c) more than 6 but less than 9 (d) more than 9 (2011) (a) 7. 4. A thin uniform rod of length l and mass m is swinging freely about a horizontal axis passing through its end. The pulley has mass m and radius R. A thin horizontal circular disc is rotating about a vertical axis passing through its centre. 6.20 JEE MAIN CHAPTERWISE EXPLORER CHAPTER 5 1. ROTATIONAL MOTION A hoop of radius r and mass m rotating with an angular velocity w 0 is placed on a rough horizontal surface. The moment of inertia of this plate about an axis perpendicular to its plane and passing through one of its corners is A thin rod of length L is lying along the xaxis with its ends at x = 0 and x = L. If the position x CM of the centre of mass of the rod is plotted against n. Its linear density (mass/length) varies with x CM x CM L L 2 O (c) (b) n L L 2 O x CM x CM L 2 L L 2 (d) O O (2008) (a) 2 2 5 2 ma (b) ma 3 6 (c) 1 ma 2 12 (d) 7 ma 2 12 (2008) For the given uniform square lamina ABCD. Its centre of mass rises to a maximum height of 1 l w 1 l 2w 2 1 l 2w 2 1 l 2w 2 (a) 3 g (b) 6 g (c) 2 g (d) 6 g (2009) 10. (a) I AC = 2 I EF (b) 2 I AC = I EF F O (c) I AD = 3 I EF (d) I AC = I EF C D A E B (2007) Angular momentum of the particle rotating with a central force is constant due to (a) constant torque (b) constant force (c) constant linear momentum (d) zero torque (2007) A round uniform body of radius R. 3. Consider a uniform square plate of side a and mass m. The initial velocity of the centre of the hoop is zero. if the string does not slip on the pulley. 5. the acceleration of the mass m. mass M and moment of inertia I rolls down (without slipping) an inclined plane making an angle q with the horizontal. The ring now rotates with an angular velocity w¢ = w m w(m + 2 M ) (a) ( m + 2 M ) (b) m w( m . A particle of mass m moves along line PC with velocity v as shown. 4 3 3 (a) l (b) l (c) l (d) l 3 4 2 (2005) 17. Let F be the r force acting on a particle having position r vector r and T be the torque of this force about the origin. the origin of the coordinate system. The torque about the point (1. A T shaped object with dimensions B shown in the figure. If its angular frequency is doubled and its kinetic energy halved. 1 2 1 2 (2006) l 16. Their moment of inertia about their diameters are respectively I A and I B such that (a) I A = I B (b) I A > I B (c) I A < I B (d) I A /I B = d A /d B where d A and d B are their densities. The moment of inertia of a uniform semicircular disc of mass M and radius r about a line perpendicular to the plane of the disc through the center is 1 2 1 2 2 2 Mr (a) Mr 2 (b) (c) Mr (d) Mr 2 4 5 (2005) 19. and another disc Y of radius 4R is made from an iron plate of thickness t/4. A force F is applied at the point P parallel to AB. What is the angular momentum of the particle about P? (a) mvL (b) mvl C L P l r O (c) mvr (d) zero.F (iˆ + ˆ j ) (d) F (iˆ + ˆ j ) . Then the relation between the moment of inertia I X and I Y is (a) I Y = 32I X (b) I Y = 16I X (c) I Y = I X (d) I Y = 64I X . The centre of mass of the new disc is a/R from the centre of the bigger disc. A force of - Fkˆ acts on O. One solid sphere A and another hollow sphere B are of same mass and same outer radii. O x y (2006) 15. The center of mass of bodies B 3 and C taken together shifts compared to that of body A towards (a) body C (b) body B (c) depends on height of breaking (d) does not shift (2005) 18. Then r r r r r r (a) rr × T = 0 and F × T ¹ 0 (b) rr × T ¹ 0 and F × T = 0 r r r r r r r (c) rr × T ¹ 0 and F × T ¹ 0 (d) r × T = 0 and F × T = 0 (2003) 22. A body A of mass M while falling vertically downwards 1 under gravity breaks into two parts; a body B of mass M 3 2 and body C of mass M . is lying on a A r smooth floor. so as to keep the centre of mass at the same position? m 2 m 1 m 1 (a) d (b) m d (c) m + m d (d) m d . (2006) 13. by what distance should the second particle be moved. A circular disc of radius R is removed from a bigger circular disc of radius 2R such that the circumferences of the discs coincide. The moment of inertia of this system about an axis through A and parallel to BD is (a) ml 2 (b) 2ml 2 (c) 3 ml 2 (d) 3ml 2 . Consider a two particle system with particles having masses m 1 and m 2. If the radius of the sphere is increased keeping mass same which one of the following will not be affected? (a) moment of inertia (b) angular momentum (c) angular velocity (d) rotational kinetic energy. A solid sphere is rotating in free space. 14. A circular disc X of radius R is made from an iron plate of thickness t. (2002) . (2003) 23. then the new angular momentum is (a) L /4 (b) 2L (c) 4L (d) L /2. If the first particle is pushed towards the centre of mass through a distance d. Four point masses. A particle performing uniform circular motion has angular momentum L. (2003) 24. Two objects each of mass M are attached gently to the opposite ends of a diameter of the ring. The value of a is (a) 1/4 (b) 1/3 (c) 1/2 (d) 1/6. such P r that the object has only the 2 l F translational motion without rotation. A thin circular ring of mass m and radius R is rotating about its axis with a constant angular velocity w. are placed at the corners of a square ABCD of side l.F (iˆ - ˆ j ) (b) F (iˆ - ˆ j ) (c) .21 Rotational Motion (a) g sin q 1 - MR 2 / I (b) g sin q 1 + I / MR 2 (c) g sin q 1 + MR 2 / I (d) g sin q 1 - I / MR 2 (2007) 11. each of value m.2 M ) wm (c) (m + 2 M ) (2006) (d) (m + M ) . –1) is z (a) . (2007) 12. (2004) 20. Find the location of P with C respect to C. (2004) r 21. 9. The velocity of centre of mass is (a) v (b) v/3 (c) v/2 (d) zero. (2002) Answer Key 1. 17. (d) (*) (d) (d) 6. 26. 28. 21. 7. 18. A solid sphere. 15. (b) (d) (d) (b) (d) 3. 16. (d) (a) (a) (c) (a) 2. 23.22 JEE MAIN CHAPTERWISE EXPLORER 25. (c) (d) (d) (d) (c) 4. 19. 22. 20. Then maximum acceleration down the plane is for (no rolling) (a) solid sphere (b) hollow sphere (c) ring (d) all same. Initial angular velocity of a circular disc of mass M is w 1 . 24. Two identical particles move towards each other with velocity 2v and v respectively. 25. What is the final angular velocity of the disc? M + m w M + m w (b) (a) 1 1 M m M M w . (2002) 27. 10. 12. (b) (d) (b) (d) . (d) (2002) (c) M + 4 m w1 M + 2 m 1 ( ( ) ) ( ( ) ) 28. 27. Then two small spheres of mass m are attached gently to two diametrically opposite points on the edge of the disc. 14. 13. 8. Moment of inertia of a circular wire of mass M and radius R about its diameter is (a) MR 2 /2 (b) MR 2 2 (c) 2MR (d) MR 2 /4. (2002) 26. (d) (b) (a) (a) (c) 5. 11. a hollow sphere and a ring are released from top of an inclined plane (frictionless) so that they slide down the plane. 6 g × dx ÷öx ø n × dx 0 3 2pn = 36 ò çè L n × x ò Ln × x t d q = 2 t - dt 3 2 ( ) 1 l m × w 2 . When the centre of mass is raised through h. (b) : Torque exerted on pulley t = FR \ a= (For circular disc) 2 mR a \ T = 2 Þ x C. R = 2 m. w = 2 t 2 - t 3 At t = 6 s.e. the increase in potential energy is mgh. As the density increases with length.(n + 1) ïü = L í ý 2 dn ( n + 2) ( n + 2) 2 ï îï þ T mg m If the rod has the same density as at x = 0 i. F = (20t – 5t 2 ).23 Rotational Motion a = 1. the centre of mass would have been at L/2.M = ) 0 L L òx 4 2 t t On integration. therefore uniform. \ Angular acceleration of disc 7. n = 0. pulley to be consider as a circular disc.(i) m As per question. (d) 5.. a = m 2. perpendicular to the plate is ..T \ a = . q = 36 rad Þ n k l 2 × w 2 \ h = . (c) : The free body diagram of pulley and mass L( n + 1) (n + 2) The variation of the centre of mass with x is given by R dx = L ïì (n + 2)1 . q = - 3 12 At.5t 2 ) ´ 2 10 a = (4t – t 2 ) (Using (ii)) mg . 2 3 L dq t 3 = 2 t 2 dt 3 3 2h 2 On integrating. (d): t I . Therefore it can only be (b). the centre of mass shifts towards the right.M = n +1 0 L ò x dx n dx n + 2 ( n + 1) = L × n +1 n + 2 L 0 n = 36 < 6 2 p Þ x C. (d) : The uniform rod of length l and mass m is swinging about an axis passing through the end... (b) : x C.(ii) Here. a mg – T = ma mg mg . t = 6 s. I = 10 kg m 2 ) or 4. mr 2w 0 = mvr + mr 2w ( ) = mvr + mr 2 v r mr 2w 0 = mvr + mvr mr 2w 0 = 2mvr rw or v = 0 2 ( or a = FR Q a = t I I Here.. 2 dw = (4t – t 2 )dt Þ mgh = 3 æ k ( 6. q This is equal to the kinetic energy h d w = (4t - t 2 ) dt dt = 1 I w 2 .M = 3.mR a 2 Therefore. w = 0 w= (Using (i)) (Q a = Ra ) ma = mg . t = T × R and I = 1 mR 2 According to law of conservation at point of contact.ma 2 2 g \ a = 3 (20t . (a) : For a rectangular sheet moment of inertia passing through O. (b) : I = (Mass of semicircular disc) ´ r 2 2 2 or I = Mr . m + 2 M r r 14.F ) = F (iˆ + ˆ = iF j ). or I z Ma 2 = 2 6 2 I By the same theorem I EF = z = Ma 2 6 \ I AC = I EF . The centre of mass is at R/3 to the left on the diameter of the original disc. The force should act at the centre of mass 2R M ¢ O ¢ x O m R 2 3 pR s × x + pR s × R = 0 M Because for the full disc. 18. I A = 2 MR 2 5 2 2 For hollow sphere. MR 11. The question should be at a distance aR and not a/R. None of the option is correct. for square plate it is 6 a 2 + a 2 = a . R x = - Þ = aR. 2 r I AC + I BD = I z Þ I AC = a = O 13. 9. or m 1 d = m 2 x or m2 16. 12 12 3 By perpendicular axes theorem. (*) :(M¢ + m) = M = p (2R) 2×s where s = mass per unit area m = sR 2×s. 2 19. the angular momentum of a particle is constant. (a) : It is a case of translation motion without rotation. I EF = M B 45° l 2 or I about B parallel to the axis through O is 2 axis A 1 l \ AO ´ = 2 2 l AO = or 2 I = I D + I B + I C a /2 B A a/2 r= 12. If no external torque is acting on a particle. r = iˆ – ˆ j ˆ iˆ ˆ j k r r \ r ´ F = 1 -1 0 0 0 .F ˆ – ˆj ( . (d) : Torque t = r ´ F r r F = . \ r 2 = a 2 4 4 2 2 \ D r C 2 2 I o + Md 2 = Ma + Ma = 4 Ma 6 2 6 2 2 I = Ma 3 8. (d) : The centre of mass of bodies B and C taken together does not shift as no external force is applied horizontally. or 2 2 a 2 + b 2 M ( a + a ) 2 a 2 = = M 12 12 12 M (2 a 2 ) M (2 a 2 ) Ma 2 + = .Fkˆ . m + 2 m 3 17. æ 2 l ö 2 ml 2 + mç ÷ 2 è 2 ø 2 2 I = 2 ml + 4 ml 2 2 I= (d) : Central forces passes through axis of rotation so torque is zero. . (d) : Let m 2 be moved by x so as to keep the centre of mass at the same position \ m 1 d + m 2 (–x) = 0 m x = 1 d . the centre of mass is at the centre O. I B = 3 MR I A 2 MR 2 3 3 = ´ = \ I B 5 2 MR 2 5 or I A < I B . (b) : Acceleration of a uniform body of radius R and mass M and moment of inertia I rolls down (without slipping) an inclined plane making an angle q with the horizontal is given by g sin q 1 + I 2 . (a) : Angular momentum is conserved \ L 1 = L 2 \ mR 2 w = (mR 2 + 2 MR 2 ) w¢ = R 2 (m + 2M) w¢ I z = 2 I= (d) : By perpendicular axes theorem. M¢ = 3pR 2s 2 C l / 2 l w¢ = m w .24 JEE MAIN CHAPTERWISE EXPLORER 2 2 I 0 = M çæ a + b ÷ö è 12 ø a O b Ma 2 . 10. (c) : For solid sphere. 3 \ | a | = D 6 ml 2 = 3 ml 2 . Y cm = ( m ´ 2l ) + (2 m ´ l ) 4 l = . (d) : AO cos 45 ° = l 2 -1 3 . 15. M + 4 m m1v1 + m2 v 2 m1 + m2 v c = m (2v ) + m ( . IY = or I Y 4 ( pR t s ) R pR st MR = = 2 2 2 2 2 (Mass)(4 R ) p(4 R ) t = s ´ 16 R 2 2 2 4 = 32pR 4 ts \ I X pR 4 st 1 = ´ = 1 I Y 2 32 pR 4 st 64 I Y = 64 I X .25 Rotational Motion 20. (d) : Q T = r ´ F r r r r r \ r × T = r × ( r ´ F ) = 0 r r r r r Also F × T = F × ( r ´ F ) = 0 .I. (d) : The bodies slide along inclined plane. (b) : Free space implies that no external torque is operating on the sphere. about its diameter = MR . (c) : Angular momentum of the system is conserved 2 23. Here the law of conservation of angular momentum applies to the system. 25. Hence angular momentum is zero. r r r 21. (a) : Angular momentum L = Iw 1 2 Rotational kinetic energy ( K ) = I w 2 L Iw´ 2 2 2 K = = Þ L = \ K w w I w2 or L1 K 1 w2 = ´ = 2 ´ 2 = 4 L2 K 2 w1 \ L L2 = 1 = L . 24. 4 4 \ I X = 2 2 Similarly. Acceleration for each body down the plane = gsinq.v ) v = . It is the same for each body. 22. (c) : v c = or M w1 . The line of motion is through P. Internal changes are responsible for increase in radius of sphere. (a) : A circular wire behaves like a ring 2 M. They do not roll. 27. (d) : Mass of disc X = ( pR t ) s where s = density 2 \ \ 1 1 MR 2 w1 = 2 mR 2 w + MR 2 w 2 2 or Mw 1 = (4m + M)w or w= 28. (d) : The particle moves with linear velocity v along line PC. m+m 2 . 2 26. Of the four choices given. When both d and h are much smaller than the radius of earth.4 × 10 8 Joules (b) 6.67 ´ 10 –9 J (b) 6.11 km s –1 (b) 1. 7. statement2 is false. R 2 R (a) Statement1 is true. statement2 is true; earth is statement2 is not a correct explanation for 1 1 statement1. (2008) (2004) . statement2 is true.67 ´ 10 –10 J –10 (c) 13. Find the work to be done against the gravitational force between them to take the particle far away from the sphere. (you may take G = 6. the flux of gravitational field passing through its sides is 12. independent of Statement2 : If the direction of a field due to a point source (a) the mass of the satellite is radial and its dependence on the distance r from the source (b) radius of its orbit 2 is given as 1/r .1 km s –1 –1 (c) 11 km s (d) 110 km s –1 (2008) Average density of the earth (a) is directly proportional to g (b) is inversely proportional to g (c) does not depend on g (d) is a complex function of g (2005) A particle of mass 10 g is kept on the surface of a uniform sphere of mass 100 kg and radius 10 cm. The time period of an earth satellite in circular orbit is 4pGM. (a) (d) ( 2 ) . The mass of a spaceship is 1000 kg. the radius of the earth is R (a) 2R (b) (c) R/2 (d) 2R 2 (2009) 10. mass m from the surface of a planet of mass M and radius R in a circular orbit at an altitude of 2R? (a) GmM 3 R (b) 5 GmM 2 GmM (c) 6 R 3 R (d) GmM 2 R (2013) 2. 3. the escape velocity from the surface of the planet would be (a) 0.67 ´ 10 –11 Nm 2 /kg 2 ) (a) 6. Then the time period of a planet in circular orbit of radius R around the sun will be proportional to Directions : The following question contains statement1 and statement2. mgR (c) mgR (a) 2 mgR (b) (d) mgR. then which of the following is correct? (a) d = 2h (b) d = h (c) d = h/2 (d) d = 3h/2 (2005) Suppose the gravitational force varies inversely as the n th power of distance. 2 4 Statement1 : For a mass M kept at the centre of a cube of side (2004) a. The required energy for this work will be (a) 6.33 ´ 10 –10 J (2005) The change in the value of g at a height h above the surface of the earth is the same as at a depth d below the surface of earth. (2004) (b) Statement1 is false. Given that the escape velocity from the earth is 11 km s –1 . the gain in the potential energy of an object of mass m raised from correct explanation for statement1.4 × 10 Joules (d) 6. The gravitational potential at a point on the line joining them where the gravitational field is zero is (a) zero 4. A planet in a distant solar system is 10 times more massive than the earth and its radius is 10 times smaller. statement2 is true; statement2 is a 11. 10 (c) 6. The value of g and R (radius of earth) are 10 m/s 2 and 6400 km respectively. its flux through a closed surface depends only (c) both the mass and radius of the orbit on the strength of the source enclosed by the surface and not (d) neither the mass of the satellite nor the radius of its orbit. (c) Statement1 is true.4 × 10 9 Joules 8. the surface of the earth to a height equal to the radius R of the (d) Statement1 is true. What is the minimum energy required to launch a satellite of 6. on the size or shape of the surface. 4Gm (b) - r 6Gm (c) - r 9Gm (d) - r (2011) 9.34 ´ 10 J (d) 3. It is to be launched from the earth’s surface out into free space. 5. If g is the acceleration due to gravity on the earth’s surface. choose the one that n - 2 ( n + 1 ) (b) R( n 2 - 1) (c) R n best describes the two statements. The height at which the acceleration due to gravity becomes g/9 (where g = the acceleration due to gravity on the surface of the earth) in terms of R.4 × 10 11 Joules (2012) Two bodies of masses m and 4m are placed at a distance r.26 JEE MAIN CHAPTERWISE EXPLORER CHAPTER GRAVITATION 6 1. (b) (a) (d) (d) 2.5R. Energy required to move a body of mass m from an orbit of radius 2R to 3R is (a) GMm/12R 2 (b) GMm/3R 2 (c) GMm/8R (d) GMm/6R. 19. then the satellite will (a) continue to move in its orbit with same velocity (b) move tangentially to the original orbit in the same velocity (c) become stationary in its orbit (d) move towards the earth. the orbital speed of the satellite is gR (a) gx (b) R - x 16. 14. (a) 10. (2003) (2004) 17. (c) . (2002) (c) 2 ö1/ 2 æ gR (d) ç R + x ÷ è ø gR 2 R + x . the new time period will become (a) 10 hour (b) 80 hour (c) 40 hour (d) 20 hour. the escape velocity will be 18. (c) (b) (c) (b) 3.27 Gravitation 13. If suddenly the gravitational force of attraction between Earth and a satellite revolving around it becomes zero.5R (d) 1. A satellite of mass m revolves around the earth of radius R at a height x from its surface. Two spherical bodies of mass M and 5M and radii R and 2R respectively are released in free space with initial separation between their centres equal to 12R. (d) 9.5R (c) 7. (d) 12. If the separation between the earth and the satellite is increased to 4 times the previous value. The escape velocity for a body projected vertically upwards from the surface of earth is 11 km/s. The kinetic energy needed to project a body of mass m from the earth surface (radius R) to infinityis (a) mgR/2 (b) 2mgR (c) mgR (d) mgR/4. If the body is projected at an angle of 45° with the vertical. (a) 11 2 km/s (b) 22 km/s (c) 11 km/s (d) 11 / 2 m/s. (2002) Answer Key 1. 8. (2002) 14. (2003) 15. (c) 5. (2002) 20. (b) 17. (c) 4. The time period of a satellite of earth is 5 hour. 7. (2003) 19. 13. (a) 18. then the distance covered by the smaller body just before collision is (a) 2. (a) 6. (a) 15. The escape velocity of a body depends upon mass as (a) m 0 (b) m 1 (c) m 2 (d) m 3 . (c) 11. 20.5R (b) 4. If they attract each other due to gravitational force only. (a) 16. If g is the acceleration due to gravity on the surface of the earth. \ v e = 2GM ´ 10 = 10 ´ 11 = 110 km s -1 R /10 7. (d) : v escape = ) If v is the velocity of the satellite at a distance 2R from the surface of the planet. has B A to pass through B. area of the Gaussian surface is 4pr 2 . GM = GM × 1 r2 R 2 9 r \ r = 3R The height above the ground is 2R. Assuming the mass is a point mass. 6.GMm = - GMm R R ( 5. \ Workdone = . all the lines passing through A are passing through B. (d) : Let x be the distance of the point P from the mass m where gravitational field is zero..67 ´ 10 –11 ) ´ (100) ´ (10 ´ 10 –3 ) 10 ´ 10 –2 –10 = 6.28 JEE MAIN CHAPTERWISE EXPLORER 1. It is only for M Gaussian surface. r Flux f g =| g × 4pr 2 | = r 2 Here B is a cube. whether B is a cube or any surface. (a) : The acceleration due to gravity at a height h from M the ground is given as g/9.x ) G (4m ) Gm =r r - r 3 3 3G (4m ) Gm = - 9 = . (b) : Energy of the satellite on the surface of the planet is E i = KE + PE = 0 + .x ) 2 or x = r 3 Gravitational potential at a point P is Gm G (4m ) =x (r . Statement2 is correct because when the shape of the earth is spherical.- GMm 6 R R GMm GMm 5 GMm =+ = 6R R 6 R = ( 2.E i = . This ensures inverse square law. . r g . r r Q Flux f = E × 4 pr 2 = e0 Every line passing through A. then total energy of the satellite is 1 GMm E f = mv 2 + 2 ( R + 2 R) ( ) 2 = 1 æ GM ö GMm m 2 çè ( R + 2 R ) ÷ø 3 R 1 GMm GMm GMm =2 3R 3R 6 R \ Minimum energy required to launch the satellite is DE = E f ..GM r 2 4 pr 2 × GM = 4 p GM . whatever be the shape. 2 Gm G (4 m ) x = or = 1 ( r - x ) 4 x2 (r . (b) : Gravitational force F = \ dW = FdR = Gm1m 2 R 2 G m1m 2 dR R 2 ¥ W ¥ Gm1m 2 dR é 1 ù \ ò dW = Gm1m2 ò 2 = Gm1m 2 ê .. (c) : Let A be the Gaussian surface enclosing a spherical charge Q..(i) g = (Using (i)) 4.R ú = ë û R R 0 R R (6. r B Q E × 4 pr 2 = A e0 Q + r Q r E = 2 4 pe0 .gravitational field = .GMm .. As explained earlier. (i) GM R 2 r 3M R 2 3 3 ´ = r= g \ g = 3 GM 4 p RG or 4 pRG 4 p R \ Average density is directly proportional to g.4 × 10 10 J 3. 2 GM for the earth R 8. the lines of field r should be normal. (ii) . v e = 11 kms –1 Mass of the planet = 10 M e . (c) : Energy required = = gR 2 ´ ) GMm R (Q g = GM R ) m R 2 = mgR = 1000 × 10 × 6400 × 10 3 = 64 × 10 9 J = 6. Radius of the planet = R/10. although all the lines are not normal.3 Gm - r r 2 r ( \ () ( ) ) Mass of earth Volume of earth M 3 M r= = (4 / 3) pR 3 4 p R 3 (a) : Density = ..67 × 10 J. the satellite has the tangential velocity and it will move in a straight line. 2 h ö æ (a) : At height. (a) : For motion of a planet in circular orbit.E. Compare Lorentzian force on charges in the cyclotron. From equation of motion. (d) : For a satellite centripetal force = Gravitational force mv 0 2 GMm \ ( R + x ) = ( R + x ) 2 gR 2 GM 2 = or v 0 = ( R + x ) ( R + x ) G ´ 5 M (12 R - x ) 2 GM Acceleration of bigger body. a1 ( x ) = 2 é GM ù êëQ g = R 2 úû .E.E.. 13. 17.. æ GM ö 1 1 = mR ç 2 ÷ = mgR 2 è R ø 2 14. (c) : Let the spheres collide after time t.R ÷ where h << R è ø 2 hg 2 hg or g – g h = or D g h = R R d æ ö At depth..) 2R = . (c) : The escape velocity of a body does not depend on the angle of projection from earth. a 2 ( x ) = (12 R - x ) 2 Acceleration of smaller body. It is 11 km/sec. (i) . \ mR w = or w = GM e R 3 = x2 x1 R 2R 9R We know that x1 + x 2 = 9 R x1 + x 1 = 9 R 5 \ x 1 = 45 R = 7. 2 GM e R Escape velocity does not depend on mass of body which escapes or it depends on m 0 . (b) : Force on object = GMm x 2 \ Work done = GMm dx x 2 2 R dx \ ò Work done = GMm ò 2 R x \ Potential energy gained 2 15.. when the smaller sphere covered distance x 1 and bigger sphere covered distance x 2 . (d) : Energy = (P. (c) : Escape velocity ve = 2 gR \ Kinetic energy = 1 mve 2 = 1 m ´ 2 gR = mgR . = R R 10.5R.çæ . F ( x ) = GM ´ 5 M 2 1 1 a ( x ) t 2 and x 2 = a 2 ( x) t 2 2 1 2 x a ( x ) Þ 1 = 1 = 5 Þ x1 = 5 x 2 x2 a 2 ( x) at x from centre of earth.29 Gravitation 9.GmM .R ÷ where d << R è ø dg or g – g d = R dg or D g d = R From (i) and (ii). ( n 2 +1 ) n +1 T = 2 p = 2 p R = 2 p R w GM GM \ T is proportional to R gR 2 . 11. g h = g ç 1 . ..5 R 6 Therefore the two spheres collide when the smaller sphere covered the distance of 7. 3R 2 R ø è 6 R 20. R + x or v 0 = é GM ù êëQ g = R 2 úû 18. (ii) 2hg dg or d = 2h. (c) : According to Kepler's law T 2 µ r 3 12. when Dg h = D g d . 3 3 T 1 1 æ T1 ö æ r 1 ö æ 1 ö 1 \ ç T ÷ = ç r ÷ = ç 4 ÷ = 64 or T = 8 è ø 2 2 2 è ø è ø or T 2 = 8T 1 = 8 × 5 = 40 hour. where R = r e + h R 2 (12 R - x) 16. x1 = ( n 2 + 1 ) . The gravitational force acting between two spheres depends on the distance which is a variable quantity. 2 2 19. Centripetal force = Gravitational force 2 \ mR w = \ GM or w = R n +1 GMm R n 2 R 1 GMm ´ 1 = GMm é . (a) : For a satellite Centripetal force = Gravitational force GmM e The gravitational force.ù = êë x úû R 2 R \ Gain in P. (a) : Escape velocity = 2 gR = GM e ( re + h ) 3 3 ( r + h ) \ T = 2 p = 2 p e w GM e \ T is independent of mass (m) of satellite.GmM ÷ö = + GmM .) 3R – (P. g d = g ç 1 . (b) : The centripetal and centrifugal forces disappear. Assume that a drop of liquid evaporates by decrease in its surface energy. it presses the semicircular parts together. The two parts are held together by a ring made of a metal R strip of cross sectional area S and length L. A uniform cylinder of length L and mass M having cross sectional area A is suspended. To fit the ring on the wheel. The length of the slider is 30 cm and its weight negligible.025 Nm –1 (d) 0. and its Young’s modulus is Y. As it cools down to surrounding temperature. it is heated so that its temperature rises by DT and it just steps over the wheel. The surface tension of the liquid film is (a) 0.1 Nm –1 (b) 0.4 m s –1 . 2T r L T T (a) r L (c) (d) r L (b) T r L (2013) 5.03 N m –1 ) (a) 4p mJ (b) 0. If the coefficient of linear expansion of the metal is a. The extension x 0 of the spring when it is in equilibrium is Mg Mg 1 + LA s (a) (b) k k M Mg Mg LA s 1 - 1 - LA s (c) (d) (2013) k M k 2 M ( ( 3. ) ) ( ) If a piece of metal is heated to temperature q and then allowed to cool in a room which is at temperature q 0 . What should be the minimum radius of the drop for this to be possible? The surface tension is T.0 × 10 –3 m (b) 7. then according to Newton’s law of cooling the correct graph between log e (q – q 0 ) and t is (a) (d) 7. such that it is half submerged in a liquid of density s at equilibrium position.30 JEE MAIN CHAPTERWISE EXPLORER 1.5 × 10 –2 N (see figure). (c) log ( e q – q 0 ) 7 log ( e q – q 0 ) PROPERTIES OF SOLIDS AND LIQUIDS CHAPTER .5 × 10 –3 m –3 (c) 9.4p mJ (2011) A liquid in a beaker has temperature q(t) at time t and q 0 is temperature of surroundings. density of liquid is r and L is its latent heat of vaporization. The water velocity as it leaves the tap is 0.0125 Nm –1 FILM w (2012) Water is flowing continuously from a tap having an internal diameter 8 × 10 –3 m. the graph between the temperature T of the metal and time t will be closed to 6. 2. L is slightly less than 2pR. The diameter of the water stream at a distance 2 × 10 –1 m below the tap is close to (a) 5.6 × 10 –3 m (2011) 8. from a fixed point by a massless spring. (2013) 4. so that its temperature remains unchanged.2p mJ (c) 2p mJ (d) 0. (a) (b) (c) (d) t log ( e q – q 0 ) log ( e q – q 0 ) (b) t t 0 t (2012) A wooden wheel of radius R is made of two semicircular parts (see figure).6 × 10 m (d) 3.05 Nm –1 (c) 0. the force that one part of the wheel applies on the other part is (a) SYaDT (b) pSYaDT (c) 2SYaDT (d) 2pSYaDT (2012) A thin liquid film formed between a Ushaped wire and a light slider supports a weight of 1. with its length vertical. Work done in increasing the size of a soap bubble from a radius of 3 cm to 5 cm is nearly (Surface tension of soap solution = 0. The temperature at the interface of the two sections is T 1 l 1 K 1 l 2 K 2 T 2 .. A long metallic bar is carrying heat from one of its ends to the other end under steadystate. r 2 and r 3? (a) r 1 < r 3 < r 2 (b) x x r 3 (b) r 3 < r 1 < r 2 (c) r 1 > r 3 > r 2 q (d) r 1 < r 2 < r 3 q (c) (d) x x (2009) 13. One end of a thermally insulated rod is kept at a temperature T 1 and the other at T 2 . If the above ball is in equilibrium in a mixture of this oil and water. A solid ball. The rod is composed of two sections of lengths l 1 and l 2 and thermal conductivities K 1 and K 2 respectively. Assume that the liquid applies a viscous force on the ball that is proportional to the square of its speed v. i. Another identical tube (B) is dipped in a soapwater solution. Two wires are made of the same material and have the same volume. It comes to equilibrium in the position shown in the figure. However wire 1 has crosssectional area A and wire 2 has crosssectional area 3A. The oil and water are immiscible.r 2 ) k (c) Vg r 1 k (d) Vg r 1 k (2008) 15. is dropped in the jar. how much force is needed to stretch wire 2 by the same amount? (a) F (b) 4F (c) 6F (d) 9F (2009) 12. A spherical solid ball of volume V is made of a material of density r 1 . D is 2 (a) b 6 a (b) b 2 2 a 2 (c) b 12 a 2 (d) b 4 a (2010) 10. F viscous = –kv 2 (k > 0). It is falling through a liquid of density r 2 (r 2 < r 1 ).31 Properties of Solids and Liquids 9. A jar is filled with two nonmixing liquids 1 and 2 having densities r 1 and r 2 respectively. A ball is made of a material of density r where r oil < r < r water with r oil and r water representing the densities of oil and water. where a and b are constants and x is the x x distance between the atoms. respectively. Which of the following is true for r 1. The variation of temperature q along the length x of the bar from its hot end is best described by which of the following figures? q q (a) 14.r 2 ) k (b) Vg (r1 . If the length of wire 1 increases by Dx on applying force F. made of a material of density r 3.e. A capillary tube (A) is dipped in water. The potential energy function for the force between two atoms in a diatomic molecule is approximately given by U ( x ) = a12 - b 6 . The terminal speed of the ball is (a) Vg (r1 . which of the following pictures represents its equilibrium position? Oil (a) Water B A B A B A B (a) (b) (c) Water (b) A Oil (d) Water Oil (c) (d) Oil Water (2008) (2010) 11. Which of the following shows the relative nature of the liquid columns in the two tubes? (2008) 16. If the dissociation energy of the molecule is D = [U(x = ¥) – U at equilibrium ]. (2004) 24.T1 ) K ö f ÷ . (2006) 23.1 m/s. (2004) 29.5 kg/m 3 ) is 0. at a distance r from the sun. (2006) 19. A wire fixed at the upper end stretches by length l by applying a force F. T 1 (2004) 27. Assuming the sun to be a spherical body of radius R at a temperature of T K. respectively. The radial rate of flow of heat in a substance between the two concentric spheres is proportional to (a) r1r 2 ( r2 - r1 ) (b) (r 2 – r 1 ) (c) r 1 r 2 T 1 T 2 ( r2 - r 1 ) r1r2 æ r 2 ö (d) ln ç ÷ è r1 ø (2005) T 2 (a) 1 (c) 2/3 K 4 x 2K (b) 1/2 (d) 1/3. If S is stress and Y is Young’s modulus of material of a wire. where r 0 is the radius of the earth and s is Stefan's constant.4 m/s (c) 0. The retarding viscous force acting on the spherical ball is (a) directly proportional to R but inversely proportional to v (b) directly proportional to both radius R and velocity v (c) inversely proportional to both radius R and velocity v (d) inversely proportional to R but directly proportional to velocity v. with f equal to slab. The rate of heat transfer through the æ A(T2 . If the terminal speed of a sphere of gold (density = 19. The figure shows a system of two concentric spheres of radii r 1 and r 2 and kept at temperatures T 1 and T 2 . The work done in stretching is (a) F/2l (b) Fl (c) 2Fl (d) Fl/2. then the ratio of the radiant energy received on earth to what it was previously will be (a) 4 (b) 16 (c) 32 (d) 64. The momentum transferred to the surface is (a) E/c (b) 2E/c (c) Ec (d) E/c 2 . the energy stored in the wire per unit volume is S 2 (a) 2Y/S (b) S/2Y (c) 2S 2 Y (d) 2 Y (2005) 21.5 kg/m 3 ) of the same size in the same liquid (a) 0. (2004) 28. (2004) 26. the elongation of the wire will be (in mm) (a) l/2 (b) l (c) 2l (d) zero.1 J. The water rises up to 8 cm.32 JEE MAIN CHAPTERWISE EXPLORER (a) ( K1l1T1 + K 2 l2T 2 ) ( K1l1 + K 2 l2 ) (b) ( K 2 l2T1 + K1l1T 2 ) ( K1l1 + K 2 l2 ) (c) ( K 2 l1T1 + K1l2T 2 ) ( K 2 l1 + K1l2 ) (d) ( K1l2T1 + K 2 l1T 2 ) ( K1l2 + K 2 l1 ) (2007) 17. If two soap bubbles of different radii are connected by a tube. A radiation of energy E falls normally on a perfectly reflecting surface. The temperature of the two outer surfaces of a composite slab. (a) R 2 s T 4 2 r (c) pr0 2 R 2s T 4 2 r (b) (d) 4 pr0 2 R 2 s T 4 r 2 r0 2 R 2sT 4 2 4 p r .5 kg/m 3 ) find the terminal speed of a sphere of silver (density 10. (2006) 18.133 m/s (d) 0. Spherical balls of radius R are falling in a viscous fluid of viscosity h with a velocity v. If the entire arrangement is put in a freely falling elevator the length of water column in the capillary tube will be (a) 4 cm (b) 20 cm (c) 8 cm (d) 10 cm (2005) 22. (2004) 25. The weight stretches the wire by 1 mm. incident on earth. respectively are T 2 and T 1 (T 2 > T 1 ). (2003) .2 m/s (b) 0. If the wire goes over a pulley and two weights W each are hung at the two ends.2 J (b) 10 J (c) 20 J (d) 0. in a steady state is ç x è ø x 20. A wire elongates by l mm when a load W is hanged from it. evaluate the total radiant power. If the temperature of the sun were to increase from T to 2T and its radius from R to 2R. Then the elastic energy stored in the wire is (a) 0. consisting of two materials having coefficients of thermal conductivity K and 2K and thickness x and 4x. (a) air flows from the bigger bubble to the smaller bubble till the sizes become equal (b) air flows from the bigger bubble to the smaller bubble till the sizes are interchanged (c) air flows from the smaller bubble to the bigger (d) there is no flow of air. A wire suspended vertically from one of its ends is stretched by attaching a weight of 200 N to the lower end.2 m/s in a viscous liquid (density = 1. A 20 cm long capillary tube is dipped in water. 26. (2002) 33. 10. Which of the following is more close to a black body? (a) black board paint (b) green leaves (c) black holes (d) red roses. (2003) 32. 23. (2002) 34. 9. 31. 17. (2002) Answer Key 1. (c) (d) (b) (c) (d) 6. the rate of cooling of a body is proportional to (Dq) n . According to Newton's law of cooling. 22. A cylinder of height 20 m is completely filled with water. 14. 12. The wavelength of the maximum intensity of the spectrum is correctly given by (a) Rayleigh Jeans law (b) Planck's law of radiation (c) Stefan's law of radiation (d) Wien's law. 15. (d) (c) (d) (a) (d) (a) 5. 33. 7. Two spheres of the same material have radii 1 m and 4 m and temperatures 4000 K and 2000 K respectively. 25. 34. The velocity of efflux of water (in ms –1 ) through a small hole on the side wall of the cylinder near its bottom is (a) 10 (b) 20 (c) 25. 8. 29. (d) (d) (a) (b) (b) (a) 4. The ratio of the energy radiated per second by the first sphere to that by the second is (a) 1 : 1 (b) 16 : 1 (c) 4 : 1 (d) 1 : 9. 27. 11. The earth radiates in the infrared region of the spectrum. (2003) 31. 28. (a) (d) (d) (c) (d) (d) 2. where Dq is the difference of the temperature of the body and the surroundings. 19. (c) (b) (d) (b) (d) . 16. 30. 21.33 Properties of Solids and Liquids 30. 32. 18. and n is equal to (a) two (b) three (c) four (d) one. 20. 13.5 (d) 5. (d) (d) (b) (d) (d) (b) 3. 24. 4 = p ´ ( d2 ) ´ 2 2 8. (a) 2.é 13 + 7 ù = é 13 . (d) : According to Newton’s law of cooling d q d q = .34 JEE MAIN CHAPTERWISE EXPLORER According to equation of motion.b = b .7 ù ú ëê x ú ëê x x û x û At equilibrium F = 0 \ 1213a .2 » 2 m s –1 \ According to equation of continuity a 1 v 1 = a 2 v 2 L Ag = Mg 2 sLAg Mg 2 x 0 = k kx0 + s ( p´ \ ) DL = aD T L 2 ´ 0. \ F = 2SYaDT 6.q = ò . r 2 = 5 cm = 5 × 10 –2 m Since bubble has two surfaces.4 m s –1 h = 0. DL = LaDT F = ( 8 ´ 10 -3 2 w ( d a b - Force. In equilibrium. So. F = . kx 0 + F B = Mg v2 = v1 2 + 2 gh = (0. (d): Let k be the spring constant of spring and it gets extended by length x 0 in equilibrium position. (d) : Here. (d) : Here.6 b = 0 or x 6 = 2 a b x x 7 a b U at equilibrium = - 2 2 a 2 a b b ( ) ( ) 2 2 2 2 2 = ab2 . S = 0.b = - b 2a 4a 2a 4 a 4 a U(x = ¥) = 0 D = [U(x = ¥) – U at equilibrium ] ( ) é b 2 ù b 2 = ê 0 .4)2 + 2 ´ 10 ´ 0.84p × 10 –4 = 4p × 10 –4 J = 0.kdt 0 log e (q – q 0 ) = – kt + C where C is a constant of integration. or F = 2T Where.03 N m –1 r 1 = 3 cm = 3 × 10 –2 m.q = .2 m ) 2 d 2 = 3. 1.03 × 128 × p × 10 –4 = 3. (d) : According to Newton’s law of cooling the option (d) represents the correct graph.6 × 10 –3 m Mg 1 - sLA k 2 M 3. the graph between log e (q – q 0 ) and t is a straight line with a negative slope. F = w F = 2 TL 2TL = w w T = 2 L Substituting the given values. surface tension.q 0 ) or q .4p mJ 9.025 Nm -1 2 ´ 30 ´ 10 -2 m 7. 4. Initial surface area of the bubble = 2 × 4pr 1 2 = 2 × 4p × (3 × 10 –2 ) 2 = 72p × 10 –4 m 2 Final surface area of the bubble = 2 × 4pr 2 2 = 2 × 4p(5 × 10 –2 ) 2 = 200p × 10 –4 m 2 Increase in surface energy = 200p × 10 –4 – 72p × 10 –4 = 128p × 10 –4 \ Work done = S × increase in surface energy = 0. Option (d) represents the correct graph.k (q . (c) : Increase in length. (d) : U = a - b x12 x 6 The thermal stress developed is T D L =Y = Y aD T S L T T or T = SYaDT From FBD of one part of the wheel.= ë 4 a ûú 4 a . we get 1. 5. we get ò q . (c) : The force due to the surface tension will balance the weight.kdt dt 0 d q Integrating both sides.5 ´ 10 -2 N T= = 0. F is the force that one part of the wheel applies on the other part. d 1 = 8 × 10 –3 m v 1 = 0.dU = dx x12 x6 dx ) -12a 6 b 12 a 6 b = . 11. so oil should be over the water.T ] K 2 A [T – T 2 ] = l1 l2 or or K 1l T 2) 2(T 1 – T) = K 2l 1(T – T(K 1 l 2 + K 2 l 1 ) = K 1 l 2 T 1 + K 2 l 1 T 2 or T = K1 l2 T1 + K 2 l1 T 2 . When the ball is moving with terminal velocity 2 a = 0. The viscous force rapidly increases with velocity. Þ v = Vg (r1 . (d) : Terminal velocity = v viscous force upwards = weight of sphere downwards 4 6 phrv = æç pr 3 ö÷ (r – s ) g è 3 ø For gold and silver spheres falling in viscous liquid. l1 / 3 l1 12. R r Sun Earth 4 2 Energy = sT ´ 4 pR 2 ´ time Area of envelope 4 p r 4 2 Energy incident per unit area on earth = sT R 2 ´ time r . mg – Vr 2g – kv 2 = ma where V is volume. in a sphere of radius r. r 3 16. the force acting down or T µ h without making finer corrections. Soap reduces the surface tension of water. according to Stefan's law. V = V1 = V2 V = A × l1 = 3A × l2 Þ l2 = l1/3 Y= D l F / A Þ F1 = YA 1 D l / l l1 D l F2 = Y 3 A 2 l2 Given Dl1 = Dl2 = Dx (for the same extension) \ F2 = Y × 3 A × ( ) Dx YADx = 9× = 9 F1 or 9 F . or 2 2 or 19. If r 3 had been greater than r 2 . (d) : For the same material. (d) : Let T be the l 1 temperature of the K 1 interface. (d) : The force acting upwards 2prT .5 9 1 v g 0. q min . v is the terminal velocity. Vmin Potential difference is proportional to resistance (thermal as well as x electric).35 Properties of Solids and Liquids 10. Therefore r 1 < r 2 . (b) : Heat flow can be compared to charges flowing in a conductor. (a) : The liquid 1 is over liquid 2. dQ Current is the same. For both sides. But it is also a wetting agent. v g r g – s 19. q 1 q 2 dt The potential difference V 1 – V V 2 rl V 1 current V at any point = I × Resistance = I ´ A Potential difference is µ l but negative. Young’s modulus is the same and it is given that the volume is the same and the area of crosssection for the wire l1 is A and that of l2 is 3A. hpr 2rg. k 15. Earth (E) in space receives part of this energy.1 m/s. the length of wire is L/2 on each side and so the elongation will be l/2. Therefore Vr 1g – Vr 2g – kv = 0.5 18 2 = = = = \ vs r s – s 10. The height of liquid supported decreases. Therefore the meniscus will not be convex as in mercury. attaining a maximum when the ball reaches the terminal velocity. 13. (c) : Energy radiated by sun. V max (temperature decreases) but it is a straight line function. thrust by Archimedes principle upwards and the force due to the force of friction also acting upwards. (c) : As r oil < r w ater . potential decreases q q max. As l increases. Then the mg = V r 1 g acceleration is zero. 14. L 2 l = L W W W 18. (b) : The forces acting on the solid Vr 2 g ball when it is falling through a viscous force liquid are mg downwards.5 – 1.5 – 1.2 vs = = = 0. so the ball will sink in the oil but r < r water so it will float in the water. it will not be partially inside but anywhere inside liquid 2 if r 3 = r 2 or it would have sunk totally if r 3 had been greater than r 2 . K1 l2 + K 2 l1 17. \ r 1 < r 3 < r 2 . (b) : Y = Force ´ L = WL A´l Al P WL \ AY Due to pulley arrangement. Therefore (d). E = sT 4 × (area 4pR 2 ) (time) This energy is spread around sun in space.r 2 ) . Since two section of rod T 1 are in series. elongation = l. rate of flow of heat in them will be equal \ l 2 K 2 T 2 K1 A [T1 . Hence option (c) is correct. As r > r oil . T )t 2 KA(T ..36 JEE MAIN CHAPTERWISE EXPLORER \ \ æ 2 4 ö Power incident per unit area on earth = ç R s2 T ÷ è r ø 2 4 2 R sT p r ´ Power incident on earth = 0 r 2 20. 25. (i) Workdone = Fl . (c) : Pressure inside the bubble = P 0 + r Smaller the radius. 22.T 1 ù ê ú f = x ê 3 ú ´ 1 x ë û ë û 1 f = . 2 29. (d) : According to Newton's law of cooling. dQ = .T1 ) t Q1 = Q 2 Þ = x 4 x or 2(T 2 – T) = (T – T 1 ) 2 T + T T = 2 1 or 3 2 T + T ù KA é Q1 = T . (d) : Energy stored per unit volume = 1 ×stress× strain 2 = Stress × stress S 2 = .. (d) : From first surface. (b) : In a freely falling elevator g = 0 Water will rise to the full length i.T2 ) r1r 2 Q = ( r2 .. (ii) E 1 (4000)4 ´ (1)2 ´ 1 ´ 4 pse 1 = = E 2 (2000) 4 ´ (4)2 ´ 1 ´ 4 ps e 1 .1 J . greater will be the pressure.. (a) : For conduction from inner sphere to outer one. rate of cooling is proportional to Dq.2 1 ú t \ x êë 2 3 û 26.T1 ) K ù KA é T2 .r ÷ . or ...T1 ) t From second surface. 2 2 30. (a) : Energy radiated E = sT 4 × (area 4pR 2 ) × time × e 2 YAl Workdone = 2 L From (i) and (ii) E 2 s (2T )4 ´ 4 p (2 R ) 2 ´ t = = 16 ´ 4 E 1 sT 4 ´ (4 pR ) 2 ´ t E 2 = 64. Q 2 = (4 x ) At steady state. Momentum transferred to surface = c Change of momentum = 28. E1 31..K ´ (4 pr 2 ) or dt dr \ Radial rate of flow Q = -4 pKr 2 dT dr r2 KA(T2 . 2Y 2 Y 21. (d) : Wien's law 32.. (b) : v = 2 gh = 2 ´ 10 ´ 20 = 20 m/s. (a) : A good absorber is a good emitter but black holes do not emit all radiations. Hence air flows from the smaller bubble to the bigger. 34.e..T 1 ] ë r1r2 û or 4 pK (T1 . Q 1 = or or 27. Air flows from higher pressure to lower pressure. \ (Dq) n = (Dq) or n = 1. .KA dT ´ (time dt ) dr dQ dT = .. YA l YAl 2 ò dW = L ò l dl = 2 L 0 or E æ E ö 2 E -ç. (d) : Elastic energy per unit volume 1 = ´ stress ´ strain 2 \ Elastic energy 1 = ´ stress ´ strain ´ volume 2 1 F DL = ´ ´ ´ ( AL ) 2 A L 1 1 = F DL = ´ 200 ´ 10 –3 = 0.. 33.÷ = c è cø c 2 E . (d) : Young's modulus Y = FL Al YAl F = \ L YAl ( dl ) or dW = F dl = L é A(T2 . KA(T2 . (d) : According to Stefan's law.. Radiant energy E = (sT 4 ) × area × time 4T 23. (b) : Initial momentum = E/c Final momentum = –E/c \ T 2 dr = -4 pK ò dT r 2 T 1 \ Qò or é r . (b) : Retarding viscous force = 6phRv obviously option (b) holds goods..T ) t x (2 K ) A(T .r1 ) \ \ æ r1r 2 ö Q is proportional to ç r . 20 cm to tube. è 2 1 ø \ r1 \ 24. 3 .r ù Q ê 1 2 ú = 4 pK [T2 . operating with an ideal monoatomic gas. The other chamber has volume V 2 and contains ideal gas at pressure P 2 and temperature T 2 . Ignoring the slight expansion of the water. the change in its internal energy is (specific heat of water is 4184 J kg –1 K –1 ) (a) 4. the intake temperature for the same exhaust (sink) temperature must be (a) 1200 K (b) 750 K (c) 600 K (d) efficiency of Carnot engine cannot be made larger than 50% (2012) Directions: Question numbers 8. If the partition is removed without doing any work on the gas.99 (2010) (2013) A Carnot engine. A Carnot engine operating between temperatures T 1 and T 2 1 . One of the chambers has volume V 1 and contains ideal gas at pressure P 1 and temperature T 1 . whose efficiency is 40%.m 2 and m 3 and the number of molecules are n 1 . extracted from the source in a single cycle is (a) 4p 0 v 0 (b) p 0 v 0 1 3 11 pv (c) (d) 2 p0v 0 2 0 0 ( ) 2. as shown in the P T diagram. The amount of heat.1 kJ (2011) 7.5 (c) 0. Two moles of helium gas are taken over the cycle ABCDA. 3.1% (b) 10. When T 2 is lowered by 62 K. Then T 1 and T 2 are. ( ) 100 g of water is heated from 30°C to 50°C. It is desired to have an engine of efficiency 60%. takes in heat from a source maintained at a temperature of 500 K. 9 and 10 are based on the following paragraph. A diatomic ideal gas is used in a Carnot engine as the working substance. 6.4 kJ (c) 84 kJ (d) 2.25 (b) 0. The masses of molecules are m 1 . Assuming the gas to be ideal the work done on the gas in taking it from A to B is (a) 200R (b) 300R (c) 400R (d) 500R 9. Then.2 kJ (b) 8. n 2 and n 3 respectively. Assuming no loss of energy. 2008) .5% (c) 12. If during the adiabatic expansion part of the cycle the volume of the gas increases from V to 32V. the efficiency of the engine is (a) 0. The work done on the gas in taking it from D to A is (a) – 414R (b) + 414R (c) – 690R (d) + 690R 10. its efficiency 6 1 increases to .5% (d) 15. the final equilibrium temperature of the gas in the container will be T1T2 ( PV 1 1 + PV 2 2 ) T T ( PV + PV ) (a) PV T + P V T (b) 1 2 1 1 2 2 1 1 1 2 2 2 PV 1 1T2 + PV 2 2T1 (c) PV 1 1T1 + PV 2 2T 2 PV 1 1 + PV 2 2 (d) PV 1 1T2 + PV 2 2T 1 PV 1 1 + PV 2 2 (2004. The above pv diagram represents the thermodynamic cycle of an engine. 5 2 × 10 B C 2P 0 Helium gas goes through a cycle ABCDA (consisting of two isochoric P 0 and two isobaric lines) as shown in A D figure.T 2 and T 3 are mixed. An insulated container of gas has two chambers separated by an insulating partition. Efficiency of this cycle is nearly (Assume the gas to be close V 0 2V 0 to ideal gas) (a) 9. The net work done on the gas in the cycle ABCDA is (a) zero (b) 276R (c) 1076R (d) 1904R (2009) 11. 4. the final temperature of the mixture is n1T1 + n2T2 + n3T 3 (T + T + T ) (a) 1 2 3 (b) n1 + n2 + n3 3 (c) n1T12 + n2T22 + n3T 3 2 n1T1 + n2T2 + n3T3 (d) n12T12 + n22T22 + n32T 3 2 n1T1 + n2T2 + n3T3 (2011) 5. respectively 3 has efficiency (a) 372 K and 310 K (c) 330 K and 268 K (b) 372 K and 330 K (d) 310 K and 248 K (2011) A B D C P(Pa) 5 1 × 10 T 500 K 300 K 8.75 (d) 0.37 Thermodynamics CHAPTER THERMODYNAMICS 8 1.4% (2012) Three perfect gases at absolute temperatures T 1 . 22. 24. Even Carnot engine cannot give 100% efficiency because we cannot (a) prevent radiation (b) find ideal sources (c) reach absolute zero temperature (d) eliminate friction. (a) (b) (b. The temperatureentropy diagram of a reversible engine cycle is given in the figure. When a system is taken from state a i to state f along the path iaf. Heat given to a body which raises its temperature by 1°C is (a) water equivalent (b) thermal capacity (c) specific heat (d) temperature gradient.4 × 10 6 J (c) 16. Its efficiency is (a) 1/3 (b) 2/3 (c) 1/2 (d) 1/4 T 2T 0 (c) It is not applicable to any cyclic process (d) It is a restatement of the principle of conservation of energy (2005) 18. A system goes from A to B via two processes I and II as shown P II in figure. W along the path ibf is (a) 14 cal (b) 6 cal (c) 16 cal f b (d) 66 cal (2007) 14. (2002) 17.2 × 10 6 J (b) 8. 10. 13. 11. 14. 25. 17. (b) (c) (b) (a) . 23. 12. The work done by the engine is (a) 4.8 × 10 6 J (d) zero. (2003) 22. it is found that Q = 50 cal and W = 20 cal. (c) (c) (b) (b) (b) 2. During an adiabatic process. 9. Which of the following is incorrect regarding the first law of thermodynamics? (a) It introduces the concept of the internal energy (b) It introduces the concept of entropy 25. (2004) 19. 15. (2002) T 0 S S 0 2 S 0 Answer Key 1. the pressure of a gas is found to be proportional to the cube of its absolute temperature. 7. 19. Which of the following statements is correct for any thermodynamic system? (a) The internal energy changes in all processes. (2002) (2005) 24. 20. A Carnot engine. A Carnot engine takes 3 × 10 6 cal of heat from a reservoir at 627°C. (2003) 21. (2003) 23. "Heat cannot by itself flow from a body at lower temperature to a body at higher temperature" is a statement or consequence of (a) second law of thermodynamics (b) conservation of momentum (c) conservation of mass (d) first law of thermodynamics. (b) (c) (b) (c) 3. c) (c) 6. Which statement is incorrect? (a) all reversible cycles have same efficiency (b) reversible cycle has more efficiency than an irreversible one (c) Carnot cycle is a reversible one (d) Carnot cycle has the maximum efficiency in all cycles. having an efficiency of h = 1/10 as heat engine. and gives it to a sink at 27°C.38 JEE MAIN CHAPTERWISE EXPLORER 12. The ratio C P /C V for the gas is (a) 4/3 (b) 2 (c) 5/3 (d) 3/2. If the work done on the system is 10 J. 18.3 J mol –1 K –1 ) (a) monoatomic (b) diatomic (c) triatomic (d) a mixture of monoatomic and diatomic. The gas is (R = 8. the amount of energy absorbed from the reservoir at lower temperature is (a) 100 J (b) 99 J (c) 90 J (d) 1 J (2007) 13. (b) (b) (a) (a) 5. (b) Internal energy and entropy are state functions. 8. is used as a refrigerator. (c) The change in entropy can never be zero. (d) The work done in an adiabatic process is always zero. 16. 21. then V (a) DU 2 > DU 1 (b) DU 2 < DU 2 (c) DU 1 = DU 2 (d) relation between DU 1 and DU 2 cannot be determined (2005) 16. Along the path ibf Q = 36 i cal. The work of 146 kJ is performed in order to compress one kilo mole of gas adiabatically and in this process the temperature of the gas increases by 7ºC. (2003) 20. (2006) 15. If DU 1 and DU 2 are the changes in internal energies A B in the processes I and II I respectively. (d) (b) (c) (d) 4. Which of the following parameters does not characterize the thermodynamic state of matter? (a) temperature (b) pressure (c) work (d) volume. (c) : Heat is extracted from the source in path DA and AB. T h = æç1 .. (b) : The final temperature of the mixture is ( ) DQ = DQ DA + DQ AB 3 3 3 R DT = V0 D P = P0V0 2 2 2 T mixture = T1n1 + n2T2 + n3T 3 n1 + n2 + n3 5.4% 13 3 PV + 5 PV 0 0 0 0 2 4. T h = 1 - 2 T1 where T 1 is the temperature of the source and T 2 is the temperature of the sink. then its efficiency becomes \ 1 3 1 = æ1 .62 2 = 6T 3 2 5 5(T 2 . h = ´ 100 Heat supplied to the gas = PV 200 0 0 ´ 100 = = 15. C p = \ CV = 3 R and C P = 5 R 2 2 5 R 2 2 p0 v 0 ù 10 5 é 2 p 2v \ DQAB = n R ê 0 0 = p v 2 ë nR nR ûú 2 0 0 \ The amount of heat extracted from the source in a single cycle is \ D Q AB = nCV D T = n Along the path BC. \ Work done by the gas W = Area of the rectangle ABCD = P 0 V 0 Helium gas is a monoatomic gas.2 Þ 2 = 1 . when T 2 is lowered by 62 K.0 0 ú = p0v 0 2 ë nR nR û 2 Along the path AB.40 = 3 100 500 500 100 5 T2 = 3 ´ 500 = 300 K 5 For 2 nd case h = 60%.. Hence DQ AB = nC pDT = nC p (T B – T A ) ( ) For monoatomic gas. T1 = 6 ´ 310 = 372 K 5 6. heat supplied to the gas at constant volume. DQ DA = nC vDT = nC v (T A – T D ) According to ideal gas equation pv pv = nRT or T = nR A B Along the path AB. (a) : The efficiency of Carnot engine. work done is equal to the area under the cycle and is taken to be positive if the cycle is clockwise. heat supplied to the gas at constant pressure. (d) : In case of a cyclic process. Cv = 3 R 2 p v ù 3 3 é 2 p v \ DQDA = n R ê 0 0 . m = 100 g = 100 × 10 –3 kg s = 4184 J kg –1 K –1 and DT = (50 – 30) = 20°C \ DQ = 100 × 10 –3 × 4184 × 20 = 8.62) 2 = 6T2 3 5T 2 – 310 = 4T 2 Þ T 2 = 310 K From equation (i). (b) : Efficiency of Carnot engine. D C For a monoatomic gas.(i) As per question.39 Thermodynamics 1. Along path DA. heat is rejected by the gas Work done by the gas Efficiency.4 × 10 3 J (Using (i)) . 5 5 R D T = (2 P0 ) DV = 5 PV 0 0 2 2 Along the path CD and DA. pressure is constant. volume is constant. \ D QBC = nC P DT = n = 3 p0v0 + 10 p0 v0 = 13 p0v0 2 2 2 2.62 1 = 1 - T1 3 T 2 . h = 1 6 çè T1 ø 6 ( T 2 5 = T1 6 Þ T1 = ) 6 T 2 5 . (b) : DQ = msDT Here. T 1 = 500 K T T 40 \ = 1 .62 ö 3 çè T1 ÷ø T 2 .T 2 . For 1 st case h = 40%.2 ö÷ Given. T 2 = 300 K \ 60 300 300 60 2 = 1 = 1 = Þ 100 T1 T1 100 5 T1 = 5 ´ 300 = 750 K 2 3.2 ö÷ è T1 ø T 1 æ \ = 1 . Hence. The total number of molecules is conserved. from C to D is equal and opposite. (c) : DU 1 = DU 2 . \ Work done along AB and CD cancel each other because pressure changes but temperature is the same.693) = –415. They cancel each other. n = P2V 2 R1T1 2 RT2 Final state = (n 1 + n 2 )RT g . DT is 200 K.693 = 277R.75 4 8. temperature remain the same. (c) : For Carnot engine efficiency h = QH . V \ Work done by the gas = W = nRT ln 2 V1 P 1 = nRT ln P2 Þ W = –600R (0. Work done on the gas from A to B = 400R. volume increases. a f Q ibf = DU ibf + W ibf Since DU iaf = DU ibf . temperature remains the same. Þ W BC + DA = (2 ln 2) × (200R) = 400R × 0. 10. (b) : As this is a simple mixing of gas. (c) : For an adiabatic process TV g–1 = constant \ T1V1g -1 = T2V2 g -1 g .1 7 For diatomic gas. P is the same. Work done on the gas in taking it from B to C. DW = 0) DU = DQ = 8. .1 T2 (32) 5 = T2 (32) 2/5 = T 2 (2 5 ) 2/5 = 4T 2 T 2 = 1 - 1 T1 4 ( ) h = 3 = 0.40 JEE MAIN CHAPTERWISE EXPLORER As DQ = DU + DW \ Change in internal energy ( .8 J mol K For diatomic gas. T 300 K ( n1 + n 2 ) = T = Efficiency of the engine. (b) : D to A.8R. Q iaf = DU iaf + W iaf or DU iaf = Q iaf – W iaf = 50 – 20 = 30 cal For the path ibf. This is the work done by the gas \ Work done on the gas = +415. 14.4 × 10 3 J = 8. Net work done on the gas of 2 moles of helium through the whole network = 277R per cycle or nearest to the answer (b). 9. even if adiabatic conditions are satisfied. . h = 1 - PV 1 1 .8R.3 = 20. pressure is decreased.1 10 = 9 b= 1/10 Q (where W is the work done) Also b = L W or Q L = b × W = 9 × 10 = 90 J. Q ibf = DU iaf + W ibf \ W ibf = Q ibf – DU iaf = 36 – 30 = 6 cal.( -146) ´ 10 or C V = -DW = n DT (1 ´ 103 ) ´ 7 –1 –1 = 20. (b) : According to first law of thermodynamics DQ = DU + DW For an adiabatic process. . change in internal energy are path i b independent. Nearest to (b). because the change in internal energy depends only upon the initial and final states A and B. (b) : According to first law of thermodynamics for the path iaf. work done on the gas = +414R.4 kJ 7. 15. g = 5 \ T1 = 7 . For taking from D to A. 13. T = T1T2 ( PV n1 + n2 T2 PV 1 1 + T1P2V2 12.8 J mol -1 K -1 2 2 Hence the gas is diatomic. 5 5 CV = R = ´ 8. P(in pascal) A (c) : 2 × 10 5 B 1 × 10 5 D C PV PV T PV + T P V 1 1 + 2 2 = 2 1 1 1 2 2 RT1 RT2 RT1T2 T1n1 + T2 n 2 1 1 + P2V 2 ) .1 = T 2 11. 500 K Path AB.1 V T1 = T 2 æç 2 ö÷ è V1 ø \ n1 = ( ) 32V V = T2 (32) g . (b) : Total work done on the gas when taking from A to B = 400R. PV = nRT for all process \ PDV = nRDT = 2R 200 = 400R. DQ = 0 \ 0 = DU + DW or DU = – DW or nC V DT = – DW 3 . Þ W BC + W DA = 2 ln 2(500R – 300R).h h 1 . PV = nRT for adiabatic as well as isothermal changes.Q L QL Coefficient of performance of a refrigerator b = 1. 25. (d) : In an adiabatic process. (c) : We cannot reach absolute zero temperature. g = 5 = 1. 0 0 17.3 = g or 2g = 3 Þ g = 3/2 5 N. 22.3 = 3 . (b) : Internal energy and entropy are state functions.67 7 For diatomic gas.5. 21.B For monoatomic gas. g = 3 = 1.1 = 3Þ 3 g . (a) : Second law of thermodynamics. (b) : Thermal capacity. (b) : Efficiency = 1 - T 2 300 1 2 = 1= 1 .2) ´ 3 J = 8. 18.2 J \ Workdone by engine = (Heat energy) × (efficiency) 2 6 = (3 ´ 10 ´ 4.3T S = 1 . 19.41 Thermodynamics Q 2 16.= T1 900 3 3 Heat energy = 3 × 10 6 cal = 3 × 10 6 × 4. T g = (constant) P g -1 or Tg/g–1 = (constant) P Given T 3 = (constant) P g \ g .4 when g = 1. (c) : The work does not characterize the thermodynamic state of matter. c) : Statements (b) and (c) are incorrect regarding the first law of thermodynamics. (a) : Efficiency h = 1 - Q 1 Q 2 = T 0 (2S 0 – S 0 ) = T 0 S 0 Q1 = T0 S0 + T0 S 0 3 = T0 S 0 2 2 T0 S 0 ´ 2 2 1 \ h = 1 . 20. the gas must be a suitable mixture of monoatomic and diatomic gases \ g = 3/2. (b. .4 × 10 6 J. 23. (a) : All reversible cycles do not have same efficiency. 24. Assuming no heat is lost to the surroundings. 3. (d) 5. One kg of a diatomic gas is at a pressure of 8 × 10 4 N/m 2 . the final temperature of the gases.42 JEE MAIN CHAPTERWISE EXPLORER CHAPTER KINETIC THEORY OF GASES 9 1. (2004) 7. 2. (a) 3/2 (b) 23/15 (c) 35/23 (d) 4/3. At what temperature is the r.1) ( g . What is the energy of the gas due to its thermal motion? (a) 3 × 10 4 J (b) 5 × 10 4 J (c) 6 × 10 4 J (d) 7 × 10 4 J (2009) If C P and C V denote the specific heats of nitrogen per unit mass at constant pressure and constant volume respectively. 9. 4. The ratio C P /C V of the mixture is (a) 1. It is moving with speed v and is suddenly brought to rest. A thermally insulated vessel contains an ideal gas of molecular mass M and ratio of specific heats g.1) 2 2 (a) 2( g + 1) R Mv K (b) 2 g R Mv K ( g - 1) 2 g Mv 2 Mv K (c) (d) (2011) K 2 R 2 R 7 (c) T f = T0 3 Two rigid boxes containing different ideal gases are placed on a table. 8. One mole of ideal monoatomic gas (g = 5/3) is mixed with one mole of diatomic gas (g = 7/5). (a) . 7. to that at constant volume. (d) 6. T f .s. (2002) 8.m. (2002) Answer Key 1. velocity of a hydrogen molecule equal to that of an oxygen molecule at 47°C? (a) 80 K (b) –73 K (c) 3 K (d) 20 K. 2 4. The temperature of the gas molecules inside will (a) increase (b) decrease (c) remain same (d) decrease for some. The boxes are then put into thermal contact with each other and heat flows between them until the gases reach a common final temperature.62 (2005) 6. while increase for others. while Box B contains one mole of helium at temperature (7/3) T 0 . (b) (c) (2006) 5. its temperature increases by ( g . then the value of g for the resulting mixture is (a) 7/5 (b) 2/5 (c) 24/16 (d) 12/7.54 (c) 1. Box A contains one mole of nitrogen at temperature T 0 . (b) (d) 3. (Ignore the heat capacity of boxes). What is g for the mixture? g denotes the ratio of specific heat at constant pressure. Cooking gas containers are kept in a lorry moving with uniform speed. 1 mole of a gas with g = 7/5 is mixed with 1 mole of a gas with g = 5/3. Then. The density of the gas is 4 kg/m 3 . in terms of T 0 is 5 3 (b) T f = T0 (a) T f = T0 2 7 A gaseous mixture consists of 16 g of helium and 16 g of oxygen. (d) (c) 2.59 (d) 1. (2002) 9.4 (b) 1. then (a) C P – C V = 28R (b) C P – C V = R/28 (c) C P – C V = R/14 (d) C P – C V = R (2007) 3 (d) T f = T0 . (d) : v rms = or T f = 3 T0 .43 Kinetic Theory of Gases 1 2 1.1 1+ 1 1 1 = + g m . 32 2 (c) : It is the relative velocities between molecules that is important. As the vessel is stopped suddenly. 28 (d) : DU = 0 3 æ 7 ö æ5 ö \ 1 ´ ç 2 R ÷ (T f .1 = 0. C V = or 8g m – 8 = 4 or 8g m = 12 12 24 or g m = 8 = 16 8. (d) : Kinetic energy of vessel = mv 2 Increase in internal energy DU = nCV D T where n is the number of moles of the gas in vessel.1 g1 .5 = 3/2. n 2 = 1/2 1 2 mv = nCV D T 2 3.62. RT M = ( vrms ) H 2 273 + 47 T = Þ T = 20 K . 2 \ ( vrm s )O 2 16 = 4 (d) : For 16 g of helium. n 1 = 4 For oxygen.T0 ) + 1´ 2 R ç T f .1 g1 .1 g 2 . f = 5.1 = 2 + 2 = 4 Þ g m . its kientic energy is used to increase the temperature of the gas \ or DT = C P = n1CP1 + n2 C P 2 n1 + n2 f where C P = çæ + 1 ÷ö R è 2 ø \ (Q n = Mm ) Mv 2 ( g . V \ U = 5 ´ 8 ´ 10 4 ´ 1 = 5 ´ 104 J 2 4 \ 28C P – 28C V = R C P = C V n1 + n2 n n = 1 + 2 or 1 + 1 = 1 + 1 g m . f = 3. n 1 = 4 16 1 = For 16 g of oxygen.5 m \ g m = 1.1 5 . (c) : For mixture of gases n1 + n2 n n = 1 + 2 g m .1) K 2R 6. (5 is the degrees of freedom as the gas is diatomic) But PV = mRT 1 kg mass 1 V= = = m 3 density 4 kg/m 3 4 P = 8 × 10 4 N/m 2 . Root mean square velocities are different from lateral translation.1 7 5 -1 - 1 5 3 2 = 5 + 3 g m - 1 2 2 ( ) ( ) (b) : Molar heat capacity = Molar mass × specific heat capacity So. or 9.1 7 - 1 3 5 (Q C = (g R . (4 ´ 32 R ) + ( 12 ´ 5 2 R) 29 (a) : For mixture of gases. the molar heat capacities at constant pressure and constant volume will be 28C P and 28C V respectively 2 8 or g - 1 = 2 m or C P . f where CV = R 2 1 2 mv = D U 2 1 2 m mv = C DT 2 M V Mv 2 DT = 2 CV 4. 2 3 5 or g . 5. 7. n1CV1 + n2 C V 2 n1 + n2 For helium. .1) ) 2.3 T0 ÷ = 0 è ø è ø or 5T f – 5T 0 + 3T f – 7T 0 = 0 or 8T f = 12T 0 (4 ´ 52 R ) + ( 12 ´ 7 2 R ) = 47 = 1. (b) : The thermal energy or internal energy is U = 5 m RT for 2 diatomic gases.1 g 2 - 1 g m . n 2 = 32 2 For mixture of gases.CV = R . t) of a wave on a string is given by 2 y ( x.2 × 10 11 N/m 2 respectively? (a) 770 Hz (b) 188. Their mean position is separated by distance X 0 (X 0 > A).729 6. If the maximum separation between them is (X 0 + A). The amplitude of a damped oscillator decreases to 0. with b as the constant of proportionality. 3. When the mass M passes through its mean position then a smaller mass m is placed over it and both of them move together with amplitude A 2 .50(m) û ëê The tension in the string is (a) 6.04(s) 0.2 Hz (d) 200. y = 0.5 N (d) 0.04 kg m –1 is given by t x ù . executes SHM with a amplitude A 1 . An ideal gas enclosed in a vertical cylindrical container supports a freely moving piston of mass M.0 N (c) 12. The tension in it produces an elastic strain of 1%.9 times its original magnitude in 5 s. has a fundamental frequency.44 JEE MAIN CHAPTERWISE EXPLORER CHAPTER OSCILLATIONS AND WAVES 10 1. The tube is dipped vertically in water so that half of it is in water. If x. the velocity and the acceleration of a particle executing simple harmonic motion of time period T. (c) (d) 0.5 N (2010) ( ) 10. the phase difference between their motion is 1/ 2 (2011) The transverse displacement y(x. f in air. then. (2013) 2. which of the following does not change with time? (a) a 2 T 2 + 4p 2 v 2 (b) aT/x (c) aT + 2pv (d) aT/v (2009) .02 (m) sin é 2p ú 0. v and a denote the displacement.( ax + bt This represents a 2 + 2 ab xt ) (a) wave moving in +xdirection with speed (b) wave moving in –xdirection with speed a b b a (c) standing wave of frequency b 1 (d) standing wave of frequency b 9. 1 b (c) 2 b 7. What is the fundamental frequency of steel if density and elasticity of steel are 7. The ratio of æ A 1 ö çè A ÷ø is 2 (2013) (2012) 5. the volume of the gas is V 0 and its pressure is P 0 . (2011) The equation of a wave on a string of linear mass density 0.6 (b) 0. attached to a horizontal spring.5 Hz (c) 178. Assuming that the system is completely isolated from its surrounding.693 b A cylindrical tube. When the spherical bob of the pendulum suffers a retardation (due to viscous drag) proportional to its velocity. The piston and the cylinder have equal cross sectional area A. p 3 (a) M M + m (c) ( MM + m ) (b) M + m M (d) ( MM+ m ) 1/ 2 2 1 A gP 0 (d) 2 p MV0 If a simple pendulum has significant amplitude (up to a factor of 1/e of original) only in the period between t = 0 s to t = t s. The fundamental frequency of the aircolumn is now f 3 f (a) (b) (c) 2f (d) f 2 4 (2012) p 4 (d) p 6 A mass M. open at both ends.5 m is made of steel. The piston is slightly displaced from the equilibrium position and released.7 (c) 0. the piston executes a simple harmonic motion with frequency 1 AgP 0 1 MV 0 (a) 2 p Ag P (b) 2 p V M 0 0 1 V0 MP 0 (c) 2 p A2 g 4. Two particles are executing simple harmonic motion of the same amplitude A and frequency w along the xaxis.5 Hz (2013) (a) b p 2 (a) 8. then t may be called the average life of the pendulum. In another 10 s it will decrease to a times its original magnitude where a equals (a) 0.25 N (b) 4.7 × 10 3 kg/m 3 and 2. t ) = e . When the piston is in equilibrium. the average life time of the pendulum is (assuming damping is small) in seconds (b) (b) (2011) A sonometer wire of length 1.81 (d) 0. then (a) A = x 0w 2 . 12 6 4 3 (2006) 23. Two springs. the maximum value of v upto which he can hear the whistle is (a) 30 ms –1 (b) 15 2 ms - 1 - 1 (c) 15 / 2 ms (d) 15 ms –1 .50 p.1 s. A whistle producing sound waves of frequencies 9500 Hz and above is approaching a stationary person with speed v ms –1 . Then. The bob of a simple pendulum is a spherical hollow ball filled with water.08 2. t) = 0.0 0. is 4. b = (b) a = 25. the lowest resonant frequency for this string is (a) 10.45 Oscillations and Waves 11. b = p 2. d = p/4 (d) A = x 0w 2 . If a simple harmonic motion is represented by its time period is (a) 2pa (b) 2p a (c) 2p/a d 2 x dt 2 + ax = 0 . The average kinetic energy during its motion from the position of equilibrium to the end is (a) 2p 2 m a 2 u 2 (b) p 2 ma 2u 2 1 2 2 (c) 4 m a u (d) 4p 2 ma 2u 2 (2007) m 18.25 s (b) 0.0 . After what time will its kinetic energy by 75% of the total energy? 1 s 1 s 1 1 (a) (b) (c) s (d) s .05 Hz (d) 1050 Hz. till water is coming out. (2006) w 21. d = 3p/4 (b) A = x 0 .0 s. The amplitude of oscillation is gradually increased.04 1. b= (d) a = (c) a = p p p p (2008) 14. respectively.005 cos(ax – bt). A sound absorber attenuates the sound level by 20 dB. (d) 2p / a (2005) . There are no other resonant frequencies between these two. A coin is placed on a horizontal platform which undergoes vertical simple harmonic motion of angular frequency w. They superpose to give beats.08 m and 2. executing simple harmonic motion with an amplitude 7 mm.4 m/s. At the starting point of the motor cycle there is a stationary electric siren. It is observed to have resonant frequencies of 420 Hz and 315 Hz. If the person can hear frequencies upto a maximum of 10000 Hz. of force k 1 k 2 constants k 1 and k 2 are connected to a mass m as shown. A point mass oscillates along the xaxis according to the law x = x 0 cos (wt – p/4). (2006) 24. u. If the wavelength and the time period of the wave are 0. A plugged hole near the bottom of the oscillating bob gets suddenly unplugged. The intensity decreases by a factor of (a) 100 (b) 1000 (c) 10000 (d) 10 (2007) 20. then a and b in appropriate units are p (a) a = 12.00 p. The speed of sound in oxygen (O 2) at a certain temperature is 460 ms –1 . The maximum velocity of a particle. (2008) 15. A motor cycle starts from rest and accelerates along a straight path at 2 m/s 2 .5 s (c) 0. d = –p/4 (2007) 16. If both k 1 and k 2 are made four times their original values. The velocity of sound in air is 300 ms –1 . (2006) 25. Three sound waves of equal amplitudes have frequencies (u – 1). A wave travelling along the xaxis is described by the equation y(x. a body oscillates simple harmonically with a period of 2 s.0 0. the frequency of oscillation becomes (a) 2f (b) f/2 (c) f/4 (d) 4f (2007) 19. (u + 1). If the acceleration of the particle is written as a = Acos(wt + d).125 s (2007) 17. The time at which the maximum speed first occurs is (a) 0. the time period of oscillation would (a) remain unchanged (b) increase towards a saturation value (c) first increase and then decrease to the original value (d) first decrease and then increase to the original value (2005) 26. The number of beats produced per second will be (a) 4 (b) 3 (c) 2 (d) 1 (2009) 12. The speed of sound in helium (He) at the same temperature will be (assume both gases to be ideal) (a) 330 ms –1 (b) 460 ms –1 (c) 500 ms –1 (d) 650 ms –1 .75 s (d) 0. How far has the motor cycle gone when the driver hears the frequency of the siren at 94% of its value when the motor cycle was at rest? (Speed of sound = 330 ms –1 ). The coin will leave contact with the platform for the first time (a) at the highest position of the platform (b) at the mean position of the platform g (c) for an amplitude of 2 w 2 g (d) for an amplitude of 2 . (a) 49 m (b) 98 m (c) 147 m (d) 196 m (2009) 13. b= .01 s (c) 10 s (d) 0. Starting from the origin. The displacement of an object attached to a spring and executing simple harmonic motion is given by x = 2 × 10 –2 cos pt metre. The frequency of oscillation of the mass is f. During observation.5 Hz (b) 105 Hz (c) 1. A particle of mass m executes simple harmonic motion with amplitude a and frequency u. d = –p/4 (c) A = x 0w 2 . (2006) 22. A string is stretched between fixed points separated by 75 cm. The period of oscillation is (a) 100 s (b) 0. E. (2003) 39. The frequency of the piano string before increasing the tension was (a) (256 + 2) Hz (b) (256 – 2) Hz (c) (256 – 5) Hz (d) (256 + 5) Hz. with a velocity onefifth of the velocity of sound. The beat frequency decreases to 2 beats per second when the tension in the piano string is slightly increased. An external force F (t) proportional to coswt (w ¹ w 0 ) is applied to the oscillator. (2003) 40. then (a) w 1 = w 2 (b) w 1 > w 2 (c) w 1 < w 2 when damping is small and w 1 > w 2 when damping is large (d) w 1 < w 2 (2004) 32. is maximum when x is maximum (d) P. but not simple harmonic motion with a period 2p/w (d) a periodic.) are measured as function of displacement x. (2003) . An observer moves towards a stationary source of sound.w 2 ) (a) w 2 . The potential energy (P. where t is in second and x in meter. The percentage increase in the time period of the pendulum of increased length is (a) 11% (b) 21% (c) 42% (d) 10%. some tape is attached on the prong of the fork 2. executing simple harmonic motion is (a) µ x (b) µ x 2 (c) independent of x (d) µ x 1/2 where x is the displacement from the mean position. then what was the original frequency of fork 2? (a) 196 Hz (b) 204 Hz (c) 200 Hz (d) 202 Hz (2005) 31. The speed of the wave is (a) 2000 m/s (b) 5 m/s (c) 20 m/s (d) 5p m/s.E. (2003) 38. Two particles A and B of equal masses are suspended from two massless springs of spring constants k 1 and k 2 . The bob of a simple pendulum executes simple harmonic motion in water with a period t. (2003) 37. What relationship between t and t 0 is true? (a) t = t 0 (b) t = t 0 /2 (c) t = 2t 0 (d) t = 4t 0 . The total energy of a particle. the ratio of amplitudes of A and B is (a) k1 / k 2 (b) k 2 /k 1 (c) k2 / k 1 (d) k 1 /k 2 . (2004) 36. What is the percentage increase in the apparent frequency? (a) 5% (b) 20% (c) zero (d) 0. 4 beats per second are heard. are equal. Now. is maximum when x = 0. but not simple harmonic motion with a period p/w (2005) 29.) and total energy (T. Two simple harmonic motions are represented pö æ by the equations y1 = 0. When the tuning forks are sounded again. 6 beats per second are heard. (2003) 41. In forced oscillation of a particle the amplitude is maximum for a frequency w 1 of the force. A tuning fork of known frequency 256 Hz makes 5 beats per second with the vibrating string of a piano. The function sin 2 (wt) represents (a) a simple harmonic motion with a period 2p/w (b) a simple harmonic motion with a period p/w (c) a periodic. If the period of oscillation with the two springs in series is T.w 2 0 0 1 m (c) m(w 2 + w 2 ) (d) w 2 + w 2 . The length of a simple pendulum executing simple harmonic motion is increased by 21%. The time displacement of the oscillator will be proportional to m 1 (b) m(w 2 . (2004) 0 0 33. while the energy is maximum for a frequency w 2 of the force. (2004) 34.E. A mass M is suspended from a spring of negligible mass.). during oscillations. then (a) T = t 1 + t 2 (b) T 2 = t 12 + t 22 –1 –1 –1 –2 (c) T = t 1 + t 2 (d) T –2 = t 1–2 (2004) + t 2 . When two tuning forks (fork 1 and fork 2) are sounded simultaneously. is zero when x = 0 (c) K.46 JEE MAIN CHAPTERWISE EXPLORER 27. The displacement y of a particle in a medium can be expressed as: y = 10 –6 sin(100t + 20x + p/4) m.1sin ç 100 pt + ÷ and y 2 = 0.E. the kinetic energy (K. If the mass is increased by m.1cosp t. respectively. Which of the following statement is true? (a) K. while the corresponding period for another spring is t 2. Then the ratio of m/M is (a) 3/5 (b) 25/9 (c) 16/9 (d) 5/3. is maximum when x = 0 (b) T.E. The spring is pulled a little and then released so that the mass executes SHM of time period T. Neglecting frictional force of water and given that the density of the bob is (4/3) × 1000 kg/m 3 .E. (2005) 28. 3 ø è The phase difference of the velocity of particle 1 with respect to the velocity of particle 2 is (a) –p/3 (b) p/6 (c) –p/6 (d) p/3. the time period becomes 5T/3. 35. if the frequency of fork 1 is 200 Hz. while the period of oscillation of the bob is t 0 in air. A particle at the end of a spring executes simple harmonic motion with a period t 1 .E. If the maximum velocities. A particle of mass m is attached to a spring (of spring constant k) and has a natural angular frequency w 0.5% (2005) 30. A body executes simple harmonic motion. The frequency of the unknown fork is (a) 286 cps (b) 292 cps (c) 294 cps (d) 288 cps. (c) (b) (b) (b) (d) (b) (c) (c) (c) 5. A tuning fork arrangement (pair) produces 4 beats/sec with one fork of frequency 288 cps. 44. The frequency u of the alternating source is (a) 50 Hz (b) 100 Hz (c) 200 Hz (d) 25 Hz. (c) (a) (a) (b) (c) (c) (c) (b) (b) 4. 15. 8. The wire passes at its middle point between the poles of a permanent magnet.47 Oscillations and Waves 42. In a simple harmonic oscillator. and is cut into n equal parts. If a spring has time period T. stands up. 23. 29. 18. (2002) 48. the frequency of a tuning fork (a) increases (b) decreases (c) remains same (d) increases or decreases depending on the material. 36. 51. potential energy is minimum (d) both kinetic and potential energies are minimum. A little wax is placed on the unknown fork and it then produces 2 beats/sec. (d) (d) (b) (a) (c) (a) (a) (c) (b) 2. 47. 31. The amplitude of the particle is (a) – 4 (b) 4 (c) 4 2 (d) 8. 25. 45. (2002) 47. 30. 11. (d) (b) (*) (c) (d) (b) (d) (a) (b) 3. A metal wire of linear mass density of 9. 40. (2002) 51. 13. 12. in ms –1 is (a) 300 (b) 600 (c) 1200 (d) 200. The displacement of a particle varies according to the relation x = 4(cospt + sinpt). 19. (2002) 50. Tube A has both ends open while tube B has one end closed. 43. and it vibrates in resonance when carrying an alternating current of frequency u. When temperature increases. A child swinging on a swing in sitting position. 50. 42. 26. 34. 14. 7. Then the equation of the unknown wave is (a) y = asin(wt + kx) (b) y = –asin(wt + kx) (c) y = asin(wt – kx) (d) y = –asin(wt – kx). 33. 32. Length of a string tied to two rigid supports is 40 cm. 27. at the mean position (a) kinetic energy is minimum. A wave y = a sin(wt – kx) on a string meets with another wave producing a node at x = 0. 10. 35. 17. 49. 46. 52. 9. 37. Maximum length (wavelength in cm) of a stationary wave produced on it is (a) 20 (b) 80 (c) 40 (d) 120. (2003) 43. (b) (b) (a) (d) (a) (c) (a) (b) . 39. (2002) Answer Key 1. 38. then the time period of each part will be (a) T n (b) T / n (c) nT (d) T.8 g/m is stretched with a tension of 10 kgwt between two rigid supports 1 metre apart. (2003) 44. potential energy is maximum (b) both kinetic and potential energies are maximum (c) kinetic energy is maximum. 21. 48. 41. 22.2 x + p metre. The speed of the wavemotion. then the time period of the swing will (a) increase (b) decrease (c) reamains same (d) increases if the child is long and decreases if the child is short. (2002) 46. (2003) ( ) 45. otherwise they are identical. The displacement y of a wave travelling in the x direction is given by y = 10-4 sin 600t . 20. 3 where x is expressed in metre and t in second. 24. The ratio of fundamental frequency of tube A and B is (a) 1 : 2 (b) 1 : 4 (c) 2 : 1 (d) 4 : 1. 16. (2002) 49. (d) (a) (b) (b) (b) (b) (c) (b) 6. 28. (2002) 52. Y = 2.48 JEE MAIN CHAPTERWISE EXPLORER 1..9 = e –5b/2m . t = 5 s) its amplitude becomes 0. t = 15 s). u = 2 w 1 gP0 A = 2p 2 p MV0 3.2 Hz 7 4. r = 7. (d) : \ u = 1 T 2 L r A FBD of piston at equilibrium Here tension is due to elasticity of wire DL ù \ T = YA é ë L û P atm A + Mg = P 0 A …(i) FBD of piston when piston is pushed down a distance x é As Y = Stress = TL ù ë Strain AD L û 1 Y DL Hence. . we get u= ( P0 + dP) A .w 2 x dt 2 We get w 2 = gP0 A2 gP0 A 2 or w = MV0 MV0 Frequency. u = 2 L r L Here.. \ f = v 2 l l .5 7.( Patm A + Mg ) = M d 2 x dt 2 …(ii) As the system is completely isolated from its surrounding therefore the change is adiabatic. (c) 5.01 1 2 ´ 1.5 m L Substituting the given values.729 d 2 x = . dV = Ax) V0 Using (i) and (iii) in (ii).2 × 10 11 N/m 2 . it behaves closed l pipe length \ f ¢ = l . For an adiabatic process PVg = constant \ Vg dP + Vg–1 PdV = 0 gPdV or dP = - V gP0 ( Ax ) ( .9) 3 (Using (i)) = 0.7 × 10 3 kg/m 3 DL = 0. \ dP = - 2. .01. 2. (d): When the tube of length l is open at both ends. 2 v v = = f (Using(i)) 2 l l 4 2 ( ) 2 l 2 .. As per question After 5 s (i.e. we get 2 2 gP A 2 gP A 2 M d 2 x = - 0 x or d 2 x = - 0 V 0 MV 0 dt dt Comparing it with standard equation of SHM. T is the tension in the wire and m is the mass per length of the wire As m = rA where r is the density of the material of the wire and A is the area of crosssection of the wire.7 ´ 103 …(iii) = 103 3 2 Hz = 178. When the tube is dipped vertically in water and half of it is in water.9A 0 = A 0 e –b(5)/2m = A 0 e –5b/2m 0.(i) where v is the speed of sound in air.e. its amplitude becomes aA 0 = A 0 e –b(15)/2m = A 0 e –15b/2m a = (e –5b/2m ) 3 = (0. (d) : The amplitude of a damped oscillator at a given instant of time t is given by A = A 0 e –bt/2m where A 0 is its amplitude in the absence of damping. (c) : Fundamental frequency of vibration of wire is u= 1 T 2 L m where L is the length of the wire.(i) After 10 more second (i.2 ´ 1011 ´ 0. b is the damping constant.. L = 1. respectively.5 Wave velocity.5 and each frequency will show 2 beats. Coefficient of t b Wave speed = Coefficient of x = a 9. (b) : The wave travelling along the xaxis is given by y(x. (d) : T 1 = 2 p (m + M ) . The intensity waxes and wanes.vsound = - v obs f f ¢ ö vsound æ . (b) : y ( x. 2 2 2 (19.25 N 2 T or T = mw 2 m k Frequencies Mean Beats u + 1 and u (u + 0. New time period of oscillation 7. linear mass density m = 0. 2a 2 ´ 2 13. Total number of beats = 4 12. t) = f(ax + bt) As there is positive sign between x and t terms. (b) : The source is at rest. a = –w 2 x where w is a constant = 2 1 2 1 11. v = w (2 p / 0.04 0.49 Oscillations and Waves 6. we get w= k ( u -2 u )t cos 2p ( u +2 u ) t. the new system is of mass = (M + m) attached to the spring.005 cos(ax – bt).v obs è f ø 330(0.04 kg m –1 The given equation of a wave is ( ) t x ù y = 0.u = = 98 m..5 and u – 0.80 ms –1 .04 ´ M ..(ii) where T is the tension in the string and m is the linear mass density Equating equations (i) and (ii).4p2 × x Þ aT = - 4 p . law of conservation of linear momentum Mv 1 = (m + M)v 2 M(A 1w 1 ) = (m + M)(A 2w 2 ) ( .(i) .u 2 is modulated by a wave of frequency 1 (rather the 2 difference of frequencies/2).50 û Compare it with the standard wave equation y = Asin(wt – kx) we get.v obs ) \ f ¢ = f sound vsound f ¢ Þ ´ vsound = vsound - v obs f f ¢ Þ ´ vsound . u + 0. The resultant frequency obtained is A 1 m + M = A2 M 8. x y = 2 A cos 2p 2 2 p . u.. the number of beats = u 1 ~ u 2 . y 1 = A cos 2pu 1 t y 2 = A cos 2pu 2 t Superposing.80) \ s = v . (u – 1). v 1 = A 1w 1 and v 2 = A 2w 2 ) T 2 = 2 p or = A 1 (m + M ) w 2 = A2 M w1 ( m M+ M ) ´ TT (Q w = 2Tp and w = 2 Tp ) T= 2 10.1 = . t) = 0. (a) : The given sources of sound produce frequencies.. . w = 2p rad s -1; k = 2 p rad m -1 0.. acceleration.04 0.( a x + bt ) Comparing equation (i) with standard equation y(x. t ) = e . hence wave travel in –x direction.94 – 1) = –vobs Þ vobs = 330 × 0.. 1 2 1 2 u1 + u 2 and this wave 2 u .5) Also v = T m . . (b) : For a simple harmonic motion.02sin éê2 p ú ë 0. T a = .. . (v .(i) k When a mass m is placed on mass M. One should detect three frequencies. For two sources of frequencies u 1 and u 2 .05 ) 2 0.04 ) 2 p ( 0. x T T The period of oscillation T is a constant. 1 2 = 6. (b) 2 p ( 0.04) = m s -1 k (2p / 0. one gets (Using (i) and (ii)) 2 + bt 2 + 2 ab xt ) 2 = e . u and (u + 1). (a) : Here. For a cosine curve (or sine curve).5 1 (u + 1) and (u – 1) u 2 Total number of beats = 4..( ax 2 \ aT is a constant.5) Hz 1 u and u – 1 u – 0. 1 beat and 1 beat per second. Using.(ii) k Consider v 1 is the velocity of mass M passing through mean position and v 2 velocity of mass (m + M) passing through mean position.06 = 19. the observer is moving away from the source. w = angular frequency.00 p .p ÷ö 4 ø è Acceleration a = A cos (wt + d) Velocity v = dx dt æ v = . 0.. Option not given.L 2 = 10 log çè I ÷ø . b = p 14. (a) : Given : x = x0 cos çæ wt . we get 3 p A = x0 w2 . force will be towards mean position and so it will be downwards.0 \ a = 25.50 JEE MAIN CHAPTERWISE EXPLORER 1 2 2 2 ma w sin w t 2 Average kinetic energy < K > 1 = < ma 2 w2 sin 2 wt > 2 2 p Therefore a = k = .. Let us choose f = 0 \ x(t) = acoswt Velocity of a particle v = dx = .. (b) : For a particle to execute simple harmonic motion its displacement at any time t is given by x(t) = a(cos wt + f) where.x0 wsin ç wt .00 = 25. sin pt = 1 p p sin pt = sin pt = or or 2 2 1 t = s = 0. maximum acceleration (aw 2 ) and so the maximum force (maw 2 ) will be at extreme positions. f = phase constant. æ 2 p ö 21.. 1 k eff 1 k1 + k 2 = 2p m 2 p m As k 1 and k 2 are increased four times New frequency. At lowest position. force will be towards mean position and so it will be upwards. (iii) éQ < sin 2 q > = êë k 2 . (i) . (b) : Given : displacement x = 2 × 10 –2 cos pt . 1 2 1 2 Kinetic energy = 2 mv = 2 m ( a w cos w t ) Total energy E = 1 ma 2 w 2 2 . I 2 100 20.5 s. (a) : In the given figure two springs are connected in parallel.08 0. æ I ö æ I ö 19. The value of the velocity given for O 2 is quite high. (b) : During simple harmonic motion.10 log çè I ÷ø 0 0 æ I ö or D L = 10log çæ I 1 ÷ö or 20 dB = 10 log ç 1 ÷ I è I 2 ø è 2 ø I I or 10 2 = 1 or I 2 = 1 .ö = x0 w 2 cos é p + ( wt . 4 16.4 –2 = 10 sec = 0. = g for O 2 = 1 + 2/5 = 1.. l 2 p p \ a= = Þ a = p ´ 100.. (iv) Velocity v = dx = -2 ´ 10-2 p sin p t dt For the first time when v = v max .08 m..a w sin w t dt Kinetic energy of a particle is K = 1 mv2 2 1 ù 2 úû [Q w = 2 pu ] 18. (a) : L 1 = 10 log ç 1 ÷ ; L 2 = 10 log ç 2 ÷ è I 0 ø è I 0 ø æ I1 ö æ I 2 ö \ L1 . At highest position. (*) : v = gP = r = 1 m w 2 a 2 < sin 2 w t > 2 1 1 = m w 2 a 2 æ ö è 2 ø 2 = gRT M 1 2 ma (2 pu ) 2 4 = p 2 ma 2 u 2 . 1 4( k1 + k 2 ) = 2 f 2 p m m k 1 f = f¢= . Therefore the effective spring constant is given by k eff = k 1 + k 2 Frequency of oscillation. (ii) . d = . 22. 3 4 7 * The value of the speed of sound in He should have been 965 m/s and that of O 2. about 320 m/s. (i) (using (i).p ÷ö 4 ø è dv Acceleration a = dt p p 2 æ = . The coin will leave contact will the platform for the first time when m(aw 2 ) ³ mg at the lowest position of the platform. (c) : In vertical simple harmonic motion.00 p.. This is opposite to weight direction of the coin.4; g for He = 1 + 2/3 = 5/3 v 2 æ g He 32 ö =ç ´ ÷ ´ 460 v1 ç 4 g O 2 ÷ è ø 5 1 32 = 460 ´ ´ ´ ´ 5 = 1420m/s. a = amplitude.. 15.. As l = 0.01 sec.x0 w cos wt .04 4 2 p w=bÞ =b Þ p 2.) ù êë è 4ø 4 úû = x0 w2 cos éwt + 3 p ù êë 4 úû Compare (iv) with (ii). or 2 17. (b) : Maximum velocity v m = a w = a ç T ÷ è ø -3 2 pa 22 (7 ´ 10 ) T = = 2 ´ ´ \ vm 7 4. .1 ´ p ) sin p t dt \ \ = (0. (c) : For a pendulum. \ T would first increase and then decrease to the original value. (b) : By Doppler's effect u¢ v s + v O = (where v s is the velocity of sound) u vs = v + (v / 5) 6 = v 5 \ Fractional increase = u ¢ . 10000 300 = 9500 300 - v Þ (300 . centre of gravity is restored to centre. w 1 = w 2 . 29.1ö = æ 6 . 5 30. The centre of gravity of system is at centre of sphere when hole is plugged.w 2 ) .= - .1 ´ p ) cos çæ pt + p ÷ö 2 ø è p p p D f = . a decrease in u 2 increased beats/sec.1 ö = 1 è u ø è5 ø 5 u 100 \ Percentage increase = = 20%. When unplugged. Centre of gravity goes on descending. 0 T= 33. (a) : In case of forced oscillations (i) The amplitude is maximum at resonance \ Natural frequency = Frequency of force = w 1 (ii) The energy is maximum at resonance \Natural frequency = Frequency of force = w 2 \ From (i) and (ii). But this is not given in question If u 2 = 196 Hz. 25. (b) : Let the successive loops formed be p and (p + 1) for frequencies 315 Hz and 420 Hz p T pv \ u = 2l m = 2 l pv ( p + 1) v = 315 Hz and = 420 Hz \ 2 l 2 l ( p + 1) v pv = 420 - 315 or 2l 2 l v 1 ´ v = 105 Þ = 105 Hz or 2l 2 l p = 1 for fundamental mode of vibration of string.cos2 wt 1 cos2 wt = - 2 2 2 It is a periodic motion but it is not SHM \ Angular speed = 2w \ Period T = 2 p p 2 p = = angular speed 2 w w Hence option (d) represents the answer. This is given in the question when beats increase to 6 \ Original frequency of second fork = 196 Hz. (b) : In case of forced oscillations. 3 2 6 2 28. water drains out. w a d pö æ 27. 32.u = æ u ¢ . total energy = 1 2 2 ma w 2 Total energy is independent of x. (c) : Under simple harmonic motion. +4 Hz 204 Hz (u2) – 4 Hz 196 Hz (u 2) 200 Hz (u 1 ) As mass of second fork increases. \ Lowest resonant frequency = 105 Hz. (d) : u = v - v s where v s is the velocity of sound in air. 31. \ Length of pendulum first increases. (d) : y = sin wt = 1 . u 2 decreases. 26. 6 w 6(2 p / T ) 6 ´ 2 p 6 (Kinetic energy) = 23.51 Oscillations and Waves Q or or \ or 75 ( E ) 100 1 75 1 2 2 ma 2 w2 cos 2 wt = ´ ma w 2 100 2 3 cos2 wt = 3 Þ cos wt = = cos p 4 2 6 p wt = 6 p p 2p 1 t = = = = sec.1 ´ 100 p ) cos ç 100 pt + 3 ÷ è ø d v2 = ( y2 ) = ( -0. a decrease in u 2 decreases beats/sec. then decreases to original value. T = 2 p l where l is measured upto g centre of gravity. (a) : Let the two frequencies be u 1 and u 2 u 2 may be either 204 Hz or 196 Hz. (d) : Standard differential equation of SHM is d 2 x + w2 x = 0 dt 2 2 Given equation is d 2 x + ax = 0 dt \ w 2 = a or w= a 2 p 2 p = .w2 1 x is proportional to m( w2 . x = asin(wt + f) where a = \ F0 / m w0 2 . If u 2 = 204 Hz. (c) : v1 = dt ( y1 ) = (0. When the bob becomes empty.v ) = 285 Þ v = 15 m /s . v s u¢ 24. T = 2 p M / k 5 T M + m Finally. 42. beats per second decrease..10 1 = = \ Fractional change = T 10 10 \ Percentage change = 10%. (c) : x = 4(cospt + sinpt) é1 ù 1 = 4 ´ 2 ê cos pt + sin p t ú 2 2 ë û p p é ù or x = 4 2 êsin 4 cos pt + cos 4 sin pt ú ë û p æ ö = 4 2 sin ç pt + ÷ 4 ø è Hence amplitude = 4 2 . vm = a 25 M M + m = 9 k k 9 m + 9 M = 25 M m 16 = . 38. 44. 39. This is not given in the question.52 JEE MAIN CHAPTERWISE EXPLORER 34.s ) l r-s = 261 Hz æ 1 k ö 2 pa 1 = (2 pa ) æç ö÷ = (2 pa ) ç ÷ T T è ø è 2 p m ø 1 10 ´ 9.8 ´ 10 -3 43. (b) : Given wave equation : y = 10-6 sin çæ 100t + 20 x + p ÷ö m 4 ø è Standard equation : y = a sin (wt + kx + f) Compare the two \ w = 100 and k = 20 w = 100 Þ 2 pn = nl = v = 5 \ k 20 2p / l \ v = 5 m/s. This is given in the question. u = 2 l m lr g r t = ´ = t0 g (r . 1 T For vibration of wire. u = 2 l m When tension T increases.8 100 = = 50 Hz. (c) : Initially. (b) : When springs are in series.c. (d) : Let the lengths of pendulum be (100l) and (121l) T ¢ = 121 = 11 \ T 100 10 T ¢ – T 11 . (c) : The possible frequencies of piano are (256 + 5) Hz and (256 – 5) Hz. (i) Due to upthrust of water on the top. (a) : Kinetic energy is maximum at x = 0.s ) \ or ( 4 ´ 1000 / 3 = 2 4000 - 1000 3 ) t = t 0 × 2 = 2t 0 .. 37. \ u = 2 ´ 1 2 9. frequency of vibration of wire become equal to frequency of a.(ii) \ g (r . (a) : At resonance...2 x + p ÷ö m 3 ø è Standard wave equation : y = asin(wt – kx + f) Compare them Angular speed = w = 600 sec –1 Propagation constant = k = 2 m –1 . 3 = 2 p k 5 M M + m ´ 2p = 2 p \ 3 k k or 35. beats/sec increase. or +5 Hz 256 Hz – 5 Hz 36.. Hence frequency of piano = (256 – 5) Hz. its apparent weight decreases upthrust = weight of liquid displaced \ Effective weight = mg – (Vsg) = Vrg – Vsg Vrg¢ = Vg(r – s). (ii) If 251 Hz increases due to tension. t 1 = 2 p m k1 For second spring t 2 = 2p m k2 2 2 æ k + k ö 4 p m 4 p m 2 2 t1 + t2 = + = 4 p 2 m ç 1 2 ÷ \ k1 k2 è k1k2 ø 2 or or é m ( k1 + k 2 ) ù t12 + t 2 2 = ê 2 p k1k 2 úû ë t12 + t2 2 = T 2 .. (c) : Maximum velocity under simple harmonic motion v m = aw \ vm = 251 Hz 1 T For piano string.. or M 9 41. (a) : Given wave equation : y = 10 -4 sin çæ 600t .. where s is density of water ær-sö g¢ = g ç ÷ or è r ø l r t = 2 p l / g ¢ = 2 p ... m m a2 k1 40. Q k m (v m ) A = (v m ) B \ a1 or k1 k a k = a 2 2 Þ 1 = 2 ... u increases (i) If 261 Hz increases... k = k1k 2 k1 + k 2 For first spring. (c) : t0 = 2 p l / g . Consider option (b) Stationary wave : 292 cps 288 cps T ¢ = 2 p m nk 48. (b) : The wax decreases the frequency of unknown fork. 49. Hence unknown fork has frequency 292 cps.53 Oscillations and Waves w 2 pu = = ul = velocity k 2 p / l w 600 = 300 m/sec . T = 2 p l / g . kinetic energy is maximum and potential energy is minimum at mean position. \ velocity = = k 2 45. T = 2 p m k For each piece. 46. It is given that the beats decrease to 2 from 4. The option is not acceptable. When the child stands up. l increases Hence frequency decreases. (c) : In tube A. Wax reduced 292 cps and so beats should decrease. k n n l max = 40 Þ l max = 80 cm. Y is not zero. Y = a sinwt – asinwt = zero This option holds good Option (c) gives Y = 2asin(wt – kx) At x = 0. (b) : Time period will decrease. 2 50. This frequency is ruled out. (b) : When temperature increases. The possible unknown frequencies are (288 + 4)cps and (288 – 4) cps. B AN l A = l 2 N l B = l 4 . 51. l A = 2l AN AN In tube B. 52. Y is not zero Option (d) gives Y = 0 Hence only option (b) holds good. (b) : For a spring. 47. the centre of gravity is shifted upwards and so length of swing decreases. Y = asin(wt – kx) – asin(wt + kx) At x = 0. (b) : + 4 cps – 4 cps 284 cps Wax reduces 284 cps and so beats should increases. l B = 4l v v \ u A = l = 2 l A N l v v u B = = l B 4 l u A 2 \ u = 1 . spring constant = nk \ \ T ¢ = 2 p m ´ 1 = T . (b) : Consider option (a) Stationary wave : Y = asin(wt + kx) + asin(wt – kx) when x = 0. (c) : In a simple harmonic oscillator. It is not given in the question. The disc rotates about an axis perpendicular to its plane and passing through its centre with an angular velocity w. (2013) Potential difference V in volts 11 E E (b) (a) R r R r . 0 Statement 2 : The electric field at a distance r(r < R) from rr the centre of the sphere is 3 e . Statement 2 is the correct explanation of Statement 1. the net force acting on the particle is proportional to 1 1 (a) - y (b) y (c) –y (d) y (2013) 3. (b) Statement 1 is false. the electric field E is plotted as a function of distance from the centre. choose the one that best describes the two statements. Statement 2. at the surface of the sphere and also at a point outside the sphere. are kept at x = – a and x = a q on the xaxis. (b) B R 5. (d) Statement 1 is true. 0 (a) Statement 1 is true. 20 15 10 5 Time t in seconds The figure shows an experimental plot for discharging of a capacitor in an RC circuit. Two charges. The electric potential at infinity is zero. Of the four choices given after the statements. A particle of mass m and charge q0 = is 2 placed at the origin. The electric potential at the point O lying at a distance L from the end A is 6. If we keep both the amount of charge placed on the disc and its angular velocity to be constant and vary the radius of the disc then the variation of the magnetic induction at the centre of the disc will be represented by the figure (a) B R (c) B R R This question has Statement 1 and Statement 2. (c) Statement 1 is true.54 JEE MAIN CHAPTERWISE EXPLORER CHAPTER ELECTROSTATICS 1. As a result of this uniform charge distribution there is a finite value of electric potential at the centre of the sphere. The graph which would correspond to the above will be (2012) (d) B 25 0 50 100 150 200 250 300 7. Q ln 2 (a) 4 p Î L 0 3 Q (c) 4 p Î L 0 4. The time constant t of this circuit lies between (a) 0 and 50 sec (b) 50 sec and 100 sec (c) 100 sec and 150 sec (d) 150 sec and 200 sec (2012) In a uniformly charged sphere of total charge Q and radius R. Statement 2 is true. is true. its potential energy changes by 3 e . Statement 1 : When a charge q is taken from the centre to the qr surface of the sphere. Q ln 2 (b) 8 p Î L 0 Q (d) 4p Î L ln 2 0 A charge Q is uniformly distributed over the surface of non conducting disc of radius R. Two capacitors C 1 and C 2 are charged to 120 V and 200 V respectively. If charge q 0 is given a small displacement (y < < a) along the yaxis. Then (a) 9C 1 = 4C 2 (b) 5C 1 = 3C 2 (c) 3C 1 = 5C 2 (d) 3C 1 + 5C 2 = 0 (2013) 2. Statement 2 is not the correct explanation of Statement 1. (2012) A charge Q is uniformly distributed over a long rod AB of length L as shown in the figure. each equal to q. It is found that by connecting them together the potential on each one can be made zero. An insulating solid sphere of radius R has a uniformly positive charge density r. Statement 2 is false. Statement 2 is true. As a result of this rotation a magnetic field of induction B is obtained at the centre of the disc. 9. Which of the following graphs most closely represents the electric field E(r) produced by the shell in the range 0 £ r < ¥.25 pF (b) 1. Two identical charged spheres are suspended by strings of equal lengths. A charge Q is placed at each of the opposite corners of a square. Let there be a spherically symmetric charge distribution with 5 r charge density varying as r(r ) = r0 - upto r = R. One of the dielectrics has dielectric constant k 1 = 3 and thickness d/3 while the other one has dielectric constant k 2 = 6 and thickness 2d/3. then the Q/q equals 1 (a) - 2 2 (b) –1 (c) 1 (d) - 2 (2009) r E(r ) (c) (2008) (d) O (2010) R R r O R r 18. Let P( r ) = 0 (2010) 0 ^ j 12.6 g cm –3 . As a result the charges approach each other with a velocity v. (d) Statement1 is false.24 × 10 –16 J (2009) E (c) (d) R r R r (2012) 8.5 pF. Statement1: For a charged particle moving from point P to point Q. If the net electrical force on Q is zero. When suspended in a liquid of density 0.8 g cm –3 .2 2 j (d) . (2008) .R 0 0 ( (c) ( ) r r 5 r 4e ( 3 R ) 0 0 ) 16. the angle remains the same.60 × 10 –17 J –16 (c) –2. The charge begins to leak from both the spheres at a constant rate. the magnitude of electric field is Q Qr 1 2 Qr 1 2 (a) 0 (b) 4 pe r 2 (c) (d) 3 pe 0 R 4 0 1 4 pe 0 R 4 (2009) 15. Two identical charged spheres suspended from a common point by two massless strings of length l are initially a distance d(d < < l) apart because of their mutual repulsion. A thin spherical shell of radius R has charge Q spread uniformly over its surface. The electric field at a distance r (r < R) from the origin is given by r0 r 5 r 4 pr0 r 5 r (a) 3e 4 . Statement2 is true. The strings make an angle of 30° with each other. Then the charge density inside the ball is (a) –24pae 0 r (b) –6ae 0 r (c) –24pae 0 (d) –6ae 0 (2011) 10. The net r O field E at the centre O is ^ ^ q q j (b) (a) 2 p 2 e r 2 j 2 2 4 p e r 0 0 ^ ^ q q (c) . (c) Statement1 is true. The separation between its plates is d.24 × 10 J (d) 2. b are constants.R (b) 3e 3 . Capacitance of the capacitor is now (a) 20. the dielectric constant of the liquid is (a) 1 (b) 4 (c) 3 (d) 2 (2010) 11. (2009) 17. This question contains Statement1 and Statement2.2 2 j 4 p e 0 r 2 p e 0 r E(r ) E(r ) (a) (b) O R r O E(r ) ^ i 13. Of the four choices given after the statements. A charge q is placed at each of the other two corners. The space between the plates is now filled with two dielectrics. Statement2 is true; Statement2 is the correct explanation of Statement1. The work done in moving 100 electrons from P to Q is (a) –9. Two points P and Q are maintained at the potentials of 10 V and –4 V respectively. and 4 R r(r) = 0 for r > R. A thin semicircular ring of radius r has a positive charge q distributed uniformly over it. choose the one that best describes the two statements. For a point ‘p’ inside the sphere at distance r 1 from the centre of the sphere. where r is the distance from the origin. A parallel plate capacitor with air between the plates has a capacitance of 9 pF. Statement2 is true; Statement2 is not the correct explanation of Statement1. Statement2 is false (b) Statement1 is true. (a) Statement1 is true. the net work done by an electrostatic field on the particle is independent of the path connecting point P to point Q. Statement2: The net work done by a conservative force on an object moving along a closed loop is zero.55 Electrostatics E 14.60 × 10 –17 J (b) 9.8 pF (c) 45 pF (d) 40. Then as a function of distance x between them (a) v µ x –1/2 (b) v µ x –1 (c) v µ x 1/2 (d) v µ x (2011) The electrostatic potential inside a charged spherical ball is given by f = ar 2 + b where r is the distance from the centre; a. where r is the distance from the centre of the shell? ( (d) ) 4 r r 5 r 3e ( 4 R ) Q r be the charge density distribution for a p R 4 solid sphere of radius R and total charge Q. If density of the material of the sphere is 1. A electric dipole is placed at an angle of 30º to a nonuniform electric field. The potential difference between the centers of the two rings is ù Q é 1 1 ú (a) zero (b) 4 pe ê R ê ú 0 ë R 2 + d 2 û QR (c) 4 pe 0 d 2 ù Q é 1 1 ú (d) 2 pe ê R 2 2 0 ê R + d úû ë (2005) 30. Two spherical conductors A and B of radii 1 mm and 2 mm are separated by a distance of 5 cm and are uniformly charged. then (a) E changes. The potential at a point x (measured in hm) due to some charges situated on the xaxis is given by V(x) = 20/(x 2 – 4) volt The electric field E at x = 4 mm is given by (a) (10/9) volt/mm and in the +ve x direction (b) (5/3) volt/mm and in the –ve x direction (c) (5/3) volt/mm and in the +ve x direction (d) (10/9) volt/mm in the –ve x direction (2007) 21. plate 2 is at a higher potential).65 × 10 6 m/s 12 (c) 7. The surface charge + q density s of the sheet is proportional to + S (a) sinq + (b) tanq + + (c) cosq B (d) cotq (2005) . If the spheres are connected by a conducting wire then in equilibrium condition. If the charges on A and B are interchanged with those on r D and C respectively. A parallel plate capacitor is made by stacking n equally spaced plates connected alternatively. Two thin wire rings each having a radius R are placed at a distance d apart with their axes coinciding. The potential situated at difference between the points A and B will be (a) 4. (2006) 26.56 JEE MAIN CHAPTERWISE EXPLORER 19. Charges are placed on r the vertices of a square as shown.02 × 10 m/s (d) 1. A parallel plate condenser with a dielectric of dielectric constant K between the plates has a capacity C and is charged to a potential V volt. 0) of X – Y coordinate system.1 m and can be treated as infinitely large. An electron is released from rest on the inner surface of plate 1. An electric charge 10 –3 mC is placed at the origin (0. The dielectric slab is slowly removed from between the plates and then reinserted. If the temperature of the block is raised by DT.0) respectively. The dipole will experience (a) a torque only (b) a translational force only in the direction of the field (c) a translational force only in a direction normal to the direction of the field (d) a torque as well as a translational force. The ratio of the energy stored in the capacitor and the work done by the battery will be (a) 1/2 (b) 1 (c) 2 (d) 1/4 (2007) 23. V changes r (c) both E and V change r (d) E and V remain unchanged q q A B D – q C – q (2007) 22. (i. The location of a point on the x axis at which the net electric field due to these two point charges is zero is (a) 8L (b) 4L (c) 2L (d) L/4 (2005) 31.11 × 10 –31 kg) (a) 32 × 10 –19 m/s (b) 2. the ratio of the magnitude of the electric fields at the surface of spheres A and B is (a) 1 : 4 (b) 4 : 1 (c) 1 : 2 (d) 2 : 1. A fully charged capacitor has a capacitance C. The Y 0. What is its speed when it hits plate 2? (e = 1. m e = 9. The net work done by the system in this process is 1 (a) zero (b) 2 (K – 1) CV 2 CV 2 ( K - 1) (c) (d) (K – 1) CV 2 (2007) K 20. It is discharged through a small coil of resistance wire embedded in a thermally insulated block of specific heat capacity s and mass m. The charges on the two rings are +Q and –Q.6 × 10 –19 C.87 × 10 6 m/s. If the capacitance between any two adjacent plates is C then the resultant capacitance is (a) C (b) nC (c) (n – 1)C (d) (n + 1)C (2005) 29. (2006) 27.e. (2006) 25.5 volt (b) 9 volt (c) zero (d) 2 volt (2007) ( ) 24. which makes an angle q with + + a large charged conducting sheet P. as P + shown in the figure. A charged ball B hangs from a silk + thread S.1 m X 1 2 plates are separated by d = 0. Let E be the electric field and V the potential at the centre. Two insulating plates are both uniformaly charged in such a way that the potential difference between them is V 2 – V 1 = 20 V. 2 and (2. Two points A and B are 2. the potential difference V across the capacitance is msD T 2msD T (b) (a) C C (c) 2mC D T s (d) mC D T s (2005) 28. V remains unchanged r (b) E remains unchanged. A battery is used to charge a parallel plate capacitor till the potential difference between the plates becomes equal to the electromotive force of the battery. Two point charges +8q and –2q are located at x = 0 and x = L respectively. (a) (d) (c) (a) (b) (d) (a) 1 2 (d) 2 n CV . 13. (2004) 34. (2003) 40. 23. with a speed v. 28. 32. If q were given a speed 2v. 40. (c) (b) (a) (a) (b) (b) (c) (*) 2. the closest distances of approach would be (a) r (b) 2r (c) r/2 (d) r/4. then brought in contact with C and finally removed away from both. (b) (c) (b) (c) (d) (b) (d) 6. then the energy stored is equal to y –q3 b2 2 Q (a) 4 pe R 0 4.2 cos q a q 2 q 3 + sin q b 2 a 2 a q 2 q 3 + 2 cos q a q 2Q 2 q (b) 4pe R . 41. (2002) 41. 36. (d) (d) (c) (a) (c) (d) (a) 5. (2003) 37.(1 + 2 2) (b) (1 + 2 2) 4 4 Q (c) . If the system is in equilibrium the value of q is Q Q (a) . (c) (d) (d) (d) (c) (d) (b) . 19. A thin spherical conducting shell of radius R has a charge q. A sheet of aluminium foil of negligible thickness is introduced between the plates of a capacitor. 4 pe 0 R 38. Two spherical conductors B and C having equal radii and carrying equal charges in them repel each other with a force F when kept apart at some distance. 21. (2002) Answer Key 1. 14. 31.2 sin q b a (2003) 36. Capacitance (in F) of a spherical conductor with radius 1 m is (a) 1. 35. 11. 30. +q 2 and –q 3 are placed as shown in the figure. 10. A charged particle q is shot towards another charged particle Q which is fixed. 2 J of work is done. 20. If the electric flux entering and leaving an enclosed surface respectively is f 1 and f 2. 22. (2004) 35. 15. the electric charge inside the surface will be (a) (f 2 – f 1 )e 0 (b) (f 1 + f 2)/e 0 (c) (f 2 – f 1 )/e 0 (d) (f 1 + f 2 )e 0 . It approaches Q upto a closest distance r and then returns. A charged particle q is placed at the centre O of cube of D O length L (ABCDEFGH).5 V. (2003) 39. (b) (a) (d) (a) (d) (b) (b) (a) 3. 38. The x component of the force on –q 1 is proportional to (a) (b) (c) q 2 q 3 . Three charges –q 1 .4 pe R 0 0 2 Q q (c) 4 pe R + 4 pe R 0 0 (d) (q + Q ) 2 .(1 + 2 2) 2 (d) Q (1 + 2 2) . (2003) 42. 12. 18. The electrostatic potential at a point P at a distance R/2 from the centre of the shell is 1 nCV 2 (c) CV 2 2 E 43.1 V (b) 8 V (c) 2 V (d) 0. (2002) (a) CV (b) b +q2 x –q1 b2 q 2 q 3 (d) 2 . Another charge Q is placed at the centre of the shell. 2 (2004) 33. If there are n capacitors in parallel connected to V volt source. (2002) F C q G B L (2002) 44.1 × 10 –10 (b) 10 –6 –9 (c) 9 × 10 (d) 10 –3 . The new force of repulsion between B and C is (a) F/4 (b) 3F/4 (c) F/8 (d) 3F/8. A third spherical conductor having same radius as that of B but uncharged is brought in contact with B. 26. 9. 16. 27. 7. 17. The capacitance of the capacitor (a) decreases (b) remains unchanged (c) becomes infinite (d) increases. If a charge q is placed at the centre of the line joining two equal charges Q such that the system is in equilibrium then the value of q is (a) Q/2 (b) –Q/2 (c) Q/4 (d) –Q/4. On moving a charge of 20 coulomb by 2 cm. then the potential difference between the points is (a) 0. q Another same charge q is H placed at a distance L from O. Then the electric flux A through ABCD is (a) q/4pe 0 L (b) zero (c) q/2pe 0L (d) q/3pe 0L. 8. The work done in placing a charge of 8 × 10 –18 coulomb on a condenser of capacity 100 microfarad is (a) 16 × 10 –32 joule (b) 3. 29. 42. Four charges equal to –Q are placed at the four corners of a square and a charge q is at its centre. 25.1 × 10 –26 joule –10 (c) 4 × 10 joule (d) 32 × 10 –32 joule. 24. 43. 44. 37. 34.57 Electrostatics 32. 39. 33. 37 × 25 V = 9. At t = t. Hence. (d): Consider a elementary ring of radius r and thickness dr of a disc as shown in figure. dQ = dV = 1 dQ 1 Q = dx 4p Î0 x 4 pÎ 0 Lx m 0Qw R m Qw R m Q w dr = 0 2 = 0 2 ò 2 pR 2pR 0 2 p R Since.58 JEE MAIN CHAPTERWISE EXPLORER 1. ( ) q When a particle of mass m and charge q0 = placed is at 2 the origin is given a small displacement along the yaxis. 5. after connection 120C 1 = 200C 2 6C 1 = 10C 2 3C 1 = 5C 2 2. R 2 r VS = 1 3 e 0 When a charge q is taken from the centre to the surface. Potential at O due to the rod is 2 L 1 Q dx V = ò dV = ò 4 pÎ L 0 Lx Q ln2 2 L 1 Q = [ln x ] L = 4 pÎ0 L 4 p Î 0 L 4. R 2 r VC = 2 e 0 Potential at the surface of the sphere. (c) : During discharging of a capacitor V = V 0 e –t/t where t is the time constant of RC circuit.37 V 0 e From the graph. .25 V This voltage occurs at time lies between 100 sec and 500 sec.VS ) q = ç q = 1 ÷ è 2e0 3 e0 ø 6 e 0 Statement 1 is false. Statement 2 is true. (c) : For potential to be made zero. (a) : Consider a small element of length dx at a distance x from O. then the situation is shown in the figure. Q dx L Potential at O due to the element is Charge on the element. Q 2 Qr dq = (2prdr ) = 2 dr p R 2 R Current due to rotation of charge on ring is O r dr R dqw Qrw dr = 2 p p R 2 Magnetic field at the centre due to the ring element m I m Qrw dr m Qw dr dB = 0 = 0 2 = 0 2 2 r 2 pR r 2 p R Magnetic field at the centre due to the whole disc I = B = ò dB = By symmetry. t = 0. V 0 = 25 V \ V = 0. Q and w are constants \ B µ 1 R Hence variation of B with R should be a rectangular hyperbola as represented in option (d). the change in potential energy is æ R 2r 1 R 2r ö R 2 rq DU = (VC . (b): Potential at the centre of the sphere. 6. Charge on the ring. time constant t of this circuit lies between 100 sec and 150 sec. \ The net force acting on the particle is qq 0 y Fnet = 2 F cos q = 2 1 2 2 4 p Î0 ( y + a 2 ) ( y 2 + a 2 ) ( ) q q y 2 æQ q = q (Given) ö 2 = ø 4p Î0 ( y 2 + a 2 )2 ( y 2 + a 2 ) è 0 2 q 2 y 1 = 2 4 p Î0 ( y + a 2 ) 3/2 As y < < a q 2 y \ F net = 1 4 p Î 0 a 3 or F net µ y 3. (b): The situation is as shown in the figure. V V = 0 = 0. the components of forces on the particle of charge q 0 due to charges at A and B along xaxis will cancel each other where along yaxis will add up. r r q inside ò E × dS = Ñ e 0 q or . 8..Rx ) 4 px dx 2 ) 3 r = 4 pr0 ò 5 x 2 . q¢ = q From equations (iv) and (iii). we get K= K= 1 1 .. 9. O l q q l T cosq q T A Tsinq B F F mg x 2 T C F d mg \ T cosq = mg 2 1 q 4 pe 0 d 2 q 2 1 \ tan q = 4 pe 0 d 2 mg When charge begins to leak from both the spheres at a constant rate. (d) : E q R r (For r > R ) The variation of E with distance r from the centre is as shown adjacent figure. i.(ii) Dividing equation (ii) by (i). dV = 4px 2 dx Let us draw a Gaussian surface of radius r(r < R) as shown in the figure above.6 = = 2 (r .(iii) mg When the balls are suspended in a liquid of density s and dielectric constant K. the forces acting on each ball are (i) Tension T (ii) Weight mg (iii) Electrostatic force of repulsion F For its equilibrium along vertical. (b): For uniformly charged sphere 1 Qr E= (For r < R ) 4 pe 0 R 3 1 Q E= (For r = R ) 4 pe 0 R 2 1 Q 4 pe 0 r 2 q ° 30 E= 10. then q 2 1 tan q = 4 pe 0 x 2 mg 1 q 2 x x = Q tan q = 2l 4 pe 0 x 2 mg 2 l 2 x q or µ 2 2 l x or q 2 µ x 3 Þ q µ x 3/2 dq 3 1/2 dx µ x dt 2 dt and T sin q = F = ( ( or v µ x -1/ 2 Q ) ) dq = constant dt T T + + mg mg F Initially. (d) : f = ar 2 + b .(s / r)] .d f = - 2 ar dr According to Gauss’s theorem..(i) O R Gaussian surface r (Using (i)) Volume of the shell. Tcosq = mg .s) (1.ù é As V = ù ú ëê ú r û ëê r û For equilibrium of balls..s r ( ) r 1.. the electrostatic force will become (1/K) times.(i) and along horizontal.e.. Tsinq = F . (c) : Consider a thin spherical shell of radius x and thickness dx as shown in the figure. we get tan q = F .(iv) According to given problem... Total charge enclosed inside the Gaussian surface is r r Qin = ò rdV = ò r0 0 0 ( ( 5 4 . E = dx x ..6 - 0. F tan q¢ = F ¢ = mg ¢ Kmg [1 ..2ar 4 pr 2 = inside e 0 3 q inside = – 8e 0apr Charge density inside the ball is q rinside = inside 4 p r 3 3 -8 e 0 a pr 3 \ rinside = 4 p r 3 3 rinside = -6ae 0 Electric field. F¢ = (F/K) while weight mg¢ = mg – Upthrust = mg – Vsg [As Upthrust = Vsg] s m mg ¢ = mg é1 .x dx R 0 4 . (a) : Figure shows equilibrium positions of the two sphere.59 Electrostatics 7.8) 11. ò dE cos q = 0 The net electric field at O is p r p ^ lrd q sin q( . (c) : Work done = potential difference × charge = (VB – VA) × q.. p R The electric field at the point p distant r 1 from the centre. R Q E = outside shell and zero inside..... C = 0 d e0 A × 3 C1 = = 9 C ; d / 3 e A × 6 C2 = 0 = 9 C 2d / 3 18.. Qq - Þ E = r 1 r 1 1 Qr × 4 pr 2 dr e 0 0 ò p R 4 R Qr 1 2 .... C r k 1 = 3 k 2 = 6 d/3 2d /3 .. If the loop is completed.. 4 pe 0 R 4 16.j ^ ) E = ò dE sin q( . • – 4 V P Q Work done in moving 100e – from P to Q. (d) : +10 V • .. (c) : If the charge density.ú ë 3 4 ëê 3 Rû R û According to Gauss’s law Q E 4 pr 2 = in e 0 pr r 4 ù 0 é 5 3 E 4 pr 2 = r e 0 ëê 3 R ûú = E = pr0 r 3 é 5 r ù 4 pr 2 e 0 ë 3 R û E = r0 r é 5 r ù 4e 0 ë 3 R û 15.. according to Gauss’s theorem is Q E∙4pr 1 2 = charge enclosed/e 0 E × 4 pr1 2 = 12.ú = pr0 ê r 3 .60 JEE MAIN CHAPTERWISE EXPLORER 14. Force of repulsion.cos q] 0 p j 2 2 4 p 2 e 0 r 2 0 4 p e 0 r q ^ =j 2 p 2 e 0 r 2 ( ) 13. r = Q 4 r . Inside the shell. 17... = 9C ´ 9C = 9 ´ C or 9 ´ 9 ´ 10-12 F 18C 2 2 Þ C total = 40.. W = (100e – )(V Q – V P) = (–100 × 1... Electrostatic force is a conservative force. No net work is done as the initial and final potentials are the same. r é 5 é 5 x 4 ù r 4 ù = 4 pr0 ê x 3 = 4 pr0 ê r 3 ú ë 12 ë 12 4 R úû 4 R û 0 4 pr0 é 5 3 r 4 ù é 5 r 4 ù r . (b) : The electric field for a E uniformly charged spherical shell is given in the figure. B dqr \ Charge on the element....24 × 10 –16 J. (d) : Linear charge density.5 pF.. (d) : C = \ C total = C1C 2 C1 + C2 as they are in series... (Work done in moving 100 negative charges from the positive to the negative potential)... the field is zero and it is maximum at the surface O and then decreases A B µ 1/r 2 .. VA and VB only depend on the initial and final positions and not on the path.. l = Þ E × 4 pr1 2 = q p r Consider a small element AB ^ j of length dl subtending an angle dq at the centre O as A dl shown in the figure.j ) = ò 2 0 4 pe 0 r 0 p qr sin qd q ^ q =-ò j Q l = 2 3 pr 0 4 p e 0 r p q q sin qd q ^ ^ =-ò j = [ . dEcosq q dq = ldl ^ i q O dl = lrdq Q d q = dE dEsinq r The electric field at the centre O due to the charge element is ( dE = ) 1 dq l rd q = 4 pe 0 r 2 4 pe 0 r 2 Resolve dE into two rectangular components By symmetry... 4 pe 0 × r 2 e 0 A = 9 ´ 10-12 F d e kA With dielectric. VA – VA = 0.6 × 10 –19 )(–14 V) = 2. (a) : The force of repulsion by Q is + q – Q cancelled by the resultant attracting A force due to q – and q – at A and B. Both the statements are true but A B statement2 is not the reason for statement1. B – 2 + q Q2 1 1 Q Q F = = × 2 2 2 4pe 0 (a + a ) 4 pe 0 2 a Total force of attraction along the diagonal (taking cosq components) ì Qq 2 ü Qq 1 Qq 1 1 = × + = í ý 4 pe 0 a 2 2 a 2 2 4 pe 0 î a 2 þ { Q 2 Qq 2 Þ = 2 2 a 2 a 1 rdV e 0 ò } Q 2 Þ = - 2 2 (a ). dV = . (b) : Energy of capacitor = Heat energy of block 1 2 CV = ms D T \ 2 2 ms D T V = . Energy stored in the capacitor U = 1 CE 2 2 1 CE 2 U = 2 \ = 1 . or C Q 1 Q .6 ´ 10-19 ´ 20 = m 9. v = 27. Therefore after it is taken out.d æç 20 ö÷ = 40 x dx dx è x 2 . Total work done = zero. \ Field is affected but not the potential.61 Electrostatics 19. 2 144 9 [16 - 4] 21. = V R A or or 2 e ´ DV 2 ´ 1. If they are exchanged. (b) : An electron on plate 1 has electrostatic potential energy. (d) : V A = 23. E B R A 1 A r 1 (4 pe0 R A ) V V RB A 1 10 -3 ´ 10 -6 = 4 pe 2 0 V B = = Y (0.4) 2 At x = 4 mm \ \ + Q X (2. 2 ) d B Potential at point B is 2 4 pe 0 RA E A R B 2 = = . \ Kinetic energy = Electrostatic potential energy 1 2 mv = e D V or 2 Positive sign indicate E is +ve x direction. the energy U i = q 2 = 2 KC Once is taken out. 22. charge is redistributed and the spheres attain a common potential V. (c) : rr1 = 2 iˆ + 2 ˆ j 2 1 ´ C A V 28. the dipole will experience a torque as well as a translational force. In the case of potential.V B = = – Q 1 ´ Q é 2 2 ù ê 2 4 pe0 ê R R + d 2 úú ë û Q é 1 1 ù 2 pe0 êê R R 2 + d 2 úú . as it is D a scalar. If a dielectric slab is slowly introduced. When it moves.65 × 10 6 m/s. ë û 30. potential energy is converted into kinetic energy. (d) : In a nonuniform electric field.4 ø ( x 2 . (a) : The potential energy of a charged capacitor 24. q B C – q r r2 = 2iˆ + 0 ˆ j r or | r2 | = r2 = 2 ( \ 1 q 1 10–3 ´10 -6 = 4 pe 0 r2 4pe0 2 V A – V B = 0. (a) : “Unit positive charge” will be repelled q by A and B and attracted by – q and – q A downwards in the same direction. (a) : Given : Potential V ( x ) = 20 x - 4 E= 40 ´ 4 160 10 = = V/m m.1 4 pe0 R 4 pe0 R 2 + d 2 Q 1 ( -Q ) 1 V B = + 4 pe0 R 4 pe0 R 2 + d 2 B 2 2 R + d R 2 .11 ´ 10 -31 v = 2. W CE 2 2 2 E A = 26. 0) \ V A . (c) : Resultant intensity = 0 O A B – 2q + 8q L d . the potential energy come back to the old value. 20. they cancel each other whatever – q may be their position. 2 q 2 C where U i is the initial potential energy. the direction of the field will be opposite. again the energy increases to the old value. (a) : Let E be emf of the battery Work done by the battery W = CE 2 ( 2 ) + ( 2 ) or 1 Q A 2 4 pe 0 R A 25. (d) : When the spherical conductors are connected by a conducting wire. (c) : n plates connected alternately give rise to (n – 1) capacitors connected in parallel \ Resultant capacitance = (n – 1)C. r | r1 | = r1 = Intensity E A = \ 2 Electric field E = . 0) r 2 1 q 4 pe 0 r1 Similarly E B = = 2 Potential at point A is V A = 4 pe0 R A2 29. . mv = .q1 )( .(i) when the third equal conductor touches B. the charge of B is shared equally between them q \ Charge on B = = charge on third conductor.... 1 4 r 4 8q 1 1 2 q = 0 4 pe0 ( L + d ) 2 4 pe0 d 2 or or \ (L + d) 2 = 4d 2 d = L Distance from origin = 2L.(i) 2 r F 1 –q 1 (90° – q ) q q 2 x F 2 Resolved part of F 2 along q 1 q 2 = F 2 cos (90° –q) = q1q 3 sin q \ 4 pe 0 a 2 along (q 1 q 2 ) Total force on (–q 1 ) é q q q q sin q ù = ê 1 2 2 + 1 3 2 ú along x axis 4 pe0 a ûú ëê 4 pe 0 b \ é q 2 q 3 ù xcomponent of force µ ê 2 + 2 sin q ú .. è ø \ Charge on q T sinq . (d) : Energy of condenser = 2 -18 2 1 Q 1 (8 ´ 10 ) = ´ = 32 ´ 10-32 J 2 C 2 (100 ´ 10 -6 ) .. 4 pe0 d 2 è 2 4 ø 8 F 2 a –Q F 1 3 q = Charge of third conductor 4 New force between B and C C= 32.q 3 ) 4 pe 0 a 2 0 F 2 makes an angle of (90° – q) with (q 1 q 2 ) y Distance CA = 2 a –q 3 2 a a = 2 2 For equilibrium. F 3 45° .(ii) –Q B A a D –Q E F 4 q a a C –Q ... a ëb û 36. (d) : Initially.q1q 2 35..2 ´ 1 = 0 2 a / 2 2 a 2 2a2 or é ù Q ê1 + 1 ú = q 2 2 2 ë û or q= Q é 2 2 + 1 ù Q = (1 + 2 2 )... 2 æqö Now this third conductor with charge ç ÷ touches C. 2 m (2 v ) = r 1 \ From (i) and (ii) 1 = r 1 Þ r = r ... (d) : Energy is conserved in the phenomenon 1 2 kqQ Initially. consider forces along DA and equate the resultant to zero 1 Q ´Q 1 Q ´ Q + cos 45 ° \ 4 pe0 ( DA)2 4 pe 0 (CA ) 2 q Distance EA = 1 Q ´ q cos 45° = 0 4 pe 0 ( EA) 2 or Q Q q + ´ 1 .. 2 êë 2 2 úû 4 33.. their è 2 ø q ö æ total charge ç q + 2 ÷ is equally shared between them... tan q = \ = 1 æ q 3 q ö 3 ç ´ ÷ = F ....62 JEE MAIN CHAPTERWISE EXPLORER kqQ 1 2 Finally. (b) : Force on (–q 1 ) due to q 2 = \ F 1 = q1q 2 4 pe 0 b 2 4 pe0 b 2 along (q 1 q 2 ) Force on (–q 1 ) due to ( . F 2 ....q 3 ) = q1q 3 F 2 = 4 pe a 2 as shown ( ... 2 1 q 34. (b) : Tsinq = sq/e 0 T cosq T qs e 0 mg T cosq = mg \ \ sq e0 mg s is proportional to tanq. (b) : Consider the four forces F 1 ... F 3 and F 4 acting on charge (–Q) placed at A. F = 4 pe 2 0 d 31..... C ¢ = 0 ( d - t ) d 39. it is 6 e . e A With air as dielectric. e0 The flux enters the enclosure if one has a negative charge (–q 2 ) and flux goes out if one has a +ve charge (+q 1 ).1 ´ 10-10 F . Q´q 4 pe 0 d 2 + Q ´ Q 4 pe 0 (2 d ) 2 = 0 d Q or d q Q Q q = - .63 Electrostatics 37. ( * ) : Electric flux through ABCD = zero for the charge placed outside the box as the charged enclosed is zero. For one surface.f1 ) e0 . (b) : Aluminium is a good conductor. (Option not given). Q = q 1 ~ q 2 40. C = 0 d e A e A = 0 = C . (a) : According to Gauss theorem. 4 42. But q for the charge inside the cube. (d) : When the system of three charges is in equilibrium. f 2 > f 1 .1 volt. With space partially filled. (b) : Total capacity = nC 1 2 \ Energy = nCV 2 43. 9 9 ´ 10 41. it is e through all the 0 q surfaces. ( f2 .f1 ) = Q Þ Q = ( f2 . (a) : W = QV \ V = W 2 = = 0. Q 20 . (a) : C = 4 pe0 R = 1 = 1. 4 pe 0 R 4pe 0 R 4pe 0 R 38. 0 44. The capacity remains unchanged. Its sheet introduced between the plates of a capacitor is of negligible thickness. (c) : Potential at any internal point of charged shell = q 4 pe 0 R 1 2 Q Potential at P due to Q at centre = 4 pe R 0 \ Total potential point q 2 Q 1 = + = ( q + 2Q ). As one does not know whether f 1 > f 2 . 1 2 1 2 This question contains Statement1 and Statement2.04 Volt (b) zero Volt (c) 2. 5. DV I I A 8.03 A P 1 to P 2 (2008) Directions : Questions 8 and 9 are based on the following paragraph. (c) Statement1 is true. The resistance of a wire changes from 100 W to 150 W when its temperature is increased from 27°C to 227°C. (iii) From the r dependence of E(r). This implies that a = 2. (2009) Shown in the figure below is a meterbridge set up with null deflection in the galvanometer. where j is the current per unit area at r. Statement2 is true; Statement2 is the correct explanation of Statement1. (a) 0. Current I enters at A and leaves from D. 4.05% (b) increase by 0. (ii) and (iii) for current I leaving D and superpose results for A and D. Statement1: The temperature dependence of resistance is usually given as R = R 0 (1 + aDt). The calculation is done in the following steps: (i) Take current I entering from A and assume it to spread over a hemispherical surface in the block. choose the one that best describes the two statements.5 × 10 –3 /°C Statement2: R = R 0 (1 + aDt) is valid only when the change in the temperature DT is small and DR = (R – R 0 ) << R 0 . We apply superposition principle to find voltage DV developed between B and C.27 A P 2 to P 1 (d) 0.64 JEE MAIN CHAPTERWISE EXPLORER CHAPTER 12 1. A 60 W bulb is already switched on. a + a (2010) (c) a1 + a 2 . 6. The resistance of the lead wires is 6 W.3 Volt (2013) Two electric bulbs marked 25 W220 V and 100 W220 V are connected in series to a 440 V supply.75 W (c) 220 W (d) 110 W. 2. What is the decrease of voltage across the bulb. Statement2 is true. (2008) A 5 V battery with internal resistance 2 W and 2 V battery with internal resistance 1 W are connected to a 10 W resistor as shown in the figure. The respective temperature coefficients of their series and parallel combinations are nearly a + a 2 a1 + a 2 a + a 2 (a) 1 . (b) 1 .2% (d) decrease by 0.27 A P 1 to P 2 (c) 0. a B b a C D DV measured between B and C is . Statement2 is false (b) Statement1 is true. (a) Statement1 is true. Consider a block of conducting material of resistivity r shown in the figure. (iv) Repeat (i).9 Volt (d) 13.1% longer. If a wire is stretched to make it 0. Of the four choices given after the statements. obtain the potential V(r) at r. 3. CURRENT ELECTRICITY The supply voltage to a room is 120 V. (d) Statement1 is false.05% (2011) Two conductors have the same resistance at 0°C but their temperature coefficients of resistance are a 1 and a 2 . Statement2 is true; Statement2 is not the correct explanation of Statement1. Which of the bulbs will fuse? (a) 100 W (b) 25 W (c) neither (d) both (2012) 55 W G 20 cm 7.2% (c) decrease by 0. The value of the unknown resistance R is R (a) 55 W (b) 13. its resistance will (a) increase by 0. when a 240 W heater is switched on in parallel to the bulb? (a) 10. a1 + a 2 2 2 2 a1a 2 a + a 2 (d) a1 + a 2 . (ii) Calculate field E(r) at distance r from A by using Ohm’s law E = rj.03 A P 2 to P 1 . The current in the 10 W resistor is P 2 5 V 2 W 2 V 1 W 10 W P 1 (b) 0. 65 Current Electricity rI (a) 2 p(a - b ) rI rI (c) a . The resistance of hot tungsten filament is about 10 times the cold resistance. an electric current will (a) flow from antimony to bismuth at the cold junction (b) flow from antimony to bismuth at the hot junction (c) flow from bismuth to antimony at the cold junction (d) not flow through the thermocouple. The internal resistances of the two sources are R 1 and R 2 (R 2 > R 1 ). The resistance of a bulb filament is 100 W at a temperature of 100ºC. An electric bulb is rated 220 volt 100 watt. antimony and bismuth.67 A. rI rI (b) pa . its resistance will become 200 W at a temperature of (a) 200ºC (b) 300ºC (c) 400ºC (d) 500ºC. If the potential difference across the source having internal resistance R 2 is zero. the ratio l B /l A of their respective lengths must be (a) 2 (b) 1 (c) 1/2 (d) 1/4. (2006) 18. If its temperature coefficient of resistance be 0.p( a + b ) rI rI (d) 2pa . the galvanometer G shows zero deflection. (2006) 16. The condition for bridge to be balanced will be P R (a) Q = S + S 1 2 P 2 R (b) Q = S + S 1 2 P R ( S1 + S 2 ) (c) Q = S S 1 2 P R ( S1 + S 2 ) (d) Q = 2 S S .( a + b ) 9.17 A (c) 0.2p(a + b) (2008) For current entering at A.33 A (d) 0. three resistance P. (2006) 17. In a potentiometer experiment the balancing with a cell is at length 240 cm. A circular wire made of B has twice the diameter of a wire made of A. are respectively based on (a) conservation of charge. An energy source will supply a constant current into the load if its internal resistance is (a) zero (b) nonzero but less than the resistance of the load (c) equal to the resistance of the load (d) very large as compared to the load resistance (2005) 22.5 W (2005) 19.5 A (b) 0. the value of the resistor R will be 10 W 500 W 5 V G (a) 0. conservation of charge. The Kirchhoff’s first law (å i = 0) and second law (å iR = å E ). where the symbols have their usual meanings. (2006) 13.005 per ºC. Two sources of equal emf are connected to an external resistance R. conservation of energy 12 V B (a) 500 W (c) 200 W R A (b) 1000 W (d) 100 W 2 V (2005) 21. The resistance of the wire at 0°C will be (a) 3 ohm (b) 2 ohm (c) 1 ohm (d) 4 ohm (2007) 11. conservation of momentum (c) conservation of energy. In the circuit. conservation of charge (d) conservation of momentum. Then for the two wires to have the same resistance. What will be the resistance of 100 W and 200 V lamp when not in use? . If the batteries A and B have negligible internal resistance. The power consumed by it when operated on 110 volt will be (a) 50 watt (b) 75 watt (c) 40 watt (d) 25 watt. On shunting the cell with a resistance of 2 W. 1 2 (2006) 15. the electric field at a distance r from A is rI rI (a) (b) 4 p r 2 8 p r 2 r I rI (c) 2 (d) (2008) r 2 p r 2 10. The internal resistance of the cell is (a) 4 W (b) 2 W (c) 1 W (d) 0. the balancing length becomes 120 cm. The resistance of a wire is 5 ohm at 50°C and 6 ohm at 100°C. Q and R connected in the three arms and the fourth arm is formed by two resistance S 1 and S 2 connected in parallel. A material B has twice the specific resistance of A. then (a) R = R1R 2 R1 + R2 (c) R = R 2 5 W 10 W I 20 W (b) R = ( R1 + R 2 ) ( R2 - R1 ) R1R 2 R2 - R1 (d) R = R 2 – R 1 (2005) 20. (2006) 12. A thermocouple is made from two metals. The current I drawn from the 5 volt source will be 10 W (b) conservation of charge. If one junction of the couple is kept hot and the other is kept cold then. (2006) 14. In a Wheatstone’s bridge. then the neutral temperature is (a) 700°C (b) 350°C (c) 1400°C (d) no neutral temperature is possible for this thermocouple. When they are joined in parallel the total resistance is P. 30 E (d) 100 . (2004) 30. The electrochemical equivalent of a metal is 1. of its standard cell is E volt. The length of a given cylindrical wire is increased by 100%. and the e. The mass of the metal liberated at the cathode when a 3 A current is passed for 2 second will be (a) 19.5 W. where i is the current in the potentiometer wire. (2003) 36. of the battery is 30 E (a) 100. When a total charge q flows through the voltmeters. A 220 volt. equal amount of metals are deposited. (2004) 32. The total current supplied to the circuit by the battery is (a) 1 A 2 W (b) 2 A 6 V (c) 4 A 6 W 3 W (d) 6 A. In a metre bridge experiment null point is obtained at 20 cm from one end of the wire when resistance X is balanced against another resistance Y. A 3 volt battery with negligible internal resistance is connected in a circuit as shown in the figure.3 × 10 –7 kg per coulomb. one of copper and another of silver. (2004) 27. I The current I. The power consumed will be (a) 750 watt (b) 500 watt (c) 250 watt (d) 1000 watt. If the electrochemical equivalent of Zn and Cu are 32.5 and 31.5 30 E (b) 100 - 0. If the balance point is obtained at l = 30 cm from the positive end.5 i . If S = nP. Two voltameters. decreases in mass by 0.m. sending a constant current through a circuit. then where will be the new position of the null point from the same end. connected in parallel.66 JEE MAIN CHAPTERWISE EXPLORER (a) 400 W (c) 40 W (b) 200 W (d) 20 W (2005) 23.5 A (c) 2 A (d) (1/3) A (2003) 35.5 W 3. If the electrochemical equivalents of copper and silver are z 1 and z 2 respectively the charge which flows through the silver voltameter is z 1 z 2 (a) q z (b) q z 2 1 q q (c) (d) z z (2005) 1 + 1 1 + 2 z2 z1 24. (2004) 29. The thermistors are usually made of (a) metals with low temperature coefficient of resistivity (b) metals with high temperature coefficient of resistivity (c) metal oxides with high temperature coefficient of resistivity (d) semiconducting materials having low temperature coefficient of resistivity.m. (2004) 33.8 × 10 –7 kg (b) 9.5 respectively. of a battery whose internal resistance is 0. if one decides to balance a resistance of 4X against Y? (a) 50 cm (b) 80 cm (c) 40 cm (d) 70 cm.5 30 E (c) 100 - 0. (2004) 28. (2004) 26. then the minimum possible value of n is (a) 4 (b) 3 (c) 2 (d) 1. The length of a wire of a potentiometer is 100 cm. A heater coil is cut into two equal parts and only one part is now used in the heater.6 × 10 kg (d) 1. the e. (2003) 37.9 × 10 –7 kg –7 (c) 6. If the lengths and radii of the wires are in the ratio of 4/3 and 2/3. The negative Zn pole of a Daniell cell. If the cold junction is kept at 0°C.f. It is employed to measure the e.13 g in 30 minutes. Due to the consequent decrease in diameter the change in the resistance of the wire will be (a) 200% (b) 100% (c) 50% (d) 300%. The resistance of the series combination of two resistances is S. 1000 watt bulb is connected across a 110 volt mains supply.1 × 10 –7 kg. the increase in the mass of the . are joined in parallel.f. If X < Y. An electric current is passed through a circuit containing two wires of the same material.m.f. then the ratio of the currents passing through the wire will be (a) 3 (b) 1/3 (c) 8/9 (d) 2. in the circuit will be 3 W 3 V 3 W (a) 1 A (b) 3W 1. The heat generated will now be (a) one fourth (b) halved (c) doubled (d) four times (2005) 25. Time taken by a 836 W heater to heat one litre of water from 10°C to 40°C is (a) 50 s (b) 100 s (c) 150 s (d) 200 s. The thermo emf of a thermocouple varies with the temperature q of the hot junction as E = aq + bq 2 in volt where the ratio a/b is 700°C. (2004) 31. (2003) 34. then R is 2 W (a) 2 W (b) 6 W 15 V (c) 5 W (d) 4 W. (a) (d) (d) (c) (a) (b) (c) 5. The mass of a product liberated on anode in an electrochemical cell depends on (a) (It) 1/2 (b) It (c) I/t (d) I 2 t. 22. (c) (c) (b) (c) (a) (c) (b) . capable of detecting current as low as 10 –5 A. (2003) 38. 35. 36. 10. If in the circuit. 12. 40. 25. 20. 11. 23. 42. The thermo e. 19. (2003) 39.qi = 2qn . 38. 13. 30.m. (where t is the time period for which the current is passed).242 g.qc = 2qn qi + qc = qn (c) (d) qc . (2002) 42. 7. 14. 34.180 g (b) 0. 16. 21. The smallest temperature difference that can be detected by this system is (a) 16°C (b) 12°C (c) 8°C (d) 20°C. 15. 37. 9. 39. (a) (a) (a) (d) (c) (c) (b) 6.f. 31. (2002) Answer Key 1. 8. 28. then (a) qi + qc = qn (b) qi . q c is the temperature of the cold junction. 18. 29. 24. A galvanometer of 40 ohm resistance. is connected with the thermocouple. A wire when connected to 220 V mains supply has power dissipation P 1 .141 g (c) 0. 27. 41. (b) (d) (a) (a) (b) (d) (b) 4. (2002) 40. 26. (a) (d) (c) (d) (c) (d) (c) 2.67 Current Electricity positive Cu pole in this time is (a) 0. Then P 2 : P 1 is (a) 1 (b) 4 (c) 2 (d) 3. (2002) 2 41. 17. Power dissipation in this case is P 2. Now the wire is cut into two equal pieces which are connected in parallel to the same supply.126 g (d) 0. (b) (d) (c) (d) (a) (c) (a) 3. q n is the neutral temperature. If q i is the inversion temperature. power R dissipation is 150 W. 33. 32. of a thermocouple is 25 mV/°C at room temperature. 07 V 240 W + 6 W Resistance of heater. RH = As bulb and heater are connected in parallel. Let R 1 and R 2 be their resistance at t°C. the supply voltage is taken as rated voltage. I = 440 2 = A (220)2 / 20 11 Potential difference across 25 W bulb . 10.. 2. Then R R R p = 1 2 R1 + R2 R (1 + a1t ) R0 (1 + a 2 t ) R p 0 (1 + a p t ) = 0 R0 (1 + a1t ) + R0 (1 + a 2 t ) R0 R0 R = 0 where.41 V . DR = 2 D l R l 120 V × 48 W = 106.1% its resistance will increase by 0. (a) : As P = V R Here. R p 0 = R0 + R0 2 . when wire is stretched by 0. Then R 1 = R 0 (1 + a 1t) R 2 = R 0 (1 + a 2t) Let R s is the resistance of the series combination of two conductors at t°C. because it can tolerate only 220 V while the voltage across it is 352 V. (b) : 25 W‑220 V 100 W‑ 220 V 440 V (Rated voltage) 2 Rated power \ Resistance of 25 W220 V bulb is (220) 2 W 25 Resistance of 100 W220 V bulb is R1 = (220) 2 W 100 When these two bulbs are connected in series.68 JEE MAIN CHAPTERWISE EXPLORER 2 1. 120 V ´ 240 W V1 = = 117. \ Resistance of bulb 120 V ´ 120 V RB = = 240 W 60 W 120 V ´ 120 V = 60 W 240 W Voltage across bulb before heater is switched on.2%. Their equivalent resistance is (240 W)(60 W) Req = = 48 W 240 W + 60 W \ Voltage across bulb after heater is switched on V2 = Potential difference across 100 W bulb (220) 2 = IR2 = 2 ´ = 88 V 11 100 Thus the bulb 25 W will be fused. (b) : Resistance of wire rl R = A On stretching. 66 V 48 W + 6 W Decrease in the voltage across the bulb is DV = V 1 – V 2 = 10. the total resistance is R2 = (220) 2 Rs = R1 + R2 = (220) 2 é 1 + 1 ù = W ë 25 100 û 20 Current. 4. (a) : Let R 0 be the resistance of both conductors at 0°C. volume (V) remains constant. R s0 = R 0 + R 0 = 2R 0 \ 2R 0 (1 + ast) = 2R 0 + R 0t(a 1 + a 2 ) 2R 0 + 2R 0ast = 2R 0 + R 0t(a 1 + a 2 ) a + a 2 \ a s = 1 2 Let R p is the resistance of the parallel combination of two conductors at t°C..04 V V As R = (220) 2 = IR1 = 2 ´ = 352 V 11 25 or (Q V and r are constants) D R % = 2 D l % R l Hence.(i) V So V = Al or A = l rl 2 (Using (i)) \ R = V Taking logarithm on both sides and differentiating we get. 3. Then Rs = R 1 + R 2 Rs0 (1 + ast) = R 0 (1 + a 1t) + R 0 (1 + a 2t) where. C E = Ir/2pr 2 . j = -1 (a + a 2 ) t ù = 1 [1 + (a1 + a 2 )t ] éê1 + 1 ú ë û 2 2 Resistance = 1 é (a + a 2 ) t ù [1 + (a1 + a 2 )t ] ê1 . If r l is the resistance per unit length (in cm) A x P 2 x – y Q R A = RB B 2 V 2 W 10 W C 11.. a = 2.(iii) = (ii) × 2 (i) – (iii) gives 32y = 1 Þ y = 1 A = 0...(ii) Þ 2x – 22y = 4 . R 100 = 6 W R t = R 0 (1 + at) where R t = resistance of a wire at t°C. a a As (a 1 + a 2 ) 2 is negligible (c) : b D rl rr = area 2 p r 2 VB ..(i) æ r ö æ 2 D A ö 4 2 = ç A ÷ç ÷ = 2 = 1 è 2r A ø è DA ø rl rl ´ 4 = 2 pr p D 2 .I r a 1 .. The resistance of a wire change from 100 W to 150 W when the temperature is increased from 27°C to 227°C. R 0 = resistance of a wire at 0°C.1 úû ë 2 2 [By binomial expansion] Þ DV = DV = (a1 + a 2 ) t 2 a + a 2 a p = 1 2 5.69 Current Electricity \ R0 R 2 (1 + a1t )(1 + a 2 t ) (1 + a p t ) = 0 2 2 R0 + R0 (a1 + a 2 ) t R0 R 2 (1 + a1t + a 2t + a1a 2 t 2 ) (1 + a p t ) = 0 2 R0 (2 + (a1 + a 2 )t ) (1 + a1t + a 2t + a1a 2 t 2 ) 1 (1 + a p t ) = 2 (2 + (a1 + a 2 )t ) 8.. a p t = ò . 20rl 80 rl = or R = 80 ´ 55 = 220 W .R 0 1 = 6 - R0 2 This is a Wheatstone bridge. As a 1 and a 2 are small quantities \ a 1a 2 is negligible + 2 – 1(x – y) + 10 ∙ y = 0 +x – 11y = 2 . 55 R 20 (d) : I 2 p r 2 a + b \ 1 (1 + a p t ) = 1 é1 + 1 (a1 + a 2 ) t ù û 2 2 ë 2 7. (i) R 100 – R 0 = R 0a (100) .03 A from P2 to P1 . The current I is a surface current. (d) : j × r = E.Edr It is true that a is small. (d) : Given : R 50 = 5 W.1 ùú .. (a) : Resistance of a wire R = y 5 V x P 1 x – y \ 1 W 4r Al A pD A2 = 4 r B l B 2 p DB 2 lB æ r A ö æ D B ö or l = ç r ÷ ç D ÷ A è B ø è A ø D Applying Kirchhoff’s law for the loops AP 2 P 1 CA and P 2 BDP 1 P 2 .VC = DV = é ( a + a 2 )t (a + a 2 ) 2 t 2 ù = 1 ê1 .I r é 1 ù a dr = ò 2 2p 2 p êë r úû a + b a + b r I r é 1 ..e. A a 1 + (a1 + a 2 ) t 1 + (a1 + a 2 ) t \ 1 (1 + a p t ) = = 2 2 + (a1 + a 2 ) t (a + a 2 ) t ù 2 éê1 + 1 ú ë û 2 Current density. one gets –10y – 2x + 5 = 0 Þ 2x + 10y = 5 or 10 – 2R 0 = 6 – R 0 or R 0 = 4 W..1 úû 2ë 2 2 6.5 × 10 –3 /°C. a = temperature coefficient of resistance. R – R0 << R0 is not true. 9. we get 20 cm 80 cm 5 . (a) : From the statement given. R . 2 p ëê a a + b û I r 2 p r 2 10. = 55 W B 2 .1 + (a1 + a 2 ) t . (ii) Divide (i) by (ii).. 32 (d) : Current is spread over an area 2pr 2 . But (150 – 100) W or 50 W is not very much less than 100 W i.. \ E = \ R 50 = R 0 [1 + a 50] and R 100 = R 0 [1 + a 100] or R 50 – R 0 = R 0a(50) . (d) : The voltameters are joined in parallel. Q S1 S2 15.l ) 4 100 .1 ÷ = R ç ÷ l è l2 ø è 2 ø é 240 .005 (100)] or 100 = R 0[1 + 0.5 A .l 2 ö r = R ç 1 .120 ù = 2 W. \ 484 17.. (d) : I = .005t ] 1 + 0. Resistance when not in use = 10 22.. It means that current flows from A to B at cold junction.005t] 200 = R 0 [1 + 0.. (d) : Resistance of the bulb V 2 (220) 2 ( R ) = = = 484 W P 100 (110) 2 Power across 110 volt = 484 110 ´ 110 Power = = 25 W . (c) : For balanced Wheatstone's bridge. 18. Hence no current flows through arm BD.. (a) : If internal resistance is zero.. I R 20.005ºC –1 R t = 200 W \ R 100 = R 0 [1 + 0. Þ 12R = 1000+ 2R Þ R = 100 W 21... (c) : The equivalent circuit is a balanced Wheatstone's bridge.. we get 100 = [1 + 0. P = R Q S S1 S 2 Q S = S1 + S 2 ( Q S 1 and S 2 are in parallel) R ( S1 + S 2 ) \ P = . 2 V 2 (200) = = 400 W P 100 400 = 40 W .005t = 2 + 1 or t = 400ºC. (c) : Resistance of hot tungsten = 23. 27.70 JEE MAIN CHAPTERWISE EXPLORER 12. (b) : Potential difference is same when the wires are put in parallel rl V = I1R1 = I 1 ´ 1 2 p r1 Again V = I 2 R2 = I 2 ´ rl 2 pr2 2 . (a) : For meter bridge experiment. 26..005t] Divide (i) by (ii).005 ´ 100] 200 [1 + 0. \ r = 2 ê ë 120 úû or E = (Given) E R 1 E R 2 2ER 2 R1 + R2 + R or R 1 + R 2 + R = 2R 2 or R = R 2 – R 1 . z 2 ö æ + 1 ç z1 ÷ø è 24.. B AB and BC are in series \ R ABC = 5 + 10 = 15 W 5 W 10 W AD and DC are in series \ R ADC = 10 + 20 = 30 W A C 10 W ABC and ADC are in parallel ( R ABC )( R ADC ) 10 W 20 W \ R eq = ( R ABC + R ADC ) D I or Req = 15 ´ 30 = 15 ´ 30 = 10 W 15 + 30 45 E 5 = = 0.l In the first case.(i) . (c) : Given : R 100 = 100 W a = 0. 16. (c) : Thermistors are made of metal oxides with high temperature coefficient of resistivity. the energy source will supply a constant current. (c) : Resistance of full coil = R Resistance of each half piece = R/2 or q 2 = \ H 2 V 2 t R 2 = ´ = H1 R / 2 V 2 t 1 \ H 2 = 2H 1 Heat generated will now be doubled. (a) : AntimonyBismuth couple is ABC couple. I 1 = I 2 12 2 or 500 + R = R .(ii) 13. 2E R1 + R2 + R Q E – IR 2 = 0 \ E = IR 2 19. 4X l 4 l = Þ = Þ l = 50 cm . (a) : Kirchhoff's first law [S i = 0] is based on conservation of charge Kirchhoff's second law (S i R = S E) is based on conservation of energy.. R1 l1 l 1 = = R2 l2 (100 – l1 ) X = 20 = 20 = 1 Y 100 - 20 80 4 In the second case. (d) : For zero deflection in galvanometer. \ Current I = Req 10 + 5 V 14. 25.. Mass deposited = z 1 q 1 = z 2 q 2 \ q1 z2 q1 + q2 z1 + z 2 = Þ = q2 z1 q2 z1 Þ z ö q æ = 1 + 2 ÷ q2 çè z1 ø q .005 × 100] R t = R 0 [1 + 0. (b) : The internal resistance of a cell is given by æ l ö æ l1 . Y (100 . (c) : qc + qi = 2 qn Þ qi + qc = qn . è øè ø 2 28.5 W 1.5 A. R 2 or 2 W 3 W 3 R ´ 100% = 300% . 2 V 2 41. (b) : Equivalent resistance = Q S = nP 6 W \ % change = 1000 ´ 30 ´ 4.5 W Þ 6 V 3 W 35. (a) : In series combination.71 Current Electricity I ´ rl I ´ rl I æ l ö æ r ö \ 1 2 1 = 2 2 2 Þ 1 = ç 2 ÷ ç 1 ÷ I 2 è l1 ø è r 2 ø pr1 pr2 2 2 l \ The new resistance R¢2 = r l ¢ 2 = r = 4 R 2 p r¢ p´ r 2 \ Change in resistance = R¢ . its radius is going to be decreased in such a way that the volume of the cylinder remains constant. + R 2 R 2 . where current i is drawn from battery 1.5 W 6 V I = 6 = 4 A. qn = - 32. (b) : P 1 = R when connected in parallel. R 48. \ 25 ´ 10 -6 39. (b) : According to Faraday's laws.5 W 1. 40. \ P1 2 2 P2 = V = 4 V = 4 P 1 R / 4 R 2 42.´ (700) = -350° C 2b 2 Neutral temperature is calculated to be –350°C Since temperature of cold junction is 0°C.3 × 10 –7 ) × (3) × (2) = 19.5 30. or K × 30 = E¢ – ir. P = R1 R 2 ( R1 + R2 ) R R \ (R 1 + R 2 ) 2 = nR 1 R 2 \ ( R1 + R2 ) = n 1 2 ( R1 + R2 ) For minimum value. (c) : Resistance of bulb = P 1000 Required power Þ 6 V (3 + 3) ´ 3 18 = = 2 W (3 + 3) + 3 9 \ Current I = V = 3 = 1. (a) : m = Z i t or m = (3. d q or 0 = a + 2bq n or E ´ 30 = E ¢ + 0. (d) : E = aq + bq 2 dE = a + 2 b q \ d q a 1 = .5 = Þ mCu 31.5 38. (c) : The equivalent circuits are shown below : 1.8 × 10 –7 kg.18 J = 1cal] or 2 V 2 (110) 110 ´ 110 = = = 250 W . R 34.18 t= = 150 sec.5 mCu = 0. no neutral temperature is possible for this thermocouple. = mZn Z Zn = mCu Z Cu when i and t are same \ dE = 0 At neutral temperature (q n). 36.5 i or E ¢ = 30 E - 0. R 1 = R 2 = R \ (R + R) 2 = n(R × R) Þ 4R 2 = nR 2 or n = 4.18) [Q 4.4 48. m µ It.13 ´ 31.5 i 100 100 2 V 2 (220) = = 48. 31. ( R / 2) ´ ( R / 2) R R eq = = R R 4 \ + 2 2 P 2 = 4. (b) : Power = V R \ 150 = (15)2 (15) 2 225 225 = + Þ R = 6 W . radius of the wire be r \ Resistance R = r l 2 r = resistivity of the wire pr 100 Now l is increased by 100% \ l ¢ = l + 100 l = 2 l As length is increased. (d) : Let the length of the wire be l.5 = 0. (c) : Electrical energy is converted into heat energy \ 836 × t = 1000 × 1 × (40 – 10) × (4.13 32. where E¢ is emf of battery. (c) : According to Faraday's laws of electrolysis.126 g 32. 29.4 W .4 37. (c) : Potential gradient along wire = E volt 100 cm E volt \ K = 100 cm For battery V = E¢ – ir.R = 3 R I 1 æ 3 öæ 2 ö2 3 ´ 4 1 or I = ç 4 ÷ç 3 ÷ = 4 ´ 9 = 3 . 1. 836 33. (a) : Let the smallest temperature be q°C \ Thermo emf = (25 × 10 –6 ) q volt Potential difference across galvanometer = IR = 10 –5 × 40 = 4 × 10 –4 volt \ (25 × 10 –6 )q = 4 × 10 –4 4 ´ 10 -4 q= = 16° C. S = (R 1 + R 2 ) In parallel combination. r2 ´ l r2 ´ l r 2 2 pr 2 ´ l = pr ¢2 ´ l¢ Þ r ¢ = l ¢ = 2l = 2 0. greater is the resistance of ammeter. deuteron and alpha particle of the same kinetic energy are moving in circular trajectories in a constant magnetic field. StatementII is the correct explanation of StatementI. MAGNETIC EFFECTS OF CURRENT AND MAGNETISM A metallic rod of length ‘l’ is tied to a string of length 2l and made to rotate with angular speed w on a horizontal table with one end of the string fixed. When a current is passed through the coil it starts oscillating; it is very difficult to stop.6 × 10 –5 Wb/m 2 ) (a) 5. The variation of the magnetic field B along the line XX¢ is given by 2 (d) 4 Bwl 2 (2013) This question has StatementI and StatementII. But if an aluminium plate is placed near to the coil. (d) StatementI is true. choose the one that best describes the two Statements. The radii of proton. (d) development of air current when the plate is placed. StatementII is true. (c) StatementI is true. StatementII is false. (2012) 6.20 Am 2 and 1. additional shunt needs to be used across it. Of the four choices given after the Statements. If there is a vertical magnetic field ‘B’ in the region. They carry steady equal currents flowing out of the plane of the paper as shown. Two long parallel wires are at a distance 2d apart.0 cm.72 JEE MAIN CHAPTERWISE EXPLORER CHAPTER 13 1.f. They are placed on a horizontal table parallel to each other with their N poles pointing towards the South. They have a common magnetic equator and are separated by a distance of 20. it stops. This is due to (a) induction of electrical charge on the plate. (b) shielding of magnetic lines of force as aluminium is a paramagnetic material. (c) electromagnetic induction in the aluminium plate giving rise to electromagnetic damping. 4. A current I flows in an infinitely long wire with crosssection in the form of a semicircular ring of radius R. The magnitude of the magnetic induction along its axis is m 0 I m 0 I m I m I (a) 2 (b) (c) 0 (d) 0 2 p R 4 p R p R 2 p 2 R (2011) 7. A coil is suspended in a uniform magnetic field.56 × 10 Wb/m (d) 3. 2 (c) 3 Bw l 2 (a) ra = r p < r d (c) ra = r d > r p Two short bar magnets of length 1 cm each have magnetic moments 1. StatementII : To increase the range of ammeter. StatementII is not the correct explanation of StatementI. with the plane of the coil parallel to the magnetic lines of force. 3. Which one of the following relation is correct? (2012) 5. (a) StatementI is false. The value of the resultant horizontal magnetic induction at the mid point O of the line joining their centres is close to (Horizontal component of earth’s magnetic induction is 3. (2013) (b) ra > r d > r p (d) ra = r p = r d B (a) X X¢ d d B (b) X X¢ d d B (c) X X¢ d d B (d) X X¢ d d (2010) . StatementII is true.m. StatementI : Higher the range.80 × 10 –4 Wb/m 2 (b) 3. induced across the ends of the rod is 2 2 (a) 5 Bw l (b) 2 Bw l 2 2 2. StatementII is true. r d and ra .00 Am 2 respectively. deuteron and alpha particle are respectively r p . the e. (b) StatementI is true.6 × 10 –5 Wb/m 2 –4 2 (c) 2.50 × 10 –4 Wb/m 2 (2013) Proton. 5 (d) e r = 0.5. (2006) 18. thin walled pipe. m r = 0. A magnet when brought close to them will (a) attract all three of them (b) attract N 1 and N 2 strongly but repel N 3 (c) attract N 1 strongly. The wire AOB carries an electric current I 1 and COD carries a current I 2 . N 2 weakly and repel N 3 weakly (d) attract N 1 strongly.5 (b) e r = 0.73 Magnetic Effects of Current and Magnetism Directions : Question numbers 8 and 9 are based on the following paragraph. The current is uniformly distributed across its cross section. (2009) 8.5 × 10 –7 T southward (c) 5 × 10 –6 T northward (d) 5 × 10 –6 T southward. Two identical conducting wires AOB and COD are placed at right angles to each other. Then (a) the magnetic field at all points inside the pipe is the same. in a direction perpendicular to the plane of the wires AOB and COD.05 × 10 –3 Wb/m 2 . r r uniform and mutually orthogonal fields E r r and B with a r velocity v perpendicular to both E and B . 24 ab 10.5 (2008) 12. A current loop ABCD is held fixed on B the plane of the paper as shown in the A a figure. A charged particle is released from rest in this region. A horizontal overhead powerline is at a height of 4 m from the ground and carries a current of 100 A from east to west. N 2 and N 3 are made of a ferromagnetic. but not zero (b) the magnetic field is zero only on the axis of the pipe (c) the magnetic field is different at different points inside the pipe (d) the magnetic field at any point inside the pipe is zero (2007) 16.a ) + (a + b) ù (c) (d) û 4 p ë ab û 4p ë 3 Due to the presence of the current I 1 at the origin (a) the forces on AB and DC are zero (b) the forces on AD and BC are zero (c) the magnitude of the net force on the loop is given by I1 I é p m 2(b . The ratio of the magnetic field at a/2 and 2a is (a) 1/2 (b) 1/4 (c) 4 (d) 1 (2007) 17. These two fields are parallel to each other. will be given by 1 m 0 ( I 2 + I 2 2 ) (a) 2 p d 1 (c) 1 m 0 2 ( I1 + I 2 2 ) 2 2 p d m æ I + I ö 2 (b) 0 ç 1 2 ÷ 2 p è d ø (d) m0 ( I + I ) 2 p d 1 2 (2007) 13.05 × 10 –5 Wb/m 2 (d) 1.05 × 10 –4 Wb/m 2 (b) 1. Angle C made by AB and CD at the origin O is 30°. (2006) 19. m r = 1. (2006) 20.a ) + ( a + b) ù û 4p 0 ë 3 (d) the magnitude of the net force on the loop is given by m 0 II 1 (b - a ).5 × 10 –7 T northward (b) 2. In a region. The arcs BC (radius = b) and DA I (radius = a) of the loop are joined by I 1 O 30° two straight wires AB and CD. A steady b D current I is flowing in the loop. A long solenoid has 200 turns per cm and carries a current i. momentum and kinetic energy of the particle are constant (2007) 14. The magnetic field on a point lying at a distance d from O.05 × 10 –2 Wb/m 2 (c) 1. The magnetic field at its centre is 6.a ù m 0 I é p 2(b . m r = 0. The value of the magnetic field at its centre is (a) 1. Needles N 1.5. If an electron is projected along the direction of the fields with a certain velocity then . respectively. The magnitude of the magnetic field (B) due to loop ABCD at the origin (O) is m I (b - a ) (a) zero (b) 0 24 ab m 0 I é b . Then r r r r r r 2 2 (a) v = B ´ E / E (b) v = E ´ B / B r r r r (c) vr = B ´ E / B 2 (d) vr = E ´ B / E 2 (2007) 15.28 × 10 –2 weber/m 2 . a paramagnetic and a diamagnetic substance respectively.5. A long straight wire of radius a carries a steady current i. Then (a) kinetic energy changes but the momentum is constant (b) the momentum changes but the kinetic energy is constant (c) both momentum and kinetic energy of the particle are not constant (d) both.5. and comes out r without any change in magnitude or direction of v . m r = 1. A uniform electric field and a uniform magnetic field are acting along the same direction in a certain region. (2008) 11. A charged particle with charge q enters a region of constant. straight. 9. The path of the particle will be a (a) circle (b) helix (c) straight line (d) ellipse. Which of the following values of these quantities are allowed for a diamagnetic mateiral? (a) e r = 1.5 (c) e r = 1. A current I flows along the length of an infinitely long. The magnetic field directly below it on the ground is (m 0 = 4p × 10 –7 T m A –1 ) (a) 2. Another long solenoid has 100 turns per cm and it carries a current i/3. steady and uniform electric and magnetic fields are present. A charged particle moves through a magnetic field perpendicular to its direction. Relative permittivity and permeability of a material are e r and m r . but repel N 2 and N 3 weakly. Another straight thin wire with steady current I 1 flowing out of the plane of the paper is kept at the origin. The length of a magnet is large compared to its width and breadth. They will (a) attract each other with a force of (b) repel each other with a force of (2p d 2 ) m 0 i 2 28. The magnetic field due to a current carrying circular loop of radius 3 cm at a point on the axis at a distance of 4 cm from the centre is 54 mT. A particle of mass M and charge Q moving with velocity v describes a circular path of radius R when subjected to a uniform transverse magnetic field of induction B. (2004) m0 i 2 (2p d ) (2005) 25. They exert a force F on each other. separated by a distance d carry a current of i A in the same direction. Now the current in one of them is increased to two times and its direction is reversed. The time period of its oscillation in a vibration magnetometer is 2 s. The materials suitable for making electromagnets should have (a) high retentivity and high coercivity (b) low retentivity and low coercivity (c) high retentivity and low coercivity (d) low retentivity and high coercivity. A particle of charge –16 × 10 –18 coulomb moving with velocity 10 ms –1 along the xaxis enters a region where a magnetic field of induction B is along the y axis. In order that each division reads 1 volt. parallel wires. Two thin long. The distance is also increased to 3d. 3 ampere and 4 ampere are the currents flowing in each coil respectively. An ammeter reads upto 1 ampere. The work done by the field when the particle completes one full circle is . A charged particle of mass m and charge q travels on a circular path of radius r that is perpendicular to a magnetic field B.74 JEE MAIN CHAPTERWISE EXPLORER (a) it will turn towards right of direction of motion (b) it will turn towards left of direction of motion (c) its velocity will decrease (d) its velocity will increase (2005) 21. (2003) r 34.09 W. then the magnetic induction at any point inside the tube is (a) infinite (b) zero m 0 2 i 2 i tesla × tesla (c) . It is bent into a circle of one turn and the magnetic field at the centre of the coil is B. separated by a distance d carry current I 1 and I 2 in the same direction. The time period of this combination will be 2 æ 2 ö (d) ç (a) 2 s (b) s (c) (2 3) s ÷ s 3 è 3 ø (2p d 2 ) (2004) m0 i 2 (2p d ) 32. It is then bent into a circular loop of n turns. and an electric field of magnitude 10 4 V/m is along the negative z axis. Two concentric coils each of radius equal to 2p cm are placed at right angles to each other. Two long conductors. the magnitude of B is (a) 10 3 Wb/m 2 (b) 10 5 Wb/m 2 16 2 (c) 10 Wb/m (d) 10 –3 Wb/m 2 . A magnetic needle is kept in a nonuniform magnetic field. The magnetic induction in weber/m 2 at the center of the coils will be (m 0 = 4p ´ 10 –7 Wb/Am) (a) 5 ´ 10 –5 (b) 7 ´ 10 –5 (c) 12 ´ 10 –5 (d) 10 –5 (2005) 23. The time taken by the particle to complete one revolution is 2 p m 2 p qB (b) (a) qB m 2 p mq (2005) (c) (d) 2pmq B qB 22. To increase the range to 10 A the value of the required shunt is (a) 0. Its internal resistance is 0.3 W (c) 0.81 ohm. (2003) (c) attract each other with a force of (d) repel each other with a force of m 0 i 2 27. The magnet is cut along its length into three equal parts and three parts are then placed on each other with their like poles together. A current i ampere flows along an infinitely long straight thin walled tube. If the charged particle continues moving along the xaxis. The new value of the force between them is (a) –2F (b) F/3 (c) –2F/3 (d) –F/3. (2004) 29. (2004) 31. (2004) (d) r 4 p r 30. It experiences (a) a force and a torque (b) a force but not a torque (c) a torque but not a force (d) neither a force nor a torque (2005) 26. Its current sensitivity is 10 divisions per milliampere and voltage sensitivity is 2 divisions per millivolt. (2004) 33. A moving coil galvanometer has 150 equal divisions.9 W (d) 0. The magnetic field at the centre of the coil will be (a) nB (b) n 2 B (c) 2nB (d) 2n 2 B. What will be its value at the centre of the loop? (a) 250 mT (b) 150 mT (c) 125 mT (d) 75 mT. A long wire carries a steady current.03 W (b) 0. the resistance in ohms needed to be connected in series with the coil will be (a) 99995 (b) 9995 (c) 10 3 (d) 10 5 (2005) 24. c) (c) (a) (b) (d) (c) 2. then (a) curved path of electron and proton will be same (ignoring the sense of revolution) (b) they will move undeflected (c) curved path of electron is more curved than that of the proton (d) path of proton is more curved. 13. 22. 20. 33. If in a circular coil A of radius R. If an electron and a proton having same momenta enter perpendicular to a magnetic field. (c) (c) (b) (b) (b) (b) (a) 6. (2002) 43. 2 2 2 4 (2003) 36. (2002) 42. 21. If its period of T ¢ oscillation is T ¢ . 29. current I is flowing and in another coil B of radius 2R a current 2I is flowing. 7. then the ratio of the magnetic fields. 31. 42. 17. A magnetic needle lying parallel to a magnetic field requires W units of work to turn it through 60°. A thin rectangular magnet suspended freely has a period of oscillation equal to T. 36. 34. (2003) 37. (2002) 40. (2003) 35. 12. 41. (c) (b) (d) (b) (a) (a) (a) 4. 40. (2003) 38. 11. 16. 27. 28. 35. The time period of a charged particle undergoing a circular motion in a uniform magnetic field is independent of its (a) speed (b) mass (c) charge (d) magnetic induction. 30. The torque needed to maintain the needle in this position will be 3 W (a) 3W (b) W (c) (d) 2W. 24. Now it is broken into two equal halves (each having half of the original length) and one piece is made to oscillate freely in the same field. 10. 39. 18. (a) (b) (b. produced by them will be (a) 1 (b) 2 (c) 1/2 (d) 4. 38. 43. 9. The magnetic lines of force inside a bar magnet (a) are from northpole to southpole of the magnet (b) do not exist (c) depend upon the area of crosssection of the bar magnet (d) are from southpole to northpole of the magnet. 8. If a current is passed through a spring then the spring will (a) expand (b) compress (c) remains same (d) none of these. 19. (a) (d) (d) (a) (b) (b) (b) 5. 15. (2002) 41. 2 ( ) 39. If an ammeter is to be used in place of a voltmeter. Curie temperature is the temperature above which (a) a ferromagnetic material becomes paramagnetic (b) a paramagnetic material becomes diamagnetic (c) a ferromagnetic material becomes diamagnetic (d) a paramagnetic material becomes ferromagnetic. 26. 32. B A and B B . 37. 25. (a) (c) (c) (c) (b) (a) (a) . 14.75 Magnetic Effects of Current and Magnetism æ Mv 2 ö (a) ç ÷ 2 pR è R ø (c) BQ 2pR (2003) (b) zero (d) BQv 2pR. (2002) Answer Key 1. the we must connect with the ammeter a (a) low resistance in parallel (b) high resistance in parallel (c) high resistance in series (d) low resistance in series. (a) (b) (b) (c) (c) (d) (a) 3. 23. the ratio is T 1 1 1 (a) (b) (c) 2 (d) . m. (b. 11. their fields are equal and opposite and their effects also cancel each other.76 JEE MAIN CHAPTERWISE EXPLORER m : 2m : 4 m = 1: 2 :1 e e 2 e Þ ra = r p < r d 1. If it is a. According to the table given.c.f. (a) A They do not contribute to the a magnetic induction at O. induced across the ends of the rod is 3 l 3 l é x 2 ù B w e = ò Bw xdx = Bw ê ú = [(3l )2 . K = 1 mv 2 or v = 2 K 2 m m 2K 2 Km = qB m qB As K and B are constants \ r = \ r µ m q rp : rd : r a = m p md m a : : qp qd qa r By Ampere’s theorem. The total magnetic field due to loop ABCD at O is B = BAB + BBC + BCD + BDA 3. Therefore.* (d) : i -7 Bnet = 4p ´ 10 [1.6 ´ 10 -5 4p ´ (10 ´ 10 -2 ) 3 -7 = 10 -3 ´ 2. F = qv × B.5 and m r = 0. When a charged particle enters a field with its velocity perpendicular to the magnetic field. Bnet = 10. 12.2 + 1] + 3.6 ´ 10 -5 10 = 2. m 0 2 ( I + I 2 2 ) 1/ 2 . the e. one takes e r = 1. (c) : The field at the same point at the same distance from the mutually perpendicular wires carrying current will be having the same magnitude but in perpendicular directions. (a) As the point O lies on broadside position with respect to both the magnets. (a) : = Consider a element of length dx at a distance x from the fixed end of the string. There will be no force.36 × 10 –4 = 2. e. Þ B = 0 - m 0 M 1 m 0 M 2 m + + B H = 0 3 ( M 1 + M 2 ) + B H 4p r 3 4 p r 3 4 p r Substituting the given values. (a) : The radius of the circular path of a charged particle in the magnetic field is given by r = mv Bq Kinetic energy of a charged particle. 4.2 + 3. The net magnetic field at point O is B net = B 1 + B 2 + B H 7. * It is assumed that this is a direct current.(2l )2 ] ë 2 û 2 l 2 2 l = 5 Bw l 2 2 5. 2. (c) 6.f.56 × 10 –4 Wb/m 2 . the current at the given instant is in the given direction. Due to parts AB and CD. c) : Due to Lorentzian force. (b) : O is along the line CD and AB. (b) Þ B= m0 I p m I p ´ + 0 + 0 ´ 4pb 6 4 pa 6 m 0 I (b - a ) . The field I I 1 30° O due to DA is positive or out of the b D paper and that due to BC is into C the paper or negative. out of the paper or positive. 24 ab 9. B × 2 pd = m 0 i r m i 4p ´ 10 -7 ´ 100A = 50 ´ 10 -7 T B = 0 = 2p ´ 4 m 2 pd Þ B = 5 × 10 –6 T southwards. induced in the element is de = B(wx)dx Hence. (b) : The straight wire is perpendicular to the segments and the fields are parallel. Then the choice (c) is correct. the motion is circular .m.2 × 10 –4 + 0.5. we get B 8. \ B = B12 + B2 2 \ B = 2 p d 1 13. (c) : The situation is as shown in the figure. (c) : The values of relative permeability of diamagnetic materials are slightly less than 1 and e r is quite high. (c) : Magnet will attract N 1 strongly. è 2 ø Therefore the answer is (b). if v is given only as E/B.(i) . perhaps by slip. (c) is the answer. a high resistance should be connected in series with the galvanometer V = I g ( G + R ) = or 5+ R = (5 + R ) 15 (5 + R ) Þ 150 = 15 1000 1000 150 ´ 1000 = 10000 \ 15 R = 9995 W. v = E .. (a) : Field along axis of coil B = At the centre of coil. (b) : T = w = v centripetal force = magnetic force Q mv 2 = qvB Þ v = qBr \ r m From (i) and (ii) 2 pr ´ m 2 pm T = = . The path of charged particle will be a straight line..(ii) 22.. 16. (Compare to electric field inside a hollow sphere). Therefore its tangential momentum changes its direction but its energy remains the same with qvB = æ 1 I w 2 = constant ö . B ¢ = m 0 i 2 R m0 iR 2 2( R 2 + x 2 ) 3/ 2 .. (a) : A force and a torque act on a magnetic needle kept in a nonuniform magnetic field. The two forces oppose B r r r E ´ B r r v = each other if v is along E ´ B i. 24. F ¢ = \ \ m0 I1 I 2 l 2 p d m 0 ( -2 I1 )( I 2 ) l 2p 3 d F ¢ -m 0 2 I1 I 2 l 2 pd 2 = ´ =F 2 p 3d m 0 I1 I 2 l 3 F¢ = – 2F/3..05 × 10 –2 Wb/m 2 ..B.. m 0 i 2 L .. (b) : Vmax = 2 = 75 m V 150 I max = = 15 mA = I g 10 Resistance of galvanometer G = 75/15 = 5 W For conversion into a voltmeter. B 2 = m 0n 2I 2 B2 n2 I 2 100 i / 3 1 = ´ = ´ = \ B1 n1 I1 200 i 6 \ B2 = B 1 6. N 2 weakly and repel N 3 weakly. B 2 r r As E and B are perpendicular to each other r r E ´ B = EB sin 90 ° = E B B2 B 2 For historic and standard experiments like Thomson’s e/m value.. (c) : Force of attraction between wires = 25. Iw is a constant. B 1 = m 0 n 1I 1 In second case. (b) : When E and B are perpendicular and velocity has no changes then qE = qvB i. 2 * The questions could have been more specific.. As 1 2 Iw is also constant. v constantly changes its direction (but r not the magnitude). \ qBr qB . -7 m 0 2 2 = 4 p ´ 10 (3)2 + (4) 2 i1 + i 2 2 ´ (2 p ´ 10 -2 ) 2 r 150 23. (d) : Magnetic field is shielded and no current is inside the pipe to apply Ampère’s law. 2 p 2 pr 21.. r r 14.e. 26. 19. (a) : Magnetic induction at centre of one coil B 1 = Similarly B 2 = m 0 i 1 2 r m 0i 2 2 r \ 2 2 m 0 2 2 2 æ m i ö æ m i ö B 2 = B12 + B 2 2 = ç 0 1 ÷ + ç 0 2 ÷ = 2 ( i1 + i 2 ) è 2 r ø è 2 r ø 4 r \ B= or B = 5 × 10 –5 Wb/m 2 .28 ´ 10 -2 = = 1. 20. The options do not mention L.. (c) : Magnetic field applied parallel to motion of electron exerts no force on it as q = 0 and force = Bevsinq = zero Electric field opposes motion of electron which carries a negative charge \ velocity of electron decreases. 2 p d N. B 2 R a 2 17.. 15..e. If angular momentum is taken.. (c) : Initially. (b) : In first case. 27. 6 6 18. although the answer is numerically correct. Current enclosed in the 1 st ampèrean path is \ I × pr12 = Ir 1 2 pR 2 R 2 m ´ current m 0 × Ir12 m 0 Ir 1 B = 0 = = path 2 pr1 R 2 2 p R 2 Magnetic induction at a distance r 2 = \ m 0 × I 2 p r2 a B1 r1r 2 2 × 2 a = 2 = = 1 .77 Magnetic Effects of Current and Magnetism mv 2 . .. whether by “momentum” it is meant tangential momentum or angular momentum.... it would have been better from the pedagogic view. (d) : Uniform current is flowing. (c) : Magnetic field exerts a force = Bevsinq = Bevsin0 = 0 Electric field exerts force along a straight line.. F = Finally. (i) Torque = MB sin 60° = (2W) sin 60° = Ig I g G S 32... (b) : The spring will compress. 35. B x = B z = 0 q \ a x = a y = 0... (a) : Particle travels along xaxis. (b) : Workdone by the field = zero.. T = 2 p I MB where I = ml 2 /12. r 1 = radius of coil = l/2p m i 2 m i p B = 0 = 0 \ 2 r1 2 l l Finally.. (b) : For an oscillating magnet.. (d) : The magnetic lines of force inside a bar magnet are from south pole to north pole of magnet.. \ Time period T¢ = 2 p 30... (b) : Materials of low retentivity and low coercivity are suitable for making electromagnets. (a) : Bqv = mv Þ r = mv = r Bq Bq r will be same for electron and proton as p.. (c) : High resistance in series with a galvanometer converts it into a voltmeter. 38. B and q are of same magnitude. vx 10 m 34.... B x = B z = 0 Electric field is along the negative zaxis.Ez + vx B y ) m . Hence v y = v z = 0 Field of induction B is along yaxis...81 0. m 0 2 pI m 0 I = 4p R 2 R B A I A R B æ 1 ö æ 2 ö = ´ = = 1 BB I B RA çè 2 ÷ø çè 1 ÷ø 42. v y = v z = 0. a z = ( . \ T ¢ = I ¢ ´ M = I ¢ ´ M T M¢ I I M ¢ \ T ¢ = 1 ´ 2 = 1 . (b) : For a vibrating magnet. Again a z = 0 as the particle traverse through the region undeflected \ E z = v x B y or B y = E z 10 4 Wb = = 10 3 2 .. (b) : Magnetic field will be zero inside the straight thin walled tube according to ampere's theorem. . r 2 = radius of coil = 2 pn m 0i ´ n nm 0i ´ 2 pn 2 m0 in 2 p = = 2 r2 2l 2 l \ B ¢ = \ 2 B ¢ 2 m 0 in p 2 l = ´ = n 2 B 2l 2 m 0 ip \ B¢= n 2 B. 2 39. It will be on account of force of attraction between two adjacent turns carrying currents in the same direction.. (a) : A ferromagnetic material becomes paramagnetic above Curie temperature. (a) : mRw2 = BqRw Þ w = Bq Þ T = 2 pm m Bq T is independent of speed.(iii) 36...81 = = 0.(i) 2 2 2 Mass× (length) I ¢ = 12 ( m / 2)( l / 2) 2 ml 2 I = = = .. B ¢ = 28. (a) : B = az = \ 43.. (b) : Initially.09 W in parallel. T = 2 p I MB where I = ml 2 /12. (d) : S + G = I Þ S = I .78 JEE MAIN CHAPTERWISE EXPLORER \ \ or 2 2 3/2 ( R 2 + x 2 ) 3 / 2 B ¢ m 0 i 2( R + x ) = ´ = B 2 R m 0 iR 2 R 3 2 2 3/ 2 B ´ ( R2 + x 2 ) 3/ 2 = 54 ´ [(3) + (4) ] = 54 ´ 125 3 3 27 (3) R B¢ = 250 mT. 40... T 8 1 2 31.. 37. M = xl. x = pole strength When the magnet is divided into 2 equal parts. M = xl. 2 p 41. (a) : W = – MB (cos q 2 – cos q 1 ) MB = – MB (cos 60° – cos 0) = 2 \ MB = 2W . 10 - 1 9 33. the magnetic dipole moment x ´ l M M ¢ = Pole strength × length = = .I g \ I¢ M ¢B 2W ´ 3 = 3 W . x = pole strength of magnet 2 m lö 3 ml 2 I I ¢ = æç öæ ´ = = (For three pieces together) ÷ç ÷ è 3 øè 3 ø 12 9 ´ 12 9 M ¢ = ( x ) æç l ö÷ ´ 3 = xl = M (For three pieces together) è 3 ø \ T ¢ = 2 p \ T¢ = I ¢ = 2 p I / 9 = 1 ´ 2 p I = T M ¢ B MB 3 MB 3 T 2 = sec. E x = E y = 0 r r r r \ Net force on particle F = q ( E + v ´ B ) Resolve the motion along the three coordinate axis F q a x = x = ( E x + v y Bz - v z B y ) \ m m S= ay = F y q = ( E + v z Bx - v x B z ) m m y F z q = ( E + v x B y - v y B x ) m m z Since E x = E y = 0.(ii) 12 12 ´ 8 8 29.. 3 3 1 ´ 0.. On taking out the inductor from the circuit the current leads the voltage by 30°. I = 2 Blv Blv (c) I1 = I 2 = (d) I1 = I 2 = I = A fully charged capacitor C with initial charge q 0 is connected to a coil of self inductance L at t = 0. On taking out the capacitance from the circuit the current lags behind the voltage by 30°. The distance between their centres is 15 cm. Which of the following statement is correct? (a) At t = 6. The centre of the small loop is on the axis of the bigger loop. Calculate the value of R to make the bulb light up 5 s after the switch has been closed.79 Electromagnetic Induction and Alternating Current CHAPTER 14 1. 3.1 × 10 –11 weber –11 (c) 6 × 10 weber (d) 3. S 2 kept open.15 mV (2011) 5. Suppose t 1 is the time taken for the energy stored in the capacitor to reduce to half its initial value and t 2 is the time taken for the charge to reduce to onefourth its initial value. The current through the battery is (a) V ( R1 + R 2 ) at t = 0 and V at t = ¥ R1R2 R2 (b) VR1R 2 2 1 2 2 at t = 0 and R + R K V L R 1 R 2 V at t = ¥ R 2 (c) V at t = 0 and V ( R1 + R 2 ) at t = ¥ R2 R1R2 LC (2011) A resistor R and 2 mF capacitor in series is connected through a switch to 200 V direct supply.4) (a) 1.7 × 10 5 W 6 (c) 2. then the flux linked with bigger loop is (a) 6. Let C be the capacitance of a capacitor discharging through a resistor R. q = CV/2 (d) At t = 2t.3 cm lies parallel to a much bigger circular loop of radius 20 cm.7 × 10 W (d) 3. the magnitude of the induced emf in the wire of aerial is (a) 1 mV (b) 0. t .50 mV (d) 0. The I 1 Q setup is placed in a uniform magnetic field going into the plane of the paper. Now switch S 1 is closed.I 2 = 3 R Blv . The three currents I 1. q = CV(1 – e –1 ) 2 (b) Work done by the battery is half of the energy dissipated in the resistor (c) At t = t. (log 10 2. If the speed of the boat is 1.6 × 10 –9 weber (b) 9. The boat carries a vertical aerial 2 m long.50 m s –1 .75 mV (c) 0. VR1R 2 R12 + R2 2 at t = ¥ (2010) In a series LCR circuit R = 200 W and the voltage and the frequency of the main supply is 220 V and 50 Hz respectively.5 = 0. A circular loop of radius 0.3 × 10 4 W (b) 1. The power dissipated in the LCR circuit is (a) 242 W (b) 305 W (c) 210 W (d) zero W (2010) . 4. Across the capacitor is a neon bulb that lights up at 120 V. the key K is closed at t = 0. I 2 and I are (d) V at t = 0 and R 2 9. (q is charge on the capacitor and t = RC is capacitive time constant). The time at which the energy is stored equally between the electric and the magnetic fields is (a) p LC P l A rectangular loop has a sliding connector PQ of length l and resistance R W R W R W v R W and it is moving with a I I 2 speed v as shown. ELECTROMAGNETIC INDUCTION AND ALTERNATING CURRENTS In an LCR circuit as shown below both switches are open initially. In the circuit shown below. Blv . I = 2 Blv (b) I1 = . If a current of 2. I = Blv (a) I1 = I 2 = 6R (b) p LC 4 (c) 2 p LC (d) 3R R R 3 R R (2010) 7.3 × 10 7 W (2011) Blv .0 × 10 –5 N A –1 m –1 due north and horizontal.0 A flows through the smaller loop.3 × 10 –11 weber (2013) A boat is moving due east in a region where the earth’s magnetic field is 5. q = CV(1 – e –2 ) (2013) 2. Then the ratio t 1 /t 2 will be (a) 2 (b) 1 (c) 1 2 (d) 1 4 (2010) 8. After a long time A B the battery is disconnected after short E circuiting the points A and B. developed between the two ends of the conductor is (a) 5 mV (b) 50 mV (c) 5 mV (d) 50 mV.t /0. If each tube D C moves towards the other ´ ´ ´ ´ ´ ´ at a constant speed v. In a series resonant LCR circuit. will be (a) zero (b) 2Blv (c) Blv (d) –Blv (2005) 23.c. The axis of rotation is perpendicular to the field.8 (d) 0.3 s (c) 0. A circuit has a resistance of 12 ohm and an impedance of 15 ohm. The ´ ´ ´ ´ ´ magnetic field B is ´ perpendicular to the plane ´ ´ ´ ´ ´ ´ of the figure. 2 second after the connection is made. The current reaches half of its steady state value in (a) 0. The switch S is closed at t = 0. the current flowing in ampere in the circuit is (a) (1 – e –1 ) (b) (1 – e) (c) e (d) e –1 (2007) 13.5 × 10 –2 V (c) 40 V (d) 250 V.4 (2005) 20. In a uniform magnetic field of induction B a wire in the form of a semicircle of radius r rotates about the diameter of the circle with angular frequency w. their mutual inductance is (m 0 = 4p × 10 –7 T mA –1 ) (a) 2. the inductance should be changed from L to (a) 4L (b) 2L (c) L/2 (d) L/4. In order to impart maximum power at 50 Hz. the voltage across R is 100 volts and R = 1 kW with C = 2 mF. L (2005) 19.25 (b) 0. circuit the voltage applied is E = E 0 sin wt. The potential drop across L as a function of time is (a) 6e –5t V (b) E R 1 S L R 2 12 - 3 t e V t (c) 6 (1 - e . l and v where l is the width of each tube. (2004) 8 R 2 R .4p × 10 –5 H –4 (c) 4. If one of the solenoids has 300 turns and the other 400 turns. One conducting U tube can slide inside another as shown in figure. For the resonant frequency to remain unchanged. In an AC generator. (2008) 12. The power factor of the circuit will be (a) 1. An inductor of inductance L = 400 mH and resistors of resistances R 1 = 2 W and R 2 = 2 W are connected to a battery of emf 12 V as shown in the figure. If the horizontal component of earth’s magnetic field is 0.8p × 10 H (d) 4. The pö æ resulting current in the circuit is I = I 0 sin ç wt . At resonance the voltage across L is (a) 4 × 10 –3 V (b) 2. Which of the following cannot be the constituent of the circuit? (a) LC (b) L alone (c) C alone (d) R. it should be connected to a capacitance of (a) 1 mF (b) 2 mF (c) 4 mF (d) 8 mF (2005) 21.80 JEE MAIN CHAPTERWISE EXPLORER 10.125 (c) 0.f.15 s (b) 0.05 s (d) 0. (2006) 16. maintaining electrical contacts ´ ´ ´ ´ ´ ´ A B between the tubes. (2006) 18.8p × 10 –5 H. The phase difference between the alternating current and emf is p/2.1 s (2005) 22.m. If the total resistance of the circuit is R the mean power generated per period of rotation is 2 ( B pr 2w ) 2 (a) Bpr w (b) 8 R 2 R 2 ( B p r w 2 ) 2 ( Bpr w ) (d) (c) . The current in the circuit 1 ms after the short circuit is (a) 1 A (b) (1/e) A (c) e A (d) 0.1 A. The self inductance of the motor of an electric fan is 10 H. then the e. a resistor L (R = 100 W) and a battery (E = 100 V) are initially connected in series as R shown in the figure. The resonant frequency w is 200 rad/s.2 ) V (d) 12e –5t V (2009) 11. The power è ø consumption in the circuit is given by E0 I 0 (a) P = 2 E0 I 0 (b) P = 2 E (c) P = zero (d) P = 0 I 0 (2007) 2 14. then the emf induced in the circuit in terms of B. Two coaxial solenoids are made by winding thin insulated wire over a pipe of crosssectional area A = 10 cm 2 and length = 20 cm. The induced emf at t = 3 s is (a) 190 V (b) –190 V (c) –10 V (d) 10 V.2 × 10 –4 T.4p × 10 –4 H (b) 2. (2004) 24. An ideal coil of 10 H is connected in series with a resistance of 5 W and a battery of 5 V. An inductor (L = 100 mH). all of the same area A and total resistance R. (2006) 15. A coil of inductance 300 mH and resistance 2 W is connected to a source of voltage 2 V. The maximum value of emf generated in the coil is (a) NABw (b) NABRw (c) NAB (d) NABR (2006) 17. The flux linked with a coil at any instant t is given by f = 10t 2 – 50t + 250. rotates with frequency w in a magnetic field B. (2004) 25. a coil with N turns.2 ÷ . A metal conductor of length 1 m rotates vertically about one of its ends at angular velocity 5 radian per second. In an a. In a LCR circuit capacitance is changed from C to 2C. The internal resistance of the battery is negligible. C. The mutual inductance of the pair of coils depends upon (a) the rates at which currents are changing in the two coils (b) relative position and orientation of the two coils (c) the materials of the wires of the coils (d) the currents in the two coils.66 H (b) 9 H D (c) 0.66 H A 3 H 3 H 3 H (d) 1 H. 20. This combination is moved in time t seconds from a magnetic field W 1 weber to W 2 weber. (2004) 29.8 H 32. ammeter because (a) A. 35. 11.m. 15. 33. of 8 V is induced in a coil. 8. Alternating current cannot be measured by D.05 second.C.C. In a transformer. L. 16. (2002) 36. (2003) 31. changes direction (c) average value of current for complete cycle is zero (d) D. (2003) 30. then that in the secondary coil is (a) 4 A (b) 2 A (c) 6 A (d) 10 A.2 H (b) 0.C. The voltage across the LC combination will be (a) 50 V (b) 50 2 V (c) 100 V (d) 0 V (zero). 21.f. A conducting square loop of side + + + + ++ L and resistance R moves in its + + + + ++ + + + + ++ plane with a uniform velocity v + + + + ++ v perpendicular to one of its sides. (c) (a) (d) (c) (c) (d) . (c) (a) (d) (b) (a) (b) 6. The induced current in the circuit is W . 22.1 H. (b) (c) (b) (a) (b) (c) 3. cannot pass through D. Two coils are placed close to each other.W 1 ) (c) - 2 Rnt n(W . the voltage across each of the components. 12. 25. 24. 13. 23. ammeter (b) A. 27. 36.c. 19.4 H (c) 0. 9. (2004) 28. 31. The core of any transformer is laminated so as to (a) reduce the energy loss due to eddy currents (b) make it light weight (c) make it robust & strong (d) increase the secondary voltage. 30. an e. 32. (2002) Answer Key 1. In an oscillating LC circuit the maximum charge on the capacitor is Q. When the current changes from +2 A to –2 A in 0. + + + + ++ A magnetic induction B constant + + + + ++ + + + + ++ in time and space. 28. A coil having n turns and resistance R W is connected with a galvanometer of resistance 4R W. (2003) 4. 14. 26. The inductance between A and D is (a) 3. The charge on the capacitor when the energy is stored equally between the electric and magnetic field is (a) Q/2 (b) Q / 3 (c) Q / 2 (d) Q. (2002) 35. (2002) 34. C and R is 50 V. If current in primary coil is 4 A. number of turns in the primary coil are 140 and that in the secondary coil are 280. (2003) 33. ammeter will get damaged. 7. Rt (2004) 27. 10. (d) (a) (c) (d) (c) (d) (d) 0. The induced emf is (a) zero (b) RvB (c) vBL/R (d) vBL. (b) (d) (a) (a) (d) (b) 5. 34. The power factor of an AC circuit having resistance (R) and inductance (L) connected in series and an angular velocity w is (a) R/wL (b) R/(R 2 + w 2 L 2 ) 1/2 (c) wL/R (d) R/(R 2 – w 2 L 2 ) 1/2 . 29.C.W 1 (a) - 2 5 Rnt n(W . 18. 17. as shown in figure. (d) (d) (c) (c) (b) (d) 2.W 1 ) (d) - 2 . circuit. The coefficient of self induction of the coil is (a) 0. In an LCR series a.81 Electromagnetic Induction and Alternating Current 26.W 1 ) (b) - 2 5 Rt (W . pointing + + + + ++ perpendicular and into the plane L at the loop exists everywhere with half the loop outside the field. e . BH = 5. (c) : Emf induced across PQ is e = Blv. V 0 = 200 V. (b): S 200 V In case charging of capacitor through the resistance is V = V0 (1 - e .15 mV Here..5 m s –1 Induced emf. Charge on a capacitor at any time t is q = q 0(1 – e –t/t ) q = CV(1 – e –t/t ) [As q 0 = CV] t At t = 2 q = CV (1 .7 × 10 6 W 6.3 ´ 10 -2 )2 ´ 2 f1 = 2[(15 ´ 10 -2 ) 2 + (20 ´ 10 -2 )2 ]3/ 2 f 1 = 9. we get 4 p ´ 10 -7 ´ (20 ´ 10 -2 )2 ´ p ´ (0.50 × 2 = 15 × 10 –5 V = 0.t / RC ) As field due to current loop 1 at an axial point m 0 I1 R 2 \ B 1 = 2(d 2 + R 2 ) 3/2 Flux linked with smaller loop 2 due to B 1 is m 0 I1 R 2 f 2 = B1 A2 = pr 2 2(d 2 + R 2 ) 3/2 The coefficient of mutual inductance between the loops is f m 0 R 2 pr 2 M = 2 = I 1 2( d 2 + R 2 ) 3/ 2 Flux linked with bigger loop 1 is m 0 R 2 pr 2 I 2 f1 = MI 2 = 2(d 2 + R 2 ) 3/2 Substituting the given values. L P O e = Blv R M R I 1 I Q R I 2 N Applying Kirchhoff’s first law at junction Q. V = 120 V.916 R ´ 2 ´ 10 -6 Þ R = 2. we get . Now. R = ? C = 2 mF and t = 5 s. q = CV (1 . capacitor is charging through a resistor R.t /2 t ) = CV (1 - e -1/ 2 ) 4.6 or e -5/ R ´ 2 ´ 10 = 80 200 Taking the natural logarithm on both sides..e ( p p = LC 4w 4 Q w = 1 LC ) 2 m F R -2 t / t .(i) -2 ) = CV (1 - e ) 5.0.0 × 10 –5 N A –1 m –1 l = 2 m and v = 1..t / t ) = CV (1 - e -1 ) t= At t = 2t. (d) : Here. we get I = I 1 + I 2 .. e = BHvl = 5 × 10 –5 × 1. (b) : Charge on the capacitor at any instant t is q = q 0 coswt Equal sharing of energy means 1 Energy of a capacitor = Total energy 2 2 2 1 q 1 æ 1 q 0 ö Þ q = q 0 = 2 C 2 è 2 C ø 2 From equation (i) q 0 = q0 cos w t 2 1 cos wt = 2 p -1 1 w t = cos = 2 4 ( ) At t = t q = CV (1 .82 JEE MAIN CHAPTERWISE EXPLORER 1. (c) : 2. (d): As switch S 1 is closed and switch S 2 is kept open. we get -5 = ln(0.1 × 10 –11 weber 3.(i) Applying Kirchhoff’s second law for the closed loop PLMQP. The equivalent circuit diagram is as shown in the figure. -6 \ 120 = 200(1 - e -5/ R ´ 2 ´10 ) .e .4) = . (c) : I 1R + IR = Blv V . the maximum energy stored in the capacitor... we get ln1 . therefore the given series LCR is in resonance.(iv) 2 1 -2 t1 / RC = e 2 Taking natural logarithms of both sides. the inductor acts as a short circuit. ..(iii) and q 0 = q0 e . ln4 = 2ln2) = 2 V rms cos f Z (Q I rms = V rms Z ) . (d) : During discharging of capacitor through a resistor. we get As XL = XC . t ln1 .2 RC t 2 = RC ln4 = 2RC ln2 \ Impedance of the circuit is Z = R = 200 W The power dissipated in the circuit is P = V rmsI rms cosf \ t 1 RC ln 2 1 1 = ´ = t2 2 2 RC ln 2 4 ( . I .(iv) Blv 3 R Blv Substituting the value of I in equation (iii). R = 200 W. we get t = 0 R 1 R 2 At time t = 0.t / RC 2 q 1 ( q0 e U = 1 = 2 C 2 C ) (Using(i)) 2 q .(i) The energy stored in the capacitor at any instant of time t is 2 K X C R tan 30° = X C or X C = 1 R R 3 Taking natural logarithms of both sides of the above equation. (a) : Here. we get 2 t 1 or t1 = RC ln 2 2 RC From equation (iv).. R 1 and R 2 are in parallel) V ( R1 + R 2 ) R1R2 9. the phase difference between the current and voltage is X tan f = L R X tan 30 ° = L or X L = 1 R R 3 When only the inductance is removed. .(iii) Adding equations (ii) and (iii).. . the phase difference between current and voltage is tan f¢ = 1 = e ... u = 50 Hz When only the capacitance is removed.(ii) L Again.t2 / RC 4 V R 1 7. V rms = 220 V.(ii) = 1 0 e -2t / RC = U 0 e -2t / RC 2 C q 2 where U 0 = 1 0 . we get I 2 = 3 R Blv 2 Blv .83 Electromagnetic Induction and Alternating Current –I 1R – IR + e = 0 8... applying Kirchhoff’s second law for the closed loop PONQP.ln 2 = - (Q ln1 = 0) R 2 (ii) \ The current through the battery is I = V = Req = V R1 R 2 R1 + R2 ( .. I1 = I 2 = 3R 3 R Substituting this value of I in equation (ii). I = Hence. The corresponding equivalent circuit diagram is as shown in the figure (i). . 2 C According to given problem U 0 = U 0 e -2 t1 / RC 2 (Using (ii)) .t / RC (Using (i)) 4 From equation (iii).. the inductor acts as an open circuit. V 2IR + I 1R + I 2R = 2Blv 2IR + R(I 1 + I 2 ) = 2Blv 2IR + IR = 2Blv I (Using (i)) R 2 3IR = 2Blv (i) I = 2 Blv 3 R . we get I 1 = The current through battery is I = V R2 At time t = ¥.. q = q 0e –t/RC . The corresponding equivalent circuit diagram is as shown in the figure (ii).ln 4 = . we get –I 2R – IR + e = 0 I 2R + IR = Blv . we get .. .. (d) : R and L cause phase difference to lie between 0 and p/2 but never 0 and p/2 at extremities. p \ Power factor cos f = cos 2 = 0 Power consumption = E rms I rms cos f = 0. l = 20 cm. maximum emf = NABw. (c) : f = 10t – 50t + 250 d f = 20t - 50 \ dt . 13. But I 0 = I 2 12 V S R 2 -3 D t = t 0 = L = 400 ´ 10 H = 0. VL = I L w = 18. N 2 = 400. (b) : Induced emf = \ 1 1 Bwl 2 = ´ (0.Rt / L ) or 2 Rt L 1 = ln 2Þ t = ln 2 e . 19. (c) : Given : E = E 0 sin wt I = I 0 sin çæ wt . 23.d f Induced emf e = dt or E 17.c. I2 = I0(1– e –t/t ) where t = mean life or L/R.6 = 50 m V.2 s R 2 W \ I2 = 6(1 – e –t/0. 22.e 10 ÷ ÷ 5 ç è ø I = (1 – e –1 ). 21. generator. I = I 0 (1 – e –Rt/L ) I 0 = I 0 (1 - e . I = I 0 e –tR/L ( -1´10-3 )´100 I = (1) e 100´10 -3 1 I = e -1 = A. 1 1 1 1 1 = Þ = = \ L1C1 L2 C2 LC L2 (2C ) 2 L2 C \ L 2 = L/2.p ÷ö 2 ø è Since the phase difference (f) between voltage and current p is 2 .e L ÷ ç ÷ è ø æ .R t ö or I = E ç 1 . From f 2 = pr 12 (m 0ni)n 2 l. 14.2 ´ 10-4 )(5)(1) 2 2 2 10 Induced emf = -4 2 = 100 ´ 10 . (d) : Current I = Z 2 2 where E = VR + (VL - VC ) Z = R 2 + ( X L - X C ) 2 At resonance. (d) : During growth of charge in an inductance. 10. 2 24. (a) : The emf induced in the circuit is zero because the two emf induced are equal and opposite when one U tube slides inside another tube.20 = 2. X L = X C \ Z = R Again at resonance. power factor cosf = 1 0 \ P = 2 rms 2 V (220 V) = = 242 W.84 JEE MAIN CHAPTERWISE EXPLORER or e = –(20t – 50) = –[(20 × 3)– 50] = –10 volt or e = – 10 volt.693) t = 2 or t = 0. A = pr 1 2 = 10 cm 2 . C 11. -7 -4 m N N A M = 0 1 2 = 4p ´ 10 ´ 300 ´ 400 ´10 ´ 10 l 0.8. N 1 = 300.4p× 10 –4 H 12. B A E (across BC) = L dI 2 + R2 I 2 dt I 1 E 12 = = 6 A. (a) : For maximum power. .2 = 12e –5t V. Z (200 W ) 16.Rt / L = = 2 .1 = C w (2 ´ 10-6 ) ´ (200) V L = 250 volt.2 ) = 12e –t/0. V L = V C \ E = V R V 100 I = R = = 0. (a) : In an a. (c) : At resonance.1 sec.5 ´ 2 ö 5 æ ç1 . (c) : Power factor cos f = = Z 15 20. R2 2 L R 1 I2 = I0(1– e –t/t ). R 12 = 0. I1 = E/R1.R t ö I = I 0 ç 1 .1 or or 2 L R 300 ´ 10 -3 ´ (0. (a) : During the growth of current in LR circuit is given by æ . e 2 15. L circuit the potential difference across AD = VBC as they are parallel.2 ) Potential drop across L = E – R2I2 = 12 – 2 × 6(1 – e –t/0.e L ÷ = ÷ Rç è ø . (d) : For the given R. L w = 1 Cw 1 1 C = = \ Lw2 10 ´ (2 p ´ 50) 2 1 = = 10 -6 F 10 ´ 104 ´ ( p )2 or C = 1 mF. w = 1 LC when w is constant. (a) : M = m 0 n 1 n 2pr 1 2 l.1 A \ R 1 ´ 10 3 \ \ I 0. (b) : Maximum current I 0 = E = 100 = 1 A R 100 The current decays for 1 millisecond = 1 × 10 –3 sec During decay. t = t0 (given). At resonance. (b) : Magnetic flux linked 2 = BA cos wt = B pr cos wt 2 . 33.e .W 1 ) 1 I = =\ . for complete cycle is zero.1 H . (d) : L = . 34. can not be measured by D. (b) : I 2 N 2 = I 1 N 1 for a transformer I N 4 ´ 140 I 2 = 1 1 = = 2 A . The voltage across LC combination will be zero. 29. (c) : Average value of A.d f . . (a) : The energy loss due to eddy currents is reduced by using laminated core in a transformer. (d) : Induced emf = vBL.n d f . (d) : In an LCR series a. 4R 2 8 R . (c) : Mutual inductance between two coils depends on the materials of the wires of the coils. (b) : Power factor = .C.05 = = 0.85 Electromagnetic Induction and Alternating Current 25.c. 1æ 1 Q ç 2 çè 2 C 31. R + L2 w 2 36. circuit.C. \ N 2 280 35.W1 ) n (W2 .1 = B p r 2 w sin w t \ Induced emf e = dt 2 e 2 B 2 p 2 r 4 w 2 sin 2 wt = \ Power = R 4 R ( B pr 2 w) 2 2 = sin wt 4 R 2 <sin wt> = 1/2 Q \ Mean power generated ( B pr 2 w) 2 1 ( Bpr 2 w) 2 = ´ = . (d) : Three inductors are in parallel 3 H 27. the voltages across components L and C are in opposite phase.n dW = 26. ( R + 4 R ) t 5 Rt 30. 3 H A 28. Let q denote charge when energy is equally shared 2 \ \ ö 1 q 2 Þ Q 2 = 2 q 2 ÷÷ = 2 C ø q = Q / 2. (b) : Induced current I = ¢ where R dt R ¢ dt f = W = flux × per unit turn of the coil n (W2 . Hence A. di / dt - 4 32.C. (c) : Let Q denote maximum charge on capacitor.8 ´ 0. R 2 D 3 H \ \ 1 = 1 + 1 + 1 = 3 = 1 L( eq ) 3 3 3 3 1 L (eq) = 1 H. ammeter. (a) Statement1 is true. choose the one that best describes the two statements. The peak value of electric field strength is (a) 12 V/m (b) 3 V/m (c) 6 V/m (d) 9 V/m (2013) 2. (c) Statement1 is true. Consider telecommunication through optical fibres. (b) Statement1 is true. An electromagnetic wave of frequency u = 3. Then r r r r r r r r r r (a) X P E and k P E ´ B (b) X P B and k P E ´ B r r r r r r r r r r (c) X P E and k P B ´ E (d) X P B and k P B ´ E (2012) This question has Statement1 and Statement2. day to day and season to season. (2002) 9. (2004) 7. (c) Optical fibres have extremely low transmission loss. which are always perpendicular to r X and each other. (b) Optical fibres are subject to electromagnetic interference from outside. Statement2 is true. (c) (2002) .3 × 10 –3 J/m 3 (b) 4. It is located on a mountain top of height 500 m. (d) (c) 3. Electromagnetic waves are transverse in nature is evident by (a) polarization (b) interference (c) reflection (d) diffraction. Statement2 is not the correct explanation of Statement1. Statement2 : The state of ionosphere varies from hour to hour. Statement2 is true. The magnetic field in a travelling electromagnetic wave has a peak value of 20 nT. 8. 9. (d) Statement1 is false. The average total energy density of the electromagnetic wave is (a) 3. (c) (b) 2.58 × 10 –6 J/m 3 –9 3 (c) 6. The rms value of the electric field of the light coming from the sun is 720 N/C. Answer Key 1. (2011) 5.35 × 10 –12 J/m 3 . Statement1 : Sky wave signals are used for long distance radio communication. Statement2 is the correct explanation of Statement1. Which of the following statements is not true? (a) Optical fibres can be of graded refractive index. 4. These signals are in general. (2006) 6. The maximum distance upto which it can detect object located on the surface of the earth (Radius of earth = 6. Then (a) wavelength is doubled and the frequency remains unchanged (b) wavelength is doubled and frequency becomes half (c) wavelength is halved and frequency remains unchanged (d) wavelength and frequency both remain unchanged. (2002) 10.4 × 10 6 m) is (a) 16 km (b) 40 km (c) 64 km (d) 80 km (2012) 3. An electromagnetic wave in vacuum has the electric and r r magnetic fields E and B . (b) 6. 7. (a) (a) 4.0 MHz passes from vacuum into a dielectric medium with permitivity e = 4. Of the four choices given after the statements. (b) 5. (b) 10. statement2 is false. The direction of polarization is given by r that of wave propagation by k . Infrared radiation is detected by (a) spectrometer (b) pyrometer (c) nanometer (d) photometer.37 × 10 J/m (d) 81. (2003) 8.0. (d) Optical fibres may have homogeneous core with a suitable cladding. Which of the following are not electromagnetic waves? (a) cosmic rays (b) gamma rays (c) brays (d) Xrays. Statement2 is true. A radar has a power of 1 kW and is operating at a frequency of 10 GHz. less stable than ground wave signals.86 JEE MAIN CHAPTERWISE EXPLORER CHAPTER 15 ELECTROMAGNETIC WAVES 1. 5. frequency remains constant and wavelength changes 7.87 Electromagnetic Waves 1. (c) : In electromagnetic wave. the peak value of electric field (E 0 ) and peak value of magnetic field (B 0 ) are related by E 0 = B 0 c E 0 = (20 × 10 –9 T) (3 × 10 8 m s –1 ) = 6 V/m 2. . 8. 9. then (h + R) 2 = d 2 + R 2 d 2 = h 2 + 2Rh Q h < < R æ E 2 ö 2 = 1 e0 E rms + 1 ç rms ÷ 2 2 m 0 ç c 2 ÷ è ø 1 1 2 2 = e0 Erms + E e m 2 2 m 0 rms 0 0 1 1 2 2 2 = e0 E rms + e0 Erms = e 0 Erms 2 2 = (8. 10. (c) : brays are not electromagnetic waves. (a) : Polarization proves the transverse nature of electromagnetic waves. (d) : Maximum distance on earth where object can be detected is d. d = 2 ´ 6. m= \ d = 2 Rh 4.58 × 10 –6 Jm –3 . r r r \ k P E ´ B (b) 1 1 2 2 (b) : u = 2 e0 Erms + 2 m Brms 0 e = 4=2 e 0 1 Since m µ l \ Wavelength is halved Hence option (c) holds good.85 × 10 –12 ) × (720) 2 = 4. (b) : Optical fibres are subject to electromagnetic intereference from outside. (a) : The direction of polarization is parallel to electric field.4 ´ 106 ´ 500 = 8 ´ 10 4 m = 80 km 3. (b) : Infrared radiation produces thermal effect and is detected by pyrometer. h d 90° R R 6. (c) : During propogation of a wave from one medium to another. r r \ X P E The direction of wave propagation is parallel to r r E ´ B. The graph between angle of deviation (d) and angle of incidence (i) for a triangular prism is represented by (c) Direction : The question has a paragraph followed by two statements. choose the one that describes the statements. Statement2 is not the correct explanation of Statement1. (b) (c) 5.4 m in front of a lens forms a sharp image on a film 12 cm behind the lens.50 is interposed between lens and film with its plane faces parallel to film. of refractive index 1. Medium 1 in z ³ 0 has a refractive index of 2 and medium 2 with z < 0 has a refractive index of 3 . this film gives an interference pattern due to light reflected from the top (convex) surface and the bottom (glass plate) surface of the film. Statement2 is true. Statement1 : When light reflects from the airglass plate interface. At what distance (from lens) should object be shifted to be in sharp focus on film? (a) 2. so that the amplitude of the light from one slit is double of that from other slit. If I m be the maximum intensity. ( ) I f 1 + 8cos ) 9 ( 2 I m f 1 + 2cos 2 3 2 m 2 ( ) (b) I m f 1 + 4cos 2 5 2 (d) I m ( 4 + 5cos f ) 9 (a) (2012) 1 m s - 1 10 (c) 10 m s –1 (b) 1 m s - 1 15 (d) 15 m s –1 (2011) .4 m (b) 3. A ray of light in medium 1 given by the vector r A = 6 3 i$ + 8 3 $j . With monochromatic light.6 m (d) 7.10 k$ is incident on the plane of separation. The intensity of the emergent light is (a) I 0 /8 (b) I 0 (c) I 0 /2 (d) I 0 /4 (2013) 4. Statement2 is the correct explanation of Statement1. An object 2. (b) Statement1 is true.8 m behind the first car is overtaking the first car at a relative speed of 15 m s –1 . (c) Statement1 is true. Statement2 : The centre of the interference pattern is dark. Statement2 is true. Statement2 is false. the reflected wave suffers a phase change of p. The fringes obtained on the screen will be (a) concentric circles (b) points (c) straight lines (d) semicircles A beam of unpolarised light of intensity I 0 is passed through a polaroid A and then through another polaroid B which is oriented so that its principal plane makes an angle of 45° relative to that of A. Of the given four alternatives after the statements. Two coherent point sources S 1 and S 2 are separated by a small distance ‘d’ as shown. Statement2 is true. The angle of refraction in medium 2 is (a) 30° (b) 45° (c) 60° (d) 75° (2011) 9. A second car 2. Let the x–z plane be the boundary between two transparent media. (2011) 8.88 JEE MAIN CHAPTERWISE EXPLORER CHAPTER OPTICS 16 1. Diameter of a planoconvex lens is 6 cm and thickness at the centre is 3 mm. the focal length of the lens is (a) 10 cm (b) 15 cm (c) 20 cm (d) 30 cm (2013) 2. the resultant intensity I when they interfere at phase difference f is given by (a) 7. A glass plate 1 cm thick.2 m (c) 5.2 m (2012) (2013) 3. Statement1 and Statement2. one of the slit is wider than other. (a) Statement1 is true. (a) 6. If speed of light in material of lens is 2 × 10 8 m/s. A thin air film is formed by putting the convex surface of a planeconvex lens over a plane glass plate. (d) Statement1 is false. The speed of the image of the second car as seen in the mirror of the first one is (2013) (d) In Young’s double slit experiment. A car is fitted with a convex sideview mirror of focal length 20 cm. 520 for red light and 1. q The incident angle q for which the light ray grazes along the wall of the rod is 1 (a) sin - 1 (b) sin - 1 3 2 2 - 1 2 - 1 1 (c) sin (d) sin (2009) 3 3 15. If the refractive C index of water is 4/3 and the fish is 12 cm below the surface. the wavelength of the unknown light is (a) 393. it will (a) travel as a cylindrical beam (b) diverge (c) converge (d) diverge near the axis and converge near the periphery ( ) v (cm) u (cm) equal to 3 (a) 4 (b) 1 2 (c) 3 2 1 (d) 2 (2007) 18. illuminates Young’s double slit and gives rise to two overlapping interference patterns on the screen. If I 0 denotes the maximum intensity. A student measures the focal length of a convex lens by putting an object pin at a distance u from the lens and measuring the distance v of the image pin. Then (a) D 1 > D 2 (b) D 1 < D 2 (c) D 1 = D 2 (d) D 1 can be less than or greater than depending upon the angle of prism. Its optical power in a liquid medium with refractive index 1. Two lenses of power –15 D and +5 D are in contact with each other. As the beam enters the medium. An initially parallel cylindrical beam travels in a medium of refractive index m(I) = m 0 + m 2I. A mixture of light. what is the maximum distance at which these dots can be resolved by the eye? [Take wavelength of light = 500 nm] (a) 6 m (b) 3 m (c) 5 m (d) 1 m (2005) 22. Two point white dots are 1 mm apart on a black paper. where m 0 and m 2 are positive constants and I is the intensity of the light beam. A thin glass (refractive index 1. consisting of wavelength 590 nm and an unknown wavelength. The initial shape of the wavefront of the beam is (a) planar (b) convex (c) concave (d) convex near the axis and concave near the periphery 11.4 nm (b) 885. Let D 1 and D 2 be angles of minimum deviation for red and blue light respectively in a prism of this glass. A transparent solid cylindrical rod has a refractive index of 2 . it is observed that the third bright fringe of known light coincides with the 4 th bright fringe of the unknown light. A light ray is incident at the mid 3 point of one end of the rod as shown in the figure. It is surrounded by air. The intensity of the beam is decreasing with increasing radius.0 nm (c) 442. I is 0 12. From this data. the radius of this circle in cm is (a) 36 5 (b) 4 5 (c) 36 7 (d) 36 / 7 12 cm (2005) 21.89 Optics Directions : Questions number 1012 are based on the following paragraph. The refractive index of glass is 1. The speed of light in the medium is (a) maximum on the axis of the beam (b) minimum on the axis of the beam (c) the same everywhere in the beam (d) directly proportional to the intensity I (2010) 13. They are viewed by eye of pupil diameter 3 mm. Approximately. The graph between u and v plotted by the student should look like ( ) ( ) ( ) v (cm) v (cm) (a) (b) O u (cm) O v (cm) (c) (d) O O u (cm) u (cm) (2008) 16. The focal length of the combination is (a) + 10 cm (b) –20 cm (c) –10 cm (d) + 20 cm (2007) 17. 10. When an unpolarized light of intensity I 0 is incident on a polarizing sheet.5 nm (d) 776. The central maximum of both lights coincide. A fish looking up through the R water sees the outside world contained in a circular q Cq horizon. (2006) 19. the intensity of the light which does not get transmitted is .8 nm (2009) 14.525 for blue light. Further. In a Young’s double slit experiment the intensity at a point l where the path difference is (l being the wavelength of 6 I light used) is I.6 will be (a) 25 D (b) –25 D (c) 1 D (d) –1 D (2005) 20.5) lens has optical power of –5 D in air. 9. Now this lens has been used to form the image of an object. (2004) 29. then the number of images formed by them is (a) 5 (b) 6 (c) 7 (d) 8. If two mirrors are kept at 60° to each other. An astronomical telescope has a large aperture to (a) reduce spherical aberration (b) have high resolution (c) increase span of observation (d) have low dispersion. 21. 2 27. (2003) 31. The image formed by an objective of a compound microscope is (a) virtual and diminished (b) real and diminished (c) real and enlarged (d) virtual and enlarged. 19. (2002) 33. 11. 26. (2004) 28. A Young’s double slit experiment uses a monochromatic source. If I 0 is the intensity of the principal maximum in the single slit diffraction pattern. 27. 34. 23. 15. (c) (c) (b) (a) (c) . 10. To demonstrate the phenomenon of interference we require two sources which emit radiation of (a) nearly the same frequency (b) the same frequency (c) different wavelength (d) the same frequency and having a definite phase relationship. Wavelength of light used in an optical instrument are l 1 = 4000 Å and l 2 = 5000 Å. one should have two plane mirrors at an angle of (a) 60° (b) 90° (c) 120° (d) 30°. 12. (2004) 26. The maximum number of possible interference maxima for slitseparation equal to twice the wavelength in Young’s doubleslit experiment is (a) infinite (b) five (c) three (d) zero. 8. 24. 16. (d) (c) (c) (*) (a) (d) 2. 30. 13. then ratio of their respective resolving powers (corresponding to l 1 and l 2 ) is (a) 16 : 25 (b) 9 : 1 (c) 4 : 5 (d) 5 : 4.90 JEE MAIN CHAPTERWISE EXPLORER (b) I 0 1 (d) I 0 4 (a) zero 1 (c) I 0 2 (2005) 23. 31. 17. is (a) sin –1 (n) (b) sin –1 (1/n) –1 (c) tan (1/n) (d) tan –1 (n). (2003) 32. Which of the following is used in optical fibres? (a) total internal reflection (b) scattering (c) diffraction (d) refraction. (2003) 30. (2002) 34. (a) (b) (d) (d) (b) (b) 3. 18. At what distance from this lens an object be placed in order to have a real image of the size of the object? (a) 20 cm (b) 30 cm (c) 60 cm (d) 80 cm. If the angle of reflection is 45°. The angle of incidence at which reflected light in totally polarized for reflection from air to glass (refractive index n). 22. (d) (b) (d) (c) (d) (a) 4. 7. (c) (b) (a) (a) (b) (a) 6. 33. (2002) Answer Key 1. To get three images of a single object. The shape of the interference fringes formed on a screen (a) straight line (b) parabola (c) hyperbola (d) circle (2005) 25. we conclude 45° that the refractive index n 1 (b) n > 2 (a) n < 2 1 (2004) (c) n > (d) n < 2 . A plano convex lens of refractive index 1. 29. 20. 35. A light ray is incident perpendicular to one face of a 90° prism and is totally internally reflected at the glassair 45° interface. 14. (2002) 35. 25. 32. 28. (d) (a) (c) (c) (b) (d) 5.5 and radius of curvature 30 cm is silvered at the curved surface. then what will be its intensity when the slit width is doubled? (a) I 0 (b) I 0/2 (c) 2I 0 (d) 4I 0 (2005) 24. (3 cm) 2 = 15 cm 2 ´ 0. so shift produced by it will be ( ) ( Shift = t 1 - ) ( ) 1 1 2 1 =1 1=1 1= cm m 1. I = I1 + I 2 + 2 I1I 2 cos f 8 \ m= c 3 ´ 10 m/s 3 = = v 2 ´ 108 m/s 2 = I1 + 4 I1 + 2 I1 (4I1 ) cos f = 5I 1 + 4I 1 cosf = I 1 + 4I 1 + 4I 1 cosf = I 1 + 4I 1 (1 + cosf) .. A 2 = 2A 1 . the radius of curvature R of its curved surface in accordance with figure will be given by = I1 + 8 I1 cos 2 R 2 = r 2 + (R – t) 2 ( 2Rt = r 2 + t 2 r 2 (Q r >> t ) 2 t Here. 12 cm According to thin lens formula 1 = 1 - 1 f v u Here.91 Optics 5. 2 f 2 Putting the value of I 1 from eqn. u = – 2. (ii) and (iii) in (i).1 = 1 + 1 f 12 ( - 240) 12 240 1 = 21 or f = 240 cm f 240 21 When a glass plate is interposed between lens and film. (i).5 3 3 To get image at film.4 m = – 240 cm. r = 3 cm.(i) Resultant intensity. lens should form image at distance v ¢ = 12 - 1 35 = cm 3 3 Again using lens formula . (c) : Here.(i) R (m - 1) or I1 = As speed of light in the medium of lens is 2 × 10 8 m/s I m 9 .4 m 0 (d) : The graph between angle of deviation(d) and angle of incidence (i) for a triangular prism is as shown in the adjacent figure.5 - 1) 2. . (d): According to lens maker’s formula 1 1 1 ù = (m . 0 ) 1 cm O 2. the fringes will be concentric circles.. we get = I1 1 + 8cos 2 R 2 = r 2 + R 2 + t 2 – 2Rt \ (Q 1 + cos f = 2cos f2 ) f 2 1 = 1 . t = 3 mm = 0. v = 12 cm \ Now this light will pass through the second polaroid B I 2 = I1 cos 2 45 ° = ( I m f 1 + 8cos 2 9 2 Film (a) : When the screen is placed perpendicular to the line joining the sources.. the inten‑ sity of light emerging from polaroid B is ( I2 )( 1 2 ) = I 4 2 4..(ii) If r is the radius and t is the thickness of lens (at the centre).3 cm I= R= R= 6. we get 15 cm f = = 30 cm. So in accordance with Malus law. R 2 = ¥ \ I 2 = 4 I1 2 Maximum intensity. . I1 = ) (c) : On substituting the values of m and R from Eqs..1) é ú f ëê R1 R2 û As the lens is planoconvex \ R 1 = R. (1. Intensity µ (Amplitude) 2 2 2 I æA ö æ 2 A ö \ 2 = ç 2 ÷ = ç 1 ÷ = 4 è A1 ø I1 è A1 ø 1. I m = ( I1 + I 2 ) 1 (m . (d) : Intensity of light after passing polaroid A is I 0 2 whose axis is inclined at an angle of 45° to the axis of polaroid A..1) = f R or f = 2 2 = ( I1 + 4 I1 ) = (3 I1 ) = 9 I1 .3 cm 3. .21 ù u ¢ 35 240 5 ë 7 48 û 240 35 u ¢ \ 1 1 é144 ...1 = sin -1 3 3 So that q c is making total internal reflection.10 k^ 10 cos i = (6 3 )2 + (8 3 ) 2 + (-10) 2 = 10 20 ( ) cos i = 1 or i = cos -1 1 = 60 ° 2 2 Using Snell’s law.92 JEE MAIN CHAPTERWISE EXPLORER 21 = 3 . (c) 13.5 nm. A = 6 3 i^ + 8 3 j^ . The point which meets the curve at u = v gives 2f. (b) : Given m = m 0 + m 2 I Speed of light in vacuum As m = Speed of light in medium c c m = c or v = m = m + m I v 0 2 As the intensity is maximum on the axis of the beam.1 = R D2 for blue m B - 1 Since m B > m R. . the path difference xd xd l = n l for bright fringes and = (2 n + 1) D D 2 for getting dark fringes. 1 1 Focal length of combination F = P = - 10 D = – 0. m (b) 10. u is negative. 11. u = f and if u = ¥. q i = q. I 0 = maximum intensity æfö I = cos 2 ç ÷ . 12. the shape of wavefront is planar. by Young’s double slits. (c) : Power of combination = P 1 + P 2 = – 15 D + 5 D = – 10 D. The central fringes when x = 0. m 1 sini = m 2 sinr 2 sin 60° = 3 sin r Þ r = 45 ° 9. both are symmetrical and this curve satisfies all the conditions for a convex lens. I 0 è ø 0 18. (ii) l 6 3 Substitute eqn. (a) : In Young’s double slit experiment intensity at a point is given by æ fö I = I 0 cos 2 ç ÷ è 2 ø where f = phase difference. (a) : As the beam is initially parallel. sin f cos qc 3 qc m2 - 1 cos qc = m2 . (d) : According to the new cartesian 1 1 1 system used in schools. u has to be negative.6 m If q c has to be the critical angle. (c) : For interference.= for v u f a convex lens. v = f. m m (b) : \ q = sin -1 r Here. 17. coincide for all wavelengths. 4 14. The third fringe of l 1 = 590 nm coincides with the fourth bright fringe of unknown wavelength l. . (d) : q f qc m = ref.1 \ sin q = m = m 2 .. (c) 8.147 ù 1 3 = or =u ¢ 5 ëê 336 ûú u ¢ 1680 u¢ = – 560 cm = – 5.1 m = – 10 cm. D 1 < 1 \ D2 \ D 1 < D 2 . the rays are parallel. qc = sin -1 1 m But q c = 90° – f.1 m 2 . index of the rod ( ) 4 1 . v is +ve f f u negative If v = ¥. Therefore v is +ve. (ii) in eqn. therefore v is minimum on the axis of the beam. 15. \ xd = 3 × 590 nm = 4 × l nm D \ l= 3 ´ 590 = 442. we get I 3 I = cos 2 æ p ö ç 6 ÷ or I = 4 . A parallel beam (u = ¥) is focussed at f and if the object is at f. (b) : Angle of minimum deviation D = A(m – 1) D1 for red m .1.6 m |u¢| = 5. 1 But 7.21 = 1 é 3 . (i) or I 0 è 2 ø Phase difference f = 2 p ´ path difference l 2p l p f= ´ f = \ or . 16.1 or 1 = 3 . (i). sin qi = m = 2 Þ sin q = m. (a) : For diffraction pattern 30.1 = -1 Pa 1. \ \ 26. (c) : Resolution limit = 1. 2 2 23.1 = 60 ° .1 D 22. (d) : For interference phenomenon. P . 34.22 l = D d yd D= 1. R . P . (a) : Straight line fringes are formed on screen. 32. (a) : Total internal reflection is used in optical fibres. sinq is never greater than (+1).22 l n > 27. (d) : According to Brewster's law of polarization.1) (m g / m l ) .1 q0 360 ° 3= .m l = 1. q° 29.22 l d y 1. ( * ) : f = ( m g .5 - 1) or P l = . (a) : n = q° .1) 1.B.5 y D 360 ° . Given : i = 45° in the medium and total internal reflection occurs at the glass air interface \ sin qc Again resolution limit = sin q = q = R 30 = = 10 cm 2m 2 ´ 1.1 = 5. æ 1 1 1 ö a 19. R 3 36 = Þ R = cm. for l 2 l 1 4000 4 360° 360 ° 35. (c) : Intensity of polarized light = I 0 /2 \ Intensity of light not transmitted I I = I 0 . P a ( -5) 5 == 8 8 8 or Optical power in liquid medium = 5 Dipotre.1 = a = fl (m g .1.1) çè R . –1.1) ( m g - 1) = m g .1) ç . : This answer is not given in the four options provided in the question.÷ fl è R1 R2 ø 24. 12 7 7 21. 31. 1. 25.6 (1. the object should be placed at (2F) of the concave mirror. 33. \ Distance of object from lens = 2 × F = 2 × 10 = 20 cm. sin C sin 45 ° 28. 8 N.R ÷ø a 1 2 æ 1 1 1 ö = (l m g .22) ´ (5 ´ 10-7 ) 6. \ F of concave mirror so formed l \ f a ( m g . 20.0 = 0 . m= \ 1 1 3 Þ sin qC = = sin qC m 4 tan qc = = 1 . (b) : For interference maxima. (a) : A planoconvex lens behaves like a concave mirror when its curved surface is silvered.0. dsinq = nl \ 2lsinq = nl n sin q = or 2 This equation is satisfied if n = –2. Hence ç f ÷ = 1 è ø \ R .6 m l (m g .sin 2 q c 3/ 4 = 3 ´ 4 = 3 4 7 7 1 . . (1.9 16 or or D= 1 > 1 > 2 . (b) : Total internal reflection occurs in a denser medium when light is incident at surface of separation at angle exceeding critical angle of the medium.6 ´ 0. (d) : For total internal reflection.93 Optics Hence intensity remains constant at I 0 I = I 0 (1) = I 0 .5 . (b) : Large aperture leads to high resolution of telescope. less than (–1) \ Maximum number of maxima can be five. 2. two sources should emit radiation of the same frequency and having a definite phase relationship.5 8 Þ Pl = - = To form an image of object size. for l1 l 2 5000 5 = = = . (d) : Resolving power is proportional to l –1 2 æ sin f ö I = I 0 ç ÷ where f denotes path difference è f ø æ sin f ö For principal maxima.1 Þ 4 q° = 360° Þ q° = 90°. f = 0. n = tan i p where i p is angle of incidence \ i p = tan –1 (n). (b) : n = \ y q (10 -3 ) ´ (3 ´ 10-3 ) 30 = » 5 m. 0. (c) : The objective of compound microscope forms a real and enlarged image. statement2 is false.94 JEE MAIN CHAPTERWISE EXPLORER CHAPTER 17 1. Statement2 is the correct explanation of Statement1. Statement 2 is true. Of the four choices given after the statements. (c) Statement 1 is true. DUAL NATURE OF MATTER AND RADIATION The anode voltage of a photocell is kept fixed. (b) O 2. Statement2 is false. choose the one that best describes the two statements. Statement2 : The maximum kinetic energy and the stopping potential of photoelectrons emitted from a surface are linearly dependent on the frequency of incident light. The plate current I of the photocell varies as follows I I (a) O l If a source of power 4 kW produces 10 20 photons/second. Statement 2 is true. The surface of a metal is illuminated with the light of 400 nm. choose the one that best describes the two statements : Statement1 : A metallic surface is irradiated by a monochromatic light of frequency u > u 0 (the threshold frequency). 8 and 9 are based on the following paragraph. they can interfere and show diffraction. Statement 1 : Davisson Germer experiment established the wave nature of electrons.41 eV (c) 1. its stopping potential is V 0 and the maximum kinetic energy of the photoelectrons is K max .09 eV (b) 1. incoming electrons outgoing electrons i d crystal plane . Statement 2 is true. (d) Statement1 is false. (d) Statement1 is false. Statement2 is true; Statement2 is not the correct explanation of Statement1. the radiation belongs to a part of the spectrum called (a) grays (b) Xrays (c) ultraviolet rays (d) microwaves (2010) l 5. Statement 2 is not the correct explanation of Statement 1. Wave property of electrons implies that they will show diffraction effects. Statement2 is true; Statement2 is the correct explanation of Statement1. The maximum kinetic energy and the stopping potential are K max and V 0 respectively. Statement2 : Photoelectrons are emitted with speeds ranging from zero to a maximum value because of the range of frequencies present in the incident light. Statement2 is true.68 eV (2009) Directions : Questions 7. (2012) This question has Statement1 and Statement2. (2008) 6. (c) Statement1 is true. Statement 2 is the correct explanation for Statement 1. When the ultraviolet light is replaced by Xrays. (a) Statement1 is true. If the frequency incident on the surface is doubled. 4. Statement1 : When ultraviolet light is incident on a photocell. The law governing the diffraction from a crystal is obtained by requiring that electron waves reflected from the planes of atoms in a crystal interfere constructively (see figure). (2010) I I (c) (2013) (d) O 3.68 eV. (a) Statement1 is true. Statement 2 is false. The kinetic energy of the ejected photoelectrons was found to be 1. both V 0 and K max increase. The wavelength l of the light falling on the cathode is gradually changed. (d) Statement 1 is false. Statement2 is not the correct explanation of Statement1. Of the four choices given after the statements. (b) Statement 1 is true. The work function of the metal is (hc = 1240 eV nm) (a) 3. (c) Statement1 is true. Statement2 is true. (b) Statement1 is true. Statement2 is true. Davisson and Germer demonstrated this by diffracting electrons from crystals. (a) Statement 1 is true.51 eV (d) 1. Statement2 is true. Statement 2 : If electrons have wave nature. (b) Statement1 is true. (2011) l O l This question has Statement 1 and Statement 2. both the K max and V 0 are also doubled. If c is the velocity of light. If g E and g M are the accelerations due to gravity on the surfaces of the earth and the moon respectively and if Millikan’s oil drop experiment could be performed on the two surfaces. (2006) 15. A charged oil drop is suspended in a uniform field of 3 × 10 4 V/m so that it neither falls nor rises.6 × 10 –19 C) (a) 1000 V (b) 2000 V (c) 50 V (d) 500 V. The charge on the drop will be (take the mass of the charge = 9. 9. D Which of the following graphs can be expected to represent the number of electrons N detected as a function of the detector position y (y = 0 corresponds to the middle of the slit)? y (a) N d d y (c) (d) N d 10. V should be about (h = 6. 8. If a strong diffraction peak is observed when electrons are incident at an angle i from the normal to the crystal planes with distance d between them (see figure). (2006) y =0 d I (a) (2007) 11. the number of electrons emitted by photocathode would (a) decrease by a factor of 2 (b) increase by a factor of 2 (c) decrease by a factor of 4 (d) increase by a factor of 4 (2005) y (b) N I (b) 14. of the incident radiation gives a straight line whose slope (a) depends on the nature of the metal used (b) depends on the intensity of the radiation (c) depends both on the intensity of the radiation and the metal used .8 × 10 –18 C. When the same source of light is placed (1/2) m away. and the stopping potential for a radiation incident on this surface 5 V. The wavelength l of the light falling on the cathode is gradually changed. its de Broglie wavelength changes by the factor (c) 1/2 (d) 2. the momentum is (a) hu/c (b) u/c (c) huc (d) hu/c 2 (2007) 12.0 eV.95 Dual Nature of Matter and Radiation 7. (2005) (a) 1/ 2 (b) 2 16. (2004) 19. e = 1. The threshold frequency for a metallic surface corresponds to an energy of 6. O (2006) l 13. de Broglie wavelength l dB of electrons can be calculated by the relationship (n is an integer) (a) d cosi = nl dB (b) d sini = nl dB (c) 2d cosi = nl dB (d) 2d sini = nl dB In an experiment. Photon of frequency u has a momentum associated with it. The work function of a substance is 4. electrons are made to pass through a narrow slit of width d comparable to their de Broglie wavelength. The time by a photoelectron to come out after the photon strikes is approximately (a) 10 –1 s (b) 10 –4 s –10 (c) 10 s (d) 10 –16 s. The longest wavelength of light that can cause photoelectron emission from this substance is approximately (a) 540 nm (b) 400 nm (c) 310 nm (d) 220 nm. (2004) 18. Electrons accelerated by potential V are diffracted from a crystal. The incident radiation lies in (a) Xray region (b) ultraviolet region (c) infrared region (d) visible region. If the kinetic energy of a free electron doubles. According to Einstein’s photoelectric equation.9 × 10 –15 kg and g = 10 m/s 2 ) (a) 3. the plot of the kinetic energy of the emitted photo electrons from a metal vs the frequency.6 × 10 –34 Js. one will find the ratio electronic charge on the moon electronic charge on the earth to be (a) g M /g E (c) 0 (b) 1 (d) g E /g M O O l l I I (c) (d) O l . A photocell is illuminated by a small bright source placed 1 m away. The plate current I of the photocell varies as follows 17. m e = 9.2 × 10 –18 C –18 (c) 1. If d = 1 Å and i = 30°.6 × 10 C (d) 4. The anode voltage of a photocell is kept fixed.3 × 10 –18 C (b) 3.1 × 10 –31 kg.2 eV. They are detected on a screen at a distance D from the slit. 7. 20. Sodium and copper have work functions 2. (a) 11.f 2 ) ùú 2 . Then the ratio of the wavelengths is nearest to (a) 1 : 2 (b) 4 : 1 (c) 2 : 1 (d) 1 : 4. (a) 17. (b) (c) (c) (a) 3. then 2 2 h 2 (a) v1 . 9.3 eV and 4. (c) . 8. 13. If the velocities of the photoelectrons (of mass m) coming out are respectively v 1 and v 2 . (a) 6. 14. 21. (2004) 20. (b) 10. (b) 12. (2002) Answer Key 1. (a) (c) (b) (d) 2. (b) 16. (c) 18.v2 = éê 2 h ( f1 . (d) 5. 15.v2 = m ( f1 - f 2 ) 1 (b) v1 + v2 = éê 2 h ( f1 + f 2 ) ùú 2 ëm û 2 2 2 h (c) v1 + v2 = m ( f1 + f 2 ) 1 (d) v1 . 19. ëm û (2003) 21.96 JEE MAIN CHAPTERWISE EXPLORER (d) is the same for all metals and independent of the intensity of the radiation. Two identical photocathodes receive light of frequencies f 1 and f 2 .5 eV respectively. (d) (a) (a) (c) 4. e.27 ´ 10 -10 ) 3 ´ 10 -10 V = 50. The stopping potential is eV 0 = K max = hu – hu 0 Therefore.1 – 1. N = 10 20 i q Energy of photon. u 0 is threshold frequency.1 × 10 –31 kg. 1240 eV nm i. 13. (a) : According to Einstein’s photoelectric equation K max = hu – f 0 where. d = 1 Å. It does not depend on g.68 eV = 1. (i) . . the work function = hu – kinetic energy = 3. 400 nm Max. 3. (ii) 12. 5. d = 4. hu = = 3. This shows the wave nature of electrons as waves can exhibit interference and diffraction. P = 4 kW = 4 × 10 3 W Number of photons emitted per second. c . (b) : DavissonGermer experiment showed that electron beams can undergo diffraction when passed through atomic crystals.6 × 10 –19 C. i = 30° i. (a) : Energy of a photon E= hu Also E = pc where p is the momentum of a photon From (i) and (ii). grazing angle q = 60°. h = 6. which is also used in electron diffraction gives nl = 2d sinq.. (c) : The graph (c) depicts the variation of l with I. 4. E = hu = hc l P . (b) : For photoelectron emission.97 Dual Nature of Matter and Radiation 1. if u is doubled K max and V 0 is not doubled. (a) : The electron diffraction pattern from a single slit will be as shown below. m e = 9. E = N hc = P l N 20 -34 8 Nhc 10 ´ 6. (a) 2. 6. . 9.68 eV. (Incident energy E) = (K. V = (12.1 eV. 7.6 × 10 –19 C) universal constant. Power of a source.. \ Electronic charge on the moon = electronic charge on the earth electronic charge on the moon = 1. (c) : For electron diffraction.63 ´ 10 ´ 3 ´ 10 or l = = 3 P 4 ´ 10 But as the angle of incidence is given.972 × 10 –9 m = 49. Statement2 is false. hu = Wf + kinetic energy \ Wf . 2p The line of maximum intensity for the zeroth order will exceed d very much.18 Volt.E. (b) : Here. or electronic charge on the earth 11.. 8.6 × 10 –34 J s. (d) : The maximum kinetic energy of the electron K max = hu – hu 0 Here.) m ax + (Work function f) .e. (b) : The wavelength of light illuminating the photoelectric surface = 400 nm. nl = 2d cosi is the formula for finding a peak... both K max and V 0 increase when ultraviolet light is replaced by Xrays.42 eV. kinetic energy of the electrons = 1. e = 1. (b) : Since electronic charge (1. \ l = ° 2 ´ 1(A) ´ sin 60 ° (assuming first order) 1 ° l = 3 A. d sin q = 10..72 Å It lies in the Xray region. u = frequency of incident light f 0 = work function of the metal Since K max = eV 0 h u f0 V 0 = - e e As u X rays > u Ultraviolet Therefore. q D x l . we get h u hu = pc or p = . Bragg’s equation for Xrays. (c) : Bragg’s relation nl = 2d sinq for having an intensity maximum for diffraction pattern. P remains constant as the source is the same. (a) : de Broglie wavelength l = h / p = h / (2mK ) h l = \ where K = kinetic energy of 2 mK particle \ l2 K1 K 1 1 = = = . or or 14. we assume light to spread uniformly in all directions.2 ´ 1. (a) : For equilibrium of charged oil drop. Option (d) represents the answer. E (3 ´ 104 ) 18.2 = 11. l m = 19. 16.f 2 ) \ 2 1 2 h v12 . Hence the slope is same for all metals and independent of the intensity of radiation. \ m 1 21. The incident radiation lies in ultra violet region.f 2 2 1 m ( v 2 . Kinetic energy of emitted electron = hf – (work function f) 1 2 mv = hf1 .63 ´ 10 ) ´ (3 ´ 10 ) = f 4.3 ´ 10-18 C.6 ´ 10-19 or l = 1110 × 10 –10 m = 1110 Å. qE = mg \ mg (9.63 ´ 10 -34 ) ´ (3 ´ 108 ) l= m or 11. n1 çè P ÷ç øè 1/ 2 ø 1 17.6 × 10 –19 ) J hc = 11. y = mx + c Kinetic energy q f f Work function \ slope of line = h h is Planck's constant.f \ 2 1 1 2 mv = hf 2 . (c) : Work function = hc/l W Na 4.v22 = ( f - f 2 ).2 × (1. l 1 K2 2 K1 2 P of source P 4 p (distance) 4 p d 2 Here.5 2 = = . according to Einstein's equation. (c) : Emission of photoelectron starts from the surface after incidence of photons in about 10 –10 sec. (d) : I = 2 = 2 \ I 2 n2 P æd ö n = Þ 2 ç 1 ÷ = 2 I1 n1 P1 è d 2 ø n1 \ n 2 æ P öæ 1 ö2 4 = ÷ = .v22 ) = h ( f1 .0 ´ 1. WCu 2.6 ´ 10 -19 l m = 310 nm. Number of photoelectrons emitted from a surface depend on intensity of light I falling on it. 15. (c) : Let l m = Longest wavelength of light hc = f (work function) \ l m \ or -34 8 hc (6.98 JEE MAIN CHAPTERWISE EXPLORER E = K m + f E = 5 + 6. Thus the number of electrons emitted n depends directly on I. (d) : According to Einstein's equation.6 ´ 10 -19 \ l (6.2 ´ 1. compare it with equation.3 1 . Kinetic energy = hf – f where kinetic energy and f (frequency) are variables.2 eV = 11. 20.9 ´ 10 -15 ) ´ 10 q = = = 3. (a) : For photoelectric effect. E.73 MeV (2012) A diatomic molecule is made of two masses m 1 and m 2 which are separated by a distance r. C.4 Z - 2 (b) A . The ratio of number of neutrons to that of protons in the final nucleus will be (a) A . B.1 eV (b) 36. (b) 14 min (d) 28 min (2011) A radioactive nucleus (initial mass number A and atomic number Z) emits 3aparticles and 2 positrons. Speed of light is c. If we calculate its rotational energy by applying Bohr’s rule of angular momentum quantization.Z .4 eV (2011) The half life of a radioactive substance is 20 minutes.Z . In a hydrogen like atom electron makes transition from an energy level with quantum number n to another with quantum number (n – 1). its energy will be given by (n is an integer) (a) n 2 h 2 2(m1 + m2 ) r 2 (b) 2 n 2 h 2 ( m1 + m2 ) r 2 (c) ( m1 + m2 ) n 2 h 2 2 m1m2 r 2 (d) ( m1 + m2 ) 2 n 2 h 2 2 m12 m2 2 r 2 (2012) Energy required for the electron excitation in Li ++ from the first to the third Bohr orbit is (a) 12.99 Atoms and Nuclei CHAPTER ATOMS AND NUCLEI 18 1. M each.8 Z - 4 (c) A . The energy released during this process is (Mass of neutron = 1. against the nuclear mass M; A. 5. the frequency of radiation emitted is proportional to 1 1 1 1 (a) 3 (b) (c) 2 (d) 3/ 2 n n n n (2013) Hydrogen atom is excited from ground state to another state with principal quantum number equal to 4.30 MeV (c) 5.Z . 6.4 Z - 8 (d) A . 4.6725 × 10 –27 kg Mass of electron = 9 × 10 –31 kg) (a) 7. 2 The speed of daughter nuclei is Dm (a) c M + D m Dm (b) c M + D m (c) c 2 D m M (d) c D m M (2010) The binding energy per nucleon for the parent nucleus is E 1 and that for the daughter nuclei is E 2 . E b F A M The above is a plot of binding energy per nucleon E b . If n > > 1. Then (a) E 1 = 2E 2 (b) E 2 = 2E 1 (c) E 1 > E 2 (d) E 2 > E 1 (2010) 10.12 Z - 4 (2010) Directions : Questions number 89 are based on the following paragraph.Infrared radiation will be obtained in the transition from (a) 2 ® 1 (b) 3 ® 2 (c) 4 ® 2 (d) 5 ® 4 (2009) B C D E 11.3 eV (c) 108. The transition from the state n = 4 to n = 3 in a hydrogen like atom results in ultraviolet radiation. The approximate time interval (t 2 – t 1 ) between the time t 2 when 2 1 of it has decayed and time t 1 when of it had decayed is 3 3 (a) 7 min (c) 20 min 7.4 MeV (d) 0. F correspond to different nuclei. D.Z .6725 × 10 –27 kg Mass of proton = 1. 3.10 MeV (b) 6. 2. A nucleus of mass M + Dm is at rest and decays into two daughter nuclei of equal mass 8. Then the number of spectral lines in the emission spectra will be (a) 3 (b) 5 (c) 6 (d) 2 (2012) Assume that a neutron breaks into a proton and an electron.8 eV (d) 122. 9. Consider four reactions: (i) A + B ® C + e (ii) C ® A + B + e (iii) D + E ® F + e (iv) F ® D + E + e where e is the energy released? In which reactions is e positive? (a) (i) and (iv) (b) (i) and (iii) (c) (ii) and (iv) (d) (ii) and (iii) (2009) . (2006) E E 0 E 1 mv 2 bombards a heavy nuclear 2 target of charge Ze. If M O is the mass of an oxygen isotope 8 O 17 . (c) Statement1 is true. The halflife period of a radioactive element X is same as the mean life time of another radioactive element Y. The corresponding halflife is 1 2 (a) 5 minutes (b) 7 minutes (c) 10 minutes (d) 14 minutes (2005) 26. The ‘rad’ is the correct unit used to report the measurement of (a) the rate of decay of radioactive source (b) the ability of a beam of gamma ray photons to produce ions in a target (c) the energy delivered by radiation to target (d) the biological effect of radiation. statement2 is true; statement2 is a correct explanation for statement1. 12. rn µ n (2008) n 14. Suppose an electron is attracted towards the origin by a force k/r where k is a constant and r is the distance of the electron from the origin. Which of the following is the nucleus of element X? (a) 9 5 B (b) 11 4 Be (c) 12 6 C (d) 10 5 B (2005) 24. statement2 is true; statement2 is not a correct explanation for statement1. Statement1 : Energy is released when heavy nuclei undergo fission or light nuclei undergo fusion. Which transition shown represents the emission of a photon with the most energy? n = 4 n = 3 n = 2 I (a) I II (b) II IV III (c) III n = 1 (d) IV (2005) 25. statement2 is true. (a) Statement1 is true. When 3 Li 7 nuclei are bombarded by protons. Starting with a sample of pure 66 Cu.100 JEE MAIN CHAPTERWISE EXPLORER Directions : Question 12 contains statement1 and statement2. the radius of the n th orbital of the electron is found to be r n and the kinetic energy of the electron to be T n . rn µ n 2 n n (c) T n independent of n. a) 3 Li . Then the distance of closest approach for the alpha nucleus with be proportional to (a) 1/Ze (b) v 2 (c) 1/m (d) 1/v 4 . (b) Statement1 is false. By applying Bohr model to this system. On passing through 36 mm of lead. The intensity of gamma radiation from a given source is I. and the resultant nuclei are 4 Be 8 . (2006) 22.28 MeV (d) 1. Then which of the following is true? 1 1 (a) Tn µ . Initially they have the same number of atoms. (2006) 21. (2006) 20. (2008) 13.60 MeV and 7.24 MeV (c) 17. choose the one that best describes the two statements. The energy spectrum of bparticles [number N(E) as a function of benergy E] emitted from a radioactive source is N (E ) N (E ) (a) (b) E E 0 (c) N (E ) E 0 (d) E 0 N (E ) E . An alpha nucleus of energy 7 23. r n µ n 1 (d) Tn µ . If the binding energy per nucleon in 3 nuclei are Li and 4 2 He 5. Then (a) X and Y decay at same rate always (b) X will decay faster than Y (c) Y will decay faster than X (d) X and Y have same decay rate initially (2007) 16. the emitted particles will be (a) neutrons (b) alpha particles (c) beta particles (d) gamma photons. statement2 is false. M P and M N are the masses of a proton and a neutron respectively. it is reduced to I/8. energy of proton must be p + 73 Li ® 2 4 2 He (a) 39. Which of the following transitions in hydrogen atoms emit photons of highest frequency ? (a) n = 1 to n = 2 (b) n = 2 to n = 6 (c) n = 6 to n = 2 (d) n = 2 to n = 1 (2007) 15. In gamma ray emission from a nucleus (a) only the proton number changes (b) both the neutron number and the proton number change (c) there is no change in the proton number and the neutron number (d) only the neutron number changes (2007) 17. then in the reaction : . (d) Statement1 is true. The diagram shows the energy levels for an electron in a certain atom. Of the four choices given. A nuclear transformation is denoted by X(n. The thickness of lead which will reduce the intensity to I/2 will be (a) 18 mm (b) 12 mm (c) 6 mm (d) 9 mm (2005) . (2006) 19. binding energy per nucleon increases with increasing Z while for light nuclei it decreases with increasing Z.46 MeV.2 MeV (b) 28. the nuclear binding energy of the isotope is 2 (a) (M O – 17 M N ) c 2 (b) (M O – 8 M P) c (c) (M O – 8 M P – 9 M N ) c 2 (d) M O c 2 (2007) 7 18.06 MeV respectively. 7/8 of it decays into Zn in 15 minutes. Statement2 : For heavy nuclei. rn µ n 2 (b) Tn µ 2 . 7. (c) (a) (c) (d) (c) (a) (c) 6. If N 0 is the original mass of the substance of halflife period t 1/2 = 5 years. If two deuteron nuclei react to form a single helium nucleus. 42. 24.7 × 10 –14 J.2 MeV. (c) (a) (c) (c) (d) (b) (a) . (2004) ( ) 29. a. (2002) 42. 40. then the energy required to remove an electron from n = 2 is (a) 10. If radius of the 13 Al nucleus is estimated to be 3. b – . The Z of the resulting nucleus is (a) 76 (b) 78 (c) 82 (d) 74. (c) (c) (d) (d) (b) (c) (b) 3. b – . 35. the rate is 1250 disintegrations per minute. The compound can emit (i) electrons (ii) protons (iii) He 2+ (iv) neutrons The emission at the instant can be (a) i. (2002) Answer Key 1. 25.6 eV. 33. 12.8 eV. then the amount of substance left after 15 years is (a) N 0/8 (b) N 0/16 (c) N 0/2 (d) N 0/4. 35. (*) (d) (b) (d) (c) (b) (a) (2003) 4. 38. If the binding energy of the electron in a hydrogen atom is 13. (c) (d) (c) (c) (c) (d) (a) 5. 28. Which of the following atoms has the lowest ionization potential? (a) 14 7 N (b) 133 55 Cs (c) 40 18 Ar (d) 16 8 O. In the nuclear fusion reaction. (2004) 31. the decay constant (per minute) is (a) 0. the temperature at which the gases must be heated to initiate the reaction is nearly [Boltzmann's constant k = 1. b + . 39.1 MeV and 7 MeV respectively. 13. 15. The ratio of their nuclear sizes will be (a) 2 1/3 : 1 (b) 1 : 3 1/2 (c) 3 1/2 : 1 (d) 1 : 2 1/3 .9 MeV (b) 26. (2003) 38. 27. 41. b – . a.101 Atoms and Nuclei 27 27. iii.1 ln 2 (d) 0. a. a. 14. A nucleus disintegrates into two nuclear parts which have their velocities in the ratio 2 : 1. ii. the recoil speed of the residual nucleus is 4 u 4 u 4 u 4 u (b) - (c) (d) - . The distance of the closest approach is of the order of (a) 1 Å (b) 10 –10 cm –12 (c) 10 cm (d) 10 –15 cm. 11. a. a.6 eV (c) 3.4 ln 2 (b) 0. A radioactive sample at any instant has its disintegration rate 5000 disintegrations per minute. (2003) 33. 32. iv (c) iv (d) ii. After 5 minutes. b + . When U 238 nucleus originally at rest. iii (b) i. The binding energy per nucleon of deuteron 1 2 H and helium nucleus 4 2 He is 1. the energy required to remove the electron from the first excited state of Li ++ is (a) 30. 2 3 4 1 H +1 H ® 2 He + n given that the repulsive potential energy between the two nuclei is ~ 7. decays by emitting an alpha particle having a speed u. 37. (2002) 41. (2003) 36.8 ln 2. 22.2 eV (b) 0 eV (c) 3. (a) (c) (c) (d) (a) (a) (a) 2. 36. a.9 MeV (c) 23. then the energy released is (a) 13. 19.2 ln 2 (c) 0. (2003) 40.4 eV. ii. At a specific instant emission of radioactive compound is deflected in a magnetic field. 29. b – .4 eV (d) 122. because (a) size of the two nuclei are different (b) nuclear forces are different in the two cases (c) masses of the two nuclei are different (d) attraction between the electron and the nucleus is different in the two cases. 30. A nucleus with Z = 92 emits the following in a sequence: a.6 eV energy is required to ionize the hydrogen atom. 18. Then. 10. 20. 9. An aparticle of energy 5 MeV is scattered through 180° by a fixed uranium nucleus.4 eV (d) 6. 16. (2003) 37. 26. 23. Which of the following cannot be emitted by radioactive substances during their decay? (a) protons (b) neutrinos (c) helium nuclei (d) electrons. 21. 34. iii. 8.6 eV (b) 13.6 MeV (d) 19.38 × 10 –23 J/K] (a) 10 7 K (b) 10 5 K (c) 10 3 K (d) 10 9 K. Which of the following radiations has the least wavelength? (a) grays (b) brays (c) arays (d) Xrays. 17. (2003) 34. (2003) (a) 238 234 234 238 39. The wavelengths involved in the spectrum of deuterium (1 2 D) are slightly different from that of hydrogen spectrum. 31. If 13. (2004) 30. (2003) ( ) 32.6 fermi then the radius of 125 52 Al nucleus be nearly (a) 4 fermi (b) 5 fermi (c) 6 fermi (d) 8 fermi (2005) 28. E = 2 I 3.1) N = 2 Here.1) = 6 2 = m1m 2 2 r m1 + m2 nh 2 p n 2 h 2 4 p 2 E = = 16 = = 0. \ N= m1 r m1 + m2 Therefore. N 0 = N 0 e -lt 3 Number of undecayed atoms after time t 1 .1) û As n > > 1 2 2 æ m2 r ö æ m1 r ö I = m1 ç + m 2 ç è m1 + m2 ÷ø è m1 + m2 ÷ø 2 RcZ 2n 2 RcZ = n4 n3 or u µ 1 3 n = L= n( n .(ii) 2 9 ´ 10 ´ 9 ´ 10 MeV 1. (c) : A diatomic molecule consists of two atoms of masses m 1 and m 2 at a distance r apart. 4.102 1.51 MeV * None of the given option is correct.(i) . m2 r m1 + m2 (c) : Number of undecayed atoms after time t 2 . m or r1 = 2 r 2 m1 m \ r1 = 2 ( r - r 1 ) m1 (Using (ii)) In the question instead of h. JEE MAIN CHAPTERWISE EXPLORER On rearranging.lt 1 0 3 0 Dividing (ii) by (i).6 ´ 10 -13 As m1r1 = m2 r2 . (*) : Mass defect.(ii) .8 eV 6.6725 × 10 –27 + 9 × 10 –31 – 1.(i) L Rotational energy.6(3) 2 E1 = - eV (1) 2 E 1 = – 122..6 eV (3) 2 DE = E 3 – E 1 = – 13. we get . h should be given.6 ´ (3) 2 = -13. Dm = m p + m e – m n = (1. Q r1 + r2 = r . E n = - 2 eV n Here.2 ù 2 ú ëê ( n . the moment of inertia can be written as \ u= 2.1) ë (n )( n .1) = RcZ 2 ê 2 = 2 ú n 2 ( n . 2 2 N = N e .4 eV \ COM r 2 r 1 r and E3 = The moment of inertia of this molecule about an axis passing through its centre of mass and perpendicular to a line joining the atoms is I = m1r12 + m2 r 2 2 -13.(n .1) 2 ù RcZ 2 (2n ... Let r 1 and r 2 be the distances of the atoms from the centre of mass.6725 × 10 –27 ) kg = 9 × 10 –31 kg Energy released = Dmc 2 = 9 × 10 –31 × (3 × 10 8 ) 2 J -31 2 According to Bohr’s quantisation condition (c) : Number of spectral lines in the emission spectra. the frequency of emitted radiation is 1 1 u = RcZ 2 é . we get (a) : In a hydrogen like atom.. r 2 = é n 2 ..6 + 122. n 2 h 2 8 p 2 I n 2 h 2 (m1 + m 2 ) 8p 2 ( m1m2 ) r 2 n 2 h 2 (m1 + m 2 ) 2 m1m2 r 2 (Using (i)) (Q h = 2 h p ) 5. n = 4 or L2 = 4(4 .4 = 108. Z = 3 (For Li ++ ) m 1 m 2 13.. 13. when an electron makes an transition from an energy level with n to n – 1..6 Z 2 (c) : Using..1) n û r 1 = Similarly. –13.t1 ) = l As per question.e. n = 2 Balmer Balmer n = 1 Lyman Lyman n = 1 – 13. When a radioactive nucleus.6 eV æ 1 1 ö hu2®1 = -13. E 2 > E 1. hence binding energy per nucleon of daughter nuclei is more than that of their parent nucleus.2 eV . Q = ( ) ( ) 1 M 2 1 M 2 1 v + v .( M +Dm) ´ (0) 2 2 2 1 2 2 2 2 = M v1 2 (Q v1 = v2 ) 2 11.. + 3a 2 e \ ZA X ¾¾ ® AZ. only A and B are light elements and D and E are heavy elements. (a) : Statement1 states that energy is released when heavy nuclei undergo fission and light nuclei undergo fusion is correct. we get ( M 2 ) v = D mc 2 1 v 1 2 = 2 Dmc 2 M = + 13. The answer is (a). B/A. (c) : When a radioactive nucleus emits an alpha particle. This is not in the ultraviolet region but this is in infrared region. Hence. (a) : When two nucleons combine to form a third one. In the possibility of fission is only for F and not C. there is no change in the proton and neutron number. DM = éêë ( M + Dm) .Z .6 ´ (d) : After decay.(i) According to law of conservation of momentum. we get ( M + Dm ) ´ 0 = M ´ v1 .4 \ = . starts at a small value. The possibility of fusion is more for light elements and fission takes place for heavy elements.-12 8 W In the final nucleus. Therefore F ® D + E + e is the correct choice. t 1/2 = half life time = 20 min éQ t = ln 2 ù 1/2 êë l úû \ t 2 – t 1 = 20 min 7. 15. Out of the choices given for fusion. 1 e 2 mv 2 4 pe 0 r would have been = r 2 \ mv 2 = e = k Þ 1 mv 2 = 1 k .. 4 pe 0 2 2 This is independent of n. r n µ n. the daughter nuclei will be more stable. (b) : T 1/2 .2 ÷ eV è 2 1 ø 2 v1 = c 2 Dm M 9. The binding energy per nucleon..6 ç 12 . Statement2 is wrong. Number of protons.5 MeV for the heavy nuclei.(ii) 13. 2 p as mv is independent of r..12 ÷ = +13. ( ) M M (c) : Mass defect. If a single nucleus splits into two.6 ç 2 . . n = 5 Pfund 10.6 eV 2 2 AX = A0 e-l X t ; AY = A0 e -lY t 16. then decreases to 7. half life of X = t Y .6 ´ 2 9 6 2 è ø This is <E n = 2 ® E n = 1 . 14. (c) : Supposing that the force of attraction in Bohr atom does not follow inverse square law but inversely proportional to r. N n = A – 12 – (Z – 8) = A – Z – 4 N n A .2 + 2 ùúû = [M + Dm – M] = Dm Energy released. Q = DMc 2 = Dmc 2 . 12. æ ö -13.-126Y ¾¾ ¾ ® A Z . N p = Z – 8 Number of neutrons.103 Atoms and Nuclei 2 = e l (t2 . its mass number decreases by 4 while the atomic number decreases by 2. higher n to lower n. (d) : n = 2 hu 2 ® 1 Equating equations (i) and (ii). Therefore A + B ® C + e is correct. rises to a maximum at 62 Ni. 4 Emission is n = 2 ® n = 1 i. one has fission. (c) : gray emission takes place due to deexcitation of the nucleus. Transition from lower to higher levels are absorption lines. or ln 2 = l(t 2 – t 1) ln 2 or (t 2 . N p Z - 8 8..t 1 ) Transition 4 ® 3 is in Paschen series. emits a b + particle (or positron (e + )) its mass number remains unchanged while the atomic number decreases by 1. From mvrn = nh . one has fusion reaction. (d) : 3 eV = 10. and energy is released.M ´ v2 or v1 = v2 2 2 Also. mean life of Y ln 2 1 = Þ l X = lY ln 2 l X lY l X > l Y n = 4 Brackett \ n = 3 Paschen X will decay faster than Y. . Transition 5 ® 4 will also be in infrared region (Brackett). Therefore during gray emission. 60 = 39..2 = 17. (c) : Kinetic energy is converted into potential energy at closest approach \ K.. T \ 1/ 3 æ m v ö = ç 2 2 ÷ .. from (i) è m2 v1 ø R 1 æ 1 ö 1/ 3 1 = = 1/ 3 . (c) : Binding energy of 7 3 Li = 7 ´ 5. kinetic energy is converted into potential energy \ 1 mv 2 = 1 q1q 2 = 1 ( Ze )(2 e ) 2 4 pe0 r0 4 pe 0 r0 or r 0 = or æ 1 ö r 0 is proportional to ç m ÷ ..6 ´ 10 -19 ) = 5 ´ 106 –14 r = 5.2 MeV Total binding energy for helium = 4 × 7 = 28 MeV \ Energy released = 28 –(2 × 2.3 × 10 m = 5.28 MeV. 18. III having maximum energy from æ 1 1 ö DE µ ç 2 .(i) In second case.2 Me V Binding energy of 4 2 He = 4 ´ 7.kx è I 0 ø In first case \ + 10 n ® 24 He + 3 7 Li According to conservation of mass number A + 1 = 4 + 7 or A = 10 According to conservation of charge number Z + 0 ® 2 + 3 or Z = 5 \ I . 22. (b) : Q I = I 0 e –kx Þ I = e 0 R1 æ A 1 ö = R2 çè A2 ÷ø 1/ 3 3 æm ö =ç 1 ÷ è m2 ø 15 / T 1 æ1ö 1 1 = Þ æç ö÷ = æç ö÷ 8 çè 2 ÷ø è 2 ø è 2 ø 15 = 3 Þ T = 5 min. 7 1 A 8 20. = P. 24.2) = 28 – 4. (c) : R is proportional to A 1/3 where A is mass number 3.06 = 28.k ´ x è 2 ø or ln(2 –1 ) = – kx or ln2 = kx From (i) and (ii) 3 × (kx) = k × 36 or x = 12 mm.. for (3 × 6) R= ´ 5 = 6 fermi .24MeV \ 4 7 Energy of proton = Energy of éë 2(2 He) -3 Li ùû = 2 × 28.(i) 1/ 3 \ t / T 15 / T \ æ I ö ln ç ÷ = .. ln æ 1 ö = . for 27 13 Al .. R2 è 2 ø 2 . (d) : Graph (d) represents the variation. 30.E. (c) : Binding energy = [ZM P + (A – Z) M N – M]c 2 = [8M P + (17 – 8) M N – M O ]c 2 = (8M P + 9M N – M O )c 2 [But the option given is negative of this].. R = R 0 A 1/3 .k ´ 36 è 8 ø ln (2 –3 ) = – k × 36 or 3ln2 = k × 36 . (d) : The 'rad' the biological effect of radiation.6 MeV..6 = R 0 (27) 1/3 = 3R 0 .E.. (c) : I is showing absorption photon..1 = 2...104 JEE MAIN CHAPTERWISE EXPLORER 17.. (a) : N = ç 2 ÷ è ø 0 1 ln æ ö = . (d) : Momentum is conserved during disintegration \ m 1 v 1 = m 2 v 2 For an atom. (d) : 3 Li + 1H ® 4 Be + Z X Z for the unknown X nucleus = (3 + 1) – 4 = 0 A for the unknown X nucleus = (7 + 1) – 8 = 0 Hence particle emitted has zero Z and zero A It is a gamma photon.. è ø 4 Ze2 = Ze 2 æ 1 ö ç ÷ 4 pe0 mv 2 pe 0 v 2 è m ø 23.24 – 39.. (c) : For closest approach.. çn n2 ÷ø è 1 N æ 1 ö 25.. 19..kx 26.. (d) : The nuclear transformation is given by A X Z So the nucleus of the element be 10 5 B.. (ii) 27. From rest of three.4 = 23.2 ÷ .. 21..3 × 10 –12 cm. . \ 3 28. (c) : Total binding energy for (each deuteron) = 2 × 1. q q 5 MeV = 1 1 2 \ 4 pe 0 r or or \ (9 ´ 109 ) ´ (92 e )(2 e ) r 9 ´ 109 ´ 92 ´ 2 ´ e r= 5 ´ 10 6 9 ´ 109 ´ 92 ´ 2 ´ (1. 5 ´ 10 6 ´ e = 29. 125 52 Al Again R = R 0 (125) 1/3 . . 4( –1b 0 ) \ Increase in Z = 4 × 1 = 4 . Electrons in the outer most orbit are at large distance from nucleus in a largesize atom.7 ´ 10-14 J 2 -14 2 = 3...(ii) 2 positrons are emitted i..4ln 2.. (b) : 133 55 Cs has the lowest ionization potential.. 5 38. 33.e.. 34. finally = 1250 \ Nl = 1250 N l 1250 1 = = \ N 0 l 5000 4 or \ \ . 234 39.. Hence (a) represents the answer. t / T 36.6 41..6 eV 2 n (2) 2 energy required = 30..4 eV n2 (2) 2 42.. 8( 2 He 4 ) \ Decrease in Z = 8 × 2 = 16 .e.. T = 7. Cs has the largest size... (i) ... 35. They are not deflected by magnetic field.38 ´ 10 Disintegration rate.. (b) : Linear momentum is conserved aparticle = 4 2 He U 238 ® X 234 + He 4 \ (238 × 0) = (238 × v) + 4u 4 u or v = - .. (ii) N e -5 l 1 N 1 = Þ 0 = Þ e -5 l = (4) -1 N0 4 N 0 4 5l = ln4 = 2ln2 2 l = ln 2 = 0. 37. molecules of a gas acquire a kinetic 3 energy = kT where k = Boltzmann's constant 2 \ To initiate the fusion reaction 3 kT = 7...105 Atoms and Nuclei \ ... initially = 5000 \ N 0l = 5000 Disintegration rate. (a) : Neutrons are electrically neutral.. (d) : At temperature T.. 31. (a) : Let decay constant per minute = l 15 / 5 N æ 1 ö æ1ö 40.6 = -30. (c) : En = 13.(iii) \ Z of resultant nucleus = 92 – 16 + 4 – 2 = 78..7 ´ 109 K.(i) Four b – particles are emitted i. (a) : Gamma rays have the least wavelength. Hence the corresponding wavelengths are different... (a) : Protons are not emitted during radioactive decay. 3 1 1 = æç ö÷ = 8 è2ø 13. Hence the ionization potential is the least....6 Þ E 2 = = 3. 2( 1b 0 ) \ Decrease in Z = 2 × 1 = 2 .(3)2 ´ 13. Of the four atoms given.Z 2 E 0 .6 eV...e. (c) : Masses of 1 H 1 and 1 D 2 are different... ... (a) : Energy E 2 = = 32.. (b) : The nucleus emits 8a particles i. (a) : N = ç 2 ÷ = ç 2 ÷ è ø è ø 0 \ N = N 0 /8......7 ´ 10 ´-23 \ 3 ´ 1. 3. The current(I) in the resistor(R) can be shown by .31 MHz (2013) Input B Output is (a) Truth table for system of four NAND gates as shown in figure is A (b) Y B (a) A B Y A B Y 0 0 0 0 0 1 0 1 0 0 1 1 1 0 1 1 0 0 1 1 1 1 1 0 (b) (c) 6. The IV characteristic of an LED is 4. The combination of gates shown below yields A X (a) (b) B (a) NAND gate (c) NOT gate 5.106 JEE MAIN CHAPTERWISE EXPLORER CHAPTER ELECTRONIC DEVICES 19 1.31 kHz (b) 10. D R V (2012) An alternating current source (V) is connected in the circuit. Pick out the correct output waveform. A Y B (2013) Input A 2. (a) 5. A diode detector is used to detect an amplitude modulated wave of 60% modulation by using a condenser of capacity 250 pico farad in parallel with a load resistance 100 kilo ohm.62 kHz (d) 5.62 MHz (c) 10. Find the maximum modulated frequency which could be detected by it. (c) (d) (b) OR gate (d) XOR gate (2010) The logic circuit shown below has the input waveforms ‘A’ and ‘B’ as shown. (c) A B Y A B Y 0 0 1 0 0 0 0 1 0 0 1 1 1 0 1 1 1 0 1 0 0 1 1 1 (d) (2009) (d) A pn junction (D) shown in the figure can act as a rectifier. (d) The number of free electrons for conduction is significant in all the three. E v decrease all E c .488 mA for an emitter current of 5.00 A (b) 1. No conduction is found between P and Q. but E g decreases E c . silicon and germanium have four valence electrons each. (b) It is an npn transistor with R as base. In the following. What is the current following in the circuit? (b) t 4 W I (c) D 1 D 2 t (2009) (d) t 7. 9. then which of the following is correct? E c Conduction band width Band gap E g Valence band width (a) (b) (c) (d) all E c . In common base mode of a transistor.71 A (d) 2. (b) The number of free conduction electrons is significant in C but small in Si and Ge. (b) NOR gate (d) NAND gate.60 mA. A and B represent two inputs and C represents the output. (d) It is a pnp transistor with R as emitter. If in a p n junction diode. The circuit represents (a) 1. E v (2006) 14. and E v decrease. and E v increase. Which of the following is true for the transistor? (a) It is an npn transistor with R as collector. (2008) A working transistor with its three legs marked P.107 Electronic Devices I (a) 10 V + 5V (b) (a) t (c) I (d) –10 V (2007) – 5V 11. E v increase E c . E g . (c) The number of free conduction electrons is negligibly small in all the three. (2008) Carbon. a square input signal of 10 V is applied as shown 5 V R L –5 V Then the output signal across R L will be – 12V (c) R – 5V R (d) . At room temperature which one of the following statements is most appropriate ? (a) The number of free electrons for conduction is significant only in Si and Ge but small in C. but E g increases. If the lattice constant of this semiconductor is decreased.31 A. Q and R is tested using a multimeter. which one of the diodes is reverse biased? + 5V A + 10V R C (a) B R (b) + 5V (a) OR gate (c) AND gate 8. some resistance is seen on the multimeter. (c) It is a pnp transistor with R as collector. In the ratio of the concentration of electrons that of holes in a semiconductor is 7/5 and the ratio of currents is 7/4 then what is the ratio of their drift velocities? (a) 4/7 (b) 5/8 (c) 4/5 (d) 5/4. E g . (2006) 12. A solid which is transparent to visible light and whose conductivity increases with temperature is formed by (a) metallic binding (b) ionic binding (c) covalent binding (d) van der Waals binding (2006) . (2006) 16. The value of the base current amplification factor (b) will be (a) 48 (b) 49 (c) 50 (d) 51. (2006) – 10V 13. (2006) 15.33 A (c) 2. By connecting the common (negative) terminal of the multimeter to R and the other (positive) terminal to P or Q. the collector current is 5. The circuit has two oppositely connect ideal diodes in parallel. (2007) 10. 2 W 3 W 12V I In the circuit below. 78. 13. 25. The manifestation of band structure in solids is due to (a) Heisenberg’s uncertainty principle (b) Pauli’s exclusion principle (c) Bohr’s correspondence principle (d) Boltzmann’s law (2004) 22.2 (b) –15. Si acts as (a) nonmetal (b) metal (c) insulator (d) none of these.1 eV (d) 2.5 eV (b) 0. 14. 9.5 eV (2005) 20. The energy band gap is maximum in (a) metals (b) superconductors (c) insulators (d) semiconductors. The resistance of (a) each of these decreases (b) copper strip increases and that of germanium decreases (c) copper strip decreases and that of germanium increases (d) each of these increases. 23. (a) (c) (a) (d) (a) 6.8 (d) – 48. the specific resistance of a conductor and a semiconductor (a) increases for both (b) decreases for both (c) increases. 26. 20. 7. 31. 22. 12. (2004) 25. 8. When pn junction diode is forward biased. then (a) the depletion region is reduced and barrier height is increased (b) the depletion region is widened and barrier height is reduced. At absolute zero.108 JEE MAIN CHAPTERWISE EXPLORER 17. In the middle of the depletion layer of a reversebiased p n junction. 24. (b) (a) (c) (c) (a) 5. (2002) 29. (2003) 27. (2004) 24. (2002) 32. 29. When npn transistor is used as an amplifier (a) electrons move from base to collector (b) holes move from emitter to base (c) electrons move from collector to base (d) holes move from base to emitter. decreases (d) decreases. (2002) Answer Key 1. Formation of covalent bonds in compounds exhibits (a) wave nature of electron (b) particle nature of electron (c) both wave and particle nature of electron (d) none of these. 11. A strip of copper and another germanium are cooled from room temperature to 80 K. 30.7 eV (c) 1. (2003) 28. 21. In a full wave rectifier circuit operating from 50 Hz mains frequency. the fundamental frequency in the ripple would be (a) 100 Hz (b) 70. 10. The difference in the variation of resistance with temperature in a metal and a semiconductor arises essentially due to the difference in the (a) crystal structure (b) variation of the number of charge carriers with temperature (c) type of bonding (d) variation of scattering mechanism with temperature. In a common base amplifier. The part of a transistor which is most heavily doped to produce large number of majority carriers is (a) emitter (b) base (c) collector (d) can be any of the above three. (c) (a) (b) (c) (b) (c) 3. 31. 16. The electrical conductivity of a semiconductor increases when electromagnetic radiation of wavelength shorter than 2480 nm is incident on it. increases.7 (c) – 24. 27. (c) (a) (a) (a) (c) . By increasing the temperature. the (a) electric field is zero (b) potential is maximum (c) electric field is maximum (d) potential is zero. 19. The band gap in (eV) for the semiconductor is (a) 0. For a transistor amplifier in common emitter configuration for load impedance of 1 kW (h fe = 50 and h oe = 25) the current gain is (a) –5. 17. 18.7 Hz (c) 50 Hz (d) 25 Hz (2005) 18. the resistance of (a) each of them increases (b) each of them decreases (c) copper decreases and germanium increases (d) copper increases and germanium decreases. A piece of copper and another of germanium are cooled from room temperature to 77 K. (d) (a) (d) (b) (c) (2002) 4. the phase difference between the input signal voltage and output voltage is (a) 0 (b) p/2 (c) p/4 (d) p (2005) 19. (b) (a) (d) (a) (a) (c) 2. 32. (2002) 30. (c) both the depletion region and barrier height are reduced (d) both the depletion region and barrier height are increased (2004) 21. 15. 28. (2004) 23. (2003) 26. 6 \ u= A B A B A × B 0 0 1 1 1 0 1 1 0 0 1 0 0 1 0 1 1 0 0 0 This is same as that of OR gate. (c) : The maximum frequency which can be detected is 1 u= 2 pma t where. If it is connected to base. the gate conducts. = 10. 11. (c) : Since diode D 1 is reverse biased. Y 0 0 1 Y 1 1 1 1 0 Y 1 1 X B B (c) : (a) is original wave (b) is a fullwave rectified (c) is the correct choice.61 kHz. 12. A 0 0 0 3. (a) : It is OR gate. 9. these electrons are in the second orbit. the option (b) represents the correct graph.109 Electronic Devices 1. 1 1 1 1 0 1 1 B 1 1 1 1 1 0 0 0 1 1 1 1 7. (a) : It is npn transistor with R as collector. ( A + B ) = A × B. 0 0 A (b) : A 6. (d) : E c and E v decrease but E g increases if the lattice constant of the semiconductor is decreased. Both are semiconductors. Alternative method The truth table of the given circuit is as shown in the table (b) : The IV characteristics of a LED is similar to that of a Si junction diode. (a) : Y B 1 2p ´ 0.61 × 10 3 Hz = 10. Si and Ge have the same lattice structure and their valence electrons are 4. 13. 8. In solid state. A 5.6 ´ 250 ´ 10 -12 ´ 100 ´ 103 By de Morgan’s theorem. Ionization energies are also less therefore Ge has more conductivity compared to Si. When either of them conducts.Carbon is an insulator. greater the possibility of overlapping of energy bands. it will be in forward bias. C = 250 pico farad = 250 × 10 –12 farad R = 100 kilo ohm = 100 × 10 3 ohm m a = 0. (d) : B 0 1 1 1 1 1 0 0 0 X = A × B 0 1 1 1 Y 1 1 A B A B A+ B A + B Verify A × B 1 1 0 0 0 1 1 0 0 1 1 1 0 0 0 1 1 0 1 0 0 1 0 0 1 1 0 0 This is the same as AND Gate of A and B. therefore it will act like an open circuit. for Si it is third and germanium it is the fourth orbit. Effective resistance of the circuit is R = 4 + 2 = 6 W. higher the orbit. For C. (a) : The current will flow through R L when diode is forward biased. This is same as the Boolean expression of OR gate. The Boolean expression of the given circuit is X = A× B = A + B (Using De Morgan theorem) = A + B (Using Boolean identity) 10. The negative waves are cut off when the diode is connected in reverse bias (d) is not the diagram for alternating current. (a) : C. (a) : Figure (a) represent a reverse biased diode. A 0 0 0 B A 1 1 1 1 1 B A 0 0 1 4. 2. . Current in the circuit is I = E/R = 12/6 = 2 A. Hence. t = CR Here. But the threshold voltages are much higher and slightly different for each colour. (b) : Pauli's exclusion principle explains band structure of solids. 31.60 - 5. like Si. I b I e . (d) : Drift velocity v d = nAe ( vd ) electron æ I e ö æ n h ö = ç ÷ ç ÷ = 7 ´ 5 = 5 . (a) : Electric field is zero in the middle of the depletion layer of a reverse baised pn junction. 27. (c) : Copper is a conductor. r decreases as temperature rises. ( vd ) hole 4 7 4 è I h ø è ne ø 14. r increases as temperature rises.488 = 49. the resistance of copper strip decreases and that of germanium increases.025 24. (b) : Variation of number of charge carriers with temperature is responsible for variation of resistance in a metal and a semiconductor. Ge. (d) : In common emitter configuration.488 = = = 5. 20. act as insulators at low temperature.110 Ic I c 5.112 I 15. 22. (a) : Electrons of ntype emitter move from emitter to base and then base to collector when npn transistor is used as an amplifier. 28. 1 + 0.I c 5. (c) : Semiconductors. 30.63 ´ 10 -34 ) ´ (3 ´ 108 ) eV \ Band gap = (2480 ´ 10 -9 ) ´ (1. 21. the resistance of copper decreases and that of germanium increases. (c) : Covalent binding. (c) : For conductor. (c) : The energy band gap is maximum in insulators. 25. (a) : In a common base amplifier. 19. 18. (a) : Frequency of full wave rectifier = 2 × input frequency = 2 × 50 = 100 Hz. Germanium is a semiconductor. 17. (c) : When pn junction diode is forward biased. When cooled. (a) : Band gap = Energy of photon of l = 2480 nm hc hc \ Energy = l J = le (eV) (6.488 0. 23. (a) : Wave nature of electron and covalent bonds are correlated. For semiconductor. (c) : Copper is conductor and germanium is semiconductor. current gain is . =- 26. 32. 29.( h fe ) -50 = A i = 1 + (25 ´ 10 -6 ) ´ (1 ´ 103 ) 1 + ( hoe )( RL ) 50 = -50 = – 48. . (a) : The emitter is most heavily doped. the phase difference between the input signal and output voltage is zero. both the depletion region and barrier height are reduced.025 1.78. (b) : b = 16.5 eV.6 ´ 10-19 ) = 0. JEE MAIN CHAPTERWISE EXPLORER When cooled. Then (a) 36 > x > 18 (b) 18 > x (c) x > 54 (d) 54 > x > 36. (c) . f) (d) (4f. she measures the column length to be x cm for the second resonance. a student notes the main scale reading of 3 mm and the number of circular scale divisions in line with the main scale as 35. The total number of divisions on the circular scale is 50. The diameter of the wire is (a) 3.73 mm (d) 3. An experiment is performed to find the refractive index of glass using a travelling microscope. 29 divisions of the main scale exactly coincide with the 30 divisions of the vernier scale.026 cm (d) 0. While measuring the diameter of a thin wire.67 mm. Total divisions on the vernier scale is 30 and match with 29 divisions of the main scale. 7. it is found that the screw gauge has a zero error of – 0. then the least count of the instrument is (a) one minute (b) half minute (c) one degree (d) half degree (2009) distance v. 4f) (2009) 5. The angle of the prism from the above data (a) 58. The coordinates of P will be (a) (2f. is plotted using the same scale for the two axes. 2. Two full turns of the circular scale of a screw gauge cover a distance of 1 mm on its main scale. Main scale reading : 58. (2008) 7. A graph between the object distance u and the image Answer Key 1. f/2) (c) (f. a student gets the first resonance condition at a column length of 18 cm during winter. the screen is adjusted to get a clear image of the object. Repeating the same experiment during summer.65 degree (c) 59 degree (d) 58. 4.052 cm (c) 0. (b) 6. 2f) (b) (f/2.77 degree (b) 58.59 degree (2012) A screw gauge gives the following reading when used to measure the diameter of a wire. If the smallest division of the main scale is halfadegree (= 0.5 degree. (b) 3.5°).111 Experimental Skills CHAPTER EXPERIMENTAL SKILLS 20 1. (2008) 6. 3. (a) 5. with the position of the object fixed. (b) (a) 2. In this experiment distances are measured by (a) a screw gauge provided on the microscope (b) a vernier scale provided on the microscope (c) a standard laboratory scale (d) a meter scale provided on the microscope. Further.5 degree Vernier scale reading : 09 divisions Given that 1 division on main scale corresponds to 0. (2008) In an optics experiment.52 cm (b) 0. A straight line passing through the origin and making an angle of 45° with the xaxis meets the experimental curve at P. The diameter of wire from the above data is : (a) 0. Main scale reading : 0 mm Circular scale reading : 52 divisions Given that 1 mm on main scale corresponds to 100 divisions of the circular scale. While measuring the speed of sound by performing a resonance column experiment.32 mm (c) 3.005 cm (2011) In an experiment the angles are required to be measured using an instrument.38 mm (b) 3. a student varies the position of a convex lens and for each position. (a) 4.03 mm. from the lens. A spectrometer gives the following reading when used to measure the angle of a prism. 30 30 2 60 (a) : According to New Cartesian coordinate system used in our 12 th classes. v u f 2 f 2 f f The answer is the same (a).5 ° + 0.5 mm = = 0.S. v and u are have the same value when the object is at the centre of curvature. = L 2 = v P u v 2 3 l = > 3 ´ 18.D L 1 = (n . n 29 L .D n L.1 M. v = 2f 1 1 1 i. (The figure given is according to New Cartesian system). The solution is (a).S.C.052 cm = 6. provided with the microscope.C. g RT assuming M is the average molar mass of M the air (i. as u is negative.S. v1 = \ 1 V.C. it 1 1 1 effectively comes to v + u = f . (a) : Least count = L 2 value of 1 main scale division The number of divisions on the vernier scale as shown below. v u f One has to take that u is negative again for calculation. 0. = n .5 ° 30 = 1 M.e.D. n 4 \ L 2 > 54 cm.D. u is +ve and v is also +ve as both are real.1 = 1 .35 = 3. = 0.5 ° = 58. v 2 > v 1 . \ 1 + 1 = 1 Þ 1 + 1 = 1 .03) = 3. 2 f + 2 f = f .S.1) M.5 ° 30 30 Reading = Main scale reading + Vernier scale reading × least count = 1- = 58.D - Þ Þ 4.01 mm 100 Diameter of wire = Main scale reading + circular scale reading × Least count = 0 + 52 × 0. 45° 1 .52 mm = 0.03 \ actual diameter of the wire = 3. (b) : Least count of screw gauge Pitch = Number of divisions on circular scale 5.5° ´ 0.65° .. summer.1 V. Here n vernier scale divisions = (n – 1) M. nitrogen) and g is also for nitrogen. 30 2.112 1. (a) : Least count of the screw gauge 0. u = 2f = radius of curvature.5 1 1 1 = ´ = ° = 1 min .e. According the real and virtual system.01 = 0.5° + 9 ´ 0. If u = radius of curvature.01 mm 50 Main scale reading = 3 mm. v 1 l = = 18 cm. (c) : v 1 = n n L 1 3.S. Vernier scale reading = 35 \ Observed reading = 3 + 0. 1 mm = 0. JEE MAIN CHAPTERWISE EXPLORER (b) : 30 VSD = 29 MSD 1 VSD = 29 MSD 30 Least count = 1 MSD – 1 VSD ( ) 29 1 MSD = ´ 0. L . At temperature T 1 n 4 At T 2 . . (b) : A travelling microscope moves horizontally on a main scale provided with a vernier scale. 2f.35 – (–0.S.D. . for a convex lens.35 zero error = –0. = 1 M.15° = 58.38 mm.D.S.. If u = v. 1 st resonance g RT1 g RT 2 ; v 2 = where T 1 and T 2 stand for winter M M and summer temperatures. the lens equation is 2 nd resonance 7. CHEMISTRY . 8) from the reduction of boron trichloride by hydrogen? (a) 89. which of these changes? (a) Molality (b) Weight fraction of solute (c) Fraction of solute present in water (d) Mole fraction (2002) Number of atoms in 558. (a) 0.023 × 10 23 (2002) Answer Key 1.25 × 10 (d) 2.2 L (c) 44. 2Al (s) + 6HCl (aq) ® 2Al 3+ (aq) + 6Cl – (aq) + 3H 2(g) (a) 11.8 L (d) 22. (b) (a) 2.6 g of elemental boron (atomic mass = 10. the mass of one mole of a substance will (a) decrease twice (b) increase two fold (c) remain unchanged (d) be a function of the molecular mass of the substance.6 L (b) 67.2 L H 2(g) at STP is produced for every mole HCl (aq) consumed (b) 6 L HCl (aq) is consumed for every 3 L H 2(g) produced (c) 33.85 g mol –1 ) is (a) twice that in 60 g carbon (b) 6. (a) 3.75 M (2013) 5.2 L H 2(g) at STP is produced for every mole Al that reacts. of Fe = 55. (b) 4. 7. 4.1 Some Basic Concepts in Chemistry SOME BASIC CONCEPTS IN CHEMISTRY CHAPTER 1 1. (a) 5.25 mole of oxygen atoms? 7. mass of carbon atom is taken to be the relative atomic mass unit.5 M HCl with 250 mL of 2 M HCl will be (a) 0.023 × 10 22 (c) half that in 8 g He (d) 558.875 M (c) 1. 3. in place of 1/12.00 M (d) 1.5 × 6. at 273 K and 1 atm. Mg 3 (PO 4 ) 2 will contain 0.975 M (b) 0. The molarity of a solution obtained by mixing 750 mL of 0. wt.5 × 10 –2 (2006) If we consider that 1/6. (2005) What volume of hydrogen gas. (2007) How many moles of magnesium phosphate.02 (b) 3.4 L (2003) With increase of temperature.125 × 10 –2 –2 (c) 1. (b) 6.6 L H 2(g) is produced regardless of temperature and 6. pressure will be consumed in obtaining 21. (c) . pressure for every mole Al that reacts (d) 67. 2.5 gram Fe (at. In the reaction. 875 M Mmix = 2. (a) : 1 atomic mass unit on the scale of 1/6 of C12 = 2 amu on the scale of 1/12 of C12.2 JEE MAIN CHAPTERWISE EXPLORER 1. – 2Al 3+ (aq ) + 6Cl (aq ) + 3H 2(g ) 6 moles of HCl produced H 2 at STP = 3 × 22. 558.25 mole of oxygen atoms are present in = 6.2 L 6 Mass of one atom of the element 1 1 amu (Here on the scale of of C - 12) 6 5.125 × 10 –2 moles of Mg 3 (PO 4 ) 2 7. Mass of one atom of the element 1 2 amu (Here on the scale of of C -12) 12 \ Numerically the mass of a substance will become half of the normal scale.8 (c) : Volume increases with rise in temperature. (b) : 1 mole of Mg 3 (PO 4 ) 2 Þ 3 moles of Mg atom + 2 moles of P atom + 8 moles of O atom 8 moles of oxygen atoms are present in = 1 mole of Mg 3 (PO 4 ) 2 1 ´ 0.4 L \ 1 mole of HCl will produce H 2 at STP 3 ´ 22. of moles) = 60/12 = 5 moles.6 = 67.4 L 2 21. atomic mass of an element M1V1 + M2 V 2 Vmix = 0.2 L = ´ 2 10. (b) : M mixV mix = M 1V 1 + M 2V 2 M mix = Now.6 g boron will require hydrogen 3 22. 4. (a) : 2Al (s) + 6HCl (aq ) = 3. (atomic weight of carbon = 12) . (b) : 2BCl 3 + 3H 2 ® 6HCl + 2B or 3 BCl 3 + H 2 ® 3HCl + B 2 3 10.5 ´ 750 + 2 ´ 250 1000 Mmix = 0.25 0.4 ´ 21.85 C (no. 8 = 3.5 = 10 moles (a) : Fe (no. of moles) = 55.8 g boron requires hydrogen = ´ 22.4 = = 11. 3 States of Matter CHAPTER 2 STATES OF MATTER 1. Experimentally it was found that a metal oxide has formula M 0.98 O. Metal M, is present as M 2+ and M 3+ in its oxide. Fraction of the metal which exists as M 3+ would be (a) 5.08% (b) 7.01% (c) 4.08% (d) 6.05% (2013) 2. For gaseous state, if most probable speed is denoted by C*, average speed by C and mean square speed by C, then for a large number of molecules the ratios of these speed are (a) C* : C : C = 1 : 1.225 : 1.128 (b) C* : C : C = 1.225 : 1.128 : 1 (c) C* : C : C = 1.128 : 1 : 1.225 (d) C* : C : C = 1 : 1.128 : 1.225 (2013) 3. 4. 5. 6. 7. 8. If 10 –4 dm 3 of water is introduced into a 1.0 dm 3 flask at 300 K, how many moles of water are in the vapour phase when equilibrium is established? (Given : Vapour pressure of H 2 O at 300 K is 3170 Pa; R = 8.314 J K –1 mol –1 ) (a) 1.27 × 10 –3 mol (b) 5.56 × 10 –3 mol –2 (c) 1.53 × 10 mol (d) 4.46 × 10 –2 mol (2010) 9. Percentages of free space in cubic close packed structure and in body centred packed structure are respectively (a) 48% and 26% (b) 30% and 26% (c) 26% and 32% (d) 32% and 48% (2010) 10. Copper crystallizes in fcc with a unit cell length of 361 pm. What is the radius of copper atom? (a) 108 pm (b) 127 pm Lithium forms body centred cubic structure. The length of (c) 157 pm (d) 181 pm (2009) the side of its unit cell is 351 pm. Atomic radius of the lithium will be 11. In a compound, atoms of element Y form ccp lattice and those (a) 300 pm (b) 240 pm of element X occupy 2/3 rd of tetrahedral voids. The formula of the compound will be (c) 152 pm (d) 75 pm (2012) (a) X 3 Y 4 (b) X 4 Y 3 The compressibility factor for a real gas at high pressure is (c) X 2 Y 3 (d) X 2 Y (2008) (a) 1 (b) 1 + Pb/RT (c) 1 – Pb/RT (d) 1 + RT/Pb (2012) 12. Equal masses of methane and oxygen are mixed in an empty container at 25°C. The fraction of the total pressure exerted by In a face centred cubic lattice, atom A occupies the corner oxygen is positions and atom B occupies the face centre positions. If (a) 1/2 (b) 2/3 one atom of B is missing from one of the face centred points, 1 273 ´ (c) (d) 1/3 (2007) the formula of the compound is 3 298 (a) A 2 B (b) AB 2 13. Total volume of atoms present in a facecentred cubic unit cell (c) A 2 B 3 (d) A 2 B 5 (2011) of a metal is (r is atomic radius) 20 3 24 3 ‘a’ and ‘b’ are van der Waals’ constants for gases. Chlorine p r p r (a) (b) 3 3 is more easily liquefied than ethane because 12 3 16 3 (a) a and b for Cl 2 > a and b for C 2 H 6 p r p r (c) (d) (2006) 3 3 (b) a and b for Cl 2 < a and b for C 2 H 6 (c) a for Cl 2 < a for C 2 H 6 but b for Cl 2 > b for C 2 H 6 14. Which one of the following statements is not true about the (d) a for Cl 2 > a for C 2 H 6 but b for Cl 2 < b for C 2 H 6 effect of an increase in temperature on the distribution of molecular speeds in a gas? (2011) (a) The most probable speed increases. The edge length of a face centred cubic cell of an ionic substance (b) The fraction of the molecules with the most probable speed is 508 pm. If the radius of the cation is 110 pm, the radius increases. of the anion is (c) The distribution becomes broader. (a) 144 pm (b) 288 pm (d) The area under the distribution curve remains the same as under the lower temperature. (c) 398 pm (d) 618 pm (2010) (2005) 4 JEE MAIN CHAPTERWISE EXPLORER 15. An ionic compound has a unit cell consisting of A ions at the 20. According to the kinetic theory of gases, in an ideal gas, between corners of a cube and B ions on the centres of the faces of the two successive collisions a gas molecule travels cube. The empirical formula for this compound would be (a) in a circular path (a) AB (b) A 2 B (b) in a wavy path (c) AB 3 (d) A 3 B (2005) (c) in a straight line path (d) with an accelerated velocity. 16. What type of crystal defect is indicated in the diagram below? (2003) Na + Cl – Na + Cl – Na + Cl – 21. How many unit cells are present in a cubeshaped ideal crystal Cl – Cl – Na + Na + of NaCl of mass 1.00 g? Na + Cl – Cl – Na + Cl – [Atomic masses : Na = 23, Cl = 35.5] + – + – + Na Cl Na Cl Na 21 (a) 2.57 × 10 (b) 5.14 × 10 21 (a) Frenkel defect 21 (c) 1.28 × 10 (d) 1.71 × 10 21 (b) Schottky defect (2003) (c) Interstitial defect (2004) 22. Na and Mg crystallize in bcc and fcc type crystals respectively, then the number of atoms of Na and Mg present in the unit cell 17. In van der Waals equation of state of the gas law, the constant of their respective crystal is b is a measure of (a) 4 and 2 (b) 9 and 14 (a) intermolecular repulsions (c) 14 and 9 (d) 2 and 4 (b) intermolecular attraction (2002) (c) volume occupied by the molecules (d) Frenkel and Schottky defects (2004) 23. For an ideal gas, number of moles per litre in terms of its pressure P, gas constant R and temperature T is 18. As the temperature is raised from 20°C to 40°C, the average (a) PT/R (b) PRT kinetic energy of neon atoms changes by a factor of which of (c) P/RT (d) RT/P the following? (2002) (a) 1/2 (b) 313/ 293 (c) 313/293 (d) 2 (2004) 24. Kinetic theory of gases proves (a) only Boyle's law 19. A pressure cooker reduces cooking time for food because (b) only Charles' law (a) heat is more evenly distributed in the cooking space (c) only Avogadro's law (b) boiling point of water involved in cooking is increased (d) all of these. (2002) (c) the higher pressure inside the cooker crushes the food (d) intermolecular collisions per unit volume. material 25. Value of gas constant R is (d) cooking involves chemical changes helped by a rise in (a) 0.082 L atm (b) 0.987 cal mol –1 K –1 –1 –1 temperature. (c) 8.3 J mol K (d) 83 erg mol –1 K –1 (2003) Answer Key 1. 7. (c) (a) 2. 8. (d) (a) 3. 9. (c) (c) 4. (b) 10. (b) 5. (d) 11. (b) 6. (d) 12. (d) 13. (d) 14. (b) 15. (c) 16. (b) 17. (c) 18. (c) 19. (b) 25. (c) 20. (c) 21. (a) 22. (d) 23. (c) 24. (d) (2002) 5 States of Matter 1. (c) : Let the fraction of metal which exists as M 3+ be x. Then the fraction of metal as M 2+ = (0.98 – x) \ 3x + 2(0.98 – x) = 2 x + 1.96 = 2 x = 0.04 \ % of M 3+ = 2. 0.04 ´ 100 = 4.08% 0.98 2 RT = M (d) : C * : C : C = 8 RT = p M 3 RT M 8 = 3 3.14 = 2: (c) : a = 351 pm For bcc unit cell, a 3 = 4r r= 4. a 3 351 ´ 3 = = 152 pm 4 4 æ 5. a ö (b) : For real gases, çè P + 2 ÷ø (V - b) = RT V At high pressure, P >> a/V 2 Thus neglecting a/V 2 gives P(V – b) = RT or PV = RT + Pb PV RT + Pb =Z= RT RT or Þ Z = 1 + Pb/RT (d) : A B 8 ´ 1 8 5 ´ 1 2 Formula of the compound is A 2 B 5 . 6. (d) : a (dm 3 atm mol –2 ) b(dm 3 mol –1 ) Cl 2 6.49 0.0562 C 2 H 6 5.49 0.0638 From the above values, a for Cl 2 > a for ethane (C 2 H 6 ) b for ethane (C 2 H 6 ) > b for Cl 2 . 7. (a) : In fcc lattice, Given, a = 508 pm r c = 110 pm \ 110 + r a = 8. 508 Þ r a = 144 pm 2 (a) : The volume occupied by water molecules in vapour phase is (1 – 10 –4 ) dm 3 , i.e., approximately 1 dm 3 . PV = nRT 3170 × 1 × 10 –3 = n H 2 O × 8.314 × 300 nH = 2 O (c) : The packing efficiency in a ccp structure = 74% \ Percentage free space = 100 – 74 = 26% Packing efficiency in a body centred structure = 68% Percentage free space = 100 – 68 = 32% 10. (b) : Since Cu crystallizes in fcc lattice, \ radius of Cu atom, r= a (a = edge length) 2 2 Given, a = 361 pm \ C * : C : C = 1 : 1.128 : 1.225 3. 9. 3170 ´ 10 -3 = 1.27 ´ 10 -3 mol 8.314 ´ 300 \ r= 361 » 127 pm 2 2 11. (b) : Number of Y atoms per unit cell in ccp lattice (N) = 4 Number of tetrahedral voids = 2N = 2 × 4 = 8 Number of tetrahedral voids occupied by X = 2/3 rd of the tetrahedral void = 2/3 × 8 = 16/3 Hence the formula of the compound will be X 16/3 Y 4 = X 4 Y 3 12. (d) : Let the mass of methane and oxygen be m g. Mole fraction of oxygen, x O 2 m m 32 1 32 = ´ = = m m 32 3m 3 + 32 16 Let the total pressure be P. \ Partial pressure of O 2 , p O 2 = P × x O 2 1 1 = P 3 3 13. (d) : In case of a facecentred cubic structure, since four atoms are present in a unit cell, hence volume = P × æ4 ö 16 V = 4 ç pr 3 ÷ = pr 3 è3 ø 3 14. (b) : Most probable velocity is defined as the speed possessed by maximum number of molecules of a gas at a given temperature. According to Maxwell's distribution curves, as temperature increases, most probable velocity increases and fraction of molecule possessing most probable velocity decreases. 1 15. (c) : Number of A ions per unit cell = ´ 8 = 1 8 1 Number of B ions per unit cell = ´ 6 = 3 2 Empirical formula = AB 3 6 JEE MAIN CHAPTERWISE EXPLORER 16. (b) : When an atom or ion is missing from its normal lattice site, 24. (d) : Explanation of the Gas Laws on the basis of Kinetic Molecular Model a lattice vacancy is created. This defect is known as Schottky One of the postulates of kinetic theory of gases is defect. + – Average K.E. µ T Here equal number of Na and Cl ions are missing from their 2 regular lattice position in the crystal. So it is Schottky defect. 1 mnC 2 µ T 1 or, or, mnCrms = kT rms 2 2 17. (c) : van der Waals constant for volume correction b is the 2 2 Now, PV = 1 mnCrms = 2 ´ 1 mnC rms = 2 kT measure of the effective volume occupied by the gas molecules. 3 3 2 3 (i) Boyle’s Law : 18. (c) : K b = 3/2 RT l Constant temperature means that the average kinetic energy K 40 T 40 273 + 40 313 of the gas molecules remains constant. = = = K 20 T20 273 + 20 293 l This means that the rms velocity of the molecules, C rms remains unchanged. 19. (b) : According to Gay Lussac's law, at constant pressure of a l If the rms velocity remains unchanged, but the volume given mass of a gas is directly proportional to the absolute increases, this means that there will be fewer collisions temperature of the gas. Hence, on increasing pressure, the with the container walls over a given time. temperature is also increased. Thus in pressure cooker due to l Therefore, the pressure will decrease increase in pressure the boiling point of water involved in cooking is also increased. i.e. P µ 1 V 20. (c) : According to the kinetic theory of gases, gas molecules are or PV = constant. always in rapid random motion colliding with each other and (ii) Charles’ Law : with the wall of the container and between two successive l An increase in temperature means an increase in the average collisions a gas molecule travels in a straight line path. kinetic energy of the gas molecules, thus an increase in C rms . l There will be more collisions per unit time, furthermore, 21. (a) : Mass (m) = density × volume = 1.00 g the momentum of each collision increases (molecules strike Mol. wt. (M) of NaCl = 23 + 35.5 = 58.5 the wall harder). l Number of unit cell present in a cube shaped crystal of NaCl of Therefore, there will be an increase in pressure. l If we allow the volume to change to maintain constant r ´ a 3 ´ N A m ´ N A mass 1.00 g = = pressure, the volume will increase with increasing M ´Z M ´ Z temperature (Charles law). 23 (iii) Avogadro’s Law = 1 ´ 6.023 ´ 10 58.5 ´ 4 It states that under similar conditions of pressure and (In NaCl each unit cell has 4 NaCl units. Hence Z = 4). temperature, equal volume of all gases contain equal number \ Number of unit cells = 0.02573 × 10 23 of molecules. Considering two gases, we have = 2.57 × 10 21 unit cells 2 2 PV 1 1 = kT1 and PV 2 2 = kT2 3 3 22. (d) : bcc Points are at corners and one in the centre of the unit Since P 1 = P 2 and T 1 = T 2 , therefore cell. 1 Number of atoms per unit cell = 8 ´ + 1 = 2 8 fcc Points are at the corners and also centre of the six faces of each cell. 1 1 Number of atoms per unit cell = 8 ´ + 6 ´ = 4 8 23. (c) : From ideal gas equation, PV = nRT \ n/V = P/RT (number of moles = n/V) 2 PV (2 / 3) kT1 V n 1 1 = Þ 1 = 1 P2V2 (2 / 3) kT2 V2 n2 If volumes are identical, obviously n 1 = n 2 . 25. (c) : Units of R (i) in L atm Þ 0.082 L atm mol –1 K –1 (ii) in C.G.S. system Þ 8.314 × 10 7 erg mol –1 K –1 (iii) in M.K.S. system Þ 8.314 J mol –1 K –1 (iv) in calories Þ 1.987 cal mol –1 K –1 7 Atomic Structure CHAPTER 3 1. ATOMIC STRUCTURE Energy of an electron is given by E = -2.178 ´ 10 -18 æ Z 2 ö J ç 2 ÷ . è n ø Wavelength of light required to excite an electron in an 8. hydrogen atom from level n = 1 to n = 2 will be (h = 6.62 × 10 –34 J s and c = 3.0 × 10 8 m s –1 ) (a) 8.500 × 10 –7 m (b) 1.214 × 10 –7 m (c) 2.816 × 10 –7 m (d) 6.500 × 10 –7 m (2013) 2. 3. 4. 5. The electrons identified by quantum numbers n and l: (1) n = 4, l = 1 (2) n = 4, l = 0 (3) n = 3, l = 2 (4) n = 3, l = 1 9. can be placed in order of increasing energy as (a) (4) < (2) < (3) < (1) (b) (2) < (4) < (1) < (3) (c) (1) < (3) < (2) < (4) (d) (3) < (4) < (2) < (1) (2012) A gas absorbs a photon of 355 nm and emits at two wavelengths. If one of the emission is at 680 nm, the other is at (a) 1035 nm (b) 325 nm (c) 743 nm (d) 518 nm (2011) (c) – 4.41 × 10 –17 J atom –1 (d) – 2.2 × 10 –15 J atom –1 7. (2009) The ionization enthalpy of hydrogen atom is 1.312 × 10 6 J mol –1 . The energy required to excite the electron in the atom from n = 1 to n = 2 is (a) 9.84 × 10 5 J mol –1 (b) 8.51 × 10 5 J mol –1 (c) 6.56 × 10 5 J mol –1 (d) 7.56 × 10 5 J mol –1 (2008) Which of the following sets of quantum numbers represents the highest energy of an atom? 1 (a) n = 3, l = 0, m = 0, s = + 2 1 (b) n = 3, l = 1, m = 1, s = + 2 1 (c) n = 3, l = 2, m = 1, s = + 2 1 (d) n = 4, l = 0, m = 0, s = + 2 (2007) The energy required to break one mole of Cl—Cl bonds in Cl 2 is 242 kJ mol –1 . The longest wavelength of light capable 10. Uncertainty in the position of an electron (mass = 9.1 × 10 –31 kg) moving with a velocity 300 m s –1 , accurate of breaking a single Cl—Cl bond is (c = 3 × 10 8 m s –1 upto 0.001% will be (h = 6.6 × 10 –34 J s) and N A = 6.02 × 10 23 mol –1 ) (a) 19.2 × 10 –2 m (b) 5.76 × 10 –2 m (a) 494 nm (b) 594 nm (c) 1.92 × 10 –2 m (d) 3.84 × 10 –2 m (c) 640 nm (d) 700 nm (2010) (2006) Ionisation energy of He + is 19.6 × 10 –18 J atom –1 . The energy of the first stationary state (n = 1) of Li 2+ is 11. According to Bohr’s theory, the angular momentum of an –17 –1 electron in 5 th orbit is (a) 8.82 × 10 J atom (b) 4.41 × 10 –16 J atom –1 6. (a) 1.52 × 10 –4 m (b) 5.10 × 10 –3 m (c) 1.92 × 10 –3 m (d) 3.84 × 10 –3 m (2010) h (a) 25 p h (b) 1.0 p h (c) 10 p h (d) 2.5 p (2006) Calculate the wavelength (in nanometre) associated with a proton moving at 1.0 × 10 3 m s –1 . 12. Which of the following statements in relation to the hydrogen (Mass of proton = 1.67 × 10 –27 kg and h = 6.63 × 10 –34 J s) atom is correct? (a) 0.032 nm (b) 0.40 nm (a) 3s orbital is lower in energy than 3p orbital. (c) 2.5 nm (d) 14.0 nm (2009) (b) 3p orbital is lower in energy than 3d orbital. In an atom, an electron is moving with a speed of 600 m/s (c) 3s and 3p orbitals are of lower energy than 3d orbital. with an accuracy of 0.005%. Certainty with which the position (d) 3s, 3p and 3d orbitals all have the same energy. of the electron can be located is (h = 6.6 × 10 –34 kg m 2 s –1 , (2005) mass of electron, e m = 9.1 × 10 –31 kg) 8 JEE MAIN CHAPTERWISE EXPLORER 13. In a multielectron atom, which of the following orbitals 17. The orbital angular momentum for an electron revolving in an described by the three quantum numbers will have the same h orbit is given by l( l + 1) × . This momentum for an energy in the absence of magnetic and electric fields? 2 p (i) n = 1, l = 0, m = 0 selectron will be given by (ii) n = 2, l = 0, m = 0 1 h (iii) n = 2, l = 1, m = 1 (a) + × (b) zero 2 2 p (iv) n = 3, l = 2, m = 1 h h (v) n = 3, l = 2, m = 0 (d) 2 × (c) 2 p 2 p (a) (i) and (ii) (b) (ii) and (iii) (2003) (c) (iii) and (iv) (d) (iv) and (v) (2005) 18. The de Broglie wavelength of a tennis ball of mass 60 g moving with a velocity of 10 metres per second is approximately 14. The wavelength of the radiation emitted, when in a hydrogen (Planck's constant, h = 6.63 × 10 –34 J s) atom electron falls from infinity to stationary state 1, would be –33 (a) 10 metres (b) 10 –31 metres (Rydberg constant = 1.097 × 10 7 m –1 ) –16 (c) 10 metres (d) 10 –25 metres. (a) 91 nm (b) 192 nm (2003) (c) 406 nm (d) 9.1 × 10 –8 nm (2004) 19. In Bohr series of lines of hydrogen spectrum, the third line from the red end corresponds to which one of the following inter 15. Consider the ground state of Cr atom (Z = 24). The numbers of orbit jumps of the electron for Bohr orbits in an atom of electrons with the azimuthal quantum numbers, l = 1 and 2 are, hydrogen? respectively (a) 3 ® 2 (b) 5 ® 2 (a) 12 and 4 (b) 12 and 5 (c) 4 ® 1 (d) 2 ® 5 (c) 16 and 4 (d) 16 and 5 (2003) (2004) 16. Which of the following sets of quantum numbers is correct for 20. Uncertainty in position of a minute particle of mass 25 g in space is 10 –5 m. What is the uncertainty in its velocity an electron in 4f orbital? (in m s –1 )? (h = 6.6 × 10 –34 J s) 1 (a) n = 4, l = 3, m = +4, s = + –34 (a) 2.1 × 10 (b) 0.5 × 10 –34 2 (c) 2.1 × 10 –28 (d) 0.5 × 10 –23 (2002) 1 (b) n = 4, l = 4, m = –4, s = - 2 21. In a hydrogen atom, if energy of an electron in ground state is 1 13.6 eV, then that in the 2 nd excited state is (c) n = 4, l = 3, m = +1, s = + 2 (a) 1.51 eV (b) 3.4 eV 1 (c) 6.04 eV (d) 13.6 eV (d) n = 3, l = 2, m = –2, s = + (2004) (2002) 2 Answer Key 1. (b) 7. (c) 13. (d) 19. (b) 2. (a) 8. (a) 14. (a) 20. (c) 3. (c) 9. (c) 15. (b) 21. (a) 4. (a) 10. (c) 16. (c) 5. (c) 11. (d) 17. (b) 6. (b) 12. (d) 18. (a) 9 Atomic Structure 1. æ 1 ö -18 2 1 (b) : E = -2.178 ´ 10 Z ç 2 - 2 ÷ è n1 n2 ø -18 E = -2.178 ´ 10 é 1 1 ù ê 2 - 2 ú (1) û ú ëê (2) E = +2.178 ´ 10 -18 ´ E= Þ Þ \ 5. 2. 3. hc 6.62 ´ 10 -34 J s ´ 3 ´ 10 8 m = E 1.6335 ´ 10 -18 J l = 1.214 × 10 –7 m 6. l l1 Þ 1 1 1 = + 355 680 l 2 l 2 h mv Given, v = 1.0 × 10 3 m s –1 355 ´ 680 = 742.769 nm » 743 nm 680 - 355 Energy required to break 1 bond = We know that, E = hc l Given, c = 3 × 10 8 m s –1 6.63 ´ 10 -34 = 3.9 ´ 10 -10 m 1.67 ´ 10 -27 ´ 1.0 ´ 103 l » 0.4 nm \ l= or 7. (c) : Given, velocity of e – , v = 600 m s –1 Accuracy of velocity = 0.005% 600 ´ 0.005 \ Dv = = 0.03 100 According to Heisenberg’s uncertainty principle, h Dx × mDv ³ 4p 6.6 ´ 10 -34 Þ Dx = 4 ´ 3.14 ´ 9.1 ´ 10 -31 ´ 0.03 8. (a) : Energy required to break 1 mol of bonds = 242 kJ mol –1 \ (b) : According to deBroglie’s equation, = 1.92 × 10 –3 m E = E 1 + E 2 or hc = hc + hc 4. -19.6 ´ 10 -18 ´ 9 = - 44.1 ´ 10 -18 J atom -1 4 l = E = hu = hc/l \ l 2 = 242 ´ 10 3 = – 4.41 × 10 –17 J atom –1 (c) : We know that 1 1 1 = + l l1 l 2 6.63 ´ 10 -34 ´ 3 ´ 108 ´ 6.02 ´ 10 23 (c) : I.E.(He + ) = 19.6 × 10 –18 J atom –1 = (a) : (1) n = 4, l = 1 Þ 4p (2) n = 4, l = 0 Þ 4s (3) n = 3, l = 2 Þ 3d (4) n = 3, l = 1 Þ 3p Increasing order of energy is 3p < 4s < 3d < 4p (4) < (2) < (3) < (1) Alternatively, for (1) n + l = 5 ; n = 4 (2) n + l = 4 ; n = 4 (3) n + l = 5 ; n = 3 (4) n + l = 4 ; n = 3 Lower n + l means less energy and if for two subshells n + l is same than lower n, lower will be the energy. Thus correct order is (4) < (2) < (3) < (1). Þ l= 6.63 ´ 10 -34 ´ 3 ´ 10 8 l E 1 (for H) × Z 2 = I.E. E 1 × 4 = –19.6 × 10 –18 E 1 (for Li 2+ ) = E 1 (for H) × 9 l = 12.14 × 10 –8 m or 6.02 ´ 10 23 = = 0.494 × 10 –6 m = 494 nm 3 = 1.6335 ´ 10 -18 J 4 hc l l= 242 ´ 10 3 242 ´ 10 3 6.02 ´ 10 23 J (a) : The ionisation of Hatom is the energy absorbed when the electron in an atom gets excited from first shell (E 1 ) to infinity (i.e., E¥ ) I.E = E¥ – E 1 1.312 × 10 6 = 0 – E 1 E 1 = –1.312 × 10 6 J mol –1 6 6 E2 = - 1.312 ´2 10 = - 1.312 ´ 10 4 (2) Energy of electron in second orbit (n = 2) \ Energy required when an electron makes transition from n = 1 to n = 2 6 DE = E 2 – E 1 = - 1.312 ´ 10 - ( -1.312 ´ 106 ) 4 105 × 10 –33 metres. Thus option (b) is = 2. total number of electrons = 12 [2p 6 and 3p 6 ] For l = 2..14 ´ 9. (c) : n = 3.1 ÷ö 2 2 l l è1 ¥ ø è n1 n2 ø \ D v = 2.6 ´ 10 -34 Dv = æ ö 4 ´ 3. (d) : For hydrogen the energy order of orbital is 20. E n = 13.1 ´ 10 -28 m s -1 \ l = 91 × 10 –9 m = 91 nm 21.. 5.5 \ Angular momentum = 2 p p correct here.6 eV or E = 13.312 ´ 10 + 5.63 ´ 10 -34 \ Dx = 4 ´ 3. The third line from the red end 2 p corresponds to yellow region i. In order to obtain less when n = 5 (given) energy electron tends to come in 1 st or 2 nd orbit. l = 2 represents 3d orbital n = 4. 15.984 × 10 4 DE = 9.92 ´ 10 -2 m 11.14 ´ 25 ´ 10 -5 (a) : 1 = R çç 12 . mvr = of orbit. 6. –1.05 × 10 –34 m = 1. h Dx ´ Dp = 4 p h h Dx × (m × Dv ) = Þ Dx = 4 p 4 pm × D v 0. l = 1 represents 3p orbital n = 3.1 2 ÷÷ Þ 1 = 1. l = 0 represents 4s orbital The order of increasing energy of the orbitals is 3s < 3p < 4s < 3d. (b) : The electron has minimum energy in the first orbit and its energy increases as n increases. (a) : l = h = 6.248 ´ 10 = 0. (b) : The value of l (azimuthal quantum number) for s electron is equal to zero. (a) : 2 nd excited state will be the 3rd energy level. (b) : 24 Cr ® 1s 2 2s 2 2p 6 3s 2 3p 6 3d 5 4s 1 we know for p. 2 nd . For l = 1. So jump 5 h h may be involved either 5 ® 1 or 5 ® 2. total number of electrons = 5 16.84 ×10 5 J mol –1 9. (c) : According to Heisenberg’s uncertainty principle.10 JEE MAIN CHAPTERWISE EXPLORER 6 6 6 = -1. 1s < 2s = 2p < 3s = 3p = 3d < 4s = 4p = 4d = 4f h Dx × mDv = (d) : Orbitals having same (n + l) value in the absence of 4 p electric and magnetic field will have same energy.1 ´ 10 -31 ´ 3 ´ 10 -3 = 1.63 ´ 10 ´ 1000 m mv 60 ´ 10 = 11. 13. l = 0 represents 3s orbital n = 3. 19. 0. 14. 10.e.6 = 1. h Orbital angular momentum = l (l + 1) × 2 p Substituting the value of l for selectron = 0(0 + 1) × h = 0 2 p -34 18. (c) : For 4 f orbital electrons. –3 s = ±1/2 17. +1.001 -3 ´ 300 = 3 ´ 10 m s -1 Here Dv = 100 6. (c) : According to Heisenberg uncertainty principle. +2. n = 4 [3d 5 ] s p d f l = 3 (because 0 1 2 3 ) m = +3. 12. 1 st .e. 3 rd . i.097 ´ 107 m -1 çæ 1 . –2. l = 1 and for d..51 eV n 2 9 . l = 2. Here n represents number nh (d) : Angular momentum of the electron. O 2 – . CO C 2 2– . C 22– . The charge/size ratio of a cation determines its polarizing power. sp 2 . Ca 2+ . NO + . Which of the following species exhibits the diamagnetic are respectively behaviour? 2– (a) sp. Li 2 and Li 2 increases in the order of (a) Li 2– < Li 2 < Li + 2 (b) Li 2 < Li + 2 < Li 2– (c) Li 2 – < Li + 2 < Li 2 11. NO – NO + . He 2 2+ (b) H 2 + . 6. is present in (a) O 2+ (b) O + 2 2 (a) benzene and ethanol (c) O – 2 (d) O 2– (2009) 2 (b) acetonitrile and acetone 10. CN . N 2 2– CN – . sp 3 .11 Chemical Bonding and Molecular Structure CHAPTER 4 1. + – Stability of the species Li 2. (2011) (d) K + < Ca 2+ < Mg 2+ < Be 2+ (2007) – + + The hybridisation of orbitals of N atom in NO 3 . 7. 3 5. C 2 2– Which one of the following pairs of species have the same bond order? (a) NO + and CN + (b) CN – and NO + (c) CN – and CN + (d) O 2 – and CN – (2008) (d) H 2 2+ . O 2 . sp 3 (a) NO (b) O 2 + (c) sp. the bond order has increased and the magnetic behaviour has changed? The molecule having smallest bond angle is (a) N 2 ® N 2+ (b) C 2 ® C 2+ (a) AsCl 3 (b) SbCl 3 (c) NO ® NO + (d) O 2 ® O 2 + (2007) (c) PCl (d) NCl (2012) (c) H 2 – . Be 2+ ? (c) AlF 6 3– and SF 6 (d) CO 3 2– and NO 3 – (2012) (a) Ca 2+ < Mg 2+ < Be + < K + The structure of IF 7 is (b) Mg 2+ < Be 2+ < K + < Ca 2+ (a) square pyramid (b) trigonal bipyramid (c) Be 2+ < K + < Ca 2+ < Mg 2+ (c) octahedral (d) pentagonal bipyramid. sp. dipoledipole as the major the shortest bond length? interaction. K + . He 2 2– 4. Which one of the following sequences represents the increasing isostructural? + order of the polarizing power of the cationic species. 9. sp (2011) (c) O (d) O (2007) 2 8. sp 2 (d) sp 2 . both the species are not likely to exist? (a) H 2 – . Which of the following hydrogen bonds is the strongest? (a) O — H F (b) O — H H Which one of the following molecules is expected to exhibit (c) F — H F diamagnetic behaviour? (d) O — H O (2007) (a) S 2 (b) C 2 (c) N 2 (d) O 2 (2013) 13. Which one of the following constitutes a group of the isoelectronic species? (c) KCl and water (d) benzene and carbon tetrachloride. In which of the following ionization processes. Among the following mixtures. (a) PCl 4 and SiCl 4 (b) PF 5 and BrF 5 Mg 2+ . NO 2 and NH 4 15. (2008) 12. (2006) . In which of the following molecules/ions are all the bonds not by the compound equal? (a) FeCl 2 (b) SnCl 2 (a) SF 4 (b) SiF 4 (c) AlCl 3 (d) MgCl 2 (2011) (c) XeF 4 (d) BF 4– (2006) Using MO theory predict which of the following species has 17. (a) (b) (c) (d) N 2 . sp 3 (b) sp 2 . CHEMICAL BONDING AND MOLECULAR STRUCTURE 2 Among the following the maximum covalent character is shown 16. He 2 2– 3. He 2 (2013) 3 In which of the following pairs the two species are not 14. N 2 . sp 3 . CO. O 2– . 2. (d) Li 2 < Li 2 – < Li + (2013) 2 In which of the following pairs of molecules/ions. 29. 15. d x 2 – y 2 Which one of the following has the regular tetrahedral structure? (c) s. Number of sigma bonds in P 4 O 10 is (a) 6 (b) 7 (2004) (c) 17 (d) 16 The bond order in NO is 2. 22. p y . d yz (b) s. p y . 20. In which of the following species is the underlined carbon having (a) Bond length in NO + is greater than in NO.12 JEE MAIN CHAPTERWISE EXPLORER 18. ClO 4 – (b) CN – . d z2 (d) s. 28. d xy (a) XeF 4 (b) SF 4 (2002) (c) BF 4– (d) [Ni(CN) 4 ] 2– (Atomic nos.: B = 5. (d) 4. 16. p x . 23. S = 16. IF 5 (2003) 19. 30. (b) (c) (c) (b) (a) 8. p z . p y . BF 3 –1 (c) NH 3. PCl 3 (d) PF 5 . 26. Ni = 28. NH . NO 3 3– 2– (d) BO 3 . Of the following sets which one does NOT contain isoelectronic species? (a) PO 4 3– . SF 4 (b) XeF 2 . (a) (b) (d) (b) 10. 27. 24. A square planar complex is formed by hybridisation of which (c) dsp 2 hybridisation atomic orbitals? (d) sp 3 d 2 hybridisation. (c) (a) (c) (b) . The pair of species having identical shapes for molecules of electrons? both species is (a) O 22– (b) B 2 (a) CF 4 . (b) (b) (b) (d) 12. 24. (b) 5. (d) 3. (c) 2. Which one of the following species is diamagnetic in nature? (a) He 2 + (b) H 2 (c) H 2+ (d) H 2– (2005) 21.5 while that in NO + is 3. (BF (b) (NH 4) 4) . Which of the following are arranged in an increasing order of their bond strengths? (a) O 2– < O 2 < O 2+ < O 22– – + (b) O 22– < O 2 < O 2 < O 2 + The maximum number of 90° angles between bond pairbond (c) O 2– < O 22– < O 2 < O 2 + – 2– pair of electrons is observed in (d) O 2 < O 2 < O 2 < O 2 (a) dsp 3 hybridisation (2002) (b) sp 3 d hybridisation 28. (a) CH 3 COOH (b) CH 3 CH 2 OH (d) Bond length is unpredictable. BF (d) (NH 2) (2002) 3 . Which of (2002) the following statements is true for these two species? 30. 20. 31. 17. (2004) (d) CH 2 CH – CH 3 (c) CH 3 COCH 3 (2002) The correct order of bond angles (smallest first) in H S. (c) (d) (b) (c) 9. Which of the following molecules/ions does not contain unpaired 25. 3 (2004) 31. (c) (a) (c) (b) 11. 19. 14. 25. C 2 2– 2– – (c) SO 32– . CO 3 . N 2 . Xe = 54) 29. (2004) (a) s. 2 BF 3 and SiH 4 is (a) H 2 S < SiH 4 < NH 3 < BF 3 (b) NH 3 < H 2 S < SiH 4 < BF 3 (c) H 2 S < NH 3 < SiH 4 < BF 3 (d) H 2 S < NH 3 < BF 3 < SiH 4 . BF 3 Answer Key 1. (b) 6. (d) 7. sp 3 hybridisation? + (c) Bond length in NO is equal to that in NO. p y . p x . SO 4 2– . Which one of the following compounds has the smallest bond angle in its molecule? (a) SO 2 (b) OH 2 (c) SH 2 (d) NH 3 (2003) 27. CO 2 (c) N 2+ (d) O 2 (2006) (c) BF 3 . + (b) Bond length in NO is greater than in NO . CO 3 . 18. 22. NO 3– (2005) 26. 13. p x . In which of the following species the interatomic bond angle is 109°28¢? –1 + (a) NH 3. 23. 21. = = (p 2py ) 2 (p * 2px ) 1 10 . extent of polarisation µ covalent character (a) : According to MOT.O. So it has more covalent character. it is paramagnetic. CO 3 2– and NO 3 – both are trigonal planar.4 = 3 2 + O2 : (s1s) 2 (s * 1s) 2 (s2s) 2 (s * 2s) 2 (s2pz) 2 (p 2px ) 2 \ B. Hence option (c) is correct. 4. N 2 and CO are isoelectronic in nature.5 2 – O2 : (s1s) 2 (s * 1s) 2 (s2s) 2 (s * 2s) 2 (s2pz) 2 (p 2px ) 2 = (p 2py) 2 (p * 2px) 2 = (p * 2py ) 1 \ B. in ionic bond. \ B. (c) : Species Bond order Li 2 1 – Li 2 0.2 = 0 2 + 4 (c) : We know that. 11.O. . (d) : The structure is pentagonal bipyramidal having sp 3 d 3 hybridisation as given below: 1 size of cation Here the AlCl 3 is satisfying the above two conditions i. lone pairbond pair repulsion also increases. (c) : Number of electrons in each species are given below N 2 = 14 CN – = 14 – O 2 = 17 C 22– = 14 + NO = 14 O 22– = 18 CO = 14 NO = 15 It is quite evident from the above that NO + .e. Bond length 2+ \ O2 has the shortest bond length. the molecular orbital electronic configuration of O2 2+ : (s 1s) 2 (s * 1s) 2 (s 2s) 2 (s * 2s) 2 (s 2pz) 2 (p 2px) 2 = (p 2py ) 2 10 . = 7. 6. (b) : As we move down the group the size of atom increases and as size of central atom increases. (b) : The structures of NO 3 – . CN . Fajan’s rule states that (i) the polarising power of cation increases.O.. = 10 . greater is the stability.5 Higher the bond order.O.5 = 2. (b) : In the given pair of species. number of electron in NO + = number of electron in CN – = 14 electrons. – C 22– . µ .5 2 O2 2– : (s 1s) 2 (s * 1s) 2 (s 2s) 2 (s * 2s) 2 (s 2pz) 2 (p2px ) 2 = (p 2py) 2 (p * 2px) 2 = (p * 2py ) 2 \ B. Increasing order of atomic radius : N < P < As < Sb Decreasing order of bond angle : NCl 3 > PCl 3 > AsCl 3 > SbCl 3 5.O.7 = 1. = 10 . Li 2+ 0.5 8. 2. (b) : PCl and SiCl 4 Þ both tetrahedral PF 5 Þ trigonal bipyramidal BrF 5 Þ square pyramidal AlF 6 3– and SF 6 both are octahedral.8 = 1. 2 . Al is in +3 oxidation state and also has small size. (d) : O 2 is expected to be diamagnetic in nature but actually 9. Thus bond angle decreases. So they are isoelectronic in nature. H 2 2+ : s(1s) 0 \ Bond order = 0 He 2 : s(1s) 2 s*(1s) 2 Bond order = \ polarising power µ 3. NO 2 + and NH 4 + is sp 2 hybridisation sp hybridisation sp 3 hybridisation 10. with increase in magnitude of positive charge on the cation \ polarising power µ charge of cation (ii) the polarising power of cation increases with the decrease in the size of a cation.13 Chemical Bonding and Molecular Structure 1.0 2 1 Q B. (d) : Species with zero bond order does not exist. The positive pole of one molecule is thus attracted by the negative pole of the other molecule. polarising power decreases as KK s(2s) 2 s*(2s) 2 s(2p z ) 2 p(2p x ) 2 p(2p y ) 2 s(2p z ) 1 K + < Ca 2+ < Mg 2+ < Be 2+ It contains one unpaired electron.5 = 2. which in turn depends upon the electronegativities of the atoms present in the molecule and the geometry of the molecule (in case of polyatomic molecules.O = 1/2 [N b – N a ] = 1/2 [10 – 4] or B. hydrogen bonding in F — H F is strongest.6 = 2 2 + 109.4 Bond order = = 3 2 + N 2 Þ s1s 2 s*1s 2 s2s 2 s*2s 2 p2p x 2 p2p y 2 p2p z 1 Þ paramagnetic 9 . The magnitude of dipoledipole forces in different polar molecules is predicted on the basis of the polarity of the molecules.5 2 NO Þ s1s 2 s*1s 2 s2s 2 s*2s 2 s2p z 2 p2p x 2 p2p y 2 p*2p x 1 Þ paramagnetic Bond order = 10 . 17. (c) : Molecular orbital configuration of O 2 Þ s1s 2 s*1s 2 s2s 2 s*2s 2 s2p z 2 p2p x 2 p2p y 2 p*2p x 1 p*2p y 1 Þ paramagnetic F 16. K + > Ca 2+ > Mg 2+ > Be 2+ The molecular orbital configuration of N + 2 ion is Hence. .O = 3 CN – ® s1s 2 s*1s 2 s2s 2 s*2s 2 p2p x2 p2p y2 s2p z 2 B.4 Bond order = = 2. NO + ® s1s 2 s*1s 2 s2s 2 s*2s 2 s2p z 2 p2p x 2 p2p y 2 B. 1 1 KK s(2s) 2 s*(2s) 2 p(2p x) p(2p y ) As the size of the given cations decreases as It contains 2 unpaired electrons.5° O 2 Þ s1s 2 s*1s 2 s2s 2 s*2s 2 s2p z 2 p2p x 2 p2p y 2 p*2p x 1 Si Þ paramagnetic Bond order = 10 .5 2 NO + Þ s1s 2 s*1s 2 s2s 2 s*2s 2 s2p z 2 p2p x 2 p2p y 2 Þ diamagnetic F F F XeF 4 : sp 3 d 2 hybridisation. The molecular orbital configuration of O 2 molecule is 15. Þ paramagnetic .14 JEE MAIN CHAPTERWISE EXPLORER Hence bond order of these two species will be also similar which is shown below.O =1/2 [10 – 4] or B. (d) : High charge and small size of the cations increases The molecular orbital configuration of B 2 molecule is polarisation. 18. Polar molecules have permanent dipoles. SiF 4 : sp 3 hybridisation and tetrahedral geometry. (a) : The molecular orbital configuration of O 2 2– ion is KK s(2s) 2 s*(2s) 2 s(2p z ) 2 p(2p x ) 2 p(2p y ) 2 p*(2p x ) 2 p*(2p y ) 2 10 4 Bond order = = 3 Here KK represents nonbonding molecular orbital of 1s orbital. Þ paramagnetic 2– O 2 Þ s1s 2 s*1s 2 s2s 2 s*2s 2 s2p z 2 p2p x 2 p2p y 2 p*2p x 1 p*2p y 1 19. 14.5 = 2. S F F SF 4 molecule shows sp 3 d hybridisation but its expected trigonal bipyramidal geometry gets distorted due to presence of a lone pair of electrons and it becomes distorted tetrahedral or seesaw with the bond angles equal to 89° and 177° instead of the expected angles of 90° and 180° respectively.4 = 1. shape is square planar instead of octahedral due to presence of two lone pair of electrons on Xe atom. 2 O 22– contains no unpaired electrons. (b) : Dipoledipole interactions occur among the polar molecules. 13. (b) : Molecular orbital configuration is KK s(2s) 2 s*(2s) 2 p(2p x ) 2 p(2p y ) 2 s(2p z ) 2 p*(2p x ) 1 p*(2p y ) 1 NO Þ s1s 2 s*1s 2 s2s 2 s*2s 2 s2p z 2 p2p x 2 p2p y 2 p*2p x 1 It contains 2 unpaired electrons.5 2 C 2 Þ s1s 2 s*1s 2 s2s 2 s*2s 2 p2p x 2 p2p y 2 Þ diamagnetic Bond order = 8 . F Bond order = 10 .5 2 N 2 Þ s1s 2 s*1s 2 s2s 2 s*2s 2 p2p x 2 p2p y 2 p2p Z 2 Þ paramagnetic 10 . (a) : F . F F Xe F BF 4 – F : sp 3 hybridisation and tetrahedral geometry. containing more than two atoms in a molecule). (c) : Because of highest electronegativity of F.4 = 2 2 C 2+ Þ s1s 2 s*1s 2 s2s 2 s*2s 2 p2p x 2 p2p y 1 Þ paramagnetic Bond order = 7 .O = 3 12. (c) : Number of electrons in SO 3 = 16 + 8 × 3 + 2 = 42 Þ paramagnetic Number of electrons in CO 3 2– = 6 + 8 × 3 + 2= 32 O 22– Þ s1s 2 s*1s 2 s2s 2 s*2s 2 s2p z 2 p2p x 2 p2p y 2 p*2p x 2 p*2p y 2 Number of electrons in NO 3 – = 7 + 8 × 3 + 1= 32 Þ diamagnetic These are not isoelectronic species as number of electrons O 2+ Þ s1s 2 s*1s 2 s2s 2 s*2s 2 s2p z 2 p2p x 2 p2p y 2 p*2p x 1 are not same. 5° Due to absence of unpaired electrons. (b) : Higher the bond order.5 2 F P O (d) : O P O O O P O O O No. (c) : The correct order of bond angle (smallest first) is H 2 S < NH 3 < SiH 4 < BF 3 and (d) (CH 2 CH — CH 3 ). . of s bonds = 16 No.15 Chemical Bonding and Molecular Structure 20. 27. The expected bond angle should be 109º28¢ but actual bond angle is less than 109º28¢ because of the repulsion between lone pair and bonded pairs due to which contraction occurs.8 = 1 2 Hence increasing order of bond order is O 2 2– < O 2 – < O 2 < O 2 + F regular tetrahedral 10 . = 10 .5 2 2 2 O 2– Þ s(1s) 2s*(1s) 2s(2s) 2s*(2s) 2s(2p z) p(2p x ) p(2p y ) 2p*(2p x ) 2p*(2p y ) 1 ; B. (b) : He 2 + ® s(1s) 2 s*(1s) 1 . the carbon atom has a multiple 92.7 = 1. (b) : Molecular orbital configuration of 21. F P O Xe F F 28. (b) : In molecules (a) CH 3 C OH . (d) : O 2 Þ s(1s) 2s*(1s) 2s(2s) 2s*(2s) 2s(2p z ) 2p(2p x ) 2 1 1 p(2p y ) 2p*(2p x) p*(2p y) ; B. no unpaired electron shape.5 = 2. B. one unpaired electron – 2 1 26.O. p y and d x 2 – y 2 orbitals with bond angles of 90°.6° S 107° N H H 109°28 ¢ H H F Si B H H H F 120° F 31. 30. one unpaired electron.5° 104. = 2 2 O 22– Þ s(1s) 2s*(1s) 2s(2s) 2s*(2s) 2s(2p z) p(2p x ) sp 3 d 2 hybridisation (twelve 90° angle between bond pair and bond pair) – F 2– p(2p y ) 2p*(2p x ) 2p*(2p y ) 2 ; B. H 2 will be diamagnetic. (c) : B Ni NC CN F square planar F . . (c) CH C CH 3 3 24. (b) : dsp 2 hybridisation gives square planar structure with s. F F F 10 .O. shorter will be the bond length. square planar seesaw shaped ×× H 92.O. (b) : Central atom in each being sp hybridised shows linear H 2 ® s(1s) 2 s*(1s) 0 .6 = 2 2 O +2 Þ s(1s) 2s*(1s) 2s(2s) 2s*(2s) 2s(2p z ) 2p(2p x ) 2 p(2p y ) 2p*(2p x ) 1.O. = CN NC 22. F S O 29. only (b) has sp 3 hybridization. . + F — Xe — F O C O 1 0 H 2 ® s(1s) s*(1s) . (a) : Both undergoes sp 3 hybridization..5° 106. Bond angle : 119. . one unpaired electron 25. (c) : SO 2 OH 2 SH 2 NH 3 H 2 ® s(1s) s*(1s) . p x . of p bonds = 4 23.5° 92..6° < 107° < 109°28¢ < 120° bond. = dsp 2 hybridisation dsp 2 sp 3 d or dsp 3 hybridisation (four 90° angles between bond pair and bond pair) hybridisation (six 90° angles between bond pair and bond pair) 10 . Thus NO + is having higher bond order than that of NO so O O NO + has shorter bond length. Assuming that water vapour is an ideal gas. wreversible = - nRT ln Vf Vi DH ° .4 kJ mol respectively. 40 and 50 J K –1 mol –1 . CHEMICAL THERMODYNAMICS A piston filled with 0. 2. If standard enthalpy of combustion of methanol is –726 kJ mol –1 . –1 H 2O (l) and CO 2(g) are –166.16 JEE MAIN CHAPTERWISE EXPLORER CHAPTER 5 1.T DS° (b) ln K = RT (c) (d) 3. w = + 208 J (b) q = + 208 J. K = e –DG°/RT DGsystem D Stotal = -T (2012) The entropy change involved in the isothermal reversible expansion of 2 moles of an ideal gas from a volume of 10 9. 4. w = – 208 J (c) q = – 208 J. the reaction would be spontaneous when (b) For a spontaneous process in an isolated system. the average bond enthalpy of N—H bond in NH 3 is (a) – 1102 kJ mol –1 (b) – 964 kJ mol –1 (c) + 352 kJ mol –1 (d) + 1056 kJ mol –1 (2010) (a) – 22. –237. 8. dm 3 to a volume of 100 dm 3 at 27°C is (a) 38. 6.88 kJ (c) + 228.3 J mol –1 K –1 (b) 35. efficiency of the fuel cell will be (a) 80 % (b) 87% (c) 90% (d) 97% (2009) Standard entropy of X 2 . D eg HCl = –349 kJ mol –1 . For the reaction.0°C.04 mol of an ideal gas expands reversibly from 50.314 J/mol K) (ln 7.3 J mol –1 K –1 (2011) is – 46 kJ mol –1 .32 kJ change (DU) when 1 mol of water is vapourised at 1 bar pressure 1 H2( g ) + O2( g ) ® H 2O( l ) ; DH = - 286. On the basis of the following thermochemical data : (D f G° (d) Exothermic processes are always spontaneous. DH and 10. – D hyd HCl – = –381 kJ mol –1 ) will be (a) +120 kJ mol –1 (b) +152 kJ mol –1 (c) –610 kJ mol –1 (d) –850 kJ mol –1 (2008) For a particular reversible reaction at temperature T. (given : molar enthalpy of vapourisation of water at 2 1 bar and 373 K = 41 kJ mol –1 and R = 8. (2007) H + (aq) = 0).3 J mol K (d) 42. If T e is the temperature at criterion for spontaneity. respectively. The reaction is 3 CH 3 OH ( l ) + O 2( g ) CO 2( g ) + 2H 2 O ( l ) . the change (a) T = T e (b) T e > T in entropy is positive. the temperature will be (a) 1000 K (b) 1250 K (c) 500 K (d) 750 K (2008) Oxidising power of chlorine in aqueous solution can be determined by the parameters indicated below: – – 1/2 Cl 2(g ) 1/2 D diss H Cl (g ) D eg H – Cl – ( g ) D hyd H Cl – (aq ) The energy involved in the conversion of 1/2 Cl 2(g) to Cl – (aq) – – (using data.8 J mol –1 K –1 –1 –1 (c) 32.2. 5. Y 2 and XY 3 are 60. (c) T > T e (d) T e is 5 times T (2010) (c) Endothermic processes are never spontaneous. 1/2 X 2 + 3/2 Y 2 ® XY 3 .0 mL to 375 mL at a constant temperature of 37.5 = 2. it absorbs 208 J of heat.2 and –394.52 kJ (2009) In a fuel cell methanol is used as fuel and oxygen gas is used as an oxidizer.88 kJ (d) – 343. w = – 208 J (d) q = – 208 J.2 kJ and 100°C. The standard enthalpy of formation of NH 3 If the enthalpy of formation of H 2 from its atoms is – 436 kJ mol –1 and that of N 2 is –712 kJ mol –1 . DH = –30 kJ. 2 At 298 K standard Gibb’s energies of formation for CH 3OH (l) .3 J mol –1 K –1 ) will be The value of enthalpy of formation of OH – ion at 25°C is .88 kJ (b) – 228. 11. w for the process will be (R = 8. to be at equilibrium. equilibrium. the internal energy + – H 2O (l) ® H (aq) + OH (aq) ; D H = 57. The values of q and 7. D diss HCl 2 = 240 kJ mol –1 .01) (a) q = + 208 J. As it does so. Identify the correct statement regarding a spontaneous process: (a) Lowering of energy in the reaction process is the only DS were found to be both +ve. w = + 208 J (2013) The incorrect expression among the following is (a) in isothermal process. 17 Chemical Thermodynamics (a) 41.00 kJ mol –1 (b) 4.100 kJ mol –1 (c) 3.7904 kJ mol –1 (d) 37.904 kJ mol –1 (a) 100 kJ mol –1 (c) 800 kJ mol –1 (2007) 12. In conversion of limestone to lime, CaCO 3(s) ® CaO (s) + CO 2(g) the values of DH° and DS° are +179.1 kJ mol –1 and 160.2 J/K respectively at 298 K and 1 bar. Assuming that DH° and DS° do not change with temperature, temperature above which conversion of limestone to lime will be spontaneous is (a) 1118 K (b) 1008 K (c) 1200 K (d) 845 K (2007) 13. (DH – DU) for the formation of carbon monoxide (CO) from its elements at 298 K is (R = 8.314 JK –1 mol –1 ) (a) –1238.78 J mol –1 (b) 1238.78 J mol –1 (c) –2477.57 J mol –1 (d) 2477.57 J mol –1 (2006) (b) 200 kJ mol –1 (d) 400 kJ mol –1 (2005) 18. A schematic plot of ln K eq versus inverse of temperature for a reaction is shown in the figure. The reaction must be 6.0 ln K eq 2.0 1.5 × 10 –3 2.0 × 10 –3 1/T (K –1 ) (a) (b) (c) (d) exothermic endothermic one with negligible enthalpy change highly spontaneous at ordinary temperature. (2005) 19. Consider the reaction: N 2 + 3H 2 ® 2NH 3 carried out at constant 14. The enthalpy changes for the following processes are listed temperature and pressure. If DH and DU are the enthalpy and below: internal energy changes for the reaction, which of the following Cl 2(g) = 2Cl (g) , 242.3 kJ mol –1 expressions is true? I 2(g) = 2I (g) , 151.0 kJ mol –1 (a) DH = 0 (b) DH = DU ICl (g) = I (g) + Cl (g) , 211.3 kJ mol –1 (c) DH < DU (d) DH > DU I 2(s) = I 2(g) , 62.76 kJ mol –1 (2005) Given that the standard states for iodine and chlorine are I 2(s) and Cl 2(g) , the standard enthalpy of formation for ICl (g) is 20. For a spontaneous reaction the DG, equilibrium constant (K) (a) – 14.6 kJ mol –1 (b) –16.8 kJ mol –1 and E° cell will be respectively (c) +16.8 kJ mol –1 (d) +244.8 kJ mol –1 (2006) (a) –ve, >1, +ve (b) +ve, >1, –ve (c) –ve, <1, –ve (d) –ve, >1, –ve (2005) 15. An ideal gas is allowed to expand both reversibly and irreversibly in an isolated system. If T i is the initial temperature and T f is the 21. The enthalpies of combustion of carbon and carbon monoxide final temperature, which of the following statements is correct? are –393.5 and –283 kJ mol –1 respectively. The enthalpy of (a) (T f ) irrev > (T f ) rev formation of carbon monoxide per mole is (b) T f > T i for reversible process but T f = T i for irreversible (a) 110.5 kJ (b) 676.5 kJ process (c) –676.5 kJ (d) –110.5 kJ (2004) (c) (T f ) rev = (T f ) irrev 22. An ideal gas expands in volume from 1 × 10 –3 m 3 to (d) T f = T i for both reversible and irreversible processes. 1 × 10 –2 m 3 at 300 K against a constant pressure of 1 × 10 5 Nm –2 . (2006) The work done is 16. The standard enthalpy of formation (DH f °) at (a) – 900 J (b) – 900 kJ 298 K for methane, CH 4(g) is –74.8 kJ mol –1 . The additional (c) 270 kJ (d) 900 kJ (2004) information required to determine the average energy for 23. The internal energy change when a system goes from state A to C – H bond formation would be B is 40 kJ/mole. If the system goes from A to B by a reversible (a) the dissociation energy of H 2 and enthalpy of sublimation path and returns to state A by an irreversible path what would of carbon be the net change in internal energy? (b) latent heat of vaporisation of methane (a) 40 kJ (b) > 40 kJ (c) the first four ionisation energies of carbon and electron (c) < 40 kJ (d) zero (2003) gain enthalpy of hydrogen 24. In an irreversible process taking place at constant T and P and (d) the dissociation energy of hydrogen molecule, H 2. in which only pressurevolume work is being done, the change (2006) in Gibbs free energy (dG) and change in entropy (dS), satisfy 17. If the bond dissociation energies of XY, X 2 and Y 2 (all diatomic the criteria molecules) are in the ratio of 1 : 1 : 0.5 and DH f for the formation (a) (dS) V, E < 0, (dG) T, P < 0 of XY is –200 kJ mol –1 . The bond dissociation energy of X 2 will (b) (dS) V, E > 0, (dG) T, P < 0 be 18 JEE MAIN CHAPTERWISE EXPLORER (c) (dS) V, E = 0, (dG) T, P = 0 (d) (dS) V, E = 0, (dG) T, P > 0 (a) (b) (c) (d) (2003) violates 1 st law of thermodynamics violates 1 st law of thermodynamics if Q 1 is –ve violates 1 st law of thermodynamics if Q 2 is –ve does not violate 1 st law of thermodynamics. (2002) 25. The enthalpy change for a reaction does not depend upon the (a) physical states of reactants and products (b) use of different reactants for the same product 28. If an endothermic reaction is nonspontaneous at freezing point (c) nature of intermediate reaction steps of water and becomes feasible at its boiling point, then (d) difference in initial or final temperatures of involved (a) DH is –ve, DS is +ve substances. (b) DH and DS both are +ve (2003) (c) DH and DS both are –ve 26. If at 298 K the bond energies of C – H, C – C, (d) DH is +ve, DS is –ve (2002) C and H – H bonds are respectively 414, 347, 615 and C 29. For the reactions, 435 kJ mol –1 , the value of enthalpy change for the reaction C + O 2 ® CO 2 ; DH = –393 J H 2 C CH 2(g) + H 2(g) ® H 3C – CH 3(g) 2Zn + O ® 2ZnO ; DH = –412 J 2 at 298 K will be (a) carbon can oxidise Zn (a) +250 kJ (b) –250 kJ (b) oxidation of carbon is not feasible (c) +125 kJ (d) –125 kJ. (2003) (c) oxidation of Zn is not feasible 27. A heat engine absorbs heat Q 1 at temperature T 1 and heat Q 2 at (d) Zn can oxidise carbon. (2002) temperature T 2 . Work done by the engine is J (Q 1 + Q 2 ). This data Answer Key 1. 7. 13. 19. 25. (b) (d) (a) (c) (c) 2. 8. 14. 20. 26. (b) (d) (c) (a) (d) 3. 9. 15. 21. 27. (a) (c) (a) (d) (d) 4. 10. 16. 22. 28. (c) (b) (a) (a) (b) 5. 11. 17. 23. 29. (c) (d) (c) (d) (d) 6. 12. 18. 24. (b) (a) (a) (b) 19 Chemical Thermodynamics 1. (b) : As it absorbs heat, \ 7. q = + 208 J (d) : For the given reaction, CH3OH ( l ) + æ V 2 ö w rev = -2.303 nRT log 10 ç V ÷ è 1 ø also, DG°f [CH3OH(l)] = –166.2 kJ mol –1 æ 375 ö w rev = -2.303 ´ (0.04) ´ 8.314 ´ 310 log 10 çè 50 ÷ø \ 2. DG°f [H2O(l)] = –237.2 kJ mol –1 and DG°f [CO2(g)] = –394.4 kJ mol –1 w rev = – 207.76 » – 208 J Now, DG°reaction = å DG of products - å DG of reactants (b) : DG° = –RT lnK DG° = DH° – TD S° = [–394.4 + 2 × (–237.2)] – (–166.2) DH° – TDS° = –RT lnK 3. = – 702.6 kJ mol –1 % Efficiency = (a) : Entropy change for an isothermal process is \ Efficiency » 97% (d) : 1 X + 3 Y ® XY 3 2 2 2 2 8. 1 DS = 2.303 ´ 2 ´ 8.314 ´ log æç 100 ö÷ è 10 ø 0 0 0 DS reaction = DS products - D S reactants = 38.294 J mol –1 K –1 » 38.3 J mol –1 K –1 (c) : 1/2N 2 + 3/2H 2 B.E. 712 0 0 \ DS reaction = DS XY - 1 DS X0 - 3 DSY 0 3 2 2 2 2 = 50 - 1 ´ 60 - 3 ´ 40 = – 40 J K –1 mol –1 2 2 NH 3 436 é1 3 ù \ ( DH °f ) NH = ê B.E.N + B.E.H - 3B.E. N - H ú 3 2 2 2 ë2 û Using equation, DG = DH – TDS We have DH = – 30 kJ, DS = – 40 J K –1 mol –1 and at equilibrium DG = 0. Therefore é1 3 ù - 46 = ê ´ 712 + ´ 436 - 3B.E. N - H ú 2 ë2 û T = – 46 = 356 + 654 – 3B.E. N—H 3B.E. N —H = 1056 B.E.N - H = 5. 9. D H -30 ´ 1000 = = 750 K DS -40 (c) : 1 Cl ® Cl ( g ) ; 2 2( g ) 1 - 240 DH1 = D diss H e = = 120 kJ mol -1 Cl 2 2 2 1056 = 352 kJ mol -1 3 e Cl( g ) ® Cl(-g ) ; DH 2 = D eg H Cl = -349 kJ mol-1 (c) : According to Gibb’s formula, DG = DH – TDS Cl(-g ) + aq ® Cl(-aq ) ; Since DH and DS, both are +ve, for DG < 0, the value of T > T e . 6. DG -702.6 ´ 100 = ´ 100 DH - 726 = 96.77% æ DH ° - T DS° ö ln K = - ç ÷ø è RT æV ö DS = 2.303nR log ç 2 ÷ è V ø 4. 3 O ¾¾ ® CO2( g ) + 2H 2O( l ) , 2 2( g ) DH = –726 kJ mol –1 – (b) : The reaction for the formation of OH (aq) is 1 H 2( g ) + O 2( g ) H +( aq ) + OH ( – aq ) 2 This is obtained by adding the two given equations. \ DH for the above reaction = 57.32 + (–286.2) = –228.88 kJ DH 3 = D hyd H e = -381 kJ mol -1 The required reaction is 1 Cl ® Cl(-aq ) ; D H 2 2( g ) 1 e e e Then DH = D diss H + D eg H + D hyd H 2 = 120 + (– 349) + (–381) = –610 kJ mol –1 20 JEE MAIN CHAPTERWISE EXPLORER 10. (b) : In an isolated system, there is neither exchange of energy 16. (a) : C + 2H 2 ® CH 4 ; DH° = –74.8 kJ mol –1 In order to calculate average energy for C – H bond formation nor matter between the system and surrounding. For a spontaneous process in an isolated system, the change in entropy we should know the following data. is positive, i.e. DS > 0. C (graphite) ® C (g) ; DH° f = enthalpy of sublimation of carbon Most of the spontaneous chemical reactions are exothermic. A H 2(g) ® 2H (g) ; DH° = bond dissociation energy of H 2 number of endothermic reactions are spontaneous e.g. melting 17. (c) : Let the bond dissociation energy of XY, X 2 and Y 2 be x kJ of ice (an endothermic process) is a spontaneous reaction. mol –1 , x kJ mol –1 and 0.5x kJ mol –1 respectively. The two factors which are responsible for the spontaneity of 1 1 a process are X 2 + Y2 ® XY ; DH f = -200 kJ mol -1 2 2 (i) tendency to acquire minimum energy DH reaction = [(sum of bond dissociation energy of all reactants) (ii) tendency to acquire maximum randomness. – (sum of bond dissociation energy of product)] 11. (d) : DU = DH – D nRT 1 é1 ù = ê DH X 2 + DHY2 - DH XY ú = 41000 – 1 × 8.314 × 373 2 ë2 û = 41000 – 3101.122 x 0.5 x = 37898.878 J mol –1 = 37.9 kJ mol –1 = + - x = -200 2 2 12. (a) : For DG° = DH° – TDS° 200 For a spontaneous process DG° < 0 x= = 800 kJ mol -1 \ 0.25 i.e. DH° – TDS° < 0 K 2 DH é 1 1 ù or DH° < TDS° or, TDS° > DH° 18. (a) : ln K = R ê T - T ú 1 ë 1 2 û DH ° 179.1 ´ 1000 i.e. T > or T > 6 DH D S ° 160.2 ln = [1.5 ´ 10-3 - 2 ´ 10 -3 ] 2 R or T > 1117.9 K » 1118 K DH 13. (a) : DH – DU = Dn g RT ´ (-0.5 ´ 10-3 ) or, ln3 = R 1 C + O 2 ® CO DH of reaction comes out to be negative. Hence reaction is 2 exothermic. 1 æ 1ö Dng = 1 - ç1 + ÷ = è 2ø 2 19. (c) : N 2 + 3H 2 ® 2NH 3 1 -1 DH – DU = - ´ 8.314 ´ 298 = - 1238.78 J mol 2 14. (c) : 1 1 I 2 (s) + Cl 2 (g) ® ICl (g) 2 2 1 1 é1 ù DH ICl (g) = ê DH I2 ( s ) ® I 2 ( g ) + DH I -I + DH Cl -Cl ú -[ D H I -Cl ] ë2 2 2 û 1 1 é1 ù = ê ´ 62.76 + ´ 151.0 + ´ 242.3ú - [211.3] ë2 2 2 û = [31.38 + 75.5 + 121.15] – 211.3 = 228.03 – 211.3 = 16.73 kJ/mol Dn = 2 – 4 = –2 DH = DU + DnRT = DU – 2RT \ DH < DU 20. (a) : For spontaneous process, DG = –ve Now DG = –RT ln K When K > 1, DG = –ve Again DGº = – nFEº When Eº = +ve, DGº = –ve 21. (d) : C (s) + O 2(g) ® CO 2(g) ; DH = –393.5 kJ mol –1 CO (g) + 1 O ® CO 2(g) ; DH = –283 kJ mol –1 2 2(g) .. (i) ... (ii) On subtraction equation (ii) from equation (i), we get 15. (a) : If a gas was to expand by a certain volume reversibly, then C (s) + O 2(g) ® CO (g) ; DH = –110.5 kJ mol –1 it would do a certain amount of work on the surroundings. If The enthalpy of formation of carbon monoxide per mole it was to expand irreversibly it would have to do the same amount = –110.5 kJ mol –1 of work on the surroundings to expand in volume, but it would 22. (a) : W = – PDV also have to do work against frictional forces. Therefore the = –1 × 10 5 (1 × 10 –2 – 1 × 10 –3 ) amount of work have greater modulus but –ve sign. = –1 × 10 5 × 9 × 10 –3 = –900 J W irrev. > W reve. ; (T f ) irrev. > (T f ) rev. 21 Chemical Thermodynamics H reversible path 26. (d) : 23. (d) : A H B irreversible path H C C +H–H H H H H–C–C–H H H DH Reaction = å BE reactant – å BE product = 4 × 414 + 615 + 435 – (6 × 414 + 347) = 2706 – 2831 = –125 kJ We know that for a cyclic process the net change in internal energy is equal to zero and change in the internal energy does 27. (d) : It does not violate first law of thermodynamics but violates not depend on the path by which the final state is reached. second law of thermodynamics. 24. (b) : For spontaneity, change in entropy (dS) must be positive, 28. (b) : For endothermic reaction, DH = +ve means it should be greater than zero. Now, DG = DH – TDS Change in Gibbs free energy (dG) must be negative means that For nonspontaneous reaction, DG should be positive Now DG is positive at low temperature if DH is positive. it should be lesser than zero. (dS) V, E > 0, (dG) T, P < 0. DG is negative at high temperature if DS is positive. 25. (c) : This is according to Hess's law. 29. (d) : DH = negative shows that the reaction is spontaneous. Higher value for DH shows that the reaction is more feasible. 22 JEE MAIN CHAPTERWISE EXPLORER CHAPTER SOLUTIONS 6 1. 2. 3. K f for water is 1.86 K kg mol –1 . If your automobile radiator 7. holds 1.0 kg of water, how many grams of ethylene glycol (C 2 H 6 O 2 ) must you add to get the freezing point of the solution lowered to –2.8°C? (a) 93 g (b) 39 g (c) 27 g (d) 72 g (2012) The density of a solution prepared by dissolving 120 g of 8. urea (mol. mass = 60 u) in 1000 g of water is 1.15 g/mL. The molarity of this solution is (a) 1.78 M (b) 1.02 M (c) 2.05 M (d) 0.50 M (2012) The degree of dissociation (a) of a weak electrolyte, A x B y is related to van’t Hoff factor (i) by the expression i - 1 i - 1 (b) a = ( x + y + 1) ( x + y - 1) ( x + y - 1) ( x + y + 1) (c) a = (d) a = i - 1 i - 1 (a) a = (2011) 9. If sodium sulphate is considered to be completely dissociated into cations and anions in aqueous solution, the change in freezing point of water (DT f), when 0.01 mol of sodium sulphate is dissolved in 1 kg of water, is (K f = 1.86 K kg mol –1 ) (a) 0.0186 K (b) 0.0372 K (c) 0.0558 K (d) 0.0744 K (2010) A binary liquid solution is prepared by mixing nheptane and ethanol. Which one of the following statements is correct regarding the behaviour of the solution? (a) The solution formed is an ideal solution. (b) The solution is nonideal, showing +ve deviation from Raoult’s law. (c) The solution is nonideal, showing –ve deviation from Raoult’s law. (d) nheptane shows +ve deviation while ethanol shows –ve deviation from Raoult’s law. (2009) Two liquids X and Y form an ideal solution. At 300 K, vapour pressure of the solution containing 1 mol of X and 3 mol of Y is 550 mm Hg. At the same temperature, if 1 mol of Y is further added to this solution, vapour pressure of the solution increases by 10 mm Hg. Vapour pressure (in mm Hg) of X and Y in their pure states will be, respectively (a) 200 and 300 (b) 300 and 400 (c) 400 and 600 (d) 500 and 600 (2009) 4. Ethylene glycol is used as an antifreeze in a cold climate. Mass of ethylene glycol which should be added to 4 kg of water to prevent it from freezing at – 6°C will be : (K f for water = 1.86 K kg mol –1 , and molar mass of ethylene glycol = 62 g mol –1 ) (a) 804.32 g (b) 204.30 g (c) 400.00 g (d) 304.60 g (2011) 5. A 5.2 molal aqueous solution of methyl alcohol, CH 3 OH, is supplied. What is the mole fraction of methyl alcohol in the 10. The vapour pressure of water at 20°C is 17.5 mm Hg. If 18 g of glucose (C 6 H 12 O 6) solution? is added to 178.2 g of water at 20°C, the vapour pressure of the resulting solution will be (a) 0.100 (b) 0.190 (a) 17.325 mm Hg (b) 17.675 mm Hg (c) 0.086 (d) 0.050 (2011) (c) 15.750 mm Hg (d) 16.500 mm Hg On mixing, heptane and octane form an ideal solution. At (2008) 373 K, the vapour pressure of the two liquid components 6. (heptane and octane) are 105 kPa and 45 kPa respectively. 11. At 80°C, the vapour pressure of pure liquid A is 520 mm of Hg and that of pure liquid B is 1000 mm of Hg. If a mixture Vapour pressure of the solution obtained by mixing 25.0 g of solution of A and B boils at 80°C and 1 atm pressure, the heptane and 35 g of octane will be (molar mass of heptane = –1 –1 100 g mol and of octane = 114 g mol ) amount of A in the mixture is (1 atm = 760 mm of Hg) (a) 50 mol percent (b) 52 mol percent (a) 144.5 kPa (b) 72.0 kPa (c) 34 mol percent (d) 48 mol percent (c) 36.1 kPa (d) 96.2 kPa (2010) (2008) Solutions 23 12. A 5.25% solution of a substance is isotonic with a 1.5% solution 21. Which one of the following statements is false? of urea (molar mass = 60 g mol –1 ) in the same solvent. If the (a) Raoult’s law states that the vapour pressure of a component densities of both the solutions are assumed to be equal to over a solution is proportional to its mole fraction. 1.0 g cm –3 , molar mass of the substance will be (b) The osmotic pressure (p) of a solution is given by the (a) 210.0 g mol –1 (b) 90.0 g mol –1 equation (p = MRT, where M is the molarity of the solution. (c) 115.0 g mol –1 (d) 105.0 g mol –1 (c) The correct order of osmotic pressure for (2007) 0.01 M aqueous solution of each compound is BaCl 2 > KCl > CH 3 COOH > sucrose. 13. A mixture of ethyl alcohol and propyl alcohol has a vapour (d) Two sucrose solutions of same molality prepared in pressure of 290 mm at 300 K. The vapour pressure of propyl different solvents will have the same freezing point alcohol is 200 nm. If the mole fraction of ethyl alcohol is 0.6, depression. (2004) its vapour pressure (in mm) at the same temperature will be (a) 360 (b) 350 22. Which of the following liquid pairs shows a positive deviation (c) 300 (d) 700 (2007) from Raoult’s law? (a) Water hydrochloric acid 14. The density (in g mL –1 ) of a 3.60 M sulphuric acid solution that (b) Benzene methanol is 29% H 2 SO 4 (molar mass = 98 g mol –1 ) by mass will be (c) Water nitric acid (a) 1.45 (b) 1.64 (d) Acetone chloroform (2004) (c) 1.88 (d) 1.22 (2007) 23. To neutralise completely 20 mL of 0.1 M aqueous solution of 15. 18 g of glucose (C 6 H 12 O 6 ) is added to 178.2 g of water. The phosphorous acid (H 3 PO 3 ), the volume of 0.1 M aqueous KOH vapour pressure of water for this aqueous solution at 100°C is solution required is (a) 759.00 torr (b) 7.60 torr (a) 10 mL (b) 20 mL (c) 76.00 torr (d) 752.40 torr (2006) (c) 40 mL (d) 60 mL (2004) 16. Density of a 2.05 M solution of acetic acid in water is 1.02 g/ 24. 6.02 × 10 20 molecules of urea are present in 100 ml of its solution. mL. The molality of the solution is The concentration of urea solution is (a) 1.14 mol kg –1 (b) 3.28 mol kg –1 (a) 0.001 M (b) 0.01 M (c) 2.28 mol kg –1 (d) 0.44 mol kg –1 (2006) (c) 0.02 M (d) 0.1 M (2004) 17. Equimolal solutions in the same solvent have 25. Which one of the following aqueous solutions will exhibit (a) same boiling point but different freezing point highest boiling point? (b) same freezing point but different boiling point (a) 0.01 M Na 2 SO 4 (b) 0.01 M KNO 3 (c) same boiling and same freezing points (c) 0.015 M urea (d) 0.015 M glucose (2004) (d) different boiling and different freezing points. (2005) 26. If liquids A and B form an ideal solution, the 18. Two solutions of a substance (non electrolyte) are mixed in the (a) enthalpy of mixing is zero following manner. 480 mL of 1.5 M first solution + 520 mL of (b) entropy of mixing is zero 1.2 M second solution. What is the molarity of the final mixture? (c) free energy of mixing is zero (a) 1.20 M (b) 1.50 M (d) free energy as well as the entropy of mixing are each zero. (c) 1.344 M (d) 2.70 M (2005) (2003) 19. Benzene and toluene form nearly ideal solutions. At 20°C, the 27. 25 mL of a solution of barium hydroxide on titration with a 0.1 vapour pressure of benzene is 75 torr and that of toluene is 22 molar solution of hydrochloric acid gave a litre value of 35 mL. torr. The partial vapour pressure of benzene at 20°C for a solution The molarity of barium hydroxide solution was containing 78 g of benzene and 46 g of toluene in torr is (a) 0.07 (b) 0.14 (a) 50 (b) 25 (c) 0.28 (d) 0.35 (2003) (c) 37.5 (d) 53.5 (2005) 28. In a 0.2 molal aqueous solution of a weak acid HX, the degree 20. If a is the degree of dissociation of Na 2 SO 4 , the vant Hoff’s of ionization is 0.3. Taking K f for water as 1.85, the freezing factor (i) used for calculating the molecular mass is point of the solution will be nearest to (a) 1 + a (b) 1 – a (a) – 0.480°C (b) – 0.360 °C (c) 1 + 2a (d) 1 – 2a (2005) (c) – 0.260 °C (d) + 0.480 °C (2003) 27.86°C.d) 6. (c) A – B interaction is weaker than A – A and (a) 0.0512°C (d) A – B interaction is stronger than A – A and (c) 0. 22. 21.24 JEE MAIN CHAPTERWISE EXPLORER 29. 8. (c) 13. find the increase in boiling point. (a) 25. 18. (a) (a) (c) (b) (a) 5.2372°C (2002) (2002) Answer Key 1. Freezing point of an aqueous solution is (–0. 29. 24. (a) 2. (b) (a) (c) (b) (b) . (a) (c) (d) (d) (b) 4. (b) DH mix < 0 K f = 1. (c) (a) (c) (c) (b. (d) 0. Elevation (a) DV mix > 0 of boiling point of the same solution is K b = 0. (b) 19. 17. (a) 7.092°C B – B interaction. 16.186°C B – B interaction (b) 0. 26. 11. 15. 9. 28.512°C. (c) (b) (d) (c) (a) 3. 14. In mixture A and B components show –ve deviation as 30. 20. 12. 30. 23. 10.186)°C. 55 » 72 kPa 7. K f = 1. 0.25 + 0.05 M 1120 / 1.25 x heptane = = 0.2 1000 5. solvent is water.15 = ´ 1000 = = 2.30 p Total = x heptane p° heptane + x octane p° octane xA y+ + yB x– i .86 5.25 + 0.8) = 2.8°C Mass of solvent = 1.8 = 1.15 1120 (a) : A x B y = 0.8 Þ w= = 93 g 62 1.2 ´ 18 93.30 114 0. P T = Total presure p° X = Vapour pressure of X in pure state p° Y = Vapour pressure of Y in pure state x X = Mole fraction of X = 1/4 x Y = Mole fraction of Y = 3/4 (i) When T = 300 K.15 g/mL Total mass of solution = 1000 + 120 = 1120 g 35 = 0.54 0.1 \ a = ( x + y - 1) 25 = 0. i = 3 m= 1 – a xa ya i = 1 – a + xa + ya = 1 + a(x + y – 1) 4.86 ´ 5. (c) : P T = p° X x X + p° Y x Y w 1 and w 2 = wt.25 Solutions 1.45 0.54 × 45 = 47. N = 1000 18 \ Mole fraction = Weight of solute Molecular mass of solute m= ´ 1000 Mass of solvent (g) m= = 6. (b) : The solution containing nheptane and ethanol shows nonideal behaviour with positive deviation from Raoult’s law.25 + 24. of solute DT f = 0 – (– 6) = 6 1.0 kg Mass of solute = ? Molecular mass of solute = 62 DT f = K f × m Here solute is methyl alcohol.6 + 1000 1093. Given n = 5.01 = 0. (b) : Given.2 + 18 where.0855 » 0.15 Mass of solute Molecular mass of solute Molarity = ´ 1000 Volume of solution 120 / 60 2 ´ 1000 ´ 1.0558 K w ´ 1000 (a) : D T f = K f × m = K f ´ 2 w1 ´ m2 \ 6 = (c) :Depression in freezing point.86 K kg mol –1 \ DT f = 3 × 1. 5.86 K kg mol –1 DT f = 0 – (–2. as a result they easily vapourise showing higher vapour presure than expected.30 = 0. (c) : Mole fraction of solute = n N + n n = number of moles of solute N = number of moles of solvent 0. DT f = i × K f × m For sodium sulphate.01 m. 8.3 = 71.86 ´ w2 ´ 1000 4000 ´ 62 6 ´ 62 ´ 4000 w 2 = = 800 g 1000 ´ 1.86 × 0.2. (a) : K f = 1. however the forces between nheptane and ethanol are not very strong. 1 kg Given. This is because the ethanol molecules are held together by strong Hbonds.01 = 0.6 p° octane = 45 kPa w / 62 w ´ 1000 = 1000 62 2. w octane = 35 g w 62 ´ 2.86 nheptane = (c) : Mass of solute taken = 120 g Molecular mass of solute = 60 u Mass of solvent = 1000 g Density of solution = 1.6 = = 0. 9. p° heptane = 105 kPa w heptane = 25 g DT f = K f × m 2.45 × 105 + 0. of solvent and solute m 2 = molecular wt.086 93.25 100 noctane = Volume of solution = 3. P T = 550 mm Hg .30 xoctane = Mass 1120 = mL Density 1. (c) : Molality. (a) : Isotonic solutions have same osmotic pressure.p s 18/180 1/10 directly proportional to its mole fraction..4 × 200 or P A o = 350 mm 19.(1) (ii) When at T = 300 K. at 1 atm pressure i..05 m= = P solution = 17.02 . 1 mole of Y is added..p s = 99 \ P sol = P°X solvent 760 ´ 99 Let A be the solute and B the solvent = 752. (d) : The extent of depression in freezing point varies with the number of solute particles for a fixed solvent only and it’s a characteristic feature of the nature of solvent also.6 M solution means 3.6 moles of H 2 SO 4 = 3. . p 2 = C 2 RT For isotonic solution. density = Volume 1000 On solving equations (1) and (2).6) + 200 ´ (1 - 0.6 × 98 g = 352.99 ps = ps Þ 760 . or.5 = 1.9 = 0. at 760 mm Hg = 480 × 1.325 1000 ´ 1.5 20.344 M Molarity of the final mixture = Þ 520 X A + 1000 – 1000 X A = 760 mm Hg 1000 1 Þ X A = or 50 mol percent 2 12.8 g \ Mass of H 2 SO 4 in 1000 mL of solution = 352. pressure is 760 .(2) 14.2 = 720 + 624 = 1344 PA° X A + PB° X B = 760 mm Hg Total volume = 480 + 520 = 1000 PA° X A + PB° (1 . (d) : p° Y = 600 mm Hg and p° X = 400 mm Hg p s N 10.2 n B 18 \ X B = = M n A + nB 18 + 178.8 g Given.25 / M = V V [Where M = molecular weight of the substance] 1. (c) : Total millimoles of solute Then.4 torr Þ 100p s = 760 × 99 Þ p s = 100 178. p1 = C1 RT .6) or 290 = 0. for two different solvents the extent of depression may vary even if number of solute particles be dissolved in them.216 g/mL Now.05 ´ 60 897 11. and mole fraction of B in solution = X B 18.26 JEE MAIN CHAPTERWISE EXPLORER \ 550 = p °X Þ ( 14 ) + p ( 43 ) ° Y p°X + 3 p Y ° = 2200 .99 where M = molarity.5/ 60 5.2.6 mole of H 2 SO 4 is present in 1000 mL of solution.a + 2 a + a = 1 + 2a 1 21. = or M = 210 60 M 13..9 P solution = vapour pressure of its pure component 1 × mole fraction in solution ps Þ 760 ´ 99 . (b) : According to Raoult’s law. DT f = k f × m For different solvents.p s n = 15. (c) : According to Raoult’s law equimolal solutions of all the and PB ° = 1000 mm Hg substances in the same solvent will show equal elevation in Let mole fraction of A in solution = X A boiling points as well as equal depression in freezing point. (d) : 3.. p1 = p 2 \ C 1 = C 2 1. M 2 = molecular mass 9.05 2. P A ° = 520 mm Hg = 2. \ Mass of 3. (c) : Na 2 SO 4 ƒ 2Na + + SO 42– 1 0 0 1 – a 2a a Vant Hoff factor (i) = 1 .5 × 0.8 = 1216 g of solution 29 Mass 1216 = = 1.2 /18 9.25 or.2 ´ 1000 16.X A ) = 760 mm Hg 1344 = 1. So. d = density. (a) : In solution containing nonvolatile solute. P = PA + PB = PA° x A + PB° x B or 290 = PA ° ´ (0.. we get p ° .8 g of H 2 SO 4 is present in 100 ´ 352.. P T = (550 + 10) mm Hg \ x X = 1/5 and x Y = 4/5 1 4 Þ 560 = p °X + p Y ° 5 5 ( ) ( ) or p°X + 4 p Y ° = 2800 .6 × PA ° + 0.28 mol kg –1 17.94 Now P solution = P°X solvent = 17.99 2. (a) : We have. value of k f is also different. 29 g of H 2 SO 4 is present in 100 g of solution \ 352. P B = P° B X B P° B = 75 torr 78 / 78 1 1 X B = (78/ 78) + (46 / 92) = 1 + 0.e. = = ps 178.5 1 P B = 75 ´ = 50 torr 1. (a) : According to Raoult's law.28 × 10 –3 mol g –1 = 2. m = 180 18 1000 d - MM 2 X B = 9.5 5.5 + 520 × 1. 23.4810 Freezing point of solution = 0 – 0. (b. Consequently the vapour pressure of the solution is greater than the vapour pressure as expected from Raoult’s law.405 × 0. i.186 1. the escaping tendency of alcohol and benzene molecules from the solution increases. M1 = CH 3 CH 3 CH 3 O —– H O —– H O —– H On adding benzene. of moles after dissociation = Kf (experimental) no.e. DT b = 0. of moles before dissociation or. higher the extent of elevation in boiling point. of moles of solute 10 -3 ´ 1000 = 0. (b) : Moles of urea = 6. p A > pº A x A; p B > p ºB x B ; p > p A + p B In solution of methanol and benzene.02 ´ 10 23 Concentration (molarity) of solution = no.481° C 29.3 = 1. methanol molecules are held together due to hydrogen bonding as shown below: 26. K f (observed) = 1.02 ´ 10 20 = 10 -3 moles 24. the observed vapour pressure of each component and total vapour pressure are greater than predicted by Raoult's law.3 K f (observed) no. N 1 V 1 (acid) = N 2 V 2 (base) 0.3 = 1. of litres of solution 100 25. DV mix < 0 and DH mix < 0.1 ´ 2 ´ 20 \ V2 = 0.1 × 2 × 20 = 0.14 25 28.3 = 2. neither heat is evolved nor absorbed during dissolution.1 × 1 × V 2 0.512 or. Here A – B attractive force is greater than A – A and B – B attractive forces.85 × 1.27 Solutions 22. Kf (observed) 1.01 M = no. (a) : For ideal solutions.2 = 0.481 = – 0.1 × 35 or. the benzene molecules get in between the molecules of methanol. (a) : Elevation in boiling point is a colligative property which depends upon the number of solute particles. thus breaking the hydrogen bonds. (a) : HX H + + X – 1 0 0 1 – 0. 27. As the resulting solution has weaker intermolecular attractions. from Raoult's law.3 + 0.3 Total number of moles after dissociation = 1 – 0.85 1 or. (b) : In solutions showing positive deviation. (c) : H 3 PO 3 is a dibasic acid.405 DT f = K f × molality = 2. (b) : DT b = K b DT f = K f W B ´ 1000 M B ´ W A W B ´ 1000 M B ´ WA DTb K b DT b = or = 0. Greater the number of solute particles in a solution. DH mix = 0.1 ´ 35 = 0. (b) : Ba(OH) 2 HCl M 1 V 1 = M 2 V 2 M 1 × 25 = 0.86 .3 0.0512°C D T f K f 0. Na 2 SO 4 ® 2Na + + SO 4 2– 0.3 + 0. d) : For negative deviation.1 ´ 1 = 40 mL 6. 30.3 0. 034M solution of the carbonic acid.3 atm (d) 0.8 × 10 –11 Select the correct statement for a saturated 0.22 (b) 9. will Mg ions start precipitating in the form of Mg(OH) 2 from a solution of 0. Four species are listed below : + (i) HCO 3– (ii) H 3O – (2010) (iii) HSO 4 (iv) HSO 3F Which one of the following is the correct sequence of their Three reactions involving H 2 PO 4 – are given below: acid strength? (i) H 3 PO 4 + H 2 O H 3 O + + H 2 PO 4 – – 2– + (a) iii < i < iv < ii (ii) H 2 PO 4 + H 2 O HPO 4 + H 3 O (b) iv < ii < iii < i (iii) H 2 PO 4 – + OH – H 3 PO 4 + O 2– – (c) ii < iii < i < iv In which of the above does H 2 PO 4 act as an acid? (d) i < iii < ii < iv (2008) (a) (i) only (b) (ii) only (2010) 12.1 × 10 –5 M (b) 5.28 JEE MAIN CHAPTERWISE EXPLORER CHAPTER EQUILIBRIUM 7 1. is 4.9 L (d) 2. the solubility product of Mg(OH) 2 is 1. (2010) The correct order of increasing basicity of the given conjugate bases (R = CH 3 ) is (a) RCOO – < HC C – < NH 2 – < R – (b) RCOO – < HC C – < R – < NH 2 – – – C < RCOO – < NH 2 – (c) R < HC – (d) RCOO < NH 2 – < HC C – < R – (2010) A vessel at 1000 K contains CO 2 with a pressure of 0. Solid Ba(NO 3 ) 2 is gradually dissolved in a 1.79 (d) 7. The value of K c for the reaction.2 × 10 –10 g (d) 6. The pK a of a weak acid.05M solution of silver nitrate to start the (c) 4. K a of this acid is 9.1 × 10 –7 M (2009) which pH. 4. 2.78.02 (2012) The pH of a 0.18 atm (2011) to form?(K sp for BaCO 3 = 5. If the total pressure at equilibrium is 0. At what concentration of Ba 2+ will a precipitate begin (c) 0.0 × 10 –4 M Na 2 CO 3 (a) 1. (c) The concentration of CO 3 2– is greater than that of HCO 3 – .8 atm (b) 3 atm solution.1 × 10 –8 M (d) 8. (d) (iii) only . BOH is 4. The pK b of a weak base. The quantity corresponding salt. (HA). 11.0 3. (b) The concentration of CO 3 2– is 0. BA. (a) 9.0 × 10 –11 . (a) The concentration of H + is double that of CO 3 2– . 1 1 N + O 2(g) at the same 2 2(g) 2 (b) 4 × 10 –4 (d) 0.5 atm.0 × 10 –13 .5 × 10 2 (c) 50. 5. How many litres of water must be added to 1 litre of an aqueous solution of HCl with a pH of 1 to create an aqueous solution with pH of 2? 8.80. Some of the CO 2 is converted into CO on the addition of graphite. The value of the ionization constant. (a) 1 × 10 –3 (b) 1 × 10 –5 (c) 1 × 10 –7 (d) 3 × 10 –1 (2012) (a) 5. (d) The concentration of H + and HCO 3 – are approximately equal.2 × 10 –5 g (2010) In aqueous solution the ionisation constants for carbonic acid are K 1 = 4.2 × 10 –7 and K 2 = 4. will be of potassium bromide (molar mass taken as 120 g mol –1 ) to be (a) 9.1 × 10 –5 M At 25°C.001 M Mg 2+ ions? (a) 8 (b) 9 (c) 10 (d) 11 6. the value of K is 10.0 × 10 –8 g (c) 1. The pH of an aqueous solution of the Solubility product of silver bromide is 5.1 × 10 –9 ) (a) 4.0 L (2013) The equilibrium constant (K c ) for the reaction N 2(g) + O 2(g) ® 2NO (g) at temperature T is 4 × 10 –4 . At 2+ (c) 8.1 molar solution of the acid HQ is 3.8 atm.1 L (c) 0.58 added to 1 litre of 0.2 × 10 –9 g (b) 1. NO (g) ® temperature is (a) 2.0 L (b) 0.01 (2008) precipitation of AgBr is (c) (i) and (ii) 7.034 M. 29 Equilibrium 13.0 × 10 –5 and 5.0 × 10 (2007) (R = 0.x ø (c) K 2 K 3 = K 1 (d) K 3 = K 1K (2008) 2 æ x ö æ x ö 14.12 (b) + 8. The pOH of an aqueous 23. equilibrium 20.2 × 10 5 (b) 5. Given the data at 25°C.84 atm. F 2 and ClF 3 ? is K c = 4. the partial pressure of PCl 3 Which of the following relation is correct? will be æ x ö æ 2 x ö (a) K 3 ∙ K 23 = K 12 (b) K1 K 2 = K3 (a) ç ÷ P (b) ç ÷ P è x + 1 ø è 1 . (a) 1 : 9 (b) 1 : 36 Ammonium hydrogen sulphide decomposes to yield NH 3 and (c) 1 : 1 (d) 1 : 3 (2008) H 2 S gases in the flask. For the following three reactions (i). Phosphorus pentachloride dissociates as follows in a closed constants are given reaction vessel. The overall dissociation 25.50 atm.800 V (b) K p is less than K c What is the value of log K sp for AgI? (c) K p = K c RT æ ö 2.5 24.98 × 10 8 (b) 3.8 × 10 c (c) 5. If degree of dissociation of X and Z be equal then the 21.0 × 10 –5 –6 at 184°C) 15 –15 (K = 1.88 × 10 6 17. The equilibrium constant for the reaction. constant of the acid will be 2NO 2(g) 2NO (g) + O 2(g) (a) 0.30 (b) 0.5. (a) – 8. the total pressure in the flask rises to 0.x ø X 2Y and Z P + Q.5 (d) 9.83 × 10 –3 g (d) 1.0831 kJ/(mol. An amount of solid NH HS is placed in a flask already containing 4 ratio of total pressures at these equilibria is ammonia gas at a certain temperature and 0.0 × 10 –10 respectively. temperature is If the solubility product constant K sp of AgIO 3 at a given (a) 0. Cl 2(g) + 3F 2(g) 2ClF 3(g) ; DH = –329 kJ 1 Which of the following will increase the quantity of ClF 3 in an SO 3(g) SO 2(g) + O 2(g) 2 equilibrium mixture of Cl 2 .9 × 10 –2 (2006) (2007) . pressure. compared at 184°C it is found that Ag + I – ® AgI + e – ; E° = 0.0 × 10 –7 g (c) HCOOH (d) CH 3 CH 2 COOH (2005) (2007) – 16. The first and second dissociation constants of an acid H 2 A are (c) 3.40 × 10 –3 (d) Adding F 2 (2005) (c) 9.5 (c) 2.18 temperature is 1. The value of K c for the reaction (a) Increasing the temperature (b) Removing Cl 2 2SO 2(g) + O 2(g) 2SO 3(g) will be (c) Increasing the volume of the container (a) 416 (b) 2. The AgIO 3 (molecular mass = 283) the equilibrium which sets in is equilibrium constant for NH 4 HS decomposition at this AgIO 3(s) Ag + (aq) + IO 3 – (aq) . What is the conjugate base of OH ? (a) O 2 (b) H 2 O bufferd solution of HA in which 50% of the acid is ionized is – (c) O (d) O 2– (2005) (a) 7. When the decomposition reaction reaches 15. The exothermic formation of ClF 3 is represented by the equation: 19.17 (d) 0. The equilibrium constants K p 1 and K P 2 for the reactions (c) ç ÷ P (d) ç ÷ P (2006) è x . less than or equal to K c depends ç ÷ è ø F upon the total gas pressure.0 × 10 –8 .9 × 10 –2 .612 (2005) (c) – 37.68 × 10 –6 (d) 3.1 ø è 1 .4 will be (a) 3.8 × 10 –2 (d) 4.059 V (d) whether K p is greater than.13 (2006) 26. Among the following acids which has the lowest pK a value? (a) 1. (ii) and (iii). what is the mass of AgIO 3 contained (c) 0.303 = 0.152 V + – (a) K p is greater than K c Ag ® Ag + e ; E° = – 0.0 × 10 –4 g (b) 28. In a saturated solution of the sparingly soluble strong electrolyte equilibrium.0 × 10 (d) 5. For the reaction. When K p and K c are 18. respectively are in the ratio of 1 : 9.83 (d) –16. Hydrogen ion concentration in mol/L in a solution of pH = 5. The pK a of a weak acid (HA) is 4.11 (2005) in 100 mL of its saturated solution? 22. (i) CO (g) + H 2O CO 2(g) + H 2(g) ; K 1 PCl 5(g) PCl 3(g) + Cl 2(g) (g) CO (g) + 3H 2(g) ; K 2 (ii) CH 4(g) + H 2 O (g) If total pressure at equilibrium of the reaction mixture is P and (iii) CH 4(g) + 2H 2 O (g) CO 2(g) + 4H 2(g) ; K 3 degree of dissociation of PCl 5 is x.3 × 10 –2 g (a) CH 3 COOH (b) (CH 3 ) 2 CH – COOH (c) 2.0 (b) 4.K)).98 × 10 –6 (2005) 1. 37. The value of K c (a) s = (K sp /128) 1/4 (b) s = (128K sp ) 1/4 for the reaction is (c) s = (256K sp ) 1/5 (d) s = (K sp /256) 1/5 (2004) (a) 3. the K p /K c is equal (a) lowering of temperature as well as pressure to (b) increasing temperature as well as pressure (a) 1/RT (b) RT (c) lowering the temperature and increasing the pressure (c) (d) 1. (b) pH + pOH = 14 for all aqueous solutions. in CuSO 4 solution deposits 1 gram equivalent of copper at –12 2+ the cathode. The corresponding solubility product is K sp . 29. (b) 3 × 10 –1 mol L –1 N 2(g) + O 2(g) 2NO (g) (c) 3 × 10 –3 mol L –1 –4 at temperature T is 4 × 10 . The equilibrium constant for the reaction. s is given in the concentrations of N 2 O 4 and NO 2 at equilibrium are terms of K sp by the relation 4. Which one of the following statements is not true? – 2– (a) The conjugate base of H 2PO 4 is HPO 4 . (2005) 38.0 × 10 –4 M and the corresponding equilibrium constant K c is (c) 1. In which of the following reactions.500 coulombs of electricity when passed through a (d) SO 2 Cl 2(g) 2NH 3(g) SO 2(g) + Cl 2(g) 42. K p /K c is (a) RT (b) (RT) –1 –1/2 (c) (RT) (d) (RT) 1/2 (2002) (2002) .6 × 10 –4 M (d) 4. increase in the volume at [P4O10 ] [P4O10 ] constant temperature does not affect the number of moles at K = K = c (a) (b) c 5[P ][O ] [P4 ][O 2 ] 5 4 2 equilibrium? 1 K = (a) 2NH 3 ® N 2 + 3H 2 5 (2004) (c) K c = [O 2] (d) c [O ] 5 2 (b) C (g) + (1/2) O 2(g) ® CO (g) The conjugate base of H 2 PO 4 – is (c) H 2(g) + O 2(g) ® H 2O 2(g) (a) PO 4 3– (b) P 2 O 5 (d) None of these. What is the equilibrium expression for the reaction (2003) P 4 (s) + 5O 2 (g) P 4 O 10 (s) ? 40. (d) 96.0 × 10 –6 M (b) 1. In general 1. Consider the reaction equilibrium: (a) (b) 50 2SO 2(g) + O 2(g) 2SO 3(g) ; DH ° = –198 kJ. The solubility product of a salt having general formula MX 2 . (2003) water is 4 × 10 . (c) uninfluenced by occurrence of thunderstorm (d) which depends on the amount of dust in air. 33. The solubility in water of a sparingly soluble salt AB 2 is respectively. CO (g) + Cl 2(g) COCl 2(g) . The correct relationship between free energy change in a reaction (a) 2. 34. (2002) 2– (c) H 3 PO 4 (d) HPO 4 (2004) 41.5 × 10 2 31. Consider an endothermic reaction X ® Y with the activation (c) DG° = RT ln K c (d) –DG° = RT ln K c (2003) energies E b and E f for the backward and forward reactions. For the reaction CO (g) + (1/2) O 2(g) = CO 2(g) .0 × 10 –10 M (2005) (a) DG = RT ln K c (b) –DG = RT ln K c 28. For the reaction equilibrium.30 JEE MAIN CHAPTERWISE EXPLORER 27.8 × 10 –2 and 1.3 × 10 2 mol L –1 30.0 (2004) RT (d) any value of temperature and pressure.02 (2004) for the forward reaction is For the reaction. 32. The concentration of M ions in the aqueous solution of the salt is 36. (c) N 2(g) + 3H 2(g) (2003) 35. Change in volume of the system does not alter the number of When rain is accompanied by a thunderstorm. On the basis of Le Chatelier's principle. the condition favourable (c) 4 × 10 –4 (d) 0. The value of K c for the reaction : (d) 3 × 10 3 mol L –1 (2003) NO (g) 1 1 N + O at the same temperature is 2 2(g) 2 2(g) 39.0 × 10 –5 mol L –1 . the collected rain moles in which of the following equilibria? water will have a pH value (a) N 2(g) + O 2(g) 2NO (g) (a) slightly lower than that of rain water without thunderstorm (b) PCl 5(g) PCl 3(g) + Cl 2(g) (b) slightly higher than that when the thunderstorm is not there 2. (c) The pH of 1 × 10 –8 M HCl is 8.2 × 10 –2 mol L –1 respectively. The molar solubility (in mol L –1 ) of a sparingly soluble salt MX 4 N 2 O 4(g) 2NO 2(g) is s. Its solubility product will be (a) E b < E f (b) E b > E f (a) 4 × 10 –15 (b) 4 × 10 –10 (c) E b = E f –15 (c) 1 × 10 (d) 1 × 10 –10 (2003) (d) there is no definite relation between E b and E f . (d) 41. 8. (a) 39. (b) 23. (a) 21. (d) 17. 1 M NaCl and 1 M HCl are present in an aqueous solution. (a) 26. (2002) (2002) Answer Key 1. (d) 18. (a) 29. (d) 34. (a) 35. (d) 37. (b) (a) 4. (d) 24. (c) (d) 3. (d) 27. (d) 33.31 Equilibrium 43. (d) 6. (a) (c) 2. (c) 44. (d) 14. (d) 22. (d) 30. (a) 20. Let the solubility of an aqueous solution of Mg(OH) 2 be x then its K sp is (a) 4x 3 (b) 108x 5 (c) 27x 4 (d) 9x (2002) 44. (a) 10. (d) 19. (c) 11. 7. (b) 28. Species acting as both Bronsted acid and base is –1 (a) (HSO 4) (b) Na 2CO 3 (c) NH 3 (d) OH –1 45. The solution is (a) not a buffer solution with pH < 7 (b) not a buffer solution with pH > 7 (c) a buffer solution with pH < 7 (d) a buffer solution with pH > 7. (a) 32. (b) 15. (c) 16. (c) 36. (a) 40. 9. (c) . (b) 5. (c) 43. (b) 12. (a) 42. (d) 13. (c) 45. (b) 31. (d) 25. (a) 38. (ii) 1 ´ 10 -11 = 8.1 × 10 –9 \ 5. (i) . KBr + AgNO 3 AgBr + KNO 3 Given. nKBr VSolution (L) Þ w KBr = 1 × 10 –11 × 120 = 120 × 10 –11 NO (g) 4. so H CO ionises more than HCO – and hence. . As sp 3 carbon is less electronegative than sp 3 nitrogen.5 atm 0. – HSO 4 ® 1.. [H + ] = K a C contribution of H + is mostly due to ionisation of carbonic acid. = 1 ´ 10 -11 M [K + ] = [Br – ] = [KBr] Molarity = By reversing the equation (ii).5 - 0.1 × 10 –5 M Thus. JEE MAIN CHAPTERWISE EXPLORER (a) : Initial concentration of aq.05 M – Þ [Ag + ] = [NO 3 ] = 0. (K sp ) AgBr = 5. (c) : N 2(g) + O 2(g) By multiplying the equation (i) by 1 1 N + O 2 2(g) 2 2(g) .]2 = Þ – 10 -11 10 -3 = 10 -8 [OH ] = 10 –4 9. K c = 4 × 10 –4 2.3) 2 (c) : (K sp ) Mg(OH) 2 = [Mg 2+ ][OH – ] 2 – 1 × 10 –11 = [0. The pK a value of other species are given below : – HCO –3 ® 10. HCl solution with pH 1 = 10 –1 M Final concentration of this solution after dilution = 10 –2 M MV = M 1 (V 1 + V 2 ) 10 –1 × 1 = 10 –2 (1 + V 2) 7. Its acidic strength is greater than any given species.5 – P + 2P = 0. K >> K . H 2 PO 4 acts as a proton donor and thus.05 × [Br – ] = 5 × 10 –13 .25 (b) : In equation (ii).1 × 10 –9 = [Ba 2+ ] × [10 –4 ] Þ [Ba 2+ ] = 5. (b) : K sp for BaCO 3 = [Ba 2+ ][CO 2– 3 ] –4 given. – thus the concentrations of H + and HCO 3 are approximately equal. [CO 2– 3 ] = 1 × 10 M (from Na 2 CO 3 ) K sp = 5. 0.P ) (0.. wt.1 = 1 + V2 0. – H + + HCO 3 ; K 1 = 4. 2 2 3 3 (a) : The order of acidity can be explained on the basis of the acidity of the acids of the given conjugate base. RCOO – is the weakest base.8 P = 0.5 . + H 3O ® –1. 5 ´ 10 -2 \ [KBr] = 1 × 10 –11 M K c¢ = K c = 4 ´ 10 -4 = 2 ´ 10 -2 1 1 N + O 2 2(g) 2 2(g) 1 1 K c¢¢ = = = 50.. . Due to sp hybridised carbon.001][OH ] 2 Þ [OH . BaCO 3 precipitate will begin to form.1 \ 1 or 10 -6 = K a ´ 0. of KBr = 120) 1 w KBr = 1. (d) : HSO 3F is the super acid.0 K c¢ 2 ´ 10 -2 5 ´ 10 -13 Þ [Br .5 – P 2P Total pressure = 0. Since RCOOH is the strongest acid amongst all. 10 -3 = K a ´ 0.1 K a = 10 –5 (a) : CO 2 (g) + C (s) 2CO (g) 0.2 × 10 –9 g H + = 10 –pH = 10 –3 5. acetylene is also acidic and hence a weak base but stronger than RCOO – .05 M – [Ag + ][Br ] = (K sp ) AgBr Þ 0. when [Ba 2+ ] = 5. Stronger is the acid. we get 3. 10.01 10 = 1 + V 2 Þ V 2 = 9 L 2NO (g) . w KBr / 120 (Mol. pH = 14 – 4 = 10 6. R – is more basic than NH 2– .8 × 10 –11 1 .6) 2 K P = CO = = = 1..2 × 10 –7 (d) : H 2 CO 3 2– – (b) : pH = 3 Molarity = 0.8 atm PCO (0. weaker is the conjugate base.] = 1 2 NO (g) (c) : Given. [AgNO 3 ] = 0. .1 M HCO 3 H + + CO 3 ; K 2 = 4.92 pOH = 4 Thus. 11.3 P 2 (2 P )2 (0. . .74 acts as an acid.0 × 10 –13 The required equation is.32 1.1 × 10 –5 M. 5 + x x Total pressure = 0..8 and pK b = 4. Hence.5 0 At equi...5 = 9..33 Equilibrium Lesser the pK a value.0 ´ 10 –4 ´ 283 ´ 100 g/100mL 1000 = 28..952 = log K sp n 1 0.(ii) 1 1 .e.059 E °cell = log K i.78 \ pH = 7 + 1/2 (pK a – pK b ) = 7 + 1/2 (4. (ii) K 1 = CH 4(g) + 2H 2 O (g) ‡ˆˆ ˆˆ† CO 2(g) + 4H 2(g) K3 = [CO 2 ][H 2 ] 4 .5 + 2x = 0. (ii) and (iii) ; K 3 = K 1 × K 2 14. (d) : For acidic buffer.a 2 K P 1 Given is 1 = .a 2 = 1 Þ 4 P1 = 1 Þ P 1 = 1 P2 9 P2 36 a 2 P2 9 1 .. -0.78) = 7. (i) [CO][H 2 O] CH 4(g) + H 2 O (g) ‡ˆˆ ˆˆ† CO (g) + 3H 2(g) K 2 = [CO][H 2 ] [CH 4 ][H 2 O] . . pH = pK a + log 13..01 1.(iii) K P 9 [H ] [A ] = K1× K 2 = 1×10 –5 × 5 ×10 –10 = 5 ×10 –15 [H 2 A ] = 5 × 10 –15 Ag + + IO 3 – [S = Solubility] S S = S 2 K sp or.800 V AgI (s) ® Ag + + I – .. (i) 1 1 . E° = – 0...a ö ç 1 + a P2 ÷ è ø 4 a 2 P 1 Þ K P = .0 × 10 –4 × 283 g/L = 1.17) = 0.9 ´ 4.059 0.5 \ pH = 4.952 V 0.5 16.84 \ x = 0.9 ´ 10 -2 For 2SO 2(g) + O 2(g) ƒ 2SO 3(g) [SO 3 ] 2 2 [SO 2 ] [O 2 ] 2 3 [HA – ] 2– [SO 2 ][O 2 ] 1/ 2 = K c = 4.a 2 4 a 2 P 1 Þ K P = . (d) : NH 4 HS ( s ) ƒ NH 3 ( g ) + H 2S ( g ) Initial pressure 0 0.9 ´ 10 -4 10000 = = 416. (b) : Higher the pK a value. 0 0.a 2 [H + ][A2– ] 18.17)(0. S = 1... strongest acid has lowest pK a value. E° = – 0.5 \ pOH = 14 – 4.0 × 10 –4 mol/L = 1. (c) : AgIO H + + A 2– ; K 2 = 5 × 10 –10 = H A – From equations (i).3 × 10 –4 g/100 mL = 2. (i) SO 3(g) + 1/2 O 2(g) ƒ SO 3(g) [SO3 ] [SO 2 ][O 2 ]1/ 2 = K c ¢ = 1 .0 × 10 –8 or.49 24.... (b) : [H + ][H A – ] = 1 × 10 –5 [H 2 A ] + 2 [CH 4 ][H 2 O] 2 H + + H A – ; 17..83 × 10 –3 g/100 mL = [A – ] [HA ] When the acid is 50% ionised.01 (a) : Given PCl 5 (g) ƒ PCl 3 (g) + Cl 2 (g) t = 0 1 0 0 t eq 1 – x x x Total number of moles = 1 – x + x + x = 1 + x æ x ö ÷ P Thus partial pressure of PCl 3 = çè 1 + x ø 21. (ii) 4. 1 At equilibrium 1 – a 0 2a 1 1–a 0 0 a a 2 KP æ 2 a ö ç 1 + a P 1 ÷ è ø = = PX æ 1 .9 ´ 10 -2 [SO 3 ] Substituting values of from equation (i) and (ii) into (iii). Hence sequence of acidic strength will be + > HSO – > HCO – HSO 3F > H 3O 4 3 12. we get 20.. (d) : Given that pK a = 4. (d) : H 2 A 3 = K c ¢ 2 = 1 4.0 ´ 10 – 4 ´ 283 g/L 1000 . weaker is the acid. (iv) ˆˆ† 2 Y ; Z ‡ˆˆ ˆˆ† P + Q X ‡ˆˆ Initial mol.135 or.11 atm 2 22.059 19. S 2 = 1..80 – 4. (d) : AgI (s) + e – ® Ag (s) + I – . higher will be its acidic strength. E° = – 0.952 = - 16. 4 a 2 P1 1 . [A – ] = [HA] or pH = pK a + log1 or pH = pK a Given pK a = 4. log K sp = 0.5 + 0..a ö ç 1 + a P1 ÷ è ø P Y 2 1 K= K P = 2 P Z æ a öæ a ö ç 1 + a P2 ÷ ç 1 + a P 2 ÷ øè ø = è K P = 2 PZ æ 1 .152 V Ag (s) ® Ag + + e – . (d) : CO (g) + H 2 O (g) ‡ˆˆ ˆˆ† CO 2(g) + H 2(g) [CO2 ][H 2 ] K1 = . (a) : SO 3(g) ƒ SO 2(g) + 1/2 O 2(g) PP P Q 15.17 atm K p = p NH 3 × p H 2 S = (0. (ii) Addition of reactants or removal of product will farour 37. (c) : The conversion of SO 2 to SO 3 is an exothermic reaction.3 × 10 –2 = 3 × 10 –3 mol L –1 39.2 ´ 10-2 mol L-1 K c = [NO 2 ] 2 1. K sp = (x) × (2x) 2 = x × 4x 2 = 4x 3 44. (c) : [N 2O4 ] = 4. As temperature increases. 1 3 1 42. (d) : M X 4 (solid) 1/ 5 \ s = 5 K sp æ K sp ö = 256 çè 256 ÷ø 30.303 RT logK c Favourable conditions: At equilibrium.2 ´ 10 -2 = [N 2 O 4 ] 4.= è 2ø 2 2 K p . – 2– + H 2PO 4 ® HPO 4 + H 24. (a) : Mg(OH) 2 ® [Mg 2+ ] + 2[OH – ] x 2x K sp = [Mg 2+ ] [OH – ] 2 or. [P4 ( s ) ] [O 2 ( g ) ] 5 – + (HSO 4) + H ® H 2SO 4 We know that concentration of a solid component is always 2– – + (HSO 4) – H ® SO 4 taken as unity. Here we should also consider [H + ] that comes from H 2 O. (a) : HCl is a strong acid and its salt do not form bufter solution. K c = 1 [O 2 ]5 45. (a) : Due to thunderstorm.æ1 + ö = 1 . increase in volume at constant temperature does not affect the number of moles at equilibrium.e. (c) : K p = K c (RT)Dn ; Dn = 1 . hence pH decreases. (a) : For endothermic reaction.0831 × 457) 1 \ K p > K c 35. As the resultant solution is acidic. S = 1. s = 1 × 10 –4 M Þ [M 2 + ] = 1 × 10 –4 M 28. . DH = +ve DH = E f – E b . where Dn = 0.34 JEE MAIN CHAPTERWISE EXPLORER 23. = 1. K c = (HSO 4 ) – can accept and donate a proton. OH – ƒ O 2– + H + 33. (a) : According to BronstedLowry concept.8 ´ 10 . (d) : Conjugate base of OH – is O 2– . hence decrease the temperature will favour the forward reaction. (d) : Cl 2(g) + 3F 2(g) ƒ 2ClF 3(g) ; DH = –329 kJ 36. (a) : CO (g) + Cl 2(g) Dn = 1 – 2 = –1 COCl 2(g) K p 1 -1 K p = K c (RT)Dn . hence decrease in DG° = –2. [H + ] also increases. –ve) hence decrease in volume K sp = [A 2+ ] [B – ] 2 = 1. (d) : pH = – log[H + ] [H + ] = antilog (–pH) = antilog(–5. it means E b < E f M 4+ (aq) + 4X – (aq) s 4s Solubility product. 40. (d) : P 4 (s) + 5O 2 (g) P 4O 10 (s) and a Bronsted base is a substance which can accept a proton [P4O10 ( s ) ] from any other substance. (d) : Conjugate base is formed by the removal of H + from acid. s 3 = 1 × 10 –12 or.98 × 10 –6 34. (d) : DG = DG° + 2. (b) : N 2(g) + O 2(g) 2NO (g) 2 [NO] K c = = 4 ´ 10 -4 [N 2 ] [O 2 ] NO (g) 1 1 N + O 2 2(g) 2 2(g) [N 2 ]1/ 2 [O 2 ] 1/2 1 = 1 = [NO] Kc 4 ´ 10 -4 1 100 = = = 50 2 ´ 10 -2 2 K c ¢ = 31. K sp = s × (4s) 4 = 256 s 5. (a) : In this reaction the ratio of number of moles of reactants to products is same i. hence pH is less than 7.4) =3. There is also a decrease in volume or moles in product side.0 × 10 –5 ) 2 or increase in pressure will favour the forward reaction. 2 : 2. temperature increases.0 × 10 –15 27.0 × 10 –5 mol L –1 (iii) Here Dn = 2 – 4 = –2 (i. hence change in volume will not alter the number of moles. 41.e. Now [H + ] = [H + ] from HCl + [H + ] from H 2 O = 10 –8 + 10 –7 = 10 –8 + 10 × 10 –8 = 11 × 10 –8 \ pH = –log(11 × 10 –8 ) = 6.1/ 2 \ K = ( RT ) c 43.9587 26. (LeChatelier's principle). (a) : K p = K c (RT)Dn Dn = 3 – 2 = 1 K p = K c (0. (c) : AB 2 A 2+ + 2B – the forward reaction. (b) : MX 2(s) ƒ M 2 + (aq) + 2X – (aq) s 2s K sp = s ∙ (2s) 2 = 4s 3 4 × 10 –12 = 4s 3 or. \ K = ( RT ) = RT c 38..8 ´ 10-2 mol L-1 [NO 2 ] = 1. DG = 0 (i) As the reaction is exothermic. Thus the reaction is favoured by low temperature and high pressure. (d) : For those reactions. 25.303 RT logK c temperature will favour the forward reaction.0 × 10 –5 × (1. (c) : pH of an acid cannot exceed 7. 29. a Bronsted acid is a substance which can donate a proton to any other substance 32.2 ´ 10 -2 ´ 1.2 = 0. 0 M (b) p(H 2) = 1 atm and [H 25°C are given below: + ] = 1.770 V (d) – 0.72 V.65 × 10 4 (c) 37.76.3 The reduction potential of hydrogen halfcell will be negative (2007) if 10. The standard reduction potentials for Zn 2+ /Zn. respectively (a) 5. E° Fe 2+ /Fe = –0.10 V) was allowed to be completely discharged at 298 K.3 (b) antilog(24. of this solution is 1.5 V (2007) (c) 3. Ni 2+ /Ni.24 × 10 –4 S m 2 mol –1 The titration gives unsatisfactory result when carried out in (d) 12.0 S cm 2 /equiv.2 M of the same solution is 520 W.26 V (b) 0. 5. D r G = + 966 kJ mol –1 .29 S m –1 . Resistance of the same cell when Fe 3+ (aq) + e – ® Fe 2+ (aq) will be filled with 0.0 V (d) 2.42 V The potential for the cell Cr | Cr 3+ (0.51 V E° Cr 2 O 7 2– /Cr 3+ = 1. 2 and 16 (c) 2. 4. The Gibb’s energy for the decomposition of Al 2 O 3 at 500°C What additional information/quantity one needs to calculate L° is as follows : of an aqueous solution of acetic acid? 4/3Al + O 2 .036 V. The reaction X + Y 2+ ® X 2+ + Y will be spontaneous when (a) X = Ni.1 M is 100 W. 5 and 16 (2013) 2. The molar (a) – 0.01 M) | Fe is (a) –0. The conductivity of standard electrode potential for the change.74 V; E° MnO 4 – /Mn 2+ = 1.0 M dilution in H 2 O (where ions move freely through a solution) at + ] = 1. 2 and 8 (b) 5.35 Redox Reactions and Electrochemistry CHAPTER 8 1.385 V conductivity of 0.44 V respectively.072 V (b) 0. 2/3Al 2 O 3 (a) L° of chloroacetic acid (ClCH 2 COOH) The potential difference needed for electrolytic reduction of (b) L° of NaCl (c) L° of CH 3 COOK Al 2 O 3 at 500°C is at least (d) The limiting equivalent conductance of H + (l° H +). The equivalent conductances of two strong electrolytes at infinite (a) p(H 2 ) = 1 atm and [H + ] = 2. Resistance of a conductivity cell filled with a solution of an electrolyte of concentration 0. y and z in the reaction are. strongest oxidising agent will be (a) MnO 4 – (b) Cl – 9.08) (d) 10 37. Y = Fe (2012) oxidises oxalic acid to carbon dioxide and water gets oxidised by oxalic acid to chlorine furnishes H + ions in addition to those from oxalic acid reduces permanganate to Mn 2+ . and Fe 2+ /Fe are – 0.2 Scm 2 /equiv. Zn | Zn 2+ (1 M) || Cu 2+ (1 M) | Cu (E° cell = 1. + ] = 2. 3+ (c) Cr (d) Mn 2+ (2013) 3. Y = Zn (b) X = Fe.36 V Based on the data given above.4 × 10 –4 S m 2 mol –1 (2006) the presence of HCl. E° Fe 2+ /Fe = – 0. (2008) Given E° Cr 3+ /Cr = –0. The relative æ [Zn 2 + ] ö concentration of Zn 2+ to Cu 2+ ç 2+ ÷ is è [Cu ] ø (a) 9.33 V; E° Cl/Cl – = 1.0 M (d) p(H 2) = 2 atm and [H (2011) L° HCl = 426.0 V (b) 4. (a) 5. The value 11. Y = Zn (c) X = Zn. Given E° Cr 3+ /Cr = – 0.0 M (c) p(H 2) = 2 atm and [H L° CH 3 COONa = 91. 7.5 V (2010) Given : E° Fe 3+ /Fe = – 0. 8.26 V (c) 0.439 V. – 0.339 V (d) –0. Y = Ni (d) X = Ni.270 V (2009) (a) 124 × 10 –4 S m 2 mol –1 Amount of oxalic acid present in a solution can be determined (b) 1240 × 10 –4 S m 2 mol –1 by its titration with KMnO 4 solution in the presence of H 2 SO 4 .23 and – 0. (c) 1. 5 and 8 (d) 2. (a) (b) (c) (d) Consider the following reaction.02 M solution of the electrolyte will be (c) 0. because HCl . REDOX REACTIONS AND ELECTROCHEMISTRY xMnO 4 – + yC 2 O 4 2– + zH + ® xMn 2 + z + 2 yCO2 + H 2 O 2 The values of x.339 V (2008) The cell.1 M) | | Fe 2+ (0. 6. 12 kg of aluminium metal by this method would (C) complexing.f.2 V respectively. (c) 10 (d) 1 × 10 10 (2003) 0.0 × 10 5 10 (b) Ca(OH) 2 + H 2SO ® CaSO + 2H O (c) 1. For the redox reaction: the change in oxidation state from +2 to +3 is easiest? Zn (s) + Cu 2+ (0.36 JEE MAIN CHAPTERWISE EXPLORER 12. A (d) A. (d) 2PCl 5 + H 2SO 4 ® 2POCl 3 + 2HCl + SO 2 Cl 2 .0 × 10 1 (b) 1. B (b) B.82 V (2003) addition of H 2SO 4 to cathode compartment.m. For which one of these metals 26. Aluminium oxide may be electrolysed at 1000°C to furnish (d) remove adsorbed oxygen from electrode surface. The equilibrium constant of the reaction at 25°C will be (2005) (a) 1 × 10 –10 (b) 29. (c) 21.1 M chloroacetic acid (c) 0.77 and +1. Fe and Co are – 0.2 (b) 552.1 M fluoroacetic acid (d) 0. Consider the following E° values.0 126.0 × 10 (d) 1. When during electrolysis of a solution of AgNO 3 .80 V Zn (s) + 2H (aq) ® Zn (aq) + H 2(g) (c) 1.63 V (2004) molar conductances of the electrolytes listed above at infinite 22.08 g (b) 10. the set of properties shown by CN – ion towards require metal species is 7 (a) A. The L° for NaBr is calculate L° HOAc .10 volt.m. R = 8.5 (b) 1.0 V and –1.83 × 10 7 C of electricity (c) 5. E cell for the cell will be (c) Fe (d) Co (2004) 2.41. Mass = 27 amu; 1 Faraday = 96.314 JK –1 mol –1 ) (a) 2HI + H 2 SO 4 ® I 2 + SO 2 + 2H 2 O (a) 1. the mass of silver deposited on the cathode will be (c) increase the E and shift the equilibrium to the right (a) 1.57. The E ° M 3 + / M 2 + values for Cr.14 V (b) 1.2 S cm 2 /mol respectively. Standard reduction electrode potentials of three metals A. Among the properties (A) reducing (B) oxidising To prepare 5.2 91. E° cell is 1. of a cell. Which of the following chemical reactions depicts the oxidising found to be 0. In a cell that utilizes the reaction.0591 F 18.0 426.f.40 V (a) 1. +1. E° 3+ 2+ = + 0. The equilibrium constant of the behaviour of H 2 SO 4 ? reaction is (F = 96. For a cell reaction involving a twoelectron change. The limiting molar conductivities L° for NaCl. –3. to (a) 517. the standard e.1 M) ® Zn + (1M) + Cu (s) (a) Cr (b) Mn taking place in a cell.5 × 10 –2 17.7 (d) 217. The molar conductivities L° NaOAc and L° HCl at infinite dilution 19. B and (d) 5. aluminium metal (At. combustion of hydrogen occurs dilution in H 2 O at 25°C. C (a) 5.9 145. 25. Electrolyte KCl KNO 3 HCl NaOAc NaCl Sn (s) + 2Fe 3+ (aq) ® 2Fe 2+ (aq) + Sn 2+ (aq) is (S cm 2 mol –1 ) 149.5 (2005) (b) create potential difference between the two electrodes (c) produce high purity water 15.91 V (d) 0.5 V. the additional value required is (a) 128 S cm 2 mol –1 (b) 176 S cm 2 mol –1 2 –1 (a) L° H 2 O (b) L° KCl (c) 278 S cm mol (d) 302 S cm 2 mol –1 (2004) (c) L° NaOH (d) L° NaCl.97 V respectively. + 2+ (a) 2. To 126.49 × 10 10 C of electricity (2005) C are +0.591 V at 25°C.500 C mol –1 . (2004) (b) 1. The reducing power of these metals are 16. (2006) 20. The cathode reaction is 3+ 0 Al + 3e ® Al 23. will (a) lower the E and shift equilibrium to the left 27.6 g (d) 108 g (2003) (2004) ( ) . involving one electron change is 13. C.1 M acetic acid (b) 0.303 RT = 0.49 × 10 C of electricity (c) C. Mn.500 (2004) Coulombs). B.14 V Fe /Fe Sn /Sn (2006) Under standard conditions the potential for the reaction 14.7 (a) generate heat (c) 390. 9650 coulombs (b) lower the E and shift the equilibrium to the right of charge pass through the electroplating bath.295 V at 25°C.0 and 426. of the cell is found to be 0. KBr and KCl are in water at 25°C are 91.8 g (d) increase the E and shift the equilibrium to the left.0 × 10 30 (2004) 4 4 2 (c) NaCl + H 2SO 4 ® NaHSO 4 + HCl 21.68 V Calculate molar conductance of acetic acid using appropriate (c) 0. The standard e.49 × 10 4 C of electricity 24.07 V (d) 0. 152 and 150 S cm 2 mol –1 respectively. The highest electrical conductivity of the following aqueous (a) B > C > A (b) A > B > C solutions is of (c) C > B > A (d) A > C > B (2003) (a) 0.77 V ; E° 2+ = - 0.1 M difluoroacetic acid. In a hydrogenoxygen fuel cell. 3. (2002) 29. 5. What will be the emf for the given cell. 7 (c) 1. (c) F P 1 33. When KMnO 4 acts as an oxidising agent and ultimately forms [MnO 4 ] –1 .37 Redox Reactions and Electrochemistry 28. 7. 5 (d) 3. 32. 26. 27. 4. MnO 2 . 21. (c) (d) (a) (c) (b) (d) 4. 31. 24. 3. EMF of a cell in terms of reduction potential of its left and right electrodes is (a) E = E left – E right (b) E = E left + E right (c) E = E right – E left (d) E = –(E right + E left ) (2002) Answer Key 1. (d) (d) (a) (a) (d) (a) 2. 16. 22. 12. If f denotes reduction potential. 3. (2002) 30. 28. 11. 23. 13. 14. 33. 20. The heat required to raise the temperature of body by 1ºC is 32. Which of the following reaction is possible at anode? (a) 2 Cr 3+ + 7H 2 O ® Cr 2 O 7 2– + 14H + (b) F 2 ® 2F – (c) (1/2) O 2 + 2H + ® H 2 O (d) None of these. 5 (b) 1. Pt | H 2 (P 1) | H + (aq) | | H 2 (P 2) | Pt RT P 1 P (b) 2 F log P (a) RT log 1 2 F P 2 RT log P 2 (d) none of these. 15. 30. 25. 1 (2002) 35. Mn 2 O 3 . 8. 19. 17. (d) (a) (a) (c) (a) (c) 6. then which is true? (a) E° cell = f right – f left (b) E° cell = f left + f right (c) E° cell = f left – f right (d) E° cell = –(f left + f right ) (2002) (a) 4. 5. Which of the following is a redox reaction? (a) NaCl + KNO 3 ® NaNO 3 + KCl (b) CaC 2 O 4 + 2HCl ® CaCl 2 + H 2 C 2 O 4 (c) Mg(OH) 2 + 2NH 4 Cl ® MgCl 2 + 2NH 4 OH (d) Zn + 2AgCN ® 2Ag + Zn(CN) 2 34. Conductivity (unit Siemen's S) is directly proportional to area of the vessel and the concentration of the solution in it and is called inversely proportional to the length of the vessel then the unit (a) specific heat of the constant of proportionality is (b) thermal capacity (a) S m mol –1 (b) S m 2 mol –1 (c) water equivalent –2 2 (c) S m mol (d) S 2 m 2 mol –2 (2002) (d) none of these. 9. 35. (a) (b) (c) (c) (c) (b) (2002) 3. 10. 18. (c) (d) (c) (a) (b) . 7. 34. Mn 2+ then the number of electrons transferred in each case respectively is (2002) 31. (c) (b) (d) (b) (b) (c) 5. 29. 1. 059 [Zn 2+ ] log 2 [Cu 2+ ] When the cell is completely discharged.1 Zn 2+ or.1 – or log 0. is the oxidising agent.2.3 - 0. (d) : The ionic reactions are : – 4/3Al 2/3Al 23+ + 4e 2– 2/3O 3 O 2 + 4e – Thus. (b) : Cr ® Cr 3+ + 3e – E° red = –0. 3.3 Cu 2+ 10. 8. E red will only be negative when p H2 > [H + ]. CH 3 COOH is or Λ°CH3COOH = Λ°CH3 COONa + Λ°HCl - Λ °NaCl So.23 – (– 0. E cell = 0 ° – Ecell = E cell 0 = 1. Y = Ni [Cr3+ ] 2 Zn + Ni 2+ ® Zn 2+ + Ni Ecell = E °cell - 0.3 0.059 log10 10 4 6 p H 2 0.059 log 10 2 + 3 ncell [Fe ] Alternatively.76) 6 (0. of electrons transferred = 4 = n DG = –nFE = – 4 × 96500 × E or 966 × 10 3 = – 4 × 96500 × E Þ 6. value of Λ ° NaCl should also be known.. 0.1) 2 Ecell = 0. (b) : According to Kohlrausch’s law. 4.059 [Zn 2+ ] log 2 [Cu 2+ ] [Zn 2+ ] 2+ [Cu ] = 2 × 1. According to Nernst equation. So option (c) is correct. JEE MAIN CHAPTERWISE EXPLORER + 2+ Titration cannot be done in the presence of HCl because KMnO 4 (d) : 2MnO 4– + 5C 2 O 42– + 16H ® 2Mn + 10CO 2 + 8H 2 O being a strong oxidizing agent oxidises HCl to Cl 2 and get \ x = 2.72) Thus X should have high negative value of standard potential E° cell = 0. COOH 2KMnO4 + 3H2SO4 + 5 COOH K2SO4 + 2MnSO4 + 10CO2 + 5H2 O 9.42 V 2+ 3+ reduction potential are good reducing agents and can be easily 2Cr + 3Fe ® 2Cr + 3Fe oxidised.01) 3 Þ E° = + 0. for a spontaneous reaction E° must be positive.039 Ered = Ered ° - log n [H + ]2 \ E c ell = 0.38 1. 7.e. no. Zn 2+ + Cu 2+ (d) : Zn + Cu 0.3 - 0. \ MnO 4 – is the strongest oxidising agent. X = Zn. the molar conductivity at infinite dilution (L°) for weak electrolyte.5 V 4 ´ 96500 (c) : Given.036) – 2 × (–0.0591 E c ell = 0. E° = E° r educed species – E° oxidised species (0. (a) : k = 1 æ l ö 1 æ l ö ç ÷ i.059 log 10 = – 0.02 M lm = k ´ 1000 1 ´ 129 1000 = ´ ´ 10-6 m 3 molarity 520 0.036 V Fe 2+ + 2e – Fe ; E° 2 = – 0.439) E° 3 = – 0. log = 37.261 V Ered = 0 - 5.77 V (d) : Oxalic acid present in a solution can be determined by its titration with KMnO 4 solution in the presence of H 2 SO 4 . E° cell = E° c athode – E° a node = –0. Fe 3+ + 3e – Fe ; E° 1 = – 0. 1.3 than Y so that it will be oxidised to X 2+ by reducing Y 2+ to Y.29 = ç ÷ R è a ø 100 è a ø l/a = 129 m –1 R = 520 W for 0.439 V Required equation is Fe 3+ + e – Fe 2+ ; E° 3 = ? Applying DG° = –nFE° \ DG° 3 = D G° 1 – D G° 2 (–n 3 FE° 3 ) = (–n 1 FE° 1 ) – (–n 2 FE° 2 ) E° 3 = 3E° 1 – 2E° 2 = 3 × (– 0.059 Cu 2+ 2+ Zn = 10 37.3 – 0.108 + 0. 11. So actual amount of oxalic acid in (a) : Greater the reduction potential of a substance.878 = 0. 2.53 V H 2 (g) (c) : 2H + (aq) + 2e – Ecell = 0.02 = 124 × 10 –4 S m 2 mol –1 . stronger solution cannot be determined.72 V 2+ + 2e – ® Fe Fe E° (c) : The elements with high negative value of standard red = –0. C = 0. z = 16 itself reduced to Mn 2+ . y = 5. for calculating the value of Λ ° CH3 COOH .0591 2 log 2 (1)2 E=- 966 ´ 10 3 = .2 M.42 – (–0. thus this is a reduction reaction and H 2SO 4 behaves as oxidising \ K c = antilog 10 = 1 × 10 10 agent.10 V its oxidation will be easiest. 13..12 ´ 103 ´ 96500 7 E ° +0.295 = log Kc Mn 2+ | Mn 3+ = –1.41 V 0.0591 log10 2 E cell = 1.0591 26. Now E = Z × F [E = equivalent weight . more is the tendency 16. L°CH3COONa + L°HCl = L°CH3 COOH + L° NaCl or. (d) : E °cell = c I st half reaction : 2HI ® I 2 –1 0 n 17. therefore 0.0591 0.e.14 + 0. Furthermore.0591 0 = 0..0705 V . (c) : L¥ AcOH = L¥ AcONa + L¥ HCl – L¥ NaCl Sn/Sn 2+ Fe /Fe = 91. The reducing power follows the following order: Difluoroacetic acid will be strongest acid due to electron B > C > A. 2 H I + H 2 S O 4 ® I 2 + S O 2 + 2 H 2 O +1–1 +1 +6 –2 0 +4 +1 –2 + On adding H 2SO 4 the [H ] will increase therefore E cell will also increase and the equilibrium will shift towards the right.e. 19.0 + 426.10 - 0. (iii) Br – Cl – Equation (i) + (ii) – (iii) L°NaBr = L° + + L° Na Br – = 126 + 152 – 150 = 128 S cm 2 mol –1 0. F = Faraday] the water obtained as a byproduct may be used for drinking by E or W = ´ Q the astronauts.. 0.5 = 0. (a) : A B C 3 ´ 5. (c) : Ecell = E °cell - n log K c In this reaction oxidation number of I increases by one. (c) : E °cell = E ° + E ° 3+ 2+ 14.0295 = 1. (a) : From Faraday's 1st law. (ii) + L° . 0.L° NaCl Thus to calculate the value of L° CH3 COOH one should know the value of L° NaCl along with L° CH3 COONa and L° HCl .0591 log K maximum electrical conductivity.57 V 2 2+ 3+ Fe | Fe = – 0.2 – 126.0591 20. (a) : Cr 2+ | Cr 3+ = +0. 21.. K c = antilog 10 or K c = 1 × 10 10 More is the value of oxidation potential more is the tendency 1 to get oxidised. (c) : Ecell = E °cell - n log As Cr will have maximum oxidation potential value. L°CH3COOH = L°CH3 COONa + L°HCl . (b) : Direct conversion of chemical energy to electric energy 15.77 = 0. Q = quantity of electricity] using the heat to form steam for driving turbines. Z = electrochemical the 40% maximum now obtainable through burning of fuel and equivalent.2 V = = 5.5 V –3. or Q = A 24. (d) : CH 3COONa + HCl ® CH 3COOH + NaCl From the reaction.39 Redox Reactions and Electrochemistry 12. (c) : Zn (s) + 2H + (aq) Zn 2+ (aq) + H 2 (g) 2 + 0. 18.0591 = 10 In this reaction oxidation number of S decreases by two.059 [Zn ] ´ p H 2 Ecell = E °cell - log 2 [H + ] 2 Ecell = 1.49 × 10 C red 27 More is the value of reduction potential. (a) : L°NaCl = L° L°KBr = L° Na + K+ L°KCl = L° K+ + L° Cl – .. thus 0.97 V or.91 V = 390. (i) + L° . (c) : CN – ions act both as reducing agent as well as good n ´ w ´ F complexing agent. can be made considerably more efficient (i.77 V or. i. upto 75%) than W = Z × Q [W = weight.591 log Kc this is an oxidation reaction and HI behaves as a reducing 1 agent. withdrawing effect of two fluorine atoms so as it will show 0.0 V –1.1 Here n = 2. (d) : Higher the acidity. (a) : In the reaction.0591 log K c II nd half reaction : H 2 SO 4 ® SO 2 +6 +4 0. higher will be the tendency to release to get reduced.295 = 0.591 Þ log K c = 0. less is the reducing power.10 – 0. Þ – 0.591 = – 0. E ° cell = 1. F At anode : 2H 2(g) + 4OH – (aq) ® 4H 2 O (l) + 4e – W ´ F W ´ F – – or Q = At cathode : O 2(g) + 2H 2O (l) + 4e ® 4OH (aq) or Q = A E 2H 2(g) + O 2(g) ® 2H 2 O (l) n [A = Atomic weight . protons and hence lighter will be the electrical conductivity.0295 logK c Co 2+ | Co 3+ = – 1.7 S cm 2 mol –1 22. n = valency of ion] 23.. 25. (d) : The oxidation states show a change only in reaction (d). (a) : E cell = E right (cathode) – E left (anode) . –2 e – H 2(P 2) P P P RT RT RT E = E° log 2 = 0 og 2 = log 1 nF P1 nF P1 nF P2 31. (c) : E cell = Reduction potential of cathode (right) – reduction potential of anode (left) = E right – E left .8 g 96500 28. (a) : Here Cr 3+ is oxidised to Cr 2O 7 . . 30. (b) : 2H + + 2e – H 2 (P 1 ) H 2 (P 2 ) 2H + + 2e – Overall reaction : H 2 (P 1 ) S × m × m 3 S × m = 2 mol m ´ mol m 2 × 3 m = S m 2 mol –1 33. (b) : The mass of silver deposited on the cathode = 108 ´ 9650 = 10. 0 +1 0 Zn + 2AgCN – +2 e 34.40 JEE MAIN CHAPTERWISE EXPLORER 27. or S = K or K = AC [K = constant of proportionality] L SL AC \ Unit of K = 2– 29. S µ L +2 2Ag + Zn(CN) 2 –5 e Mn 2+ [MnO 4 ] –1 – –3 e +4 MnO 2 35. (c) : +3 Mn 2 O 3 –4 e – +7 e – [KMnO 4 ] – – 32. (b) : It is also known as heat capacity. (b) : S µ A (A = area) S µ C (C = concentration) 1 S µ ( L = length) L AC Combining we get. (2010) (log 2 = 0. Cl 2 + H 2 S – – Cl + + HS H + + Cl + S (fast) – + B. In this equation.25 h (2010) (a) 1 (b) 0 (c) 3 (d) 2 (2006) The halflife period of a first order chemical reaction is 6. H 2S Û H + HS (fast equilibrium) – – Cl 2 + HS 2Cl + H + + S (slow) (a) A only (b) B only (c) Both A and B (d) Neither A nor B 5. The time required for the completion of 99% 11.6 minutes (2009) For a reaction 1 A ® 2 B rate of disappearance of A is related 2 to the rate of appearance of B by the expression (a) - d [ A] d [ B ] = 4 dt dt (b) . The rate of a reaction doubles when its temperature changes from 300 K to 310 K.301) (a) 60. the rate increased by two times.314 J K –1 mol –1 and log 2 = 0.5 kJ mol –1 (2013) For a first order reaction.5 h (d) 0. equation? – – + + H + Cl + Cl + HS (slow) A. When the initial concentration of NO with Br 2 to form NOBr.025 M in 40 minutes. 2. Consider the reaction : + – Cl 2(aq) + H 2 S (aq) S (s) + 2H (aq) + 2Cl (aq) The rate of reaction for this reaction is rate = k[Cl 2 ][H 2 S] Which of these mechanism is/are consistent with this rate 9.3 minutes (b) 23. (A) ® products.0 mol L –1 . the halflife did not change. Rate of a reaction can be expressed by Arrhenius equation as : of the chemical reaction will be k = Ae –E/RT .25 mol L –1 if it NOBr 2(g) + NO (g) ® 2NOBr (g) is a zero order reaction? If the second step is the rate determining step.d [ A] = d [ B ] dt dt (2008) Consider the reaction. 2A + B ® products. If the temperature is raised by 50°C.301) (a) 230. The rate of reaction when the concentration of A is 0. . (2007) The energies of activation for forward and reverse reactions for A 2 + B 2 ƒ 2AB are 180 kJ mol –1 and 200 kJ mol –1 respectively. the concentration of A changes from 0. E represents .47 × 10 –5 M/min (c) 1. of the reactant A is 2. The presence of a catalyst lowers the activation energy of both (forward and reverse) reactions by 100 kJ mol –1 . how much time does it take NO (g) + Br 2(g) NOBr 2(g) for its concentration to come from 0. 4.d [ A] = 1 d [ B ] dt 2 dt (c) . 6.73 × 10 –4 M/min (d) 1. the order of the (a) 1 h (b) 4 h reaction with respect to NO (g) is (c) 0.93 minutes.73 × 10 –5 M/min (2012) 3.5 kJ mol –1 (b) 53.6 kJ mol –1 7. When the concentration of A alone was doubled. Activation energy of such a reaction will be (R = 8.41 Chemical Kinetics CHAPTER CHEMICAL KINETICS 9 1.03 minutes (c) 46. The unit of rate constant for this reaction is (a) s –1 (b) L mol –1 s –1 (c) no unit (d) mol L –1 s –1 .06 minutes (d) 460. When concentration of B alone was doubled. the rate of the reaction increases by about (a) 10 times (b) 24 times (c) 32 times (d) 64 times (2011) 8. The following mechanism has been proposed for the reaction A Products is 1 hour.d [ A] = 1 d [ B ] dt 4 dt (d) . (c) 48.6 kJ mol –1 (d) 58.1 M to 0.01 M is (a) 3.47 × 10 –4 M/min (b) 3. The rate of a chemical reaction doubles for every 10°C rise of temperature. The enthalpy change of the reaction (A 2 + B 2 ® 2AB) in the presence of a catalyst will be (in kJ mol –1 ) (a) 20 (b) 300 (c) 120 (d) 280 (2007) The time for half life period of a certain reaction 10.50 to 0. (d) 16. (a) s –1 . the order of the reaction is decreases from 0. (2003) For the reaction system: 2NO ( g) + O 2(g) ® 2NO 2(g) . R is Rydberg constant.4 M in 15 minutes. A is adsorption factor. In a first order reaction. (c) 4. (c) 18. (b) 12.42 JEE MAIN CHAPTERWISE EXPLORER (a) the energy above which all the colliding molecules will react (b) the energy below which colliding molecules will not react (c) the total energy of the reacting molecules at a temperature. For the reaction A + 2B ® C. (2006) 19.1 M to 0. volume is suddenly reduced to half its value by increasing the pressure on it. (b) 2.8 M to 0. (a) 6. (d) 13. (a) 21. E a is energy of activation. (a) 10. In the respect of the equation k = Ae –E a /RT in chemical kinetics. (d) 20. the rate of reaction will (a) diminish to onefourth of its initial value (b) diminish to oneeighth of its initial value (c) increase to eight times of its initial value (d) increase to four times of its initial value. (a) 15. (a) 23. (2002) 21. (c) 5. Units of rate constant of first and zero order reactions in terms of molarity M unit are respectively 17. If the reaction is of first order with respect to O 2 and second order with respect to NO 2 . (c) .025 M is (c) 5 (d) 7 (2002) (a) 30 minutes (b) 15 minutes (c) 7. H 2 + I 2 ® 2HI is 15. rate is given by R = [A] [B] 2 then 16. (a) 7. s –1 (d) M. (a) 17. the ratio of the new rate reactant to drop to 3/4 of its initial value. (c) 8. M which one of the following statements is correct? (c) Ms –1 . A reaction was found to be second order with respect to the concentration of carbon monoxide.10/k (b) 0. the concentration of the reactant. If the rate constant for to the earlier rate of the reaction will be as a first order reaction is k. (2006) (a) (b) (c) (d) k is equilibrium constant. t 1/4 can be taken as the time taken for the concentration of a A and halving the concentration of B. (d) 22. (b) 14. the t 1/4 can be written as 1 (a) m + n (b) (m + n) (a) 0. The formation of gas at the surface of tungsten due to adsorption is the reaction of order (a) 0 (b) 1 (c) 2 (d) insufficient data. The rate law for a reaction between the substances A and B is given by rate = k [A] n [B] m . (d) -2 = -2 2 = (2002) dt dt dt (2004) 22. On doubling the concentration of 13. (c) 19. (2005) 20. (a) 3.69/k (d) 0. The time taken (a) 3 (b) 6 for the concentration to change from 0.75/k (2005) 14. (2004) 23. The correct statement in relation to this reaction == - (a) dt dt dt is that the –1 (a) unit of k must be s d [H 2 ] d [I2 ] 1 d [HI] (b) = = (b) t 1/2 is a constant dt dt 2 dt (c) rate of formation of C is twice the rate of disappearance d [H 2 ] 1 d [I2 ] d [HI] (c) 1 = = - of A 2 dt 2 dt dt (d) value of k is independent of the initial concentrations of d [H 2 ] d [I ] d [HI] A and B. The differential rate law for the reaction. with everything else kept the same. The rate equation for the reaction 2A + B ® C is found to be : d [H 2 ] d [I2 ] d [HI] rate = k[A] [B]. (2003) 12. If the concentration of carbon monoxide is doubled. A reaction involving two different reactants can never be (a) unimolecular reaction (b) first order reaction (c) second order reaction (d) bimolecular reaction. the rate of reaction will be (a) remain unchanged (b) tripled (c) increased by a factor of 4 (d) doubled.29/k 2 (c) (n – m) (d) 2 (n – m) (2003) (c) 0. s –1 (2002) Answer Key 1. Ms –1 (b) s –1 . (b) 9. T (d) the fraction of molecules with energy greater than the 18.5 minutes (d) 60 minutes. activation energy of the reaction. (d) 11. 693 Now.693 6. t1/2 = 6. (c) : Rate at 50°C = 2 10 = 2 5 = 32 times. i. (d) : NO (g) + Br 2(g) ƒ NOBr 2(g) NOBr 2(g) + NO (g) ® 2NOBr (g) [rate determining step] Rate of the reaction (r) = K [NOBr 2] [NO] where [NOBr 2] = K C [NO][Br 2] r = K ∙ K C ∙ [NO][Br 2][NO] r = K¢ [NO] 2 [Br 2 ] The order of the reaction with respect to NO (g) = 2 11.693 l = (for 1 st order reaction) t 1/ 2 - 1 d [ A] 1 d [ B ] = 1/ 2 dt 2 dt d [ A] 1 d [ B ] = dt 4 dt (b) : Rate = k [A] x [B] y When [B] is doubled.200 = - 20 kJ mol –1 The correct answer for this question should be –20 kJ mol –1 . Hence the reaction is first order with respect to A.303 a log t a - x Þ a = 0. Rate = - (a) : For the first order reaction k= 2. unit of rate constant is L mol –1 s –1 . .93 ´ log(100) 0.0347 × 0. keeping [A] constant halflife of the reaction does not change.43 Chemical Kinetics 1.303 ´ 8.25 hours 6.e. \ Rate = [ A]1 [ B ] 1 \ order = 2 Now for a nth order reaction. (c) : Given. the rate also doubles.06 minutes.314 ´ 10 ë û Ea = 0.303 ´ 6. Rate at T1 °C t1/ 2 = (c) : For this reaction. But no option given is correct. unit of rate constant is (L) n–1 (mol) 1–n s –1 when n = 2.1 M.0347 min –1 40 0.303 [ A 0 ] log l [ A ] (a) : DH R = E f .025 M. keeping [B] constant.693 = 23. The given rate equation is only consistent with the mechanism A.303 R ë T1T2 û Ea é 310 .300 ù log 2 = -3 ê 310 ´ 300 ú 2. for a first order reaction t 1/ 2 = k i. 9.25 M Þ 0. t 1/2 = 1hr.303 0.47 × 10 –4 M min –1 50 3. (a) : The rate equation depends upon the rate determining step.T 1 ù k log 10 2 = ê ú k1 2.93 Since reaction follows 1st order kinetics. 5. t = 40 min 8. Hence the reaction is first order with respect to B.025 40 [A] product Thus. 0. t 1/2 is independent of the concentration of the reactant. or t = 7.303 k= log = log 4 = 0. (b) : k = Ae –E/RT where E = activation energy. E a = 53. a – x = 0. [ A 0 ] 100 Q Reaction is 99% complete.25 = –t + 0. [A 0 ] = 2M \ k= 2 = 1 mol L-1 hr -1 2 ´ 1 Integrated rate law for zero order reaction is [A] = –kt + [A 0 ] Here.03 × 2 log(10) = 46.5 M and [A] = 0.e. the minimum amount of energy required by reactant molecules to participate in a reaction. rate = k[A] rate = 0. 0. t 1/2 is given as [ A0 ] 2k or k = [ A0 ] 2t1/2 Given.Eb = 180 . 2.303 ´ 8. 2. Hence we can ignore sign and select option (a).5 Þ t = 0.01 M min –1 = 3.93 min 0.3010 ´ 2. 10. [A 0 ] = 0. \ = [ A] 1 E a é T2 . (d) : For a zero order reaction. Now when [A] is doubled.1 2.314 ´ 10 -3 ´ 93 ´ 10 3 10 2. 4.6 kJ mol –1 2. (b) : As r = k[A] n r2 r1 = k 2 = k1 r 2 Since r = 2 1 \ t = (Given) k 2 = 2 k1 where [A0] = initial concentration and [A] = concentration of A at time t. hence its concentration may be Ratio between new and earlier rate taken constant.44 JEE MAIN CHAPTERWISE EXPLORER dx = k [CO] 2 dt r 2 = k[2CO] 2 = 4k[CO] 2 Thus. concentration becomes two the balanced stoichiometric equation. (d) : Rate 1 = k [A] n [B] m Rate = k[RCl][H 2O] expected order = 1 + 1 = 2 On doubling the concentration of A and halving the But actual rate law : concentration of B Rate = k¢[RCl] actual order = 1 Rate 2 = k [2A] n [B/2] m Here water is taken in excess. (a) : Generally. A = frequency factor T = temperature. where n = order of reaction 16.29 log = k 3 k Therefore.1 M to 0. 20. R = [A] × [B] 2 Thus.e. Thus the rate may be expressed as -d [H 2 ] . according to the rate law expression doubling the concentration of CO increases the rate by a factor of 4. For a second order reaction. (c) : Given r1 = 13.d [ I 2 ] 1 d [ HI ] = = dt dt 2 dt The negative sign signifies a decrease in concentration of the reactant with increase of time. (c) : In Arrhenius equation.303 4 0. (a) : Order is the sum of the power of the concentrations terms in rate law expression. i. (b) : t 1/ 4 = 2. 22.e. 2A + B ® C 1 d [ A] d [ B ] d [C ] == Rate = 2 dt dt dt i. Here the molecularity of the reaction = 2 and the order of the k[2 A]n [ B / 2] m n 1 m n . rate of formation of C is half the rate of disappearance of B. (a) 21. 2t 1/2 = 2 × 15 = 30 minutes.4 M in 15 minutes. k is L mol –1 sec –1 for a second order reaction. Rate 2 = k [2NO] 2 [2O 2 ] But a reaction involving two different reactants can a first order Rate1 k [NO]2 [O 2 ] Rate 1 1 = = reaction. E a = energy of activation.e. 2 moles of HI are formed in the same time interval. (d) : H 2 + I 2 ® 2HI When 1 mole of H 2 and 1 mole of I 2 reacts. For example. 14. This equation can be used for calculation of energy of activation. the concentration of reactant will fall from 0. n = 1 Þ first order reaction i. 1 t 1/ 2 = ka i. molecularity of simple reactions is equal to 18. .025 M in two half lives.m =2 ´ = 2 = 2 reaction = 1. k = Ae –E a /RT k = rate constant. 12. R = gas constant.e. order of reaction = 1 + 2 = 3 23. (d) : 2 A + B ® C rate = k [A] [B] The value of k (velocity constant) is always independent of the concentration of reactant and it is a function of temperature only. Thus. k [ A]n [ B ]m ( ) 15. (c) : Rate 1 = k [NO] 2 [O 2 ] the sum of the number of molecules of reactants involved in When volume is reduced to 1/2. t 1/2 is inversely proportional to initial concentration. t 1/2 = 15 minute. (a) : Unit of K = (mol L –1 ) 1 – n s –1 .8 M n = 0 Þ zero order reaction to 0. (a) : The concentration of the reactant decreases from 0. 17. involving two different reactants can never be unimolecular. unit of rate constant. a reaction times. for the following reaction Rate 2 k[2NO]2 [2O 2 ] or Rate 2 8 RCl + H 2O ® ROH + HCl \ Rate 2 = 8 Rate 1 Expected rate law : 19. C and D are 0. (2009) Gold numbers of protective colloids A. 0. Which of the following statements is NOT correct? (a) Magnesium chloride solution coagulates. the gold sol more readily than the iron (III) hydroxide sol (b) Sodium sulphate solution causes coagulation in both sols (c) Mixing of the sols has no effect (d) Coagulation in both sols can be brought about by electrophoresis (2005) (2013) 2.01. respectively. (b) More easily liquefiable gases are adsorbed readily. V c as compared to the volume of a solute particle in a true solution V s could be (a) ~ 1 (b) ~ 10 23 –3 (c) ~ 10 (d) ~ 10 3 (2005) Which one of the following characteristics is not correct for physical adsorption? (a) Adsorption on solids is reversible (b) Adsorption increases with increase in temperature (c) Adsorption is spontaneous (d) Both enthalpy and entropy of adsorption are negative. 4. protective powers is (a) B < D < A < C (b) D < A < C < B (c) C < B < D < A (d) A < C < B < D (2008) The volume of a colloidal particle. (2006) 6. In Langmuir’s model of adsorption of a gas on a solid surface (a) the rate of dissociation of adsorbed molecules from the surface does not depend on the surface covered (b) the adsorption at a single site on the surface may involve multiple molecules at the same time (c) the mass of gas striking a given area of surface is proportional to the pressure of the gas (d) the mass of gas striking a given area of surface is independent of the pressure of the gas. (c) 6. (2012) 3. which of the following is correct? (a) x µ p1 m (c) x µ p 0 m (b) x µ p1/ n m (d) All the above are correct for different ranges of pressure.50. 7. The correct order of their 8. (c) Under high pressure it results into multi molecular layer on adsorbent surface. (d) 5. (2003) Answer Key 1. (c) . (c) (d) 2.10 and 0. B. Al 3+ and Ba 2+ for arsenic sulphide sol increases in the order : (a) Al 3+ < Na + < Ba 2+ (b) Al 3+ < Ba 2+ < Na + (c) Na + < Ba 2+ < Al 3+ (d) Ba 2+ < Na + < Al 3+ 5. Which of the following statements is incorrect regarding physisorption? (a) It occurs because of van der Waals forces.005. (d) (b) 3. (d) 4. The coagulating power of electrolytes having ions Na + . (d) Enthalpy of adsorption (DH a dsorption ) is low and positive. 8.45 Surface Chemistry CHAPTER SURFACE CHEMISTRY 10 1. The d+isperse phase in colloidal iron (III) hydroxide and colloidal gold is positively and negatively charged. 7. According to Freundlich adsorption isotherm. 0. respectively. e. pressure. (c) : For a negatively charged sol. Therefore adsorption always takes place with evolution of heat. Smaller the value of gold number greater will be protecting power of 8. (d) : For true solution the diameter range is 1 to 10 Å and for colloidal solution diameter range is 10 to 1000 Å. (d) : The different protecting colloids differ in their protecting powers. it was derived by Langmuir that the mass of the gas adsorbed per gram of the adsorbent is related to the equilibrium pressure according to the equation: x aP = m 1 + bP where x is the mass of the gas adsorbed on m gram of the adsorbent. the protective colloid. 3. P is the pressure and a. greater the positive charge on cations. 7. .46 JEE MAIN CHAPTERWISE EXPLORER 1. Thus 1 protective power of colloid µ Gold number 5. i. 4. 10 3 (b) : During adsorption. it is an exothermic process and since the adsorption process is exothermic. 3 Vc (4 / 3) prc 3 æ r c ö = = Vs (4 / 3)p r s 3 çè r s ÷ø Ratio of diameters = (10/1) 3 = 10 3 V c /V s . like As 2S 3. DH = –ve) but its value is quite low because the attraction of gas molecules and solid surface is weak van der Waals forces. Zsigmondy introduced a term called Gold number to describe the protective power of different colloids. (c) : Assuming the formation of a monolayer of the adsorbate on the surface of the adsorbent. greater is the coagulating power.e. the physical adsorption occurs readily at low temperature and decreases with increasing temperature. Hence on mixing coagulation of two sols may be take place. 2. (Le Chatelier's principle). b are constants. (c) : Opposite charges attract each other. there is always decrease in surface energy which appears as heat.. (d) : Physical adsorption is an exothermic process (i. (d) : According to Freundlich adsorption isotherm x = kp1/ n m 1/n can have values between 0 to 1 over different ranges of 6. (2004) (c) 5. 7. (b) 8. Its halflife period is 30 days. the mass of it remaining undecayed after 18 hours would be (a) 4.084 g (c) 3. when initial amount is 64 grams is (a) 16 g (b) 2 g (c) 32 g (d) 8 g (2002) Answer Key 1. If halflife of a substance is 5 yrs. If the initial mass of the isotope was 200 g. (c) 10. bparticle is emitted in radioactivity by (a) conversion of proton to neutron (b) form outermost orbit (c) conversion of neutron to proton (d) bparticle is not emitted. (2005) 9.0 g (2003) The radionucleide 234 90 Th undergoes two successive bdecays followed by one adecay. (a) 3. (c) 11.47 CHEMISTRY CHAPTER NUCLEAR CHEMISTRY* 11 1. The atomic number and the mass number respectively of the resulting radionucleide are (a) 92 and 234 (b) 94 and 230 (c) 90 and 230 (d) 92 and 230 (2003) 10. (c) . if one emission is an a particle. isotope? (a) bparticle emission (b) Neutron particle emission (c) Positron emission (d) aparticle emission (2007) 7.125 g (d) 4. the mass remaining after 24 hours undecayed is (a) 1. (a) 100 days (b) 1000 days (c) 300 days (d) 10 days (2007) 234 In the transformation of 238 92 U to 92 U . Which of the following nuclear reactions will generate an 6.0 g (d) 16.167 g (2004) Consider the following nuclear reactions: X A + ® Y X N + 2 4 2 He ; Y N ® B L + 2 b The number of neutrons in the element L is (a) 142 (b) 144 (c) 140 (d) 146 238 92 M The halflife of a radioactive isotope is three hours. (2002) 11.0 g (b) 8. after how many days will it be safe to enter the room? 8. 2. (b) 2. The halflife of a radioisotope is four hours. 23 (d) the isobar of 11 Na (2005) Hydrogen bomb is based on the principle of (a) nuclear fission (b) natural radioactivity (c) nuclear fusion (d) artificial radioactivity.042 g (b) 2.0 g (c) 12. If the initial velocity is ten times the permissible value. (a) 9. what should be the other emission(s)? (a) Two b – (b) Two b – and one b + (c) One b – and one g (d) One b + and one b – (2006) 24 A photon of hard gamma radiation knocks a proton out of 12 Mg nucleus to form (a) the isotope of parent nucleus (b) the isobar of parent nucleus 23 (c) the nuclide 11 Na 5. (a) 4. A radioactive element gets spilled over the floor of a room. (d) 6. (c) * Not included in the syllabus of JEE Main since 2008. then the total amount of substance left after 15 years. 3. If the initial mass of the iosotope were 256 g. 4. – 1 n 0 – a A – 4 Z – 2 Y; – b A X Z +b+ A X Z A – 1 Z X A Y Z + 1 6. A X Z A X Z 2. = 3. (a) : U 92 –a 234 2.693 238 ( 1 2 ) N = 200 ´ 230 88 N 8.303 A t = log 0 = log 0 l N l A 6 or. 234 92 U Thus in order to get 92 U 234 as end product 1a and 2b particles should be emitted. ( ) A Z – 1 Y (a) : Let A be the activity for safe working. it is produced by the conversion of a neutron to a proton at the moment of emission. JEE MAIN CHAPTERWISE EXPLORER (b) : The atoms of the some elements having same atomic number but different mass numbers are called isotopes.69 days » 100 days 0.303 10 A = log 0.303 N 2. 24 (c) : 12 Mg + g ® 23 1 11 Na + 1 p 5. (c) : Hydrogen bomb is based on the principal of nuclear fusion. The temperature produced by the explosion of n 6 ( 2 ) ( ) N = N 0 1 2 Þ 2.303 ´ 30 = 99. Given A 0 = 10 A A 0 × N 0 and A × N 2. 2. 4. 1 H and 1 H releasing huge amount of energy. number of neutrons in element L = 230 – 86 = 144 (a) : t 1/2 = 3 hours. (c) : Since the nucleus does not contain bparticles. (c) : t 1/2 = 4 hours n n = T = 24 = 6 ; N = N 0 1 t1/ 2 4 2 1 Þ N = 256 N = 4.0 g -2 b -a ® Th 230 (c) : 90 Th 234 ¾¾¾ ® 92 X 234 ¾¾¾ 90 Elimination of 1a and 2b particles give isotope. 10. In hydrogen bomb. n = T/t 1/2 = 18/3 = 6 9.48 1.125 g (b) : 238 92 M ® 230 4 88 N + 2 2 He 230 + 86 L + 2 b ® Therefore. (d) : t 1/2 = 5 years. a mixture of deuterium oxide and tritium oxide is enclosed in a space surrounding an ordinary atomic bomb. 7.693 90 A – b 234 B 91 –b = 3. 1 1 0 0 n ® +1 p + –1 e 11.693/ 30 A the atomic bomb initiates the fusion reaction between 3 2 . n n = 3 T t1/ 2 = 1 1 n = N 0 æ ö = 64 æ ö = 8 g è 2ø è 2 ø 15 = 3 5 .303 ´ 30 = log 10 0. S 2– (d) Li + . Based on lattice energy and other considerations which one of (c) Na + > Mg 2+ > Al 3+ > O 2– > F – + – 2+ 2– 3+ the following alkali metal chlorides is expected to have the (d) Na > F > Mg > O > Al (2010) highest melting point? Following statements regarding the periodic trends of chemical (a) LiCl (b) NaCl reactivity of the alkali metals and the halogens are given. Lattice energy of an ionic compound depends upon (a) charge on the ion only in the halogens with increase in atomic number down the (b) size of the ion only group. Ca 2+ . electron gain enthalpy of Na + will be (a) + 2. Se and Ar? (a) Ca < Ba < S < Se < Ar 8. K + . Na + . S 2– (c) N 3– . Ca 2+ .55 eV (c) – 5. (2006) . (2006) The increasing order of the first ionisation enthalpies of the elements B.1 eV (d) – 10. Which among the following factors is the most important in (c) Chemical reactivity increases with increase in atomic making fluorine the strongest oxidising agent? number down the group in both the alkali metals and (a) Electron affinity halogens. Which of the following represents the correct order of increasing first ionization enthalpy for Ca. 13. K + (b) Ca 2+ . The first ionisation potential of Na is 5. species is (a) S 2– . Sr 2+ . Ba. Cl – . 49 The decreasing values of bond angles from NH 3 (106°) to SbH 3 (101°) down group15 of the periodic table is due to (a) increasing bondbond pair repulsion (b) increasing porbital character in sp 3 (c) decreasing lone pairbond pair repulsion (d) decreasing electronegativity. S 2– (2012) 3. (b) Ionization energy (d) In alkali metals the reactivity increases but in the halogens (c) Hydration enthalpy it decreases with increase in atomic number down the (d) Bond dissociation energy (2004) group. In which of the following arrangements the order is NOT (a) Al 2O 3 < MgO < Na 2 O < K 2 O according to the property indicated against it? (b) MgO < K 2O < Al 2 O 3 < Na 2 O (a) Al 3+ < Mg 2+ < Na + < F increasing ionic size (c) Na 2 O < K 2 O < MgO < Al 2O 3 (b) B < C < N < O increasing first ionisation enthalpy (d) K 2O < Na (2011) 2 O < Al 2O 3 < MgO (c) I < Br < F < Cl increasing electron gain enthalpy The correct sequence which shows decreasing order of the (with negative sign) ionic radii of the element is (d) Li < Na < K < Rb increasing metallic radius (a) O 2– > F – > Na + > Mg 2+ > Al 3+ (2005) (b) Al 3+ > Mg 2+ > Na + > F – > O 2– 11. Ca 2+ . S. K + . 5. 4. S 2– (c) K + . Ca 2+ . O 2– . 6. Cl – .2 eV (2013) 2. Cl – (d) Cl – . Ca 2+ (2006) Which one of the following orders presents the correct sequence of the increasing basic nature of the given oxides? 10. Mg 2+ .1 eV. S 2– . K + . P. Which (c) KCl (d) RbCl (2005) of these statements gives the correct picture? (a) The reactivity decreases in the alkali metals but increases 12. S and F (lowest first) is (a) F < S < P < B (b) P < S < B < F (c) B < P < S < F (d) B < S < P < F (2006) Which one of the following sets of ions represents a collection of isoelectronic species? (a) K + . (2005) the group. F – .55 eV (b) – 2. (b) Ca < S < Ba < Se < Ar (c) S < Se < Ca < Ba < Ar (d) Ba < Ca < Se < S < Ar (2013) The increasing order of the ionic radii of the given isoelectronic 9. (c) packing of the ion only (b) In both the alkali metals and the halogens the chemical reactivity decreases with increase in atomic number down (d) charge and size of the ion. The value of 7. Sc 3+ (b) Ba 2+ . Cl – .Classification of Elements and Periodicity in Properties CHAPTER 12 CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES 1. (a) 10. nos. (d) 18. Cl = 17. According to the periodic law of elements. (a) 4. : Ce = 58. 19. Which one of the following ions has the highest value of ionic radius? and then an endothermic step as shown below. Al = 13. K = 19. (c) . Mg = 12. Yb = 70 and Lu = 71) (2004) (2002) Answer Key 1. Br35) (b) oxygen has high electron affinity + . (b) 9. Cl – . (a) 15. (d) 12. (a) Li + (b) B 3+ – – –1 O (g) + e = O (g) ; DH° = –142 kJmol 2– (c) O (d) F – (2004) O – (g) + e – = O 2– (g) ; DH° = 844 kJmol –1 17. (d) Sn > Yb > Ce > Lu. Cl – (d) Ca 2+ . (a) 11. Sc 3+ (a) Ce > Sn > Yb > Lu (b) Sn > Ce > Lu > Yb (d) Na + . 13. Ca = 20. Ca 2+ . (c) (c) (d) (a) 2. Which is the correct order of atomic sizes? (c) K + . (b) 17. (2003) (b) Na + . (b) 16. the variation in (2004) properties of elements is related to their (a) atomic masses 15. Mg 2+ (b) N 3– . Al 3+ . Sc 3+ . Cs + . Na = 11. Cl – . The formation of the oxide ion O 2– (g) requires first an exothermic 16. (d) 14. Na + (a) Na (c) O – ion will tend to resist the addition of another electron (c) Be. (b) 6. Sc = 21) (At. Which one of the following groupings represents a collection This is because of isoelectronic species? (a) oxygen is more electronegative (At. F – 19. Br (2003) (d) O – ion has comparatively larger size than oxygen atom. Mg 2+ . (d) 8. Nos. Sn = 50. (c) 5.: Cs55. 18. (c) 3. Cl – (d) nuclear neutronproton number ratios. Which one of the following sets of ions represents the collection (b) nuclear masses of isoelectronic species? (c) atomic numbers (a) K + . Al 3+ . Ca 2+ .: F = 9. Sc 3+ . Mg 2+ . F – . Ca 2+ .50 JEE MAIN CHAPTERWISE EXPLORER 14. (c) Lu > Yb > Sn > Ce (Atomic nos. 7. For this reason. So.E. (c) : NH 3 PH 3 AsH 3 SbH 3 15. in atomic number due to following reasons: (a) As the size increases.51 Classification of Elements and Periodicity in Properties 1. The ionisation enthalpy decreases as we move down a group.4 11. ionization enthalpy of nitrogen is greater Al 2O 3 < MgO < Na 2 O < K 2 O than oxygen. melting point decreases size. P(1s 2 2s 2 2p 6 3s 2 3p 3 ) has a stable half filled electronic configuration than S (1s 2 2s 2 2p 6 3s 2 3p 4 ).5° 93. (a) : Isoelectronic species are those which have same number Bond angle 106. the Thus all these ions have 18 electrons in them. ionisation enthalpy of P is higher than S. (b) Due to decrease in electronegativity from F to I.0 17. ionic radii decreases. 12. they have a strong tendency to lose the single valence s NaCl is expected to have the highest melting point among electron to form unipositive ions having inert gas configuration. This reactivity arises due to their low ionisation enthalpies And while descending in the group basic character of corresponding oxides increases. (b) : As we move from left to right across a period. Order of nuclear charge: Ca 2+ > K + > Cl – > S 2– Protons : 20 19 17 16 Electrons : 18 18 18 18 9.5° 91. smaller the 11. But LiCl has lower melting (d) : All the alkali metals are highly reactive elements since point in comparison to NaCl due to covalent nature. (d) : Element: B S P F I.5° 91. So the order of increasing ionisation enthalpy should be \ B < C < N < O. The bond angle in ammonia is less than 109° 28¢ due to K + = 19 – 1 = 18 ; Ca 2+ = 20 – 2 = 18 repulsion between lone pairs present on nitrogen atom and Sc 3+ = 21 – 3 = 18 ; Cl – = 17 + 1 = 18 bonded pairs of electrons. Thus increasing order of ionic radii 2+ + – 2– Ca < K < Cl < S (a) : While moving from left to right in periodic table basic character of oxide of elements will decrease. Nuclear charge is maximum of the specie with maximum protons. anions generally have greater size than cations.1 eV 2. The weak F – F bond makes fluorine the strongest oxidising electron by the nucleus becomes less. the bond between halogen and other elements becomes 14. (c) : Electron gain enthalpy = – Ionisation potential = – 5. and high negative values of their standard electrode potentials. 8. . the ionisation enthalpy increases with increasing atomic number. Also greater. \ 10. the attraction for an additional 13. But N(1s 2 2s 2 2p 3 ) has a stable half filled electronic \ Correct order is configuration. (b) : For isoelectronic species as effective nuclear charge increases. (a) : K + = 19 – 1 = 18 e – Cl – = 17 + 1 = 18 e – Ca 2+ = 20 – 2 = 18 e – Sc 3+ = 21 – 3 = 18 e – Thus all the species are isoelectronic. (d) : The bond dissociation energy of F – F bond is very low. ionisation enthalpy increases with increasing atomic number. 4. bond angles gradually decrease due to decrease in bond pair lone pair repulsion. 7. the nuclear charge (Z) of the ion. is B < C < O < N. (d) : The value of lattice energy depends on the charges present However. the reactivity of halogens decreases with increase on the two ions and the distance between them. halogen. 3. Thus the order is : going down the group because lattice enthalpies decreases 2– – + 2+ 3+ O > F > Na > Mg > Al as size of alkali metal increases. the correct order of increasing the first ionization enthalpy isoelectronic species. (c) : The addition of second electron in an atom or ion is weaker and weaker. Thus. always endothermic.3° of electrons. 5. (b) : In case of halides of alkali metals. Among So. given halides. (a) : All the given species are isoelectronic.3 10. 6.4 In general as we move from left to right in a period. As we move down the group. (eV): 8. (d) : Ionization enthalpy decreases from top to bottom in a group while it increases from left to right in a period. O 2– has highest value of ionic radius. (b) : Isoelectronic species are the neutral atoms. 19.8 e 10 9 – z For F . the size decreases and when z/e ratio decreases the size increases. Number of electrons in N 3– = 7 + 3 = 10. (c) : This can be explained on the basis of z/e nuclear charge . Number of electrons in F – = 9 + 1 = 10 Number of electrons in Na + = 11 – 1 = 10 18.9 e 10 For Li + .5 e 2 z 8 For O 2– . = = 0. in case of lanthanides Hence. { } z 3 = = 1. (c) : According to modified modern periodic law. the properties of elements are periodic functions of their atomic numbers.52 JEE MAIN CHAPTERWISE EXPLORER 16. But due to lanthanide contraction. La > Sn. (a) : Generally as we move from left to right in a period. no. Thus the atomic radius of Sn should be less than lanthanides.5 e 2 z 5 For B 3+ . cations or Ce > Sn > Yb > Lu. anions of different elements which have the same number of . electrons but different nuclear charge. there is regular decrease in atomic radii and in a group as the atomic number increases the atomic radii also increases. = = 0. there is a continuous decrease in size with increase in atomic number. = = 2. Hence the atomic radius follow the given trend : 17. of electrons whereas z/e ratio increases. 8. (2002) Answer Key 1.53 General Principles and Processes of Isolation of Metals GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF METALS CHAPTER 13 (a) Magnetite (c) Galena 1. (d) 6. (2012) Which of the following factors is of no significance for roasting sulphide ores to the oxides and not subjecting the sulphide ores to carbon reduction directly? (a) CO 2 is more volatile than CS 2 . (d) . (c) Ag and Au (d) Fe and Ni. (b) Metal sulphides are thermodynamically more stable than CS 2 . 6. (c) CO 2 is thermodynamically more stable than CS 2 . 7. (2002) The metal extracted by leaching with a cyanide is (a) Mg (b) Ag (c) Cu (d) Na. (2002) Cyanide process is used for the extraction of (a) barium (b) aluminium (c) boron (d) silver. the appropriate electrodes are cathode anode (a) pure zinc pure copper (b) impure sample pure copper (c) impure zinc impure sample (d) pure copper impure sample. Which method of purification is represented by the following equation? Ti (s) + 2I 2(g) TiI 4(g) Ti (s) + 2I 2(g) 2. During the process of electrolytic refining of copper. (d) Metal sulphides are less stable than the corresponding oxides. Aluminium is extracted by the electrolysis of (a) bauxite (b) alumina (c) alumina mixed with molten cryolite (d) molten cryolite. (c) (b) 2. Which one of the following ores is best concentrated by froth flotation method? When the sample of copper wth zinc impurity is to be purified by electrolysis. (a) Cupellation (b) Poling (c) Van Arkel (d) Zone refining (2004) 5. These are (a) Sn and Ag (b) Pb and Zn 8. (c) (2002) 4. (a) (c) 3. (2005) 4. (b) Cassiterite (d) Malachite. (2008) 7. some metals present as impurity settle as ‘anode mud’. (c) 5. 3. Zn ® Ag 2 S + 2NaCN ® Na[Ag(CN) 2 ] ¾¾¾ Na 2 [Zn(CN) 4 ] + Ag ¯ (c) : Frothflotation method is used for the concentration of sulphide ores. Ni.CS 2 is more volatile than CO 2 . The method is based on the preferential wetting (d) : The impure metal is made anode while a thin sheet of pure metal acts as cathode. thorium and uranium. 4. The reduction process is on the thermodynamic stability of the products and not on their 7. ores to their oxides. zirconium. volatility. .54 1. dissolve in the solution and less electropositive impurities such as Ag. (b) : Silver ore forms a soluble complex with NaCN from which silver is precipitated using scrap zinc. So option (a) is of no significance for roasting sulphide 6. argentocyanide (soluble) 8. Au and Pt collect below the anode in the form of anodic mud. Here galena (PbS) is the only sulphide ore. 5. (c) : In the electrolytic refining of copper the more electropositive impurities like Fe. Co. JEE MAIN CHAPTERWISE EXPLORER (c) : Van Arkel method which is also called as vapourphase refining is used for preparing ultrapure metals like titanium. (c) : Aluminium is obtained by the electrolysis of the pure alumina (20 parts) dissolved in a bath of fused cryolite (60 parts) and fluorspar (20 parts). Zn. (d) : Gold and silver are extracted from their native ores by MacArthur forrest cyanide process. the pure metal is deposited on the cathode and equivalent amount of the metal gets dissolved from the anode. 3. properties with the frothing agent and water. etc. sod. 2. On passing the current. (a) : Oxidising roasting is a very common type of roasting in metallurgy and is carried out to remove sulphur and arsenic in the form of their volatile oxides. (b) Saturation of water with MgCO 3 . (2003) . (2008) 3. (b) Electrolysis of water. In context with the industrial preparation of hydrogen from water gas (CO + H 2 ). (d) Addition of Na 2 SO 4 to water. Answer Key 1.9%) can be made by which of the following processes? (a) Mixing natural hydrocarbons of high molecular weight. (a) 3. (d) H 2 is removed through occlusion with Pd. (b) CO and H 2 are fractionally separated using differences in their densities. (2012) 2. (b) 2. (c) Which one of the following processes will produce hard water? (a) Saturation of water with CaCO 3 . (c) CO is removed by absorption in aqueous Cu 2Cl 2 solution. (c) Saturation of water with CaSO 4 . which of the following is the correct statement? (a) CO is oxidised to CO 2 with steam in the presence of a catalyst followed by absorption of CO 2 in alkali.55 Hydrogen CHAPTER HYDROGEN 14 1. (d) Reaction of methane with steam. Very pure hydrogen (99. (c) Reaction of salt like hydrides with water. (c) : Permanent hardness is introduced when water passes over rocks containing the sulphates or chlorides of both of calcium and magnesium. the gases and steam over an iron oxide or cobalt oxide or chromium oxide catalyst at 673 K resulting in the production of more H 2 . . The entire reaction is called water gas shift reaction.56 1. JEE MAIN CHAPTERWISE EXPLORER (b) : Dihydrogen of high purity is usually prepared by the electrolysis of water using platinum electrodes in presence of small amount of acid or alkali. Dihydrogen is collected at cathode. (a) : Carbon monoxide is oxidised to carbon dioxide by passing 3. CO + H2 O Fe2O 3 . CoO 673 K CO2 + H 2 NaOH CO 2 alkali Na2CO 3 CO 2 is absorbed in alkali (NaOH). 2. It forms an insoluble hydroxide M(OH) 2 which is soluble in NaOH solution. (a) exhibiting maximum covalency in compounds (b) forming polymeric hydrides (c) forming covalent halides (d) exhibiting amphoteric nature in their oxides. It forms its oxide MO which becomes inert on heating. (c) NH 4 NO 3 (d) NaNO 3 (2012) 2. (b) (b) 2. (2002) Answer Key 1. (c) prevent action of water and salt (d) prevent puncturing by undersea rocks. (b) (b) 4. the two elements differ in 9. (2004) 11. (c) 11. (b) . 5. 6. Then M is (a) Mg (b) Ba (c) Ca (d) Be. The set representing the correct order of ionic radius is (a) Li + > Be 2+ > Na + > Mg 2+ (b) Na + > Li + > Mg 2+ > Be 2+ (c) Li + > Na + > Mg 2+ > Be 2+ (d) Mg 2+ > Be 2+ > Li + > Na + (2009) 3. Beryllium and aluminium exhibit many properties which are similar. 7.57 sBlock Elements CHAPTER sBLOCK ELEMENTS 15 1. (2003) The solubilities of carbonates decrease down the magnesium group due to a decrease in (a) lattice energies of solids (b) hydration energies of cations (c) interionic attraction (d) entropy of solution formation. 8. (2003) In curing cement plasters water is sprinkled from time to time. (2003) The substance not likely to contain CaCO 3 is (a) a marble statue (b) calcined gypsum (c) sea shells (d) dolomite. (2003) A metal M readily forms its sulphate MSO 4 which is water soluble. KO 2 (potassium super oxide) is used in oxygen cylinders in space and submarines because it Several blocks of magnesium are fixed to the bottom of a ship (a) absorbs CO 2 and increases O 2 content to (b) eliminates moisture (a) keep away the sharks (c) absorbs CO 2 (b) make the ship lighter (d) produces ozone. Which of the following on thermal decomposition yields a basic as well as an acidic oxide? (a) KClO 3 (b) CaCO 3 7. (2004) 10. (a) 6. (d) 5. 8. 9. (2002) One mole of magnesium nitride on the reaction with an excess of water gives (a) one mole of ammonia (b) one mole of nitric acid (c) two moles of ammonia (d) two moles of nitric acid. But. (a) 10. This helps in (a) keeping it cool (b) developing interlocking needlelike crystals of hydrated silicates (c) hydrating sand and gravel mixed with cement (d) converting sand into silicic acid. (b) (b) 3. The ionic mobility of alkali metal ions in aqueous solution is maximum for (a) K + (b) Rb + + (c) Li (d) Na + (2006) 4. Being a lighter element. hence the increasing order of ionic mobility is Li + < Na + < K + < Rb + 8. water insoluble Be(OH) 2 and BeO. (b) : The composition of gypsum is CaSO 4∙2H 2O. 2. 3. (b) : Water develops interlocking needlelike crystals of hydrated silicates.g. Since the mobility of ions is inversely proportional to the size of the their hydrated ions. (b) : 7. 9. 11.58 JEE MAIN CHAPTERWISE EXPLORER magnesium makes the ship lighter when it is fixed to the bottom of the ship. The solubility of carbonate of metals in water is generally low. 4. Also Li and Mg are diagonally related and hence the order is Na + > Li + > Mg 2+ > Be 2+ . the ionic radii decrease due to increase in effective nuclear charge as the additional electrons are added to the same shell. (d) : Be forms water soluble BeSO 4 . However they dissolve in water containing CO 2 yielding bicarbonates. (a) : Beryllium has the valency +2 while aluminium exhibits its valency as +3. The degree of hydration. and this solubility decreases on going down in a group with the increase in stability of carbonates of metals. (a) : 4KO 2 + 2CO 2 ® 2K 2 CO 3 + 3O 2 . 10. The reactions involved are the hydration of calcium aluminates and calcium silicates which change into their colloidal gels. (b) : The alkali metal ion exist as hydrated ions M + (H 2O) n in the aqueous solution. and decrease in hydration energy of the cations. (b) : Moving from left to right in a period. Calcium hydroxide binds the particles of calcium silicate together while aluminium hydroxide fills the interstices rendering the mass impervious. (c) : Mg 3 N 2 + 6H 2 O ® 3Mg(OH) 2 + 2NH 3 6. Hence Li + ion is mostly hydrated e. (b) : The stability of the carbonates of the alkaline earth metals increases on moving down the group. At the same time. on account of its lightness. great affinity for oxygen and toughness is used in ship. 1. however from top to bottom the ionic radii increase with increasing atomic number and presence of additional shells. [Li(H 2 O) 6 ] + . decreases with ionic size as we go down the group. Be(OH) 2 is insoluble in NaOH giving sodium beryllate Na 2BeO 2. some calcium hydroxide and aluminium hydroxides are formed as precipitates due to hydrolysis. It does not have CaCO 3 . 5. (b) : Magnesium. (b) The vapour at 200°C consists mostly of S 8 rings. (d) The oxidation state of sulphur is never less than +4 in 11. In which of the following arrangements.5O2 (c) 2XeF2 + 2H2O ® 2Xe + 4HF + O2 (d) XeF6 + RbF ® Rb[XeF7] (2009) 8. (b) ONCl and ONO – are not isoelectronic. Ge. whereas there is no possibility of such interaction between (c) Silicon reacts with NaOH (aq) in the presence of air to give C and F in CF4. Na 2 SiO 3 and H 2 O. Identify the incorrect statement among the following. (2011) Which of the following statements regarding sulphur is 10. Which one of the following reactions of xenon compounds is not feasible? (a) XeO3 + 6HF ® XeF6 + 3H2O (b) 3XeF4 + 6H2O ® 2Xe + XeO3 + 12HF + 1. 7. (c) O 3 molecule is bent. 9. (b) Nitrogen cannot form dppp bond. (d) Chlorides of both beryllium and aluminium have bridged chloride structures in solid phase. (c) At 600°C the gas mainly consists of S 2 molecules. Boron cannot form which one of the following anions? 3– – (a) BF 6 (b) BH 4 – – (c) B(OH) 4 (d) BO 2 (2011) 4. (2011) 5. Sn and Pb increases steadily in the sequence (a) PbX 2 = SnX 2 = GeX 2 = SiX 2 (b) GeX 2 = SiX 2 = SnX 2 = PbX 2 (c) SiX 2 = GeX 2 = PbX 2 = SnX 2 (2007) (d) SiX 2 = GeX 2 = SnX 2 = PbX 2 . (c) Single N — N bond is weaker than the single P – P bond. The correct reason for higher B – F bond dissociation energy as compared to that of C – F is (a) smaller size of Batom as compared to that of Catom 12. (2007) (2009) . (b) stronger s bond between B and F in BF3 as compared to (a) Br 2 reacts with hot and strong NaOH solution to give NaBr that between C and F in CF4 and H 2O. (b) Boric acid is a protonic acid.59 pBlock Elements CHAPTER 16 pBLOCK ELEMENTS 1. Which of the following is the wrong statement? (a) Ozone is diamagnetic gas. (c) Beryllium exhibits coordination number of six. BF3 than that between C and F in CF4. Which of the following exists as covalent crystals in the solid state? (a) Phosphorus (b) Iodine (c) Silicon (d) Sulphur (2013) 2. (2008) Among the following substituted silanes the one which will give rise to cross linked silicone polymer on hydrolysis is (a) R 3 SiCl (b) R 4 Si (c) RSiCl 3 (d) R 2 SiCl 2 (2008) The stability of dihalides of Si. The bond dissociation energy of B – F in BF3 is 646 kJ mol –1 whereas that of C – F in CF4 is 515 kJ mol –1 . (d) lower degree of pppp interaction between B and F in (d) Cl 2 reacts with excess of NH 3 to give N 2 and HCl. Which of the following statement is wrong? (a) The stability of hydrides increases from NH 3 to BiH 3 in group 15 of the periodic table. 6. the sequence is not strictly according to the property written against it? (a) CO2 < SiO2 < SnO2 < PbO2 : increasing oxidising power (b) HF < HCl < HBr < HI : increasing acid strength (c) NH3 < PH3 < AsH3 < SbH3 :increasing basic strength (d) B < C < O < N : increasing first ionization enthalpy (2009) (2013) 3. (d) Ozone is violetblack in solid state. incorrect? (a) S 2 molecule is paramagnetic. (d) N 2 O 4 has two resonance structure. Which one of the following is the correct statement? (a) B 2 H 6 ∙2NH 3 is known as ‘inorganic benzene’. its compounds. (c) significant pp pp interaction between B and F in BF3 (b) Ozone reacts with SO 2 to give SO 3. 1 and 2 lone pairs of electrons on the (a) It is used to fill gas in balloons instead of hydrogen because central atom respectively it is lighter and noninflammable.60 JEE MAIN CHAPTERWISE EXPLORER 13. (2005) 15. it gives (a) one sigma. two pi. SiO . Aluminium chloride exists as dimer. Which of the following statements is true? 25. This transformation is related to (c) HClO 4 is a weaker acid than HClO 3 . (2004) (c) two sigma. Which (a) each silicon atom is surrounded by four oxygen atoms and of the following statements about these chlorides is correct? each oxygen atom is bonded to two silicon atoms (a) MCl 2 is more volatile than. The molecular shapes of SF 4 . 0 and 1 lone pairs of electrons on the (a) sp 2 and sp 2 (b) sp 2 and sp 3 central atom respectively 3 2 (c) sp and sp (d) sp 3 and sp 3 . grey powder. (a) an interaction with nitrogen of the air at very low (d) HNO 3 is a stronger acid than HNO 2 . The structure of diborane (B 2 H 6 ) contains (a) SO 2 < P 2 O 3 < SiO 2 < Al 2 O 3 (a) four 2c2e bonds and two 3c2e bonds (b) SiO 2 < SO 2 < Al 2 O 3 < P 2 O 3 (b) two 2c2e bonds and four 3c2e bonds (c) Al 2 O 3 < SiO 2 < SO 2 < P 2 O 3 (c) two 2c2e bonds and two 3c3e bonds (d) Al 2 O 3 < SiO 2 < P 2 O 3 < SO 2 . The states of hybridisation of boron and oxygen atoms in boric 20. each oxygen atom is bonded to two silicon atoms (c) MCl 2 is more ionic than MCl 4 . Al 2 Cl 6 in solid state as well as in solution of nonpolar solvents such as benzene. The correct order of the thermal stability of hydrogen halides (H – X) is soil? (a) HI > HBr > HCl > HF (a) Ammonium sulphate (b) HF > HCl > HBr > HI (b) Potassium nitrate (c) HCl < HF > HBr < HI (c) Urea (d) HI > HCl < HF > HBr (2005) (d) Superphosphate of lime (2007) 23. M forms chlorides in +2 and +4 oxidation states. (2005) 27. (2006) temperatures 17. White metallic tin buttons got converted to (b) In aqueous medium HF is a stronger acid than HCl. 0 and 2 lone pairs of electrons on the (b) It is used as a cryogenic agent for carrying out experiments central atom respectively (2005) at low temperatures. 1 and 1 lone pair of electrons on the central atoms respectively 29. Which one of the following statements regarding helium is incorrect? (c) different with 0. (b) each silicon atom is surrounded by two oxygen atoms and (b) MCl 2 is more soluble in anhydrous ethanol than MCl 4 . (2004) (d) four 2c2e bonds and four 3c2e bonds (2005) 28. Which of the following oxides is amphoteric in character? (a) HClO 3 and Cl 2O (b) HClO 2 and HClO 4 (a) CaO (b) CO 2 (c) HCl and Cl 2 O (d) HCl and HClO 3 (2006) (c) SiO 2 (d) SnO 2 (2005) 16. (a) zero (b) two (d) It is used in gascooled nuclear reactors. A metal. The number of hydrogen atom(s) attached to phosphorus atom (c) It is used to produce and sustain powerful superconducting in hypophosphorous acid is magnets. Regular use of the following fertilizers increases the acidity of 22. CF 4 and XeF 4 are acid (H 3 BO 3 ) are respectively (a) the same with 2. Among Al O . 21. When calcium carbide are dissolved in water. (d) different with 1. of their uniforms. The number and type of bonds between two carbon atoms in 26. (2004) (b) the same with 1. In silicon dioxide 14. (c) silicon atom is bonded to two oxygen atoms (d) MCl 2 is more easily hydrolysed than MCl 4 . one pi (d) two sigma. (2005) . P O and SO the correct order of acid 2 3 2 2 3 2 strength is 19. (2004) (c) one (d) three. The soldiers of Napolean army while at Alps during freezing winter suffered a serious problem as regards to the tin buttons (a) H 3 PO 3 is a stronger acid than H 2 SO 3 . one pi 3+ + 3Cl – (a) Al 3+ + 3Cl – (b) [Al(H 2O) (b) one sigma. Heating an aqueous solution of aluminium chloride to dryness (b) a change in the crystalline structure of tin will give (c) a change in the partial pressure of oxygen in the air (a) AlCl 3 (b) Al 2 Cl 6 (d) an interaction with water vapour contained in the humid (c) Al 2 O 3 (d) Al(OH)Cl 2 (2005) air. two pi 6] 3– (c) [Al(OH) 6 ] + 3HCl (d) Al 2 O 3 + 6HCl. (2006) (d) there are double bonds between silicon and oxygen atoms. What products are expected from the disproportionation reaction of hypochlorous acid? 24. MCl 4 . (2004) 18. (None) 3. 33. (a) 13. The explanation for it is that (a) concentrated hydrochloric acid emits strongly smelling HCl gas all the time 40. In case of nitrogen. 3. Graphite is a soft solid lubricant extremely difficult to melt. (a) 5. 35. 26. (a) 4. It is due to 32. (b) 31. In XeF 2 . (c) (b) (a) (a) (c) (c) 12. 41. (c) 7. XeF 4 . Which one of the following statements is correct? (a) Manganese salts give a violet borax test in the reducing flame. PCl as well as PCl are possible. (a) 25. 29. 27. (2003) 31. ammonia solution dissolves only AgCl. XeF 6 the number of lone pairs on Xe are (b) oxygen in air reacts with the emitted HCl gas to form a respectively (a) 2. (2003) 37. 17. concentrated hydrochloric (b) Absolutely pure water does not contain any ions. The reason for this anomalous behaviour is that graphite (a) is a noncrystalline substance (b) is an allotropic form of diamond (c) has molecules of variable molecular masses like polymers (d) has carbon atoms arranged in large plates of rings of strongly bound carbon atoms with weak interplate bonds. When H 2 S is passed through Hg 2 S we get (a) The mixture only cools down (a) HgS (b) HgS + Hg 2 S (b) PCl 3 and HCl are formed and the mixture warms up (c) Hg 2 S + Hg (d) Hg 2 S. Alum helps in purifying water by (a) forming Si complex with clay particles (b) sulphate part which combines with the dirt and removes it (c) coagulating the mud particles (d) making mud water soluble. 3 cloud of chlorine gas (c) strong affinity of HCl gas for moisture in air results in (c) 4. Which one of the following is an amphoteric oxide? (a) ZnO (b) Na 2O (c) SO 2 (d) B 2O (2003) 3 5 availability of vacant d orbitals in P but not in N lower electronegativity of P than N lower tendency of Hbond formation in P than N occurrence of P in solid while N in gaseous state at room temperature. (d) (d) (b) (d) (c) (a) 10. (b) From a mixed precipitate of AgCl and AgI. 2. acid pulls moisture of air towards itself. 15. 22. 1. (c) (c) (d) (b) (c) (c) 9. 14. 30. 24. (d) On boiling a solution having K + . (d) due to strong affinity for water. Ca 2+ and HCO 3 – ions we get a precipitate of K 2 Ca(CO 3 ) 2 . This moisture (c) Chemical bond formation takes place when forces of forms droplets of water and hence the cloud. (d) 6. (d) 37. 42. 38. (c) 2. What may be expected to happen when phosphine gas is mixed (2002) with chlorine gas? 42. 28. 36. Which one of the following substances has the highest proton (2002) affinity? (a) H 2 O (b) H 2 S 39. (c) PCl 5 and HCl are formed and the mixture cools down (d) PH 3∙Cl (2003) 2 is formed with warming up. 40. 34. 39. 33. 1. 36. Concentrated hydrochloric acid when kept in open air sometimes produces a cloud of white fumes. (2002) 35. (c) (d) (b) (b) (a) (d) 11. 1 (b) 1. (2002) Answer Key 1. (2003) (d) In covalency transference of electron takes place. 2 (d) 3. (2003) 38. NCl 3 is possible but not NCl 5 while in case (c) NH 3 (d) PH 3 (2003) of phosphorus. (c) Ferric ions give a deep green precipitate on adding potassium ferrocyanide solution. attraction overcome the forces of repulsion. 21. (2002) forming of droplets of liquid solution which appears like 41. 2. (d) (b) (d) (b) (c) (c) . Which one of the following pairs of molecules will have permanent dipole moments for both members? (a) SiF 4 and NO 2 (b) NO 2 and CO 2 (c) NO 2 and O 3 (d) SiF 4 and CO 2 (2003) 3 (a) (b) (c) (d) 34. Glass is a (a) microcrystalline solid (b) supercooled liquid (c) gel (d) polymeric mixture. 16. Which of the following statements is true? a cloudy smoke (a) HF is less polar than HBr. 32. (a) 19. 20.61 pBlock Elements 30. 23. 18. (b) 8. and +6 are more common. the basic character decreases on going down the group due to decrease in the availability of the lone pair of electrons because of the increase in size of elements from N to Bi. 6. 3. (d) : 3Br 2 + 6NaOH ® 5NaBr + NaBrO 3 + 3H 2 O O 3 + SO 2 ® O 2 + SO 3 Si + 2NaOH + O 2 ® Na 2SiO 3 + H 2O . proton donor) but behaves as a Lewis acid by accepting a pair of electrons from OH – ions. Borazine The size of the central atom increases from N to Bi therefore. the tendency to form a stable covalent bond with small atom like hydrogen decreases and therefore. it is called inorganic benzene. 2B3N3H6 + 12H 2 Borazine has structure similar to benzene and therefore. (c) : RSiCl 3 on hydrolysis gives a cross linked silicone.C – F. (None) : All the statements are correct. Hence option (d) is correct. stability decreases. (a) : Due to nonavailability of dorbitals. – + B(OH) 3 + 2H 2 O ® [B(OH) 4] + H 3O BeCl 2 like Al 2Cl 6 has a bridged polymeric structure in solid phase generally as shown below. +4. It is a notable part that boric acid does not act as a protonic acid (i. +6 oxidation states but +4 Cl . 3B2H6 + 6NH 3 3[BH2(NH3)2 ]+ BH 4 – or B2H6 ∙ 2NH 3 or Cl 11. +2. So the stability also decreases.E. 450 K room temp. (c) : In group15 hydrides. (c) : Due to the inert pair effect (the reluctance of ns 2 electrons of outermost shell to participate in bonding) the stability of M 2+ ions (of group IV elements) increases as we go down the group. Also.e. B is sp 2 hybridised and has a vacant 2porbital (c) : In BF3 which overlaps laterally with a filled 2porbital of F forming strong pppp bond. 8. 12. the maximum covalency of boron cannot exceed 4. Beryllium exhibits coordination number of four as it has only four available orbitals in its valency shell.62 JEE MAIN CHAPTERWISE EXPLORER 1. 1300°C 440°C 280°C 150°C 7. Cl Cl – Be Cl Cl Cl Cl Cl Al Al Cl Polymeric structure of Al2 Cl 6 OH R (ii) HO – Si – OH + H O – Si – OH + H O – Si – OH OH OH R –H2 O R HO – Si – O – Si – O – Si – OH OH R R OH R OH OH R (iii) HO – Si – O – Si – O – Si – OH –3H2 O OH OH OH H O H O H O HO – Si – O – Si – O – Si – OH R R R R R R – O – Si – O – Si – O – Si – O – O O O – O – Si – O – Si – O – Si – O – R R R Cross linked silicone Be – Cl Polymeric structure of BeCl 2 OH R – Si – OH R (a) : The reaction is not feasible because XeF6 formed will XeF6 + 3H2O 9. 3H2 O (i) R – Si – Cl –3HCl Cl R XeO3 + 3H2F2.B – F > B. Thus. Therefore. Decomposition temperature 5. (a) : Thermal stability decreases gradually from NH 3 to BiH 3 . NH 3 PH 3 AsH 3 SbH 3 BiH 3 4. 10.. (d) : Boric acid is a weak monobasic acid (K a = 1. (c) 2. C does not have any vacant porbitals to undergo pbonding. Thus B. However in CF4. boron is unable to expand its octet. correct order of basicity is NH3 > PH3 > AsH3 > SbH3. further produce XeO3 by getting hydrolysed.0 × 10 –9 ). B2H6 ∙ 2NH 3 Heat. The formation can be explained in three steps (d) : Sulphur exhibits –2.E. thermal stability of HX The crystallisation of AlCl 3 from aqueous solution. the four orbitals of each of the boron atom overlap with two terminal hydrogen atoms forming two normal B – H sbonds. Due to successive decrease in the strength + – 2AlCl 3 + aq. H . (b) : Hypophosphorous acid (in ClO 3 – ) and –1 (in Cl – ). 2– . SiCl 4. XeF 4 (sp 3 d 2 . (b) : As the size of the halogen atom increases from F to I. H – F to H – I (H – F < H – Cl < H – Br < H – I). 2NH 3 + 3Cl 2 N 2 + 6HCl 6NH 3 + 6HCl 6NH 4 Cl 8NH 3 + 3Cl 2 N 2 + 6NH 4 Cl 13. O O 18. Now. H H . . F (XeF 4 ) O 16. molecules also decreases from HF to HI (HF > HCl > HBr > yields an ionic solid of composition [AlCl 2 (H 2 O) 4 ] + HI). state are ionic in nature and behave as reducing agent. (d): CaObasic. e. HO H H Hence. CO 2 and SiO 2acidic. > H – I). F F Further as we move down the group. trigonal bipyramidal with one equatorial position SnCl 4. . . 1s orbital of hydrogen atoms (bridge atom) and one of hybrid orbitals of the other boron atom overlap to form a delocalised orbital covering the three nuclei with a pair of electrons. . Number of hydrogen atom(s) attached to phosphorus atom = 2. B B B B . SnCl 2 . two lone pairs). . Such a bond is known as three centre two electron (3c – 2e) bonds. H . (a) : (NH 4 ) 2 SO 4 + 2H 2 O (2H + + SO 4 2– ) + 2NH 4 OH Strong acid Weak base (NH 4) 2SO 4 on hydrolysis produces strong acid H 2SO 4. Hence H 2 SO 3 22. whereas those which show +2 oxidation occupied by 1 lone pair). no lone pair). which increases the acidity of the soil. (a) : According to molecular orbital theory. (c) : The elements of group 14 show an oxidation state of +4 Structure of diborane and +2.g. H H 14. therefore. 20. H . 2– . (d) : 3HClO 4(aq) ® HClO 3(aq) + 2HCl (aq) (CF 4 ) (SF 4 ) It is a disproportionation reaction of hypochlorous acid where – the oxidation number of Cl changes from +1 (in ClO ) to +5 21. (d) : SF 4 (sp 3 d. SnO 2amphoteric. H – X bond length in HX molecules also increases from is a stronger acid than H 3 PO 3 . . Al 2 Cl 6 + H 2 O . . greater P is the acidity.g. 1 s . (b) : Aluminium chloride in aqueous solution exists as ion from the fact that H – X bond dissociation energies decrease pair. Two of reacts both with acids and bases. PbCl 2 . H . from H – F to H – I. .63 pBlock Elements Cl 2 reacts with excess of ammonia to produce ammonium chloride and nitrogen. One of the remaining hybrid orbital (either filled or empty) of one of the boron atoms. tetrahedral. Due to higher dissociation energy of H – F bond and molecular The increase in H – X bond length decreases the strength of association due to hydrogen bonding in HF. It is a network 190°C to give the nonionic dimer Al 2Cl 6. as it three have one electron each while the fourth is empty. square planar. etc. e. . 24. . HNO 3 is a stronger acid than HNO 2 . greater is the electronegativity and higher is the oxidation state of the central atom. H . CF 4 (sp 3 . heat solid in which each Si atom is surrounded tetrahedrally by four + – [AlCl 2(H 2 O) 4 ] [AlCl 4 (H 2 O) 2 ] ∙ xH 2 O 190°C oxygen atoms. 23. PbCl 4. H H H . the tendency of the element F F F to form covalent compound decreases but the tendency to C S Xe form ionic compound increases. The compounds showing an oxidation state of +4 are covalent compound and have tetrahedral structures. greater is the acidity. Ca 2+ [ C C ] 2 p – O – Si – O – Si – O – O O 19. (b) : Calcium carbide is ionic carbide having [ C C ] . each of the two boron atoms is in sp 3 hybrid state. etc. HF is a weaker H – X bond from H – F to H – I (H – F > H – Cl > H – Br acid than HCl. (d) : Higher is the oxidation state of the central atom. . The decrease in the strength of H – X bond is evident 17. (a) : Silicon dioxide exhibits polymorphism. . HClO 4 is a stronger acid than HClO 3 . – [AlCl 4 (H 2O) This compound decomposes at about 2] ∙ xH 2O. . Of the four hybrid orbitals. F F F F F F 15. ® [AlCl 2(H 2O) 4] (aq) + [AlCl 4(H 2O) 2] (aq) of H – X bond from H – F to H – I. accepting proton to form ammonium ion as it has tendency to donate an electron pair. be decanted off. Grey tin White tin (cubic) (tetragonal) 33. AgBr and AgI at the room temperature are 2.0 × 10 –13 and 8. each carbon atom makes a use of sp 2 hybridisation. since the bonding between the layers XeF 6 sp 3 d 3 1 lone pair involving only weak van der Waal's forces. (a) : Helium is twice as heavy as hydrogen. 3 O . + H – N + H . its lifting power is 92 percent of that of hydrogen. AgCl + 2NH 4OH ® [Ag(NH 3) 2]Cl + 2H 2O Diammine silver chloride 30. (b) : Grey tin is very brittle and easily crumbles down to a powder in very cold climates. . sp . 26. . This is called tin disease or tin plague. 34. SiO 2 is slightly acidic whereas P 2O 3 and 35. sharing of electrons between two nonmetal atoms takes place. Ba. (c) : Ammonia is a Lewis base. and cryogenic research where temperatures close to absolute zero are needed.8 × 10 –10 . . (b) : Al 2 Cl 6 + 12H 2 O 2[Al(H 2 O) 6 ] + 6Cl ZnO + H 2 SO 4 ® ZnSO 4 + H 2 O 27. (c) : Due to the higher electronegativity of F. H O . H . 36. In covalency. having a general formula R 2 O∙MO∙6SiO 2 where R = Na or K 39. This gives softness. H H + H – N ® H H The change of white tin to grey tin is accompanied by increase in volume.5 × 10 –17 respectively. Zn or Pb. (b) : H . sp 3 29. (c) : HCl gas in presence of moisture in air forms droplets of SO 2 are the anhydrides of the acids H 3 PO 3 and H 2 SO 3 . 37. (c) : The negatively charged colloidal particles of impurities get neutralised by the Al 3+ ions and settle down and pure water can solid solution (supercooled liquid) of silicates and borates. PH 3 + 4Cl 2 ® PCl 5 + 3HCl 2 B sp 28. Further. 2 2 6 2 3 15 P = 1s 2s 2p 3s 3p 31. 32. they are even less soluble than AgCl. (c) have permanent dipole moment. Thus.. (d) : Acidity of the oxides of non metals increases with the zinc sulphate electronegativity and oxidation number of the element. ZnO + 2NaOH ® Na 2 ZnO 2 + H 2 O Al 2O 3 < SiO 2 < P 2O 3 < SO 2 sodium zincate Al 2O 3 is amphoteric.. phosphorus pentachloride. 5.64 JEE MAIN CHAPTERWISE EXPLORER SnO 2 + 4HCl ® SnCl 4 + 2H 2O SnO 2 + 2NaOH ® Na 2 SnO 3 + H 2 O 25... (b) : Glass is a transparent or translucent amorphous supercooled 38. (c) : Phosphine burns in the atmosphere of chlorine and forms sp 3 O . (d) : Graphite has a twodimensional sheet like structure and In phosphorus the 3dorbitals are available. Due to greater solubility of AgCl than AgI. character of graphite. (a) : ZnO is an amphoteric oxide and dissolves readily in acids 3+ – forming corresponding zinc salts and alkalies forming zincates. (a) : 7 N = 1s 2 2s 2 3p 3 and M = Ca. 40. (c) : NO 2 and O 3 both have unsymmetrical structures. H . (d) : XeF 2 sp 3 d 3 lone pairs The above layer structure of graphite is less compact than that XeF 4 sp 3 d 2 2 lone pairs of diamond. HF is more polar than HBr pure water contains H + and OH – ions. Consequently. so they 42. liquid solution in the form of cloudy smoke. these layers can slide over each other. greasiness and lubricating 41. ammonia solution dissolves only AgCl and forms a complex. Helium has the lowest melting and boiling points of any element which makes liquid helium an ideal coolant for many extremely lowtemperature applications such as superconducting magnets. (b) : The solubility product of AgCl. AgI is the least soluble silver halide. The lattice energies of AgBr and AgI are even higher because of greater number of electrons in their anions. Four successive members of the first row transition elements are listed below with atomic numbers.65 d and fBlock Elements CHAPTER 17 d and f BLOCK ELEMENTS 1. (2011) (d) more energy difference between 5f and 6d than between 4f and 5dorbitals. Which of the following arrangements does not represent the 7. Fe and Co is (a) Cr > Mn > Fe > Co (b) Mn > Cr > Fe > Co 11. Which of the following statements about iron is incorrect? (a) Ferrous compounds are relatively more ionic than the 8. The actinoids exhibit more number of oxidation states in general (c) Cr > Fe > Mn > Co (d) Fe > Mn > Cr > Co than the lanthanoids. the main reason being (b) All the members exhibit +3 oxidation state. will the magnitude of D oct be the highest? (a) [Co(NH 3 ) 6 ] 3+ (b) [Co(CN) 6 ] 3– 3– 3+ (c) [Co(C 2O (d) [Co(H 2O) (2008) 4) 3] 6] In context of the lanthanoids. (2013) 2. all the 4s and 3d electrons are used for bonding. (2009) In which of the following octahedral complexes of Co (At. Mn. the transition metals show basic character and form cationic complexes. (d) Ferrous oxide is more basic in nature than the ferric oxide. (a) more reactive nature of the actinoids than the lanthanoids (c) Because of similar properties the separation of lanthanoids (b) 4f orbitals more diffused than the 5f orbitals is not easy. (b) The ionic sizes of Ln(III) decrease in general with increasing atomic number. (d) Once the d 5 configuration is exceeded. The correct order of E° M2+ values with negative sign for /M (2008) the four successive elements Cr. Larger number of oxidation states are exhibited by the actinoids with increasing atomic number in the series. (c) In the highest oxidation states of the first five transition elements (Sc to Mn). (2012) 3. (2009) Knowing that the chemistry of lanthanoids (Ln) is dominated by its +3 oxidation state. correct order of the property stated against it? (a) Sc < Ti < Cr < Mn : number of oxidation states (b) V 2+ < Cr 2+ < Mn 2+ < Fe 2+ : paramagnetic behaviour (c) Ni 2+ < Co 2+ < Fe 2+ < Mn 2+ : ionic size (d) Co 3+ < Fe 3+ < Cr 3+ < Sc 3+ : stability in aqueous solution. (d) Ln(III) hydroxides are mainly basic in character. In context with the transition elements. Which one of them is expected to have the highest E° M3+ /M 2+ value? (a) Co (Z = 27) (b) Cr (Z = 24) (c) Mn (Z = 25) (d) Fe (Z = 26) (2013) Iron exhibits +2 and +3 oxidation states. which of the following statement is not correct? (a) There is a gradual decrease in the radii of the members 10. which of the following statements is incorrect? (a) In addition to the normal oxidation states. corresponding ferric compounds. the tendency to involve all the 3d electrons in bonding decreases. This is because (2010) . than those by the lanthanoids. (c) Ferrous compounds are more easily hydrolysed than the corresponding ferric compounds. 27). (c) Ln(III) compounds are generally colourless. 6. (c) lesser energy difference between 5f and 6d than between (d) Availability of 4f electrons results in the formation of 4f and 5d orbitals compounds in +4 state for all the members of the series. The outer electronic configuration of Gd (Atomic No : 64) is (a) 4f 3 5d 5 6s 2 (b) 4f 8 5d 0 6s 2 4 4 2 (c) 4f 5d 6s (d) 4f 7 5d 1 6s 2 (2011) 9. (b) Ferrous compounds are less volatile than the corresponding ferric compounds. no. (b) In the highest oxidation states. the zero oxidation state is also shown by these elements in complexes. 4. which of the following statements is incorrect? (a) Because of the large size of the Ln(III) ions the bonding in its compounds is predominantly ionic in character. 5. (c) sodium (d) magnesium. (d) Cerium (IV) acts as an oxidising agent. The lanthanide contraction is responsible for the fact that (c) Manganese nitrate (a) Zr and Y have about the same radius (d) Silver nitrate. (a) Poor shielding of one of 4felectron by another in the These are obtained by floating molten glass over a liquid metal subshell. on strong heating? (2006) (a) Ferric nitrate (b) Copper nitrate 16. (2007) (b) The +3 oxidation state of cerium is more stable than +4 13. oxidation state.) among is (a) [MnCl 4 ] 2– > [CoCl 4 ] 2– > [Fe(CN) 6 ] 4– (b) [MnCl 4 ] 2– > [Fe(CN) 6 ] 4– > [CoCl 4 ] 2– 12. Identify the incorrect statement among the following: (c) [Fe(CN) 6 ] 4– > [MnCl 4 ] 2– > [CoCl 4 ] 2– (a) 4fand 5forbitals are equally shielded. The oxidation state of chromium in the final product formed by 27. (d) Evolved I 2 is reduced. (c) +2 (d) +3 (d) CrO 42– (2003) is oxidised to +7 state of Cr. (2004) 15. (a) The common oxidation states of cerium are +3 and +4.40 Å (c) 1. The correct order of magnetic moments (spin only values in B. Which of the following statements about cerium is incorrect? (d) The chemistry of various lanthanoids is very similar. respectively (a) one.66 JEE MAIN CHAPTERWISE EXPLORER (a) the 5f orbitals extend further from the nucleus than the 4f orbitals (b) the 5f orbitals are more buried than the 4f orbitals (c) there is a similarity between 4f and 5f orbitals in their angular part of the wave function (d) the actinoids are more reactive than the lanthanoids. Nickel (Z = 28) combines with a uninegative monodentate ligand – 2– (a) Cu 2I (b) CuI 2 is formed. The number of 2 is formed.: Mn = 25. Of the following outer electronic configurations of atoms.73 (2006) solution is added to it. The metal used can be (b) Effective shielding of one of 4felectrons by another in the (a) mercury (b) tin subshell. (m B )] of Ni 2+ in aqueous solution would be (atomic number of (c) The +4 oxidation state of cerium is not known in solutions. Heating mixture of Cu 2 O and Cu 2 S will give (a) Cu + SO 2 (b) Cu + SO 3 (c) CuO + CuS (d) Cu 2SO 3 (2005) 20. Which one of the following nitrates will leave behind a metal (d) Greater shielding of 5d electrons by 4felectrons. Calomel (Hg 2 Cl 2 ) on reaction with ammonium hydroxide gives (a) HgNH 2 Cl (b) NH 2 – Hg – Hg – Cl (c) Hg 2 O (d) HgO (2005) of Lu 3+ (Atomic number of Lu = 71)? (a) 1. What would happen when a solution of potassium chromate is treated with an excess of dilute nitric acid? the reaction between KI and acidified potassium dichromate 2– (a) Cr 3+ and Cr 2O 7 are formed. partially filled orbitals. solution is 2– (b) Cr 2 O 7 and H 2 O are formed. (b) dBlock elements show irregular and erratic chemical (Atomic nos. Excess of KI reacts with CuSO 4 solution and then Na 2 S 2 O 3 (c) 0 (d) 1. For making good quality mirrors. Fe = 26. plates of float glass are used. the (c) one. (2005) .90 22. Cerium (Z = 58) is an important member of the lanthanoids.84 (b) 4. tetrahedral 23. (2003) (c) Poorer shielding of 5d electrons by 4felectrons. X to form a paramagnetic complex [NiX 4 ] . (2004) (c) La and Lu have partially filled dorbitals and no other 21. (2007) 19. Which of the statements is incorrect for this reaction? 14.06 Å. (2003) (b) Zr and Nb have similar oxidation state (c) Zr and Hf have about the same radius 26. The radius of La 3+ (Atomic number of La = 57) is 1. which does not solidify before glass. (2004) Ni = 28) (a) 2. 25.06 Å (d) 0. square planar (d) two. (a) +4 (b) +6 (c) CrO 42– is reduced to +3 state of Cr.M. (d) [Fe(CN) 6 ] 4– > [CoCl 4 ] 2– > [MnCl 4 ] 2– . square planar.85 Å (2003) 18. unpaired electron(s) in the nickel and geometry of this complex 2 2 3 (2004) ion are.60 Å (b) 1. (c) Na S O is oxidised. Which (d) Zr and Zn have the same oxidation state. tetrahedral (b) two. The “spinonly” magnetic moment [in units of Bohr magneton. highest oxidation state is achieved by which one of them? (2006) (a) (n – 1)d 8 ns 2 (b) (n – 1)d 5 ns 1 3 2 (c) (n – 1)d ns (d) (n – 1)d 5 ns 2 . Co = 27) properties among themselves. Which of the following factors may be regarded as the main cause of lanthanide contraction? 24. (2005) one of the following given values will be closest to the radius 17. 34. 13. (2003) Answer Key 1. Fe = 26) ions is (a) 3 (b) 4 (c) 5 (d) 6 (2003) 32. How do we differentiate between Fe 3+ and Cr 3+ in group III? (a) By taking excess of NH 4 OH solution. 35. 29. 23. 9. 14. 17. 32. +5 (2002) results in liberation of some violet coloured fumes and droplets of a metal appear on the cooler parts of the test tube. 10. 12. (2003) (2002) 35. 25. La 3+ . (d) Both (b) and (c). chromium (Cr). 20. The number of delectrons retained in Fe 2+ (At. (b) (a) (d) (a) (c) (a) . 33. 24. Pm 3+ and Yb 3+ in increasing order of their ionic radii. (b) (b) (a) (a) (d) (b) 2. +3 (b) +2. 22. The atomic numbers of vanadium (V).67 d and fBlock Elements 28. However it becomes soluble (a) +2. Which of the following ions has the maximum magnetic moment? solid is (a) Mn 2+ (b) Fe 2+ (a) (NH 4 ) 2 Cr 2 O 7 (b) HgI 2 2+ (c) Ti (d) Cr 2+ (2002) (c) HgO (d) Pb 3 O 4 . 8. +4 (d) +3. (d) (c) (c) (b) (d) (a) 5. A red solid is insoluble in water. 16. (a) Yb 3+ < Pm 3+ < Ce 3+ < La 3+ (b) Ce 3+ < Yb 3+ < Pm 3+ < La 3+ (c) Yb 3+ < Pm 3+ < La 3+ < Ce 3+ (d) Pm 3+ < La 3+ < Ce 3+ < Yb 3+ 30. (c) By decreasing OH – ion concentration. 24. 18. Heating the red solid in a test tube (c) +3. The most stable ion is (a) [Fe(OH) 3 ] 3– (b) [Fe(Cl) 6 ] 3– second ionisation enthalpy? 3– 3+ (c) [Fe(CN) 6] (d) [Fe(H 2O) (2002) 6] (a) V (b) Cr (c) Mn (d) Fe (2003) 34. (a) (c) (b) (a) (d) (d) 3. 31. 27. 11. A reduction in atomic size with increase in atomic number is a characteristic of elements of (a) high atomic masses (b) d block (c) f block (d) radioactive series. +4 if some KI is added to water. 28. no. Arrange Ce 3+ . (2002) manganese (Mn) and iron (Fe) are respectively 23. (c) (b) (a) (c) (b) (b) 4. (d) (a) (a) (b) (b) (c) 6. 19. The red 36. 29. 36. 21. 7. 25 and 26. 15. Which one of these may be expected to have the highest 33. Most common oxidation states of Ce (cerium) are 31. 30. 26. (b) By increasing NH 4+ ion concentration. Hence shielding of 4f is Cl { more than 5f. the total lanthanide contraction. 5. etc; possess almost the same electron decreases in actinides. Therefore all these three subshells can participate. 17. An 4f orbital is NH 2 Hg 2Cl ® Hg + Hg + NH 4 Cl + 2H 2 O 2 + 2NH 4OH nearer to the nucleus than 5f orbitals. The black substance is a mixture of mercury and mercuric nucleus on the valency electrons due to presence of electrons amino chloride. atomic number weaker field ligand than NH 3 and C 2 O 42– therefore 3+ 3– 3+ increases by one unit and the addition of one electron occurs D oct [Co(H 2 O) 6 ] < D oct [Co(C 2 O 4 )] < [Co(NH 3 ) 6 ] at the same time in 4fenergy shell. (a) : Calomel on reaction with ammonium hydroxide turns 12.91 V E° Fe 2+ /Fe = – 0. 4. Thus.28 V (b) : When the transition metals are in their highest oxidation state. [NiX 4 ] 2– ® 4p × × × × × × × × sp 3 hybridisation Number of unpaired electrons = 2 Geometry = tetrahedral. 6d and 7s energy levels are of comparable 16. +4. +5. +6 and +7. (a) : The decrease in the force of attraction exerted by the black. 3. This brings the valence shell nearer to the nucleus and 10. The pairs of elements 4f orbitals (lanthanides) hence the hold of nucleus on valence such as ZrHf. Colour appears due to presence of unpaired felectrons which undergo ff transition. of second and third transition series resemble each other more 11. 2. 6. the elements energies. to the fact that the 5f. (c) : Actinoids show different oxidation states such as +2. Black .18 V E° Cr 2+ /Cr = – 0.e. NbTa. hence the size of atom or ion goes on decreasing as we move +3. For this reason the actinoides properties. the 4felectrons shield each other are given below : quite poorly from the nuclear charge. they no longer have tendency to give away electrons. (a) : As we proceed from one element to the next element in the lanthanide series. However +3 oxidation state is most in the series. On account of the very Common ligands in order of increasing crystal field strength diffused shapes of forbitals. i. in the inner shells is called shielding effect. The wide range of oxidation states of actinoids is attributed 9. (b) : Strong field ligand such as CN – . 14. 3d 4p 4s (a) (c) : Ferrous oxide is more basic. The sum of the successive reactions is equal to common among all the actinoids. the nuclear charge. (c) : In each vertical column of transition elements. H 2 O is a 15. 7.44 V Ni 2+ ® E° Co 2+ /Co = – 0. thus they are not basic but show acidic character and form anionic complexes. the effect of nuclear I – < Br – < Cl – < F – < OH – < C 2 O 4 2– < H 2 O < NH 3 < en < NO 2 – charge increase is somewhat more than the changed shielding < CN – effect. exhibit more number of oxidation states in general.68 1. 8. less volatile and less easily hydrolysed than ferric oxide. (b) : Ni ® (b) : E° Mn 2+ /Mn = – 1. (c) : Ln 3+ compounds are generally coloured in the solid state as well as in aqueous solution.84 54 7 1 2 64 Gd = [Xe] 4f 5d 6s (d) : Availability of 4f electrons does not result in the formation 3d 4s of compounds in +4 state for all the members of the series. JEE MAIN CHAPTERWISE EXPLORER (b) : Number of unpaired electrons in Fe 2+ is less than Mn 2+ . more ionic. (a) : As the distance between the nucleus and 5f orbitals closely than the elements of first and second transition series (actinides) is more than the distance between the nucleus and on account of lanthanide contraction. Number of unpaired electrons (n) = 2 (d) : The electronic configuration of m = n(n + 2) = 2(2 + 2) = 8 » 2. 13. usually produce low spin complexes and large crystal field splittings. MoW. (a) : 28 Ni ® [Ar] 3d 8 4s 2 so Fe 2+ is less paramagnetic than Mn 2+ . Magnetic moment = n n + 2 31. . i. 21. 2K 2CrO 4 + 2HNO 3 ® K 2 Cr 2O 7 + 2KNO 3 + H 2 O or. This causes a reduction in the size of the entire 4f n shell. . . . 36. As we move through the lanthanide series. 19. [Fe(CN) 6 ] 4– ® . (d) : 26 Fe = 1s 2 2s 2 2p 6 3s 2 3p 6 3d 6 4s 2 Fe 2+ = 1s 2 2s 2 2p 6 3s 2 3p 6 3d 6 The number of d electrons retained in Fe 2+ = 6. hence solubility product of Fe(OH) 3 is attained. potassium iodide forming a complex. (c) : With increase in atomic number i.85 Å) is the ionic radius of Lu 3+ . . . . (d) : When heated at red heat.69 d and fBlock Elements 28.e. The mutual shielding effect of f electrons is very little. . (a) : Mercury is such a metal which exists as liquid at room temperature. 144244 3 sp 3 Number of unpaired electrons = 3 . Cl – is most basic amongst all. Nos. Ce 4+ is stable due to 4 f 0 configuration. . . . Hg + I 2 HgI 2 –1 22. This is because of the electronic configuration of Cu + which is 3d 10 (completely filled) and of Cr + which is 3d 5 (halffilled). (b) : The second ionisation potential values of Cu and Cr are sufficiently higher than those of neighbouring elements. (d) : +6 +3 2– Cr 2 O 7 + 14H + + 6I – 2Cr 3+ + 7H 2 O + 3I 2 29. 33. (a) : According to their positions in the periods.e. (d) : NH 4 + ions are increased to suppress release of OH – ions. Thus the lowest 35. (b) : 4KI + 2CuSO 4 0 0 I 2 + Cu 2 I 2 + 2K 2 SO 4 +2. (a) : [MnCl 4 ] 2– ® - 3d 4s - . greater the number of unpaired electrons. 2CrO 4 2– yellow H + 2– Cr 2 O 7 + H 2 O orange these oxidation states are only stable in those cases where stable 4 f 0 . . 144244 3 sp 3 Number of unpaired electrons = 5 . 70 61 58 57 silver. (b) : violet vapours 32. (c) : The common stable oxidation state of all the lanthanides value (here 0. . . . the electron is to be removed from very stable configurations. . (n – 1)d 5 ns 2 can achieve the maximum oxidation state of +7. . - - - 4p . [CoCl 4 ] 2–® . (a) : Mn 2+ (3s 2 3p 6 3d 5 ) has the maximum number of unpaired electrons (5) and therefore has maximum moment.5 +2 I 2 + 2Na 2 S 2 O 3 –1 Na 2 S 4 O 6 + 2NaI (n – 1)d ns 23. the number of the principal shell increases and therefore. (lanthanide ions) decreases from La 3+ to Lu 3+ .. however increases by one at each step. (b) : A more basic ligand forms stable bond with metal ion. greater will be the HgI 2 + 2KI ® K 2 HgI 4 paramagnetic character. i. . The oxidation states of +2 and +4 are also exhibited and 27. K 2HgI 4 . . 4f electrons are being added one at each step. 4 f 7 or 4 f 14 configurations are achieved. . for the second ionisation potentials. Yb 3+ < Pm 3+ < Ce 3+ < La 3+ 25. This is due to the shape of the f orbitals. Colour of precipitate is different. the inward pull experienced by the 4f electrons increases. . AgNO 3 decomposes to metallic At. The nuclear charge. the ionic radii of Ln 3+ contraction. 18. in moving down a group. is +3. . (c) : +4 oxidation state of cerium is also known in solutions. (b) : Dilute nitric acid converts chromate into dichromate and H 2O. Ionic radii (pm) 86 98 103 106 2AgNO 3 ® 2Ag + 2NO 2 + O 2 Ionic size decreases from La 3+ to Lu 3+ due to lanthanide 26. 34. (d) : Due to lanthanide contraction. these values are in the order: 24. But in case of f block elements there is a steady decrease in atomic size with increase in atomic number due to lanthanide contraction. 20. HgI 2 on heating liberates I 2 gas. .e. the size of the atom increases. 144244 3 d 2 sp 3 Number of unpaired electrons = 0 30. Hence. (a) : Cu 2S + 2Cu 2O ® 6Cu + SO 2 This is an example of autoreduction. (b) : The precipitate of mercuric iodide dissolves in excess of where n = number of unpaired electrons. 5 g mol –1 ) is passed through a cation exchanger. Pt = 78) The magnetic moment (spin only) of [NiCl 4] (2007) (a) 1. The IUPAC name for the complex [Co(NO 2 )(NH 3 ) 5 ]Cl 2 is (a) nitritoNpentaamminecobalt(III) chloride Which one of the following has an optical isomer? (b) nitritoNpentaamminecobalt(II) chloride (a) [Zn(en) 2 ] 2+ (b) [Zn(en)(NH 3 ) 2 ] 2+ (c) pentaammine nitritoNcobalt(II) chloride (c) [Co(en) 3 ] 3+ (d) [Co(H 2 O) 4 (en)] 3+ (2010) (d) pentaammine nitritoNcobalt(III) chloride.82 BM 4. exhibit optical isomerism? (a) [Co(en)(NH 3 ) 2 Cl 2 ] + (b) [Co(en) 3 ] 3+ (c) [Co(en) 2 Cl 2 ] + (d) [Co(NH 3 ) 3 Cl 3 ] (2013) 2. respectively (a) 6 and 3 (b) 6 and 2 (c) 4 and 2 (d) 4 and 3 (2008) (2011) 11. (c) 2. Which among the following will be named as dibromidobis (ethylene diamine) chromium(III bromide? 10. (2011) 13. In Fe(CO) 5 .5 g mol ).: Fe = 26.84 BM. Which of the following has a square planar geometry? 2– 2– (a) [Cr(en) 2 Br 2]Br (b) [Cr(en)Br 4 ] – (a) [PtCl 4] (b) [CoCl 4] (c) [Cr(en)Br 2 ]Br (d) [Cr(en) 3 ]Br 3 (2012) (c) [FeCl 4 ] 2– (d) [NiCl 4 ] 2– . the Fe – C bond possesses (b) The complex is paramagnetic. 5.70 JEE MAIN CHAPTERWISE EXPLORER CHAPTER 18 COORDINATION COMPOUNDS 1. Which of the following compounds shows optical isomerism? Which of the following has an optical isomer? 2+ 2– (a) [Cu(NH 3) (b) [ZnCl 4] 4] (a) [Co(NH 3 ) 3 Cl] + (b) [Co(en)(NH 3 ) 2 ] 2+ (c) [Cr(C 2 O 4 ) 3 ] 3– (d) [Co(CN) 6 ] 3– (2005) 3+ 3+ (c) [Co(H 2O) (d) [Co(en) 2(NH (2009) 4 (en)] 3) 2] .78 g of AgCl (molar mass = 143. 6. Ni = 28.675 g of CoCl 3 ∙6NH 3 (molar mass = 267. (2006) (d) The complex gives white precipitate with silver nitrate solution. 7. The 14. How many EDTA (ethylenediaminetetraacetic acid) molecules Which of the following facts about the complex [Cr(NH 3 ) 6 ]Cl 3 are required to make an octahedral complex with a Ca 2+ ion? is wrong? (a) Six (b) Three (a) The complex involves d 2 sp 3 hybridisation and is (2006) (c) One (d) Two octahedral in shape. 2– is (At. Which of the following complex species is not expected to 9. 12.82 BM (b) 5. (d) 1.46 BM 3. The value of the ‘spin only’ magnetic moment for one of the following configurations is 2. (b) d 4 ( in weak ligand field) The formula of the complex is (At. 8. (2006) A solution contains 2. mass of Ag = 108 u) (c) d 3 ( in weak as well as in strong fields) (a) [CoCl(NH 3 ) 5 ]Cl 2 (b) [Co(NH 3 ) 6 ]Cl 3 (d) d 5 (in strong ligand field) (2005) (c) [CoCl 2 (NH 3) (d) [CoCl 3 (NH 3 ) 3 ] (2010) 4 ]Cl Which of the following pairs represents linkage isomers? 15. (c) ionic character (d) scharacter only. (a) pcharacter only (b) both s and p characters (c) The complex is an outer orbital complex.41 BM The coordination number and the oxidation state of the element E in the complex [E(en) 2 (C 2 O 4 )]NO 2 (where (en) is ethylene diamine) are. Which one of the following cyano complexes would exhibit the (a) [Cu(NH 3 ) 4][PtCl lowest value of paramagnetic behaviour? 4 ] and [Pt(NH 3 ) 4 ][CuCl 4 ] 3– 3– (b) [Pd(PPh 3 ) 2 (NCS) 2 ] and [Pd(PPh 3 ) 2 (SCN) 2 ] (a) [Cr(CN) 6] (b) [Mn(CN) 6] 3– 3– (c) [Co(NH 3 ) 5 (NO 3 )]SO 4 and [Co(NH 3 ) 5 (SO 4 )]NO 3 (c) [Fe(CN) 6 ] (d) [Co(CN) 6 ] (2005) (d) [PtCl 2 (NH 3 ) 4 ]Br 2 and [PtBr 2 (NH 3 ) 4 ]Cl 2 (2009) 16. The correct one is chloride ions obtained in solution were treated with excess of –1 (a) d 4 ( in strong ligand field) AgNO 3 to give 4. Co = 27. nos. 11.71 Coordination Compounds 17. (c) (c) (b) (a) 6. 9. One mole of the same complex reacts with two moles of AgNO 3 solution to yield two moles of (2005) AgCl (s). Answer Key 1. the oxidation state complex? of nickel is 4– 4– (a) [Fe(CN) 6] (b) [Mn(CN) 6] (a) –1 (b) 0 (c) [Co(NH 3 ) 6 ] 3+ (d) [Ni(NH 3 ) 6 ] 2+ (c) +1 (d) +2 (2003) [Atomic nos. In this context which of the following statements is (a) In acidic solutions hydration protects copper ions. 21. (b) Haemoglobin is the red pigment of blood and contains iron. (d) Copper hydroxide is an amphoteric substance. 24. 14. 8. 19. 21. (2002) (a) the number of ligands around a metal ion bonded by sigma bonds 27. 13. The oxidation state of Cr in [Cr(NH 3 ) 4 Cl 2 ] + is (a) +3 (b) +2 (c) +1 (d) 0 19. (c) In alkaline solutions insoluble Cu(OH) 2 is precipitated (c) Cyanocobalamine is B 12 and contains cobalt. 15. 22. incorrect? (b) In acidic solutions protons coordinate with ammonia (a) Chlorophylls are green pigments in plants and contain molecules forming NH 4+ ions and NH 3 molecules are not calcium. 7. (d) (b) (d) (d) (b) 2. available. (2004) 23. Co = 27. The structure of the complex is (a) [Co(NH 3 ) 5 Cl]Cl 2 Which one of the following has largest number of isomers? (b) [Co(NH 3 ) 3 Cl 2 ]∙2NH 3 + 2+ (a) [Ru(NH 3) (b) [Co(NH 3) 4Cl 2] 5Cl] (c) [Co(NH 3 ) 4 Cl 2 ]Cl∙NH 3 2+ + (c) [Ir(PR 3) (d) [Co(en) 2Cl 2H(CO)] 2] (d) [Co(NH 3 ) 4 Cl]Cl 2 ∙NH 3 (2003) (R = alkyl group. en = ethylenediamine) (2004) 24. 27. 17. (2002) 18. (c) (a) (c) (a) 5. 22. gives 3 moles of ions on dissolution in water. Fe = 26. (b) (b) (a) (b) . 12. 26. The IUPAC name of the coordination compound K 3 [Fe(CN) 6 ] is (a) potassium hexacyanoferrate (II) (b) potassium hexacyanoferrate (III) (c) potassium hexacyanoiron (II) (d) tripotassium hexacyanoiron (II) (2005) (c) the number of ligands around a metal ion bonded by sigma and pibonds both (d) the number of only anionic ligands bonded to the metal ion. 10. 25. One mole of the complex compound Co(NH 3 ) 5 Cl 3 . In the coordination compound. Ammonia forms the complex ion [Cu(NH 3 ) 4 ] 2+ with copper ions in alkaline solutions but not in acidic solutions. CH 3 – Mg – Br is an organometallic compound due to (b) the number of ligands around a metal ion bonded by pi (a) Mg – Br bond (b) C – Mg bond bonds (c) C – Br bond (d) C – H bond. 23. Ni = 28] (2004) 26. (d) CarboxypeptidaseA is an enzyme and contains zinc. 16. (a) (d) (a) (a) (b) 3. 20. What is the Coordination compounds have great importance in biological reason for it? systems. The type of isomerism present in nitropentamine chromium (III) chloride is The coordination number of a central metal atom in a complex (a) optical (b) linkage is determined by (c) ionization (d) polymerisation. which is soluble in excess of any alkali. K 4 [Ni(CN) 4 ]. 18. (2004) (2003) Which one of the following complexes is an outer orbital 25.: Mn = 25. 20. (c) (a) (d) (d) (b) 4. in the complex. 5. . This overlap is also called back donation of electrons In[Pd(PPh3)2(NCS)2]. (b) : In a metal carbonyl. en en Ni 2+ . [Co(en)2(NH3)2] 3+ exhibits optical isomerism. The hybridisation scheme is as shown in figure. about the central 2valent platinum ion in a square planar configuration.03 moles atom is formed when a vacant hybrid bond of the metal atom of AgCl on treatment with AgNO 3. (b) : No. the nickel is in +2 oxidation state and the ion has the electronic configuration 3d 8 . 2. 2. Formation of pbond is caused when a filled orbital of the metal (b) : Linkage isomerism is exhibited by compounds containing atom overlaps with a vacant antibonding p* orbital of C atom ambidentate ligand.01 267. it is an octahedral complex. (c) : EDTA. [M(AA) 3 ] n± . the metal carbon bond possesses both the s and pcharacter. Optical isomerism is common in octahedral complexes of the general formula. by metal atom to carbon. where (AA) is a symmetrical bidentate ligand. the formula of the containing a lone pair of electrons. It is an inner orbital complex. Thus among the given options. [M(AA) 2 a 2 ] n± . (c) : In the paramagnetic and tetrahedral complex 2– [NiCl 4] . (c) : The complex [Cr(NH 3 ) 6 ]Cl 3 involves d 2 sp 3 hybridization 9. the linkage of NCS and Pd is through N. complex is [Co(NH 3 ) 6 ]Cl 3 . Ca 2+ + [H 2 EDTA] 2– ® [Ca(EDTA)] 2– + 2H + 12. Thus. it implies that 3 chloride overlaps with an orbital of C atom of carbon monoxide ions are ionisable. In [Pd(PPh3)2(SCN)2].78 = 0.e.03 143. [Ar]3d 8 : NH3 3+ Co [NiCl 4 ] 2– : 3+ Co NH3 m= H3N H3N en en n( n + 2) BM = (2(2 + 2) = 8 Nonsuperimposable mirror images = 2. of moles of AgCl = 7. of CO. Therefore coordination number of the complex is 6 i. (c) : Optical isomers rarely occur in square planar complexes due to the presence of axis of symmetry.5 11. (d) : Optical isomerism is usually exhibited by octahedral compounds of the type [M(AA)2B2]. (a) : In 4coordinate complexes Pt. A sbond between metal and carbon Since 0. Oxidation number of E in the given complex is x + 2 × 0 + 1 × (–2) = +1 \ x = 3 4. the linkage of SCN and Pd is through S.82 BM (a) : In the given complex [E(en) 2 (C 2 O 4 )] + NO 2 – ethylene diamine is a bidentate ligand and (C 2 O 4 2– ) oxalate ion is also bidentate ligand. of moles of CoCl 3 ∙6NH 3 = No. The free acid H 4 EDTA is insoluble and the disodium salt Na 2 H 2 EDTA is the most used reagent.01 moles of the complex CoCl 3∙6NH 3 gives 0. 6. presence of plane of symmetry. [Ma 2 b 2 c 2 ] n± .675 = 0. Square planar complexes rarely show optical isomerism on accout of presence of axis of symmetry.5 4. only [Co(en) 3 ] 3+ shows optical isomerism. among the 10. Thus. as it involves (n – 1)d orbitals for hybridization.. [Mabcdef] n± .72 JEE MAIN CHAPTERWISE EXPLORER 1. [M(AA) 2 ab] n± and [M(AB) 3 ] n± . (d) : [Co(NH 3 ) 3 Cl 3] will not exhibit optical isomerism due to 8. (a) 3. which has four donor oxygen atoms and two donor nitrogen atoms in each molecule forms complex with Ca 2+ ion. the four ligands are arranged given options. 73 Coordination Compounds – M + + + C C M O O ox 3– (a) + – . . C – + – + M ox . O . – + Cr Cr ox (b) 3– ox ox ox – – M C . O . + + (a) The formation of the metal ¬ carbon sbond using an unshared 17. (b) : K 3 [Fe(CN) 6 ] Potassium hexacyanoferrate(III) 18. (a) : Let the oxidation state of Cr in [Cr(NH 3 ) 4 Cl 2 ] + = x The poverlap is perpendicular to the nodal plane of sbond. x + 4(0) + 2(–1) = +1 13. (d) : [Co(NO 2 )(NH 3 ) 5 ]Cl 2 x – 2 = +1 or, x = +1 + 2 = +3 pentaaminenitritoNcobalt(III) chloride 19. (d) : [Co(en) 2 Cl 2 ] + shows geometrical as well as optical 14. (a) : Spin only magnetic moment = n(n + 2) B.M. isomerism. Where n = no. of unpaired electron. 20. (a) : Chlorophyll are green pigments in plants and contains Given, n(n + 2) = 2.84 magnesium instead of calcium. or, n (n + 2) = 8.0656 pair of the C atom. (b) The formation of the metal ® carbon pbond. or, n = 2 21. (d) : Complex ion Hybridization of central ion In an octahedral complex, for a d 4 configuration in a strong field [Fe(CN) 6 ] 4– d 2 sp 3 (inner) 4– ligand, number of unpaired electrons = 2 [Mn(CN) 6 ] d 2 sp 3 (inner) 3+ [Co(NH 3 ) 6 ] d 2 sp 3 (inner) 15. (d) : [Co(CN) 6 ] 3– 2+ [Ni(NH 3) sp 3 d 2 (outer) Co ® [Ar] 3d 7 4s 2 6] 3+ 6 0 Co ® [Ar] 3d 4s 22. (a) : The number of atoms of the ligands that are directly bound 3d 4s 4p In presence of strong field ligand CN – pairing of electrons takes place. ×× ×× ×× ×× ×× ×× 14444444444442 444444444444434 d 2 sp 3 to the central metal atom or ion by coordinate bonds is known as the coordination number of the metal atom or ion. Coordination number of metal = number of s bonds formed by metal with ligands. 23. (a) : Given reactions can be explained as follows: [Co(NH 3 ) 5 Cl] 2+ + 2Cl – Þ 3 ions. [Co(NH 3 ) 5 Cl]Cl 2 [Co(NH 3 ) 5 Cl]Cl 2 + 2AgNO 3 ® [Co(NH 3 ) 5 Cl](NO 3 ) 2 + 2AgCl There is no unpaired electron, so the lowest value of 24. (b) : In acidic solution, NH forms a bond with H + to give NH + 3 4 paramagnetic behaviour is observed. ion which does not have a lone pair on N atom. Hence it cannot act as a ligand. 16. (c) : Optical isomers rarely occur in square planar complexes on account of the presence of axis of symmetry. Optical 25. (b) : Let the oxidation number of Ni in K [Ni(CN) ] = x 4 4 isomerism is very common in octahedral complexes having 1 × 4 + x × (–1) × 4 = 0 Þ 4 + x – 4 = 0 Þ x = 0 general formulae: [Ma 2 b 2 c 2 ] n± , [Mabc def] n± , [M(AA) 3 ] n± , [M(AA) 2 a 2 ] n± , 26. (b) : The nitro group can attach to metal through nitrogen n± n± as (–NO 2 ) or through oxygen as nitrito (–ONO). [M(AA) 2ab] and [M(AB) 3] (where AA = symmetrical bidentate ligand and AB = 27. (b) : Compounds that contain at least one carbonmetal bond unsymmetrical bidentate ligand). are called organometallic compounds. 74 JEE MAIN CHAPTERWISE EXPLORER CHAPTER 19 ENVIRONMENTAL CHEMISTRY 1. The gas leaked from a storage tank of the Union Carbide plant in Bhopal gas tragedy was (a) phosgene (b) methylisocyanate (c) methylamine (d) ammonia (2013) 2. Identify the wrong statement in the following. (a) Acid rain is mostly because of oxides of nitrogen and sulphur. (b) Chlorofluorocarbons are responsible for ozone layer depletion. (c) Greenhouse effect is responsible for global warming. (d) Ozone layer does not permit infrared radiation from the sun to reach the earth. (2008) 3. Answer Key 1. (b) 2. (d) 3. (c) The smog is essentially caused by the presence of (a) O 2 and O 3 (b) O 2 and N 2 (c) oxides of sulphur and nitrogen (d) O 3 and N 2 . (2004) 75 Environmental Chemistry 1. (b) violet rays from sun to reach the earth. 2. (d) : The thick layer of ozone called ozoneplanket which is 3. effective in absorbing harmful ultraviolet rays given out by the sun acts as a protective shield. It does not permit the ultra (c) : Photochemical smog is caused by oxides of sulphur and nitrogen. 76 JEE MAIN CHAPTERWISE EXPLORER PURIFICATION AND CHARACTERISATION OF ORGANIC COMPOUNDS CHAPTER 20 1. 2. 29.5 mg of an organic compound containing nitrogen was digested according to Kjeldahl’s method and the evolved ammonia was absorbed in 20 mL of 0.1 M HCl solution. The excess of the acid required 15 mL of 0.1 M NaOH solution for complete neutralization. The percentage of nitrogen in the compound is 3. (a) 29.5 (b) 59.0 (c) 47.4 (d) 23.7 (2010) The ammonia evolved from the treatment of 0.30 g of an organic compound for the estimation of nitrogen was passed in 100 mL of 0.1 M sulphuric acid. The excess of Answer Key 1. (d) 2. (c) 3. (c) acid required 20 mL of 0.5 M sodium hydroxide solution for complete neutralization. The organic compound is (a) acetamide (b) benzamide (c) urea (d) thiourea. (2004) In a compound C, H and N atoms are present in 9 : 1 : 3.5 by weight. Molecular weight of compound is 108. Molecular formula of compound is (a) C 2H (b) C 3H 6N 2 4N (c) C 6 H 8 N 2 (d) C 9 H 12 N 3 . (2002) 77 Purification and Characterisation of Organic Compounds 1. (d) : The % of N according to Kjeldahl’s method = w N 1 = Normality of the standard acid = 0.1 N w = Mass of the organic compound taken = 29.5 mg = 29.5 × 10 –3 g V = Volume of N 1 acid neutralised by ammonia = (20 – 15) = 5 mL. Þ 2. %N = 1.4 ´ 0.1 ´ 5 29.5 ´ 10 -3 Percentage of nitrogen in urea (NH 2 ) 2 CO 14 ´ 2 = 60 ´100 = 46.6 \ The compound must be urea. 1.4 ´ N1 ´ V = 23.7 (c) : Equivalents of NH 3 evolved 100 ´ 0.1´ 2 - 20 ´ 0.5 = 1 = 1000 1000 100 Percent of nitrogen in the unknown organic compound = 1 ´ 14 ´ 100 = 46.6 100 0.3 3. (c) : C H N 9 : 1 : 3.5 9 1 3.5 : : 12 1 14 3 1 1 : : 4 1 4 3 : 4 : 1 Empirical formula = C 3 H 4 N (C 3 H 4 N) n = 108 (12 ´ 3 + 1 ´ 4 + 14) n = 108 54n = 108 Þ n = 108/54 = 2 Molecular formula = C 6 H 8 N 2 78 JEE MAIN CHAPTERWISE EXPLORER CHAPTER 21 1. SOME BASIC PRINCIPLES OF ORGANIC CHEMISTRY The order of stability of the following carbocations is CH 2 CH 2 II III (a) III > I > II (c) II > III > I 2. 3. 4. 5. (b) III > II > I (d) I > II > III The correct decreasing order of priority for the functional groups of organic compounds in the IUPAC system of nomenclature is (a) – CONH 2 , – CHO, – SO 3 H, – COOH (b) – COOH, – SO 3 H, – CONH 2 , – CHO (c) – SO 3 H, –COOH, –CONH 2 , – CHO CH—CH 2 ; CH 3 —CH 2 —CH 2 ; I 9. (2013) How many chiral compounds are possible on monochlorination 10. The electrophile, EÅ attacks the benzene ring to generate the of 2methyl butane? intermediate scomplex. Of the following, which scomplex (a) 2 (b) 4 is of lowest energy? (c) 6 (d) 8 (2012) NO 2 NO 2 Identify the compound that exhibits tautomerism. + (a) (b) (a) 2Butene (b) Lactic acid + H (c) 2Pentanone (d) Phenol (2011) E H E Out of the following, the alkene that exhibits optical isomerism is NO 2 H (a) 2methyl2pentene (b) 3methyl2pentene H (c) 4methyl1pentene (d) 3methyl1pentene (2010) (c) (d) + E + E The IUPAC name of neopentane is (a) 2methylbutane (b) 2,2dimethylpropane (c) 2methylpropane (d) 2,2dimethylbutane (2008) 11. The absolute configuration of HO2 C (2009) 6. 7. 8. (2008) (d) – CHO, –COOH, – SO 3 H, –CONH 2 The number of stereoisomers possible for a compound of the molecular formula CH 3 – CH CH – CH(OH) – Me is (a) 3 (b) 2 (c) 4 (d) 6 (2009) CO2 H is HO H H (a) S, R (c) R, R (b) S, S (d) R, S OH (2008) The alkene that exhibits geometrical isomerism is 12. Which one of the following conformations of cyclohexane is (a) propene (b) 2methylpropene chiral? (c) 2butene (d) 2methyl2butene (2009) (a) Boat (b) Twist boat Arrange the carbanions, (c) Rigid (d) Chair (2007) (CH3 )3 C , CCl3 , (CH 3 ) 2CH , C6 H 5CH 2 13. Increasing order of stability among the three main conformations in order of their decreasing stability (i.e. eclipse, anti, gauche) of 2fluoroethanol is (a) eclipse, gauche, anti (a) C6 H5 CH 2 > CCl3 > (CH3 )3C > (CH 3 ) 2 CH (b) gauche, eclipse, anti (c) eclipse, anti, gauche (b) (CH3 ) 2 CH > CCl3 > C6 H5 CH 2 > (CH3 )3 C (d) anti, gauche, eclipse. (2006) (c) CCl3 > C6 H 5 CH 2 > (CH 3 ) 2CH > (CH 3 )3 C 14. The increasing order of stability of the following free radicals (d) (CH3 )3C > (CH 3 )2 CH > C6 H 5 CH 2 > CCl3 (2009) is 79 Some Basic Principles of Organic Chemistry . . . . (a) (CH 3 ) 2 CH < (CH 3 ) 3 C < (C 6 H 5 ) 2 CH < (C CH < (C 6 H 5 ) 3 C . . . . (b) (C 6 H 5 ) 3 C < (C 6 H 5 ) 2 CH < (CH 3 ) 3 C < (CH 3 ) 2 CH . . . . (c) (C 6 H 5 ) 2 CH < (C 6 H 5 ) 3 C < (CH 3 ) 3 C < (CH C < (CH 3 ) 2 CH . . . . C < (C 6 H 5 ) 2 CH (d) (CH 3 ) 2 CH < (CH 3 ) 3 C < (C 6 H 5 ) 3 C < (C H CH 3 H – C Å C 2 H 5 – CH – C 2 H 5 H (III) (2006) (IV) it is true that (a) all four are chiral compounds (b) only I and II are chiral compounds (c) only III is a chiral compound (d) only II and IV are chiral compounds. 15. CH 3 Br + Nu ® CH 3 – Nu + Br The decreasing order of the rate of the above reaction with nucleophiles (Nu – ) A to D is [Nu – = (A) PhO – , (B) AcO – , (C) HO – , (D) CH 3 O – ] (a) D > C > A > B (b) D > C > B > A (c) A > B > C > D (d) B > D > C > A. (2006) 20. The reaction : – – 16. The IUPAC name of the compound shown below is Cl (a) (b) (c) (d) Br (a) (b) (c) (d) (CH 3 ) 3 C – Br 2bromo6chlorocyclohex1ene 6bromo2chlorocyclohexene 3bromo1chlorocyclohexene 1bromo3chlorocyclohexene. H 2 O (2003) (CH 3 ) 3 C – OH elimination reaction substitution reaction free radical reaction displacement reaction. (2002) 21. Which of the following does not show geometrical isomerism? (a) 1,2dichloro1pentene (2006) (b) 1,3dichloro2pentene 17. The decreasing order of nucleophilicity among the nucleophiles (c) 1,1dichloro1pentene is (d) 1,4dichloro2pentene. (2002) (1) CH 3 C – O – (2) CH 3 O – 22. A similarity between optical and geometrical isomerism is that O (a) each forms equal number of isomers for a given compound O (b) if in a compound one is present then so is the other – H C S – O – (c) both are included in stereoisomerism (3) CN (4) 3 (d) they have no similarity. (2002) O (a) 1, 2, 3, 4 (b) 4, 3, 2, 1 23. Racemic mixture is formed by mixing two (c) 2, 3, 1, 4 (d) 3, 2, 1, 4 (a) isomeric compounds (2005) (b) chiral compounds 18. Due to the presence of an unpaired electron, free radicals are (c) meso compounds (a) chemically reactive (d) optical isomers. (2002) (b) chemically inactive (c) anions 24. Arrangement of (CH 3 ) 3 C–, (CH 3 ) 2 CH–, CH 3 CH 2 – when attached (d) cations (2005) to benzyl or an unsaturated group in increasing order of inductive effect is 19. Among the following four structures I to IV, (a) (CH 3 ) 3 C – < (CH 3 ) 2 CH – < CH 3 CH 2 – CH 3 O CH 3 (b) CH 3 CH 2 – < (CH 3 ) 2 CH – < (CH 3 ) 3 C – C 2 H 5 – CH – C 3 H 7 CH 3 – C – CH – C 2 H 5 (c) (CH 3) 2CH – < (CH 3) 3C – < CH 3CH 2 – (I) (II) (d) (CH 3) C – < CH CH – < (CH ) CH –. (2002) 3 3 2 3 2 Answer Key 1. (a) 7. (c) 13. (a) 2. (a) 8. (c) 14. (a) 3. 9. 15. (c) (c) (a) 4. (d) 10. (c) 16. (c) 5. 11. 17. (b) (c) (d) 6. (c) 12. (b) 18. (a) 19. (b) 20. (b) 21. (c) 22. (c) 23. (d) 24. (b) (d) : 3Methyl1pentene exhibits optical isomerism as it has an asymmetric Catom in the molecule. only 2butene exhibits cistrans isomerism. (c) : 5. (c) : The given compound has a C (*) carbon. (b) : neopentane or 2. Also. exhibit geometrical isomerism. Out of four possible isomers only I and III are chiral. greater is its stability. (a) : * H3 C – HC CH – C H(OH) – Me H C H3 C C group and one chiral H H C C CH(OH)Me Me(OH)CH C H CH3 d. 2pentanone exhibits tautomerism. 7.2dimethylpropane + E This structure will be of lowest energy due to resonance stabilisation of +ve charge. 2. (a) : Greater the number of resonating structures a carbocation 6. possess. the presence of electronwithdrawing NO 2 group will destabilize the +ve charge and hence they will have greater energy. dynamic equilibrium is known as tautomerism. 2° carbanions are more stable. 3. . In all other three structures. (c) : The groups having +I effect decrease the stability while groups having –I effect increase the stability of carbanions. Compounds in which the two groups attached to a double bonded carbon are different. out of 2° and 3° carbanions. thus. thus the decreasing order of stability is : CCl3 > C6 H5 CH 2 > (CH3 ) 2 CH > (CH 3 )3 C. 4.80 JEE MAIN CHAPTERWISE EXPLORER 1. (c) : When two groups attached to a double bonded carbon atom are same. 9. the compound does not exhibit geometrical isomerism. (c) : The type of isomerism in which a substance exists in two readily interconvertible different structures leading to a 8. (c) : The order of preference of functional groups is as follows: –SO 3 H > –COOH > – COOR > – COX > –COCl > –CONH 2 > –CHO > –CN > C O > –OH > –SH > –NH 2 > C C > –C C– > –NO 2 > –NO > –X H 10. Benzyl carbanion is stabilized due to resonance. l isomers of cis form \ Total stereoisomers = 4. more basic the species increases the inductive effect increases. . but different spatial arrangement of atoms phenyl group attached to the carbon atom holding the odd or groups. (d) : Strong bases are generally good nucleophile. 2 (1) 16. Hence basicity as well as nucleophilicity order is 2 O 5 1 CH 3 O – > CH 3 4 6 C O – > H 3 C S O O 3 O – – Now CN – is a better nucleophile than CH 3O . (c) : Both involves compounds having the same molecular and structural formulae. dextro and laevorotatory accordingly. If the nucleophilic atom or the centre is the same. Further as the number of 22.e. (c) : C C – CH 2 – CH 2 – CH 3 that the steric strain is maximum. Benzyl free radicals are stabilised by resonance and hence are more stable than alkyl free radicals. H 19.e. electron increases.e. 15. the order of stability of free radical is as follows: Identical groups (Cl) on C l will give only one compound. . In fully eclipsed conformation F and OH groups are so close 21. Only I and II are chiral compounds. as number of – CH 3 group nucleophilicity parallels basicity. 3 2 3 3 6 5 2 6 5 3 mixture.e. stronger is the active as it does not have any plane of symmetry. (b) : –CH 3 group has +I effect. R ) 17. (a) : If the nucleophilic atom or the centre is same. optical isomers is termed as racemic mixture or dl form or (±) (CH ) CH < (CH ) C < (C H ) CH < (C H ) C i. This order can be easily explained on the general concept that a weaker acid has a stronger conjugate base. (b) : A chiral object or compound can be defined as the one that is not superimposable on its mirror image.. . 13. the stability of a free radical increases 23. (a) : Free radicals are highly reactive due to presence of an unpaired electron. They readily try to pairup the odd electrons. nucleophile. i. or we can say that all the four groups attached to a carbon atom must be different. Hence it does not show geometrical isomerism. (b) : This is an example of nucleophilic substitution reaction. (d) : An equimolar mixture of two i. H skew or gauche CH 3 CH 3 * H * H (II) CH CO – C – C (I) C 2 H 5 – C – C 3 2 5 3 7 H H 20. tertiary > secondary > primary. (a) : Hence decreasing order of nucleophilicity is CH 2 – CH 2 F CN – > CH 3 O – > CH 3 OH C O – > H 3 C O S O – 2fluoroethanol F H F H H H H HO HO H OH H H H eclipsed anti or staggered F OH H H H H O O F H 18. (c) : COOH COOH HO2 C CO2 H HO HO H H H OH C C 2 HO H H 1 OH Cl 3 OH H 1 3 COOH COOH – > HO – > PhO – > AcO – CH 3O Here. – Br + OH – (CH 3 ) 3 C OH + Br (CH 3 ) 3 C The anti conformation is most stable in which F and OH groups Nucleophile Leaving group Substrate are far apart as possible and minimum repulsion between two Cl H groups occurs. more basic the species. O is the same in all these species. fully eclipsed . stronger is the nucleophile. (c) : (6) (2) (5) (3) (4) 3bromo1chlorocyclohexene Br (R . nucleophilicity 12.81 Some Basic Principles of Organic Chemistry 11. . i. 24.e. the nucleophilic atom i. hence this conformation is Cl most unstable. (a) : On the basis of hyperconjugation effect of the alkyl groups. bond. The order of stability of these conformations is Condition for geometrical isomerism is presence of two different anti > gauche > partially eclipsed > fully eclipsed atoms of groups attached to each carbon atom containing double 14. (b) : The twist boat conformation of cyclohexane is optically parallels basicity. (2007) One mole of a symmetrical alkene on ozonolysis gives two moles of an aldehyde having a molecular mass of 44 u. A gaseous hydrocarbon gives upon combustion 0. Which branched chain isomer of the hydrocarbon with molecular mass 72 u gives only one isomer of mono substituted alkyl halide? (a) Neopentane (b) Isohexane 9. 5. Which of the following molecules is expected to rotate the (b) CH 3 CH 2CH C CCH 2CH planepolarised light? 2 2 CH 3 (c) CH 3 CH 2C CH CHO COOH CHCH 3 (d) CH 3 CH (2008) H (b) HO H (a) H 2 N In the following sequence of reactions. The empirical formula of the hydrocarbon is (a) C 7 H 8 (b) C 2 H 4 (c) C 3 H 4 (d) C 6 H 5 (2013) 2.2dibromopropane? H 6. (2007) (a) CH 4 (b) CH 3 – CH CH 2 11. the alkene affords CH 2 OH H the compound B O H O B. (2012) Toluene is nitrated and the resulting product is reduced with tin and hydrochloric acid. Presence of a nitro group in a benzene ring (a) deactivates the ring towards electrophilic substitution (b) activates the ring towards electrophilic substitution (c) renders the ring basic (d) deactivates the ring towards nucleophilic substitution.82 JEE MAIN CHAPTERWISE EXPLORER CHAPTER HYDROCARBONS 22 1.72 g of water 8. (c) Neohexane (d) Tertiary butyl chloride (2012) 3. and 3. 4. The reaction mixture so formed contains (a) mixture of o and mbromotoluenes (b) mixture of o and pbromotoluenes (c) mixture of o and pdibromobenzenes (d) mixture of o and pbromoanilines. (2010) (c) benzyl chloride C – H produces The treatment of CH 3MgX with CH 3C (d) o and pchlorotoluene.08 g of CO 2. The reaction of toluene with Cl 2 in presence of FeCl 3 gives predominantly alkene is (a) mchlorobenzene (a) ethene (b) propene (b) benzoyl chloride (c) 1butene (d) 2butene. (2008) (a) CH 3 – CH H (2008) C – CH 3 (b) CH 3 – C CH 2 + HBr ® CH + 2HBr ® (c) CH 3 CH CHBr + HBr ® The hydrocarbon which can react with sodium in liquid (d) CH CH + 2HBr ® (2007) ammonia is (a) CH 3 CH 2 C CCH 2 CH 3 12. The 10. Which of the following reactions will yield (c) CH 3 C C – CH 3 2. The product so obtained is diazotised and then heated with cuprous bromide. 2Hexyne gives trans2hexene on treatment with (a) Li/NH 3 (b) Pd/BaSO 4 (c) LiAlH 4 (d) Pt/H 2 (d) CH3 – C 7. CH3 CH CHCH 3 3 A 2 Zn The compound B is (a) CH 3CHO (b) CH 3CH 2CHO (c) CH 3 COCH 3 (d) CH 3 CH 2COCH 3 (c) (2008) (d) SH H 2 N NH 2 H Ph H Ph (2007) . (2003) monochlorinated compounds is (a) nhexane (b) 2.2dimethylpentane 4.4dimethyl5. The IUPAC name of (a) (b) (c) (d) 19. (b) 18. (2007) 14.3dimethylbutane 24. Which of these will not react with acetylene? conditions (a) NaOH (b) ammonical AgNO 3 (d) 1bromo2butene under kinetically controlled conditions. (2002) conditions (c) 3bromobutene under thermodynamically controlled 26. (d) 16. (b) 17. Which one of the following has the minimum boiling point? (a) nButane (b) 1Butyne formation of (c) 1Butene (d) Isobutene (2004) (a) primary alcohol (b) secondary or tertiary alcohol 23. 25. 19. 13. 14. (b) (d) (d) (a) 9. 26.5diethyl4. (a) 15.83 Hydrocarbons 13. This (d) mixture of secondary and tertiary alcohols. On mixing a certain alkane with chlorine and irradiating it with (c) mixture of primary and secondary alcohols ultraviolet light.5diethylpentane 5. c) 7. (d) 10. What is the product when acetylene reacts with hypochlorous (a) 3bromobutene under kinetically controlled conditions acid? (b) 1bromo2butene under thermodynamically controlled (a) CH 3 COCl (b) ClCH 2 CHO (c) Cl 2 CHCHO (d) ClCHCOOH. (a) 2.4dimethylpentane. Acid catalyzed hydration of alkenes except ethene leads to the 22. (c) 12. (d) 21. it forms only one monochloroalkane. (b) 24. (2007) 15. Which one of the following is reduced with zinc and hydrochloric acid to give the corresponding hydrocarbon? (a) Ethyl acetate (b) Acetic acid (c) Acetamide (d) Butan2one (2004) 21. (a) (a) (b) (c) 8. 20. The compound formed as a result of oxidation of ethyl benzene by KMnO 4 is (a) benzyl alcohol (b) benzophenone (c) acetophenone (d) benzoic acid. Reaction of one molecule of HBr with one molecule of 1. (2005) alkane could be (a) propane (b) pentane 17. (b) 22. Amongst the following compounds. (a) 11. (a. the isomer which can give two (c) isopentane (d) neopentane. (2005) is 3ethyl44dimethylheptane 1. Butene1 may be converted to butane by reaction with (c) 2. 3 butadiene at 40°C gives predominantly 25. Me Å Me N Et OH nBu D 20. the optically active alkane having lowest molecular mass is (a) CH 3 – CH 2 – CH 2 – CH 3 CH 3 (b) CH 3 – CH 2 – CH – CH 3 The alkene formed as a major product in the above elimination reaction is (a) (b) CH 2 Me Me (c) CH 3 – C – CH 2 C 2 H 5 Me (d) CH 3 – CH 2 – C (2006) (d) (c) H CH (2004) 16. 2Methylbutane on reacting with bromine in the presence of sunlight gives mainly (a) 1bromo2methylbutane (b) 2bromo2methylbutane (c) 2bromo3methylbutane (d) 1bromo3methylbutane. (a) 3. (d) . Of the five isomeric hexanes. (c) 4. (2003) 18.1diethyl2. (d) 5. (c) Na (d) HCl (2002) (2005) Answer Key 1. (2005) (a) Zn HCl (b) Sn HCl (c) Zn Hg (d) Pd/H 2 . (d) 6.2dimethylbutane (d) 2methylpentane. (b) 23. 08 CH 3 O O Monoozonide x 0. mass of R = 44 – 29 = 15 CH 3 CH 3 CH 3 This is possible. (a) : The complete reaction sequence is as follows H H y = 0. JEE MAIN CHAPTERWISE EXPLORER (a) : Moles of water produced = Moles of CO 2 produced = 0.07 7 = = y 0.04 18 5. FeCl 3 Cl \ The aldehyde is CH 3 CHO and the symmetrical alkene is + Cl 2 + CH 3 HCCHCH 3 . CuBr/D NH 3 or EtNH 2 at low temperature or LiAlH 4 as reducing agent (antiaddition). sodium alkylides. CH 3 HNO3/H 2SO 4 CH 3 Sn/HCl Nitration Toluene NO 2 NH 2 p nitrotoluene CH 3 3. only when R is —CH 3 group.07 44 Equation for combustion of an unknown hydrocarbon. is CMgX (c) : Terminal alkynes react with sodium in liquid ammonia to yield ionic compounds i.07 and 2 H O 3 C C CH 3 H3 C H O C C H3 C \ y = 0. 3. (a) : As the molecular mass indicates it should be pentane 8. Þ R + 12 + 1 + 16 = 44 10. 2. æ yö y C x H y + ç x + ÷ O 2 ® xCO 2 + H 2 O è 4ø 2 7. (b) : The reaction sequence is as follows : and neopentane can only form one mono substituted alkyl CH 3 halide as all the hydrogens are equivalent in neopentane.08 = 0. (d) : Mol. NaNO2 /HCl 2. Br + O – O N N – O O N – O O O N + From the resonating structures + of it can be seen that the Molecular mass of RCHO = 44 nitrogroup withdrawn electrons from the rings and hence it deactivates the benzene ring for further electophilic substitution.04 Þ x = 0. C x H y (a) : Grignard reagent reacts with compounds having active C–H ® CH 4 + CH 3 C CH 3 MgX + CH 3 C 6. O N O (d) : o bromotoluene (a) : O 4. or acidic hydrogen atom to give alkane.72 = 0.84 1. CH 3 Br p bromotoluene 9. (a.e. ochlorotoluene Cl pchlorotoluene .08 8 Zn–H2 O Cleavage 2CH3 CHO + ZnO \ The empirical formula of the hydrocarbon is C 7 H 8 . c) : For trans products we should take Na or Li metal in 1. it is the less sterically hindered bhydrogen that is removed and hence less substituted alkene is the major product. (a) : CH 3 – CH 2 – CH 2 – C – C – CH 2 – CH 3 H 3 C – C – C – CH 3 CH 3 CH 2 – CH 3 3ethyl4. 2dibromopropane 7 C CH 3 OH (secondary alcohol) HBr CH 3 CH 3 CH 3 (3º) – HBr CH 1 CH 3 CH 3 13. regardless of length is oxidised to –COOH. the 1.(CH 3) 2CHCH 2CH 3 127°C 2methylbutane (CH 3) 2C(Br)CH 2CH 3 2bromo2methylbutane . 2dibromopropane CH 3 CH CHBr OH – CH 3 CH 3 C OH CH CH 3 CH 3 Br CH 3 CH 3 CH 3 (tertiary alcohol) 17. (d) : H 3 C – C – C – CH 3 Ethyl benzene Benzoic acid Cl H When oxidises with alkaline KMnO 4 or acidic Na 2Cr 18.85 Hydrocarbons 11. CHO 12. 6 CH CH 3 – CH – CH 2 Br 1. at acarbon. Br 2 /hu So. (b) : The reactivity order of abstraction of H atoms towards bromination of alkane is 3°H > 2°H > 1°H. CH 3 5 4 3 2 H 2.3dimethylbutane Due to the presence of chiral carbon atom. it is optically active. 1dibromoethene CH 3 CH 3 – CH CH 2 HBr CH 3 – CH – CH 3 CH 3 Br + C CH CH 3 migration + CH 3 C (2º) CH 3 2bromopropane CH 3 OH Br CH 3 – C CH HBr CH 3 – C CH 2 CH 3 – C – CH 3 Br Br 2. (b) : CH 3 C H + CH 2 CH CH 3 Br 1. 4dimethylheptane H Cl 2 CH 2 Cl CH 3 H 3 C – C — C – CH 3 H H COOH CH 2 – CH 3 H + CH 3 CH 3 KMnO 4 14. (b) : HO – C* – H CH 3 CH 3 H 3 C – C – C – CH 3 CH 2 OH H 2. (d) : CH 3 D CH – CH CH 2 HBr 40° CH 3 Me N Et – OH nBu In Hofmann elimination reaction.2addition Therefore. CH 2 more hindered bH 15.3dimethylbutane has two types of hydrogen atoms so on monochlorination gives only two monochlorinated compounds.4addition product is thermodynamically controlled product entire side chain (in benzene homologues) with atleast one H and formed at comparatively higher temperature. 19. (b) : CH 3 Br HC CH HBr HBr H – C CH 2 H – C – CH 3 Br 16. (b) : 1. (b) : The number of monohalogenation products obtained from any alkane depends upon the number of different types of hydrogen it contains.4addition + CH 3 CH(Br) – CH CH 2 + less hindered bH CH 2 (Br)CH CHCH 3 1. hence it is expected to rotate plane of polarized light. 1bromo2butene will be the main product under thermodynamically controlled conditions.2addition product is kinetically controlled product while 2O 7. 1. C 2 H 5 Optically active due to presence of chiral carbon atom.86 JEE MAIN CHAPTERWISE EXPLORER 20. the lower is the boiling point. CH(OH) CHOH 2 HOCl HOCl Thenalkanes have larger surface area in comparison to branched 25. (c) : CH 3 – C * three types of hydrogen (three monohalogenation product) one type of hydrogen (one monohalogenation product) neopentane Thus the given alkane should be neopentane. Thus. Compound containing only one type of hydrogen gives only one monohalogenation product. . Acetylene reacts with the other three as: CH CNa Na NH 3(l) CH HCl CH CH 2 HCl CHCl CH 3 CHCl 2 [AgNO 3 + NH 4 OH] CAg CAg + NH 4 NO 3 white ppt. (d) : The number of monohalogenation products obtained from 26. CH 3 CH 2 CH 3 two types of hydrogen (two monohalogenation product) propane CH 3 CH 2 CH 2 CH 2 CH 3 pentane two types of hydrogen (two monohalogenation product) be the stronger acid H 2 O and the stronger base CH 3 C C . (d) : Among the isomeric alkanes. O CH 3 – CH 2 – C – CH 3 Zn Hg Butan2one HCl CH 3 – CH 2 CH 2 – CH 3 Butane CH 3 CH 3 – CH – CH 2 – CH 3 isopentane CH 3 H 3 C – C – CH 3 CH 3 H 21. therefore. (c) : CH CH CHCl CHCl 2 chain isomers (as the shape approaches that of a sphere in the CHO branched chain isomers). they have lower boiling CHCl 2 points in comparison to straight chain isomers. intermolecular forces are weaker –H 2 O in branched chain isomers. acid and amide. 24. (d) : Butan2one will get reduced into butane when treated with zinc and hydrochloric acid following Clemmensen reaction whereas Zn/HCl do not reduce ester. The greater the Butane branching of the chain. 23. (d) : H 3 C – CH 2 – CH CH 2 Pd/H 2 Butene 1 22. the normal isomer has a higher H 3 C – CH 2 – CH 2 – CH 3 boiling point than the branched chain isomer. (a) : Acetylene does not react with NaOH because product would any alkene depends upon the number of different types of hydrogen it contains. Oxidation of (A) gives an acid (B). What is DDT among the following? (2006) (a) A fertilizer (b) Biodegradable pollutant 12. secondary or tertiary. The structure of the compound that gives a tribromoderivative on (c) Nonbiodegradable pollutant treatment with bromine water is (d) Greenhouse gas (2012) CH 3 Consider the following bromides : CH 2 OH (a) (b) OH The correct order of S N 1 reactivity is (a) A > B > C (b) B > C > A (c) B > A > C (d) C > B > A CH 3 CH 3 OH (c) (2010) (d) OH (2006) . C 8H 7.87 Organic Compounds Containing Halogens ORGANIC COMPOUNDS CONTAINING HALOGENS CHAPTER 23 1. 4. A solution of (–) – 1 – chloro – 1 – phenylethane in toluene racemises slowly in the presence of a small amount of SbCl 5 . is (a) CH 3 Cl (b) (C 2 H 5 ) 2 CHCl (c) (CH 3 ) 3 CCl (d) (CH 3 ) 2 CHCl (2008) 9. (c) 1phenylcyclopentene Which alcohol reacts fastest and by what mechanism? (d) 3phenylcyclopentene. Which of the following is the correct order of decreasing S N 2 reactivity? (a) R 2CH X > R 3C X > RCH 2 X (b) RCH X > R 3 C X > R 2 CH X (c) RCH 2 X > R 2CH X > R 3C X (d) R 3 C X > R 2 CH X > RCH 2 X (X is a halogen) (2007) CH 2 Br CH 2 Br (a) (b) CH 3 CH 3 CH 2 Br C 2 H 5 (c) (2013) (d) Br CH 3 2. 5. The organic chloro compound. C 8 H 6 O 4 . Compound (A). (2006) (a) Tertiary alcohol by S N 2. Identify the compound (A). (a) by heating phenol with HF and KF (d) Secondary alcohol by S N 2. Fluorobenzene (C 6 H 5 F) can be synthesised in the laboratory (c) Tertiary alcohol by S N 1. gives a white precipitate when warmed with alcoholic AgNO 3 . (b) Secondary alcohol by S N 1. 6. Which of the following on heating with aqueous KOH produces acetaldehyde? (a) CH 3COCl (b) CH 3CH 2 Cl (c) CH 2 ClCH 2Cl (d) CH CHCl (2009) 3 2 8. (2013) (b) from aniline by diazotization followed by heating the Iodoform can be prepared from all except diazonium salt with HBF 4 (a) isopropyl alcohol (b) 3methyl2butanone (c) by direct fluorination of benzene with F 2 gas (c) isobutyl alcohol (d) ethyl methyl ketone. (2012) (d) by reacting bromobenzene with NaF solution. which shows complete stereochemical inversion during a S N 2 reaction. due to the formation of 10. Reaction of trans2phenyl1bromocyclopentane on reaction (a) free radical (b) carbanion with alcoholic KOH produces (c) carbene (d) carbocation. 9Br. (B) easily forms anhydride on heating. 11. (2013) (a) 4phenylcyclopentene An unknown alcohol is treated with the “Lucas reagent” to (b) 2phenylcyclopentene determine whether the alcohol is primary. 3. (d) 13. The structure of the major product formed in the following reaction is NaCN DMF Cl CN (b) Cl NC CN I Cl (c) CN (d) (2006) I CN (2004) 18. Bottles containing C 6 H 5 I and C 6 H 5 CH 2 I lost their original labels. The end (a) alkenes (b) alkyl copper halides solution in each tube was made acidic with dilute HNO 3 and (c) alkanes (d) alkenyl halides.88 JEE MAIN CHAPTERWISE EXPLORER 13. Alkyl halides react with dialkyl copper reagents to give taken in a test tube and boiled with NaOH solution. (d) (d) 11.3dichloropentane (d) 2hydroxypropanoic acid (2004) 14. (2003) (a) 1chloropentane (b) 2chloropentane Answer Key 1. A and B were separately 15. (d) 19. (c) 5. (c) . 15. (2005) then some AgNO 3 solution was added. Tertiary alkyl halides are practically inert to substitution by yellow precipitate. 21. (c) (c) (a) 10. Acetyl bromide reacts with excess of CH 3 MgI followed by treatment with a saturated solution of NH 4 Cl gives (a) acetone (b) acetamide (c) 2methyl2propanol (d) acetyl iodide. Substance B gave a 16. Which of the following will have a mesoisomer also? (a) 2chlorobutane (b) 2. 17. (c) 4. (2004) (c) predominantly 1butene (d) predominantly 2butyne. (2005) 21. Which of the following compounds is not chiral? (d) Addition of HNO 3 was unnecessary. (b) 3. 16. (a) 18. (b) 9. (b) 20. The compound formed on heating chlorobenzene with chloral formation of in the presence of concentrated sulphuric acid is (a) equimolar mixture of 1 and 2butene (a) gammexene (b) DDT (b) predominantly 2butene (c) freon (d) hexachloroethane. They were labelled A and B for testing. (2005) (b) A was C 6 H 5 CH 2 I (c) B was C 6 H 5 I 17. Which one of the following statements is S N 2 mechanism because of true for this experiment? (a) insolubility (b) instability (a) A was C 6 H 5 I (c) inductive effect (d) steric hindrance.3dichlorobutane (c) 2. (a) 14. (d) 8. (2004) I (a) (c) 1chloro2methylpentane (d) 3chloro2methylpentane 19. Elimination of bromine from 2bromobutane results in the 20. (b) 2. (b) (a) 12. (c) 6. (a) 7. in the transition state there will be five groups attached to the carbon atom at which reaction occurs. and hence can give tribromoderivative. (d) : KOH 5. Compound (B). step. (b) : S N 1 reaction rate depends upon the stability of the 3phenylcyclopentene carbocation. turbidity appears immediately with tertiary alcohol by S N 1 mechanism. and the bulkier the group. as we move from 1° alkyl halide to 3° alkyl halide. 4. (c) : In Lucas test. a d– HO – or d d– X C X b d (transition state) a C + X – b d HO group In S N 2 reaction. . (c) H Ph Ph 6. (c) : The compounds with form iodoform. the order of reactivity is thus. H 10. the more the reaction will be hindered sterically. because it has +I effect increases which makes the carbon bearing halogen less positively polarised and hence less readily attacked by the two ortho and one para position free with respect to OH group nucleophile. the nucleophile attacks from back side resulting in the inversion of molecule. (c) : S N 2 mechanism occurs as a 1. Thus there will be crowding in the transition state. as carbocation formation is the rate determining It follows E2 mechanism. Also. So the decreasing order of reactivity of halide is RCH 2 X > R 2 CHX > R 3 CX (primary) (secondary) (tertiary) Thus all the compounds except isobutyl alcohol will form Br H iodoform. (a) : Since the compound on treatment with Br 2 water gives a tribromoderivative. (d) : + N NCl – NH 2 0°C HBF 4 NaNO 2 + HCl –HCl aniline diazotization benzene diazonium chloride + – C 6 H 5 N 2 BF 4 benzene diazonium tetra fluoroborate D C 6 H 5 F + BF 3 + N 2 fluorobenzene 8. the next stable carbocation is formed from reactions take place when the two groups to be eliminated are (C).89 Organic Compounds Containing Halogens 9. it is a 2° carbocation. (b) : 7. 11. they are attached i. (a) : In S N 2 reactions. (a) : OH + C b (nucleophile) 2. therefore it must be mcresol.e. (d) : A carbocation intermediate is formed during racemisation. Hence S N 2 reaction is favoured by small groups on the carbon atom attached to halogens. (A) forms the least stable 1° trans and lie in one plane with the two carbon atoms to which carbocation. E2 reactions are stereoselectively trans. forms a 2° allylic carbocation which is Hughes and Ingold proposed that bimolecular elimination the most stable. 3. the crowding increases and 12. alkyl halide reacts with lithium dialkyl cuprate. Nu C X – H – C – Cl 3 OH H – C – Cl 4 CH 3 – C – COOH H – C – H H CH 3 2.3dichlorobutane have meso isomer due to the presence of plane of symmetry.3dichloropentane 2hydroxypropanoic acid 2.2 bis (pchlorophenyl) ethane or DDT b Cl H H H H O Mg I CH 3 OH CH 3 1 1butene (20%) C C H – C – Cl CH 3 CH CHCH 3 a CH 3 NH 4 Cl H – C – Cl 2butene (80%) d CH 3 CH 3 a CH 3 C 2methyl2propanol 19.1trichloro2. 17. 20. (d) : Nu + O OH Br I – C Cl 2. R 2 CuLi + R¢X ® RR¢ + RCu + LiX X 1 2 2 In elimination reaction of alkyl halide major product is obtained according to Saytzeff’s rule. B C 6 H 5 CH 2 I NaOH C 6 H 5 CH 2 ONa HNO 3 /H + C 6 H 5 CH 2 OH AgNO 3 Yellow precipitate .90 JEE MAIN CHAPTERWISE EXPLORER CH 3 CH 3 Br 2 H 2 O Br OH mcresol 18. Thus there will be crowding in the transition state. (b) : 14. (d) : CH 3 KOH (alc. the alkene which is most substituted one predominates. in the transition state. d H a C Nu + X – d Cl CCl 3 CH O + H Cl In an S N 2 reaction. (a) : To be optically active the compound or structure should possess chiral or asymmetric centre.6tribromo3methylphenol CH 2 CH 3 No yellow precipitate Thus A must be C 6 H 5 I. (b) : DDT is prepared by heating chlorobenzene and chloral with concentrated sulphuric acid. there will be five groups attached to the carbon atom at which reaction occurs. H Cl H H H * H – C – C – C – C – C – H H – C – C – C – C – C – H H H H H H H H H H H 1chloropentane 2chloropentane A C 6 H 5 I NaOH C 6 H 5 ONa HNO 3 /H + C 6 H 5 OH AgNO 3 Cl CH 3 H H H H CH 3 Cl H H * * * H – C – C – C – C – C – H H – C – C – C – C – C – H H H H H H H H H H H 1chloro2methylpentane 3chloro2methylpentane Cl H 2 SO 4 – H 2 O CCl 3 CH Cl 1. (c) : In Corey House synthesis of alkane. (b) : CH 3 CH 2 CHCH 3 C CH 3 CH 2 CN NaCN Cl DMF Br b CH 3 CH 3 Mg I CH 3 Mg I I 16.3dichlorobutane CH 3 15.) 2bromobutane Cl O Mg I + CH 3 CH 2 CH CH 2 b 1 2 CH 3 H – C – H H – C – Cl 3 CH 3 CH 3 2chlorobutane 2.4. (a) : of bulky groups make the reaction sterically hindered. (c) : Br 13.1. which states that when two alkenes may be formed. and presence 21. 4. (2010) 7. 4. The (c) nbutyl alcohol (d) npropyl alcohol. HBr reacts with CH 2 CH – OCH 3 under anhydrous conditions (c) ethyl chloride (d) ethyl ethanoate. 4. . The major product obtained on interaction of phenol with sodium hydroxide and carbon dioxide is (a) benzoic acid (b) salicylaldehyde (c) salicylic acid (d) phthalic acid. (2012) Phenol. (d) H 3 C – CHBr – OCH 3 .91 Alcohols. 6tribromophenol. OH NO 2 (III) OCH 3 (IV) (b) II > IV > I > III (d) III > I > II > IV (2013) In the following sequence of reactions. CH 3 CH 2 OH P + I 2 A Mg ether B HCHO C H 2 O D the compound D is (a) propanal (b) butanal Sodium ethoxide has reacted with ethanoyl chloride. Orthonitrophenol is less soluble in water than p and mnitrophenols because (a) onitrophenol shows intramolecular Hbonding (b) onitrophenol shows intermolecular Hbonding (c) melting point of onitrophenol is lower than those of 9. Arrange the following compounds in order of decreasing acidity. Phenyl magnesium bromide reacts with methanol to give (a) a mixture of anisole and Mg(OH)Br (b) a mixture of benzene and Mg(OMe)Br (c) a mixture of toluene and Mg(OH)Br (d) a mixture of phenol and Mg(Me)Br. 6. is (a) 1Butanol (b) 2Butanol (c) 2Methylpropan2ol (d) 2Methylpropanol. 3. (2011) at room temperature to give Phenol is heated with a solution of mixture of KBr and KBrO 3 . OH OH . OH . + CHCl 3 + NaOH CHO The electrophile involved in the above reaction is (a) Å (a) dichloromethyl cation (CHCl 2 ) (b) dichlorocarbene (: CCl 2) (b) (c) trichloromethyl anion (CCl 3 ) Å (d) formyl cation (CHO) (c) (d) (2010) (2006) 12. (2006) .HCl and anhydrous ZnCl 2 . 5. (a) CH 3 CHO and CH 3 Br The major product obtained in the above reaction is (b) BrCH 2 CHO and CH 3 OH (a) 2bromophenol (b) 3bromophenol (c) BrCH 2 – CH 2 – OCH 3 (c) 4bromophenol (d) 2. From amongst the following alcohols the one that would react fastest with conc. (2008) Cl CH 3 (I) (II) (a) IV > III > I > II (c) I > II > III > IV 2. Phenols and Ethers ALCOHOLS. (2009) 8. 6trinitrobenzene (c) onitrophenol (d) pnitrophenol. when it first reacts with concentrated sulphuric acid and then with concentrated nitric acid. (2006) (2011) The main product of the following reaction is – ONa + OH 11. PHENOLS AND ETHERS CHAPTER 24 1. m and pisomers (d) onitrophenol is more volatile in steam than those of m and pisomers. (2007) compound that is produced in this reaction is (a) diethyl ether (b) 2butanone 10. gives (a) nitrobenzene (b) 2. (None of the option is correct) (b) 14. This is due to (a) dipolar character of ethers (b) alcohols having resonance structures (c) intermolecular hydrogen bonding in ethers (d) intermolecular hydrogen bonding in alcohols. (d) 4. For which of the following parameters the structural isomers C 2 H 5 OH and CH 3 OCH 3 would be expected to have the same (d) (c) values? (Assume ideal behaviour) CH 2 COOH (2005) OH (a) Heat of vaporisation OH (b) Vapour pressure at the same temperature 14.92 JEE MAIN CHAPTERWISE EXPLORER 13. (a) 8. The latter on acidic hydrolysis gives chiral carboxylic acid. (a) 7. During dehydration of alcohols to alkenes by heating with concentrated H 2 SO 4 the initiation step is (a) protonation of alcohol molecule 15. (d) 5. (2003) (b) CH 3 CH 2 CH 2 CHCH 3 CH 3 (c) CH 3 CH 2 CCH 2 CH 3 OH Answer Key 1.3dimethyl1hydroxy cyclohexane 1. (b) 18. Among the following compounds which can be dehydrated very (b) formation of carbocation easily? (c) elimination of water (a) CH 3 CH 2 CH 2 CH 2 CH 2 OH (d) formation of an ester. 15. (d) 2. (b) 6. (c) 9.1dimethyl3hydroxy cyclohexane 3. 13. An ether is more volatile than an alcohol having the same molecular formula. 18. (d) (d) 3. (d) (c) 11. 16. The structure of the carboxylic acid is (d) CH 3 CH 2 CHCH 2 CH 2 OH 16. 19. (c) 12. pcresol reacts with chloroform in alkaline medium to give the compound A which adds hydrogen cyanide to form the compound B.1dimethyl3cyclohexanol. (d) . The IUPAC name of the compound (a) (b) (c) (d) CH(OH)COOH (a) (b) CH(OH)COOH OH OH is HO CH 3 CH 3 (2004) CH 3 3. (b) 17. (2004) CH 3 17.3dimethyl1cyclohexanol 1. (d) (c) 10. (2003) (2005) OH 19. The best reagent to convert pent3en2ol into (c) Boiling points pent3en2one is (d) Gaseous densities at the same temperature and pressure (a) acidic permanganate (2004) (b) acidic dichromate CH 3 CH 2 COOH (c) chromic anhydride in glacial acetic acid (d) pyridinium chlorochromate. 93 Alcohols. 7.HCl and anhydrous ZnCl 2 is Lucas reagent. 1° alcohol + Lucas reagent No reaction. Mg H – C – OH CH 2 CH 3 n propyl alcohol (D) H 2 O HCHO H – C – OMgI CH 2 CH 3 (C) . Thus. conc. (d) : CH 3 CH 2 OH P CH 3 CH 2 I I 2 CH 3 CH 2 MgI (B ) (A ) H H The preferential formation of this compound is due to conjugation in the compound.6tribromophenol. 2° alcohol + Lucas reagent Turbidity after 5 mins. H2 SO 4 373 K Phenol 5. : (None of the option is correct) OH OH SO3 H + Conc. 2° and 3° alcohols. Phenols and Ethers 1.e. (d) : Electron donating groups (—CH 3 and —OCH 3 ) decrease while electron withdrawing groups (—NO 2 and —Cl) increase the acidity.4.. therefore order of decreasing acidity is III > I > II > IV. The product formed is salicylic acid. (d) : KBr (aq) + KBrO 3(aq) Br 2(aq) This bromine reacts with phenol gives 2. . (b) : OH 2Phenol sulphonic acid SO3 H 4Phenol sulphonic acid OH Conc. 8. 6. i. (a) : oNitrophenol is stable due to intramolecular hydrogen bonding. (d) : 4. (c) : The reaction of phenol with NaOH and CO 2 is known as KolbeSchmidt or Kolbe’s reaction. Since —OCH 3 is a stronger electron donating group than —CH 3 and —NO 2 is stronger electron withdrawing group than —Cl. which is used to distinguish between 1°. 3. 2. HNO 3 D NO 2 O2 N NO 2 Picric acid 9. 3° alcohol + Lucas reagent Immediate turbidity. It is difficult to break the Hbonding when dissolved in water thus less soluble. (c) : The reagent. the required alcohol is 2methylpropan2ol. 19.3dimethyl1cyclohexanol Molecular weight 2 As both the compounds have same molecular weights. the initiation step is protonation of alcohol. (b) : CH 3OH + C ® C 6H 6H 5MgBr 6 + Mg(OCH 3)Br CH 3 13. . It is hydrolysed rapidly by dilute acids at room temperature to give methanol and aldehyde. Hence. (d) : The reason for the lesser volatility of alcohols than ethers is the intermolecular association of a large number of molecules due to hydrogen bonding as – OH group is highly polarised. (c) : Br CH 3 CH 3 CH(OH)COOH OH 14. (d) : Vapour density = 18. hydrogen bonding No such hydrogen bonding is present in ethers. boiling point and heat of vaporization will differ as ethanol has hydrogen bonding whereas ether does not. OH – + CHCl 3 ƒ HOH + : CCl 3– ® Cl – + : CCl 2 12. : CCl 2 generated from chloroform by the action of a base. R R R R O – H O – H O – H O – H 15. (d) : Pyridinium chlorochromate oxidises an alcoholic group selectively in the presence of carboncarbon double bond. (b) : + ONa + CHCl 3 + NaOH CHO (ReimerTiemann reaction) The electrophile is dichlorocarbene. 17. (a) : Dehydration of alcohol to alkene in presence of concentrated H 2 SO 4 involves following steps : H + – C – C – –H 2 O H OH H OH 2 alcohol protonated alcohol – C – C – + + H carbonium ion –H + – C C – alkene CH(CN)OH OH 2 CH 3 H + /H 2 O HCN 1 OH CH 3 Å The more stable carbocation is generated thus more easily it will be dehydrated. The rest of these three properties; vapour pressure. (b) : CH 3 CHO OH 3 ® 3. However. Thus. (c) : The ease of dehydration of alcohols is tertiary > secondary > primary according to the order of stability of the carbocations. both will have the same vapour density. it undergoes many addition reactions at the double bond. – C – C – + CHCl 3 + KOH CH 3 – CH 2 – C – CH 2 – CH 3 OH HO H 3 C – CH – OCH 3 H + CH 3 – CH 2 – C – CH 2 – CH 3 16. H 2 C CH – OCH 3 H + Å H 2 C – CH – OCH 3 H Br – OH 11. gaseous density of both ethanol and dimethyl ether would be same under identical conditions of temperature and pressure.94 JEE MAIN CHAPTERWISE EXPLORER 10. under anhydrous conditions at room temperature. (d) : Methyl vinyl ether is a very reactive gas. A liquid was mixed with ethanol and a drop of concentrated H 2 SO 4 was added. The compound formed as a result of oxidation of ethyl benzene by KMnO 4 is (a) benzyl alcohol (b) benzophenone (c) acetophenone (d) benzoic acid. (a) A < B < C < D (b) D < B < C < A trichloroacetate ion and another compound. Ketones and Carboxylic Acids ALDEHYDES. (a) (b) (c) (d) CO 2 H is (C) CF 3 CO 2 H (D) Me Ozonolysis of an organic compound gives formaldehyde as one of the products. PhCOCH 3 D. (2006) (b) a vinyl group 11. which of the following is the most appropriate reagent? (a) ZnHg/HCl (c) NaBH 4 3. (2007) Silver mirror test is given by which one of the following compounds? 10.2trichloroethanol trichloromethanol 2. KETONES AND CARBOXYLIC ACIDS CHAPTER 25 1. 5. CH 3COCH 3 by using NaOH. (b) Na. 6. An organic compound A upon reacting with NH 3 gives B. The mixture of the products contains sodium C.2trichloropropanol chloroform In Cannizzaro reaction given below – 2 PhCHO :OH PhCH 2 OH + PhCOO – 8. OH – (2012) 2.95 Aldehydes. The liquid was (a) CH 3 OH (b) HCHO (c) CH 3 COCH 3 (d) CH 3 COOH (2009) 9. A compound with a fruity smell was formed. Among the following the one that gives positive iodoform test (c) an isopropyl group upon reaction with I 2 and NaOH is (d) an acetylenic triple bond.2. (2011) (a) CH 3 CH 2 CH(OH)CH 2 CH 3 (b) C 6 H 5 CH 2 CH 2 OH The strongest acid amongst the following compounds is CH 3 (a) CH 3 COOH (c) H 3 C (d) PhCHOHCH 3 (2006) (b) HCOOH OH (c) CH 3 CH 2 CH(Cl)CO 2 H 12. (2009) CH 3 2. (c) CH 3 CH 2 CH 2 COOH (d) CH 3 —CH—COOH (2013) In the given transformation. The correct order of increasing acid strength of the compounds (a) Acetaldehyde (b) Acetone (A) CH 3 CO 2H (B) MeOCH 2 CO 2H (c) Formaldehyde (d) Benzophenone (2011) Me 4. The increasing order of the rate of HCN addition to compounds (d) ClCH 2 CH 2 CH 2 COOH (2011) A D is Trichloroacetaldehyde was subjected to Cannizzaro’s reaction A. NH 3 (d) NH 2 – NH 2 . The other (c) D < C < B < A (d) C < D < B < A (2006) compound is . This confirms the presence of (a) B < D < A < C (b) D < A < C < B (a) two ethylenic double bonds (c) D < A < B < C (d) A < D < C < B.2. HCHO B. A is (a) CH 3 CH 2 COOH (b) CH 3COOH 7. (2011) the slowest step is (a) the attack of :OH – at the carboxyl group (b) the transfer of hydride to the carbonyl group (c) the abstraction of proton from the carboxylic group (d) the deprotonation of PhCH 2 OH. C in presence of KOH reacts with Br 2 to give CH 3CH 2 NH 2. liq. On heating. PhCOPh. B gives C. (a. When CH 2 CH – COOH is reduced with LiAlH 4 . The IUPAC name of CH 3 COCH(CH 3 ) 2 is (c) Butanol (d) Benzoic acid (2004) (a) isopropylmethyl ketone 14. CH 3 O O – R – C CH 3 R – C + Nu Z – (d) + Z Nu is fastest when Z is (a) Cl (b) NH 2 (c) OC 2 H 5 (d) OCOCH 3 . 14. In the anion HCOO the two carbonoxygen bonds are found to (2002) be of equal length. 15. 10. The general formula C n H 2n O 2 could be for open chain (a) diketones (b) carboxylic acids (c) diols (d) dialdehydes. (c) (d) (c) (c) 6. (d) The anion is obtained by removal of a proton from the ® ethyl butanoate acid molecule. (2003) (b) CH 3 COONa + C 2 H 5 OH 21. What is the reason for it? (a) Electronic orbitals of carbon atom are hybridised. 8. the (b) 2methyl3butanone composition of the resultant solution is (c) 4methylisopropyl ketone (a) CH 3 COOC 2 H 5 + NaCl (d) 3methyl2butanone. Cl 2 22. Which of the following compounds has wrong IUPAC name? O bond is weaker than the C – O bond. (b) (c) (a) (c) 5. Rate of the reaction. 19. (a) 12. KOH (2002) B What is B ? (a) CH 3 CH 2 COCl (b) CH 3 CH 2 CHO (c) CH 2 CHCOOH (d) ClCH 2 CH 2 COOH (2004) – 17. (b) CH 3 – CH – CH 2 – CHO ® 3methylbutanal (2003) CH 3 18. (2003) (c) CH 3 – CH – CH – CH 3 ® 2methyl3butanol OH CH 3 O 19. 22. 13. 17. (d) (d) (a) (d) 3. (c) CH 3 COCl + C 2 H 5 OH + NaOH (CH 3 ) 2 C CH – CH 2 – CHO gives (d) CH 3 Cl + C 2 H 5 COONa. (a) (b) (b) (b) 2. CH 3 CH 2 COOH red P A alc. 20. 23. On vigorous oxidation by permanganate solution. CH 3 COOH + CH 3 CH 2 COOH CH – OH + CH 3 CH 2 CH 2 OH C O + CH 3 CH 2 CHO . Consider the acidity of the carboxylic acids: (i) PhCOOH (ii) oNO 2C 6H 4COOH (iii) pNO 2C 6H 4COOH (iv) mNO 2C 6H 4COOH Which of the following order is correct? (a) i > ii > iii > iv (b) ii > iv > iii > i (c) ii > iv > i > iii (d) ii > iii > iv > i (a) CH 3 – C – CH – CH 2 CH 3 CH 3 CH 3 (b) CH 3 (2004) (c) CH 3 16. 21. (b) The C (a) CH 3 – CH 2 – CH 2 – COO – CH 2 CH 3 (c) The anion HCOO – has two resonating structures. the compound obtained will be (a) CH 3 – CH 2 – COOH (b) CH 2 CH – CH 2 OH (d) CH – CH – C – CH – CH ® 2methyl3pentanone 3 2 3 CH 3 (2002) Answer Key 1. (b) . (c) 18. (2004) OH OH 15. Which one of the following undergoes reaction with 50% sodium (c) CH 3 – CH 2 – CH 2 OH hydroxide solution to give the corresponding alcohol and acid? (d) CH 3 – CH 2 – CHO. c) (d) (d) (b) 4. 16.96 JEE MAIN CHAPTERWISE EXPLORER 13. 7. (2003) (a) Phenol (b) Benzaldehyde 20. 11. 23. On mixing ethyl acetate with aqueous sodium chloride. 9. H2SO4 are added must be an acid. Ketones and Carboxylic Acids O O 1. R R 10. tend to increase the acid strength of substituted acid. Mechanism : – C6 H 5 – C O + OH CH 3 —CH 2 —C—NH 2 ( C ) Fast H 2. 3. c) : Formaldehyde and acetaldehyde can be oxidised by Tollen’s reagent to give silver mirror. (c) : Effect of substituent on the acid strength of aliphatic acids: (i) Acidity decreases as the +Ieffect of the alkyl group increases. \ CH 3 COOH + C 2 H 5 OH conc. (ii) Acidity decreases as the –Ieffect as well as number of 6. Htransfer to the carbonyl group is the rate determining step and hence the slowest. (d) : H H – C6 H 5 – C – O + O C – C6 H 5 Slow. etc. regardless of length is oxidised to – COOH.97 Aldehydes. – NH 2 etc. (d) : Ethyl benzene 4. 5. (a) : In Cannizzaro’s reaction one molecule is oxidised to halogen atoms decreases. it is an ester. – OH. On the basis of given information the relative order of increasing acid strength of the given compounds is (CH 3 ) 2 COOH < CH 3 COOH < CH 3 OCH 2 COOH < CF 3 COOH Vinyl group (CH 2 CH –) on ozonolysis gives formaldehyde. COOH CH 2 – CH 3 KMnO 4 9. Hydride transfer OH H C6 H 5 – C O + O – C – C6 H 5 OH to – CHOH only. (a) : CH 3 —CH 2—C—OH NH 3 ( A ) 7. Rearrangement H – – OH group and alkene are acid‑sensitive groups so Clemmensen reduction cannot be used and NaBH 4 reduces – 8. In case of Cannizzaro reaction. (c) : CH 3 CH 2 CH(Cl)COOH is the strongest acid due to –I effect HCOOH > CH 3COOH > (CH 3) 2CHCOOH > (CH 3) 3CCOOH of Cl. tend to decrease while electron withdrawing substituents like –NO 2 . –CHO. . the entire side chain (in benzene homologues) with atleast one H at acarbon. C6 H 5 – CH 2 – OH + C6 H5 COO (d) : Since the compound formed has a fruity smell. (a. FCH 2COOH > ClCH 2COOH > BrCH 2COOH > ICH 2COOH > CH 3COOH F 3 CCOOH > F 2 CHCOOH > FCH 2 COOH > CH 3 COOH (iii) Electron donating substituents like – R. (b) : Benzoic acid When oxidises with alkaline KMnO 4 or acidic Na 2 Cr 2 O 7 .H 2 SO 4 CH 3 COOC 2 H 5 + H 2 O. – + CH 3 —CH 2 —C—ONH 4 ( B ) D O CH 3 —CH 2 —NH 2 Br 2 /KOH (b) : Rate determining step is always the slowest step. thus the liquid to which ethanol and conc. carboxylate ion and the other is reduced to alcohol. (d) : H – C – C – C – C – H H CH 3 H 3methyl2butanone 21. (d) : NO 2 1 2 3 Cl 2 red P –HCl CH 3 CHClCOOH (A) alc. From the above information. (c) : HCOO – exists as C O O– O H–C O H – C – O– The aromatic aldehydes and ketones are less reactive than their So. it reduces carboxylic the rate of HCN addition to compounds HCHO. NO 2 ) at pposition have more pronounced electron withdrawing effect than as – NO 2 group at mposition (–I effect) \ Correct order of acidity is ii > iii > iv > i. D PhCHOHCH 3 + 4I 2 + 6NaOH ¾¾ ® CHI 3 + PhCOONa + 5NaI + 5H 2 O p isomer due to orthoeffect. 13.e. (d) : Iodoform test is given by only the compounds containing CH 3 CO – or CH 3 CHOH – group. (b) : Aldehydic group gets oxidised to carboxylic group. (b) : LiAlH 4 is a strong reducing agent. This is due to the +R effect of the benzene ring. (c) : CH 3 CH 2 COOH chloride. decreases. KOH –HCl CH 2 (B) 4 NO 2 23. the two carbonoxygen bonds are found to be of equal length. the reactivity of the ketones further overlapping between the 3p orbital of Cl and 2p orbital of carbon.98 JEE MAIN CHAPTERWISE EXPLORER 11. aliphatic analogous. 18. OH CH 3 oisomer will have higher acidity then corresponding m and 3methyl2butanol CHCOOH . Double bond breaks and carbon gets oxidised to carboxylic group. group into primary alcoholic group without affecting the basic PhCOCH 3 and PhCOPh is skeleton of compound. 14. PhCOPh < PhCOCH 3 < CH 3 COCH 3 < HCHO. The basicity of the leaving group H R H increases as The lower reactivity of ketones over aldehydes is due to +I – Cl – < RCOO – < RO – < NH 2 effect of the alkyl (R) group and steric hindrance. CHO CH 2 OH 50% NaOH Benzaldehyde Benzyl alcohol COONa + Sodium benzoate CH 2 CH – COOH H O H 1 2 3 LiAlH 4 [H + ] CH 2 CH – CH 2 OH H 4 20. As the size Secondly least stabilization by resonance due to ineffective of the alkyl group increases. it is clear that increasing order of 19.e. As M group (i. CH 3 CH 3 (CH 3 ) 3 (CH 3 ) 2 CH C O > C O > (CH 3 ) 3 (CH 3 ) 2 CH 17. Dialdehydes C n H n O 2 . 12. (a) : R – C R – C + Z + Nu Order of reactivity: Nu Z H R Reactivity of the acid derivatives decreases as the basicity of R C O > C O C O > the leaving group increases. (b) : Diketones C nH 2n – 2 O 2.. (a) : CH 3COOC 2H 5 + NaCl (aq) ® no reaction i. 16. (c) : CH 3 – CH – CH – CH 3 Electron withdrawing group increases the acidity of benzoic acid. the resultant solution contains ethyl acetate and sodium 22. (b) : Benzaldehyde will undergo Cannizzaro reaction on treatment with 50% NaOH to produce benzyl alcohol and benzoic acid as it does not contain ahydrogen. (c) : Addition of HCN to carbonyl compounds is a characteristic O O – – nucleophilic addition reaction of carbonyl compounds. CH 3 COCH 3 . Carboxylic acid C nH 2n O 2 PhCHO > PhCOCH 3 > PhCOPh Diols C n H 3n O 2. COOH COOH COOH COOH NO 2 < < < 15. When primary amine reacts with chloroform in ethanolic KOH synthesis nor for separation of amines? then the product is (a) Hinsberg method (b) Hofmann method (a) an isocyanide (b) an aldehyde (c) Wurtz reaction (d) Curtius reaction (2005) (c) a cyanide (d) an alcohol. CH 3 CH 2 NH 2 + CHCl 3 + 3KOH ® (A) + (B) + 3H 2 O. 8. On heating it gives NH 3 alongwith a solid residue. 9. (b) 11. (2002) Answer Key 1. (a) (d) 3. (2003) the reaction is continuously removed. (a) 6. CH 3 NH 2 and (CH 3 ) 2 NH is (2005) (a) CH 3 NH 2 < NH 3 < (CH 3) 2 NH (b) (CH 3 ) 2 NH < NH 3 < CH 3 NH 2 Amongst the following the most basic compound is (a) benzylamine (b) aniline (c) NH 3 < CH 3NH 2 < (CH 3) 2NH (c) acetanilide (d) pnitroaniline (2005) (d) CH 3 NH 2 < (CH 3 ) 2 NH < NH 3 (2003) Which one of the following methods is neither meant for the 13. Ethyl isocyanide on hydrolysis in acidic medium generates (c) (NH 2 ) 2 CO (d) CH 3 CH 2 CONH 2 (2005) (a) ethylamine salt and methanoic acid (b) propanoic acid and ammonium salt Reaction of cyclohexanone with dimethylamine in the presence (c) ethanoic acid and ammonium salt of catalytic amount of an acid forms a compound if water during (d) methylamine salt and ethanoic acid. the compounds (A) and (B) are respectively (a) C 2 H 5 NC and 3KCl (b) C 2 H 5 CN and 3KCl 10. 12.67% and N = 46. The compound is (a) CH 3 NCO (b) CH 3 CONH 2 11. The compound formed is generally known as (a) a Schiff’s base (b) an enamine (c) an imine (d) an amine 6. 5. A compound with molecular mass 180 is acylated with 8. The number of amino groups present per molecule of the former compound is (a) 6 (b) 2 (c) 5 (d) 4 (2013) Which of the following is the strongest base? (a) NH 2 (b) (c) NH 2 (d) CH 3 NHCH 3 CH 2 NH 2 (2004) In the chemical reaction. 3.67% while rest is oxygen. 2. (c) 13. (c) 10. 9.99 CHEMISTRY ORGANIC COMPOUNDS CONTAINING NITROGEN CHAPTER 26 1. CH 3 COCl to get a compound with molecular mass 390. The correct order of increasing basic nature for the bases NH 3 . (a) 12. 7. (d) 5. The solid residue gives violet colour with alkaline copper sulphate NC (d) H 3 C (2003) solution. (c) CH 3 CH 2 CONH 2 and 3KCl (d) C 2 H 5 NC and K 2 CO 3 . (a) 2. (2007) Which one of the following does not have sp 2 hybridized carbon? (a) Acetone (b) Acetic acid (c) Acetonitrile (d) Acetamide (2004) Which one of the following is the strongest base in aqueous solution? (a) Methylamine (b) Trimethylamine (c) Aniline (d) Dimethylamine (2007) (a) H 3 C CN (b) H 3 C N 2 Cl The reaction of chloroform with alcoholic KOH and p toluidine forms An organic compound having molecular mass 60 is found to NHCHCl 2 (c) H 3 C contain C = 20%. 4. (c) 7. (d) (c) 4. (c) . H = 6. 66 1.00 6. the nonbonding electron pair of nitrogen is delocalized into benzene ring by resonance. (a) : Due to resonance of electron pair in aniline. In benzylamine electron pair is not involved in resonance. This anomalous behaviour of tertiary amines is due to steric factors i. crowding of alkyl groups cover nitrogen atom from all sides and thus makes it unable for protonation. NHCOCH 3 NH 2 CH 2 NH 2 NH 2 ethanolic KOH.. (c) : In Wurtz reaction alkyl halide reacts with sodium metal in the presence of dry ether to give alkane. 8. the covering of alkyl groups 3 2 3 3 Acetonitrile Acetamide over nitrogen atom from all sides makes the approach and bonding by a proton relatively difficult. Electron withdrawing (C 6 H 5 –) groups decreases electron density on nitrogen atom and thereby decreasing basicity.67 6. (c) : CH 3 – CO – CH 3 ; CH 3 – COOH Acetic acid electrons to proton increases and hence the basicity of amines Acetone 3 3 also increases. Thus the relative strength is in order (CH 3 ) 2 NH > CH 3 NH 2 > NH 3 . then a bad smell compound isocyanide is formed. (c) : Except the amines containing tertiary butyl group. 4. . This is called carbylamine reaction and this reaction is used as a test of primary amines. 3. H 3 C (c) : Element Simplest Percentage Relative no. 13. due to which basicity decreases. (d) : The increasing order of basicity of the given compounds is (CH 3 ) 2 NH > CH 3 NH 2 > (CH 3 ) 3 N > C 6 H 5 NH 2 Due to the +I effect of alkyl groups. (b) : 6. Further the presence of electron donating groups in the benzene ring increase the basic strength while electron withdrawing group decrease the basic strength of substituted aniline.e.67 D C 2 H 5 N C + 2H 2 O Ethylisocyanide NH 2 CONHCONH 2 + NH 3- biuret Biuret gives violet colour with alkaline copper sulphate solution.100 JEE MAIN CHAPTERWISE EXPLORER 390 - 180 = 5 42 1.. (a) : When a primary amine reacts with chloroform with . the nonbonding electron pair of nitrogen does not take part in resonance. The alkyl groups. basic strength decreases. . So aliphatic amines are more basic than aniline. sp sp 2 sp sp ; CH – CONH CH – C N In case of tertiary anine (CH 3) N. . hence the basicity 10. which are electron releasing groups. as a result the electron density on the N atom decreases. (d) : H 3 C NH 2 + CHCl 3 + 3KOH decreases. .33 1. Primary amine Isocyanide . 11. 5.67 46. N(CH 3 ) 2 O + (CH 3 ) 2 NH The above reaction is known as carbylamine reaction and is generally used to convert primary amine into isocyanide. of amino groups = 2. 7. D RNH 2 + CHCl 3 + 3KOH ¾¾ ® RNC + 3KCl + 3H 2O NO (ii) dehydration enamine 2 I II III IV Decreasing order of basic strength is II > I > IV > III. increase the electron density around the nitrogen thereby increasing the availability of the lone pair of electrons to proton or Lewis acids and making the amine more basic. the compound is H 2 NCONH 2 . all lower aliphatic amines are stronger bases than ammonia because of +I (inductive) effect.67 3. In other three compounds. (d) : In this compound. 2NH 2 CONH 2 NC + 3KCl + 3H 2 O (i) H + H + C 2 H 5 NH 2 + HCOOH Ethylamine Methanoic acid 12. So. 1 4 2 1 The molecular formula is CH 4 N 2 O. ratio of atom C H N O 20. (a) : CH 3CH 2NH 2 + CHCl 3 + 3KOH C 2H 5NC + 3KCl + 3H 2O This is called carbylamine reaction. (a) : Alkyl isocyanides are hydrolysed by dilute mineral acids to form primary amines. The observed order in the case of lower members is found to be as secondary > primary > tertiary.67 26.. (c) : No. the electron density on sp 3 sp 3 sp 2 sp 3 sp 2 nitrogen increases and thus the availability of the lone pair of 9. . (a) 4. 2. (b) 6. polymerization is (a) HNO 3 (b) AlCl 3 (c) BuLi (d) LiAlH 4 (2012) 6. (b) (a) 2.6 (d) polystyrene. (2002) Answer Key 1. hydrogen bonding is (a) natural rubber (b) teflon (c) nylon6.101 Polymers CHAPTER POLYMERS 27 1. (c) . (c) 3.6 (c) Terylene (d) Bakelite (2005) Nylon threads are made of (a) polyvinyl polymer (b) polyester polymer (c) polyamide polymer (d) polyethylene polymer. Bakelite is obtained from phenol by reaction with (a) HCHO (b) (CH 2 OH) 2 (c) CH 3 CHO (d) CH 3 COCH 3 (2008) 4. (2010) 3. (b) 5. Which of the following is fully fluorinated polymer? (a) Neoprene (b) Teflon (c) Thiokol (d) PVC (2005) 7.g. The polymer containing strong intermolecular forces e. 7. Which of the following is a polyamide? (a) Teflon (b) Nylon6. (2003) Polymer formation from monomers starts by (a) condensation reaction between monomers (b) coordinate reaction between monomers (c) conversion of monomer to monomer ions by protons (d) hydrolysis of monomers. The species which can best serve as an initiator for the cationic 5. Cl 4. acids such as H 2SO 4.6 involves amide (CONH) linkage therefore. n (a) : Polymerisation takes place either by condensation or addition reactions. (b) : Neoprene : — CH 2 – CH C – CH 2 — (c) : Nylon threads are polyamides.102 JEE MAIN CHAPTERWISE EXPLORER 1. n n HOOC(CH 2 ) 4 COOH + n H 2 N(CH 2 ) 6 NH 2 Teflon : — CF 2 – CF 2 — Adipic acid n Thiokol : Nylon (polyamide) n Cl n Hexamethylene diamine HO – OC – (CH 2 ) 4 – CONH(CH 2 ) 6 NH – CH 2 — CH 2 – S – S – CH 2 — CH 2 – S – S – CH 2 CH 2 – PVC : — CH 2 – CH — 280°C high pressure 7. (c) : Nylon6. Nylon6. SnCl 4 or BF 3 in H 2O. 6 6. ( HN – (CH 2 ) 6 – NHCO – (CH 2 ) 4 – CO ) n (a) : Bakelite is a thermosetting polymer which is made by reaction between phenol and HCHO. AlCl 3 . HF. 3. . They are the condensation polymers of diamines and dibasic acids. 2. it will also have very strong inter molecular hydrogen bonding between (b) : Polymers having amide linkages (– CONH) are known as polyamides. (b) : Cationic polymerisation is initiated by use of strong Lewis 5. n(H 2 N(CH 2 ) 6 NH 2 ) + n(HOOC(CH 2 ) 4 COOH) Hexamethylene diamine Adipic acid amide linkage group of two polyamide chains. Insulin production and its action in human body are responsible for the level of diabetes. heterocyclic base and phosphate ester (a) Sugars (b) Amines linkages are at (c) Primary alcohols (d) Nitro compounds (2012) (a) C 5¢ and C 2¢ respectively of the sugar molecule Which one of the following statements is correct? (b) C 2¢ and C 5¢ respectively of the sugar molecule (a) All amino acids are optically active. aD(+)glucose and bD(+)glucose are (b) Enzymes are normally heterogeneous catalysts that are very (a) enantiomers (b) conformers specific in action. (2006) 14. (2003) . (a) fixed configuration of the polypeptide backbone (d) Enzymes are specific biological catalysts that possess well (b) ahelical backbone defined active sites. 6. Which base is present in RNA but not in DNA? are (a) Uracil (b) Cytosine (a) –OH and –COOH (c) Guanine (d) Thymine (2004) (b) –CHO and –COOH 13.103 Biomolecules CHAPTER 28 BIOMOLECULES The term anomers of glucose refers to (a) isomers of glucose that differ in configurations at carbons one and four (C1 and C4) (b) a mixture of (D)glucose and (L)glucose (c) enantiomers of glucose (d) isomers of glucose that differ in configuration at carbon one (C1). 5. In both DNA and RNA. (2012) 11. 7. 4. (2007) The pyrimidine bases present in DNA are (a) cytosine and adenine (b) cytosine and guanine (c) cytosine and thymine (d) cytosine and uracil. (d) C 5¢ and C 1¢ respectively of the sugar molecule (c) All amino acids except glutamic acid are optically active. (c) epimers (d) anomers. 8. Synthesis of each molecule of glucose in photosynthesis 9. (c) hydrophobic interactions (2004) (d) sequence of aamino acids. The reason for double helical structure of DNA is operation of (a) van der Waal's forces (b) dipoledipole interaction (c) hydrogen bonding (d) electrostatic attractions. This compound belongs to which of The presence or absence of hydroxy group on which carbon the following categories? atom of sugar differentiates RNA and DNA. The two functional groups present in a typical carbohydrate 12. (c) C 1¢ and C 5¢ respectively of the sugar molecule (b) All amino acids except glycine are optically active. (c) C O and –OH (a) Enzymes are specific biological catalysts that can normally (d) –OH and –CHO (2009) function at very high temperatures (T ~ 1000 K). (a) 1 st (b) 2 nd (a) A coenzyme (b) A hormone rd th (c) 3 (d) 4 (2011) (c) An enzyme (d) An antibiotic (2004) 3. Identify the correct statement regarding enzymes. Which of the following compounds can be detected by Molisch’s test? 10. (2006) 1. (2008) (c) Enzymes are specific biological catalysts that cannot be The secondary structure of a protein refers to poisoned. (2005) (d) All amino acids except lysine are optically active. involves (a) 6 molecules of ATP (b) 18 molecules of ATP (c) 10 molecules of ATP (d) 8 molecules of ATP (2013) 2. (c) 3. which is found in amino acid is (a) – COOH group (b) – NH 2 group (c) – CH 3 group (d) both (a) and (b). (b) 10. (c) 16. The functional group. (c) 14. (b) 9. (b) 7. (d) 2. (b) 6. (d) 5. (d) 12. (2003) 16. Complete hydrolysis of cellulose gives (a) Dfructose (b) Dribose (c) Dglucose (d) Lglucose. (b) 17. (b) 13. (c) 11.104 JEE MAIN CHAPTERWISE EXPLORER 15. RNA is different from DNA because RNA contains (a) ribose sugar and thymine (b) ribose sugar and uracil (c) deoxyribose sugar and thymine (d) deoxyribose sugar and uracil. (2002) Answer Key 1. (2002) 17. (a) 8. (c) 4. (d) 15. (a) . Thus the two functional groups present are C O (aldehyde or ketone) and –OH. (b) : The sugar molecule found in RNA is Dribose while the sugar in DNA is D2deoxyribose. a form in which the OH at C 1 in aldoses and C 2 in ketoses lies towards the right and b form in which it lies towards left. (c) : NH 2 N OH a D(+)glucopyranose H OH b D(+) glucopyranose A pair of stereoisomers which differ in configuration at C1 are known as anomers. 6. fructose. (b) : 6CO 2 + 18ATP + 12NADPH + 6RuBP ® 6RuBP + Glucose + 18ADP + 18P + 12NADP + One molecule of glucose is formed from 6CO 2 by utilising 18ATP and 12NADPH. 4. Thus glucose.. The sugar D2deoxyribose differs from ribose only in the substitution of hydrogen for an – OH group at 2position as shown in figure. aD(+) glucose ƒ equilibrium mixture ƒ b(D) (+) glucose As a result of cyclization the anomeric (C1) becomes asymmetric and the newly formed – OH group may be either on left or on right in Fischer projection thus resulting in the formation of two isomers (anomers). 9. (c) : DNA contains cytosine and thymine as pyrimidine bases and guanine and adenine as purine bases. all exist in a and b form. 3. based on the dehydration of carbohydrate by sulphuric acid to produce an aldehyde. (b) : Secondary structure of proteins is mainly of two types. . (b) : Insulin is a proteinaceous hormone secreted by bcells by islet of Langerhans of pancreas in our body. (d) : Due to cyclic hemiacetal or cyclic hemiketal structures. ribose. 7. amino acid coils as a right handed screw (called ahelix) because of the formation of hydrogen bonds between amide groups of the same peptide chain. with two molecules of phenol resulting in red or purple coloured compound. (d) : Structures of aD(+)glucose and bD(+)glucose are : 6 5 H 4 HO H OH 3 O H H a 4 H 1 OH HO 2 H OH 3 H – C – OH H – C – OH H – C H – C CH 2 OH bDglucose aDglucose O OH H 1 OH 2 10. HO – C – H H – C – OH H – C – OH H – C – OH 5. which condenses 8.e. all the pentoses and hexoses exist in two stereoisomeric forms i. Glucose exists in two forms aDglucose and bD glucose. The isomers having – OH group to the left of the C1 is designated bDglucose and other having – OH group on the right as aDglucose.105 Biomolecules 1. etc. (i) N OH 5¢ H O HO – C – H CH 2 OH 6 5 O and HO – P – O – CH 2 O 4¢ C H O H 3¢ N N 1¢ H C C C 2¢ HO OH H ahelix : This structure is formed when the chain of a 11. (b) : Glycine is optically inactive while all other amino acids are optically active. (c) : Carbohydrates are essentially polyhydroxy aldehydes HO – C – H and polyhydroxy ketones. (a) : Molisch’s test is a sensitive chemical test for the presence of carbohydrates. 2. (ii) bplated sheet : In this structure the chains are held together by a very large number of hydrogen bonds between C O and NH of different chains. –NH 2 . Cellulose is a straight chain polysaccharide composed of Dglucose units which are joined by b glycosidic linkages. guanine and adenine. 14. (c) : (C 6H 10 O 5) n + nH 2O Cellulose three hydrogen bonds H + nC 6H 12 O 6 Dglucose derivatives (b) Purine derivatives (c) Sugar Thymine Adenine Guanine Deoxyribose Uracil Adenine Guanine Ribose . as well as an amino group. 13.e.106 JEE MAIN CHAPTERWISE EXPLORER 12. (a) Pyrimidine Cytosine Cytosine Adenine Thymine Cytosine Guanine two hydrogen bonds 15. Both have the same purine bases i. Hence cellulose on hydrolysis produces only Dglucose units. (d) : An amino acid is a bifunctional organic molecule that which normally functions effectively at body temperature. (b) : DNA RNA of their nucleotide monomers. (a) : RNA contains cytosine and uracil as pyrimidine bases while DNA has cytosine and thymine. contains both a carboxyl group. –COOH. (d) : Enzymes are shape selective specific biological catalysts 16. (c) : The two polynucleotide chains or strands of DNA are linked up by hydrogen bonding between the nitrogenous base molecules 17. Which one of the following types of drugs reduces fever? (a) Analgesic (b) Antipyretic (c) Antibiotic (d) Tranquiliser (2005) 4. (c) 5. Aspirin in known as (a) phenyl salicylate (b) acetyl salicylate (c) methyl salicylic acid (d) acetyl salicylic acid (2012) 2. (2003) (b) OCOCH3 (c) H2 C CH – CN and H2 C CH – CH CH2 5. is used as (c) (2002) . Which of the following could act as a propellant for rockets? (a) Liquid hydrogen + liquid nitrogen (b) Liquid oxygen + liquid argon (c) Liquid hydrogen + liquid oxygen (d) Liquid nitrogen + liquid oxygen. (d) 2. Cl Answer Key 1. (c) 3. (b) 4.107 Chemistry in Everyday Life CHEMISTRY IN EVERYDAY LIFE CHAPTER 29 1. BunaN synthetic rubber is a copolymer of (a) 3. and (d) H2 C CH – CN H2 C CH – C CH2 (a) antiseptic (c) analgesic (2009) COOH The compound (b) antibiotic (d) pesticide. (c) : BunaN is a copolymer of butadiene and acrylonitrile. . (b) : An antipyretic is a drug which is responsible for lowering temperature of the feverish organism to normal but has no effect 1. (c) : The compound is acetyl salicylic acid (Aspirin). 4. (c) : Liquid hydrogen (because of its low mass and high enthalpy of combustion) and liquid oxygen (as it is a strong supporter of combustion) are used as an excellent fuel for rockets.108 JEE MAIN CHAPTERWISE EXPLORER 3. Drugs which relieve or decrease pain are termed analgesics. 5. (d) : Aspirin 2. on normal temperature states. PRINCIPLES RELATED TO PRACTICAL CHEMISTRY Which of the following reagents may be used to distinguish 3.109 Principles Related to Practical Chemistry CHAPTER 30 1. (a) The compound formed in the positive test for nitrogen with the Lassaigne solution of an organic compound is (a) Fe 4 [Fe(CN) 6 ] 3 (b) Na 3[Fe(CN) 6] (c) Fe(CN) 3 (d) Na 4[Fe(CN) (2004) 5NOS. (d) 2. 2. (b) 3. between phenol and benzoic acid? (a) Aqueous NaOH (b) Tollen’s reagent (c) Molisch reagent (d) Neutral FeCl 3 (2011) Biuret test is not given by (a) proteins (b) carbohydrates (c) polypeptide (d) urea (2010) Answer Key 1. . (a) : 3Na 4 [Fe(CN) 6 ] + 4Fe 3+ ® Fe 4 [Fe(CN) 6 ] 3 + 12Na + Prussian blue . solution. Benzoic acid gives buff coloured (pale dull yellow) precipitate 3. (b) : Biuret test is used to characterise the presence of —CONH group in a compound.110 1. JEE MAIN CHAPTERWISE EXPLORER (d) : Phenol gives violet colouration with neutral ferric chloride 2. with neutral ferric chloride solution. MATHEMATICS . Z Í X and Y Ç Z is empty is (a) 2 5 (b) 5 3 (c) 5 2 (d) 3 5 (2012) 3. y) : y = x + 1 and 0 < x < 2} T = {(x. let f (x) = x + 5x + 1. B and C are three sets such that A Ç B = A Ç C and A È B = A È C. 4. then (a) A = C (c) A Ç B = f (b) B = C (d) A = B (2009) 3 For real x. (2011) (a) (– ¥. x ³ –1. (c) Statement1 is true. Relations and Functions SETS. Statement2 : f is bijection. Then (a) R is an equivalence relation but S is not an equivalence relation (b) neither R nor S is an equivalence relation (c) S is an equivalence relation but R is not an equivalence relation (d) R and S both are equivalence relations (2010) Let f (x) = (x + 1) 2 – 1. Show that f is invertible and its inverse is (a) g ( y ) = y . y) : x – y is an integer}. Statement2 : B = {(x. n. ¥) 5. Statement2 is true; Statement2 is a correct explanation for Statement1. Which one of the following is true? (a) T is an equivalence relation on R but S is not (b) Neither S nor T is an equivalence relation on R (c) Both S and T are equivalence relations on R (d) S is an equivalence relation on R but T is not (2008) Let f : N ® Y be a function defined as f (x) = 4x + 3 where Y = {y Î N : y = 4x + 3 for some x Î N}. y)|x. 2. Statement2 is false. Statement2 is true. Statement1 : The set {x : f (x) = f –1 (x)} = {0. ÷ m. 5}. y) Î R × R : x = ay for some rational number a} is an equivalence relation on R. n. Let R be the set of real numbers. (b) Statement1 is false. –1}. 7. ¥) – {0} (d) (0. y are real numbers and x = wy for some rational number w}; ì æ m p ö S = í ç . 3. (c) Statement1 is true. q ¹ 0 and qm = pn}. Statement2 is true; Statement2 is not a correct explanation for Statement1. Statement2 is true; Statement2 is a correct explanation for Statement1. The number of subsets of A × B having 3 or more elements is (a) 211 (b) 256 (c) 220 (d) 219 (2013) Let X = {1. (a) Statement1 is true. respectively. ¥) The domain of the function f ( x ) = (2011) Consider the following relations: R = {(x. 2. then (a) (b) (c) (d) f is onto R but not oneone f is oneone and onto R f is neither oneone nor onto R f is oneone but not onto R (2009) Let R be the real line. 1 is | x | - x (b) (– ¥. 0) (c) (– ¥. p and q are integers such that î è n q ø 10. RELATIONS AND FUNCTIONS CHAPTER 1 1.1 Sets. Statement2 is true; Statement2 is not a correct explanation for Statement1.3 4 (c) g ( y ) = 4 + y + 3 4 (b) g ( y ) = 3 y + 4 3 (d) g ( y ) = y + 3 4 (2008) . Z) that can be formed such that Y Í X. (2009) If A. The number of different ordered pairs (Y. y) Î R × R : y – x is an integer} is an equivalence relation on R. 9. 4. (b) Statement1 is false. Consider the following subsets of the plane R × R : S = {(x. Let A and B be two sets containing 2 elements and 4 elements 6. (d) Statement1 is true. Statement2 is true. (a) Statement1 is true. Statement2 is false. Statement1 : A = {(x . (d) Statement1 is true. 8. 2. not symmetric and transitive.n .. ¥) x 3 – 3x 2 + 3x + 3 24. The function f ( x) = log( x + x 2 + 1) is 14. 5. 5}. (2003) é0. (4. defined by f ( x ) = sin x . (2006) (a) f (x) = f (–x) (b) f (2 + x) = f (2 – x) (c) f (x + 2) = f (x – 2) (d) f (x) = –f (–x). Let R = {(3. (2004) 13. 4).. (6. p ö p (b) 0. 2). (2. è 1 - x 2 ÷ø (c) neither an even nor an odd function then f is both oneone and onto when B is the interval (d) an even function. 3) (6. (2004) 12! (c) 3!(4!) 3 12! (d) 3!(4!) 4 . –4] x 3 + 6x 2 + 6 f ( x ) = 3 2 + log10 ( x3 . ¥) (d) (1. 2) ( ) ( ) 16. 3. (3. The number of ways to partition 17. 2). 3. 0) È (1. 2). A function is matched below against an interval where it is supposed to be increasing. then the interval of S is letter in common}. 4}. The domain of the function f ( x ) = sin ( x . 9). 3) (c) [2.2 JEE MAIN CHAPTERWISE EXPLORER 11. The relation . Let R = {(1. 2. The relation R is 12! 12! (a) not symmetric (b) transitive (a) (4!) 3 (b) (4!) 4 (c) a function (d) reflexive. (2007) 18. ¥) (c) (–1. ¥) 2x 3 – 3x 2 – 12x + 6 (d) oneone but not onto. 2.x ). (12. R = {(x.3 cos x + 1. 6) (9. Which of the following pairs is (a) onto but not oneone incorrectly matched? (b) oneone and onto both Interval Function (c) neither oneone nor onto (a) [2. 3]. 9).. (2005) 22. symmetric and transitive (b) reflexive. 3] (d) [1. 1)} be a relation on S is the set A = {1. . Let f : (–1. is 1 ù 4 - x æ (d) ç -¥ . 1] (b) [–1. 3). (3. 2. 2 û (2005) is by f (n ) = í 2 ú ï . 12). 3} (d) {1. (b) an equivalence relation (2004) (c) reflexive only (d) reflexive and transitive only. C of equal size. B. (3. (c) f (–x) (d) f (a) + f (a – x). (3. 12}. 9. Thus A È B È C = S.3) is 9 - x 2 (a) reflexive and symmetric only (a) [1. (2003) . 3. (2004) relation R by : 19. 0) È (1. (2005) A Ç B = B Ç C = A Ç C = f. symmetric and transitive line x = 2. Let W denote the words in the English dictionary. y) Î W × W | the words x and y have at least one is onto. when n is even î 2 15. when n is odd p p é p p ù . 1] Then R is (c) [0. 3. If f : R ® S.2 . 2. (2003) (b) (–¥. 12). 2 (a) ê 2 ÷ 23.12} is to be partitioned into three (a) f (x) (b) –f (x) sets A. 3). 2. then (d) reflexive. A function f from the set of natural numbers to integers defined ë ø ì n . 2] (b) [2. The set S = {1. 4. (2. 6} 12. 3 ú 3x 2 – 2x + 1 (2005) è û (a) (–1. 12). A real valued function f (x) satisfies the functional equation f (x – y) = f (x) f (y) – f (a – x) f (a + y) where a is a given constant and f (0) = 1. f (2a – x) is equal to (b) (1. 6. 2) È (2. Define the (c) {1. be a function defined by (a) an odd function æ 2 x ö (b) a periodic function f ( x ) = tan -1 ç . 2 (d) ëê . symmetric and not transitive 20. 4. The range of the function F(x) = 7 – x P x – 3 is (a) {1. . 3] (d) [–1.. (2004) (a) not reflexive. (a) [0. The graph of the function y = f (x) is symmetrical about the (c) reflexive.1 .1 is 21. 2) È (2. 3. Domain of definition of the function (c) (–¥. 4} (b) {1. ï (c) - 2 . 6)} be a relation on the set A = {3. 1) ® B. 21. (d) 2 (2003) 26. (b) (b) (c) (c) . 1] (d) [–9. Which one is not periodic? (a) |sin3x| + sin 2 x (b) cos x + cos 2 x 2 (c) cos4x + tan x (d) cos2x + sinx. (d) (b) (d) (d) (c) 2. 10. 7. 18. (a) (a) (b) (a) (a) 5. 14. 13. then å f ( r ) is r = 1 7( n + 1) (a) 2 7n(n + 1) (c) 2 ( ) (b) 7n(n + 1) x ù 28. 15. The period of sin 2q is (a) p 2 (b) p (c) p 3 (d) p/2. 12. (a) [1. 22. (2002) and f (1) = 7. 9] (c) [–9. 26. 19. 27. 28. 20. The domain of sin -1 éê log 3 is ë 3 úû 7 n . 23. 17. (c) (a) (a) (b) 6. If f : R ® R satisfies f (x + y) = f (x) + f (y). 8. 9. 11. (2002) (2002) Answer Key 1. y Î R n 27. 9] (b) [–1. for all x. –1]. 25. (d) (b) (c) (c) (b) 3. (a) (a) (d) (b) (b) 4. 16. 24. Relations and Functions 25.3 Sets. Y Ç Z = f Þ x Î B È A Þ x Î B Number of ways = 3 5 . y) Î Z and (y. ÷ø Î S n b k Î N. ÷ø Î S since mn = mn. 4. z) Î A Þ (x. y) Î A and (y. x – x = 0 Î Z n n symmetric – for (x. 5. 2. ÷ Î S . f ¢(x) = 3x 2 + 5 > 0 \ f is strictly increasing (c) : We have (x. The number of subsets having atleast 3 elements = 8 C 3 + 8 C 4 + 8 C 5 + 8 C 6 + 8 C 7 + 8 C 8 = 2 8 – ( 8 C 0 + 8 C 1 + 8 C 2 ) = 256 – 1 – 8 – 28 = 219 (b) : Let x Î C Suppose x Î A Þ x Î A Ç C Þ x Î A Ç B (Q A Ç C = A Ç B) Thus x Î B Again suppose x Î/ A Þ x Î C È A (d) : X = {1.e.(2) \ (x. which gives (x + 1) 2 – 1 = x The function. 0 < x < 2} is not æm pö æ p a ö reflexive for (x. then y = f (x) Þ 4k + 3 = 4x + 3 Þ x = k Î N This means S is transitive. because it’s a polynomial function Thus x must be –ve. Thus R is not Hence f is oneone also. x) Î Z i. corresponding to any y Î codomain there $ some x Î (a) : f ( x ) = | x | - x domain such that f (x) = y. Thus S is symmetric. n = q and q = b . symmetric and transitive. (x. Thus corresponding to any y Î Y we have x Î N. The function (b) : The solution of f (x) = f –1 (x) are given by then is onto. Z Í X. |x| – x > 0 Þ |x| > x Also f is continuous on R.4 1.3 y = 4x + 3 Þ x = \ x = –1. y) Î Z Þ x – y Î Z æ m p ö Þ y – x Î Z i. z) Combining (1) and (2) we get B = C. x) Î S would imply x = x + 1 Again.3 or g ( y ) = is the inverse of the function. w) Î Z. 0) Ï R for any w but (0. 10. being both oneone and onto is invertible. x 1. . = Þ 4x 1 + 3 = 4x 2 + 3 Þ x 1 = x 2 n b Thus f (x 1 ) = f (x 2) Þ x 1 = x 2 . x) Î R for w = 1 implying that R is reflexive.e. For a ¹ 0. y) is in B but (y.(1) Þ z – x = integer Similarly we can show that B Í C . (b) : The function is f : R ® R Also (x. As a polynomial of odd degree has always at least one real 1 root. x) Þ Symmetric Let y Î R then y = x 3 + 5x + 1 Hence A is a equivalence relation but B is not. q ÷ø Î S Þ qm = pn transitive – for (x. about f being ONTO or not. giving ç . 0) is not in B. ç .3 x . nothing can be said y . y) : x – y Î Z} is æ m m ö As çè . 0 \ f -1 ( x ) = 4 4 But as no codomain of f is specified. giving p m ö æ x – w Î Z i. 4. Hence f is ONTO.. 3. y) : y = x + 1. a) Î R. S is reflexive. symmetric. x) Î A is true Þ Reflexive f (x) = x 3 + 5x + 1 As (x. ÷ Î S èn qø è q b ø Þ 0 = 1 (impossible) Thus S is not an equivalence relation means qm = pn and bp = aq. w) Î Z But this can be written as np = mq. 5}; Y Í X .. Þ x 3 + 5x + 1 – y = 0 (0. Thus in both cases x Î C Þ x Î B (a) : y – x = integer and z – y = integer Hence C Í B . 6. x 2 Î N m p p a m a i.e. \ T is an equivalence relation on R. 3. i.e. T = {(x. (y. reflexive – for (x. reflexive. \ x Î (– ¥. f (x) = x. 4 7. Hence the function is one æm aö one.. y) Î A Þ (y. Þ Transitive 8. ÷ Î S and ç .. (a.. (a) : Let f (x 1 ) = f (x 2 ). 2. for some Thus çè . Þ x – y Î Z and y – w Î Z. 9. Let y Î Y be a number of the form y = 4k + 3.e. (a) : To be an equivalence relation the relation must be all – Hence R is not an equivalence relation. 0). x) Î Z çè n . è q n ø S = {(x.. JEE MAIN CHAPTERWISE EXPLORER (d) : A × B will have 2 × 4 = 8 elements. Þ (x + 1) 2 – (x + 1) = 0 Þ (x + 1)x = 0 y . But (x. 1) we have 2 – x x = 2 2 + x f ( x ) = tan -1 çæ 2 x 2 ÷ö è 1 . b) Î R . (9. 3) Î R but (3. (c): If y = f (x) is symmetrical about the line x = a then Þ (a. 2 ÷ 2 è 1 . 3) Ï R 2 2 ú ëê û \ relation is not symmetric which means our choice (a) and = 2 sin (x – 60°) (b) are out of court. 3} is required range \ Let x = DON. q(x) ¹ 0 = tan –1 tan2q = 2 tan –1 x Þ - f (x + 2) = f (x – 2) p p £ sin –1 (x – 3) £ 2 2 p p Þ – sin £ x – 3 £ sin 2 2 now D f of p(x) is - p 2 x ö p < tan -1 æç < . (d) : For (3. (iii) Let (x. Relations and Functions R is not reflexive as (1. 5 letter in common. f (0) = 1 ) 17. (b) : Given relation R such that R = {(x. 9) Î R. a) Î R which is hold i. 16. 3) Î R and (3. (a. Þ R is symmetric.e. (a) : Number of ways = 12 C4 ´ 8 C4 ´ 4 C4 = 12! . 6)}. x) has every letter common \ R is reflexive \ from (i). y) Î R and (y. 4. 12. (d): Let f (x) = g(x) + 1 \ R is reflexive. 12). 14. z = SHE then (x. R is reflexive. y = NEST. (9. 6).e. é1 ù 3 where g(x) = 2 ê sin x cos x ú 13. \ F(3) = 4 P 0 . symmetric but not transitive. 9). F(4) = 3 P 1 .x ù úû ëê é 1 + x 2 + é ù 1 = –log ê ú = – log ê 1 ê êë 1 + x 2 . y) Î W × W | the word x and y have atleast 18. 4} and (2.(i) Þ x 2 < 9 |x| < 3 i.(*) f ( a ) = 0. \ –2 £ 2 sin (x – 60°) £ 2 For reflexivity a Î R. R is not symmetric as (2.e (–¥.. z) Î R. 4) Î R and (2. (3. We need to prove reflexivity and transitivity. (b) : f (x) = (By x = tan q ) p ( x ) (say) q ( x ) then Domain of f (x) is D f p(x) Ç D f q(x). z) Ï R. 3)(6. (b.. –1 £ 2 sin (x – 60°) + 1 £ 3 Again. (a) : R is a function as A = {1. (d) : f (x) = x 3 + 6x 2 + 6 f ¢( x ) = 3 x 2 + 12 x = 3 x ( x + 4) –3 f ¢( x ) > 0 Þ x < -4 È x > 0 the interval x < – 4 i. 12). (6. 2. (ii). 2) Ï R (4!) 3 R is not transitive as (1. F(5) = 2 P 2 But R is not transitive \ {1. –3 < x < 3 . 1) Ï R.tan q ø \ 21. 19. x ) Î R " x Î W (iii) x – 3 £ 7 – x Þ x £ 5 \ (x. y) Î R then (y.x ø 2 tan q ö \ f (tan q ) = tan -1 æç 2 ÷ è 1 . c) Î R 20. (b) : Given f (x – y) = f (x) f (y) – f (a – x) f (a + y) let x = 0 = y f (0) = ( f (0)) 2 – ( f (a)) 2 1 = 1 – ( f (a)) 2 Þ f (a) = 0 \ f (2a – x) = f (a – (x – a)) = f (a) f (x – a) – f (a + x – a)f (0) By using (*) = 0 – f (x)(1) = – f (x) ( Q .. one letter in common} (i) \ 7 – x > 0 Þ x < 7 where W denotes set of words in English dictionary (ii) x – 3 ³ 0 Þ x ³ 3 Clearly ( x . (12. (c) : F(x) to be defined for x Î N.x úû ë xù ú ú û . – 4] matched correctly and after checking others we find that f (x) = 3x 2 – 2x + 1 Þ f ¢(x) > 0 for x > 1/3 which is not given in the choice.5 Sets. (3. 2. (c) : For x Î (–1. (3. x) Î R as x and y have atleast one x = 3.. 9).x ø 2 15. 3.. (a) : f (x) = log é x 2 + 1 + êë xù úû 2 \ f (–x) = log é 1 + x . 12).. 1) Î R but (1. 1) Ï R 11. 3) Î R 2 3 4 Þ 2 £ x £ 4 Again 9 – x 2 > 0 .(ii) From (i) and (ii) we have \ 2 £ x < 3 is correct Domain 22. for transitivity of (a. c) Î R f (x + a) = f (x – a) which is also true in R = {(3. (a) : If y = sin –1 a.1 - 1 Þ 1 = 2 2 – + f (3) = f (1) + f (2) \ x ¹ ± 2 cos x let us take. (b) : If n is odd. f (x) = cos x Let f (x + T) = f (x) Þ cos T + x = cos x Þ T + x = 2np ± x which gives no value of T independent of x \ f (x) cannot be periodic Now say g(x) = cos 2 x + cos x which is sum of a periodic and non periodic function and such function have no period.. 24.. d) Similarly we can say that cos 4x + tan 2 x and cos2x + sin x are periodic function. Finally f (n) is oneone onto function. n is even) Let f (2k 1 ) = f (2k 2 ) 2 k1 2 k = ..6 JEE MAIN CHAPTERWISE EXPLORER = – f (x) Þ f (x) + f (–x) = 0 Þ f (x) is an odd function. then –1 £ a £ 1 é é æ x ö ù ù æxö \ –1 £ log 3 ç ÷ £ 1 ê as y = sin -1 ê log 3 ç ÷ ú ú è 3 ø û û è 3 ø ë ë Þ x 1 £ £ 31 3 3 Þ 1 £ x £ 9 . y = 1 \ f (1 + 1) = 2f (1) = 2(7) \ f (2) = 2(7) 27.2 2 2 Þ k 1 = k 2 Þ f (n) is oneone if n is even Þ - Again f (n) = n - 1 2 (b) Now cos 2 x is periodic with period p and for period of 1 f ¢ (n) = > 0 [n Î N if n is odd 2 - 1 and f ¢ (n) = < 0 [n Î N if n is even 2 Now all such function which are either increasing or decreasing in the stated domain are said to be onto function.C.M of their periods LCM (p.p ) æp ö \ L. (c) : Let x = 0 = y \ f (0) = 0 and x = 1. Þ k 1 = k 2 26.C. 0) È (1. So.. let n = 2k – 1 Let n f (2k 1 – 1) = f (2k 2 – 1) \ = 7(1 + 2 + 3 + .1 . cos x + cos 2 x is non periodic function.x 2 \ D f g(x) = R – {–2.e. (c) : Let g(x) = 3 4 ..+ n) = 7n(n + 1) 2 p and period of sin 2 x is p 3 1 - cos 2 x (a) Same as the period of |sin x| or whose period 2 is p Now period of |sin 3x| + sin 2 x is the L. ¥ ) 25. 0) È (1.1 2k . 23. x = 1. 2) È (2. (b) : Period of |sin 3x| is Þ f (n) is oneone functions if n is odd Again. ¥ ) –1 0 1 \ Domain of f (x) is (–1.M of ç . If n = 2k (i. + f (n) r = 1 2 k 2 . \ x 3 – x > 0 + \ x Î (–1. 2} f (x) = log 10 (x 3 – x) x(x + 1) (x – 1) > 0 – å f (r ) = f (1) + f (2) + f (3) + . (b) : Let f (q) = sin 2q = |sin q| Period of |sin q| is p 28. p ÷ = = p HCF (3. y = 2 \ = 7 + 2(7) = 3(7) and so on.1) è 3 ø (c. y = 0 \ f (1 + 0) = f (1) + f (0) = 7 (given) x = 1. B) equals (a) (1. If the cube roots of unity are 1. + ç z + 6 ÷ is z ø è z ø z ø è zø è è (a) 18 (a) –p (b) 54 (b) p/2 (c) 6 (d) 12. 1) 12. –1 + 2w. 1 – 2w. then (1 - i ) (a) x = 2n. w. æ 1 + z ö ÷ equals then arg çè 1 + z ø p . 0) (b) (–1. 1) (c) (0. (2005) 13.7 Complex Numbers CHAPTER COMPLEX NUMBERS 2 1. and Arg(z) – Arg(w) = p/2. where n is any positive integer (b) x = 4n + 1. then the maximum value of | Z | is equal to Z (c) 2 + 2 (d) 3 + 1 5 + 1 (b) 2 (2009) If Z (a) 6. then the value If z is a complex number of unit modulus and argument q. 3. 1 + 2w 2 (d) –1. 1) (d) (1. where n is any positive integer (c) x = 2n + 1. where z is a complex number. Then arg z equals (a) 3p/4 (b) p/2 (c) p/4 (d) 5p/4. then (a) 2 (b) –1 (c) 1 ( ) is equal to x y + p q 2 2 ( p + q ) (d) –2. 2.. are (a) –1. where n is any positive integer (d) x = 4n. If z 2 + z + 1 = 0. (2007) 8. (2003) . (2012) 10. (2004) 15. then arg z 1 – argz 2 is equal to If w(¹ 1) is a cube root of unity. If z and w are two nonzero complex numbers such that |zw| = 1. Then (A. If w = z - (1/ 3) i (a) a circle (b) an ellipse (c) a parabola (d) a straight line. If |z + 4| £ 3.. then zw is equal to (a) –1 (b) i (c) –i (d) 1. then z lies on 11. (2003) 17. (2006) 2 2 2 2 æ 1 ö æ 2 1 ö æ 3 1 ö æ 6 1 ö of ç z + ÷ + ç z + 2 ÷ + ç z + 3 ÷ + . –1. then the point represented by the z - 1 complex number z lies (a) either on the real axis or on a circle not passing through the origin. w be complex numbers such that z + i w = 0 and zw = p. If | z 2 – 1 | = | z | 2 + 1. and (1 + w) 7 = A + Bw. 1 + 2w.q (b) q (a) (c) p – q (d) – q 2 (2013) The conjugate of a complex number is number is 1 (a) i - 1 (b) -1 i - 1 (c) 1 . (c) either on the real axis or on a circle passing through the origin. 9. (2004) 1/3 14. then the maximum value of |z + 1| is (a) 6 (b) 0 (c) 4 (d) 10. (2005) (2011) The number of complex numbers z such that |z – 1| = |z + 1| = |z – i| equals (a) 0 (b) 1 (c) 2 (d) ¥ (2010) 5. Then that complex i - 1 1 i + 1 (d) -1 i + 1 (2008) 7. 1 – 2w 2 . w 2 then the roots of the equation (x – 1) 3 + 8 = 0. –1 (b) –1. (2006) (c) –p/2 (d) 0. 4 = 2. (2005) z and |w| = 1. Let z. If z 1 and z 2 are two nonzero complex numbers such that |z 1 + z 2 | = |z 1 | + |z 2 |. (b) on the imaginary axis. then z lies on (a) a circle (b) the imaginary axis (c) the real axis (d) an ellipse. z 2 If z ¹ 1 and is real. If 1 + i = 1 .. (d) on a circle with centre at the origin. (2004) x 16. If z = x – iy and z = p + iq. The value of 2k p 2 k p ö æ + i cos ÷ is 11 11 ø k =1 è (b) 1 (c) –1 10 (a) i å ç sin (d) –i. 4. where n is any positive integer. –1 – 2w 2 (c) –1. then (c) Re(z) > 3 (d) Re(z) > 2. (d) 5. (b) 14.4 < z - 2 . (c) 7. (b) . z 20. The locus of the centre of a circle which touches the circle (c) a 2 = 4b (d) a 2 = b. z and w are two nonzero complex number such that | z | = | w | numbers) will be and Arg z + Arg w = p then z equals (a) an ellipse (b) a hyperbola (a) w (b) – w (c) w (d) – w (c) a circle (d) none of these (2002) (2002) Answer Key 1.8 JEE MAIN CHAPTERWISE EXPLORER 18.z1 = a and z . its solution is given by being complex further. (a) 11. (c) 6. If z . (2003) z . z 1 and z 2 form (a) Re(z) > 0 (b) Re(z) < 0 an equilateral triangle. (d) 20. (a) 8. (d) 12. (2002) (a) a 2 = 2b (b) a 2 = 3b 21. (c) 3. z 1 & z 2 are complex 19. (d) 16. 9. (d) 17. (d) 18. (b) 10. 15. assume that the origin. Let z 1 and z 2 be two roots of the equation z 2 + az + b = 0. (d) (d) (b) (b) 4. (b) 2. 21. (d) 13 (a) 19.z 2 = b externally (z. 2 £ r - z 2 is real. 4.. 0) and radius 3 units. 1).(ii) 6.2r . 1 1 -1 = = i - 1 ..5 £ r £ 1 + 5 (i) and (ii) gives Þ ( z . 3. These points are noncollinear.1) 2 + y 2 Þ – y(x 2 – y 2 ) + 2xy(x – 1) = 0 Þ y(x 2 + y 2 – 2x) = 0 Þ y = 0 or x 2 + y 2 – 2x = 0 \ \ z lies on real axis or on a circle passing through origin. If z is a real number. (a) : We have for any two complex numbers a and b ||a| – |b|| £ |a – b| 5 ..£2 Þ r 2 . 1.4 £ 0 r The corresponding roots are 2 ± 20 r= = 1 ± 5 2 Thus 1 .1 Let z = x + iy 2 4 £ 2 r 10 = å æç sin 2 k p + i cos 2 k p ö÷ = . hence r £ 1 + 5 . 2 nd solution : We resort to definition of modulus.1)( z .z .1 i + 1 Im (–1. then r - 4 £ 2 r The left inequality gives r 2 + 2r – 4 ³ 0 The corresponding roots are Þ . (b) : Note that Set |Z| = r > 0. hence the desired number is the centre of the (unique) circle passing through these three noncollinear points. 0) is 6 units. Þ zz .1 £ r £ 5 + 1 We have z = ( z ) giving z = Þ z + z = 0 (z being purely imaginary) Þ 1 + y 2 = y 2 – 2y + 1 \ y = 0 r £ -1 - 5 So. 1 and i. Hence maximum distance of z from (–1. the greatest value is 5 + 1.. (d) : (1 + w) 7 = (–w 2 ) 7 = –w 14 = –w 12 w 2 = –w 2 = 1 + w = A + Bw given Hence on comparison.iy ) is real ( x . 9. 5. z - 1 2 ( x . |z – 1| 2 = |z – i| 2 Thus. w 2 \ æ z + 1 ö + æ z 2 + 1 ö + . (b) : 1 st solution : |z – 1| = |z + 1| = |z – i| reads that the distance of desired complex number z is same from three points in the complex plane –1. we have (A.9 Complex Numbers 1 + z 1 + z = = z 1 + z 1 + 1 z 2 Observe that | z | = 1 = zz 1 + z Then the arg of the number is just the argument of z 1 + z and that’s q. 0) n 2 k p ö æ 2kp (d) : å ç sin n + 1 + i cos n + 1 ÷ ø k =1 è \ Þ (x – 1) 2 + y 2 = x 2 + (y – 1) 2 4 4 4 Now | Z | £ ZÞ | Z | £ 2 |z| Z | Z | 1 i - 1 (a) : z lies on or inside the circle with centre (–4. (c) : z ¹ 1. Again. |z – 1| = |z + 1| Þ |z – 1| 2 = |z + 1| 2 . 0) (–4..1) = ( z + 1)( z + 1) Thus x = 0 -2 ± 20 = ± 5 2 Thus r ³ 5 . 0) is the desired point. B) = (1.(i) But r > 0.y + 2 xiy ) (( x .1 (As r > 0) Again consider the right inequality 4 r..1) . 8...z + 1 = zz + z + z + 1 Þ 1 + y 2 = (y – 1) 2 (because x = 0) or implies that r ³ 5 . z - 1 2.i .i 11 11 ø k =1 è (d) : z 2 + z + 1 = 0 Þ z = w. then r= z 2 is real. (d) : z = 7. æ z 6 + 1 ö ç ÷ ç ÷ z ÷ø çè è z2 ø z 6 ø è 2 2 = 4 (w + w 2 ) 2 + 2(w 3 + w 3 ) 2 = 4 (– 1) 2 + 2(2 2 ) = 4 + 8 = 12 2 O Real . (0. =(x 2 + y 2 ) + 1 Again consider x = –1. 2. 1 – 2w 2 are roots of (x – 1) 3 + 8 = 0 but on the other hand the other roots does not satisfies the equation (x – 1) 3 + 8 = 0. –1 cannot be the proper choice. 1 – 2w 2 (i = 1.i ø Þ (x + 1) 3 + 2(x + 1) 2 (w 2 – w) – 4(x + 1) = 0 x æ 2 i ö which cannot be expressed in the form of given equation (x – Þ ç ÷ = 1 1) 3 + 8 = 0. (d) : Method I : Let z 1 = cosq 1 + i sinq 1 .. z 1 z 2 = b Þ x 3 – 3x 2 + 3x – 1 + 8 = 0 Þ (x – 1) 3 + 8 = 0 which matched Again 0. 13 (a) : Þ Þ Þ Þ 3 z 3 z \ w = 3 z . 11. (d) : Method (1) : (By making the equation from the given x y roots) + = –2(p 2 + q 2 ) Let us consider x = –1. –1 –2w 2 Þ 2x = 0 \ Required equation from given roots is x = 0 Þ (x + 1)(x + 1 – 2w)(x + 1 + 2w 2 ) = 0 Þ z lies on imaginary axis. p q Adding the equations of (*) we get 12. Now consider the roots è 2 ø x i = –1. z 2 = cosq 2 + isinq 2 \ z 1 + z 2 = (cosq 1 + cosq 2 ) + i(sinq 1 + sinq 2 ) Now |z 1 + z 2 | = |z 1 | + |z 2 | Þ (cos q1 + cos q2 )2 + (sin q1 + sin q2 )2 = 1 + 1 Þ 2(1 + cos(q 1 – q 2 )) = 4 (by squaring) Þ cos(q 1 – q 2 ) = 1 Þ q 1 – q 2 = 0 (Q cos0° = 1) Þ Arg z 1 – Arg z 2 = 0. Method 2 (by taking cross checking) As (x – 1) 3 + 8 = 0 . –1. z 1 .i = 3 z Þ x – iy = p 3 – 3pq 2 + i(3p 2 q – q 3 ) Þ 3( x ) + i (3 y . (c) : | zw | = 1 Þ | z || w | = 1 So | z | = – Product of roots taken all at a time = 0 | w | Now sum of roots x 1 + x 2 + x 3 p Again Arg ( z ) – Arg (w) = 2 = –1 + 1 – 2w + 1 – 2w 2 = 3 z 2 z Product of roots taken two at a time \ = i = z i from (1) w w = –1 + 2w – 1 + 2w 2 + 1 + 2(w 2 + w) + 4w 3 = 3 z 1 Product of roots taken all at a time = z z i Þ z w = = - i .. For real axis y = 0 Þ (x + 1)[(x + 1) 2 + (x + 1)(2w 2 – 2w) – 4w 3 ] = 0 x 1 + i æ ö Þ (x + 1)[(x + 1) 2 + 2(x + 1)(w 2 – w) – 4 ] = 0 16.10 JEE MAIN CHAPTERWISE EXPLORER 10. 1 – 2w.i 3 z - i z + iw = 0 z = -iw Þ z = iw w = –iz \ arg (–iz 2 ) = p arg (–i) + 2arg (z)= p 2arg(z) = p + p/2 = 3p/2 arg(z) = 3p/4 x = 0 Imaginary axis 14. (–2) 3 + 8 = 0 Þ 0 = 0 Similarly for 1 – 2w we have (x – 1) 3 + 8 = 0 Þ (1 – 2w – 1) 3 + 8 = 0 Þ (–2w) 3 + 8 = 0 Þ –8 + 8 = 0 and for 1 – 2w 2 we have (1 – 2w 2 – 1) 3 + 8 = 0 0 z 1 z 2 \ 0 2 + z 1 2 + z 2 2 = 0z 1 + z 1 z 2 + z 2 0 = 0 Þ z 1 2 + z 2 2 = z 1 z 2 (z 1 + z 2 ) 2 = 3z 1 z 2 a 2 = 3b . n Î I + and the equation with these roots is given by 3 2 x – (sum of the roots)x + x(Product of roots taken two at a time) 1 .(*) and x = –1 satisfies (x – 1) 3 + 8 = 0 i. –1 + 2w. (b) : |z – 1| = |z| + 1 Þ Let z = x + iy (x + 1) 3 = 0 which does not match with the given equation 2 + y 2 Þ (x – 1) (x – 1) 3 + 8 = 0 so x = –1. 1 – 2w. z 2 are vertices of an equilateral triangle with given equation. \ w i 2 = (–1)[(1 – 2w)(1 – 2w )] = –7 3 2 \ Required equation is x – 3x + 3x + 7 = 0 18.e. z 2 are roots of z 2 + az + b = 0 \ z 1 + z 2 = –a. (1) 17. –1 p q \ Required equation from given roots is 2 2 (x + 1)(x + 1)(x + 1) = 0 15. –1.. 3) Þ i x = 1 Þ i x = (i) 4n Þ x = 4n.1) = 3( x + iy ) ( z = x + iy ) Þ x = p 3 – 3pq 2 and y = –(3p 2 q – q 3 ) x y Þ (3x) 2 + (3y – 1) 2 = 9(x 2 + y 2 ) Þ 6y – 1 = 0 which is straight = p 2 – 3q 2 and = –(3p 2 – q 2 ) (*) line. (d) : z 1/3 = p + iq Þ x – iy = (p + iq) 3 Þ 3 z . (b) : As z 1 . (d) : Given w = Þ w 6 (–8) + (8) = 0 Þ 0 = 0 \ – 1.. (d) : Given ç ÷ = 1 è 1 . (c) : |z – 4| < |z – 2| or |a – 4 + ib| < |(a – 2) + ib| by taking z = a+ib Þ (a – 4) 2 + b 2 < (a – 2) 2 + b 2 Þ –8a + 4a < –16 + 4 Þ 4a > 12 Þ a > 3 Þ Re(z) > 3 21. . A hyperbola is the locus of a point which moves in such a way that the difference of its distances from two fixed points (foci) is always constant. (b) : Let |z| = |w| = r \ z = re ia and w = re ib where a + b = p (given) - = re Now Z e ia = re i(p – b) = re ip × e –ib = – re –ib = – w 20. which represent a hyperbola Since.11 Complex Numbers 19. (b) : r a z 1 z 3 b z 2 z 1 z 3 – z 3 z 2 = (a + r) – (b + r) = a - b = a constant. 6. where I is 2 × 2 identity matrix. Statement2 is true; Statement2 is (a) Statement1 is true. Statement2 is true; Statement2 is not a correct explanation for Statement1. Statement2: |adj A| = |A| (d) Statement1 is true.1 ö ç 1 ÷ ç 1 ÷ (c) ç ÷ (d) ç ÷ è 0 ø è -1ø 2 2 (b) – 1 (c) – 2 2 (d) 1 (2012) Let A and B be two symmetric matrices of order 3. Statement2 is false. b.12 JEE MAIN CHAPTERWISE EXPLORER MATRICES AND DETERMINANTS CHAPTER 3 1. If P 3 = Q 3 and 2 5.1 ÷ (a) ç ÷ è 0 ø æ 1 ö ç . (2010) The number of 3 × 3 non‑singular matrices. then u 1 + u 2 is equal to ç ÷ ç ÷ è0ø è 0 ø 4. The system has (2013) The number of values of k. with four entries as 1 and all other entries as 0. Let a. If . (2011) not a correct explanation for Statement1 (b) Statement1 is true. has no solution. for which the system of equations (k + 1)x + 8y = 4k. Statement2 is true.1 ÷ (b) ç ÷ è -1ø æ . Define Tr(A) = sum matrix A. is (a) 1 3. é1 a 3 ù If P = ê1 3 3 ú is the adjoint of a 3 × 3 matrix A and ê ú êë 2 4 4 úû 7. then a is equal to (a) 11 2. Statement‑1 : Tr(A) = 0. kx + (k + 3)y = 3k – 1. The number of values of k for which the linear equations (c) Statement1 is false. Statement2 is true. (a) (b) (c) (d) (2013) æ1 ö æ 0 ö that Au1 = ç 0 ÷ and Au 2 = ç 1 ÷ . (b) 5 (d) at least 7 (2010) Let A be a 2 × 2 matrix with non‑zero entries and let A 2 = I. Statement2 is true; Statement2 is not a correct explanation for Statement1 (c) Statement1 is true. c be such that b(a + c) ¹ 0. is (a) less than 4 (c) 6 9. P Q = Q P. (b) 5 (d) 4 (b) 2 (c) 3 (d) infinite æ 1 0 0 ö Let A = ç 2 1 0 ÷ . Statement2 is false. Let A be a 2 × 2 matrix (c) Statement1 is true. (d) Statement1 is false. Statement2 is true; Statement2 is a kx + 4y + z = 0 correct explanation for Statement1. (c) 0 Consider the system of linear equations x 1 + 2x 2 + x 3 = 3 2x 1 + 3x 2 + x 3 = 3 3x 1 + 5x 2 + 2x 3 = 1 Statement‑2 : |A| = 1. Statement 2 : AB is symmetric matrix if matrix multiplication of A with B is commutative. Statement2 is true; Statement2 is Statement1: adj (adj A) = A a correct explanation for Statement1.1 ö æ . (b) Statement1 is false.1 ö ç . |A| = 4. (a) Statement1 is true. 2x + 2y + z = 0 (2009) possess a non‑zero solution is (a) 1 (b) zero (c) 3 (d) 2 (2011) 11. Statement2 is false. then determinant of (P + Q ) is equal to (a) 0 8. (2010) (a) Statement1 is true. (b) Statement1 is true. 10. Statement2 is true; Statement2 is a correct explanation for Statement1. Statement‑1 : A (BA ) and (AB )A are symmetric matrices. If u 1 and u 2 are column matrices such ç ÷ è 3 2 1 ø æ . Statement2 is true. infinite number of solutions exactly 3 solutions a unique solution no solution of diagonal elements of A and |A| = determinant of (2012) Let P and Q be 3 × 3 matrices with P ¹ Q. 4x + ky + 2z = 0 (d) Statement1 is true. a 3 . . y ¹ 0 then D is 1 1 1 + y (a) (b) (c) (d) divisible by x but not y divisible by y but not x divisible by neither x nor y divisible by both x and y. If a + b + c = – 2 and cx and z = bx + ay. (c) If det A ¹ ± 1. c be any real numbers. Let I be the 2 × 2 (2006) identity matrix. Let A be a 2 × 2 matrix with real entries. y = az + 21. Statement2 : If A ¹ I and A ¹ – I.. then the value of the log a n log a n +1 log a n + 2 log a n +3 log a n + 4 log a n + 5 . then A –1 exists and all its entries are non x + ay + z = a – 1. by the principle of mathematical (c) Statement1 is true.1 0 ÷÷ . Then -b b + 1 b . then tr(A) ¹ 0. then A –1 exists but all its entries are not (a) 0 (b) 1 (c) 2 (d) 3. a... z not all zero such that x = cy + bz.13 Matrices and Determinants æ1 2ö æ a 0 ö 18. (2005) Statement1 : If A ¹ I and A ¹ –I. Let A = ê0 a ê êë0 0 (a) 1/5 (b) (2007) aù A 2 | = 25. Assume that A 2 = I. Suppose that there are real 2 2 2 numbers x. (a) A (b) A + I (c) I – A (d) A – I. then which of the following will be always true? (a) A = B (b) AB = BA (c) either A or B is a zero matrix (d) either A or B is an identity matrix. (2004) . If A and B are square matrices of size n × n such that A 2 – B 2 = (A – B)(A + B). The only correct statement about the ç .1 0 0 ÷ è ø matrix A is (a) A –1 does not exist (b) A = (–1)I. then which one of the (a) Statement1 is true. Statement2 is false 1 1 ë û ë û (b) Statemen1 is false. integers if a is (d) If det A = ± 1. é1 0 ù é1 0 ù 20. If D = 1 1 + x 1 for x ¹ 0. where I is a unit matrix (c) A is a zero matrix (d) A 2 = I..then det A =–1. the sum of diagonal entries 2 19. then A –1 exists and all its entries are integers (a) either –2 or 1 (b) –2 (2008) (c) 1 (d) not –2. are G. Denote by tr(A).1 1 15. x + y + az = a – 1 has no solutions. Statement2 is true; Statement2 is induction a correct explanation for Statement1 (a) A n = 2 n – 1 A – (n – 1)I (d) Statement1 is true. log a n + 8 (c) 0 (d) –2. è 3 4ø è 0 b ø (a) there cannot exist any B such that AB = BA c c . If | A ú 5 úû 5 (c) 5 2 (d) 1. a 2 . b Î N. then |a | equals 5a ú . then the inverse of A is of A. (2005) a a +1 a -1 1 1 a +1 b +1 c . then A need not exist then f (x) is a polynomial of degree (b) If det A = ± 1.1 c + 1 (-1)n + 2 a (-1)n +1 b (- 1) n c (b) there exist more than one but finite number B’s such that then the value of n is AB = BA (a) any even integer (b) any odd integer (c) there exists exactly one B such that AB = BA (c) any integer (d) zero (2009) (d) there exist infinitely many B’s such that AB = BA. Let A be a square matrix all of whose entries are integers. is (a) 2 log a n + 6 (a) 2 (b) –1 log a n + 7 (b) 1 (c) –2 (d) 5. then a is (2004) determinant 25. (2007) 17. y. (2005) 13. Statement2 is true following holds for all n ³ 1.P. Let A = ç 0 .1 ö ç 23. é5 5 a 16. Let A = 2 1 -3 and 10( B ) = -5 0 a ÷ .. Let a. Let A = ç ÷ and B = ç ÷ . (1 + a 2 ) x (1 + b 2 ) x 1 + c 2 x Then which one of the following is true? –1 (a) If det A = ± 1. (2005) necessarily integers 22. 12. Statement2 is true; Statement2 is (b) A n = nA – (n – 1)I (2008) not a correct explanation for Statement1 (c) A n = 2 n – 1 A + (n – 1)I (d) A n = nA + (n – 1)I. If B is the ç ÷ ç ÷ è 1 -2 3 ø 1 ø è1 1 inverse of matrix A. The system of equations a x + y + z = a – 1. . b. (2004) 2 2 ö æ 4 æ 1 -1 1 ö ç ç ÷ 24. If a 1 . a n . If A = ê ú and I = ê0 1 ú ... (2006) æ 0 0 .1 b -1 c + 1 = 0. Then a 2 + b 2 + c 2 + 2abc is equal to 1 + a 2 x (1 + b 2 ) x (1 + c 2 ) x (a) 1 (b) 2 (c) – 1 (d) 0 (2008) f ( x ) = (1 + a 2 ) x 1 + b 2 x (1 + c 2 ) x 14.1 + a . If A – A + I = 0. (a) 5. (d) 29. then (2003) log l log m p 1 q 1 log n r 1 (a) –1 equals (b) 2 (c) 1 (d) 0. (d) 21. then a b ax + b bx + c is ax + b bx + c 0 is equal to b c w (c) w 2 (2003) (d) 0. 11.14 JEE MAIN CHAPTERWISE EXPLORER 26. 7. If 1. (2002) (a) +ve (c) –ve 29. (a) 19. 9. (d) 16. (d) 15. (c) 20. then b a ë û ë û (a) a = a 2 + b 2 . 8. (a) 10. (b) 28. (d) 24. (b) (ac – b 2 )(ax 2 + 2bx + c) (d) 0. (d) (2002) . (a) (d) 3. Answer Key 1. all positive. (c) (d) 6. (a) 22. (b) (c) 4. (c) 27. q th and r th term of a GP. (c) 25. (b) 26. b = a 2 – b 2 (c) a = 2ab. b = ab. (b) 12. (d) 13. é a b ù 2 é a b ù 27. If A = ê ú and A = ê b a ú . (a) (d) 2. (a) 14. m. 1 n then D = w w 2 n (a) 1 wn w2 n w2 n 1 1 n (b) w 28. (a) 17. w 2 are the cube roots of unity. n are the p th . (b) 23. w. If l. If a > 0 and discriminant of ax 2 + 2bx + c is –ve. b = 2ab (b) a = a 2 + b 2 . b = a 2 + b 2 (d) a = a 2 + b 2 . (d) 18. \ k = 2. Number of nonsingular matrices = 6 C 1 × 6 C 1 = 36 We can also exhibit more than 6 matrices to pick the right choice. 3 For k = 1. P 2 Q = Q 2 P. 3. Au = ê0 ú . Hence. A(BA) = (AB)A by associativity. \ a = 11 Remark : det(adj A) = (det A) n – 1. 4 k 2 we have k 4 2 = 0 2 2 1 é1 0 0 ù é1 ù é0 ù ê 2 1 0 ú . 3x + 6y = 8 which are parallel lines.1û ë For no solution of AX = B a necessary condition is det A = 0. é a ù Let u1 = êb ú ê ú êë c úû é1 ù Þ a = 1. 3x 1 + 5x 2 + 2x 3 = 1 As such the system is inconsistent and hence it has no solution. which on expansion gives k 2 – 6k + 8 = 0 Þ 7. (k – 2)(k – 4) = 0. Thus P is symmetric. x + 4y = 2 which is just a single equation in two variables. |P 2 + Q 2 | = 0 5. where A is a matrix of order n. Au 2 = ê1 ú ê ú 3 p + 2q + r = 0 Þ r = -2 êë0 úû é 1ù é 0 ù é 1 ù u1 + u 2 = ê -2ú + ê 1ú = ê -1 ú ê ú ê ú ê ú êë 1úû êë -2 úû êë -1 úû (d) : For the system to possess nonzero solution. Also (AB) T = B T A T = BA = AB (Q A and B are commutative) Þ 6. For k = 3. . 8. So no solution in this case. (a) : P 3 = Q 3 .e. 2a + b = 0 Au 1 = ê 0 ú ê ú Þ b = -2. then det A will be nonzero and all the elements of that term will be 1 each. x + 4y = 2 It has infinite solutions.15 Matrices and Determinants 1. Au = ê1 ú A = (b) : 1 ê ú 2 ê ú ê ú êë3 2 1 úû êë0 úû êë0 úû é p ù Let u2 = ê q ú ê ú êë r úû AB is also symmetric. det(adj A) = (det A) 2 Þ 2a – 6 = 16 Þ 2a = 22. 3a + 2b + c = 0 Þ c = 1 êë 0 úû é0 ù Þ p = 0. (d) : Let A(BA) = P Then P T = (ABA) T = A T B T A T (Transversal rule) = ABA = P ék + 1 8 ù éxù é 4 k ù (a) : The equation is ê úê ú= ê ú k k + 3û ëyû ë3k . 4. k + 1 8 Þ = 0 k k + 3 Þ (k + 1)(k + 3) – 8k = 0 Þ k 2 + 4k + 3 – 8k = 0 Þ k 2 – 4k + 3 = 0 Þ (k – 1)(k – 3) = 0 \ k = 1. 2 p + q = 1 Þ q = 1. Again. 4 (d) : x 1 + 2x 2 + x 3 = 3 2x 1 + 3x 2 + x 3 = 3 3x 1 + 5x 2 + 2x 3 = 1 A quick observation tells us that the sum of first two equations yields (x 1 + 2x 2 + x 3 ) + (2x 1 + 3x 2 + x 3 ) = 3 + 3 Þ 3x 1 + 5x 2 + 2x 3 = 6 But this contradicts the third equation. é1 a 3 ù (a) : P = ê1 3 3 ú ê ú ëê 2 4 4 úû Let P = 1(12 – 12) – a(4 – 6) + 3(4 – 6) = 2a – 6 Also. é a1 a2 A = ê b1 b2 ê (d) : êë c1 c2 a 3 ù b 3 ú ú c3 úû Let A = (a 1b 2c 3 + a 2 c 1 b 3 + a 3 b 1 c 2 ) – (a 1 c 2 b 3 + a 2 b 1 c 3 + a 3c 1b 2 ) If any of the terms be nonzero. PQ 2 = P 2 Q Þ P(P 2 + Q 2 ) = (Q 2 + P 2 )Q Þ P(P 2 + Q 2 ) = (P 2 + Q 2 )Q P ¹ Q Þ P 2 + Q 2 is singular. 2. the equation becomes 2x + 8y = 4. the equation becomes 4x + 8y = 12. i.. 1 a Þ D+ ( -1)n b + 1 b . (d) : D = 1 1 + x 1 = .1 -b = 0 c é a 2 + bg b ( a + d ) ù A2 = ê ú 2 ú ëê g ( a + d ) d + bg û é a 2 + bg b ( a + d )ù é1 A2 = I = ê ú=ê 2 ú ë0 ëê g ( a + d ) d + bg û 2 2 giving a + bg = 1 = d + bg and g(a + d) = b(a + d) = 0 As A ¹ I. (d) : Each entry of A is an integer.c .1 15.1 + c c -1 c +1 a a +1 a -1 a +1 b +1 c . tr (A) = a + d = 0 {a = – d} Statement2 is false because tr (A) = 0 13.1 = 0 14.1 = 0 c c .1 + ( -1) a .bg b g .bg = -1 Statement1 is therefore true. so the cofactor of every c c + 1 c - 1 entry is an integer.1 c + 1 a - b a + 1 a . which gives |adj A| = |A| Thus statement2 is also true. we have a = – d det A = 1 .1 n .b Þ c -1 a = 0 b a -1 c + 1 c - 1 c (Changing rows to columns) Þ 1(1 – a 2 ) + c(– c – ab) – b(ca + b) = 0 a a + 1 a . A ¹ – I. (a) : A2 = êê0 a 5a úú êê0 a 5 a úú êë0 0 5 úû êë0 0 5 úû . C 3 ® C 3 – C 1 ) a+c 0 a + c 1 0 0 = -b + c 0 b + c R 1 ® R 1 + R 3 .b b + 1 b . And then each entry of adjoint is integer.b 2 b . But statement2 doesn’t explain statement1.1 b .1 a -1 b -1 c .1 Þ a 2 + b 2 + c 2 + 2abc = 1 n Þ D + ( -1) -b b + 1 b . a 11. 10.1 = 0 c c . (a) : System of equations x – cy – bz = 0 cx – y + az = 0 bx + ay – z = 0 has non trivial solution if the determinant of coefficient matrix is zero 1 .bg 0 ù 1 úû = -1 + bg .0) = xy c -2 c + 1 1 0 y Expanding along 2nd column Hence D is divisible by both x and y.1 1 1 1 + y C 2 ® C 2 – C 3 c -2 c + 1 (Apply C 2 ® C 2 – C 1 .1 This means all entries in A –1 are integers.1 c . (a) : Let A = éê ú . (Changing columns in cyclic order doesn’t change the deter Also det A = ± 1 and we know that minant) n n Þ D + (–1) D = 0 Þ {1 + (–1) }D = 0 1 A -1 = ( adj A ) det A a a + 1 a . (a) : We have adj(adj A) = |A| n–2 A Here n = 2. JEE MAIN CHAPTERWISE EXPLORER éa b ù (c) : Let A = ê ú ë g d û a bù 12. (b) : Þ a +1 a -1 -b b + 1 b . é5 5a a ù é5 5 a a ù 16.1 .1 = 0 ( -1) n + 2 a ( -1)n +1 b (- 1) n c a + 1 b + 1 c . which gives adj(adj A) = A The statement1 is true. Now D = -b b + 1 b . R 2 ® R 2 + R 3 = 1 x 0 = 1( xy . Again |adj A| = |A| n – 1 Here n = 2. D = 2{(a + c)(b + c) – (a + c) (c – b)} = 2(a + c)2b = 4b(a + c) ¹ 0 (By hypothesis) Now {1 + (–1) n }D = 0 Þ 1 + (–1) n = 0 Which mean n = odd integer. We have ë g d û é a 2 + bg b (a + d) ù é1 0 ù A2 = ê ú=ê ú ú ë0 1 û ëê g (a + d ) d + bg û Which gives a + d = 0 and a 2 + bg = 1 So we have Tr(A) = 0 det A = ad – bg = –a 2 – bg = – (a 2 + bg) = –1 Thus statement1 is true but statement2 is false.16 9.1 c + 1 1 1 1 a 2 a . 17 Matrices and Determinants 22. (b) : For no solution | A | = 0 and (adj A)(B) ¹ 0 é 25 25a + 5a 2 5a + 5a + 25 a 2 ù ê ú A2 = ê 0 a2 25a + 5 a 2 ú ê ú 0 0 25 ú ëê û 1 2 2 Given |A | = 25, 625a = 25 Þ | a |= . 5 a 1 1 Now | A | = 0 Þ 1 a 1 = 0 1 1 a Þ a 3 – 3a + 2 = 0 Þ (a – 1) 2 (a + 2) = 0 Þ a = 1, –2. But for a = 1, | A | = 0 and (adj A)(B) = 0 Þ for a = 1 there exist infintiely many solution. Also the each equation becomes x + y + z = 0 again for a = –2 |A| = 0 but (adj A)(B) ¹ 0 Þ $ no solution. 17. (b) : Give A 2 – B 2 = (A + B)(A – B) Þ 0 = BA – AB Þ BA = AB æ 1 2ö æ a 0 ö 18. (d) : A = ç and B = ç ÷ è 3 4ø è 0 b ÷ø 23. (d) : (i) |A| = 1 \ A –1 does not exist is wrong statement æ 1 2ö æ a 0ö æ a 2 b ö = Now AB = ç è 3 4÷ø çè 0 b÷ø çè 3a 4b ÷ø ... (i) and æ a 0ö æ 1 2ö æ a 2 a ö BA = ç = è 0 b÷ø çè 3 4÷ø çè 3b 4b ÷ø ... (ii) As AB = BA Þ 2a = 2b Þ a = b \ æ a 0 ö B=ç = aI 2 Þ $ infinite value of a = b Î N è 0 a ÷ø æ -1 0 0 ö ç (ii) (–1) I = 0 -1 0 ÷ ¹ A Þ (b) is false ç ÷ è 0 0 -1 ø (iii) A is clearly a non zero matrix \ (c) is false We left with (d) only. 24. (d) : Given A –1 = B = 10 A –1 = 10 B 19. (c) : A 2 – A + I = 0 Þ I = A – A × A IA –1 = AA –1 – A(AA –1 ) , A –1 = I – A. æ 1 0 ö 20. (b) : A = ç ÷ è 1 1 ø æ1 0ö 3 æ 1 0 ö n æ 1 0 ö \ A2 = ç ÷, A = ç ÷ so A = ç ÷ 2 1 3 1 è ø è ø è n 1 ø æ n 0 ö æ n - 1 0 ö and nA – (n – 1)I = ç ÷-ç ÷ n - 1 ø è n nø è 0 æ 1 0 ö n =ç ÷ = A . n 1 è ø æ ç Þ ç è æ Þ ç ç è æ Þ ç ç è 4 -5 1 4 -5 1 2 2 ö 0 a ÷ = 10 A –1 . ÷ -2 3 ø 2 2 ö 0 a ÷ (A) = 10I ÷ -2 3 ø 4 -5 1 2 2 öæ 1 0 a ÷ç 2 ÷ç -2 3 ø è 1 æ 10 0 0 ö = ç 0 10 0 ÷ ç ÷ è 0 0 10 ø -1 1 ö 1 -3 ÷ ÷ 1 1 ø ...(*) Þ –5 + a = 0 (equating A 21 entry both sides of (*)) Þ a = 5 a2 a3 a = = ... n = r a1 a2 an - 1 which means a n , a n + 1 , a n + 2 Î G.P. Þ a n + 1 2 = a n a n + 2 Þ 2 log a n + 1 – log a n – log a n + 2 = 0...(i) Similarly 2 log a n + 4 – log a n + 3 – log a n + 5 = 0 ...(ii) and 2 log a n + 7 – log a n + 6 – log a n + 8 = 0 ...(iii) Using C 1 ® C 1 + C 3 – 2C 2 we get D = 0 26. (d) : As w is cube root of unity \ w 3 = w 3n = 1 25. (c) : 21. (c) : Applying C 2 ® C 2 + C 3 + C 1 f ( x ) = 1 + 2 x + x ( a 2 + b2 + c 2 ) 1 + a2 x 1 (1 + c 2 ) x (1 + a 2 ) x 1 (1 + c 2 ) x (1 + a 2 ) x 1 1 + c 2 x Applying R 1 ® R 1 – R 2 , R 2 ® R 2 – R 3 and using a 2 + b 2 + c 2 = – 2 we have (1 + 2 x - 2 x ) 1 - x 0 0 0 0 x - 1 = (1 – x) 2 (1 + a 2 ) x 1 1 + c 2 x = x 2 – 2x + 1 \ degree of f (x) is 2. 1 n w \ w2 n wn w2 n w2 n 1 1 w n = (w 3n – 1) – w n (w 2n – w 2n ) + w 2n (w n – w n ) = 0 18 JEE MAIN CHAPTERWISE EXPLORER æ a b ö æ a b ö 27. (a) : A = AA = ç ÷ç ÷ è b a ø è b a ø 2 æ a 2 + b2 2 ab ö æ a b ö =ç ÷= ç 2 ab a 2 + b2 ÷ çè b a ÷ø è ø \ 28. (c) : C 1 ® xC 1 + C 2 – C 3 = 1 x 0 0 b c ax 2 + 2bx + c bx + c l = t p = AR p – 1 Þ log l = log A + (p – 1) log R Similarly, log m = log A + (q – 1) log R and log n = log A + (r – 1) log R \ ax + b bx + c 0 (ax 2 + 2bx + c ) 2 [b x + bc – acx – bc] x = (b 2 – ac) (ax 2 + 2bx + c) = (+ve) (–ve) < 0 = 29. (d) : Let A be the first term and R be the common ratio of G. P. log l log m log n p 1 q 1 r 1 = log A + ( p - 1) log R log A + ( q - 1) log R log A + ( r - 1) log R = log A - log R log A - log R log A - log R c 1 µ c 3 = 0 + 0 = 0 p 1 q 1 r 1 p 1 p log R q 1 + q log R r 1 r log R p 1 q 1 r 1 c 1 µ c 2 19 Quadratic Equations CHAPTER 4 1. 2. 3. 4. 5. QUADRATIC EQUATIONS The real number k for which the equation 2x 3 + 3x + k = 0 has two distinct real roots in [0, 1] (a) lies between 2 and 3 (b) lies between –1 and 0 (c) does not exist (d) lies between 1 and 2 (2013) If the equations x 2 + 2x + 3 = 0 and ax 2 + bx + c = 0, a, b, c Î R have a common root, then a : b : c is (a) 3 : 2 : 1 (b) 1 : 3 : 2 (c) 3 : 1 : 2 (d) 1 : 2 : 3 The equation e sinx – e –sinx – 4 = 0 has (a) exactly one real root. (b) exactly four real roots. (c) infinite number of real roots. (d) no real roots. (2013) (2012) 9. If the roots of the quadratic equation x 2 + px + q = 0 are tan30° and tan15°, respectively then the value of 2 + q – p is (a) 2 (b) 3 (c) 0 (d) 1. (2006) 10. All the values of m for which both roots of the equation x 2 – 2mx + m 2 – 1 = 0 are greater than –2 but less than 4, lie in the interval (a) –2 < m < 0 (b) m > 3 (c) –1 < m < 3 (d) 1 < m < 4. (2006) 11. The value of a for which the sum of the squares of the roots of the equation x 2 – (a – 2)x – a – 1 = 0 assume the least value is (a) 0 (b) 1 (c) 2 (d) 3. (2005) 12. If the roots of the equation x 2 – bx + c = 0 be two consecutive integers, then b 2 – 4c equals (a) 3 (b) –2 (c) 1 (d) 2. (2005) Let a, b be real and z be a complex number. 13. If both the roots of the quadratic equation If z 2 + az + b = 0 has two distinct roots on the line x 2 – 2kx + k 2 + k – 5 = 0 Re z = 1, then it is necessary that are less than 5, then k lies in the interval (a) |b| = 1 (b) b Î (1, ¥) (a) (6, ¥) (b) (5, 6] (c) b Î (0, 1) (d) b Î (–1, 0) (2011) (c) [4, 5] (d) (– ¥, 4). b are the roots of the equation x 2 – x + 1 = 0, then If a and a 2009 + b 2009 = (a) –2 (b) –1 (c) 1 (d) 2 (2010) a n x n (2005) a n – 1 x n – 1 14. If the equation + + ... + a 1 x = 0, a 1 ¹ 0, n ³ 2, has a positive root x = a, then the equation na n x n – 1 + (n – 1)a n – 1 x n – 2 + ... + a 1 = 0 has a positive root, which is (a) smaller than a (b) greater than a (c) equal to a (d) greater than or equal to a. (2005) 15. Let two numbers have arithmetic mean 9 and geometric mean 4. Then these numbers are the roots of the quadratic equation (a) x 2 + 18x – 16 = 0 (b) x 2 – 18x + 16 = 0 (c) x 2 + 18x + 16 = 0 (d) x 2 – 18x – 16 = 0. (2004) 6. If the roots of the equation bx 2 + cx + a = 0 be imaginary, then for all real values of x. The expression 3b 2 x 2 + 6bcx + 2c 2 is (a) less than 4ab (b) greater than – 4ab (c) less than – 4ab (d) greater than 4ab (2009) 7. The quadratic equations x 2 – 6x + a = 0 and x 2 – cx + 6 = 0 have one root in common. The other roots of the first and second equations are integers in the ratio 4 : 3. Then the common root is 2 (a) 2 (b) 1 (c) 4 (d) 3 (2008) 16. If (1 – p) is a root of quadratic equation x + px + (1 – p) = 0 then its roots are If the difference between the roots of the equation (a) 0, –1 (b) –1, 1 (c) 0, 1 (d) –1, 2. (2004) x 2 + ax + 1 = 0 is less than 5 , then the set of possible 2 17. If one root of the equation x + px + 12 = 0 is 4, while the values of a is equation x 2 + px + q = 0 has equal roots, then the value of (a) (3, ¥ ) (b) (– ¥ , –3) q is (c) (–3, 3) (d) (–3, ¥ ). (2007) (a) 3 (b) 12 (c) 49/4 (d) 4. (2004) 8. 20 JEE MAIN CHAPTERWISE EXPLORER (a) 3x 2 – 25x + 3 = 0 (b) x 2 + 5x – 3 = 0 18. The value of a for which one root of the quadratic equation 2 2 (c) x 2 – 5x + 3 = 0 (d) 3x 2 – 19x + 3 = 0. (2002) (a – 5a + 3)x + (3a – 1)x + 2 = 0 is twice as large as the other is 22. Difference between the corresponding roots of (a) –2/3 (b) 1/3 (c) –1/3 (d) 2/3. (2003) x 2 + ax + b = 0 and x 2 + bx + a = 0 is same and a ¹ b, then 19. If the sum of the roots of the quadratic equation (a) a + b + 4 = 0 (b) a + b – 4 = 0 ax 2 + bx + c = 0 is equal to the sum of the squares of their (c) a – b – 4 = 0 (d) a – b + 4 = 0. (2002) a , b and c 2 23. Product of real roots of the equation x + | x | + 9 = 0 reciprocals, then c a b are in (a) is always positive (b) is always negative (a) geometric progression (c) does not exist (d) none of these. (2002) (b) harmonic progression 2 (c) arithmeticgeometric progression 24. If p and q are the roots of the equation x + px + q = 0, then (d) arithmetic progression. (2003) (a) p = 1, q = –2 (b) p = 0, q = 1 (c) p = –2, q = 0 (d) p = –2, q = 1. (2002) 20. The number of real solutions of the equation 2 x – 3|x| + 2 = 0 is 25. If a, b, c are distinct +ve real numbers and a 2 + b 2 + c 2 = 1 (a) 4 (b) 1 (c) 3 (d) 2. (2003) then ab + bc + ca is (a) less than 1 (c) greater than 1 21. If a ¹ b but a 2 = 5a – 3 and b 2 = 5b – 3 then the equation whose roots are a/b and b/a is (b) equal to 1 (d) any real no. Answer Key 1. (c) 2. (d) 7. (a) 8. (c) 13. (d) 19. (b) 25. (a) 14. (a) 20. (a) 3. 9. 15. 21. (d) (b) (b) (d) 4. 10. 16. 22. (b) (c) (a) (a) 5. 11. 17. 23. (b) (b) (c) (c) 6. 12. 18. 24 (b) (c) (d) (a) (2002) 21 Quadratic Equations 7. (a) : Let a and 4b be the root of 1. (c) : Let f (x) = 2x 3 + 3x + k , f ¢(x) = 6x 2 + 3 > 0 Thus f is strictly increasing. Hence it has atmost one real root. x 2 – 6x + a = 0 But a polynomial equation of odd degree has atleast one root. and a and 3b be those of the equation x 2 – cx + 6 = 0 Thus the equation has exactly one root. Then the two distinct From the relation between roots and coefficients roots, in any interval whatsoever is an impossibility. No such a + 4b = 6 and 4ab = a (c) exists. 2 a + 3b = c and 3ab = 6 2. (d) : In the equation x + 2x + 3 = 0, both the roots are imaginary. we obtain ab = 2 giving a = 8 a b c Since a, b, c Î R, we have = = The first equation is x 2 – 6x + 8 = 0 Þ x = 2, 4 1 2 3 For a = 2, 4b = 4 Þ 3b = 3 Hence a : b : c : : 1 : 2 : 3 For a = 4, 4b = 2 Þ 3b = 3/2 (not an integer) sinx –sinx 3. (d) :e – e – 4 = 0 So the common root is a = 2. Þ (e sinx ) 2 – 4e sinx – 1 = 0 Þ t 2 – 4t – 1 = 0 4 ± 16 + 4 Þ t = = 2 ± 5 2 8. i.e., e sin x = 2 + 5 or 14 2 4244 - 53 (neglected) | a - b | = (a + b )2 - 4ab , | a - b | = a 2 - 4 - ve Since, | a - b | < 5 Þ a 2 - 4 < 5 Þ a 2 – 4 < 5 Þ a 2 < 9 Þ – 3 < a < 3. sin x = ln (2 + 5) > 1 \ No real roots. 4. (b) : Let roots be 1 + ai, 1 + bi, then we have, (a Î R) (1 + ai) + (1 + bi) = – a Þ 2 + (a + b)i = – a (1 + ai)(1 + bi) = b Comparing we have, a = – 2 and a = – b Now (1 + ai)(1 – ai) = b Þ 1 + a 2 = b Þ b = 1 + a 2 As a 2 ³ 0 we have b Î (1, ¥) 1 ± i 3 5. (b) : We have x 2 – x + 1 = 0 giving x = . 2 Identifying these roots as w and w 2 , we have a = w, b = w 2 . We can also take the other way round that would not affect the result. Now a 2009 + b 2009 = w 2009 + w 4018 = w 3k + 2 + w 3m + 1 (k, m Î N) = w 2 + w = –1. ( . . . w 3k = 1) 6. (b) : The roots of bx 2 + cx + a = 0 are imaginary means c 2 – 4ab < 0 Þ c 2 < 4ab Again the coeff. of x 2 in (c) : x 2 + ax + 1 = 0 Let roots be a and b, then a + b =–a and ab = 1 9. (b) : a = tan 30°, b = tan 15° are roots of the equation x 2 + px + q = 0 \ tan a + tan b = – p and tan a ∙ tan b = q using tan a + tan b = tan (a + b) (1 – tan a tan b) Þ – p = 1 – q Þ q – p = 1 Þ 2 + q – p = 3 10. (c) : Let a, b are roots of the equation (x 2 – 2mx + m 2 ) = 1 –3 –1 3 5 Þ x = m ± 1 = m + 1, m – 1 Now –2 < m + 1 < 4 ...... (i) and –2 < m – 1 < 4 ....... (ii) ......(A) ì Þ -3 < m < 3 í and 1 < m < 5 .......(B) î By (A) & (B) we get –1 < m < 3 as shown by the number line. 3b 2 x 2 + 6bcx + 2c 2 is +ve, so the minimum value of the 11. (b) : Let f (a) = a 2 + b 2 = (a + b) 2 – 2ab expression = (a – 2) 2 + 2(a + 1) 2 2 2 2 2 2 36b c - 4(3b )(2c ) 12 b c \ f ¢(a) = 2(a – 2) + 2 == = - c 2 For Maxima | Minima f ¢(a) = 0 (3b 2 ) 12b 2 Þ 2[a – 2 + 1] = 0 Þ a = 1 2 2 As c < 4ab we have –c > – 4ab Again f ¢¢(a) = 2, Thus the minimum value is – 4ab. f ¢¢(1) = 2 > 0 Þ at a = 1, f (a) will be least. 22 JEE MAIN CHAPTERWISE EXPLORER By (i) and (ii) we have 12. (c) : Let a, a + 1 are consecutive integer \ (x + a)(x + a + 1) = x 2 – bx + c Comparing both sides we get Þ –b = 2a + 1 c = a 2 + a \ b 2 – 4c = (2a + 1) 2 – 4(a 2 + a) = 1. 9 a 2 13. (d) : Given x 2 – 2kx + k 2 + k – 5 = 0 Roots are less than 5 Þ D ³ 0 Þ (–2k) 2 ³ 4(k 2 + k – 5) Þ k £ 5 ...(A) Again f (5) > 0 Þ 25 – 10k + k 2 + k – 5 > 0 Þ k 2 – 9k + 20 > 0 Þ (k – 4)(k – 5) > 0 Þ k < 4 È k > 5 ...(B) ...(C) Also sum of roots < 5 Þ k < 5 2 from (A), (B), (C) we have k Î (–¥, 4) as the choice gives number k < 5 is (d). 14. (a) : If possible say f ( x ) = a0 x n + a1 x n -1 + ... + an x a 2 - 5 a + 3 ´ 2 2a 2 ( a 2 - 5a + 3) 2 2 2 Þ 9(a – 5a + 3) = (1 – 3a) 2 Þ a = 3 19. (b) : Given a + b = 1 (1 - 3a ) 2 = + 1 = ( a + b ) 2 - 2 ab a 2 b2 a 2b 2 Þ 2a 2 c = bc 2 + ab 2 2a c b Þ c , a , b Î A.P = + Þ a b c b a c Þ reciprocals are in H.P 20. (a) : Given x 2 – 3|x| + 2 = 0 If x ³ 0 i.e. |x| = x \ The given equation can be written as x 2 – 3x + 2 = 0 \ f (0) = 0 (x – 1) (x – 2) = 0 x = 1, 2 Similarly for x < 0, x 2 – 3|x| + 2 = 0 Þ x 2 + 3x + 2 = 0 Þ Now f ( a ) = 0 (\ x = a is root of given equation) Þ n -1 n - 2 + ... + a1 = 0 has at least \ f ¢( x ) = nan x + ( n - 1) an -1 x one root in ]0, a [ Þ na n x n–1 + (n – 1) a n–1 x n–2 + .... + a 1 = 0 has a +ve root smaller than a. Þ Hence 1, –1, 2, –2 are four solutions of the given equation. a b and which 21. (d) : We need the equation whose roots are b a are reciprocal of each other, which means product of roots is a b = 1. In our choice (a) and (d) have product of roots 1, b a so choices (b) and (d) are out of court. In the problem choice, None of these is not given. If out of four choices only one choice satisfies that product of root is 1 then you select that choice for correct answer. Now for proper choice we proceed as, a ¹ b, but a 2 = 5a – 3 and b 2 = 5b – 3, Changing a, b by x \ a, b are roots of x 2 – 5x + 3 = 0 Þ a + b = 5, ab = 3 a b a 2 + b 2 19 a b + = and product . = 1 now, S = = ab 3 b a b a \ Required equation, x 2 – (sum of roots) x + product of roots = 0 19 Þ x 2 – x + 1 = 0 3 2 Þ 3x – 19x + 3 = 0 is correct answer. 15. (b) : Let the two number be a, b a + b = 9 and ab = 4 \ 2 \ Required equation 2 x 2 – 2(Average value of a, b)x + G. M = 0 x 2 – 2(9)x + 16 = 0 16. (a) : As 1 – p is root of x 2 + px + 1 – p = 0 Þ (1 – p) 2 + p(1 – p) + (1 – p) = 0 (1 – p) [1 – p + p + 1] = 0 Þ p = 1 \ Given equation becomes x 2 + x = 0 Þ x = 0, – 1 17. (c) : As x 2 + px + q = 0 has equal roots \ p 2 = 4q and one root of x 2 + px + 12 = 0 is 4. \ 16 + 4p + 12 = 0 \ p = –7 49 \ p 2 = 4q Þ q = 4 18. (d) : Let a, 2a are roots of the given equation \ sum of the roots 1 - 3 a a + 2a = 3a = 2 a - 5a + 3 and product of roots 2 a(2a) = 2a 2 = 2 a - 5a + 3 x = –1, –2 ...(i) ...(ii) 22. (a) : Let a, b are roots of x 2 + bx + a = 0 \ a + b = –b and ab = a again let g, d are roots of x 2 + ax + b = 0 \ g + d = –a and gd = b Now given 23 Quadratic Equations Now if q = 0 then p = 0 Þ p = q If p = 1, then p + q = – p q = – 2p q = – 2(1) q = – 2 Þ p = 1 and q = – 2 a – b = g – d Þ (a – b) 2 = (g – d) 2 Þ (a + b) 2 – 4ab = (g + d) 2 – 4gd Þ b 2 – 4a = a 2 – 4b Þ b 2 – a 2 = –4(b – a) Þ (b – a) (b + a + 4) = 0 Þ b + a + 4 = 0 as (a ¹ b) 23. (c) : x 2 + |x| + 9 = 0 Þ |x| 2 + |x| + 9 = 0 Þ \ no real roots (³ D < 0) 24 (a) : Given S = p + q = – p and product pq = q Þ q(p – 1) = 0 Þ q = 0, p = 1 25. (a) : In such type of problem if sum of the squares of number is known and we needed product of numbers taken two at a time or needed range of the product of numbers taken two at a time. We start square of the sum of the numbers like (a + b + c) 2 = a 2 + b 2 + c 2 + 2(ab + bc + ca) Þ 2(ab + bc + ca) = (a + b + c) 2 – (a 2 + b 2 + c 2 ) ( a + b + c) 2 - 1 Þ ab + bc + ca = < 1 2 (b) Statement1 is false. (2010) 10. Let T n be the number of all possible triangles formed by joining 6. (d) Statement1 is false. Statement1 : The number of different ways the child can buy the six icecreams is 10 C 5 . 3.24 JEE MAIN CHAPTERWISE EXPLORER PERMUTATIONS AND COMBINATIONS CHAPTER 5 1. 20 C – 20 C + 20 C – 20 C + .. then the value of n is (a) 5 (b) 10 (c) 8 (d) 7 (2013) 2. If a voter votes for at least has 9 distinct blue balls. Statement2 is false. then the number of ways in which he can vote out at random and then transferred to the other.. statement2 is true; statement2 is a 0 1 2 3 10 correct explanation of statement1. the number of ways in which one or more balls can be selected from 10 white. (a) 0 (b) 20 C 10 (c) – 20 C 10 (d) 1 20 C 10 .1) 10C j . (2011) 4. The number is of ways in which this can be done is (a) 5040 (b) 6210 (c) 385 (d) 1110. (a) 3 (b) 36 (c) 66 (d) 108 (2006) (2010) . 9 green and 7 black balls is (a) 630 (b) 879 (c) 880 (d) 629 (2012) 7. At an election. The sum of the series Statement2 : S 1 = 90 × 2 8 and S 2 = 10 × 2 8 . Statement2 is true; Statement2 is a correct explanation for Statement1.. not greater than the number to be elected.. – . a voter may vote for any number of candidates. Then the number of such arrangements is (a) at least 500 but less than 750 (b) at least 750 but less than 1000 (c) at least 1000 (d) less than 500 (2009) In a shop there are five types of icecreams available. There are 10 There are two urns. 4 novels and 1 dictionary are to be selected and arranged in a row on a shelf so that the dictionary is always in the middle. Statement2 is true (c) Statement1 is true.. A child buys six icecreams. (2007) (c) Statement1 is true. Statement2 : The number of different ways the child can buy the six icecreams is equal to the number of different ways of arranging 6 A’s and 4 B’s in a row. (a) Statement1 is true. å j 10 C j j =1 8. (b) Statement1 is true. vertices of an nsided regular polygon. Urn A has 3 distinct red balls and urn B candidates and 4 are to be elected. Statement2 is true; Statement2 is not a correct explanation for Statement1 (2008) How many different words can be formed by jumbling the letters in the word MISSISSIPPI in which no two S are adjacent? (a) 7 ∙ 6 C 4 ∙ 8 C 4 (b) 8 ∙ 6 C 4 ∙ 7 C 4 j =1 (c) 6 ∙ 7 ∙ 8 C 4 (d) 6 ∙ 8 ∙ 7 C 4 (2008) Statement1 : S 3 = 55 × 2 9 . statement2 is true; statement2 is not 2 a correct explanation for statement1. Statement2 : The number of ways of choosing any 3 places from 9 different places is 9 C 3 . From each urn two balls are taken one candidate. From 6 different novels and 3 different dictionaries. statement2 is true. (a) Statement1 is true.. (c) Statement1 is true. If T n + 1 – T n = 10. Let S1 = 10 å 10 j ( j . statement2 is false. å j 2 10 C j . Statement2 is true. Statement2 is false (b) Statemen1 is false. Statement2 is true; Statement2 is a correct explanation for Statement1 (d) Statement1 is true. Statement1 : The number of ways of distributing 10 identical balls in 4 distinct boxes such that no box is empty is 9 C 3 . 9. (d) Statement1 is true.. S 2 = j =1 10 and S3 = 5.. Assuming the balls to be identical except for difference in colours.. + 20 C is (a) Statement1 is true. Statement2 is true; Statement2 is not a correct explanation for Statement1. If C r denotes the number of combinations of n things taken (a) 602 (b) 603 (c) 600 (d) 601. Total number of such numbers are 14. by (2002) (a) 30 (b) 5 ! ´ 4 ! (c) 7 ! ´ 5 ! (d) 6 ! ´ 5!. (2005) r at a time. (a) 5. (b) (a) (b) (b) 3. The sum of integers from 1 to 100 that are divisible by 2 or 15. (c) (d) (d) (d) 4. 2. 3. 13. 6 50 56 . 13. If the letters of the word SACHIN are arranged in all possible (a) 196 (b) 280 (c) 346 (d) 140. 5.25 Permutations and Combinations 11. The number of ways in which 6 men and 5 women can dine 5 is at a round table if no two women are to sit together is given (a) 3000 (b) 3050 (c) 3600 (d) 3250. (a) 18. 15. (2002) (2004) 20. 4 repetition allowed is (2005) (a) 125 (b) 105 (c) 128 (d) 625. (a) 6. How many ways are there to arrange the letters in the word (2002) GARDEN with the vowels in alphabetical order? (a) 360 (b) 240 (c) 120 (d) 480. 7. 1. The value of C4 + å (2003) r =1 18. 1. 3. (c) . Total number of four digit odd numbers that can be formed using 0. (2003) 21. 2. A student is to answer 10 out of 13 questions in an examination (a) 216 (b) 375 (c) 400 (d) 720. the digits 0. 2. (2002) The number of choices available to him is Answer Key 1. 14. (a) (b) (a) (d) 2.r C3 is 12. 8. then (2003) the word SACHIN appears at serial number n 17. then the expression n C r + 1 + n C r – 1 + 2 × n C r equals (a) n + 2 C r + 1 (b) n + 1 C r (c) n + 1 C r + 1 (d) n + 2 C r . 19. 9. such that he must choose at least 4 from the first five questions. (a) 16. 21. 20. Number greater than 1000 but less than 4000 is formed using (a) 56 C 4 (b) 56 C 3 (c) 55 C 3 (d) 55 C 4 . ways and these words are written out as in dictionary. (c) 10. 3. 19. (d) 17. 7 are 16. Five digit number divisible by 3 is formed using 0. (c) 12. Then number of ways of distributing 8 identical balls in 3 distinct boxes so that none of the boxes is empty is (a) 312 (b) 3125 (c) 120 (d) 216. (2004) 4. (d) 11. (a) 3 8 (b) 21 (c) 5 (d) 8 C 3 . 6 and 7 without repetition. .2) = 10 6 6 Þ Then the number of ways = 6 C4 ∙ 3 C1 ∙ 4! 3n(n – 1) = 60 Þ n(n – 1) = 20 Þ n 2 – n – 20 = Þ Þ (n – 5)(n + 4) = 0 \ n = 5 7. ..26 1. 1 dictionary can be selected in 3 C 1 ways. P.20 C11 . we have 7 letters M. 10(10 .20 C1 + 20 C 2 . Statement1 is false. + 20 C10 . I. 6.. P... \ Statement2 is true So..20 C3 + . 7 black balls is = (10 + 1) (9 + 1) (7 + 1) – 1 = 880 – 1 = 879 ways (c) : x 1 + x 2 + x 3 + x 4 = 10 The number of positive integral solution is 6 + 4 – 1 C 4 – 1 = 9 C 3 8.. (a) : A voter can vote one candidate or two or three or four candidates \ Required number of ways = 10 c 1 + 10 c 2 + 10 c 3 + 10 c 4 = 385 Fixed 11. 3.. 9 green.1) (d) : Q 20 C0 + 20 C1 x + . Statement2 is true. I. of word start with H = 5! No. we only have to arrange 4 novels which can be done in 4! ways.. Also out of 3 dictionaries. Then statement1 is true and statement2 is false..1) = 10 2 2 Þ n – n – 20 = 0.1)(n . \ n = 5 Here we have used n C r + n C r + 1 = n + 1 C r + 1 n Þ 2.1) 10 C j 9.1) n(n ....20 C9 ) + 20 C10 = 0 20 C0 ..1) + j ) × 10 C j j =1 10 = å j =1 9 10 j ( j . 5. of word start with I = 5! No. I.1) 8 j ( j .. C2 = 10 Þ (b) : Number of ways in which one or more balls can be selected from 10 white.. + 20 C20 x 20 = (1 + x ) 20 After putting x = –1. å j( j .20 C12 + .20 C 9 + 20 C10 = 1 C10 2 20 10... + 20 C10 x10 + å 10 = 9 ´ 10 .. + 20 C20 = 0 j = 2 10 S2 = 10 å j× 10 C j = 10 j =1 10 S3 = å j =1 9 å C j -1 = 10 ´ 2 j =1 10 j 2 × 10C j = å ( j ( j ...1) 10C j + å j × 10 C j j =1 = 90 ∙ 2 8 + 10 ∙ 2 9 = (45 + 10)2 9 = 55 ∙ 2 9 . (d) : S A C H I N No.. of word start with C = 5! No.. of word start with A = 5! No. (d) : The number of ways = ( 3 C 2 ) × ( 9 C 2 ) = 3 ´ 2(20 C0 . The required number of ways = 7 ∙ 5 ∙ 3 ∙ 8 C 4 = 7 ∙ 6 C 4 ∙ 8 C 4 .. 2 nd solution : n + 1 C 3 – n C 3 = 10 n (n .2 j ( j . JEE MAIN CHAPTERWISE EXPLORER (a) 1 st solution : n + 1 C 3 – n C 3 = 10 : Since the dictionary is fixed in the middle. 4 novels can be selected in 6 C 4 ways.. 7 = 7 × 5 × 3 way of arranging them = 24 And four S can be put in 8 places in 8 C 4 ways..2 = 90 ´ 2 8 C0 . of word start with N = 5! Total words = 5! + 5! + 5! + 5! + 5! = 5(5 !) = 600 . we get 20 å 8C j . Number of different ways of arranging 6A’s and 4B’s in a row 10 10 = = C4 = Number of different ways the child can buy 6 ´ 4 the six icecreams. . (c) : S1 = = 6∙5 0 ∙ 3∙ 24 = 1080 = 1080 2 (b) : We have to find the number of integral solutions if x 1 + x 2 + x 3 + x 4 + x 5 = 6 and that equals 5+6–1 C 5–1 = 10 C 4 Thus Statement1 is false. 9 ´ 8 = 108 2 (c) : Out of 6 novels.20 C1 + 20 C2 .20 C1 + 20 C2 -20 C3 + .20 C 3 + .1) × × C j . ( n + 1) n(n ... 4. = 0 (a) : Leaving S.. I.. It is the same as the number of ways of choosing any 3 balls from 9 different places. 100 = 9 + 6 × 2 = 21 Set of numbers divisible by 10 are 10.. S 20 = [5 + 100] = 10 × 105 = 1050 2 m = y – 1. (b) : (i) Each box must contain at least one ball since no = n + 2 C r + 1 box remains empty so we have the following cases 18. 10. .27 Permutations and Combinations Now add the rank of SACHIN so required rank of SACHIN = 600 + 1 = 601. n = z – 1 10 i. of question 8 No. 2. 4. of question 5 No.. where H = Hundred place = 2 C 1 × 4 × 4 × 4 T = Ten’s place III 6. . 5.e. 15. I 1.100 As 1. 1. (b) : Set of numbers divisible by 2 are 2.E these alphabets can be arrange themselves by 2! ways 17. (d) 1 ´ 3! 3 × + 3! × 2 20.. 7 Number of ways of 6 men at a M We have to fill up four places like TH H T U round Table is n – 1! = (6 – 1)!= 5! (Case: If repetition is allow) Now we left with six places between the men and there are 5 C 6 2 4 C = 5 × 6 2 × 4 6 5 women. (c) : Let number of digits formed x. .. (a) : Case (i) : Required ways for first case = 5 C 4 × 8 C 6 = 140 Case (ii): No. 3. 5 1. 2. 3. = 128 U = Unit place \ Number of ways 19. which means left extreme digit will be either 2 or 3. 1 we get 50 C4 + 50 C3 + 51 C3 + 52 C3 + 53C3 + 54 C3 + 55 C3 ( Q nCr + n Cr +1 = n + 1 Cr +1 ) = 51 C 4 + 51 C 3 + 52 C 3 + 53 C 3 + 54 C 3 + 55 C 3 = 52 C 4 + 52 C 3 + 53 C 3 + 54 C 3 + 55 C 3 = 53 C 4 + 53 C 3 + 54 C 3 + 55 C 3 = 54 C 4 + 54 C 3 + 55 C 3 = 55 C 4 + 55 C 3 = 56 C 4 16. Similarly. y. 1. 2. Box Number of balls \ 1000 < x < 4000. z respectively S 50 = 25[102] then x + y + z = 8 and no box is empty so each x.. \ Required numbers = 2 C 1 × H T U I 1. 1. y. 3. 3. 2. 4 have case ways and Now sum of numbers divisible by 2 is given by 1... 4 2. 2. 3. of question 5 No. of question 4 No. 6 2.. (a) 50 6 C4 + å 56 . 4.. 6. 2. (a) : Consider n C r – 1 + 2 n C r + n C r + 1 6! = ( n C r – 1 + n C r ) + ( n C r + n C r + 1 ) \ Number of words = = 360 2! = n + 1 C r + n + 1 C r + 1 14. 4. 4. 30. (d) : Number of women = 5 M M = 25 × 122 = 3050 Number of men = 6 M M 21. 2. 5.100 2! Set of numbers divisible by 5 are 5. (l + 1) + (m + 1) + (n + 1) = 8 are non negative integers and S 10 = [10 + 100] = 5 × 110 2 \ Required number of ways = n + r – 1 C r \ Required sum = 25 × 102 + 1050 – 550 = 3 + 5 – 1 C 5 = 7 C 5 = 7 C 2 = 25[102 + 42 – 22] M 15. 3. 3.r C3 r = 1 Putting r = 6. 4 have equal number of ways of arranging the 50 n S 50 = [2 + 100] using S n = [a + l] balls in the different boxes. 20. 2 2 (ii): Let the number of balls in the boxes are x. z ³ 1 20 Þ l + m + n + 3 = 8 where l = x – 1. = 720 \ Required number of ways = 5! × 6 P 5 = 5! × 6! . of question 8 5 8 C 5 C 5 \ Required ways for case (ii) = 5 C 5 × 8 C 5 8 ´ 7 ´ 6 = 56 3 ´ 2 ´ 1 Total number of ways = 140 + 56 = 196 = 13. 5. of question 6 No. 12. these 5 women can be arranged themselves by P 5 1 1 = 5 × 36 × 4 way. (a) : Number of letters = 6 Number of vowels = 2 namely A. (d) : Odd numbers are 1. MATHEMATICAL INDUCTION AND ITS APPLICATION 2. having n radical signs then by methods of mathematical induction which is true (a) a n > 7. " n ³ 1 (b) a n > 3.. Statement2 is true; Statement2 is not a correct explanation for Statement1 (2008) Answer Key 1. (b) Let S(k) = 1 + 3 + 5 + . Then which of the following is true? (a) S(k) Þ S(k – 1) (b) S(k) Þ S(k + 1) (c) S(1) is correct (d) principle of mathematical induction can be used to prove the formula. Statement2 is true; Statement2 is a correct explanation for Statement1 (d) Statement1 is true.. (a) Statement1 is true. Statement2 is true 3.28 JEE MAIN CHAPTERWISE EXPLORER CHAPTER 6 1.. 1 1 1 + + .. + (2k – 1) = 3 + k 2 . n ( n + 1) < n + 1.. " n ³ 1 (c) a n < 4.. 1 2 n Statement2 : For every natural number n ³ 2. (d) 2. (c) Statement1 is true.. (b) 3. + > n . " n ³ 1 (d) a n < 3. Statement1 : For every natural number n ³ 2. (2002) . Statement2 is false (b) Statemen1 is false. " n ³ 1. (2004) If a n = 7 + 7 + 7 + . S of S(k) So S(1) is not true. we get 1 1 1 + + .. i.(i) ..(iii) 1 1 1 1 1 + + + .... L. + > k 1 2 k Þ k> Þ k+ ( k + 1) ....1 Þ k +1 Step 1 : For n = 2... + > k 1 2 k 1 Adding on both sides. (b) : S(k) = 1 + 3 + .(i) When k = 1.. + + > k + 1 1 2 3 k k + 1 By Assumption step..e.........S of S(k) ¹ R...(ii) Let S(k) is true \ 1 + 3 + 5 + . Now S(k + 1); 1 + 3 + 5 + . P (2) : Statement2 For n = k k ( k + 1) < k + 1 Þ Q k k + 1 > k k < k + 1 k k + 1 . + > n 1 2 n 1 1 + > 2 is true 1 2 Step 2 : Assume P(n) is true for n = k. + + > k + 1 1 2 3 k k + 1 hence (ii) is true for n = k + 1 hence P(n) is true for n ³ 2 So. + (2k – 1) = k 2 + 3 Þ 1 + 3 + 5 + . + + > k + 1 2 k k +1 k + 1 k > k + 1 - \ a n = = 1± 1 + 28 2 1 ± 29 > 3 2 ..... + (2k – 1) + (2k +1) = 3 + (k + 1) 2 . (b) : a n = k + 1 < k + 1 k + 1 Þ 1 k + 1 From (iii) & (iv) Step 3 : For n = k + 1.. we get k + 1 1 1 1 1 1 + + .(iv) .....H. Statement1 and Statement2 are correct but Statement2 is not explanation of Statement1 2. (d) : Statement1 1 1 1 Let P ( n ) : + + .. 1 1 1 + + .29 Mathematical Induction and Its Application 1..H. + (2k – 1) = 3 + k 2 .. we have to show that 1 1 1 1 1 + + + .(ii) . + (2k – 1) + (2k + 1) = 3 + k 2 + 2k + 1 = (k + 1) 2 + 3 Þ S(k + 1) true \ S(k) Þ S(k + 1) 7 + a n Þ a n 2 – a n – 7 = 0 For k ³ 2 k Þ k > k +1 Multiplying by k Þ 1 > 1 > k + 1 k + 1 3.. If x is so small that x 3 and higher powers of x may be neglected. (b) 132 (c) 144 (d) –132 (2011) The remainder left out when 8 2n – (62) 2n +1 is divided by 9 is (a) 2 (b) 7 (c) 8 (d) 0 n Statement1 : å ( r + 1) n Cr = ( n + 2)2 n -1 r = 0 n Statement2 : ( ) (2006) (2006) 11 é 1 ù If the coefficient of x 7 in ê ax 2 + equals the coefficient bx úû ë 11 é æ 1 ö ù in ê ax . (2005) (c) ab = 1 (d) b of x –7 10. (2004) n - 5 n - 4 5 6 n (a) 6 (b) (c) n - 4 (d) n - 5 . (2009) 5. 2. then a n is 1 (1 . 7.ax )(1 - bx) (a) n – 1 (b) 1 n - 1 2 (c) 1 n 2 (d) 2n - 1 .1 ö æ çè 2/3 1/3 ÷ is x . (a) å ( r + 1) n Cr xr = (1 + x ) n + nx (1 + x ) n -1 ( 3 ) (1 + x )3/ 2 . Statement2 is false (d) - x 2 . n) is (a) (20.. The term independent of x in expansion of 10 x +1 x . 20) (c) (45.a n +1 (c) (d) .. then (m. then a and b satisfy the relation è bx ø û ë (a) a + b = 1 (b) a – b = 1 a = 1.ç 2 ÷ ú . 3. n if (1 – y) m (1 + y) n = 1 + a 1 y + a 2 y 2 + . the sum of 5 th n – 1 2 n (a) (–1) (n – 1) (b) (–1) (1 – n) and 6 th terms is zero. 3 (c) x - 3 x 2 (a) Statement1 is true. Statement2 is true; Statement2 is a 11.b n (b) b - a b - a a n +1 . (d) an odd positive integer.b n +1 b n +1 . n 1 r t 5 and t n = å n . The coefficient of the middle term in the binomial expansion in powers of x of (1 + ax) 4 and of (1 – ax) 6 is the same if correct explanation for Statement1 a equals (d) Statement1 is true. The coefficient of x in expansion of (1 + x)(1 – x) is In the binomial expansion of (a – b) n . 45) (b) (35. n ³ 5. (c) an irrational number..1 + 1 x 2 then 1/ 2 (1 - x ) 3 2 (a) 3 x + x 8 may be approximated as 3 2 (b) 1 - x 8 r = 0 6.. 7 The coefficient of x in the expansion of (1 – x – x 2 + x 3 ) 6 is (a) –144 4.. (2005) 2 8 8 (b) Statemen1 is false. then n is equal to 13. 2 (2004) . b - a b - a For natural numbers m. then ( 3 + 1)2 n . Statement2 is true; Statement2 is (a) –3/10 (b) 10/3 (c) –5/3 (d) 3/5. If s n = å n s n r = 0 C r r = 0 C r (2007) If the expansion in powers of x of the function is a 0 + a 1 x + a 2 x 2 + a 3 x 3 + . 35) (d) (35. If n is a positive integer. not a correct explanation for Statement1 (2004) (2008) n n 12. (2012) 9.x + 1 x - x1/2 ø (a) 120 (b) 210 (c) 310 (d) 4 (2013) 8. (b) a rational number other than positive integers..( 3 - 1) 2 n is (a) an even positive integer. then a/b equals (c) (n – 1) (d) (–1) n – 1 n. 45). and a 1 = a 2 = 10. b n . Statement2 is true (c) Statement1 is true.30 JEE MAIN CHAPTERWISE EXPLORER CHAPTER BINOMIAL THEOREM 7 1.a n a n . (a) . The number of integral terms in the expansion of ( 3 + 8 5) is (a) 33 (b) 34 (c) 35 (d) 32. (c) 6. 19. If x is positive. the first negative term in the expansion of (1 + x) 27/5 is (a) 5 th term (b) 8 th term (c) 6 th term (d) 7 th term. 13. (b) 12. (b) (d) (c) (c) 2. r and n are positive integers r > 1. (a) 9. (c) 11. (a) 17. (c) 15. (d) 16. The positive integer just greater than (1 + . If the sum of the coefficients in the expansion of (a + b) n is 4096. (d) 14. n > 2 and coefficient of (2002) th th 2n (r + 2) term and 3r term in the expansion of (1 + x) are Answer Key 1. then n equals (a) 3r (b) 3r + 1 (c) 2r (d) 2r + 1. (c) 5. (a) 4. (a) 10. (2003) 256 15. then the greatest coefficient in the expansion is (a) 1594 (b) 792 (c) 924 (d) 2924.31 Binomial Theorem 14. The coefficients of x p and x q in the expansion of (1 + x) p + q are (a) equal (b) equal with opposite signs (c) reciprocals of each other (d) none of these. (2002) 19.0001) 1000 is (a) 4 (b) 5 (c) 2 (d) 3. (2002) equal. 17. (2003) 16. (d) 3. (b) 18. (2002) 18. 7. (c) 8. .ax )(1 - bx ) = (a 0 + a 1 x + ...x -1/2 ) 10 \ 20 .x1/3 + 1) ( x + 1)( x . ù ....( 3 - 1) 2 n = 2 é 2n C1∙( 3) 2n -1 + 2 nC3 ∙( 3) 2n .) Þ (a 0 + a 1x + ...2 b 2 + . (c) : n n å ( r + 1) n Cr = å r × n Cr + å n Cr r =0 r =0 r = 0 n n n = å r × × n -1 Cr -1 + å n C r r r =0 r = 0 = n × 2n -1 + 2n = 2n -1 ( n + 2) Thus Statement1 is true... + a nx = 1 + x (a + b) + x 2 (a 2 + ab + b 2 ) + x 3 (a 3 + a 2 b + ab 2 + b 3 ) + ..... 45) ( (c) : Tr + 1 of ax 2 + 1 bx ) 11 1 Tr + 1 of æç ax .....r 1 Cr ( ax ) r æç .4 = . + a n–1 x n–1 + a n x n + .. (***) + .3 + .. + x n n (a + a n–1 b + a n–2 b 2 + .b ) = - Cr -1 x r + å n Cr x r = nx (1 + x )n -1 + (1 + x ) n Substitute x = 1 in the above identity to get å ( r + 1) n Cr = n × 2n -1 + 2 n Statement2 is also true & explains Statement1 also. + a b n–1 + b n (a) : (1 – x – x 2 + x 3 ) 6 = ((1 – x)(1 – x 2 )) 6 8 = –1 (modulo 9) ( On comparing the coefficient of x n both sides we have an irrational number........ n n n r n r n r Again å ( r + 1) Cr x = å r × Cr x + å Cr x r =0 r = 0 n = nå r=0 a n .bx ) -1 (1 .r..... ìï ( x1/3 + 1)( x 2/3 .. (a) : Using Modulo Arithmetic Also 62 = –1 (modulo 9) = [(–1) 2n – (–1) 2n + 1 ]mod 9 = (1 + 1)mod 9 = 2 mod 9 Þ Remainder = 2 n 5.. x +1 x . n) = (35.. (m.. 3..a n +1 ......... (*) Differentiating w.... Þ n(1 + y) n–1 (1 – y) m – m(1– y) m–1 (1 + y) n = a 1 + 2a 2 y + 3a 2 y 2 + 4a 4 y 3 + ...t.1 ö æ (b) : ç 2/3 1/3 è x ..(**) Again differentiating (**) with respect to y we have [n(n – 1)(1 + y) n–2 (1 – y) m + n(1 + y) n–1 (–m) (1 – y) m–1 ] –[m(1 + y) n (m – 1)(1 – y) m–2 (1–y) m–1 n(1 + y)+n–1] = 2a 2 + 6a 3y ...ax ) -1 (1 .) (1 + bx + b 2 x 2 .5 r 6 = ( x1/3 ......r ( ) 1 bx 11 ) 1 bx = 11 C 5 a 6 b5 . y both sides of (*) we have –m(1 – y) m–1 (1 + y) n + (1 – y) m n(1 + y) n–1 = a 1 + 2a 2 y + 3a 3 y 2 + 4a 4 y 3 + .. n n -4 4 (b) : C4 a ( .........1) üï =í ý x ( x ... + ...... ë û = (1 – 6 C 1 x + 6 C 2 x 2 – 6 C 3 x 3 + 6 C 4 x 4 – 6 C 5 x 5 + 6 C 6 x 6 ) (1 – 6 C 1 x 2 + 6 C 2 x 4 – 6 C 3 x 6 + 6 C 4 x 8 – 6 C 5 x 10 + 6 C 6 x 12 ) Coeff. b - a (d) : (1 – y) m (1 + y) n = 1 + a 1 y + a 2 y 2 + a 3 y 3 + ... + a n x n + ....... r = 4 (c) : ( 3 + 1)2 n ..x + 1 x - x1/2 ÷ø 6.2 ö÷ è bx ø 11 2 \ Coeff. + ab n–1 + b n )+ .(B) Solving (A) and (B) n = 45......32 JEE MAIN CHAPTERWISE EXPLORER 10 1.2 ö÷ bx ø è ( 11 = = 11Cr (ax 2 ) r 11 ..1) þ x 2/3 .. m = 35 \ n n -1 r = 0 ( a n + a n -1b + a n . = 6 ∙ 20 – 20 ∙ 15 + 6 ∙ 6 = 120 – 300 + 36 = –144 4. 10 7.. of x 7 = (– 6 C 1 )(– 6 C 3 ) + (– 6 C 3 )( 6 C 2 ) + (– 6 C 5 )(– 6 C 1 ) = 8. Now putting y = 0 in (**) and (***) we get n – m = a 1 = 10 (A) and m 2 + n 2 –(m + n) –2 mn = 2a 2 = 20 .) (1 – bx) –1 = (1 + ax + a 2 x 2 + . + ......5 r =0 6 Thus 2...... = = (1 – x) 6 (1 – x 2 ) 6 Þ 8 2n – (62) 2n + 1 n a n = a n + a n–1 b + a n–2 b 2 + . . + .) n + ...... + b n x n + . of x 7 in ax + 11 ....x1/3 + 1 ï îï 20 . + abn -1 + b n )( b . b 5 ) C5 a n -5 ( - b ) 5 Þ (d) : From given 1 = (1 ... b n +1 . Tr + 1 = ( - 1)r 10 Cr x Þ \ Term = 10 C 4 = 210.a ) b - a (Multiplying and dividing by b – a) 9.. 16.3r + 1 = r + 1 ï Þ 2 n = 4 r Þ 3r – 1 = r + 1 and í ï n = 2 r î 2r = 2 n .. = 11 × 3×4 × 7 = 924 6 5 4 3 2 1 3 . 8 7 11 ´ 9 . 8. (a) : Coefficient of middle term in (1 + ax) 4 = coefficient of middle term in (1 – ax) 6 3 10 12.å n C r t n = n å 1 n Cr n - 10 9 10 \ The positive integer just greater than is 2...... (c) : Consider (a + b) n = C 0 a n + C 1 a n – 1 b n + C 2 a n– 2 b 2 + .1 + 3 × 1 x + 3 × 2 × 1 x 2 + .33 Binomial Theorem 11 1 ö æ and coefficient of x –7 in ç ax .3 x 2 é1 + 1 x + 1 × 3 × 1 x 2 + . 8 11. + ç 4 ÷ ç ÷ 4 è 10 ø è 10 ø 2 è 10 ø \ ab = 1.(ii) \ By (i) and (ii) Coefficient of x p in (1 + x) p + q = Coefficient of x q in (1 + x) p + q 19. 9 17. (n .. (a) : (3 1/2 + 5 1/8 ) 256 r T r + 1 = 256 C r ( 3) 256 - r 5 8 256 - r r ... (b): (1 + x) (1 – x) n = (1 – x) n + x(1 – x) n \ Coefficient of x n is = (–1) n + (–1) n – 1 n C 1 = (–1) n [1 – n] \ 4 C 2a 2 = 6 C 3 (– a) 3 Þ a = - n 13. (d) : 1/ 2 (1 . 7 .r ) n C n . (a) : In the expansion of (1 + x) p + q T r + 1 = p + q C r x r \ Coefficient of x p = p + q C p å n C r n < 1 + . n > 2 and Coefficient of T r + 2 = Coefficient of T 3r in (1 + x) 2n Þ 2n C r + 1 = 2n C 3r – 1 R < \ ì 2 n .. 2 . 3 ( ) (1 + x )3/ 2 . . 8 ..1) .x 2 (1 - x ) -1/ 2 = . 1 . (c) : Given r > 1. .. = ´ .. are both positive integer 2 8 \ r = 0.....( n .r r = 0 1 n n Þ t n = n r =0 r å nC r .r = r = 0 replacing n – r by r r = 0 t n = ns n – t n t n n \ = s n 2 = ( p + q )! q !( p + q - q )! = ( p + q )! q ! p ! 27 n( n . (c): t n = r r = 0 t n = å n .. (d) : General term in the expansion of (1 + x ) 5 For integral terms 10 1 1 1 + + 3 + .r å n C n .x ) 3 1 + x + 3 1 1 x 2 + .2 ÷ = bx ø è 11 Now a6 = b5 C5 11 C 6 a 5 b 6 11 C 6 a 5 b6 1000 1 ö æ 16.. ù 8 úû 8 êë 2 2 2 2! 3 = .r + 1) r x r ! 32 27 \ n – r + 1 < 0 Þ + 1 < r Þ r > 5 5 T r + 1 = Þ r > 6 15.¥ = 9 10 10 2 10 ìQ nCx = n C y ï r = 1 í Þ x + y = n ï or x = y î 18. 2 2 2 2! 2 2! 4 = (1 - x ) 1/ 2 ( )( ) 3 = .. + C n 2 n = 4096 = 2 12 Þ n = 12 (even) Now (a + b) n = (a + b) 12 as n = 12 is even so coefficient of greatest term is C n = 12C12 = 12 C6 n 2 = 2 12 11 10 9 .x 2 + higher powers of x 2 .(i) Also coefficient of x q in (1 + x) p + q is = p + q C q 14. (c) : Let R = ç 1 + 4 ÷ 10 ø è 10 3 1 2 999 æ 1 ö æ 1 ö æ 1 ö = 1 + 1000 ç 4 ÷ + 1000 + .256 \ 256 = 0 + (n – 1)8 using t n = a + (n – 1)d 256 256 \ = n – 1 \ n = + 1 8 8 n = 32 + 1 Þ n = 33 ( p + q)! ( p + q )! = p !( p + q - p)! p ! q ! .1 + 1 x 2 10...... + C nb Putting a = b = 1 \ 2 n = C 0 + C 1 + C 2 + ..... . n Statement 2 : å ( k 3 . (2012) 3.. is (a) 150 (b) zero (c) – 150 (d) 150 times its 50 th term (2012) 10. .1 + 1 - . Statement 2 is false. Statement 2 is true....( k . 12.77.777...P.... If the coefficients of r th .. . = a 10 = 150 and a 10 .. If the terms of the geometric progression are alternately positive and negative... p ¹ q ..P. If a1 + a2 + .P. (a) Statement 1 is true. Then the common ratio of this progression is equals (a) (b) 1 5 - 1 2 ( 5 1 (c) 2 1 - 5 ( ) ) 1 (d) 2 5.34 JEE MAIN CHAPTERWISE EXPLORER CHAPTER 8 1. + a p p 2 a = . The sum of the third and the fourth terms is 48. each term equals the sum of the next two terms. a n are in H.. a 2 .. (2005) .. + (361 + 380 + 400) is 8000. then the expression a 1a 2 + a 2a 3 + . . correct explanation for Statement 1. then 6 equals a1 + a2 + . SEQUENCES AND SERIES The sum of first 20 terms of the sequence 0. then the first term is (a) 4 (b) – 4 (c) – 12 (d) 12 (2008) 1 .P. If 100 times the 100 th term of an A... + a n – 1 a n is equal to (a) n(a 1 – a n) (b) (n – 1)(a 1 – a n) (c) na 1 a n (d) (n – 1)a 1 a n . Statement 2 is true; Statement 2 is not a correct explanation for Statement 1. then the 150 th term of this A. 4.P. 2. be terms of an A.. (r + 1) th and (r + 2) th terms in the If a 1 = a 2 = . a 3 . with binomial expansion of (1 + y) m are in A.. is (a) 7 (99 - 10 -20 ) 9 (b) 7 (179 + 10-20 ) 81 (c) 7 (99 + 10 -20 ) 9 (d) 7 (179 - 10-20 ) 81 7. + aq q 2 a 21 (a) 41/11 (b) 7/2 (c) 2/7 (d) 11/41. (2013) Statement 1 : The sum of the series 1 + (1 + 2 + 4) + (4 + 6 + 9) + (9 + 12 + 16) + . His total saving from the start of service will be ` 11040 after 11. (2006) If a 1. 6.. (c) Statement 1 is false.1)3 ) = n 3 for any natural number n . k = 1 8. then m and r satisfy common difference –2.. 0. A man saves ` 200 in each of the first three months of his service. upto infinity is The sum of the series 2! 3! 4! 1 (a) e - 2 (b) e + 1 2 (c) e –2 (d) e –1 . a 2..7... (b) Statement 1 is true... . 0.. (d) Statement 1 is true. (a) 20 months (b) 21 months (c) 18 months (d) 19 months (2011) The sum to infinity of the series 1 + 2 + 6 + 10 + 14 + . a 11 . is 3 32 33 34 (a) 3 (b) 4 (c) 6 (d) 2 (2009) The first two terms of a geometric progression add up to 12. are in an A. Let a n denote the number of notes he counts in the n th minute. Statement 2 is true; Statement 2 is a 9. (2007) Let a 1 . (2006) A person is to count 4500 currency notes.. In each of the subsequent months his saving increases by ` 40 more than the saving of immediately previous month. 5. then the time taken by him to count the equation all notes is (a) m 2 – m(4r – 1) + 4r 2 + 2 = 0 (a) 24 minutes (b) 34 minutes (b) m 2 – m(4r + 1) + 4r 2 – 2 = 0 (c) 125 minutes (d) 135 minutes (c) m 2 – m(4r + 1) + 4r 2 + 2 = 0 (2010) (d) m 2 – m(4r – 1) + 4r 2 – 2 = 0. with nonzero common difference equals the 50 times its 50 th term. (2007) In a geometric progression consisting of positive terms.P. . log 3[4 × 3 x – 1] are in A. are in G. 1 1 (b) G. 13. R 1 . y 1 ).. 21. (d) e - 1 ..P. (b) H. m and T n = . (2005) 14. n n n 13. then the points (x 1 .. 25. (2003) 21. (c) ArithmeticGeometric Progression (d) A. then a – d equals n m (c) H.).P. 8. 24. 27. a 3 .P. 14. b. x 3 and y 1 . R. is equal to log an + 8 (a) 0 (b) 1 (c) 2 (a) log e 2 – 1 (c) log e (4/e) (2005) 15. n. 19. c are in A. y 2 . If f (1) = f (–1) and a.G. a n . When n is odd. (d) (d) (a) (d) (b) 3. with the same common ratio. Let R 1 and R 2 respectively be the maximum ranges up and down on an inclined plane and R be the maximum range on 16. (d) G. then f ¢(a). If x 1 . 26.P. y 3 ) (a) lie on an ellipse (b) lie on a circle (c) are vertices of a triangle (d) lie on a straight line. (2003) 23. (b) (d) (b) (c) 6.P. whose first term is a and the horizontal plane. (2003) 22. x 2 . upto ¥ is equal to 1 × 2 2 × 3 3 × 4 (b) log e 2 (d) 2log e 2. (b) are in H. 17. (c) satisfy a + 2b + 3c = 0 (d) are in A. 2 2 n ( n + 1) n ( n + 1) (2002) (a) (b) 4 2 26. and |a| < 1.. (x 2 ..1 + 1 . 18. (a) 256 (b) 512 (c) 1024 (d) none. + 9 = n is even. 23.. (b) (b) (c) (c) (b) 4. (2002) 2 4 1 2 + 2 × 2 2 + 3 2 + 2 × 4 2 + 5 2 + 2 × 6 2 + . 15.. Then.. . (c) (d) (2004) e 2 28. 1 1 + .....P.P. (b) ArithmeticGeometric progression (c) A. then x equals (a) log 3 4 (b) 1 – log 3 4 17. then the determinant log an log an + 1 log a n + 2 D = log an + 3 log an + 4 log an + 6 log an + 7 log a n + 5 . (2002) then a. The sum of first n terms of the series (c) 1 – log 3 (d) log 4 3.P. x + 4cy + cz = 0 has a nonzero solution. 22. 10. |b| < 1. (b) (d) (c) (d) (b) 5.P.35 Sequences and Series ¥ ¥ ¥ n=0 n=0 n = 0 (a) are in G. Let f (x) be a polynomial function of second degree. then the product of its 9 terms is 19. 7. z are in (a) H.P. Fifth term of a GP is 2. ¥ is 4 × 2! 16 × 4! 64 × 6! e - 1 (a) e + 1 (b) (c) e + 1 e e 2 e 20.P. 12.P. y 3 are both in G.. The sum of series 2! 4! 6! 27. If x = å a . x + 3by + bz = 0. (2004) (c) (d) éê square is 100. c are in A. The common ratio of GP is ú ë 2 û 2 (a) 5 (b) 3/5 (c) 8/5 (d) 1/5. z = å c where a. The sum of the series 1 + 1 + 1 + 1 + . 1 1 1 (2002) + + + . 11. b. y = å b . y... Let T r be the r th term of an A.P. The value of 21/ 4 × 41/8 × 81/6 . If a 1 ..P.. Sum of infinite number of terms in GP is 20 and sum of their 2 3n(n + 1) n(n + 1) ù . R 2 are in common difference is d.P. f ¢(b) and f ¢(c) are in (a) G. is n( n + 1) when 3 3 3 3 3 2 25. T = . 9. 28. 20.. If the system of linear equations x + 2ay + az = 0. log 9 (3 1 – x + 2). ¥ is (e - 1) 2 (e 2 - 1) (a) (b) 2 e 2 e (a) 1 (b) 2 (c) 3/2 (d) 4.P.. . If for some positive integers m. |c| < 1 then x. 1 – 2 + 3 – 4 + . (b) (c) (a) (b) (a) 2. 2 e (2005) (2003) 1 . 2 - 2) 2 ( e (2002) (e - 1) . a 2 . If 1. c Answer Key 1.. y 2 ) and (x 3 .. the sum is (a) 425 (b) – 425 (c) 475 (d) – 475.P. (2003) (a) 1/mn (b) 1 (c) 0 (d) m n (2004) 24.P. The sum of the series (d) 4. is 18.P. b.. 16. (d) ArithmeticGeometric Progression (A. . (a) (b) (a) (c) . m ¹ n. (a) A. 3 3 3 3 3 7 (179 + 10-20 ) 81 2.1)( -21)) = 3000 2 Let n – 10 = m Þ m × 148 –m(m – 1) = 3000 Þ m 2 – 149m + 3000 = 0 Þ (m – 24)(m – 125) = 0 9.(1) .600 = 10440 è 2 ÷ø Þ n 2 + 5n – 546 = 0 Þ (n + 26)(n – 21) = 0 8.3 ö Þ ç (480 + 40n .. (d) : Given a 1 . n ..7 14444244443 Hence.4)40} = 11040 2 Now a = æ n ...to ¥ ú = ∙ = ∙ = 2 3 êë 3 û 3 1 - 1 3 2 / 3 3 2 Þ S = 2 \ S = 3 3 7. n = 34 is the only answer.. a1 + a2 + .1) d q \ m = 24..160) = 11040 ... (b) : We have a 1 + a 2 + . ar 3 . + r = (1 ... T 3 = 19 = 27 – 8 Þ T n = n 3 – (n – 1) 3 4é 1 4 1 ù 4 1 1 + + .(1) .P.. + a n = 4500 – 10 × 150 = 3000 Þ 148 + 146 + ..10-20 ) ý 9 íî 9 þ 1 + (1 + 2 + 4) + (4 + 6 + 9) + .. ..77777. 12 12 12 = = = -12 From (1) 1 + r 1 .1) d p = 2 a1 + ( q .. we have a 135 = 148 + (135 – 1)(–2) = 148 – 268 < 0 But a 34 is positive....e.x = 1 .10 . a = ar + ar 2 Þ r 2 + r – 1 = 0 -1 + 5 .36 JEE MAIN CHAPTERWISE EXPLORER 1.. (a) : Let S = 1 + 7ì 1 ü 20 .. .. (b) : tr = 0.. ar 2 ...... be a. we have a + ar = 12 ar 2 + ar 3 = 48 on division we have = \ Statement 2 is a correct explanation of statement 1..+ - . 2! 3! 4! upto infinity \ n = 21. (b) :100 (a + 99d) = 50 (a + 49d) Þ a + 149d = 0 i. r terms 7 7 7 7 = + + . T 150 = 0 3 ´ 200 + ..... (b) : Let it happens after n months.x + x . . upto infinity.+ ..x + x . 3 32 33 3 4 1 1 2 6 10 S = + + + + . \ r = – 2 3. 3 3 31 3 1 3 Subtracting (2) from (1).(2) ar 2 (1 + r ) 48 = Þ r 2 = 4 a (1 + r ) 12 \ r = ± 2 But the terms are alternately positive and negative. giving n = 34. (d) :Statement 1 : . a p p 2 = 2 a1 + a2 + .... ar...3 {2 ´ 240 + ( n . a 3 .. 5... T 2 = 7 = 8 – 1... a 2 ... 135 But for n = 135. = 3000 n ......2 - 1 2 3 4 (d) : Q e .(2) 2 1 4 4 4 S = 1 + + 2 + 3 + 4 + . Then put x = 1.. a q q Þ r = Þ p [ 2 a1 + ( p .10...r ) 10 102 10 9 20 S 20 = å r =1 = 7æ tr = ç 20 9è 20 ö 10.. + (361 + 380 + 400) is 8000 n 3 3 3 Statement 2 : å ( k ..1) ) = n k = 1 Statement 1 : T 1 = 1. be terms of A.1) d ] 2 Þ 2 a 1 + ( p . we get 6... 2 10. 4.. 1! 2! 3! 4! (b) : Given..10 Þ × (2 ´ 148 + (n ..(k .(1 . 125...... .1) d ] p 2 2 = 2 q q [ 2 a1 + ( q .r ÷ = ø r =1 å 2 6 10 14 + + + + . (c) : Let the G. we get 1 1 1 1 e -1 = 1 .P.. + a n = 4500 Þ a 11 + a 12 + . 1 . c = x y z log b + ( r + 7)log R Þ using (applying C 2 ® 2C 2 – (C 1 + C 3 )) log b + ( r . a n – 1 a n = (n – 1)a 1a n ... a 2... å bn = ..(ii) ..1 d - a n d .37 Sequences and Series Þ [2a 1 + (p – 1)d]q = p[2a 1 + (q – 1)d] as a. b. z = 1. (b) : T r + 1 = m C r y r ...P.r + 1)! Þ log b + ( r + 1)log R = M an -1 a n = log b + r log R log b + ( r + 2)log R log b + ( r + 3)log R log b + ( r + 4)log R 1 n 1 m 1 1 Now T m – T n = - = (m – n) d n m 1 1 Þ d = and a = mn mn \ a – d = 0 T n = a + (n – 1) d = 17.r ) + ( r + 1)!( m . ¥ ÷ ú ç ÷ 2 ê ø úû ë è 2 2! 2 (4!) 2 (6!) = öù 1 é æ x 2 x 4 x 6 + + + . y . a.. ¥ 4(2!) 16(4!) 64(6!) 1 1 1 = 1+ 2 + 4 + + . \ log r = log b + ( r - 1)log R 11. y= . | b | < 1.. å c n = 1...x ùû 2 where x = 1/2 = 1 [ e1/ 2 + e -1/ 2 ] = e + 1 2 2 e 16.. (b): As S n is needed for n is odd let n = 2k + 1 \ S n = S 2k + 1 = Sum up to 2k terms + (2k + 1) th term 2k (2k + 1) 2 + last term 2 ( n - 1) n 2 n 2 ( n + 1) = + n 2 as n = 2k + 1 = 2 2 = . . ..r + 1)! ( r + 1)!( m . (n) Adding (i).r + 1)( m .a n = 0 1 ...b 1 - c Þa = log b +( r + 6)log R y ..a2 a a a1 a 2 = 1 = 1 - 2 d d d a 2 a 3 a2 a 3 = - d d Þ .r + 1)! ( r + 1)!( m .P. (a) : Let t r denote the r th term of G.1)!( m . b.. (i) .P. ¥ and n å a = n=0 ¥ ¥ 1 1 1 .r - 1)! r !( m - r )! Þ r ( r + 1) ( m ..1 ö 2 1 1 x . (d) : Given a 1.. \ 12....(i) .. 13.a 1.P. \ m C r – 1 + m C r + 1 = 2 × m C r Þ m! m ! m ! + = 2 ( r . a n are in H..1)log R log b + ( r + 5)log R a .b n = 0 1 - c 1 1 1 \ x = ...r + 1)! r (r + 1) + (m – r + 1)(m – r) = 2(r + 1)(m – r + 1) Þ r (r + 1) + (m – r) 2 + m – r – 2(r + 1) (m – (r – 1))=0 Þ r (r + 1) + m 2 + r 2 – 2mr + m – r + 2 (r 2 – 1) – 2m(r + 1) = 0 Þ m 2 – m (4r + 1) + 4r 2 – 2 = 0. (a) : Given | a | < 1. b= . c Î AP \ 2b = a + c Þ 2a 1 (q – p) = d[(q – 1)p – (p – 1)q] Þ 2a 1 (q – p) = d(q – p) Þ 2a 1 = d æ y .1 x -1 z . z Î H. Î A. 2 1 1 1 + + + .r +1) = ( r + 1)!( m ... (n) equations we get a 1 a n - a 1a 2 + a 2 a 3 + a 3 a 4 + .1) log R 0 log b + log R ( r + 1) 1 log b + ( r + 2) log R 0 log b + ( r + 4) log R 2 log b + ( r + 5) log R 0 log b + ( r + 7) log R 1 = ´ 0 = 0. (ii) .1 z . Þ 1 1 1 ..P. (ii) a n . | c | < 1... 14. a n – 1 a n = d d 1 1 Also = + ( n ..1 .a n = ( n . (c) : 1 + = öù 1é æ 1 1 1 ê 2 ç1 + 2 + 4 + 6 + .. ¥ ÷ ú ê 2 ç1 + 2ë è 2! 4! 6! ø û = 1 éë e x + e . (c) : T m = a + (m – 1) d = 2( r + 1)( m ..1 2 ç ÷ = x + z Þ y = x + z è y ø Þ a 6 a + 5d a + 10 a 1 = 1 = 1 a21 a1 + 20 d a1 + 40 a1 Þ a 6 = 11 a21 41 Þ x .1) a1 a n d Þ a 1a 2 + a 2 a 3 + ... a1 a 2 a n Þ 1 1 = d a2 a1 Now from given determinant we have log b + ( r . ¥ 2 2! 2 (4!) 26 (6!) 15..1) d an a1 a1 ..... with first term b and common ratio R \ tr = bR r .. c Î A. 1 2 n 1 1 ö å çè 2 n .1 . 1 1 1 + – ... R 2 Î H...P. 2 Þ log 3 (3 1 – x + 2) = log 3 (4∙3 x – 1) + 1 Þ 3 1 – x + 2 = (4∙3 x – 1) × 3 Q log 3 3 = 1. + (2n – 1) 3 ] n = odd = 5 – 2 3 [1 3 + 2 3 + . (c) : Let q be the angle of inclination of plane to horizontal and u be the velocity of projection of the projectile u 2 u 2 .. 1 3 b b = 0 1 4 c c Þ (3bc – 4bc) – 2a(c – b) + a(4c – 3b) = 0 2ac = b a + c Þ a. Þ f ¢ (a)... + 9 3 ) – (2 3 + 4 3 + 6 3 + 8 3 ) = (1 3 + 3 3 + 5 3 + .38 JEE MAIN CHAPTERWISE EXPLORER = –[log e 2 – 1] = 1 – log e 2 Now s = s 1 – s 2 = (A) – (B) = log e 2 – 1 + log e 2 = log(4/e) é ù x 2 x 4 + + . t = 3/4 Þ 3 x = –1/3 which is not possible 3 3 Þ 3 x = and t = 4 4 Þ x log 3 3 = log 3 3 – log 3 4 ........ C are collinear.. B(ar.P.÷ + ç .r ) b = and a (1 - r ) a br (1 .. 24.+ .(B) 25... (d) : Let x 1 = a \ x 2 = ar. (a) : e x + e –x = 2 ê1 + 2! 4! úû ëê -1 1 1 e + e – 1 = + + .. 2ab.. Þ a 2 . C(ar 2 ..¥ 2 × 3 4 × 5 6 × 7 å 1 log 3 (3 1 – x + 2). c Î A. f ¢ (b). ¥ 2 2! 4! 1 1 1 (e - 1) 2 + + + ..(A) (By taking logarithm at the base 3 both sides) Þ x = 1 – log 3 4 1 (2n) (2n + 1) s 2 = å t n¢ = t¢ n = 1 = (2n )(2 n + 1) R 1 .¥ 1 × 2 2 × 3 3 × 4 1 1 1 + + Let s 1 = + .P.. ac Î A. f ¢ (c) = 2ac as a.... ¥ ú 2 3 4 5 ë û = [1 3 + 3 3 + .P. (c): s = \ s n = å t n = 22. B. 1 – x Þ 3 + 2 = 12∙3 x – 3 x 1 Þ 3 [(3 – x ) + 2] = 12∙3 2x – 3∙3 x (multiplying 3 x both side) 2 x Þ 12t – 5t – 3 = 0 where t = 3 Þ (3t + 1) (4t – 3) = 0 Þ t = –1/3.. R. ¥ ú 18. y 3 = br 2 Now A(a..1)(2n) 2n . f ¢ (b) = 2ab... . but point B is common so A.r ) b = slope of BC = ar (1 - r ) a as slope of AB = slope of BC \ AB || BC... br). Now slope of AB = Þ 21. b. + n 3 ] n = even = 4 = [2n(n + 1) (n + 2) (n + 3) – 12n(n + 1) (n + 2) + 13n(n é n 2 ( n + 1) 2 ù + 1) – n] n = 5 (odd) – 2 3 ê ú 4 êë úû n = 4 (even) (Remember this result) .. ¥ è 2 3 ø è 3 4 ø é 1 1 1 1 ù = . R 2 = \ R 1 = g (1 . br 2 ) 23.sin q) g (1 + sin q ) 2 g 2 1 1 + \ = 2 = R u R1 R2 20. x 3 = ar 2 and y 1 = b \ y 2 = br.. f ¢ (c) Î A. log 3 (4∙3 x – 1) Î A. \ t n = . (b) : For non trivial solution the determinant of the coefficient of various term vanish 1 2 a a i.2 n ÷ø 1 1 1 1 + .÷ + .. c Î H.P.¥ 1 × 2 3 × 4 5 × 6 æ b(1 . Þ 2a 2 . Þ 1 1 1 = (2n ...ê . 2ac Î A. ab.P.. b. (d) : Let the polynomial be f (x) = ax 2 + bx + c given f (1) = f (–1) Þ b = 0 \ f (x) = ax 2 + c now f ¢ (x) = 2ax \ f ¢ (a) = 2a 2 . (c) : As 1.+ + ... b). (a) : (1 3 + 3 3 + 5 3 + ..¥ 2 3 4 5 = log e 2 = 1 - Again s 2 = 1 1 1 + + + . + 9 3 ) – 2 3 (1 3 + 2 3 + 3 3 + 4 3 ) 1 ö æ 1 ç ÷ è 2n 2 n + 1 ø å æ 1 1 ö æ 1 1 ö = ç . ¥ = 2! 4! 6! 2 e 19..P.+ .e. .. r = common ratio 1 - r S¥ = 20 a According to question = 20 1 - r Þ a = 20(1 – r) .(*) ...(i) 2 Also a 1 - r 2 = 100 Solving (i) and (ii) we have r = 3/5 = 2l (say) 1 2 3 4 + + + + .r 1 + r Þ a = 5(1 + r) 27.ar 8 i = 1 8 ´ 9 = a9 r 2 .P is a and common ratio r \ t 5 = ar 4 = 2 9 \ a a ...P.. (b) : Let terms of G. ar..... (b) : Let first term of a G.r 4 1 2 so S¥ = 2 1 Where l = 28.. Now (B) – (A) Þ = + + 2 4 8 16 32 64 a 1 2 l = ´ = \ l = 1 1 . Þ = 100 1 .¥ 4 8 16 32 l 1 2 3 4 + + + ... are a.. ar 2 ......39 Sequences and Series 1 2 3 4 + + + + ... ¥ 8 16 32 = [2 × 5 × 6 × 7 × 8 – 12 × 5 × 6 × 7 + 13 3 æ 16 ´ × 5 × 6 – 5] – 2 ç è = [3750 – 5(505)] – 2 × 16 × 25 25 ö ÷ 4 ø = 1225 – 800 = 425 26. (b) : S¥ = 2 4 Õ ai = a × ar ar 2 ..¥ = 0 + + 2 8 16 32 64 l 1 1 1 1 1 + + + ..(ii) = a 9 r 36 = (ar 4 ) 9 = 2 9 = 512 .. (A) . a \ S¥ = where a = first term. (B) . .. 1 ö f(x) = [x] cos ç p .r. differentiable in (2. (b) Statement 1 is true. ì sin( p + 1) x + sin x . Statement 2 is true; Statement 2 is a correct explanation for Statement 1. a firm is manufacturing 2000 items. Statement 2 is false. DIFFERENTIAL CALCULUS (b) 1 (c) 2 Statement 2 : a = (d) –1/4 (2013) At present. Statement 2 is true; Statement 2 is not a correct explanation for Statement 1. x>0 x = – 1 and x = 2. 5. x Î R 8. Consider the function. x ¹ 0 has extreme values at ï x+x . then the new level of production of items is 7. (d) Statement 1 is true.cos2 x )(3 + cos x ) lim is equal to x ® 0 x tan4 x (a) 1/2 2. 5) and f(2) = f(5). If f : R ® R is a function defined by æ 2 x . The values of p and q for which the function continuous for every real x. (2012) A spherical balloon is filled with 4500 p cubic metres of helium gas. Statement 2 is true; Statement 2 is a correct explanation for Statement 1. (a) 3000 (b) 3500 (c) 4500 (d) 2500 (2013) 3. continuous only at x = 0. It is estimated that the rate of change of production P w. (2012) x ï f (x) = ï q . then f is dy 2 equals to æ d 2 y ö æ dy ö -2 (a) çç 2 ÷÷ ç ÷ è dx ø è dx ø æ d 2 y ö æ dy ö-3 (b) çç 2 ÷÷ ç ÷ è dx ø è dx ø -1 æ d 2 y ö (c) çç 2 ÷÷ è dx ø -1 (d) æ d 2 y ö æ dy ö-3 .40 JEE MAIN CHAPTERWISE EXPLORER CHAPTER 9 1.çç ÷÷ ç ÷ (2011) è dx 2 ø è dx ø æ 1 . is continuous for all x in R . (b) Statement 1 is true.2)} ö ÷ lim çç ÷ x ® 2 è x-2 ø (a) equals - 2 (b) equals 1 2 equals 2 (c) does not exist (d) discontinuous only at nonzero integral values of x. 10. (d) Statement 1 is true. ïî x 3/2 Statement 1 : f has local maximum at x = – 1 and at x = 2. (2012) 9. f(x ) = |x – 2| + |x – 5|. Statement 2 is false. are (a) (b) (c) (d) 6. Statement 2 is true; Statement 2 is not a correct explanation for Statement 1. 5]. then (a) 1 2 (b) 1 dy at x = 1 is equal to dx (c) 2 (d) 1 (a) Statement 1 is true. Statement 1 : f ¢(4) = 0 Statement 2 : f is continuous in [2. If y = sec(tan –1x).t. b Î R be such that the function f given by í ï 2 f ( x ) = ln | x | + bx 2 + ax . x < 0 ï discontinuous only at x = 0.x . (c) Statement 1 is false. d 2 x (2011) . (1 . (c) Statement 1 is false. Statement 2 is true. If the firm employs dx 25 more workers. additional number of workers x is given by dP = 100 - 12 x .cos {2( x . Statement 2 is true. (a) Statement 1 is true. x = 0 Let a. where [x ] denotes the greatest integer è 2 ÷ø function. then the rate (in metres per minute) at which the radius of the balloon decreases 49 minutes after the leakage began is (a) 2/9 2 1 -1 and b = 2 4 (b) 9/2 (c) 9/7 (d) 7/9 (2012) (2013) 4. If a leak in the balloon causes the gas to escape at the rate of 72p cubic metres per minute. Then which one of the following holds? p p (a) The cubic has maxima at both and – 3 3 p p (b) The cubic has minima at and maxima at – 3 3 p (c) The cubic has minima at – and maxima 3 p at 3 p p (d) The cubic has minima at both and – 3 3 (2008) If f has a local minimum at x = –1. (a) 5 (b) 7 (c) Statement1 is true. 1) ® R be a differentiable function with f (0) = –1 and f ¢(0) = 1. then in the interval [–1. Statement2 is true. if x > -1 (a) 1 (b) log 2 (c) –log 2 (d) –1 (2009) 18. q = 2 2 17. (2009) (c) 2 (d) –1. Statement2 is true; Statement2 is 22. 1] : (a) (b) (c) (d) . (c) f is differentiable at x = 0 and at x = 1 1 (d) f is differentiable at x = 0 but not at x = 1 Statement2 : 0 < f ( x ) £ .1)sin if x ¹ 1 (2010) 19. Let f : R ® R be a positive increasing function with lim x ®¥ (a) 1 f (3 x ) f (2 x ) = 1. (2010) 2 2 21. 1 2 (b) Statement1 is true. Statement2 is true; Statement2 is x 7 + 14x 5 + 16x 3 + 30x – 560 = 0 have? not a correct explanation for Statement1. for all x Î R. Let f : R ® R be defined by ì k . (2007) (2009) real root of P¢(x) = 0.1 ïî 0 if x = 1 14. q = - 2 2 (a) p = . (2007) 4 3 2 16. Statement2 is true. q = 1 3 2 2 5 1 (d) p = . Statement2 is false. Let f (x) = x|x| and g(x) = sin x. then a possible value of k is (a) 1 (b) 0 (c) –1/2 (d) –1 1 ì ï( x .41 Differential Calculus 3 2 1 2 1 3 (c) p = . e x + 2 e . Let y be an implicit function of x defined by x 2x – 2x x cot y – (b) p = . 2 2 (2008) (a) Statement1 is true. If p and q are positive real numbers such that p + q = 1. if x £ -1 f ( x ) = í î 2 x + 3. 2 2 Statement2 : gof is twice differentiable at x = 0. Given P(x) = x + ax + bx + cx + d such that x = 0 is the only 23. Then lim = f (x) x ®¥ f ( x) (b) 2/3 (c) 3/2 (d) 3 (2010) 13. q = 1= 0. (2007) (a) Statement1 is true. Statement2 is true; Statement2 is a 20. Suppose the cubic x 3 – px + q has three distinct real roots where p > 0 and q > 0. (c) 1 (d) 3 (2008) (d) Statement1 is false. Then y¢(1) equals (2011) 11. the maximum value of (p + q) is Statement1 : gof is differentiable at x = 0 and its 1 1 (a) (b) derivative is continuous at that point. Let f : (–1. then 15. g (x ) = [f (2f (x ) + 2)] 2 . f ( x ) = .x (b) f is neither differentiable at x = 0 nor at x = 1 Statement1 : f (c) = 1/3.2 x . Then which of the following is P(–1) is not minimum but P(1) is the maximum of P true ? P(–1) is the minimum but P(1) is not the maximum of P neither P(–1) is the minimum nor P(1) is the maximum (a) f (x) is differentiable everywhere of P (b) f (x) is not differentiable at x = 0 P(–1) is the minimum and P(1) is the maximum of P (c) f (x) ³ 1 for all x Î R (d) f (x) is not differentiable at x = 1. (c) 2 (d) 2. for some c Î R . Let f : R ® R be a continuous function defined by Then which one of the following is true? 1 (a) f is differentiable at x = 1 but not at x = 0 f (x) = . If P(–1) < P(1).. The function f : R – {0} ® R given by not a correct explanation for Statement1. How many real solutions does the equation correct explanation of Statement1. Let f ( x ) = í x . |x| + 1}. (b) Statement1 is true. Then g¢(0) = (a) 4 (b) – 4 (c) 0 (d) –2 (2010) 12. Let f : R ® R be a function defined by f (x) = min {x + 1. Statement2 is false.2 x x e - 1 (c) Statement1 is false. can be made continuous at x = 0 by defining f (0) as (d) Statement1 is true; Statement2 is true; Statement2 is a (a) 0 (b) 1 correct explanation for Statement1. (d) (2005) (c) ( a . then function in 1 . (c) (d) (2005) 2 2 6p 54p (2004) 3 x 2 + 9 x + 17 is 3 x 2 + 9 x + 7 (c) 1 (d) 17/7. 1) and at that point the tangent to the graph is y = 3x – 5. (2005) fence are of same length x. x ¹ . 3). –1) È (–1. (2007) 2 2 (b) it passes through the origin m n m + n 26 If x ∙ y = (x + y) .b )2 . 2 (c) ç 4 2 ÷ (d) ç 2 4 ÷ (2007) è ø è ø 1 2 . A function y = f (x) has a second order derivative f ¢¢(x) = 6(x – 1). If 2a + 3b + 6c = 0.2 . Let a and b be the distinct roots of ax 2 + bx + c = 0. If f is a realvalued differentiable function satisfying y | f ( x ) . (2006) ) 2 x 2 y + = 1 is a 2 b 2 (a) ab (b) 2ab (c) a/b (d) ab (2005) 39. p . 0) and (3. then f (1) 27.tan x .b ) 2 (a) 0 (b) æp. (2006) 31. (2006) 2 37. ¥) (b) (– ¥. The function f (x) = tan –1 (sin x + cos x) is an increasing 33. 6].b ) 2 2 25. The set of points where f ( x ) = is differentiable.cos( ax 2 + bx + c ) lim is equal to æ pö æ p pö x ®a ( x . If x is real. The normal to the curve x = a(cos q + q sin q ). A triangular park is enclosed on two sides by a fence and on equals the third side by a straight river bank. then at least one root of the equation ax 2 + bx + c = 0 lies in the interval (a) (2. (2004) 3 of ice of uniform thickness that melts at a rate of 50 cm /min. (2006) 38. pö æ . (2004) 40. 2 ÷ è ø è ø a 2 ( a . A value of c for which conclusion of Mean Value Theorem holds for the function f (x) = log e x on the interval [1. then (b) (a) x (a) f (6) < 8 (b) f (6) ³ 8 2 8 (c) f (6) = 5 (d) f (6) < 5. - a . The two sides having (a) 1 (b) 2 (c) 0 (d) – 1. If f (1) = –2 and f ¢(x) ³ 2 3 2 x 3 for [1. y Î R and f (0) = 0. 3) (b) (1. p When the thickness of ice is 5 cm. Let f ( x ) = 1 .42 JEE MAIN CHAPTERWISE EXPLORER 24. is p p 1 cm/min 1 cm/min f ( x ) is continuous in éê 0. x Î éëê0. 3] is 34. Angle between the tangents to the curve y = x 2 – 5x + 6 at the points (2. The maximum area enclosed by the park is 36. Suppose f (x) is differentiable at x = 1 and x 28.p thickness of ice decreases. The function g ( x ) = (a) x = 2 (c) x = 0 x 2 + has a local minimum at 2 x (b) x = –2 (d) x = 1. 1) (d) (1. then f ¢(1) equals 1+ | x | h ® 0 h (a) 4 (b) 3 (c) 6 (d) 5. ùú . then the function (a) 1/4 (b) 41 is (a) (x + 1) 3 (b) (x – 1) 3 2 32. (2005) 1 2 (c) x (d) px 2 .a 2 ( a . (2005) (a) (– ¥. then f is ë 2 û 4 (b) (a) 18p 36p 1 1 5 cm/min 1 cm/min. (2006) 30. the maximum value of ( ) . ùû 2 ú 4 4 x . ¥). 2 ÷ (a) ç 0. then dy/dx is (c) it is at a constant distance from the origin x + y y p (a) (b) xy (2005) (d) it passes through a 2 . 0) È (0. is lim 1 f (1 + h ) = 5. Let f be the differentiable for " x. y = a(sinq – qcosq) at any point q is such that (a) log 3 e (b) log e 3 p 1 (a) it makes angle + q with xaxis (c) 2 log 3 e (d) log e 3. then the rate at which the 41.p . ¥) (d) (0. (a) - (b) (c) 1 (d) –1. x. Area of the greatest rectangle that can be inscribed in the ellipse ( 29. ¥) (c) (– ¥. 0) is (a) p/2 (b) p/3 (c) p/6 (c) p/4. x x (c) xy (d) . 2) (c) (0. p ö .f ( y ) | £ ( x - y )2 .a ) 2 (b) ç . A spherical iron ball 10 cm in radius is coated with a layer (c) (x – 1) (d) (x + 1) 2 . (2006) 35. If its graph passes through the point (2. If lim ç1 + x + 2 ÷ = e . (c) f(x) is continuous everywhere (2003) (d) f(x) is discontinuous everywhere. æ at ö y = a cos t . ì x. If f (x) = x n . (2003) 53.g ¢¢ ( x ) = 0. (2003) 48.. then f ¢(5) is (a) 0 (b) 1 (c) 6 (d) 2. The maximum distance from origin of a point on the curve value of the sum at x equal to (a) 1 (b) –1 (c) –2 (d) 2. æ at ö x = a sint – b sin è ø (2003) b ì -æç 1 + 1 ö÷ | x| x ø ï .. 2] such (a) 1 (b) 2 (c) 1/2 (d) 3.2 f ( x ) 52. x = 0 (a) continuous for all x. + n is n®¥ n ®¥ n n (a) zero (b) 1/4 (c) 1/5 (d) 1/30. (2002) 1 - cos 2 x is 2 x (a) 1 (c) 0 55. y and f (5) = 2. If 2a + 3b + 6c = 0 (a. Let f (2) = 4 and f ¢(2) = 4 then lim equals x - 2 x ® 2 f (a ) g ( x ) . b..tan( x / 2) ][1 . If f (1) = 1. then f (x) is 47. (2003) 54. (2002) . both a..g (a ) f ( x ) + g (a ) lim = 4. b > 0 is b (a) a – b (c) 2 a + b (b) a + b 2 (d) 57. except x = 0 50. 1) (a) a Î R.b cos è ø .. If f ( x ) = í xe è ïî 0 . f ( x ) . (2002) 46. where a > 0. (2002) attains its maximum and minumum at p and q respectively 2 such that p = q. f ¢(1) f ¢¢(1) f ¢¢¢ (1) (-1)n f n (1) (2002) is f (1) + + . b Î R(d) a = 1 and b = 2. 5] (2004) (d) none of these (2002) 43.f ( a ) .43 Differential Calculus 2 x 51. but not differentiable at x = 0 (b) neither differentiable not continuous at x = 0 (c) discontinuous everywhere (d) continuous as well as differentiable for all x. then a equals 58. g n (a) exist and are not equal for some n. lim æ x + 5 x + 3 ö ç ÷ 2 x ®¥ è x + x + 3 ø 1 .x ) (a) f(x) is continuous at every x. if x is rational and f ( x ) = í (2003) î . (2003) 45. is 3 3 x ®p / 2 [1 + tan( x / 2) ] [ p - 2 x ] (c) e (d) 1.. The value of 4 4 4 3 3 3 lim 1 + 2 + 3 5+ . If f (x + y) = f (x) × f (y) " x. are ax 2 + bx + c = 0 has x ®¥ è x ø (a) At least one in (0.log(3 . c Î R) then the quadratic equation æ a b ö 2 42. Further if x f (2) .. if x is rational. If the function f (x) = 2x 3 – 9ax 2 + 12a 2 x + 1.. lim x ® 0 (b) –1 (d) does not exist. (2002) (a) 0 (b) 1/32 (c) ¥ (d) 1/8. Let f (a) = g(a) = k and their n th derivatives f n (a). b = 2 (b) a = 1. f (2) = 3g (2) = 9 (2003) then f (x)– g(x) at x = 3/2 is (a) 0 (b) 2 49. The real number x when added to its inverse gives the minimum 56. f (x) and g (x) are two differentiable function on [0. 5] as (a) 2 n – 1 (b) 0 (c) 1 (d) 2 n . except x = 0 (a) –1/3 (b) 2/3 (c) –2/3 (d) 0. f ¢(0) = 3. g ( x ) - f ( x ) (a) 2 (b) – 2 x ® a (c) – 4 (d) 3. then the values of a and b. f ¢ (1) = 2. x ¹ 0 . the value of k is x x ® 0 (b) f(x) is discontinuous at every x.1 x - 1 is (b) 4 (d) 1/2. f ¢(1) = 2 g ¢(1) = 4 . b Î R (b) At least one root in [2. then Lt x ®1 (a) 2 (c) 1 (2002) a 2 - b 2 . + 1! 2! 3! n ! 59. Then log(3 + x) . + n - lim 1 + 2 + 3 5 + . If lim = k . (2002) then the value of k is 1 (a) 2 (b) 1 2 x (c) 0 (d) 4. t h a t f ¢¢ ( x ) . 3] (c) a Î R. then the value of (c) 10 (d) 5.. (c) At least one root in [4.x.sin x ] [ (a) e 4 (b) e 2 lim 44. f is defined in [–5. (b) (b) (a) (c) (a) (a) (b) (b) (b) (a) 3. 45. 16. 37. 13. 44. 56. 54. 34. 32. ([x] denotes greatest integer less than [ x ] x ® 0 or equal to x) (a) has value – 1 (b) has value 0 (c) has value 1 (d) does not exist (2002) 60. 43. 42. 29. 25. 26. 35. 19. 15. (d) (c) (b) (c) (c) (b) (c) (c) (a) (a) 4. 20. 58. 9. 27. 52. Answer Key 1. If y = ( x + 1 + x 2 ) n . 48. 14. 18. 47. 8. lim 2 61. 59. 60. 50. 33. 57. 11. 39. 40. 31. (a) (a) (a) (b) (c) (a). 28. 41. (c) (b) (a) (c) (d) 5. 61. 22. (a) (a) (b) (d) (a) (b) (b) (b) (c) (d) (2002) . then (1 + x ) (a) n 2 y (c) –y d 2 y dy + x is dx dx 2 (b) –n 2 y (d) 2x 2 y. 51. 12. 17. 49. 23. 21. 24. (c) (b) (d) (a) (a) (c) (a) (a) (d) (b) 6. 36. 46.44 JEE MAIN CHAPTERWISE EXPLORER log x n - [ x ] . 10. 30. (c) (a) (d) (b) (c) (b) (d) (d) (b) (a) (a) 2. 55. 7. n Î N . 38. 53. (c) : Let x = 2 + h lim h®0 x ® n+ è dx ø 1 . (a) : P(0) = 2000 = l. x = 2 \ Statement 1 : f has local maxima at x = –1. True íç ÷ ý 8. \ l = 2000 P(25) =100 × 25 – 8 × 25 3/2 + 2000 = 3500.1 ö p 2 ÷ø 5. LHL = –1. radius decreases at a rate of m/min 9 Þ 972 p = 2 ì -1 ü d x d æ dx ö d ï æ dy ö ï Statement1 : f ¢(4) = 0.45 Differential Calculus (1 . x = 0 ï 2 x+x . x>0 ïî x 3/2 . íç ÷ ý × = =-ç ÷ ×ç ÷ ï dy dx ïî è dx ø þ æ 2 x . dP = (100 - 12 x ) dx 2 3 / 2 ×x + l 3 and 1 1 + 4 b + a = 0 Þ a = 2 2 1 x2 + 2b = - 1 x2 - æ 1 1 ö 1 = -ç + < 0 è x 2 2 ÷ø 2 1 2 + l dy 1 = sec(tan -1 x ) × tan(tan -1 x ) × dx 1 + x2 dy 1 1 = 2 × 1 × = dx x = 1 2 2 4.7.72 p m 3 / min. lim f ( x ) = 0 x ® n- \ f(n) = 0 Þ f(x) is continuous for every real x. x>5 î 1 4 \ Statement 2 : a = . f ( x ) = [ x ] cos çè æ pö = [ x ] cos ç px .2 b + a = 0 Þ b = - 4 æ 2sin 2 x ö æ x ö = lim ç ÷ (3 + cos x ) ÷×ç x ® 0 è x2 ø è tan4 x ø P = 100 x .x ï . ì ï sin( p + 1)x + sin x . b = - 7. (a) : f(x) = |x – 2| + |x – 5| ì7 .8 x f ¢(– 1) = 0 and f¢(2) = 0 f ¢¢( x) = - 1 = 2 ´ ´ 4 = 2 4 3 / 2 1 + 2bx + a x 1 Þ . (a) : f ( x ) = í q. x ¹ 0 has extreme values at x = – 1. dx 2 è dx ø -3 æ dy ö d 2 y =-ç ÷ è dx ø dx 2 9. v = v 0 + 49∙ = 4500p – 49 × 72p dt v= = 4500p – 3528p = 972p 4 3 pr Þ r 3 = 243 × 3 = 3 6 Þ r = 9 3 dr dr 18 2 \ .72 p = 4 p ´ 81 ´ Þ == - dt dt 81 9 2 Thus. 2 £ x £ 5 ï 2 x .12 × Þ f ¢ ( x ) = dv = .1 . differentiable in (2. we have. 6. (c) : xlim ® 0 1 . v0 = 4500 p dt 4 3 dv 4 dr pr \ = p ´ 3 r 2 ´ 3 dt 3 dt dv After 49 min. lim f ( x) = 0.2 x . Thus limit doesn’t exist.cos 2 h |sin h | = lim h h h ® 0 RHL = 1. 5]. (d) : y = sec (tan –1 x) Þ [Given] for all x Î R – {0} Þ f has a local maximum at x = – 1. True -1 -2 -1 d ïì æ dy ö ïü dx æ dy ö d 2 y æ dy ö But Statement 2 is not a correct explanation for statement 1.÷ = [x] sin px è 2 ø Let n be an integer. x < 0 x ï ï 10.cos2 x )(3 + cos x ) x(tan4 x) 1. (a) : f(x) = ln|x| + bx 2 + ax. (c) : f : R ® R. 5) and f(2) = f(5). (b) : dx P = 100 x . x = 2 dP = 100 - 12 x 2.cos2 x = lim (3 + cos x ) x ® 0 2 æ tan4 x ö x ×ç ÷ è x ø Integrating. x = 2. 3. (b) : 2 = dy çè dy ÷ø = dy ï dy îï è dx ø þ Statement2 : f is continuous in [2. x < 2 ï f ( x ) = í 3. we have D of 4x 2 + 3ax + 2b is less than zero i. 1 – 2 cot y – 1 = 0 15. of P(x) = P(1) But minimum of P(x) doesn’t occur at x = –1. e + 2e ³ 2 2 2 1 1 £ Then x e + 2 e . = lim h ® 0 11. x = 0 ï 2 x cos x 2 .sin x . x. 12.cosec 2 y ) + cot y × x x (1 + ln x )] = 0 dx Let the composite function gof (x) be denoted by H(x). x < 0 Again H ¢¢ ( x ) = í 2 2 2 . \ . x > 0 î H¢(x) is continuous at x = 0 for H¢(0) = LH¢(0) = RH¢(0) ì-2 cos x 2 + 4 x 2 sin x 2 . (a) : Using A.M..e. P(–1) is f (c) = 1/3. b 2 – 4ac < 0 then ax 2 + bx + c > 0 " x Î R) So P¢(x) < 0 if x Î [–1. x ³ 0 î 2 cos x . we have f (x) < f (2x) < f (3x) Dividing by f (x) leads to 1 < As lim x ®¥ f (2 x ) f (3 x ) < f ( x) f ( x) f (3 x ) = 1. (b) : gof (x) = g(f (x)) = sin(x|x|) Þ cot y = 0 \ y = p/2 2 Differentiating (i) w.e.. 17. Let x = 0 g¢(0) = 2f (2f (0) + 2) ∙ f ¢(2f (0) + 2) ∙ 2f ¢(0) = 2f (0) ∙ f ¢(0) ∙ 2f ¢(0) = (–2)(1)(2) = – 4. x ³ 0 Again. we have by Squeez theorem f ( x) f (2 x ) or Sandwich theorem.h) .x x -x ³ e x × 2 e .× 2 × h = 1 × 0 = 0 - h h h ® 0 RH ¢ (0) = lim g¢(x) = 2f (2f (x) + 2) ∙ f ¢(2f (x) + 2) ∙ 2f ¢(x) H (0 .x .e.G. (d) : lim f ( x) = 1 x ®1 + h ® 0 + As f has a local miminum at x = –1 f (–1 + ) ³ f (–1) ³ f (–1 – ) Þ 1 ³ k + 2 k £ –1 Thus k = –1 is a possible value. Thus such that Thus H(x) is NOT twice differentiable at x = 0 16. lim = 1. x < 0 Then H ( x ) = ìí 2 î sin x . 14. p + 2 = q = 1/2 \ p = –3/2.. Thus. decreasing and P¢(x) > 0 if x Î (0. inequality.sin h 2 sin h 2 = lim. 0) i.H (0) - h 1 2 2 Observe that f (0) = 1/3..2[ x x ( . not the minimum. (a) : P(x) = x 4 + ax 3 + bx 2 + cx + d P¢(x) = 4x 3 + 3ax 2 + 2bx + c P¢(0) = 0 Þ c = 0 Also P¢(x) = x(4x 2 + 3ax + 2b) As P¢(x) = 0 has no real roots except x = 0. x < 0 =í 2 dy î sin x . we can say that as f is continuous.H (0) h æ sin h 2 ö sin h 2 . Here we are able to At x = 1 we have identify the point as well. (3a) 2 – 4 ∙ 4 ∙ 2b < 0 then 4x 2 + 3ax + 2b > 0 " x Î R (If a > 0.. (d) : x 2x – 2x x cot y – 1 = 0 .x 1 x x e + 2 e -x is always positive.e.0 = lim ç 2 ÷ h ø h h ® 0 + h ® 0 + è h = lim = 1 ∙ 0 = 0 Thus H(x) is differentiable at x = 0 ì-2 x cos x 2 . lim f ( x ) = - x ® 0 sin( p + 1) + sin x = p + 2 x LH ¢ (0) = lim h ® 0 - Now. x ³ 0 2 x 2 x (1 + ln x ) . p/2) we have .t. increasing Max. e x + 2 e .46 JEE MAIN CHAPTERWISE EXPLORER lim f ( x ) = + x ® 0 1 2 sin x 2 . x ®¥ f ( x ) 13. (a) : As f is a positive increasing function. i. Using extremevalue theorem. (b) : g(x) = {f (2f (x) + 2} 2 We have on differentiation with respect to x..M. x < 0 ï Also H ¢ ( x) = í 0 .. q = 1/2. At P(1.(i) f will attain a value 1/3 at some point..x 2 2 As 0 < 1 e + 2 e .r. we have ì. 1] i. we have £ H (0 + h) .4 x sin x LH¢¢(0) = –2 and RH¢¢(0) = 2 As f (–1) = k + 2 Þ k + 2 £ 1. M.ny ö mx + nx .f ( a ) f (3) .e.2 æ 0 ö ç form ÷ x ®0 ( e ø . f (x) is an strictly increasing function. + 0] = 0 P = 0 2 ù é1 22.47 Differential Calculus 2(1 + ln1) .my ..M ³ G. (b) : Denote x 3 – px + q by f (x) i.1 . Hence it is not differentiable at x = 0.1) sin .my ÷= dx çè y(x + y) x ( x + y ) ø Þ dy æ nx . (c) : Let 2x 24. x Î R Hence f (x) is differentiable for all x Î R. i.0 1 (1 + h .2 x æ 0 ö ç form ÷ x ®0 x ( e2 x - 1) è 0 ø = lim P 18.1) = lim = lim sin h h h®0 h ® 0 As the limit dosen’t exist. x we get i. (c) : Let p = cos q. f ¢( c ) = f (b ) . f (x) = x 3 – px + q Now for expression.f (1) f ¢ (1) = lim . 4 ÷ .2 .e.r. (d) : f ¢( x ) = f ¢( x ) = 1 .(cos x .f (0) Again f ¢ (0) = lim . By using A.cos x > sin x 2 4 æ p pö Hence y = f (x) is increasing in ç . 3 f (1 + h ) .2[1(-1) Þ 2+ 2 \ ( ) dy dx ( ) dy dx ( ) dy dx 1 p 2 + q 2 ³ pq Þ pq £ 2 2 (p + q) 2 = p 2 + q 2 + 2pq Þ ( p + q ) £ 2.t.sin x 2 + sin 2 x If f ¢(x) > 0 then f (x) is increasing function For . 21. 19. è ø 25. if the limit exists h h ® 0 1 ( h . (a) : f (x) = min {x + 1. f ¢(x) = 0. 3 3 f ¢¢(x) = 6x By using L’ Hospital rule 2 e 2 x . c 2 2log 3 e 26.log e 1 1 = log e 3 2 2 Þ 1 1 1 = log e 3 = \ c = 2 log 3 e.e. |x| + 1} f (0) = lim p ö ÷ > 0 3 ø p and minima at 3 p . (b) : f (0) = lim ê x .my ö y y dy y = = Þ = . x ®0 4 e + 4 xe2 x 23.1 û P = - 1 e2 x . q = sin q Þ dy æ n m + n ö m + n m = dx çè y x + y ÷ø x + y x Þ dy æ nx + ny . dx çè nx . f (x) = x 7 + 14x 5 + 16x 3 + 30x – 560 2x Þ f (x) = x + 1.mx .1) sin . 3x 2 – p = 0 p p . if the limit exists. m n dy m + n æ dy ö + = 1 + x y dx x + y çè dx ÷ø so it can have at the most one solution. h h ® 0 f (1 + h ) - f (1) \ lim h h ® 0 1 (1 + h . (c) : By LMVT.p < x < p .sin1 f ( h ) f (0) h .1) + 2 xe2 x è 0 f (0) = lim x = - Again use L’ Hospital rule æ p ö æ f ¢¢ ç ÷ < 0 Þ f ¢¢ ç è è 3 ø Thus maxima at - 4 e 2 x = 1. \it is not diffentiable at x = 1 f ( h ) . It can be shown that it has exactly one solution. (a) : x m × y n = (x + y) m + n Taking log both sides we get \ f ¢(x) = 7x 6 + 70x 4 + 48x 2 + 30 m log x + n log y = (m + n) log(x + y) Þ f ¢(x) > 0 " x Î R Differentiating w.sin x ) 1 + (sin x + cos x ) 2 cos x .1 \ lim = lim h h h® 0 h ® 0 But this limit dosen’t exist.f (1) = b-a 3 - 1 f ¢( c ) = log e 3 . (b) : By definition 20.2 x ú x ® 0 ë e .my ÷ø x x dx x 0 £ q £ p/2 p + q = cos q + sin q Þ maximum value of (p + q) = 2 Second method . b ) ö ù ÷ú ê sin ç 2 2 2 ø ú ´ a ( x . 3 x 2 + 9 x + 7 px 2 + qx + r [find the solution of the inequality A y 2 + B y + K ³ 0] where A = q 2 – 4pr = –3. x < 0 ï1 .b ) 2 .b ) ù a ê ú 4 ú ëê û 2 30.a )( x .5 = 1 è dx ø y = 0 since m 1 m 2 = –1 Þ tangents are at right angle i. (c) : [ y . p 2 x 2 + 2 x 1 2 g ¢( x ) = - 2 \ 2 x for maxima and minima g¢(x) = 0 Þ x = ± 2 4 Again g ¢¢( x ) = 3 > 0 for x = 2 x < 0 for x = –2 \ x = 2 is point of minima 31.b) ù 2sin 2 ê úû a 2 ( x .a ) 2 é a ( x .a )( x . (b) : As a is root of ax 2 + bx + c = 0 \ aa 2 + ba + c = 0. Now equation of normal to the curve 34. B = 4ar + 4PC – 2bq = 126 é æ a ( x . (b) : For the range of the expression = y = ax 2 + bx + c æ 2 ö 2sin 2 ç ax + bx + c ÷ 2 è ø = lim x ® a ( x . x ³ 0 î 1 + x A T C ì 1 .a (sin q .b ) = lim ê è a ( x . (a). solve –3y 2 + 126 + y – 123 ³ 0 Þ 3y 2 – 126y + 123 £ 0 Þ y 2 – 42y + 41 £ 0 Þ (y – 1)(y – 42) £ 0 Þ 1 £ y £ 42 Þ maximum value of y is 42 4 3 32.b ) ú 2 x ® a ê ú 2 ëê û =1´ a 2 ( a .1 £ sin 2 a £ 1 2 2 1 2 \ Maximum are of DABC = x 2 x 28. 0). .a ) 2 x ® a 29. (c) : AT = x sin a BT = x cos a Area of triangle 1 ABC = base × height 2 1 = (2 BT )( AT ) 2 a 1 2 = (2 x cos a sin a ) B 2 1 2 1 2 = x sin 2 a £ x as .b ) 2 2 ë = lim ´ 2 2 4 x ®a 2 é ( x .a (cos q + a sin q)) sin q Þ x cos q + y sin q = a (1) Now distance from (0.. 18p 33. (c) : Given f ( x ) = 1+ | x | ì x . x ³ 0 ïî (1 + x ) 2 f ¢ (x ) is finite quantity " x Î R \ f ¢ (x ) is differentiable " x Î ( -¥ . x < 0 ï (1 .e. (3.q cos q ) ] =- cos q ( x . Now ( lim 1 . given points (2. (a) : Let g ( x ) = 3 x 2 + 9 x + 17 ) . (a) : Given equation y = x 2 – 5x + 6.a )( x . 2 dy dy d q = × = tan q = slope of tangent dx d q dx \ Slope of normal to the curve = – cotq = tan (90 + q). 0) dy = 2 x - 5 dx æ dy ö = 4 .cos( ax 2 + bx + c ) ( x .a ) distance (d) = 1 \ distance is constant = a.x ) 2 ï Þ f ¢ ( x ) = í ï 1 . è dt øat y = 5 4 p (15) 2 dt = 1 cm/min.a ) ( x .5 = -1 say m 1 = ç dx ÷ è ø x = 2 \ y = 0 and m2 = æç dy ö÷ at x = 3 = 6 .48 JEE MAIN CHAPTERWISE EXPLORER 27. ¥ ) K = b 2 – 4ac = –123 i. (a) : v = p( y + 10) where y is thickness of ice 3 10 10 + y Þ dy dv = 4 p (10 + y ) 2 dt dt æ dy ö 50 dv = ç ÷ as = 50 cm 3 / min.e. 0) to x cosq + y sinq = a is (0 + 0 .x Þ f ( x ) = í ï x . 2y (1 . (b) : Any point on the ellipse 2 x 2 y + 2 = 1 2 a b is (acos q .p 41. 1) 1 = (2 – 1) 3 + c 1 Þ c 1 = 0 f (x) = (x – 1) 3 Þ Lt Lt x ® p/4 2x æ d 2 A ö dA p = 0 Þ q = and ç 2 ÷ = -8 ab sin 2 q d q 4 è d q ø q = p / 4 2 as d A < 0. (b) : Let if possible f ¢(x) = 2 for Þ f (x) = 2x + c (Integrating both side w. \ sides of rectangle are 2 2 Required area = 2ab.tan x putting 4x – p = t x ® p/4 4 x . \ Area is maximum for q = p/4.f ( y ) £ ( x - y ) 2 lim x® y = f ( x ) .f ( x )] ax 3 bx 2 Þ Lt = 4 Þ + + cx 39.H. 1) 40.x ÷ ú ø û ë è 4 æp ö tan ç . (b) : Given f ² (x) = 6(x – 1) f (x) = (by integration) Q f (0) = 0 Þ f ( x ) (" x Î R ) = 0 \ f (1) = 0. bsinq ) so the area of rectangle inscribed in the ellipse is given by A = (2acosq) (2bsinq) \ A = 2 ab sin 2 q Þ dA = 4 ab cos 2 q d q Now for maximum area 6( x - 1) 2 + c 2 é Q f ( x ) = y = 3 x + 5 ù 3 = 3 + c ê f ¢ ( x ) = 3 " x Î R ú c = 0 ë û 2 f ¢ (x) = 3(x – 1) f (x) = (x – 1) 3 + c 1 as curve passes through (2.x ÷ è 4 ø 2 x a b ö æ 42.÷ a b è 4 2 ø f (0) = 0 and f (1) = + + c 3 2 Lt f (a) = 4 x ® a . æ p x ö ax 3 bx 2 tan ç . f (x) = 0 ( f ¢ ( x ) < 0.sin x ) + + cx called Here integration of ax 2 + bx + c is è 4 2 ø 3 2 44. not possible ) Þ f ( x ) = k Þ f (x) = 0 ax 3 bx 2 + + cx = 0 3 2 \ at least one root of the equation ax 2 + bx + c = 0 lies in (0.cos ç .f ( y ) £ lim x .2 x ö 2 4. (c) : Given f ( x ) . d q 2 2 a 2 b .2 x ) given condition then at least one root of the equation ax 2 + è 4 ø bx + c = 0 must lies in that interval. f ¢ (x) = Þ Þ so Þ Þ \ 37. (d) : As f (x) is differenatiable at x = 1 f (1 + h ) 5 = lim assumes 0/0 form h h ® 0 f ¢ (1) \ f ¢(1) = 5. –2 = 2 + c Þ c = – 4 \ f (x) = 2x – 4 \ f (6) = 2 × 6 – 4 = 8 \ f (6) ³ 8.f (a) . Now use the intervals in f (x) if f (x) satisfies the x ®p/2 æ p .t.x ÷ ax 3 bx 2 è 4 2ø è 2 ø Now f (x) = + + cx Lt 3 2 x ®p/2 æ p x ö ( p 2 x ) 2 4.÷ (1 . ç .49 Differential Calculus 2a + 3b + 6 c 6 = 0 given 2a + 3b + 6c = 0 \ x = 0 and x = 1 are roots of 35. (c) : Let f (x) = x ® a g ( x ) - f ( x ) 3 2 Note : In such type of problems we always consider f (x) as k = 4 the integration of L.x ÷ ´ (1 + tan x ) 4 ø è Lt = –1/2 x ® p/4 æp ö 4 ç . x) \ f (1) = 2 + c.r. (a) : \ 38.tan x ) ´ (1 + tan x ) é æp öù (1 + tan x) ê -4 ç . (b) : Lt it by f (x). 36. æ p x ö æp ö tan ç . 5 = lim h ®0 1 1 . (d) : Lt x ® a f (a) g ( x ) .g (a) f ( x ) + g (a ) = 4 g ( x ) - f ( x ) f (a)[ g ( x) .y x - y x ® y Þ f ¢( x ) £ 0 .S of the given equation without constant. ç ÷ ( p .÷ 1 . (b) : e 2 = ç 1 + + 2 ÷ x x ø è e 2 = e (1¥ form) é a b ù Lt ê1+ + -1ú (2 x ) x x 2 û x ®¥ ë e 2 = e 2a Þ 2a = 2 \ a = 1 and b Î R 43. log(3 - x ) = k x xö x ö æ æ log ç 1 + ÷ .. (b) : = \ f ¢ (x) = 1 – 1/x 2 and f ² (x) = 2 x 3 now f ¢ (x) = 0 Þ x = ± 1 \ f ²(1) > 0 Þ x = 1 is point of minima. 6 As f (0) = f (1) = 0 and f (x) is continuous and differentiable . + n 3 n5 2 n( n + 1)(2 n + 1)(3n 2 + 3n . ì -æ 1 + çè | x | ï 47.÷ 4 2 æ p . + C n x n .2 f (2) x - 2 ( x .. + 1! 2! 3! n ! n C – n C + n C . ç .(i) Putting x = –1 in both side of (i) we get 0 = C 0 – C 1 + C 2 – C 3 + .(A) 2a + 3b + 6 c = = 0 given.1) n = f ² (x) = n(n – 1)x n – 2 so = C 2 2! 2! f n (x) = n(n – 1) .1 \ 2 = a 2 = 2a Þ a 2 – 2a = 0 a(a – 2) = 0 Þ a = 0.2) f (2) . x = 0 = 0 é 1 1 ù ê .H.÷ 3ø 3 ø è è k = Lt x ® 0 x xö x ö æ æ log ç 1 + ÷ log ç 1 .2 x ö ç ÷ è 4 2 ø è 4 ø 14 + 24 + 34 + .50 JEE MAIN CHAPTERWISE EXPLORER æp xö æ p x ö tan ç .h )e è h h ø .log ç 1 .+ (–1) n C 0 1 2 n Now (1 + x) n = C 0 + C 1 x + C 2 x 2 + ....1) ù = ú 30 úû 50.2[ f ( x ) .D at x = 0 Lt (0 + æ 1 1 ö -ç + ÷ h) e è h h ø x ® 0 + æ 1 1 ö -ç .x + x ú û ë and Lt .. 2a and f ² (a) < 0 and f ² (2a) > 0 Now p = a and q = 2a and p 2 = q f n (1)(- 1) n = (–1) n n C n n! f ¢(1) f ¢¢(1) f ¢¢¢(1) (- 1) n f n (1) + - = + . x ® 0 = 0 As LHL = RHL \ f (x) is continuous at x = 0 x ® 0 x ® 0 Again RHD at x = 0 is - 0 = 0 h also we have L. (b) : For maximum and minima f ¢ (x) = 0 Þ 6x 2 – 18ax + 12a 2 = 0 and f ² (x) = 12x – 18a f ¢ (x) = 0 Þ x = a. (a) : Given f (x) = í x e ï0 î Lt x ® 0+ f ( x ) = Lt xe . xf (2) ... \ By Rolle’s theorem f ¢ (x) = 0 Þ ax 2 + bx + c = 0 has at least one root in the interval (0. (c) : Lt n ® ¥ n ® ¥ n5 Þ Þ 13 + 23 + .÷ 4 2 1 2 1 è ø è 4 2 ø Lt = ´ = 2 p x x ®p/2 æ ö 4 16 32 4.. (a) : Let us consider f (x) = + cx 3 2 a b \ f (0) = 0 and f (1) = + + c 3 2 ..(B) also in [0.÷ 3ø 3 ø è è + Lt k = Lt x x x®0 x ® 0 ´3 . 1]. (b) : f (x) = x n \ f (1) = 1 = n C 0 \ f ¢ (x) = nx n – 1 so –f ¢ (1) = – n = – n C 1 f ¢¢(1) n(n .2/ x x ® 0 + 1 ö x ÷ø .0 is = 1 - h so L.H.. (a) : f (x) = x + 1/x 49... \ é .. 1). (d) : We have lim ç 2 ÷ x ®¥ è x + x + 3 ø 1 / x æ 1 + 5 + 3 ö ç x x 2 ÷ = lim ç 1 3 ÷ x ®¥ ç 1 + + 2 ÷ è x x ø = 10 = 1 ..D at x = 0 \ f (x) is non differentiable at x = 0 48..f ( x ) = Lt ..0 êUsing 1 4 + 2 4 + . + n 4 - Lt 45.H.÷ 2 sin 2 ç .Lt n (n + 1) n ® ¥ 4 ´ n 5 n ® ¥ 30 × n 5 = Lt n( n + 1)(2 n + 1)(3n 2 + 3n - 1) 30 3 n( n + 1)(6n + 9 n 2 + n .D ¹ R.f (2)] Lt x ® 2 x - 2 52...÷ (0 ....´ 3 3 3 1 1 2 k = + = 3 3 3 ax 3 bx 2 + 51. a = 2 f (1) – Lt x ® 0 log(3 + x ) .e x®0 x ¹ 0 ..2 f ( x ) + 2 f (2) .1) .. + n 4 ë 6 - 0 30 1 = 5 46. (c) : Lt x ® 2 lim [f (2) – 2 f ¢(x)] = x® 2 = 4 – 2 × 4 = – 4 1/ x æ x 2 + 5 x + 3 ö 53. ÷ b bø è ø ø è = a 2 + b 2 . x at ì ï a sin t .b cos = y b î x 2 + y 2 = AB = at ö æ at ö ö æ + cos2 ç ÷ ÷ . (d) : As f ¢¢ (x) – g ¢¢ (x) = 0 Þ f ¢ (x) – g¢(x) = k f ¢ (1) – g¢(1) = k \ k = 2 So f ¢ (x) – g¢(x) = 2 Þ f (x) – g(x) = 2x + k 1 f (2) – g(2) = 4 + k 1 k 1 = 2 So f (x) – g(x) = 2x + 2 [ f ( x) - g ( x)] 3 = 2 ´ 3 + 2 = 5 \ x = 2 2 59. B(x. (a) : Lt x ®0 56. (c) : Given f (x + y) = f (x) f (y) \ f (0 + 0) = (f (0)) 2 Þ f (0) = 0 or f (0) = 1 but f (0) ¹ 0 f ( x ) f (h ) - f ( x ) f ( x + h ) - f ( x ) Now f ¢(x) = Lt = h Lt ® 0 h h ® 0 h f (h ) .1 é x ù ê1 + ú êë 1 + x 2 úû n 1 2 y 1 = n éê x + 1 + x ùú . (a) : Lt x ® 1 f ( x ) .1 x - 1 1 ´ 2 ´ 1 ´ 2 = 2 2 58. (a) : Let 2 sin 2 x x 2 = Lt sin x = 1.2 ab cos ç t .1 h f ¢(5) = f (5) × 3 = 2 × 3 = 6 Now f ¢(x) = f (x) Lt h ® 0 \ 55. ë û 1 + x 2 ny dy ö æ y 1 = ç y 1 = ÷ dx ø 1 + x 2 è Þ y 1 2 (1 + x 2 ) = n 2 y 2 (0/0 form) Þ y 1 2 (2x) + (1 + x 2 ) (2y 1 y 2 ) = 2yy 1 n 2 Þ y 2 (1 + x 2 ) + xy 1 = n 2 y .2 ab cos a (since |cos a| £ 1) £ a 2 + b 2 - 2 ab = a – b.0). 57. (d) : n log x log x n - [ x ] - 1 = x Lt ® 0 [ x ] x ® 0 [ x ] Lt log x does not exist x ® 0 [ x ] which does not exist as Lt 2 61.51 Differential Calculus 54. y) = í at ï a cos t . (a) : y 1 = n éê x + 1 + x ùú ë û n .1 f ¢(x) = f (x) Lt h ® 0 h f (h ) - 1 \ f ¢(0) = f (0) Lt h ® 0 h f (h ) - 1 3 = Lt (³ f (0) = 1) h ® 0 h f (h ) .b sin b = x A(0. (b) = x ® 0 æ æ at ö a 2 (sin 2 t + cos2 t ) + b 2 ç sin 2 ç ÷ è b ø è \ 2 x × f ¢ (x) x ® 1 2 f ( x) 1 Lt 60. x axis and lying in the first quadrant is (a) 36 (b) 18 (c) 27 4 9.5p(t ) - 450 .ò x3 y( x 3 ) dxùû + C 3ë 7.k (T - t).t ) 2 2 I 2 (c) T - k (d) – 2 (2012) (b) e –kT (a) I - 10.3 ò x3 y( x3 ) dx + C 3 (a) g(x) – g(p) (b) 1 3 x y( x3 ) . StatementII is true. x Î R . (b) Statement1 is true. The area (in sq. dt then the time at which the population becomes zero is 3. (c) StatementI is false. StatementII is false.ò x y( x ) dxùû + C 3 2. then y(ln 2) is equal to dx (a) 13 (c) 7 (b) – 2 (d) 1 11. The value V (t ) depreciates at a rate given by differential equation k (T . If p (0) = 850. 6. a (b) ln 18 dV ( t ) = . y = ò | t | dt .2 cos x | + k . b 1 ln18 2 (c) 2 ln 18 (d) 9 (2013) 4. units) bounded by the curves y = x . StatementII is a correct explanation for StatementI. The value of (b) 2 (d) ln 9 the equipment is StatementII : ò f ( x )dx = ò f ( a + b . which are parallel to the line y = 2x.ò x2 y( x 3 ) dx + C 3 (c) g ( p ) (c) 1 é 3 x y( x3 ) . 2y – x + 3 = 0. Then the scrap value V(T) of is equal to p/6.x )dx. then g(x + p) equals 0 (a) 1 3 x y( x3 ) . (d) 10 2 3 (2012) The population p(t) at time t of a certain mouse species satisfies the differential equation (2013) dp(t ) = 0.52 JEE MAIN CHAPTERWISE EXPLORER CHAPTER INTEGRAL CALCULUS 10 1. then tan x . StatementII is true. (d) StatementI is true StatementII is true. StatementII is not a correct explanation for StatementI. (a) p (a) log 2 2 ò 0 8 log (1 + x ) 1 + x 2 dx is (b) log2 5 (2011) . If g( x ) = ò cos 4t dt . If the integral ò 5 tan x dx = x + a ln|sin x . (2013) 5. are 0 (b) ± 3 (d) ± 1 20 2 3 (b) 10 2 (c) 20 2 x equal to (a) ± 2 (c) ± 4 (b) g(x)∙g(p) g( x ) 1 3 3 2 3 (d) éëx y( x ) .2 a is equal to (a) 1 (c) – 1 (2012) Let I be the purchase value of an equipment and V(t) be the value after it has been used for t years. If (d) I - kT 2 2 (2011) dy = y + 3 > 0 and y(0) = 2. (d) g(x) + g(p) (2012) 2 The area bounded between the parabolas x = y and 4 x 2 = 9y and the straight line y = 2 is (2013) (a) The intercepts on x axis made by tangents to the curve. x If ò f ( x )dx = y( x ) then ò x 5 f ( x 3 ) dx is equal to 8. StatementI : The value of the integral p/3 ò 1 + p / 6 dx tan x b a (a) StatementI is true. where k > 0 is a constant and T is the total dt life in years of the equipment. where [x] denotes the greatest (2. (2006) 0 6 28. (2007) x log t 1 dt . local minimum at p and 2p. (2011) 1 sin x cos x dx and J = ò dx .. (2006) . ò [( x + p)3 + cos 2 ( x + 3p)] dx is equal to . + f (a)} a f ([a]) – {f (1) + f (2) + .÷ + C è 2 12 ø (d) Then ò p( x ) dx equals 1 log tan æ x .ø + c (d) x + log sin è x . the tangent to the parabola at the point a 25.... The value of the integral. The area of the region enclosed by the curves y = x .log cos è x . y = 1/x and the positive xaxis is dx equals 22.. x = e. + f ([a])} [a] f (a) – {f (1) + f (2) + .. ç 2 12 ÷ 2 è ø (2007) x 0 (a) 41 (b) 21 (2008) (c) 41 dt = p is 2 t t . + f (a)}.ø + c 4 4 18. ò [cot x]dx. (a) (b) (c) (d) (d) p log 2 8 (2011) 1 x 5 p ö ÷ . (c) log tan æç + ö÷ + C 2 è 2 12 ø 1 æ x p ö (b) log tan ç . The solution for x of the equation ò 15. 24. The area of the region bounded by the parabola (2007) (y – 2) 2 = x – 1. Let I = ò Then which one of the following is true? 2 2 (a) I > and J < 2 (b) I > and J > 2 3 3 2 2 (d) I < and J > 2 (c) I < and J < 2 3 3 21. 16. where [. Let F ( x ) = f ( x ) + f çæ ÷ö . ò x f (sin x ) dx is equal to 0 pö pö æ æ (c) x .3 p / 2 (a) p 4 32 (b) p 4 p + 32 2 (c) p 2 (d) p - 1 . 2 (2006) p 27. is 0 equal to (a) 1 (c) – p /2 (b) –1 (d) p/2 (2009) sin x dx 2 ò p sin æ x . p (0) = 1 and p (1) = 41. 1]. a > 1. The area enclosed between the curves y 2 = x and y = |x| is (a) 1/6 (b) 1/3 (c) 2/3 (d) 1.ø + c 4 4 (2008) 19..p ö + C .x + x (c) 2 dx is (d) 1.p / 2 26.] denotes the greatest integer function.ø + c (b) x + log cos è x . local maximum at p and 2p. where f ( x ) = ò 1 +t è x ø 1 (c) 4 2 - 1 (d) 4 2 + 1 (2010) Then F(e) equals (a) 1 (b) 2 (c) 1/2 (d) 0. ò (a) 3/2 square units (b) 5/2 square units cos x + 3 sin x (c) 1/2 square units (d) 1 square units (2011) (a) log tan æç x + p ö÷ + C è 2 12 ø 14. The area bounded by the curves y = cosx and y = sinx between 3 2 (a) 3 p the ordinates x = 0 and x = is 2 (b) 2 2 (c) 2 (d) p . For x Î ç 0. x x 0 0 20. Let p ( x ) be a function defined on R such that 1 x p p¢(x) = p¢(1 – x ). for all x Î [0. local maximum at p and local minimum at 2p. The value of is a f (a) – {f (1) + f (2) + . 3) and the xaxis is (a) 4 2 - 2 (b) 4 2 + 2 (a) 6 (c) 12 (b) 9 (d) 3 1 integer not exceeding x is (2009) (a) (b) (c) (d) p 17.. The area of the plane region bounded by the curves x + 2y 2 = 0 and x + 3y 2 = 1 is equal to 4 (a) 3 5 (b) 3 1 (c) 3 p p (a) p ò f (cos x ) dx 0 0 p/2 p p/2 (c) ò f (sin x ) dx 2 0 (d) p ò 3 (2008) (a) 1/2 (b) 3/2 ò f (cos x ) dx .ö è 4 ø pö pö æ æ (a) x . (2006) . 2 (d) 3 (b) p ò f (sin x ) dx x 9 .log sin è x . define f ( x ) = ò t sin t dt Then f has 2 ø 0 local minimum at p and local maximum at 2p. + f ([a])} [a] f ([a]) – {f (1) + f (2) + . The value of ò [ x ] f ¢( x) dx.1 2 2 (d) 42 (2010) 23. (2007) 13.53 Integral Calculus (c) p log2 æ è 12. The area enclosed between the curve y = log e (x + e) and the .F (1). lim ú equals n ® ¥ ê n ë n2 û n2 n2 n2 1 1 (b) sec1 (a) cosec1 2 2 1 tan1 (c) (d) tan1. (a) 2 (b) 1 (c) 4 (d) 3. then value of p p (b) 4 + 2 - 1 (a) 4 . If ò sin( x .2 coordinate axes is (a) 7/3 (b) 14/3 (c) 28/3 (d) 1/3... (2004) 31. The parabolas y 2 = 4x and x 2 = 4y divide the square region 2 p / 2 (sin x + cos x ) bounded by the lines x = 4. If f ( x ) = f ( a ) ò xg{ x(1 - x )} dx and f ( . (c) (log x ) 2 + 1 + C (d) (2005) (c) 64 (d) 15. I3 = ò 2 dx and I 4 = ò 2 dx 1 0 0 1 1 log tan x . x > 0. I 1 = 1 + e x 38. a > 0.2 + 1 (A.a ) f ( a ) I 2 = (2005) I 2 g{ x(1 - x)} dx. Let be a nonnegative continuous function such that the area bounded by the curve y = f ( x ). then the value of I is 1 f ( . then one of the possible values If ò x e log x 1 x + C (a) 2 + C (b) of k is (log x ) 2 + 1 x + 1 (a) 16 (b) 63 x xe x + C . ò 2 3 1 1 3 2 2 cos x - sin x x x2 x x 33. (2005) 3 2 41. If 40. (2005) (c) (sina. The value of ò (a) p/2 p p / 2 39. B) is p p (a) (–sina . (2004) (2005) dx is equal to 42. cosa ) (d) (–cosa . The area of the region bounded by the curves 0 46. (a) 1 : 2 : 3 (b) 1 : 2 : 1 (2004) (c) 1 : 1 : 1 (d) 2 : 1 : 2. Let F ( x) = ç ÷ .54 JEE MAIN CHAPTERWISE EXPLORER é 1 sec2 1 + 2 sec2 4 + . sina ) (c) 1 . The value of ò |1 - x | dx is 32. (d) (2004) p 2 8 2 f is 2 sin x 43. xaxis and the ordinates 1 log tan x - p + C (c) p 2 8 p 2 p x = b > x = and is b sin b + cos b + 2b . S 2 . If I1 = ò 2 dx .3 p + C then (a) 2 8 2 (a) I 1 > I 2 (b) I 2 > I 1 (c) I 3 > I 4 (d) I 3 = I 4 . sin x (2005) æ e ö d 45. y = 4 and the coordinate axes. n lim is f ( x ) 3 n ®¥ r = 1 4 t 1 Then lim dt equals f (2) = 6. ò í 4 2 ý dx 3 sin x îï 1 + (log x ) þï dx = F (k ) . I 2 = ò 2 dx . sina ) (2004) n 1 35. ò (a) 1 – e (b) e – 1 (c) e (d) e + 1. x = 3 and the xaxis is x ®0 x sin x (a) 3 (b) 2 (c) 1 (d) 4. (2004) p 30. 1 (2005) log cot x + C (b) 2 2 f ( x ) 34.1) ïü is equal to 36. 2 cos 2 x dx .p 1 + a (b) ap (c) 2p e x .4 - 2 . (2003) 1 + x 2 ( () ( ( ) ) ( ) ( ) ( ) ( ) ) ( ( ) ) ( ) 3 x 2 2 ò sec tdt 37.a ) dx = Ax + B logsin( x . 2 dx è x ø ïì (log x .a ) ò (a) –1 (b) –3 (c) 2 (d) 1. is x .. The value of I = ò dx is S 1 . + 1 sec2 1 ù 29. If ò x f (sin x) dx = A ò f (sin x )dx.4 + 2 (d) 1 . then A is (d) p/a 0 (2005) 0 (a) p/4 (b) p (c) 0 (d) 2p.a ) + C . f ¢ (2) = . cosa ) (b) (cosa. Let F : R ® R be a differentiable function having r / n å e 44. The value of lim is y = |x – 2|. (a) 2 (b) 1 (c) 0 (2004) (d) 3 (2003) . S 3 are respectively the areas of these parts numbered 1 + sin 2 x 0 from top to bottom; then S 1 : S 2 : S 3 is (a) 2 (b) 1 (c) 0 (d) 3. x = 1. 48 x ® 2 6 x - 2 (2004) (a) 36 (b) 24 (c) 18 (d) 12. Then 4 4 4 1 log tan x + 3 p + C . If f (y) = e y .p - 1 (2003) 48. 2 x (1 + sin x ) dx is 1 + cos 2 x (a) p 2 /4 (b) p 2 ò -p (2003) (c) 0 (d) p/2. If y = f (x) makes +ve intercept of 2 and 0 unit x and y and encloses an area of 3/4 square unit with the axes then t F( t ) = ò f ( t . Then 2 (a) e + e - 3 2 2 57. units (d) none of these (2002) a a + b b ò f ( x ) dx 2 a is 1 (a) p + 1 1 1 (c) p . units.e - 5 .a b ò f ( x) dx 2 a p 54. then lim n [ I n + I n . The area bounded by the curves y = lnx. 40. 41. 16. 46. I n = ò tan n x dx . 24. 38. 37. 56. 2 a 1 (b) 1 - p 1 (d) p + 2 . y = |ln x| and y = |ln|x|| is (a) 4 sq. 29. (2003) is 0 (a) 3/2 (b) 1 ò sin x dx (a) 20 1 2 (d) e . 11. then ò x f ( x) dx is equal to (a) n p + 1 n ®¥ 0 1 (a) n + 2 1 1 (c) n + 1 + n + 2 1 p + 2 p + 3 p + . c) (b) (2002) . 57. 43. units (c) 6 sq.x ) n dx is 52. 20. 19. (b) (b) (c) (*) (c) (c) (c) (b) (a) 6. 48. 26.(1 + t ) (c) F (t) = te –t ò x f ¢ ( x ) dx (b) F (t) = t e t (d) F (t) = 1 – e t (1 + t). 50. 58.. (b) (a) (a) (a) (b) (c) (c) (b) (a) (d) 2. (2002) (2003) 51.x ) dx 2 a (b) b . 54. ò [ x 2 ] dx is 0 (a) 2 - 2 (b) 2 + 2 (c) (d) 2 - 1 2 - 2. b (d) a + b ò f (b . (2002) p/ 4 0 2 (c) e + e + 5 2 2 (d) –3/4. 22. units (b) 4 sq. units (c) 10 sq. (2003) 2 58. b (c) a + b b ò f (a + b . 53. 28. 44. 34. y = ln |x|. 30. 10. 39. + n p 3. If f (a + b – x) = f (x). 8. The value of the integral I = ò x(1 . 23. 45. 51. (2002) 49. 21.2 ] equals n ®¥ 0 (a) 1/2 (b) 1 (c) ¥ (d) 0. 36. 15. 12. 47. Answer Key 1. (c) (c) (b) (c) (b) (c) (a) (b) (a) (c) 5. 17. 9. The area of the region bounded by the curves y = |x – 1| and y = 3 – |x| is (a) 3 sq. g(y) = y ; y > 0 and 55. (a. 52. 31. 7. (d) (d) (a) (a) (d) (a) (b) (c) (b) (b) 4. 25. then 2 0 (a) F (t ) = et .55 Integral Calculus 1 47. 14. (2002) is p the value of the integral ò f ( x ) g ( x) dx is 2 (b) e . 49. units (b) 6 sq.x) dx. 32. units (d) 2 sq. (d) (c) (b) (c) (c) (b) (c) (b) (b) (d) (2002) 53.n + 2 1 (d) n + 1 .y ) g( y )dy . 18. 33.. 55. d) (b) (d) (c) (a) (c) (d) (a. and g(x) be a function that satisfies f (x) + g(x) = x 2 . lim 1 1 (b) n + 1 . 13. 27. 2 2 (b) 8 (c) 10 (d) 18. 35. Let f (x) be a function satisfying f ¢(x) = f (x) with f (0) = 1 56.e - 3 2 2 (c) 5/4 10 p 50. 42. we have 2 x .56 JEE MAIN CHAPTERWISE EXPLORER 1. (c) : p ò 850 3 2 Area = ò [(2 y + 3) .ò x 2 y( x 3 ) dx + C 3 Þ Þ a = 2 dy = | x | = 2.900 ò - 50 0 3 Þ p = 900 - 50∙ e t /2 0 If p = 0. 1 3 x y( x3 ) .2) 5sin x = sin x . (b) : Let x 3 = u. x 2 = 9y Area bounded by the parabolas and y = 2 2æ 2 yö = 2 ´ ò ç3 y dy = 5 ò ydy ÷ è 2 ø 0 0 = 5´ \ x = 1. (d) : Solving y = x with 2y – x + 3 = 0. 9. ò | t | dt = . \ x = ± 2 (d) : dx x 6. (a) : x 2 = .2 sin x - 2 cos x 1 = éëu ò f ( u)du . we get 1 u f (u) du 3 ò Now ò x 5 f ( x 3 )dx = a (cos x + 2 sin x ) 5 tan x = 1 + tan x .5 p( t) - 450 dt t 2 dp p .3)( x + 1) = 0 y 4 7. (d) : 1 900 t /2 =e Þ t = 2 ln 18 50 dV = .x + 3 = 0 Þ ( x . y1 = ò| t | dt = ò tdt = 0 0 -2 and y2 = t2 = 2 2 0 Þ g( x + p) = .900 = dt Þ 2ln =t p . then 3x 2 dx = du 5 tan x dx = x + a ln|sin x . then = 9 + 9 – 9 = 9 p/ 3 4.ò tdt = .y2 ] dy = y + 3 y - 0 3 y 3 ( y )3/2 10 20 2 = ´ 2 2 = 3/2 3 3 d( p(t )) = 0.2 cos x p 6 Again StatementII is true.2 cos x | + k tan x .2 5. We then get x = ±1 3. (b) : ò Also suppose ò f ( x )dx = y( x ) Differentiating both sides.ò ( ò f ( u)) duùû 3 = 2.2 0 Tangents are y – 2 = 2(x – 2) and y + 2 = 2(x + 2) Then the x intercepts are obtained by putting y = 0. we have V(T) = a = I 2 2 . d) : g( x ) = ò cos 4t dt 0 We can solve for y to get 2 sin x (1 + 2 a) + cos x ( a .k(T - t) dt + cot x On integration. 2 I = tan x p / 3æ ò çè 1 + p / 6 p/ 3 = æp pö = ò p / 6 1 1 tan x ò 1 × dx = çè 3 . V = + tan x ö ÷ dx 1 + tan x ø \ I = p 12 At t = 0.2cos x sin x .2 0 x é sin 4t ù sin 4 x Þ g ( x) = ê = ë 4 úû 0 4 2 2 (a. V(t) = I \ a = I - kT 2 k(T . 9 8.t) 2 + a 2 kT 2 + a Þ I = 2 2 kT As t = T.6 ÷ø = p / 6 sin 4( x + p) sin 4 x = 4 4 Þ g(p) = 0 Þ g(x + p) = g(x) + g(p) or g(x) – g(p). (c) : I = ò p / 3 dx p/ 61 + Adding. sin x )dx æ 1 1 ö æ 1 1 1 1 ö = 2ç + . (cos x .cos x - sin x ] p / 4 8 2 . 3). I = ò p( x )dx = ò p(1 .x )]dx = ò [ . we have ln5 = C Now ln(y + 3) = x + ln5 As x = ln2 we have ln(y + 3) = ln2 + ln5 = ln10 Þ y + 3 = 10 Þ y = 7.y ) 2 dy = ò t 2 dt = ê ú = = 9 3 ë û 0 3 0 0 p 17.x )dx Adding we have On adding we get 2 I = ò {[cot x ] + [ .q ÷ø ÷ø dq 0 ì æ æp ö öü = ò ln í(1 + tan q) ç 1 + tan ç . (a) : Area = 1 + ò dx = 1 + ln x 1 = 3 2 1 x 2 2 14. x 12. Set x = 0 to obtain p(0) = –p(1) + k Þ 1 = – 41 + k.t.1 ÷ + ç + + + ÷ è 2 ø è 2 2 2 2 2 ø ln(1 + tan q)dq 0 p/ 4 p /4 ò 5 p /4 (As the first and third integrals are equal in magnitude) Let x = tan q Þ J = Adding 2J = ò 3 p /2 (sin x . (a) : dy 10.cot x] dx 0 Now. 3) is y – 3 = 1 (x –2) Þ x – 2y + 4 = 0 2 Þ I = 8 J = p ln 2.6 y + 9) dy = ò ( y .57 Integral Calculus dy 15. (b) : I = ò 2 0 1 + x 1 Let J = ln (1 + x ) ò 1 + x 2 0 ò p/4 = 2 [ sin x + cos x ]0 dx ò æp ö ò ln(1 + tan çè 4 .2 = 4 2 - 2 16.3) 2 dy 0 0 (Let 3 – y = t) 3 3 3 é t 3 ù 3 3 = ò (3 . (b) : f ( x ) = ò t sin tdt 0 f ¢ ( x) = x sin x f ¢¢ ( x ) = x cos x + 1 -1/ 2 x sin x 2 f ¢¢ ( p) = . x Ï Z. p(x) = –p(1 – x) + k. y¢ = 1/2 The equation of the tangent to the parabola at (2. k being the constant of integration. we have 2(y –2)y¢ = 1 y ¢ = Þ 1 2( y - 2) at (2.q) ÷ø dq = 0 p/ 4 æ æp öö ò ln(1 + tan q) + ln çè 1 + tan çè 4 .4)] dy 0 3 e e 13.sin x )dx + 0 0 0 p 2 I = ò ( -1) dx = -p \ I = –p/2 0 Note that [x] + [–x] = 0. (c) : dx = y + 3 Þ y + 3 = dx As y(0) = 2. x Î Z = –1.p < 0 ; f ¢¢ (2 p) = 2 p > 0 Thus at p maximum and at 2p minimum.(2 y .2) 2 + 1 . (b) : (y – 2) 2 = x –1 Differentiating w. The desired area = 1 p/ 4 8 ln (1 + x ) dx 11. (b) : p¢(x) = p¢(1 – x) On integration.r.q ÷ ÷ ý dq è4 ø øþ è î 0 p /4 2J = ò (ln 2)dq = 0 p p ln 2 Þ 8 J = 4 ln 2 4 4 5 p / 4 + [ . \ k = 42 1 3 = ò ( y 2 .x ) dx = ò kdx = ò 42 dx = 42. (c) : I = ò [cot x] dx 0 p 0 1 p I = ò [cot(p .cot x ]} dx 0 1 p 0 1 1 2 I = ò p( x) + p(1 . The area of the bounded region 3 = ò [( y .cos x )dx + p /4 0 p /4 Now J = 5 p /4 (cos x . 0 Thus I = 21. . x. (c) : 1 ò = log tan æç + ö÷ + c ...ö + c è 4 ø a 25.2 ï 2 2 =í ý n 1 n 3 2 ï .. (d) : 2 ò p sin æ x .. if n = 2 m + 1 ï î n n-2 2 þ p = 2 p 27.x ) dx 0 1 é2 x 2 ù 2 1 1 = ê x3/ 2 .. There is no correct option.. (c) : F ( x ) = ò 1 1 p = 2× x+ 2× ln sin æ x . 2 (a) : Solution x + 2y 2 = 0 and x + 3y 2 = 1 we have 1 – 3y 2 = – 2y 2 2 (– 2.. (a) : Required area Y y = |x | x = y 2 (1.ú = . + f ([a])] -p 2 26.. 2 2 2 12 ø p è sin x + 6 ( ) X 1 p = 0 + 2 ´ 2 2 p ì 2 ï n í using fact ò sin x dx 0 ï î ì n .1) f ¢( x ) dx + ò Kf ¢( x ) dx = f (2) – f (1) + 2[ f (3) – f (2)] + 3[ f (4) – f (3)] + ..... (c) : Let I = ò éë( x + p )3 + cos 2 ( x + 3p ) ùû dx -3 p 2 Putting 0 x + p = z -p p -3 p -p Þ z = and x = Þz= 2 2 2 2 \ dx = dz and x + 3p = z + 2p also 3 ù1 é y = 2 ò (1 .ö + c è 4 ø 2 2 19.58 JEE MAIN CHAPTERWISE EXPLORER x p -1 23.. – 1) x + 3y 2 = 1 x + 2y 2 = 0 y¢ 1 3 = ò 1 f ¢( x ) dx + ò 2 f ¢( x ) dx + . (*) : éësec t ùû = 2 2 sin x dx 18.... p = x + ln sin æ x . (c) : In the interval of integration sin x < x 1 K = – [ f (1) + f (2) + . 0) dx 1 x p 22.3 y ) . + y 2 = 1 Þ \ y = ± 1 y = – 1 Þ x = – 2 y = 1 Þ x = – 2 The bounded region is as under x¢ 1/ x ln t ln t dt + ò dt 1 + t 1 +t 1 1 24. + f (K)] + K f (a) O 1 1 a ò ( K . 1 ´ p if n = 2 m ü ï n n ...ö sin ú dx è 4 4 û 4 ø ë x = - 2.sec -1 2 = p p sin æ x . (d) : Let I = ò xf (sin x) dx 0 .ö è 4 ø sec -1 x .= 3 2 ë û 0 3 2 6.+ ö è 4 4 ø = 2 ò dx p æ sin x . (i) .ú 3 û 0 ë 0 2 4 = sq... (b) : ò [ x ] f ¢( x ) dx . p p p 3 p . 1) A 1 = [a] f (a) – [ f (1) + f (2) + ...1 × n .ö è 4 ø p pù é p = 2 ò ê cos + cot æ x . say [a] = K such that a > 1 c being a constant of integration. x x x 1 æ ln t 2 ln t ö ln t F ( x ) = ò ç + dt = ò dt = (ln x ) ÷ 2 1 + t (1 + t ) t t è ø 1 1 F(e) = 1/2... 0) x B (– 2. 1) 1 = ò ( x .y 2 ) dy = 2 ê y .1 Þ sec x = + = 2 2 4 4 l= ò [z3 + cos 2 (2p + z )] dz = ò -p 2 -p 2 1 1 sin x x 2 2 I=ò dx < ò dx = ò xdx = é x 3/ 2 ù = êë 3 ú 0 3 û x x 0 0 0 p 2 z 3dz p 2 + ò cos 2 z dz -p 2 p 2 = 0 (an odd function) +2 ò cos 2 z dz 2 \ I < 3 0 1 1 cos x 1 dx < ò dx = [2 x ]1 0 = 2 x x 0 0 Also J = ò \ J < 2 21. (K – 1) [ f (K) – f (K –1)] + K[ f (a) – f (K)] y 2 2 The desired area = 2 ò [(1 ..3 ... (0. units 3 3 x= p 2 \ 20.( -2 y )] dy = 2 ´ K 2 K -1 (1...... .p sin B + 2 4 p p \ f = 1 .. (ii) 0 a 32.x )ö dx = p ò f (cos x ) dx è ø 2 \ e éë z (log e z .. 1) = ò log z dz 0 2 a y = loge (x + e ) e using ò f ( x ) dx = ò f ( a . (b) : Required area = ò log e ( x + e ) dx a p = pp p 2 \ I = ò f (sin x )dx = 2 ò f (sin x ) dx 20 2 0 0 0 p 0 2 2 2 x > 2 x \ 3 2 b f ( x ) = ò f ( x ) dx = b . x 3 > x 2 28... (c) : Total area = 4 × 4 = 16 sq. B p \ f ( x ) = ò 1 2 34.3 1 p ú = × . (a) : For 0 < x < 1. + sec 2 1 29.log x ) 2 ö \ òç dx = ò f ¢( x ) dx = f ( x ) ç 1 + (log x ) 2 ÷÷ è ø 2 C y = 4 2 x 16 dx = = S1 4 3 0 16 16 \ S2 = 16 ´2 = .p 1 + a a a p 3 I 1 > I 2 . ´ if n is even ú n n-2 2 2 ëê û f ( x ) = p 2 31. (c) : Method by cross check x Consider f ( x ) = (log x )2 + 1 2 x log x 1 + (log x ) 2 x \ f ¢ ( x ) = (1 + (log x )2 ) 2 \ f ¢ ( x ) = 1 + (log x ) 2 .1)ùû 1 33... 48 36. = 4 f ¢(2) ´ ( f (2))3 = 1 ´ 4 ´ 6 ´ 6 ´ 6 = 18.e O A n = 4 æ 1 . [using ò f ( x ) dx = 2 ò f ( x ) dx if f (2a – x) = f (x)] b O 1 – e 1 By (i) & (ii) on adding p 2 (0.. (c) : n lim n ®¥ n 2 n n è n ø 2 n 2 æ n ö 1 1 2 4 = lim 2 sec 2 2 + 2 sec 2 çæ 2 ÷ö + . (2) 2 f ( x ) = ò cos 2 x dx = 2 ò cos 2 x dx = 2 ´ 2 ò cos 2 x dx . 2 = lim x ® 2 -p p p -p 0 .+ 2.3 = 3 ò 2 2 3 a .log x ) ö = ç ç (1 + log x ) 2 ÷÷ è ø æ (1 . 2 f ( x ) = 4 ´ 1 ´ p 2 2 0 4 (0/0) form. (a) : Let f ( x ) = ò cos x x dx ( a > 0) .p 1 + a p b b cos 2 x \ f ( x ) = ò dx \ ò f ( x ) dx = ò f ( a + b . (c) : lim ò x ® 2 6 1 tan1.x ) dx 0 Y 1 .59 Integral Calculus p 0 I = ò ( p - x) f (sin x ) dx .2 if x > 2 ï y = if x = 0 37. S3 = ò 4 t 3 dt x - 2 f ¢ ( x ) ´ 4( f ( x )) 3 1 p/2 p/2 é ù n ê By using ò sin x dx ú 0 ê ú ê n ..e. (c) : According to question r = n 2 1 r sec 2 r å = lim å æç r2 ö÷ sec 2 r = n lim n n n ®¥ r = 0 n n ®¥ r = 1 è n ø 1 3 0 \ 1 1 2 4 1 sec 2 2 + 2 sec 2 æç 2 ö÷ + .2 log x (1 + log 2 x ) 2 2 æ (1 . 3 3 \ S 1 : S 2 : S 3 is 1 : 1 : 1. (c) : í 0 ï 2 .log x ö x \ òç dx = 2 ÷ 1 + (log x ) 2 è 1 + (log x ) ø Hence (c) is correct answer and we can check the other choices by the similar argument.(1) .x ..1 n . and for 1 < x < 2. units ( B sin B + p4 cos B + B 2 ) ( ) 2 ( ) ( ) a x cos 2 x dx 1 + a x ò f (b) = sin B + B cos B .. 2 x < 2 x Þ I 3 < I 4 0 a = 1. (b) : Using fact ò X a 2 p = p ò f æ sin( . ì x .x + x r = n 3 \ 2 x > 2 x 3 as 2 x > 2 x 6 Area of x 2 > x 3 i. 2 4 2 30....x ) ..a f ( a + b + x ) + f ( x ) a 2 1 = ò x sec 2 ( x 2 ) dx = 0 0 B ( > p / 4) ò f ( x ) dx = p/ 4 p / 4 f ( x ) 35..x if x < 2 î Required area = Area of DLAB + Area of DMBC . + 2 sec ç 2 ÷ n n ®¥ n èn ø n n è n ø ( ) 2 \ ò 2 x dx > ò 2 x dx x dx = 6 . 2 -1 ò = 1 ( x 2 .x \ 2 ) dx .a) dx = Ax + B log sin (x – a) + C p Þ Differentiating w. (c) : M(3. –1.x)} dx = if | x |< 1 2 ïî(1 .a and f (–a) = \ f (a) = 1 + e a 1 + e . (d) : f ( .x 2 ìï \ |1 – x 2 | = í 1 1 [AL × AB + BC × CM] = [1 × 1 + 1 × 1] = 1 2 2 e x 38.x )) if x < -1 and x ³ 1 = ò ò (1 . (a) : ò 0 = (sin x + cos x ) 2 (sin x + cos x) 2 cos x - 1 sin x 2 dx 1 1 dx 2 cos( x + p/4) ò 1 1 dx 2 sin æ p + x + p ö ç ÷ 4 ø è2 ò 1 1 dx 2 2 sin æ x + 3p ö cos æ x + 3 p ö ç ÷ ç ÷ è2 8 ø è 2 8 ø ò æ p x ö sec 2 ç 3 + ÷ 1 è 8 2 ø dx = æ x 3 p ö 2 2 tan ç + ÷ è 2 8 ø 0 p 2 ò 1 ò p 2 Þ A f (sin x ) dx = 0 0 ò 1 2 f ( .2 A(1. x = 3 Now 1 . 0) 3 3 2 x – y = 3 ò (1 . x > x 2 – 41.a ) b using b f ( x)dx = ò f (a + b .a) = A + sin( x .a ) sin ( x .x) g {(1 .60 JEE MAIN CHAPTERWISE EXPLORER 2 . 3 y = L(1. 1) (0.1) dx ò -1 1 4 28 æ 2 ö 20 = = + 2 ç ÷ + 3 3 3 è ø 3 42. x = 2. (c) : As f (x) = 1 + e x e a e .a ) 1 ò a cos x . 0) C(3.a ) p 1 2 ò g {(1 . 0) B(2.a \ f (–a) + f (a) = 1 f ( a ) x g {x(1 .x ) dx ò a f ( a ) f ( a ) ò 2 Þ Þ 2I 1 = I 2 \ I 2 2 = I1 1 = = p 2 ò x f (sin x ) dx ò = A f (sin x ) dx p 2 ò or A f (sin x ) dx = 0 p 2 p p ò ò 0 0 p f (sin x) dx = xf (sin x) dx 2 p = ´ 2 f (sin x ) dx 2 ò ò 0 p 2 ò 0 0 = Þ A f (sin x ) = p f (sin x ) dx Þ A = p p/2 40.r.(1 . 1.b sin x dx where a = b = 1 a = r cos q = 1 b = r sin q = 1 \ r = 2 q = tan –1 (b/a) let f ( a ) ò (1 . 0) x = 1. (b) : 0 sin x ò sin( x .x)( x) } dx f ( .x 2 ) dx + ( x 2 .x) x} dx x g { x (1 . x both sides æ cos x ö 2 + sin x ÷ = ç è -1 ø 0 Þ = 1 – (–1) = 2 Þ sin x B cos ( x .x 2 dx = Putting 1 – x 2 = 0 \ x = ± 1 Points –2.x )(1 + x ) dx -2 .a) sin x = A sin (x – a) + B cos (x – a) . 1) ò 1 . (b) : = a 1 æ x 3 p ö ´ 2 log tan ç + ÷ +c 8 ø 2 2 è2 1 æ x 3 p ö log tan ç + ÷ +c 8 ø 2 è2 p/2 dx = ò (sin x + cos x) dx 43.a ) 39.x ) } dx = f ( .1) dx + -2 ò 3 (1 .t. .2][(2n )(2n . (b) : ò x (1 ..x) f (a + b – x) dx t x sin x tan x 2 = Lt x ® 0 x sin x tan x 2 = Lt x ® 0 2 sin x x x tan x 2 1 = 1 × 1 = 1 = Lt sin x x ® 0 x 2 Lt x ® 0 x Putting ò x f ( x) dx dx .... (c) : Let I = a b 3 esin x dx = F(k) – F(1) I = 1 64 sin z Þ ò e z 1 dz = F(k) – F(1) where (x 3 = z) 64 F ( z ) 1 ] Þ [ Þ Þ F(64) – F(1) = F(k) – F(1) k = 64 = F(k) – F(1) 2 46.2 2 ´ 2 ´ 1 (2n + 4)(2n + 2) 1 = ( n + 2)(n + 1) 1 1 = (by partial fraction) n + 1 n + 2 = b 4 3 sin x 3 ò x e dx = F(k) – F(1) 1 45.y y dy (By replacing y ® t – y in f (y)) 0 t 0 q q F(t) = . B) = (cos a .2)... q = p/2 and 1 n p/2 ò x (1 . q = 0 x = 1.2)(2n .2] ù = ú (4 n + 2)(4 n)(4 n .. (a) : From given F(t) = 0 = e ò t .2] = (2n + 4)(2n + 2)(2n)(2n - 2)..e x ) dx 0 1 2 x òx e 0 [(2n )(2n .2). (b) : As f (x) = f ¢ (x) and f (0) = 1 f ¢ ( x ) = 1 Þ f ( x ) Þ log(f (x)) = x Þ f (x) = e x + k Þ f (x) = e x as f (0) = 1 Now g(x) = x 2 – e x \ 1 ò f ( x ) g ( x ) dx = 0 1 q cos 2 n + 1 q d q 0 = ò e x ( x 2 .2).q) e d q ò ò t 0 = (t eq ) 0 t – [(q – 1) eq ] 0 t = t(e t – 1) – (t – 1)e t – 1 = e t (t – t + 1) – t – 1 = e t – (t + 1) 50.e 2 x dx ò 0 1 – = [(x 2 – 2x + 2)e x ] 0 1 æ e 2 x ö çç ÷÷ è 2 ø 0 . 4.q)e d q = (t .4) .61 Integral Calculus sin x = A (sin x cosa – cos x sina) + B (cos x cos a + sin x sin a) sin x = sin x (A cos a + B sin a) + cos x(B cos a – A sin a) Now solving A cos a + B sin a = 1 and B cos a – A sin a = 0 (A.x) n dx 0 x = sin 2q dx = 2 sin q cos q dq x = 0.x ) dx = ò sin 2 q cos 2 n q (2 sin q cos q)dq \ 0 p 2 I = 0 I = ò x f (a + b . (b) : n Lt ®¥ 1 x = e dx = e – 1 ò 0 p/2 \ 2 ò sin 3 q cos 2 n + 1 q d q 0 2[2 ´ (2n)(2n . (c): Given 4 Þ 2 3 x ò x3 48.x ) dx \ I = ò 2 a 2 ò a 49. sin a) r r = n 1 n e n r = 1 å 44.ò xf ( x) dx a a b b a+b a + b f ( x ) dx = f (a + b .. (b) : ( tan t ) 0 x Lt x ® 0 1 47.2 û ò f (t .4.(t .x) dx a b ò (a + b) f ( x)dx .y) g ( y) dy t 0 2n + 1 b ò (a + b) f (a + b – x) dx – a b p/2 ò sin a b 1 3 2 n + 1 qdq = 2 ò sin q cos é ê Using êë ò (a + b .. (a). . (b) : I n = n ò tan x dx 0 1 (–1.. 0 ò 51..+ (–1) f n (x)] 55.x) .. ( ò I n – 2 = = 0 + ò 1 dx 1 = 2 -1 ò [ x 1 2 ] dx x dx .62 JEE MAIN CHAPTERWISE EXPLORER æp ö æ e 2 . (a) : Required Area = dx p ò p sin x dx - 0 ò sin x dx 0 = 18 (Using period of |sin x| = p) = 10 × 2 – 1 × 2 f(x) = |log|x|| p/4 57. 0) ò 2(1 + x ) dx + ò 2 dx + ò (4 .0 ÷ = p 2 - = (e – 2) – çç ÷÷ = e – 4 è ø 2 2 è 2 ø 2 n x x n n n n Using f (x)e dx = e [f (x) – f 1 (x) + f 2 (x) + . units..(.ò ç ..x + 1) dx + (3 .( x .... (c) : sin x cos x ) 0 p n . units \ 52..] = 0 – 10 p 56.. f n are derivatives of first.÷ . 0) 2 3 [\ f(2) = 0. (d) : Given f ( x ) dx = 3/4 where f 1 . f 2 .x + 1) + (3 . second .2 0 I n + I n – 2 = 1 n + 1 \ n(I n + I n – 2 ) = 1 1 + 1/n \ n Lt ® ¥ n(I n + I n – 2 ) = 1 1 2 ò 0 2 x sec 2 x dx [ x 2 ] dx = ò 2 [ x 2 ]dx + 0 0 2 (By putting cos x = t) n . 0) p/4 (1. x dx 0 è xø 0 1 = 2[(1 – 0) + (x) 0 ] = 4 sq..2 ò tan p/4 = x dx 0 0 58. + ç ÷ ú ènø è n ø úû êë è n ø r=n = n Lt ®¥ 1 ´ n 2 ò x f ¢( x) dx = xò f ¢( x) dx – 0 1 p + 2 p + . (d) : ò sin x 10 p 53. curve having intercept 4 = –3/4 2 units on xaxis.(.2 x ) dx -1 0 2 1 = 4 sq. (b) : Required area 0 = ò 2 ò ò (3 + x) .2 x dx 0 1 æ 1 ö = 2 éë x log x ùû .. (a) : Lt n p n ® ¥ 1 = n Lt ®¥ n 1 æ r ö 1 p å n çè n ÷ø = r = 1 x p dx = ò 0 0 1 p + 1 ò f ( x) dx 0 3 = 2f(2) – 4 2 ( x ))0 - 3/4 = ( x × f p é æ 1 ö p æ 2 ö p æ n ö ù ê ç ÷ + ç ÷ + .2 ò tan tan n .2 x × (sec 2 x –1) dx + p p ò p/4 x sin x p p/4 tan n x dx + 0 -p p p/4 \ I n + I n – 2 = ò 1 + cos 2 x dx + 2 ò 1 + cos 2 x dx 54..n th order..x) . 0) = 2ò log x dx 0 ) p 1 p x -p x sin x x sin x ò 1 + cos 2 x dx = 2 × 2 ò 1 + cos 2 x = 0 + 2 -p 0 = p sin x x sin x ò 1 + cos 2 x dx = 4 × 2 ò 0 1 + cos 2 x dx 0 p p æ ö p by using xf (sin x ) dx = f (sin x) dx ÷ ç ò ò 2 0 è ø 0 = 4 p = 4 ´ 2 × 2 = 4p (tan -1 p/2 ò p/4 ò 1 + cos 2 x dx 0 ò tan n . + n p (2.1 ö e 2 3 = 4p × ç .1) dx -1 0 = 1 1 0 1 2 (0. (b) : 2 tan n . 2) = ke (2003) . x > 0 then dx dx dy 1 - x 1 (c) x 2 = y 2 + xy (a) (b) dx x x x 1 + x dy 2 2 . where A and B are arbitrary constants is of (a) second order and second degree (b) first order and second degree (c) first order and first degree (d) second order and first degree. (2006) 12. degree as follows (a) order 1. 6.e tan y ) = 0 is dx (a) 2 xe tan y = e 2 tan y + k -1 -1 tan (b) xe -1 (c) xe 2 tan -1 y -1 = tan -1 y + k y -1 = e tan y + k -1 . (2004) dx dy y + . If x = e y + e ..tan y . then the solution of the equation dx is 8. degree 3.log x + 1). The differential equation for the family of curves x 2 + y 2 – 2ay The differential equation of all circles passing through the = 0. (d) ( x . (b) order 1. (d) x = y + 3 xy . is (a) y¢¢ = y¢y (b) yy¢¢ = y¢ (c) yy¢¢ = (y¢) 2 (d) y¢ = y 2 (2009) The solution of the differential equation 5. 3. degree 2 (d) order 1. 7. 0 < x < p/2 is (a) sec x = (tanx + c)y (b) y secx = tanx + c (c) y tanx = secx + c (d) tanx = (secx + c)y (2010) The differential equation which represents the family of curves y = c 1 e c 2 x . where c > 0. (2005) The solution of the differential equation ydx + (x + x 2 y)dy = 0 is 1 1 (a) xy + log y = C (b) .xy = C (d) logy = Cx. to ¥ dy (b) y 2 = x 2 - 2 xy is 11. (2004) 10. y (a) x log æç ö÷ = cy è x ø æ x ö (b) y log ç ÷ = cx è y ø æ x ö (c) log ç ÷ = cy è y ø y (d) log æç ö÷ = cx . DIFFERENTIAL EQUATIONS Solution of the differential equation cos xdy = y(sinx – y )dx . degree 1 (c) order 2. degree 2 (2008) The differential equation of the family of circles with fixed radius 5 units and centre on the line y = 2 is (a) (x – 2) 2 y¢ 2 = 25 – ( y – 2) 2 (b) (x – 2) y¢ 2 = 25 – ( y – 2) 2 (c) ( y – 2) y¢ 2 = 25 – ( y – 2) 2 (d) ( y – 2) 2 y¢ 2 = 25 – ( y – 2) 2 (2008) 9. where c 1 and c 2 are arbitrary constants. è x ø (2005) The differential equation representing the family of curves y 2 = 2 c ( x + c ). is of order and dy x + y = dx x satisfying the condition y(1) = 1 is (a) y = x ln x + x (b) y = ln x + x (c) y = x ln x + x 2 (d) y = x e (x – 1) 4.xy + log y = C 1 (c) .63 Differential Equations CHAPTER 11 1. If x dy = y (log y . where a is an arbitrary constant is origin and having their centres on the xaxis is (a) (x 2 – y 2 )y¢ = 2xy (b) 2(x 2 + y 2 )y¢ = xy dy (a) y 2 = x 2 + 2 xy (c) 2(x 2 – y 2 )y¢ = xy (d) (x 2 + y 2 )y¢ = 2xy. is a parameter. The solution of the differential equation dy (1 + y 2 ) + ( x .. 2. (c) (d) (2004) (2007) 1 + x x dx The differential equation whose solution is Ax 2 + By 2 = 1. The solution of the equation (c) 2. (a) 5. (a) (2002) (2002) .2 x e + c + d . (2003) dx 2 14. (d) 14. 1 (c) 3. (c) 8. 4. are respectively (a) 1. (a) 2.64 JEE MAIN CHAPTERWISE EXPLORER 13. 2 (b) 3.2 x e + cx 2 + d 4 (d) 1 .2 x 15. (a) 6. (d) 13. (d) 12. (c) 3. 3 (d) 2. 1. (a) 11. (b) 3. (a) 7. The degree and order of the differential equation of the family 2 (a) 1. (b) (d) 1. 2 d 2 y = e .2 x e + cx + d 4 (c) 1 . (d) 10. 3 3 of all parabolas whose axis is xaxis. 4 Answer Key 1. (b) 15. 2. (a) 9. The order and degree of the differential equation 2 3 (1 + 3 dydx ) = 4 d 3 y are dx 3 (a) 1 - 2 x e 4 (b) 1 . dx 2 2 (d) : Given A x + B y = 1 As solution having two constants.sec 2 x dx + k Þ v sec x = .t..( y - 2) dx From (1) and (2) on eliminate ‘a’ . - y = ln x + k x Þ y = x ln x + kx As before. so that dx dx dv x + vx = 1 + v We have v + x = dx x dv dx Þ x = 1 Þ dv = dx x dx x dy ( y . . x.. Then y = x ln x + x 2nd Method (Inspection) : Rewriting the equation dy x + y = as dx x xdy – ydx = xdx dx = - cos 2 x On integration. we get 2( x ... (i) B x dx Differentiating (*) w. = e ò xdy .tan x = - sec x Þ y 2 dx y 4.y ) = y tan x .r. y ¢ = c1c2ec2 x = c2 y .r. (i) .. y = 1 giving 1 = 0 + (ln k) \ ln k = 1..(1) ...tan x + k Þ 2..t. only choices (a) and (b) are possible Again A x 2 + B y 2 = 1 . which is linear in v . curves. dx = e ln sec x = sec x The solution is v ´ sec x = ò ..h ) + 2 y dy = 0 dx dy h = x + y to eliminate h.ydx (y – 2) 2 (y¢) 2 = 25 – (y – 2) 2 (a) : General equation of all such circles is (x – h) 2 + (y – 0) 2 = h 2 .a ) + ( y ..a = .2) 2 = 25 è dx ø sec x = . (a) : 1st solution: cos x dy = y (sin x – y )dx Þ cos x dy = y sin x dx – y 2 dx Þ cos x dy – y sin x dx = –y 2 dx Þ d(y cos x) = –y 2 dx Þ d( y cos x) ( y cos x )2 Þ v = ln x + ln k As v = y/x we have y = x ln x + (ln k)x At x = 1.(ii) y¢ y = Þ y ¢¢y = ( y ¢ )2 y ¢¢ y ¢ Which is the desired differential equation of the family of 6.y 2 sec x = Þ dx dx cos x dy .C Þ sec x = y(tan x + C ) y Differentiating w. we get 3. (*) A y dy ..y tan x = ... x y¢¢ = c 2 y¢ From (i) and (ii) upon division .t. y = x ln x + x (d) : The equation to circle is (x – a) 2 + (y – 2) 2 = 25 Differentiation w. putting value of h in dx equation .2) 2 æ ö + ( y ..y 2 sec x Þ dx 1 dy 1 . evaluating constant. \ order of differential equation is 2 so our choices (b) & (c) are discarded from the list. dv + (tan x )v = - sec x .r..= Þ .t.tan x .65 Differential Equations 1. we have I.F..(2) 2 Þ (c) : y = c1 e c2 x 5. = On integration 1 tan x dx x 2 y dx d æ ö = è x ø x Setting.2) = 0 dx dy Þ x . x dy ( x . (a) : 1st Method (Homogeneous equation): dy dv = v + x Let y = vx.. dy \ we get y 2 = x 2 + 2 xy .. x Again on differentiating .(i) Again differentiating w.y = v . (i) where h is parameter Þ (x – h) 2 + y 2 = h 2 Differentiating.. we have Þ – sec x = –y tan x + yk Þ sec x = y (tanx + C) where C is a constant 2nd solution: We have dy dy y(sin x .r. 66 JEE MAIN CHAPTERWISE EXPLORER y . (a) : x = e Þ x = e y + x Differentiate w. dy 1 æ y ö dv dx = Þ log çè ÷ø = cx .. to x) Þ order 2 and degree 1 (Concept: Exponent of highest order derivative is called degree and order of that derivative is called order of the differential equation.F. (ii) æ dy ö + ç ÷ = 0 Þ 2 è dx ø dx (by differentiating (ii) w. v log v x x 2 dx tan -1 y + 1x = e dy e tan y dx 1 + x Þ x ∙ I. (a) : Given family of curve is x 2 + y 2 – 2ay = 0 Þ 2a = Þ e 2 tan y +c 2 -1 = e 2 tan y + k =- y 2 x (a . (a) : From the given equation (1 + y 2 ) dy = y (log y .. x) dy y Þ = 2a dx - 1 Þ ò 1 + y 2 dy -1 - 1 tan Þ x e 2 2 ò y × I..t. = Þ = dy 1 + y 2 1 + y 2 ö dy yæ æ y ö y \ = ç log ç ÷ + 1 ÷ Now put = v dx xè è xø x ø \ v logv dx = x dv Þ 11.r.(1) 2 x + y y Also from (1). (c) : ç 1 + 3 ÷ = 4 çç 3 ÷÷ dx ø è è dx ø 3 2 é d 3 y ù dy ö æ Þ çè 1 + 3 ÷ø = ê 4 dx3 ú dx ë û \ highest order is 3 whose exponent is also 3.t. (b) : Given \ \ dx 2 = e –2x dy e -2 x = + c dx - 2 e - 2 x y = + cx + d 4 . (d) : x 12..÷ = è xy ø y Þ 1 - = –log y + C xy y -1 y tan Þ x e - d ( xy ) 2 ( xy ) 10.r.1 = e tan 13. degree 3. 2x + 2yy¢ – 2a y¢ = 0 Þ y = d 2 y 1 + log y = C xy 2 .. xy d 2 y æ dy ö æ dy ö + x ç ÷ = y ç ÷ 2 dx dx è ø è dx ø Þ order 2 degree 1. Þ order 1.dy Þ y Þ dy æ 1 ö d ç . (i) ..r..(i) \ yy 1 = c 2 3 2 y = 2 × yy1 x + yy1 Þ ( y .) dy y tan Þ 2 x e y - 1 2 tan = e y 2 .2 xyy1 ) = 4 ( yy1 ) 2 3 3 1 Þ (y – 2xyy 1 ) = 4y y 9.. x after taking logarithm both sides dy 1 dy 1 - x Þ \ = 1 + = dx x dx x æ x 2 + y 2 ö ç ÷÷ y¢ = 0 2x + 2yy¢ – ç y è ø 2xy + y¢(2y 2 – x 2 – y 2 ) = 0 Þ y¢(x 2 – y 2 ) = 2xy + k æ d 3 y ö dy ö 3 æ 14.log x + 1) dx 8. d 2 y 15..= B . (a): As axis of parabola is xaxis which means focus lies on xaxis.(ii) 2 7.....F = e (d) : y = 2 c ( x + c ) \ 2yy 1 = 2c Now putting c = yy 1 in (i) we get ( ) . Equation of such parabola’s is given by y 2 = 4a(x – k) Þ 2yy 1 = 4a (by differentiating (i) w. (b) : y dx = –(x 2 y + x) dy Þ ydx + xdy = –x 2 y dy Þ ydx + xdy 2 ( xy ) = .... -1 where I.F. y + e y + e 2 æ d 2 y ö æ dy ö y ç 2 ÷ + ç ÷ è dx ø è dx ø By (i) and (ii) we get A . 0) . –2) and touching the axis of x at 7. Statement2 is not a correct explanation for Statement1. Statement2 is a correct explanation for Statement1. If the centre of the ellipse is at the origin and its axes are the coordinate axes. Statement 2 is false. Statement2 : If the line. then the (2013) slope of the line PQ is The xcoordinate of the incentre of the triangle that has the (a) – 2 (b) – 1/2 coordinates of mid points of its sides as (0. 3y = x - 3 3y = x - 1 (b) y = 3x - 3 (d) y = x + 3 (2013) The equation of the circle passing through the focii of the x 2 y2 + = 1 and having centre at (0. Statement2 is true. tangent to the parabola y 2 = 16 3 x the ellipse 2x 2 + y 2 = 4. 0) (c) – 1/4 (d) – 4 (2012) is 2 2 2 2 2 11. 4) in the 12. (d) Statement 1 is true.67 Two Dimensional Geometry CHAPTER 12 1. 5 (m ¹ 0) is their m common tangent. (c) Statement1 is false. (2013) 3. y 2 = 16 3 x and the ellipse 2 x 2 + y 2 = 4 is y = 2 x + 2 3 Statement 2 : If the line y = mx + 4 3 . then the equation of the ellipse is (a) 4x 2 + y 2 = 8 (b) x 2 + 4y 2 = 16 (c) 4x 2 + y 2 = 4 (d) x 2 + 4y 2 = 8 (2012) 10. The bisector of the ratio 3 : 2. (m ¹ 0) is a common m The length of the diameter of the circle which touches the xaxis at the point (1. (3. Statement2 is true. The lines L 1 : y – x = 0 and L 2 : 2x + y = 0 intersect the line L 3 : y + 2 = 0 at P and Q respectively. 3) is 16 9 2 2 (a) x + y – 6y + 7 = 0 (b) x 2 + y 2 – 6y – 5 = 0 ellipse 5. 0) also passes through the point (a) (2. (1. (d) Statement1 is true. then m satisfies m 4 – 3m 2 + 2 = 0. the equation of the reflected ray is (a) (c) 4. Statement2 is true. Statement 2 is true; Statement 2 is a correct explanation for Statement 1. where O (c) x 2 + y 2 – 6y + 5 = 0 (d) x 2 + y 2 – 6y – 7 = 0 is the origin. 2x 2 + 2y 2 = 5 and a parabola y2 = 4 5x . (b) Statement 1 is true. –2) (c) (–2. (b) Statement1 is true. 9. 5) (d) (–5. (a) 6 (b) 11/5 (c) 29/5 (d) 5 Statement1 : The ratio PR : RQ equals (2012) . 0) and passes through the point (2. 2. (a) Statement 1 is true. 1) and (1. (2012) curves is y = x + 5. y = mx + (a) Statement1 is true. Statement 2 is true. 2) (2013) Statement 1 : An equation of a common tangent to the parabola Given : A circle. TWO DIMENSIONAL GEOMETRY The circle passi ough (1. Statement1 : An equation of a common tangent to these then m satisfies m 4 + 2m 2 = 24. 8. 3) is (a) 6/5 (b) 5/3 (c) 10/3 (d) 3/5 (2012) An ellipse is drawn by taking a diameter of the circle (x – 1) 2 + y 2 = 1 as its semiminor axis and a diameter of the circle x 2 + (y – 2) 2 = 4 as its semimajor axis. (c) Statement 1 is false. then k equals acute angle between L 1 and L 2 intersects L 3 at R . 2) to meet the coordinate axes at P and Q such that it forms a triangle OPQ. Statement 2 is true; Statement 2 is not a correct explanation for Statement 1. 1) and (2. If the area of the triangle OPQ is least. 1). A line is drawn through the point (1. 6. –5) (b) (5. Statement2 is false. A ray of light along x + 3y = 3 gets reflected upon reaching x-axis. The two circles x + y = ax and x + y = c (c > 0) touch each (a) 2 - 2 (b) 1 + 2 other if (c) 1 - 2 (d) 2 + 2 (2013) (a) a = 2c (b) |a| = 2c (c) 2 |a| = c (d) |a| = c (2011) If the line 2 x + y = k passes through the point which divides the line segment joining the points (1. The point diametrically opposite to the point P(1. – 4) (2008) 16. is (c) exactly one value of p (a) y = 0 (b) y = 1 (d) all values of p (2009) (c) y = 2 (d) y = 3 (2010) (a) 3x 2 + 5y 2 – 15 = 0 (c) 3x 2 + 5y 2 – 32 = 0 24. then 20. Then a possible value 19. 0) (d) (0. 4) and Q(k. Statement2 is true; Statement2 is a triangle ABC is at the point correct explanation for Statement1. 0 . Equation of the ellipse whose axes are the axes of coordinates of the ellipse is and which passes through the point (–3. 0) is equal to 1/3. 0 3 ( ) (d) (0. The shortest distance between line y – x = 1 and curve x = y 2 is (a) 8 (b) 4 (c) 3 4 (d) (c) ( ) 5 . The equation of the tangent to the curve x 2 (b) all except two values of p parallel to the x axis. The line L given by + = 1 passes through the point (2008) 5 b (13. which in turn is inscribed in another (2011) ellipse that passes through the point (4. y) meets the xaxis at G. If P and Q are the points of intersection of the circles (b) 5x 2 + 3y 2 – 32 = 0 x 2 + y 2 + 3x + 7y + 2p – 5 = 0 and x 2 + y 2 + 2x + 2y – p 2 = 0. 32). 3) has yintercept – 4. – 4) 23 17 23 (a) (b) 17 (c) (d) (c) (– 3. (2011) ( ) 13. 2 2 (d) 5x + 3y – 48 = 0 (2011) then there is a circle passing through P. then the locus of P is (a) x = 1 (b) 2x + 1 = 0 (c) x = –1 (d) 2x – 1 = 0 (2010) 15 15 x 2 + 17 (2010) 26. A focus of an ellipse is at the origin. Then the vertex of the parabola is at m (a) (2. 0) to the distance from the (a) Statement1 is true. The shortest distance between the line y – x = 1 and the curve of k is x = y 2 is (a) – 4 (b) 1 (c) 2 (d) – 2 2 3 3 2 3 3 2 (a) (b) (c) (d) (2008) 8 5 4 8 (2009) 28. bisector of an angle divides 21.68 JEE MAIN CHAPTERWISE EXPLORER Statement2 : In any triangle. Then the length of the semi 2 major axis is 5 8 2 4 (b) (c) (d) (a) x y 3 3 3 3 17. The directrix is the line 1 x = 4 and the eccentricity is . 4) (b) (3. The circle at two distinct points if (a) –85 < m < –35 (c) 15 < m < 65 (b) –35 < m < 15 (d) 35 < m < 85 (2010) 27. where c1 and c2 are arbitrary constants. Statement2 is true; Statement2 is . If the distance of G from the origin is twice the abscissa of P. Q and (1. 0) (2009) 3 2 8 22. 0). 0 (a) (b) 4 2 not a correct explanation for Statement1. 0) on the x y + = 1 . The normal to a curve at P(x. 2) (c) (1. (2007) (c) yy ¢¢ = (y¢ ) 2 (d) y¢ = y 2 (2009) . Then the equation 14. 0) (b) (0. is (a) circle (b) hyperbola (a) y¢¢ = y ¢ y (b) yy ¢¢ = y ¢ (c) ellipse (d) parabola. one of them from the point (1. 5 5 (d) Statement1 is true. Then the distance between L and K is circle x 2 + y 2 + 2x + 4y – 3 = 0 is c 3 (a) (3. Statement2 is true. If two tangents drawn from a point P to the parabola y 2 = 4x are at right angles. Three distinct points A. 1) (2008) y 2 = 4x + 8y + 5 intersects the line 3x – 4y = 18. 1) and has (a) x 2 + 12y 2 = 16 (b) 4x 2 + 48y 2 = 48 2 (c) 4x 2 + 64y 2 = 48 (d) x 2 + 16y 2 = 16 (2009) eccentricity is 3 2 3 5 23. point ( –1. 1) for 4 y = x + . (b) Statement1 is false. The differential equation which represents the family of curves the curve is a y = c1 ec2 x . Then the circumcentre of the (c) Statement1 is true. (a) all except one value of p that is 15. The ellipse x 2 + 4y 2 = 4 is inscribed in a rectangle aligned with the coordinate axes. Statement2 is false. The perpendicular bisector of the line segment joining P(1. 4) (d) (– 3. A parabola has the origin as its focus and the line x = 2 as the directrix. B and C are given in the 2dimensional coordinate plane such that the ratio of the distance of any the triangle into two similar triangles. The line K is parallel to L and has the equation 25. 31. The locus of a point P (a. which of the following 2 cos a sin a remains constant when a varies ? (a) abscissae of vertices (b) abscissae of foci (c) eccentricity (d) directrix. then the equation of the locus of è 2 ø its centre is 1 ö æ 1 ö æ (c) ç . 3 ÷ (d) ç -3. 2). 0) and R = (3. Its equation is angled triangle with AC as its hypotenuse. For the hyperbola 2 y 2 line y = ax + b is a tangent to the hyperbola x 2 . If a circle passes through the point (a. 1) and are tangent to xaxis.69 Two Dimensional Geometry 29. b) and cuts the circle æ 1 ö (a) ç 0. (a) (2. The locus of the centre of the circle is 2 (a) a circle (b) an ellipse 27 9 (c) x 2 + y 2 = (d) x 2 + y 2 = . x > 0. k). 0). 2}. Let C be the circle with centre (0. An ellipse has OB as semi minor axis.2 = 1. (2007) such that the area of one of the sectors is thrice the area of 33. 1) be the vertices of a right between the axes is bisected at A. ¥) 2 2 2 x + y = p orthogonally. then m is (a) 1 (b) 2 (c) –1/2 (d) –2. 3} (b) {–3. (d) xy = 105 (b) xy = (2006) 40.2 = 1 is a b (a) a circle (b) an ellipse (c) a hyperbola (d) a parabola. (2007) 34. 3) and radius 2. (2006) (c) 0 £ k £ 1 (d) k ³ 1 . 0) (2005) (c) (–1. Then the eccentricity of the x ellipse is 2 35. then a belongs to 44. If the area of the (a) x + y = 7 (b) 3x – 4y + 7 = 0 triangle is 1 square unit. (2006) (a) 2ax + 2by – (a 2 – b 2 + p 2 ) = 0 è2 ø è 2 ø (b) x 2 + y 2 – 3ax – 4by + (a 2 + b 2 – p 2 ) = 0 36. (d) x + 2 3 x + y = 0 (2007) is 8. then the set of values which ‘k’ can (c) 4x + 3y = 24 (d) 3x + 4y = 25. then the equation of the of the centre of the circles. Let P = (–1. If (h. Q = (0. x > 0 1 1 2 1 1 (b) (c) (d) . The equation of a tangent to the parabola y 2 = 8x is another sector then y = x + 2. b) moving under the condition that the y 2 x 2 . (2005) C that subtend an angle of 2p/3 at its centre is 45. k) are the coordinate of a circle of area 49p square units. (2007) 2 2 38. A circle touches the xaxis and also touches the circle with centre 3 2 2 (a) x + y = (b) x 2 + y 2 = 1 at (0. (2007) 42. . Consider a family of circles which are passing through the 37. (a) 3/5 (b) 1/2 (c) 4/5 (d) 1/ 5 . ÷ (b) (3. If one of the lines of my 2 + (1 – m 2 )xy – mx 2 = 0 is a bisector of the angle between the lines xy = 0. Let A(h. (2005) (a) 2 4 2 3 and y = 3x. In an ellipse. 1) and C(2. then the set of values of k is circle is given by the interval (a) x 2 + y 2 + 2x – 2y – 47 = 0 (b) x 2 + y 2 + 2x – 2y – 62 = 0 1 1 1 k £ (a) . Then its eccentricity is. 0) and radius 3 units. The (c) 2ax + 2by – (a 2 + b 2 + p 2 ) = 0 equation of the locus of the mid points of chord of the circle (d) x 2 + y 2 – 2ax – 3by + (a 2 – b 2 – p 2 ) = 0. If the lines 3x – 4y – 7 = 0 and 2x – 3y – 5 = 0 are two diameters point (–1. 3} (d) {0. –2} diameters of a circle and divide the circle into four sectors (c) {1. (2005) 43. A straight line through the point A(3.£ k £ (b) 2 2 2 (c) x 2 + y 2 – 2x + 2y – 62 = 0 (d) x 2 + y 2 – 2x + 2y – 47 = 0. (2006) take is given by 2 2 41. 4) (b) (–2. the distance between its focii is 6 and minor axis 30. (2006) The locus of the vertices of the family of parabolas a 3 x 2 a 2 x + - 2 a is 3 2 105 (a) xy = 64 35 (c) xy = 16 y= 3 4 64 . (2007) 39. 3 3) be three points. F and F ¢ its focii and the angle FBF ¢ is a right angle. 1) (d) (0. The point on this line from which the other tangent (a) 3a 2 – 2ab + 3b 2 = 0 (b) 3a 2 – 10ab + 3b 2 = 0 to the parabola is perpendicular to the given tangent is (c) 3a 2 + 2ab + 3b 2 = 0 (d) 3a 2 + 10ab + 3b 2 = 0. a ) falls inside the angle made by the lines y = . If the pair of lines ax + 2(a + b)xy + by = 0 lie along (a) {–1. The equation of the bisector of the angle PQR is 3 x + y = 0 (a) 2 (b) x + 3 y = 0 (c) 3 y = 0. (c) a parabola (d) a hyperbola. 4) is such that its intercept 32. (2005) (2006) 4 4 . B(1. If (a.÷ . b) and cuts the circle x 2 + y 2 = 4 orthogonally. then the centroid A(p. If nonzero numbers a. If the sum of the slopes of the lines given by x 2 – 2cxy – 7y 2 = 0 is four times their product. (2004) 1 7 7 (c) 3 . (2004) (2004) . then the equation of the ellipse is (a) 4x 2 + 3y 2 = 12 (b) 3x 2 + 4y 2 = 12 (c) 3x 2 + 4y 2 = 1 (d) 4x 2 + 3y 2 = 1. 3) and making intercepts on the coordinate axes whose sum is –1 is x y y + = 1 and x + = 1 2 3 2 1 x y y x + = 1 (b) . is 1/2. y = a sinq at q always is AB. (2005) 51. 1) and the mid points of two 55.P. (a) (b) - 1. (2004) 59. If the circles x 2 + y 2 + 2ax + cy + a = 0 and 54. then (a) d 2 + (2b – 3c) 2 = 0 (b) d 2 + (3b + 2c) 2 = 0 the locus of the vertex C is the line (c) d 2 + (2b + 3c) 2 = 0 (d) d 2 + (3b – 2c) 2 = 0. The locus of the other end of the of the triangle is diameter through A is - 1 7 7 (a) (y – p) 2 = 4qx (b) (x – q) 2 = 4py . The equation of the straight line passing through the point (4. b.= 1 and 2 3 - 2 1 (a) (2004) 52. If the lines 2x + 3y + 1 = 0 and 3x – y – 4 = 0 lie along diameters of a circle of circumference 10p. –2) (b) (–1. with its centre at the origin. Equation of the circle on AB as a diameter is 2 2 2 2 passes through the fixed point (a) x + y + x + y = 0 (b) x + y – x + y = 0 (a) (0.1 (c) 2 3 - 2 1 x y y x + = 1. (2004) 53. then the locus of its centre is 48. Let A(2. (2005) 57. (d) . If one of the lines given by 6x 2 – xy + 4cy 2 = 0 is 3x + 4y = 0. (2004) 58. The normal to the curve x = a(1 + cosq). If a vertex of a triangle is (1. - 2 (d) (1. then the straight line (a) 2ax – 2by + (a 2 + b 2 + 4) = 0 x y 1 + + = 0 always passes through a fixed point. That point a b c (b) 2ax + 2by – (a 2 + b 2 + 4) = 0 is (c) 2ax + 2by + (a 2 + b 2 + 4) = 0 (a) (–1. (c) (a. ( ) ( ) ) below the xaxis at a distance of 2/3 from it below the xaxis at a distance of 3/2 from it above the xaxis at a distance of 2/3 from it above the xaxis at a distance of 3/2 from it. 0) and Q a point on the locus y 2 = 8x. a) (c) x 2 + y 2 – x – y = 0 (d) x 2 + y 2 + x – y = 0. The eccentricity of an ellipse. 2) and (3. then c equals 49. A variable circle passes through the fixed point sides through this vertex are (–1. a). (2004) 47. 2) (d) 2ax – 2by – (a 2 + b 2 + 4) = 0. The intercept on the line y = x by the circle x 2 + y 2 – 2x = 0 61. –2). of the lines ax + 2by + 3b = 0 and bx – 2ay – 3a = 0.= - 1 and 2 3 - 2 1 x y y + = - 1 and x + = . 0) (b) (0.70 JEE MAIN CHAPTERWISE EXPLORER 46. q) and touches xaxis. –3) and B(–2. The locus of mid point of PQ is (a) x 2 – 4y + 2 = 0 (b) x 2 + 4y + 2 = 0 2 (c) y + 4x + 2 = 0 (d) y 2 – 4x + 2 = 0. (2004) 1 (c) 1. then the equation x 2 + y 2 – 3ax + dy – 1 = 0 intersect in two distinct points P of the circle is and Q then the line 5x + by – a = 0 passes through P and Q for (a) no value of a (a) x 2 + y 2 + 2x + 2y – 23 = 0 (b) exactly one value of a (b) x 2 + y 2 – 2x – 2y – 23 = 0 (c) exactly two values of a (c) x 2 + y 2 – 2x + 2y – 23 = 0 (d) infinitely many values of a. 3 (d) 1. (2005) 56. 60. 2). (2005) (d) x 2 + y 2 + 2x – 2y – 23 = 0. b) ¹ (0. 0) is ( ) ( ) ( (a) (b) (c) (d) 50. The line parallel to the xaxis and passing through the intersection (a) 3 (b) – 1 (c) 1 (d) – 3. c are in H. If a circle passes through the point (a. 1) be vertices of a triangle ABC. 3 . 0) (d) (a. If the then centroid of this triangle moves on the line 2x + 3y = 1. If one of the directrices is x = 4.. where (2004) (a. then c has the value (a) 2 (b) –1 (c) 1 (d) –2. (a) 3x + 2y = 5 (b) 2x – 3y = 7 (2004) (c) 2x + 3y = 9 (d) 3x – 2y = 3. 3 3 3 2 (c) (x – p) = 4qy (d) (y – q) 2 = 4px. (2005) Let P be the point (1. If a ¹ 0 and the line 2bx + 3cy + 4d = 0 passes through the 2 2 points of intersection of the parabolas y = 4ax and x = 4ay. A point on the parabola y 2 = 18x at which the ordinate increases 69. units. A triangle with vertices (4. The normal at the point (bt 12 . The centre of the circle passing through (0. 0). then - 9 9 2 . 0) and (1. (3. The foci of the ellipse x + 2 = 1 and the hyperbola 16 b (a 1 – a 2 )x + (b 1 – b 2 )y + c = 0. (b sint. 72. each of radius 3. b 2) is 70. (–1. If the pair of lines 68. where t is a parameter. If the pairs of straight lines x 2 – 2pxy – y 2 = 0 and 2 2 2 3 1 1 3 x 2 – 2qxy – y 2 = 0 be such that each pair bisects the angle . Then the equation of the circle is 76.a22 + b12 - b2 2 144 81 25 1 (a) 5 (b) 7 (c) 9 (d) 1. then intersect on the yaxis then (a) r < 2 (b) r = 2 (a) 2fgh = bg 2 + ch 2 (b) bg 2 ¹ ch 2 (c) r > 2 (d) 2 < r < 8. If the two circles (x – 1) 2 + (y – 3) 2 = r 2 and ax 2 + 2hxy + by 2 + 2gx + 2fy + c = 0 x 2 + y 2 – 8x + 2y + 8 = 0 intersect in two distinct points. The lines 2x – 3y = 5 and 3x – 4y = 7 are diameters of a circle (2002) having area as 154 sq. (2003) 1 1 1 .71 Two Dimensional Geometry 62. (b) ( a12 + a22 + b12 + b2 2 ) 2 (2003) ( ) ( (c) (a12 + b12 . (d) (2002) (c) 2 2 2 2 between the other pair. 5) is (a) isosceles and right angled (b) isosceles but not right angled (c) right angled but not isosceles (d) neither right angled nor isosceles (2002) 64. –b cost) and (1. . Then the value of b 2 is (a) a12 . lie on the circle (a) x 2 + y 2 + 2x – 2y = 47 x 2 + y 2 = 25. - 2 (b) (a) 65. The locus of any point in the set is (b) x 2 + y 2 – 2x + 2y = 47 (a) 4 £ x 2 + y 2 £ 64 (b) x 2 + y 2 £ 25 (c) x 2 + y 2 – 2x + 2y = 62 2 2 (c) x + y ³ 25 (d) 3 £ x 2 + y 2 £ 9 (2002) (d) x 2 + y 2 + 2x – 2y = 62. If the equation of the locus of point equidistant from the points 2 y 2 (a 1 .a12 - b1 2 ). x y p x y p The equation of its diagonal not passing through the origin is 75. 2bt 1) on a parabola meets the 2 at twice the rate of the abscissa is parabola again in the point (bt 2 . then 74. (2004) (c) (2. (2003) 4 (a) x 2 + y 2 = 2 (b) x 2 + y 2 = 4p 2 66. (2003) (c) abc = 2fgh (d) none of these (2002) ( ( ) ) ( ( ) ) . 0) and (c) (3x + 1) 2 + (3y) 2 = a 2 – b 2 touching the circle x 2 + y 2 = 9 is (d) (3x – 1) 2 + (3y) 2 = a 2 – b 2 . b 1) and (a 2.a22 - b2 2 ) (d) 1 2 ( a + b22 . –4) (a) t2 = -t 1 + 2 (b) t2 = t 1 - 8 2 t1 t1 9 9 . –1). The equation of a circle with origin as a centre and passing through equilateral triangle whose median is of length 3a is is (a) x 2 + y 2 = 9a 2 (b) x 2 + y 2 = 16a 2 2 2 2 2 (a) (3x – 1) + (3y) = a + b 2 2 2 (c) x + y = 4a (d) x 2 + y 2 = a 2 . (2002) (b) (3x + 1) 2 + (3y) 2 = a 2 + b 2 73. The centres of a set of circles. (2003) 77. A square of side a lies above the x axis and has one vertex p at the origin. The point of lines represented by (a) y (cosa + sina) + x (sina – cosa) = a 3ax 2 + 5xy + (a 2 – 2)y 2 = 0 (b) y (cosa + sina) + x (sina + cosa) = a and ^ to each other for (c) y (cosa + sina) + x (cosa – sina) = a (a) two values of a (b) " a (d) y (cosa – sina) – x (sina – cosa) = a. 2bt 2 ). Locus of mid point of the portion between the axes of (a) p = –q (b) pq = 1 x cosa + y sina = p where p is constant is (c) pq = –1 (d) p = q. 2 2 ) (2003) 71. (a) (b) (2. . a sint). 4) (d) 8 2 (c) t2 = t 1 + 2 (d) t2 = -t 1 - 2 . Locus of centroid of the triangle whose vertices are (a cost. (2002) angle a (0 < a < p/4) with the positive direction of x axis. The side passing through the origin makes an 1 1 2 (c) 2 + 2 = 2 (d) 12 + 12 = 4 2 . (2003) (c) for one value of a (d) for no values of a. then c = 2 2 y x 1 = coincide. 67. 0). (2003) t1 t1 63. 48. 68. 47. 21. 16. 39. 36. 20. 54. 51. 71. 30. 8. 56. 69. 57. 59. 17. 40. 43. 67. 63. 41. 61. 26. 31. 18. 64. 14. 70. 24. 74. 12. 10. c) (b) (c) (a) (c) (a) (a) (b) (a) 5. 79. 25. 29. 42. 77. 13. 72. 78. (a) (d) (d) (a) (d) (c) (c) (d) (c) (d) (c) (a) (a) 6. 60. 19. 9. 62. 75. 52. (b) (d) (c) (d) (d) (c) (d) (b) (b) (c) (c) (b) (b) (b) 2. 46. (a) (c) (c) (c) (c) (a) (a) (c) (d) (b) (d) (d) (d) 3. 58. 35. 50. (a) (a) (b) (b) (a) (d) (c) (d) (c) (c) (c) (c) (c) . 22. (a) (b) (d) (a) (a) (b) (a) (c) (b) (d) (d) (d) (a) 4. 49. 65. 38. 53. 23. 66. 7. 27. (2002) (d) none of these (2002) (c) -1 ± 2 Answer Key 1. 15. 34. 44. 32. 28.72 JEE MAIN CHAPTERWISE EXPLORER 78. 37. 55. Two common tangents to the circle x 2 + y 2 = 2a 2 and parabola y 2 = 8ax are angle of measure 45° at the major segment of the circle then value of m is (a) x = ±(y + 2a) (b) y = ±(x + 2a) (a) 2 ± 2 (b) -2 ± 2 (c) x = ±(y + a) (d) y = ±(x + a). 11. If the chord y = mx + 1 of the circle x 2 + y 2 = 1 subtends an 79. 33. 73. 45. (d) (a) (c) (a) (b. 76. 3 \ x . 3) \ 1 + 9 + k 2 – 6k = k 2 Þ k= 5 10 Þ diameter = 3 3 9. 2. 0) As a 2e 2 = a 2 – b 2 = 7 we have equation of circle as (x – 0) 2 + (y – 3) 2 = ( 7 ... r = 2 Þ b = 4 Þ x 2 y 2 + = 1 Þ x 2 + 4 y 2 = 16 16 4 10. ç m ÷ = çç .3y . then m satisfies m 4 + 2m 2 – 24 = 0 From (1).4 + 2 è ø m è 1 Þ 1 ( x - 3) .2 è ø 2 4 x 2 1 . 8 14 + = k Þ k = 6 5 5 (d) : Statement 1 : y 2 = 16 3x . (b) :(x – 1) 2 + y 2 = 1. 1 st solution : x coordinate of incentre ax + bx2 + cx3 2 ´ 2 + 2 2 ´ 0 + 2 ´ 0 = 1 = a+b+c 2 + 2 + 2 2 4 2 = = = 2 . y = m x + x 2 y 2 2 + = 1 . 1) ; B (2. y 1) divides line segment AB in the ratio 3 : 2 1 + m2 Þ (1 + m2 )m2 = 2 which gives m 4 + m 2 – 2 = 0 Þ (m 2 – 1)(m 2 + 2) = 0 As m Î R. 1). (3.0)2 + (0 . 0) is given by {(x – 3) 2 + y 2 )} + ly = 0 As the circle passes through (1. we have 5 2 x1 = 3(2) + 2(1) 8 3(4) + 2(1) 14 = y1 = = 5 5 5 5 2x + y = k passes through P(x 1 .3 = 0 4 3 m 48 m2 = 4+ 2 m1 2 2 ö ÷ ÷ø = 4 + 2m 2 Þ 24 = 2 + m 2 m 2 m 4 + 2m 2 – 24 = 0 .2 2 ÷ tan = 2 . (d) : Foci are given by (± ae. (a) : y = mx + c Þ 2 = m + c Coordinates of P & Q : P (0. (a) : Let a tangent to the parabola be As it is a tangent to the circle x 2 + y 2 = 5/2. –2). r = 1 Þ a = 2 and x 2 + (y – 2) 2 = 4.73 Two Dimensional Geometry 1.(1) (m 2 + 6) (m 2 – 4) = 0 Þ m = ± 2 Þ Þ Statement 2 : If y = mx + 4 3 is a common tangent to m y 2 = 16 3 x and ellipse 2 x 2 + y 2 = 4. (1.3)2 \ x 2 + y 2 – 6y – 7 = 0 5.x = .x - 5 3. 4) P(x 1. 1) happen to be a right angled triangle with vertices as shown. 8.3. c). m 2 = 1 \ m = ± 1 Also m = ± 1 does satisfy m 4 – 3m 2 + 2 = 0 Hence common tangents are 7. 0) and (1. –2) lies on the circle. y 1 ) \ 2´ 5 y = mx + ( m ¹ 0) m æ 5ö çè ÷= m ø (a) : A (1. (b) : The system of circles touches the line y = 0 at the point 6. 0) . 4. A simple calculation shows that (5. The point of incidence is ( 3 . we can determine l which gives 4 + 4 – 2l = 0 \ l = 4 The circle is (x – 3) 2 + y 2 + 4y = 0. Hence the equation of reflected ray is y = \ 3y . Q(–c/m . x = m1y + 4 m1 + 2 2 4 Þ y= y = x + 5 and y = . (c) : Let the equation of the circle is (x – 1) 2 + (y – k) 2 = k 2 It passes through (2.2 4+2 2 2 + 2 A 2 nd solution : r = ( s - a)tan 2 æ 4 + 2 2 ö p =ç . statement 2 is a correct explanation for statement 1.4 + m = m1 m1 m1 2 2 æ æ4 3ö 2 Now. (a) : As the slope of incident ray is - ray has to be 1 3 1 so the slope of reflected 3 . 0) 2 1 c ´ | c | ´ = A Þ c = A 2 m 2m . (a) : The triangle whose sides midpoints are given to be (0. (c) : Let P be (y 2 . 3) is the point of contact and equation of the tangent is y = 3. 1 1 . (b) : The circle is x 2 + y 2 – 4x – 8y – 5 = 0 Þ (x – 2) 2 + (y – 4) 2 = 5 2 Length of perpendicular from centre (2.0) = 2 2 Þ Þ a a = ±c 2 2 4 2 + 12 12. Þ 2 3 4 Then the equation to line k is 4x – y = –3 The distance between lines k and c is 20 + 3 a = c . dx x 3 As the tangent is parallel to xaxis. The equation of directrix is x = –1.2 + = A 2m 2m 2 m dA =0 dm 1 2 1 2 Þ =0Þ = Þ m 2 = 4 Þ m = ± 2 2 m2 2 m 2 Q 16. 1) lies on it Þ 2 x 2 y + = 1 2 2 a b 9 + 1 = 1 a 2 b 2 Also b 2 = a2 æç 1 . 4 x .m | < 5 Þ (10 + m) < 25 5 Þ –25 < 10 + m < 25 Þ –35 < m < 15 19..= 1. x 2 + y 2 = c 2 çè x . (d) : y = x + Let P(t 2 . 17... we have 8 1= 0 Þ x 3 = 8. 0) & radius = a and c è 2 ø 2 Thus .1 / 2 t = - 2 t The shortest distance between the curves occur along the common normal. 4) on the line 3x – 4y – m = 0 should be less than radius. O divides PQ in the ratio of OP : OQ which is 2 2 : 5 but it fails to divide triangle into two similar triangles. (d) : The centres and radii are 2 2 æ aö a ±c çè ÷ø + (0 .(i) Again differentiating w. we get y ¢ = c1c2 ec2 x = c2 y . t) be a point on y 2 = x The slope of normal at P = - 1 1 the slope of tangent at P = . y = 2 + 2 = 2 + 1 = 3 2 Thus (2. i.t. x.= 4. 0 ö÷ and (0. hence this is the locus of point P. \ c = - .74 JEE MAIN CHAPTERWISE EXPLORER Þ (2 . (c) : y = c1 ec2 x Differentiating w. \ | a| = c.b 2 ) Minimum value = 1 × = 23 17 18. (d) : 4 a 2 1 4-1 3 = × = 4 × 1 2 4 2 14. Thus –2t = –1 Þ t = 1/2 4 Thus the point P is (1/4.÷ø + y = 2 4 13 32 32 13 8 + =1 Þ = 1=Þ b = - 20 5 b b 5 5 x y Thus the line is .(ii) .+ 1 ) ( = 4 2 = 2 3 3 2 = 8 4 2 20..4 m + 4 m 2 = A Þ =A Þ .y = 20 5 20 The equation of line parallel to 4x – y = 20 has slope 4.r.. we have æ aö a 2 2 . the perpendicular tangents intersect at the directrix. Centre æç a . (d) : As the line passes through (13.e.a . (c) : From a property of the parabola. 11. (c) : Let the ellipse be (–3. b 2 = 32 7 5 2 The equation to ellipse becomes 3x 2 + 5y 2 = 32 15. 13. 1/2) x 2 The required shortest distance dy 8 = 1 - On differentiation.16 . 32). \ x = 2 3 x 4 So.t.2 ö÷ Þ 5b 2 = 3 a 2 è 5 ø Upon solving we get a = 32 .y + 1| y + 1| = y 2 – y + 1 (Q y 2 – y + 1 > 0 ) (4 ac . 2 2 As |y 2 – 3 c |6 .r.m) 2 m 2 . y ) Perpendicular distance from P to x – y + 1 = 0 is | y 2 . (a) : In triangle OPQ.. x y¢¢ = c2y¢ . k) then P¢ º (2h – a. 0) As points A. 1) lies on the above ellipse 4 1 1 1 3 + = 1 Þ = 1 . is (2. (c) : The vertex is the mid point of FN. Q and (1. 1). we have 23. the corner of the Þ x + 5y + 2p – 5 + p 2 = 0 a > ae ) e Remark : The question should have read “The corresponding directrix” in place of “the directrix”.e ö = 4 Þ a æ 2 . of the circles S1 : x 2 + y 2 + 3x + 7y + 2p – 5 = 0 and 7ö k + 1 ö æ 15 k 2 .ö = 4 è e ø è 2 ø Þ a× Þ 9[(x – 1) 2 + y 2 ] = (x + 1) 2 + y 2 Þ a ..y = . 1) doesn’t lie on PQ. 0) F (0.. the coordinate of P¢ is given by P¢ º (2 × – 1 – 1.. 1 + 5 + 2p – 5 + p 2 ¹ 0 x+ y = | 2 x | Þ y dy = x or y dy = - 3 x dx dx dx 2 2 Þ p + 2p + 1 ¹ 0 Þ (p + 1) ¹ 0 \ p ¹ –1 ydy = xdx or ydy = –3xdx Thus for all values of p except ‘–1’ there is a circle passing through After integrating. the circumcentre of triangle ABC is (5/4. C lie on this circle. so its equation is x 2 y 2 + = 1 16 b2 x = 2 1 the slope of the original line PQ l Bisector 1 = ( k .ae = 4 e y2 x2 y 2 3 x 2 = + c or =+c 2 2 2 2 Þ x 2 – y 2 = –2c or 3x 2 + y 2 = 2c Þ x 2 – y 2 = c 1 or 3x 2 + y 2 = c 2 . The midpoint = è 2 2 ø A(2. 3) . (a) : The radical axis. b) and C(h.4 k - 1 P(1.(i) P. –2) Remark : If P be (a. 0) V l = - rectangle in 1st quadrant.1) æ 0 4 Þ = è è 2ø 2 ø 2 2 Þ k 2 – 1 = 15 Þ k 2 = 16 \ k = ± 4.75 Two Dimensional Geometry From (i) and (ii) upon division y¢ y = y ¢¢ y ¢ Þ Þ y ¢¢y = ( y ¢ ) 2 which is the desired differential equation of the family of curves. As C is the mid point P and P¢. 22. Q and (1. – 2).. 21. 1).1 = ( k .e. 1) it’s necessary Y .( X - x ) Þ G º ç x + y × . (b. we get . (a) : The slope of i. S2 : x 2 + y 2 + 2x + 2y – p 2 = 0 is S1 – S2 = 0 28. y) is dy ö æ dx If there is a circle passing through P. 4) æ k + 1 7ö . which reduces to x 2 + y 2 - (note that 1 1 Þ a æ . that is. 0) N 27.1) æ x è y . y) such that PM 1 = where M º (1. 0) (–1. (a) : The given ellipse is 3 8 = 4 \ a = 2 3 25. y = – 4 satisfies it. Again the ellipse circumscribing the rectangle passess =- through the point (4. vertex = (1. 0). (a) : Let P be a general point (x.2ø è 2 ø x 2 3 2 + y = 1 Þ x 2 + 12y 2 = 16 16 4 As x = 0. B. 0). dy i. 2 × – 2 – 0) C P¢ P º (– 3.e.0 ÷ dy dx ø è and sufficient that (1.1) 2 + y 2 1 = we have ( x + 1) 2 + y 2 3 Þ x 2 + y 2 - 5 x + 1 = 0 2 The locus is a circle with centre (5/4. 0) and N º (–1.= 16 b 2 4 4 b 2 Þ b 2 = 4/3 Thus the equation to the desired ellipse is The equation to the bisector l is Þ 7ö k + 1 ö æ = ( k .1) 3 . (b) : Obviously the major axis is along the xaxis The distance between the focus and the corresponding directrix a = - ae = 4 e (2. 24. (d) : The centre C of the circle is seen to be (– 1. the point A. Q (k . 0) PN 3 ( x . 2k – b) 8x 2 + 8y 2 – 20x + 8 = 0 10 x + 1 = 0 4 x 2 y 2 + = 1 4 1 26. – 4) (1. c) : Equation of normal at P(x. which in the case of intersection of the circles is the common chord. (i) . k) be mid point of AB Ð AOM = 60° Now OA = OB = 3 and OM ^ AB (By properties of circle) Now OA = h 2 + k 2 . B (1. 1) 33.> 0 . 3 3) 120° 60° Q(0.0). 16 ÷ è ø \ M(h. (d) : Let AB is chord of circle and M(h.0) Þ if a varies then the abscissa of foci remain constant. we get R (3.. (ii) Solving (i) and (ii) we get 1 < a < 3 2 36.. (ii) (o. 1) C (2.. k = y) . 0) 2 x 2 + y = 1 ( a > b ) a 2 b2 Given 2b = 8 and 2ae = 6 38. –1) \ equation of circle is (x – 1) 2 + (y + 1) 2 = 7 2 Þ x 2 + y 2 – 2x + 2y – 47 = 0 Þ Þ x 2 = } 3 3 35 a y . o ) b 4 By (i) and (ii) we have ae = 3 b 2 = 16 e 2 Þ a 2 9 16 1 . then P lies on directrix x = –2 of y 2 = 8x Hence P º ( - 2. k ) Þ 3 x + y = 0 2 Q pr 2 = 49 p r = 7 Point of intersection of AB and PQ is (1. (a) : Let (1 – m 2 )x 2 = 0 Þ m = ± 1. (c) : Given lines are y = 2 ( x > 0) and y = 3x (x > 0) using (a. k be the locus of the vertex of family of parabola 3 2 2 y = a x + a x . (b) : Q b 2 = a 2 (e 2 – 1) Þ sin 2 a = cos 2 a (e 2 – 1) Þ tan 2 a + 1 = e 2 Þ e 2 = sec 2 a Vertices º ( ± cos a ..e. where x = h + ...0) º ( ±1.0) Coordinate of focii are ( ± ae. 0) xy = 0 are y = ± x Put y = ± x in my 2 + (1 – m 2 )xy – mx 2 = 0. y = k + 4 a 16 a 3 hk = çæ -3 ÷ö çæ -35 a ÷ö è 4 a ø è 16 ø 105 hk = 64 105 xy = 64 2 (taking h = x. 3.1 | ´ 1 = 1 2 k = –1. 34. k ) A F¢ (a.76 JEE MAIN CHAPTERWISE EXPLORER 29. (i) 2 2 and a – 3a < 0 .. Angle bisector Second method Equation of bisectors of lines P (–1. a 2 ) in these lines a a 2 . 32.e2 = e 2 ( Q b 2 = a 2 (1 – e 2 ) as a > b) Þ 9 3 e = Þ 5 Þ O (o.1 ³ 0 Þ k ³ .. D ³ 0 1 2 k . 2 30. o ) 31. 1) then (–1 – h) 2 + (1 – k) 2 = k 2 Þ h 2 + 2h – 2k + 2 = 0. (a) : Let h.m ) = 0 m B Þ m = ± 1. (b) : Let P is the required point. { Þ æ -3 -35 a ö vertex is ç 4 a .2 a 3 2 \ Þ a 3h 2 a 2 h + = 2 a 3 2 3k 3h 6 = h 2 + 2 a a 2 a3 k= 2 Þ 3æ 35a ö æ 3 k+ = h + ö 16 ø è 4 a ø a 3 è i. (d) : Equation of circle (x – h) 2 + (y – k) 2 = k 2 It is passing through (–1. o ) 2 . x 35. (d) : Let OA = r Given area = 49 p A P co efficient of xy co efficient of y 2 3x – 4y = 7 2x – 3y = 5 O 2 \ Sum of slopes = - (1 . b ) (o . (c) : Slope of the required angle bisector is tan120° = - 3 Hence equation of the angle bisector is y = . (a) : Sum of the slopes = - Þ h 2 + k 2 = 9 9 2 2 Þ x + y = 4 4 37. OM = r cos q h + k = 3cos60 ° 3 h 2 + k 2 = 2 A¢ B F A 39..3( x - 0) A (1. (a) : 1 ´ | k . 2 ÷ a ø è B (a. if r 1 . 0) b c . 2 44.1 = m 1 (say) Slope of BF = Þ a 2 + a + 1 = 0 and ae b a b 2 = m 2 (say) Slope of BF ¢ = 1 3 Þ a+ + = 0 and –(c – d + b) = b/a - ae 2 4 Now m 1 × m 2 = –1 Þ b ´ b = . (c) : Let locus of the centre of circle be (a.. (c) : Given y = ax + b and 1 .a2 a 2 )( -b2 - b 2 ) Now 5ax + (c – d)y + a + 1 = 0 and Þ a 2 a 2b2 = . It means given line is the equation if (*) are equal of common chord and the equation of common chord of two \ keeping D = 0 in (*) we have intersecting circle is S 1 – S 2 = 0 4a 2 a 4b2 = 4( b2 . (c) : Let locus of centre be a. b) is given by x + y = 1 a b where a = 2a = 6 and b = 2b = 8 x y \ required equation be + = y 6 8 Þ 4x + 3y = 24 tan q = If C 1 . 0) Þ 1 – e 2 = e 2 . 3) and radius is 2. 41.. b ) B Mid point (a. b then according to given. e = ± at A(a. 0).b 4 + a 2 a 2 b 2 5x + by – a = 0 represent same equation. (c) : ax 2 + 2( a + b ) xy + by 2 = 0 (*) Let q be the angle between the lines represent by * \ tan q = F¢ y 2 x = 1 a 2 b 2 Þ a 2 + b 2 – 6b + 9 = b 2 + 4 + 4b and a 2 + b 2 – 6b + 9 = b 2 – 4b + 4 Þ a 2 – 10b + 5 = 0 and x 2 = 2b + 5 \ b2 x 2 .d + 1 = . ae . 0) B (0.y = b . 2e 2 = 1 . \ 5 a = c . Þ a2 a 2 b2 = b2 (b 2 + b2 ) Þ a 2 a2 = b2 + b 2 2 2 2 2 Þ a x . 0) 45. 0) x Internal touch.d = a + 1 5 b -a 43. F(ae.77 Two Dimensional Geometry 40. b ) (a. A C 1 C 2 = r 2 ± r 1 Þ (a – 0) 2 + (b – 3) 2 = |b ± 2| 2 42.ab a + b y 2 ( a + b ) 2 . (a. r 2 are radii of circles then 2 h 2 . Þ a 2 a 2b2 = ( b2 . (a) : As the line passes through P and Q which are the point Now the line y = ax + b will be tangent to circle if both roots of intersection of two circles. b ) æ b 2 ö 2 ç Q e = 1 . (0.a 2 y 2 = a 2 b2 Þ x 2 = 10y – 5 and x 2 = 2y – 5 Þ b2 x 2 . C 2 are centres of the circles with radii r 1 . b). This case does not exist as centre of circle is (0.(*) 46. 4) o F x A (a . 5) C ¼ = Area of DBC ¼ } {3 area OAB (a.a2 ( ax + b ) 2 = a 2 b2 Both are parabolas but x 2 = 2y – 5 does not exist.a 2 a 2 )( -b2 a 2 - a 2 b2 ) = 5ax + (c – d)y + a + 1 = 0. (b) : F ¢(–ae. \ No value of a exist. b ) (0. b) = M (3.ae ( ) . by using y = aa + b 2 2 2 2 2 2 2 2 2 Þ x (b – a a ) – 2a abx + (–b a – a b ) = 0 .b2b2 + b2 a 2 a 2 .ab \ tan45 ° = a+b Þ 3a 2 + 2ab + 3b 2 = 0. r 2 respectively then (C 1 C 2 ) 2 = r 1 2 + r 2 2 Þ a 2 + b 2 = p 2 + (a – a) 2 + (b – b) 2 Þ p 2 + a 2 + b 2 – 2aa – 2bb = 0 Þ 2ax + 2by – (a 2 + b 2 + p 2 ) = 0. b ) B Case (i) Now q = 45 {Q area of one sector = 3 time the area of another sector} q D 2 ( a + b) 2 . (c) : Now the equation of line which meet the xaxis and y axis y Þ b 2 = a 2 e 2 Þ a 2 – a 2 e 2 = a 2 e 2 (0.ab a +b (0.1 d – b – c = +b/a has no solution.. so \ equation of directrix x = e a 2 + b 2 + 2ga + 2fb + 4 = 0 a Now replacing g. (b) : Intercepts made by the lines with coordinate axis is y (–3b/a .. y¢) be point of locus mid point of PQ.. 0) and Common intercept is (0. y 3 ) æ x1 + x2 + x3 y1 + y2 + y3 ö . 2) x 3 = 3 × 2 – 1 = 5 G y 3 = 2 × 2 – 1 = 3 P(1. –1) Also circumference of the circle be given 2pr = 10p \ r = 5 \ Required equation of circle is (x – 1) 2 + (y + 1) 2 = 5 2 or x 2 + y 2 – 2x + 2y – 23 = 0 (0.. 0) 52.. (d) : The equation ax 2 + 2hxy + by 2 = 0 a 2 ´ 3 2 Further b = by (*) = (y – m 1 x) (y – m 2 x) 4 1 2h 4 ´ 3 Þ m 1 + m 2 = - = . (–3b/a.33+ 5 . 0) Þ = h. 37 ) . –3/2) 55.(*) Þ ç ÷ = 1 – 1/4 = 3/4 \ 2g 1 g 2 + 2f 1 f 2 = c 1 c 2 (where (–g. –3/2) and (3a/b.. 1 \ A(0. c = 1/3 and a = 1/2. 1 + 33 + 3 ) = (1.. (2) x B Solving (1) and (2) we get y = C A x = 0.. B(4a.2 a + y 2 = 4 is 2 æ b ö x 2 + y 2 + 2gx + 2fy + c = 0 ... (c) : As per given condition centre of the circle is the point of intersection of the 2x + 3y + 1 = 0 and 3x – y – 4 = 0 \ centre is (1.. k ) Þ (2k) 2 = 8(2h – 1) 2 2 Þ y = 2(2x – 1) Þ y – 4x + 2 = 0.. (c) : The point of intersection of parabola’s y 2 = 4ax and x 2 = 4ay are A(0. 1 \ y = 0. y ¢ ) . 0). –3/2). (1) and line be y = x . y 2 ) 47. –3/2) and (0. 0). (b) : Equation of directrix x = 4 which is parallel to yaxis xaxis so axis of the ellipse is xaxis. b ® y \ (p – x) 2 = 4q(y) which required locus of one end point of x 2 y 2 + = 1 (a > b) the diameter. 0). y = –2 (1. b = 1/2. (c) : Given circle x 2 + y 2 – 2x = 0 .. 1) and equation of circle in the diameter form is (x – 0) (x – 1) + (y – 0) (y – 1) = 0 Þ x 2 + y 2 – (x + y) = 0 54. q) Þ (p – a) 2 = 4qb 51. 2 2 \ x¢ = 2h – 1. ÷ \ Centroid G = ç 3 3 è ø = (1 .. b). (0.. y respectively \ 4 = e \ 2ax + 2by – (a 2 + b 2 + 4) = 0 Þ a = 2 (³ e = 1/2) 57. B(1. (b) : Let the equation of circle cuts orthogonally the circle x 2 Again e = 1/2 and e 2 = 1 . (c) : Equation of circle AB as diameter is given by (x – p) (x – a) + (y – q) (y – b) = 0 Þ x 2 + y 2 – x(p + a) – y (q + b) + pa + qb = 0 .(1) Now (1) touches axis of x so put y = 0 in (1) we have x 2 – x(p + a) + pa + qb = 0 .. 1) M 2 (3. 0) (–3a/b. (d) : \ x 2 = 2(– 1) – 1 = – 3 y 2 = 2 × 2 – 1 = 3 M(–1. f by x.(2) and D = 0 in equation (2) B(a. 2) R(x3. c = 1/4 Putting these value in the given equation we get x + 2y + 3 = 0 and 2x + 3y + 4 = 0 .. y ¢ + 0 x ¢ + 1 P(1. Let equation of ellipse be Now a ® x. = k .(i) . b = 1/3.78 JEE MAIN CHAPTERWISE EXPLORER Q(x2. –f) are point of locus) è a ø Also the equation of one directrix is x = 4 Þ c = – 4 a Again circle (i) passes through (a. a 2 b 2 2 b 56. (d) : Let M(x¢.(*) b 2 = = 3 4c b 4 3 x 2 y 2 m 1 m 2 = c Hence equation of ellipse is 2 + 2 = 1 2 a b and 3x + 4y = 0 Þ m 1 = –3/4 2 2 x y 2 + Þ = 1 or 3x 2 + 4y 2 = 12 \ m 2 = - 4 3 c 50. b) \ (p + a) 2 = 4[pa + qb] A(p. Q(x¢. 49. 4a) as the line 2bx + 3cy + 4d = 0 passes through these points \ d = 0 and 2b(4a) + 3c(4a) = 0 Þ 2b + 3c = 0 Þ (2b + 3c) 2 + d 2 = 0 53. y¢) lies on y 2 = 8x M( h. 48. –2) is required point on the line.(*) Solving the equations of (*) we have x = 1. y¢ = 2k Now (x¢. (d) : Let us take the two set of values of a = 1. (c) : According to the problem square lies above xaxis y – y 1 = - (x – x 1 ) dy Now equation of AB using two point form.. k) by (x. (c) : Given equations are 2 3 - 2 1 x 2 – 2qxy – y 2 = 0 . b) lies on 2x + 3y = 1 \ 2a + 3b = 1 Joint equation of angle bisector of the line (i) and (ii) are same Now centroid of ABC is x 2 . Þ 2y = dt 2 dt Again pR 2 = 154. (b) : Coordinate of centre may be (1.(i) x 4 3 O A(a. y) we get choice (a) is correct. 61.79 Two Dimensional Geometry Now by (*) we have æ 3 2 ö 1 .b cos t + 0 B (b.2 ö æ 2 + (-2) + h -3 + 1 + k ö \ = . y 2 81 9 = = 18 72 8 9 So the required point is x = .(1) (c) : Let locus of point C(h.sin a) Þ y – a sin q = (x – a(1 – cos q)) (y – a sin a) = [x – a cos a] cos q a (cos a + sin a ) which passes through (a...(ii) Þ = 1 a 1 + a By squaring (i) and (ii) then adding we get Þ a = ± 2 (as a = 2 b = –3 (3h – 1) 2 + (3k) 2 = a 2 (cos 2 t + sin 2 t) + b 2 (cos 2 t + sin 2 t) and a = –2 b = 1) Replacing (h. equidistant from (a 1 . 0) 3 k = a sin t – b cos t . taking ratio of their coefficients 3 è 3 ø 1 -p Þ 2h + 3k = 9 Þ 2x + 3y = 9 \ = q 1 (c) : The equation of normal at q is Þ pq = –1 1 66. (a) : Let (h.. 62. \ h = (4. so dy 18 dy choices (a) and (d) are out of court. b 2 ) is given by (a – a 1 ) 2 + (b – b 1 ) 2 = (a – a 2 ) 2 + (b – b 2 ) 2 (a) : If m 1 and m 2 are slope of the lines then by given condition Þ a 1 2 + b 1 2 – 2a 1a – 2b 1b – a 2 2 – b 2 2 + 2a 2a + 2b 2b = 0 m 1 + m 2 = 4m 1 m 2 Þ 2(a 2 – a 1 )a + 2(b 2 – b 1 )b + a 1 2 + b 1 2 – b 2 2 – a 2 2 = 0 2 c 4 - = Þ Þ (a 2 – a 1 )x + (b 2 – b 1 ) y + 7 7 1 Þ c = 2 (a 2 + b 1 2 – a 2 2 – b 2 2 ) = 0 2 2 2 1 By using ax + 2hxy + by = 0 1 - 2h a Þ c = - [a 1 2 + b 1 2 – a 2 2 – b 2 2 ] Þ m 1 + m 2 = and m 1 m 2 = 2 b b 64. We get dx y – y 1 = m(x – x 1 ) sin q a (cos a . 60..(2) As (a. a + b = –1 . –1 satisfies the given equations of diameter..y 2 xy æ h k . (d) : Let a. R 2 = 49 \ R = 7 . k) be the coordinate of centroid (d) : Given OA + OB = –1 y a cos t + b sin t + 1 i.. b is the point of locus. k) and centroid (a. –1) or dt dt (–1.ç + ÷ = è 4 c ø 4c 1 2 1 8 3 3 + + - = Þ Þ - = 4c c 4c 4 c 4 4 9 3 - = Þ \ c = –3 4c 4 58.. x y x y + = 1 so equation are - = 1 and 65. 1 + 1 -q ç ÷ or çè 3 3 ÷ø 3 3 è ø Þ qx 2 + 2xy – qy 2 = 0 . b) x 2 – 2pxy – y 2 = 0 .e.. b 1 ) and (a 2 . 3) 3 \ equation of the line be a sin t . 0) k = x y 3 = 1 a 1 + a Þ 3h – 1 = a cos t + b sin t . y = 9/2 8 Þ y = 9/2 \ x = 63. 1) but 1.. 59. 0) Þ y(cos a + sin a) + x(cos a – sin a) = a sin a(cos a + sin a) + a cos a (cos a – sin a) dy dx (d) : Given y 2 = 18x and = 2 = a(sin 2a + cos 2a) dt dt = a(1) dy dx \ 2y = 18 67.(3) æ h ö 3( k - 2) \ 2 ç ÷ + = 1 Now (2) and (3) are same. –1) and r 2 =t 2 = 3 Again product of the slope of AC and AB 1 = × (–5) = –1 5 Þ AC b AB Þ right at A 72. ç . (a) : AB = 26 .0) æ 2 ö = 4 ç + f ÷ 12 15 ´ è 4 ø = 3 Again foci = a 1 e 1 = x 2 + y 2 = 9 x 2 + y 2 + 2gx + 2fy = 0 5 12 2 9 = 1 + 4f Common tangent \ focus of hyperbola is (3. 5) A O C(–1. 2 2 x y \ equation of common tangent is 2gx + 2fy –9 = 0 and distance - and eccentricity for = 1 is 144 81 from the centre of circle x 2 + y 2 = 9 to the common tangent is 25 25 equal to the radius of the circle x 2 + y 2 = 9 2 2 ì a1 + b 1 0 + 0 . 2bt 1 ). k) p sin a = 2 k sin 2a + cos 2a (4. –1) \ ABC is isosceles X Þ = p2 4h 2 + p 2 4 k 2 y 0. 74.80 JEE MAIN CHAPTERWISE EXPLORER \ Required equation of circle be (x – 1) 2 + (y + 1) 2 = 49 Þ x 2 + y 2 – 2x + 2y = 47 68. (c) : Given median of the equilateral triangle is 3a. (d) : (x – 1) 2 + (y – 3) 2 = r 2 \ C 1 (1. AC = –f = ± 2 26 Y B(3. (d) : Since the normal at (bt 1 2 . 2 h \ b 2 9 = 1 – e 2 = 1 – or 16 16 Þ b 2 = 7 71. (b) : Let equation of circle be x 2 + y 2 + 2gx + 2fy + c = 0.9 ï e 1 = \ = 3 ï a 1 2 4 g 2 + 4 f 2 í 81 15 ï Þ 3 2 = 4(g 2 + f 2 ) 9 3 1 + = 3 ïî \ e1 = 144 12 1 (0. 0) As their foci are same \ 4e = 3 \ e = 3/4 f = ± 2 2 æ b ö b 2 = 1 – ÷ è a ø 16 \ \ e 2 = 1 – ç æ1 Centre of the required circle be ç . t1 As it passes through (0.1) 2 + (3 + 1)2 = 5 In D LMD. 0) so –g = . (LM) 2 = (LD) 2 + (MD) 2 Now for intersecting circles 2 r 1 + r 2 > C 1 C 2 and |r 1 – r 2 | < C 1 C 2 LM ö 2 = 9a 2 + æ (LM) ç ÷ \ Þ r + 3 > 5 and |r – 3| < 5 è 2 ø Þ r > 2 and –5 < r – 3 < 5 3 Þ r > 2 and –2 < r < 8 \ (LM) 2 = 12a 2 Þ (LM) 2 = 9a 2 4 Þ r Î (2. 0 ÷ ç è cos a ø = 1 x 2 + 1 y 2 = 4 p 2 x . 2bt 2 ) L 2 = (3a – R) 2 + R ç ÷ M R \ t 1 x + y = 2bt 1 + bt 1 3 passes through (bt 2 2 . (b) : Eccentricity for 2 + 2 = 1 1 a b and as it passes (1. 3) and r 1 = t 1 = r (x – 4) 2 + 1(y + 1) 2 = 9 \ C 2 (4. on parabola y 2 = 4bx 2 æ LM ö meet the parabola again at (bt 2 2 . so C 1 C 2 = (4 . k) is ç ÷ 2 cos a 2 sin aø è p \ cos a = . 0) = (4e.0) æ p ö . (OM) 2 = (OD) 2 + (MD) 2 69. 0) = (ae. 2bt 2 ) è 2 ø O Þ R 2 = 9a 2 + R 2 – 6aR + 3a 2 Þ t 1 bt 2 2 + 2bt 2 = 2bt 1 + bt 1 3 D Þ 6aR = 12a 2 Þ t 1 (t 2 2 – t 1 2 ) = 2(t 1 – t 2 ) R = 2a N Þ t 1 (t 2 + t 1 ) = –2 So equation of circle be 2 (x – 0) 2 + (y – 0) 2 = R 2 = (2a) 2 Þ t 2 + t 1 = Þ x 2 + y 2 = 4a 2 t1 2 Þ t 2 = - – t 1 73. . 8) Again in triangle OMD. 0) so c = 0 x 2 y 2 70. è2 p p ö æ . 0) 2gx + 2fy = 9 \ f 2 = 2 so focus of ellipse (ae. 0) ö æ 1 ö 2 ÷ . p sin a Midpoint (h.2 ÷ ø è2 ø (0. (d) : \ (h. 2 is b 2 = a 2 (1 –e 2 ) Now x 2 + y 2 + 2gx + 2fy = 0 and x 2 + y 2 = 9 touches each other. 0) to the tangent line = Radius of circle î by1 + f = 0 a h g h b f g f c On compairing the equation given in (*) we get bg = fh and bg 2 = fgh .(1) Þ m = 1 ± 2 and –1 ± 2 Now say point of intersection on y axis be (0..(1) a ¶s ¶s = mx + = 0 = m ¶ x ¶ y and \ y 2 = 8ax = 4(2a)x ¶s \ Equation of tangent to parabola \ = 0 Þ ax + by + g = 0 Þ ¶ x 2a ... b = 0 a + b tan 45° = ± 2 m 2 ... Þ x 2 (1 – m 2 ) + 2mxy = 0 Now tan q = ± 77.. (a) : As s = ax 2 + 2hxy + by 2 + 2gx + 2fy + c = 0 represent a pair of line ìï a = 1 ....(3) Now using (2) and (3) in equation (1) we have 1(1 – m 2 ) \ Þ Þ Þ 2 a = ± 2 a ´ m 1 (0 0) 1 + m 2 2 2 m (1 + m ) – 2 = 0 (m 2 – 1) (m 2 + 2) = 0 m = ± 1 Required equation of tangent y = mx + = ± (x + 2a) y = mx + 2a m 2a m . (a) : Using fact : Pair of lines Ax 2 + 2hxy + By 2 = 0 are b to each other if A + B = 0 Þ 3a + a 2 – 2 = 0 Þ a 2 + 3a – 2 = 0 Þ There exist two value of a as D > 0 -3 ± 17 2 76. \ Equation of such circle be (x – a) 2 + (y – b) 2 = 3 2 Þ a 2 + b 2 – 2ax – 2by + 25 = 9 Þ x 2 + y 2 – 2x 2 – 2y 2 + 25 = 9 Þ x 2 + y 2 = 25 – 9 Þ x 2 + y 2 = 16 and x 2 + y 2 = 25 Þ 4 £ x 2 + y 2 £ 64 \ a = abc + 2fgh – af 2 – bg 2 – ch 2 = 0 Þ (abc – af 2 ) + (fgh – bg 2 ) + fgh – ch 2 = 0 Þ 0 + 0 + fgh – ch 2 = 0 \ ch 2 = fgh Now adding (2) and (4) 2fgh = ch 2 + bg 2 .81 Two Dimensional Geometry 75. (b) : Let common tangent to the curves be intersection of pair of line be obtained by solving the equations y = mx + c . b) is the centre of the circle whose radius is 3. (c) : Equation of chord y = mx + 1 Equation of circle x 2 + y 2 = 1 Joint equation of the curve through the intersection of line and circle be given by x 2 + y 2 = (y – mx) 2 .m 2 2 h 2 .. y 1 ) and point of 79.(2) 2 2 Again ax + 2hxy + by + 2gx + 2fy + c = 0 meet at yaxis \ x = 0 Þ by 2 + 2fy + c = 0 whose roots must be equal \ f 2 = bc af 2 = abc ..(2) y = mx + ¶s m and = 0 Þ hx + by + f = 0 Þ ¶ y which is also tangent to the circle 2 x 2 + y 2 = 2a 2 = ( 2a ) ì hy1 + g = 0 (*) í Now Distance from (0.ab where í ïî h = m.0 1 - m2 Þ = ± 2m 2 = 0 \ Þ m ± 2m – 1 = 0 Þ m = ± 1 ± 2 or abc + 2fgh – af 2 – bg 2 – ch 2 = 0 .. (a) : Let (a.(4) 78. If the straight lines x -1 y . (a) 2/5 (c) 2/3 x . 3. then the integer k is equal to (a) – 2 (b) – 5 (c) 5 (d) 2 (2008) .2 z . . 7) (c) (–5. 6. 6. Statement2 : The plane x – y + z = 5 bisects the line segment joining A (3. 5 5 5 (b) 6 - 3 2 . 5) (b) (5. If the angle between the line x = y . –3. Statement2 is true; Statement 2 is not a correct explanation for Statement1. If the lines x . 4) in the plane x – y + z = 5. (b) 5/3 (d) 3/2 (2011) A line AB in threedimensional space makes angles 45° and 120° with the positive xaxis and the positive yaxis respectively. 6) and B(1. 3). –15) (d) (6. Statement2 is true; Statement2 is not a correct explanation for Statement1. (a) Statement1 is true. Statement2 is true. (c) Statement1 is true. 1. 6) is the mirror image of the point B (1.5 = = If the lines x .1 z . 7 7 7 (c) -6 - 3 2 . Statement2 is false. b) equals (a) (–6.1 y . If AB makes an acute angle q with the positive zaxis. Statement2 is true. Then (a.2 y . 7 7 7 (d) 6.1 = = and = = k 2 3 3 k 2 intersect at a point.2 = = bisects the line 1 2 3 segment joining A(1.1 z . (c) – 1 (d) 2/9 (2012) Statement1 : The point A(1. (d) Statement1 is true. 5 7 9 3 (b) (c) (d) (a) 2 2 2 2 (2013) 3. coplanar. (c) Statement1 is true.1 y + 1 z . (b) Statement1 is false. . 0. –3.3 z .2 y . then k can have (a) exactly three values (b) any value (c) exactly one value (d) exactly two values (2013) 2. (b) Statement1 is true. 0.1 z + 2 = = line lie in the plane 3 - 5 2 x + 3y – az + b = 0. An equation of a plane parallel to the plane x – 2y + 2z – 5 = 0 and at a unit distance from the origin is (a) x – 2y + 2z – 1 = 0 (b) x – 2y + 2z + 5 = 0 (c) x – 2y + 2z – 3 = 0 (d) x – 2y + 2z + 1 = 0 (2012) 4. 1. . 4). (2010) Statement1 : The point A(3. The direction cosines of the vector are (a) 6 - 3 2 . 3) in the line x y . then q equals (a) 30º (b) 45º (c) 60º (d) 75º.k z = = = = intersect. 3. 2 respectively. 2 (2009) 11.4 and are 1 1 2 k - k 1 7.4 z .3 y . 7) is the mirror image of the point B(1. (2011) 6.2 = = 1 2 3 Statement2 : The line : x y . and 2 3 4 1 2 1 then k is equal to (a) 9/2 (b) 0 5. Statement2 is true; Statement2 is a correct explanation for Statement1. The projections of a vector on the three coordinate axis are 6. (d) Statement1 is false. –17) (2009) 10.82 JEE MAIN CHAPTERWISE EXPLORER THREE DIMENSIONAL GEOMETRY CHAPTER 13 1. (2010) Let the x .3 x . (a) Statement 1 is true.1 z .2 = y . Distance between two parallel planes 2x + y + 2z = 8 and 4x + 2y + 4z + 5 = 0 is 8.1 x . 7) and B(1.3 and the plane = 2 l æ 5 ö then l equals x + 2 y + 3 z = 4 is cos -1 ç è 14 ÷ø 9.3 = z . Statement2 is true; Statement2 is a correct explanation of Statement 1. Statement2 is false. a) p p p p (b) 2 (c) 6 (d) 3 .4 x . Then 2 2 ø 14. (2005) 28. z = c ¢y + d will be perpendicular. then cos 2q equals 18. The angle between the lines 2x = 3y = –z and 6x = –y = –4z is (a) 90º (b) 0º (c) 30º (d) 45º. a.kˆ . then the angle that the line intersection are given by makes with the positive direction of the zaxis is (a) (3a. z = cy + d and x = a ¢y + b ¢. –3) (c) x – y – z = 1 (d) 2x – y – z = 1.1 y . (a.. Distance between two parallel planes x – 2y = 0 is 2x + y + 2z = 8 and 4x + 2y + 4z + 5 = 0 is (a) 7/2 (b) 5/2 19 ö æ 17 (a) ç . with each of the x and z axis. If the vectors c lies in the plane of a and x 2 + y 2 + z 2 – 10x + 4y – 2z = 8 then a equals r b . is such that è 3 ø 3 sin 2b = 3sin 2q. . 9. 2a. If L makes an angle a (b) (a) 9 3 3 with the positive xaxis. The coordinates of each of the points of of each of xaxis and yaxis. 15.5 = = = = and 1 1 - k k 2 1 are coplanar if (a) k = 1 or –1 (b) k = 0 or –3 (c) k = 3 or –3 (d) k = 0 or –1. z = cy + d and (a) 3/5 (b) 1/5 x = a¢y + b¢. The two lines x = ay + b. z = c¢y + d¢ are perpendicular to each other if (c) 2/3 (d) 2/5. (2007) (2004) (a) a = 8.1 z . a. b = 2 (c) a = 4. then x equals (a) 1 (b) –1 (c) 2 (d) –2. The distance between the line r r = 2iˆ . a). 3. 1 ÷ (d) ç 5 . 2a) (b) (3a. 3a).2 = = and the 1 2 2 1 plane 2 x . 1. then the coordinates of one of the sphere and the plane of the other end of the diameter are (a) x – y – 2z = 1 (b) x – 2y – z = 1 (a) (4. 3..ˆ j + 4 kˆ ) and the plane (2008) r ˆ r × ( i + 5 ˆ j + kˆ ) = 5 is Let L be the line of intersection of the planes 10 10 2x + 3y + z = 1 and x + 3y + 2z = 2. The two lines x = ay + b. 2 are coplanar. 2a. –3) (d) (4. (2004) (a) aa¢ + cc¢ = –1 (b) aa¢ + cc¢ = 1 a c a c + = -1 + = 1 . (2007) 25. 2 meets each of the lines x = y + a = z and x + a = 2y = 2z. The image of the point (–1. . 4 ÷ . 3).5 .j + 2 k and spheres x 2 + y 2 + z 2 + 6x – 8y – 2z = 13 and r ˆ r r c = xi + ( x . which it makes with yaxis. 16. if and only if . (b) 5 3 (c) - 4 3 (d) 3 .y + l z + 4 = 0 is such that sin q = . 3.. b = 4 17. z = 1 + ls and t x = . (2006) è ø If the angle b.2 y . (2004) (c) (4. The line passing through the points (5. b. a. a. 3.4 z . the value 3 of l is 21. 3a). If the angle q between the line x . (a. a) (d) (2a. 3a).. y = 1 + t. 4) (c) 3/2 (d) 9/2 (2004) è 3 ø 3 19 ö 13 ö æ 17 æ9 27. b = 6 13. A line makes the same angle q. The x 2 + y 2 + z 2 + 7x – 2y – z = 13 and If (2. If the plane 2ax – 3ay + 4az + 6 = 0 passes through the r ˆ ˆ ˆ r ˆ ˆ ˆ midpoint of the line joining the centres of the Let a = i + j + k . (d) (2006) ¢ ¢ a c a¢ c¢ 19. then cos a equals 10 3 . (2005) (a) – 4 (b) –2 (c) 0 (d) 1. (2003) 30. (2a. 5) (b) (4. (a. 3 (a) - 5 (b) a = 2. 1) æ 17 -13 ö .83 Three Dimensional Geometry 12. . intersection of the spheres (2007) 24. (c) (d) (2005) 1 1 1 3 10 . 3a).3 z . (a) 4 (c) (3a. (c) ç .2 ˆj + 3kˆ + l (iˆ . –3. a) and (3. . with parameters s and t respectively. 4 ÷ (b) (15. 4 (2005) 22.2) ˆ j . 1. 4) in the 3 plane 26. 3a. then l equals (a) –1/2 (b) –1 (c) –2 (d) 0. 11. If the straight lines x = 1 + s. The plane x + 2y – z = 4 cuts the sphere x 2 + y 2 + z 2 – x + z – 2 = 0 in a circle of radius (a) 1 (b) 3 (c) 2 (d) 2. The lines (c) x + 1 y . (2004) 20. 3a. . y = –3 – ls. b = 8 (d) a = 6. A line with direction cosines proportional to If a line makes an angle of p /4 with the positive directions 2. crosses the yzplane at the point è 0. 5) is one end of a diameter of the sphere x 2 + y 2 + z 2 – 3x + 3y + 4z = 8 is the same as the intersection x 2 + y 2 + z 2 – 6x – 12y – 2z + 20 = 0. z = 2 – t. b = i . (a) 1 (b) (c) (d) 2 2 3 (2007) 23. (2005) 29. 0). A tetrahedron has vertices at O (0. 36. (c) (a) (c) (b) (a) (b) 4. 2 (c) 4 (d) 1.84 JEE MAIN CHAPTERWISE EXPLORER (a) (b) (c) (d) aa¢ + bb¢ + cc¢ = 0 (a + a¢) (b + b¢) + (c + c ¢) = 0 aa¢ + cc¢ + 1 = 0 aa¢ + bb¢ + cc¢ + 1 = 0.1 + 1 . b ¢.1 . a 2 b 2 c 2 a¢2 b¢2 c¢ 2 (2003) 33. 27. c . 0. 1.1 . 21. 23. 8.1 = 0 a 2 b 2 c 2 a¢2 b¢ 2 c¢ 2 (b) 1 .1 = 0 a 2 b 2 c 2 a¢2 b¢2 c¢ 2 (c) 1 + 1 + 1 . 7. b. If a 1 + b 3 = 0 and vectors (1. 1. Then the angle between the faces OAB and ABC will be (a) cos –1 (17/31) (b) 30° (c) 90° (d) cos –1 (19/35). A (1. (0. 2 (d) 2.1 . 18. 26. 9. (b) (b) (b) (a) (a) (d) 3. c 2 ) are noncoplanar. 3) and C (–1. (2003) 32. 1. b . (2002) (d) Answer Key 1. 1 (b) 1. 35. 11. 12. 1. 1. 29. c ¢ from the 1 + c3 origin. 14. 1. 31. 13. Two systems of rectangular axes have the same origin. then the product abc equals (a) –1 (b) 1 (c) 0 (d) 2. 0. (c) (d) (a) (d) (c) (b) . The d. a 2 ) .r. (1. 33. 24. (d) (c) (c) (a) (b) (a) 2. 20.1 + 1 + 1 . 0). a . 2).1 . 30. (a) (b) (b) (a) (c) (b) 5. B (2.1 = 0 a 2 b 2 c 2 a¢2 b¢2 c¢ 2 1 + 1 + 1 + 1 + 1 + 1 = 0. If b b 2 c c2 34. 32. 25. b 2 ) and plane cuts them at distance a. c and a ¢. 0) x + 2y + 2z + 7 = 0 is which makes an angle p/4 with plane x + y = 3 are (a) 2 (b) 3 (a) 1. 15. (2003) (c) 1. 17. 2. 1). then a a 2 1 + a 3 31. 19. 28. The radius of the circle in which the sphere x 2 + y 2 + z 2 + 2x – 2y – 4z – 19 = 0 is cut by the plane 36. (d) (b) (d) (d) (b) (c) 6. 2. The shortest distance from the plane 12x + 4y + 3z = 327 to the sphere x 2 + y 2 + z 2 + 4x – 2y – 6z = 155 is 3 (b) 13 (a) 11 4 (c) 39 (d) 26. (2003) (2003) r r r r 35. of normal to the plane through (1. 10. 22. 1. 34. (2003) (a) 1 + 1 . 16. r r (1. 3 = 0 we have the lines as perpendicular ö æ 5 + 3 l ö -1 ÷ = sin ç ÷ ø è 14 5 + l 2 ø 2 So. 2. 3. 2. so statement 1 and statement 2 are true but statement 2 is not a correct explanation of statement 1.3 y . Again (2. 2. 4x + 2y + 4z = –5 Distance between planes = 16 . x .1 . (c) : Equation of a plane parallel to x – 2y + 2z – 5 = 0 and at a unit distance from origin is x – 2y + 2z + k = 0 | k | = 1 Þ | k |= 3 3 Þ \ x – 2y + 2z – 3 = 0 or x – 2y + 2z + 3 = 0 4.1 = = = r1 2 3 4 x . 6. –2) lies on the plane x .5 ö = = = -2 ç ÷ø è 1 -1 1 3 Þ a .(.1 = b . As 1.3 c.0 + 6. 4r 1 + 1 = r 2 Þ 2r 1 – r 2 = 2. we have a-1 b. 4) is (2. 2 8. But it does not explain statement1. we get 1(1 + 2k) + 1(1 + k 2 ) – 1(2 – k) = 0 Þ k 2 + 1 + 2k + 1 – 2 + k = 0 Þ k 2 + 3k = 0 Þ k(k + 3) = 0 \ k = 0.4 æ 1 .0 y . . Also the midpoint of AB lies on the given line. (c) : 1 2 l x + 2y + 3z = 4 we have (3)(1) + (–5)(3) + 2(–a) = 0 Þ –12 – 2a = 0.1 = . 3r 1 – 1 = 2r 2 + k.1 z + 2 9. Statement ‘2’ holds even if the line is not perpendicular. This situation is possible. (b) : The planes are 4x + 2y + 4z = 16.5) 2 2 4 + 2 + 4 2 = 21 7 = 6 2 3. 1. (a) : The line is = = 3 - 5 2 The direction ratios of the line are (3. As the line lies in the plane x + 3y – az + b = 0. 1. 6) Again the midpoint of A(3. 6) and B(1. so the plane bisects the segment AB. As the point lies on the plane. 3).10 + k Þ k = 10 = 2 2 2 5.1 1 1 .3 + 4 . – 4). 7) and B(1.85 Three Dimensional Geometry 1. 6.3 and r2 = - 5 2 9 11 9 .k z = = = r2 1 2 1 or 2r 1 + 1 = r 2 + 3. 7). c) Thus by image formula. (5 + 3 l ) + 5 = 1 (sin 2 q + cos 2 q = 1) 2 14(5 + l ) Þ (5 + 3 l ) 2 5 + l 2 14 + 5 = 14 Þ (5 + 3l) 2 + 5(5 + l 2 ) = 14(5 + l 2 ) Þ 25 + 30l + 9l 2 + 25 + 5l 2 = 70 + 14l 2 Þ 30l + 50 = 70 Þ 30l = 20 \ l = 2/3 7.1 z . (b) : Let the image be (a.4 = 2 1 -1 1 \ (a. –5. 0.2 y . (d) : For the lines to be coplanar Angle between line and plane (by definition) æ = sin -1 ç 1 × 1 + 2 × 2 + l × 3 è 1 + 4 + 9 1 + 4 + l2 1 . so n = Þ c os q = 1 1 Þ n = 4 2 1 2 1 . 5) & the equation of the plane is x – y + z =5. m = - 2 As l 2 + m 2 + n 2 = 1. (a) : and x .1 y + 1 z . \ q = 60° . –3 So there are two values of k. 3) is (0.k = 0 k 2 1 Expanding. 2). 1. (d) : The direction ratios of the line segment joining A(1. and 4r 1 – r 2 = – 1 – 2r 1 = 3 Þ r1 = \ - . b) is (–6. b.2 – 4. c) = (3. b.3 = c . we have n 2 = We take positive values.3 = = 6. \ a = –6 Þ 2 + 3 + 2a + b = 0 Þ b = –2a – 5 = 12 – 5 = 7 Hence (a. (c) : We have l = 1 2 1 . The direction ratios of the given line is (1. 5 . 1) and (5. (b) : a = i + j + k .y2 )2 + ( z1 - z2 ) 2 62 + 32 + 2 2 = 36 + 9 + 4 = 7 uuur The direction cosines of PQ are = x1 .z 1 = = a1 b1 c 1 x . z¢) in ax + by + cz + d = 0 is given by x .a = -9 Þ a = 6 2 \ 9 -13 x= .e. 3 1 = .b ) = -3 \ b = 4 Þ 1(1 – 2x + 4) –1 (–1 – 2x) + 1 (x – 2 + x) = 0 Þ 5 – 2x + 1 + 2x + 2x – 2 = 0 Þ x = – 2. we have which on solving given 2k 2 + 5k – 25 = 0 Þ 2k 2 + 10k – 5k – 25 = 0 Þ 2k(k + 5) – 5(k + 5) = 0 Þ (2k – 5) (k + 5) = 0 \ k = - 5. l = m = n 3 - 3 3 1 1 1 1 \ l:m:n = : .5 y .y2 z1 - z 2 . (d) : Image of point (x¢.2) ˆ j .(1 . y¢. 1) Let the other end of diameter is (a. (b) : As the lines intersect.j + 2 k and r c = xiˆ + ( x . 17.x2 ) 2 + ( y1 .z ¢ -2( ax ¢ + by ¢ + cz ¢ + d ) = = = a b c a 2 + b 2 + c 2 13 Again z = m ( a .1 z .: Þ cos a = . y1 – y2.b a - 1 The line crosses the yz plane where x = 0. (d) : The equation of the line passing through (3.kˆ r r r [a b c ] = 0 Then cos a = 5 2 12.x2 y . (b) : Let required angle is q p p Q l = cos .a + = - 2 2 2 2 2 Þ x + 1 y . we get.3 ˆ j + 3 kˆ .6) = = = 1 -2 0 5 3 Þ . . L = 2 3 1 1 3 2 = 3iˆ .4 -2( -1 . (b) : Let the vector PQ be (x1 – x2. i. 6 3 2 .1) + a = - 2 5 13 3 5 13 Þ .b ) = 2 2 2 2 Þ (1 . b = i . 9 + 9 + 9 3 Second method If direction cosines of L be l.y 2 z .86 JEE MAIN CHAPTERWISE EXPLORER uuur 10.x2 y1 . b. n. 15.y1 z . g) b + 3 3 = a + 2 Þ a = 4 . 7 7 7 1 1 1 1 -1 2 = 0 x x .5 = 2 m \ m = - 2 17 Again y = m (1 . z1 – z2) we have x1 – x2 = 6 y1 – y2 = –3 z1 – z2 = 2 Length of PQ PQ = ( x1 . (a) : Fact : Two lines and x ..a = = = m (say) 2 1 . (c) : Direction of the line. 2 16.y ¢ z .( a . a) is x . 6 = Þ b = 9 2 2 g + 5 1= Þ g = -3. m = cos then n = cos q 4 4 We know that l 2 + m 2 + n 2 = 1 p p cos2 + cos 2 + cos2 q = 1 Þ 4 4 Þ 1 1 + + cos 2 q = 1 2 2 Þ cos 2 q = 0 Þ q = p/2 Thus required angle is p/2. 1. z = 4 5 5 iˆ ˆ ˆ j k 13.x1 x .x ¢ y . 18.3 z . 6.1) + a = - Þ . then 2l + 3m + n = 0. m.b ) + 1 = 2 5 17 5 15 Þ . b.2 -1 11.z 2 = = a2 b2 c2 are ^ if a 1 a 2 + b 1 b 2 + c 1 c 2 = 0 . 3 3 3 3 r ˆ ˆ ˆ r ˆ ˆ ˆ 14.y = . - .b ) + 1 = Þ . l + 3m + 2n = 0 After solving.e. PQ PQ PQ i. (c) : Centre of sphere º (3.(1 . 4. –1/2 y z x+a = = 2 1 1 Let P º (r. 0. b.( -5) 12 + 52 + 1 2 10 .. d = 2 2 2 6 1 + 2 + 1 (ii ) (iii) (ii) and (iii) Þ r – l = 2a . (a) : Let x 1 . 1.2 j + 3 k ) × ( i + 5 j + k ) . 1) and (5.9 = 1. 6 6 + 2x 21. –2. CD : 20. –2. a).(ii) 2 3 + 2 + ( -6) 2 2 2 2 + ( -12) + ( -3) 2 = 0 Þ 10x – 5y – 5z = 5 Þ 2x – y – z = 1 y + a z x = = 1 1 1 25. r – l According to question direction ratios of PQ are (2. 1 y 1 r .. (a) : d = \ d = d = r r a × n .(i) M (1... 1. y + z = 2 r r a × n1 = d1 ) z . z axis respectively then l 2 + m 2 + n 2 = 1 Þ 2l 2 + m 2 = 1 or 2n 2 + m 2 = 1 Þ 2 cos 2q = 1 – cos 2b (a = g = q) 2 cos 2q = sin 2b Þ 2 cos 2q = 3 sin 2q (given sin 2b = 3 sin 2q) Þ 5 cos 2q = 3 28.. 2 d = 2 . r – l – a. 1) 1 1 5 + + 2 \ R = 4 4 2 d = ^ distance from centre to the plane is equal to \ r .d = = a 1 c .4 3 . z 1 be any point on the plane \ Radius of the circle 2x + y + 2z – 8 = 0 = Radius of sphere – perpendicular distance from centre of sphere to plane \ 2x 1 + y 1 + 2z 1 – 8 = 0 2 ( ) æ ö = ç 5÷ .9 6 è 2 ø = 15 .d ¢ = = a¢ 1 c ¢ As lines are perpendicular \ a a¢ + 1 + c c¢ = 0 Þ a a¢ + c c¢ = –1 and . (d) : Centre of spheres are (–3. 1) using mid point in the equation 2ax – 3ay + 4az + 6 = 0 Þ 2a – 3a + 4a + 6 = 0 Þ a = –2. y 1 . (d) : Equation of the plane of intersection of two spheres S 1 = 0 = S 2 is given by S 1 – S 2 = 0 6 . (b) : Angle between the line and plane 90– q is same as the angle between the line line q and normal to the plane a1 a2 + b1b2 + c1c 2 Plane \ cos(90 . r – a.d n (2 i .1 y + 3 z . r) and Q = (2l – a.(1) (i) and (iii) Þ l = a r = 3a. l. 2) R = Radius of sphere is g 2 + f 2 + w 2 - c = C 2 (5. l = a \ p º (3a.. 3 22. 1) x .87 Three Dimensional Geometry Given lines can be written as x . (c) : From the given lines we have x . (A) 1 l l .2 l + a r -l-a r-l = = 2 1 2 (i ) 1 + 0 + 1 . 26.24 + 18 2 C 1 (–3.. a × n2 = d 2 ( x 1 8 Normal of plane + 4x 2 y + + 4z 5 = 0 2(2 x + y + 2 z . 3a) and Q º (a.8) + 21 \ d = 2 2 2 = 21 7 = 6 2 4 + 2 + 4 27. 19. 4.. (a) : From given line y x y z x z = = and = = 3 2 -6 2 -12 - 3 a1a2 + b1b2 + c1c 2 cos q = 2 2 a1 + b12 + c12 a22 + b22 + c2 cos q = 23. y. 3 3 . g with x.b¢ y z . l ) Direction ratios of PQ are r – 2l + a. (a) : Centre of sphere is 1/2. (a) : If a line makes the angle a. 2a. a. (b) : Given AB = \ q = 90º.b y z . 1) 24.1 = = = s .q ) = 2 a12 + b12 + c12 a22 + b22 + c2 1 (1 ´ 2 + 2 ´ ( -1)) + 2 l Þ = 3 12 + 22 + 22 22 + 12 + l 5 Þ l = . we get 1 a Þ (1 + abc) 1 b 1 c a 2 b 2 = 0 c 2 Þ (1 + abc) = 0 by using (A) \ Centre of sphere is (–2. b .. 3 12x + 4y + 3z – 327 = 0 aa¢ + cc¢ + 1 = 0 r r r r r r 31..d = = and a 1 c x .x1 a1 a2 y2 . 1.. c .j . 2) PB ^ from centre to the plane z2 - z 1 c 1 = 0 c2 . a 2 ) . two lines cos q = x ..(*) xIntercept y Intercept z Intercept x y z + + = 1 Þ a b c So the distance from (0.5 j - 3 k 12 + 12 + 4 + 19 = 5 and centre (–1. (c) : Given lines can be written as x . b 2 ) . (i .z 2 = = are coplanar if a2 b2 c2 x2 . 0. (b) : The radius and centre of sphere x 2 + y 2 + z 2 + 2x – 2y – 4z – 19 = 0 is 29. (b) : Using fact.1 1 = d 1 = 1 1 1 1 1 1 + 2 + 2 + 2 + 2 2 2 a b c a b c Þ . P 2 1 a a 1 b b 2 1 c c 2 A \ a 2 a + 1 b b 2 b 3 + 1 = 0 c c2 a now ¹ 0 3 c 3 + 1 On solving. (1. 35. (A) 2 12 + 4 + 3 2 .1 z-2 = = 1 2 - 2 As lines (A) and (B) are coplanar and 32.b ¢ y . (1. b.. 3) Required distance is .0 y . 0.88 JEE MAIN CHAPTERWISE EXPLORER x .3k i . 0) to this plane to the plane (*) is given by 0 + 0 + 0 .x1 y .(2) 2 + 12 + 32 + 155 = 26 – 13 = 13 units.j . we find the distance from the centre of sphere to the planeRadius of sphere k = 0 or k = –3 30.5 j .327 2 B .3k ) . c 2 ) are non coplanar. 1. (d) : Concept using angle between the faces is equal to the angle between their normals. c) is x y z + + = 1 .y2 z .y1 z .24 + 4 + 9 . 0..y1 b1 b2 1 -1 -1 1 - k = 0 2 1 Þ 1 k Þ (5i .(B) = 5i – j – 3k and vector b to the face ABC is uuur uuur AB ´ AC = i –5j – 3k Þ 1 -4 -1 1 -l l = 0 1 2 - 2 (2l – 2l) + 4(–2 –l) –1(2 + l) = 0 Þ 5l = –10 \ l = –2 \ cos q = \ \ Let q be the angle between the faces OAB and ABC 19 19 \q = cos -1 æ ö è 35 ø 35 33.z 1 = = and a1 b1 c1 x .1 + 2 + 4 + 7 = 4 1 + 2 2 + 2 2 Now (AB) 2 = AP 2 – PB 2 = 25 – 16 = 9 \ AB = 3 k 2 + 3k = 0 34.0 z . (c) : Now equation of the plane through (a. 1.x2 y .b y . uuur uuur \ Vector b to the face OAB is OA ´ OB . a . 0) (0.0 z . (a) : As vectors (1.3k ) 5i .d = = a¢ 1 c¢ \ Required condition of perpendicularity is –2. (b) : In order to determine the shortest distance between the plane and sphere. 0) (0. c ) or (0.89 Three Dimensional Geometry Similarly. 0) or (0. 1. b. 0) \ Also angle between (*) and x + y + 0z = 3 is p/4 p \ cos = 4 a + a 2 a = 2 2 2 2 2 1 +1 a + b + c 2 2 a 2 + c 2 Þ 2a 2 + c 2 = 4a 2 Þ c = ± \ 2 a DR’s a.2 . c 1 + 2 1 b¢ 2 + 1 2 c ¢ (0. a. a. ± 2a \ Required DR’s are 1.2 . 1.(*) (by using (0. 0) d = ? k y x y z + + = 1 a b c Now d 1 = d 2 given (as origin is same) Þ 1 1 1 1 = 1 1 1 1 + + a 2 b 2 c 2 a ¢ 2 b ¢ 2 c ¢ 2 1 1 1 1 1 1 Þ 2 + 2 + 2 . 0) (0. 2 match with choice (b) . 0) in (*)) (0. 0. c¢ ) (a . (b) : Let DR’s of normal to plane are a. b . 0.b¢ . 1. 2 or 1. c i. 0.e.2 = 0 a b c a¢ b¢ c ¢ + + a(x – 1) + b(y) + c(z) = 0 Þ a(0 – 1) + b(1) + c(0) = 0 Þ – a + b = 0 Þ a = b . d 2 = 1 a¢ z 36. 0) or (a¢ . b. – 2 Hence 1.. 1. 0. 0.. v .j .k .k^ and c = i^ . 33 ® ^ ^ Let AB = 3 i + 4 k be two unit vectors. r r r r The vectors a and b are not perpendicular and c and d are r r r r r r (a) p (b) 0 (c) p 4 (d) two vectors satisfying : b ´ c = b ´ d and a × d = 0.j^ + 2 k^ (a) . –3) r r r r (c) (–2. Then the angle between a and c is (2011) with the altitude directed from the vertex B to the side AD.[ pv w qu ] . q) If a = çè 3 i + k ÷ø and b = 7 çè 2 i + 3 j . –2) (2010) r r 3( p × q ) r r r æ p×q ö r (b) r = -3 q + r r p (a) r = q . q) r r r é r r r (d) exactly one value of (p.j^ + 2 k^ (c) i^ . then the angle between is (a) p 3 (b) p 4 (c) p 6 (d) 8. r r æ a×c ö r a × b r r r æ a×c ö r (d) c + ç r r ÷ b è a × b ø r (b) c . b = 2 i + 4 j + 4 k and c = l i + j + m k are mutually orthogonal.q + ç r r ÷ p (2012) r r r r r r r r r ( p × p) [ 3u pv pw ] . (b) 2 i^ . m) = (a) (–3. If r is the vector that coincides r (d) i^ + j^ - 2 k^ ^ ^ ^ r ^ ^ ^ (2010) r ^ ^ ^ If the vectors a = i . If u . b = 1 (c) a = 1. r (2010) r Let a = j^ . p 2 (2012) 3.j^ . The vector a = a i^ + 2 j + b k lies in the plane of the vectors r ^ ^ r ^ ^ b = i + j and c = j + k and bisects the angle between r r b and c . then the length of the median through A is (a) (b) 45 (c) 18 (d) 72 7. b = 2 i + 4 j + 4 k and c = l i + j + m k are mutually orthogonal.6 k ÷ø .ç r r ÷ c (2011) è a × b ø r r ^ ^ r ^ ^ ^ Let a = j . ^ ® ^ ^ ^ ^ sides of a triangle ABC. b = 2 (b) a = 2. then the value 10 (c) all values of (p. 3) (d) (3. r r r r then the equality r r 3( p × q ) r r r æ p×qö r (c) r = 3 q . uuur r uuuur r Let ABCD be a parallelogram such that AB = q . –2) (2013) 2.2 j + 4 k are the 9.90 JEE MAIN CHAPTERWISE EXPLORER CHAPTER VECTOR ALGEBRA 14 ® 1. Then the vector b satisfying r r r r r a ´ b + c = 0 and a × b = 3 is p 2 (2008) ^ ^ r 12.i^ + j^ - 2 k^ (b) 2 i^ . AD = p and r ÐBAD be an acute angle. –3) (c) (–2.k and c = i . q) r 1 æ ^ r 1 æ ^ ^ ö ^ ^ ö (b) more than two but not all values of (p . then (l. c = a^ + 2 b and d = 5 a^ + 4 b are perpendicular to each other.j + 2 k .i^ + j^ - 2 k^ (c) i^ .r r p (d) r = . b = 2 (d) a = 2. q) (2009) of ( 2 a . 5. w are noncoplanar vectors and p. Then which one of the following gives possible values of a and b? (a) a = 1. 2) (b) (2. Then the r vector d is equal to r r r æ b ×c ö r (a) b + ç r r ÷ c è ø 6. Then the vector satisfying r r r r r a ´ b + c = 0 and a × b = 3 is (a) .[ 2w qv qu ] = 0 holds for è p × p ø (a) exactly two values of (p.j^ - 2 k^ If the vectors AB = 3 i + 4 k and AC = 5 i .k^ . 3) (d) (3. If the vectors r r ^ ^ . The nonzero vectors a . b = 1 (2008) .b ) × ë( a ´ b ) ´ ( a + 2b ) ùû is r r r 11.ç r r ÷ p r r r ( p × p) è p × p ø 10. m) = (a) (–3. b and c are related by (a) 5 (b) 3 (c) –5 (d) –3 r r r r r r a = 8b and c = -7b .ç r r ÷ b è ø a × b r r æ b × cr ö r (c) b . q are real numbers.j^ - 2 k^ (d) i^ + j^ - 2 k^ (2010) r ^ ^ ^ r ^ ^ ^ r ^ ^ ^ If the vectors a = i . 2) (b) (2. then (l. r then r is given by 4.j + 2 k . C with position (a) 14 (b) 7 (c) 2 (d) 14. The values of a.k . (a) a (b) 3a r 2 r 2 (2003) (c) 4a (d) 2 a .v ) ´ (v . For any vector a . b = xiˆ + ˆ j + (1 .3 ˆ j - 5 kˆ and aiˆ . If a . w (d) more than two values of q (2007) r r r are perpendicular to each other then u . . w be such that u = 1. b . r r r r r r r r r r r r c = yiˆ + xjˆ + (1 + x . a . If the vectors of these are collinear. then r r r r r r r r r r 1 r r r (u + v . (2006) then the vectors a + 2b + 3c .w) × (u . The length of the median through A is is equal to r 2 r 2 (a) 72 (b) 33 (c) 288 (d) 18. (d) equal to zero. Let u . v = iˆ . for which the points A. Let a . iˆ . v and w are three noncoplanar vectors. (2005) (2003) uuur uuur r r ˆ2 r ˆ 2 r ˆ 2 19. (2005) r r r ˆ is a unit j and w = iˆ + 2 ˆ j + 3kˆ . b and c be distinct nonnegative numbers. b × c ¹ 0. then a × b + b × c + c × a is equal to (a) only x (b) only y (a) –7 (b) 7 (c) 1 (d) 0. b and c be nonzero vectors such that 29. 28.x ) k and 26. c are three vectors. c are noncoplanar vectors and l is a real number. c are noncoplanar vector and l is a real number 3iˆ + ˆ j - kˆ which displace it from a point iˆ + 2 ˆj + 3 kˆ to r r r r r r r r then éë l ( a + b ) l 2 b lc ùû = éë a b + c b ùû for j + kˆ . j + kˆ respectively are vectors 2iˆ . If C is the mid point of AB and P is any point outside AB. If a . Let a = i . then 2uˆ ´ 3 vˆ is a unit vector for 1 2 2 (a) no value of q .3 ˆ (2004) the vertices of a rightangled triangle at c are r r r 23. b . l b + 4 c and (2l . If u ˆ and v ˆ are unit vectors and q is the acute angle between (a) (b) 3 3 them. (b) exactly one value of l (2004) (c) exactly two values of l r r r (d) exactly three values of l. Then éë a . (2005) (2004) r ˆ ˆ r r r ˆ r r r r 18.ˆ r r r then vector such that u × nˆ = 0 and v × nˆ = 0 . Let a . v . Let a. then | w × nˆ | is equal uur uur uuur uur uur uuur r to (a) PA + PB + PC = 0 (b) PA + PB + 2 PC = 0 uur uur uuur uur uur uuur (a) 1 (b) 2 (c) 3 (d) 0. (2005) 25. the value of ( a ´ i ) + ( a ´ j ) + ( a ´ k ) 27. (2006) 24. The work done in standard units by the point 5iˆ + 4 ˆ (a) no value of l the forces is given by (a) 25 (b) 30 (c) 40 (d) 15. B. (2003) . iˆ + kˆ and ciˆ + cjˆ + bkˆ lie in a plane. such that a + b + c = 0. A particle is acted upon by constant forces 4iˆ + ˆ j - 3 kˆ and r r r 16. where a . w = 3. (c) (2004) (d) 3 3 (b) exactly one value of q r r r r r r 22.1)c are non r r r r r r r r r coplanar for 15. The vectors AB = 3iˆ + 4 kˆ and AC = 5iˆ . (a) 2 and 1 (b) –2 and –1 r r r r r r (c) –2 and 1 (d) 2 and –1. If the (c) exactly two values of q r r r r r r projection v along u is equal to that of w along u and v . b . If u . v = 2. (c) PA + PB = PC (d) PA + PB = 2 PC (2005) (2003) r r r r r r 21.ˆj + kˆ. If the vector a + 2 b is collinear with r r aiˆ + ajˆ + ckˆ .y ) kˆ . then sinq equals (c) 3u × u ´ w (d) 0. (2004) (c) perpendicular (d) parallel. b . If q is the acute angle between the 3 (a) ur × vr ´ wr (b) ur × wr ´ vr r r r r vectors b and c . b and c be three nonzero vectors such that no two r r 17. then c is r r c and b + 3 c is collinear with a (l being some nonzero (a) the arithmetic mean of a and b r r r scalar) then a + 2 b + 6 c equals (b) the geometric mean of a and b r r r (c) the harmonic mean of a and b (a) l c (b) l b (c) l a (d) 0. then a and c are (b) all except one value of l (a) inclined at an angle of p/3 between them (c) all values of l (b) inclined at an angle of p/6 between them (d) no value of l. | c | = 3. If n 20.v + w equals 14. (c) neither x nor y (d) both x and y. b and c are any three r r r r r r (a) all except two values of l vectors such that a × b ¹ 0. Let u = iˆ + ˆj.2 ˆ j + 4 kˆ are the sides of a triangle ABC. c ùû depends on | b | = 2. If ( a ´ b ) ´ c = a ´ (b ´ c ) .w) equals ( a ´ b ) ´ c = b c a .Vector Algebra 91 2 2 13. 9. b = 2 and the angle between a and b is 6 then r 2 ( ar ´ b ) is equal to (a) 48 (b) 16 r (c) a (d) none of these (2002) Answer Key 1. 19.ˆ j + 5 kˆ respectively. (2002) r r r r r r 36. iˆ . 22. If a. (d) (a) (b) (d) 25. (2002) r r r r r r 35. 8. (a) (d) (a) (d) 26. 17. (d) 31. (b) (a) (b) (a) 29. If a. Consider A. c = 3 then angle between vector b and c is (a) 60º (b) 30º (c) 45º (d) 90º. 7. (a) (a) (c) (c) 30. (*) 36. 3l c + 2 m ( a ´ b ) = 0 then (a) 3l + 2m = 0 (b) 3l = 2m (c) l = m (d) l + m = 0. -iˆ .5 ˆ j and b = 6iˆ + 3 ˆ j are two vectors and c is a vector r r r r r r such that c = a ´ b then a : b : c = (a) 34 : 45 : (c) 34 : 39 : 45 39 (b) 34 : 45 : 39 (d) 39 : 35 : 34. (b) 33. given that a + b + c = 0 (a) 25 (b) 50 (c) –25 (d) –50. b . (c) 34. c are vectors such that [ a b c ] = 4 then r r [ar ´ b b ´ cr cr ´ ar ] = (a) 16 (b) 64 (c) 4 (d) 8. (a) 35. 13. (b) 3. Then ABCD is a (a) rhombus (b) rectangle (c) parallelogram but not a rhombus (d) square. If a ´ b = b ´ c = c ´ a then a + b + c = (a) abc (b) –1 (c) 0 (2003) (d) 2. (a) 32.92 JEE MAIN CHAPTERWISE EXPLORER 30. (c) (a) (a) (a) 28. (b) 4. a = 3i . C and D with position vectors 7iˆ . 20.3 ˆ j + 4 kˆ and 5iˆ . (a) . 12. 10. (c) 37. b = 5. (b) 2. | c | = 3 thus what will be the value of r r r r r r r r r a × b + b × c + c × a . b . 15. c are vectors show that a + b + c = 0 and r r r r r a = 7. (d) (a) (d) (d) 27. r r r 33. 24. 21. B. | b | = 4.6 ˆ j + 10kˆ . 14.4 ˆ j + 7kˆ . r r r r r r r r r 31. (a) 5. (2002) (2002) r r r 34. (2002) r r ˆ r 32. (a) 6. 23. 16. If a = 4. If | a | = 5. (2002) r r r p r 37. 11. 18. m.r r + c ( a . m = 2.pq[u v w] + 2 q 2 [u v w] = 0 r r r Þ (3 p 2 .k = 0 r ^ ^ ^ \ b = .4b$ ) = 5 .b )d rr r r r r r r r Þ ( a.b )d = -( a. r r Thus a and c are antiparallel. 4) + (5.c )b + ( a.^j + 2 k^ . b = 2 i^ + 4 j^ + 4 k^ .j - k r Thus. rrr r rr r r r Also [ a b c ] = [b c a ] = [c a b ] (cyclic) rrr r r r And [a b c ] = -[ a c b ] ( Interchange of any two vectors) r r r r r r r r r [3 u pv pw] . r 9.2b .b × b = - 5 r r (b) : a × b ¹ 0 (given) r r a×d = 0 r r r r r r r r r r Now.3 = 0 Þ b$ ∙ a$ = \q = 2 3 r uuur uuuur (d) : r = BA + AQ r uuur uuuur = .k ) = -2 i .k^ ) .b ) r (a) : We have ar ´ b + cr = 0 Multiplying vectorially with ar . - 1. v .q + projection of BA across AD rr r r ( p∙q ) p = .( a × a)b + a ´ c = 0 r r ^ ^ ^ ^ ^ ^ ^ ^ a ´ c = ( j .q + r r ( p∙p ) r r r r r r (c) : (2a .j + 2 k . 4) 2 uuuur \ AM = 42 + 12 + 4 2 = 33 r 2. 4)} 2 1 = (8.( a × b )a + 2( a × b )b . (a) : ar = 8 b r r c = -7 b r r r r a and b are parallel and b and c are antiparallel.2 i . w are noncoplanar.4 b$ r r \ c ∙d = 0 Þ ( a$ + 2b$ ) ∙ (5 a$ . (a) : We have ar ´ b + cr = 0 Multiplying vectorially with .4 a × a .j . 3( j^. (a) : a = i .d )b . q) º (0. (a) : ar = i^ . 1 p Þ 6 b$ ∙a$ . m = 2. 4.( a.( a.k^ = 0 r ^ ^ ^ \ b = . we have r r r r r a ´ (a ´ b ) + a ´ c = 0 r r r r r r r r Þ ( a × b )a . .2.2 i^ . b = 2 i + 4 j + 4 k . Thus a = l ( b + c ) ^ ö ^ ö ^ ^ æ^ ^ æ^ ^ ^ ^ a i + 2 j + b k = l ç i + j + j + k ÷ = l ç i + 2 j + k ÷ è 2 è 2 ø 2 ø .b )c = ( a.i+ j.2b . 0) is the only possibility. (a) : a lies in the plane of b and c . 0. (a) : cr = a$ + 2b$ .k ) .j - k r ^ ^ ^ ^ ^ Thus. 5.4b$ ∙a$ + 10b$ ∙a$ - 8 3. (d) : AM = 1 ( AB + AC ) 2 1 = {(3. 6.k ) ´ ( i . roots are real Þ q 2 – 24q 2 ³ 0 Þ –23q 2 ³ 0 Þ q 2 £ 0 Þ q = 0 And thus p = 0 Thus (p. . Also a bisects the r r r r r angle b and c .j . d = 5 a$ . (d) : We have [la mb nc ] = lmn[a b c ] for scalars l.2k r 10.k ) = -2 i . p. - 2. 3( j . r r r r 12. 8) = (4.b ) × {( a × a ) b .( a × a)b + a ´ c = 0 r r ^ ^ ^ ^ ^ ^ ^ ^ a ´ c = ( j .2(b × b )a } r r r r r r r r = (2a .k ) ´ ( i .b )c r r r r ( a × c ) b r Þ d = .j . r r Hence the angle between a and c is p. r ^ ^ ^ c = li + j+mk r r r r a and c are orthogonal Þ a × c = 0 giving l – 1 + 2m = 0 r r Also b and c are orthogonal Þ 2l + 4 + 4m = 0 Solving the equation we get l = –3.pq + 2 q 2 )[u v w] = 0 r r r r r r As u . r 11. r r r r r r 8.[2w qv qu ] = 0 r r r r r r r r r Þ 3 p 2 [u v w] .c )b . q Î R As a quadratic in p.i+ j.93 Vector Algebra uuuur uuur uuur 1.b ) × (b - 2a) = .^j . [u v w] ¹ 0 Hence 3p 2 – pq + 2q 2 = 0.b ) × {( a ´ b ) ´ a + 2( a ´ b ) ´ b } r r r r r r r r r r r r r r = (2 a .b ) × {( a ´ b ) ´ ( a + 2b )} r r r r r r r r = (2 a . we have r r r r r a ´ (a ´ b ) + a ´ c = 0 r r r r r r r r Þ ( a × b )a . b ´ c = b ´ d Þ a ´ (b ´ c ) = a ´ (b ´ d ) rr r rr r r r r r r Þ ( a. n. cr = l i^ + j^ + m k^ r r r r a and c are orthogonal Þ a × c = 0 giving l – 1 + 2m = 0 r r Also b and c are orthogonal Þ 2l + 4 + 4m = 0 Solving the equation we get l = –3.[ pv w qu ] .2k r r ^ ^ ^ ^ ^ ^ 7. = u 2 + v 2 + w2 .94 JEE MAIN CHAPTERWISE EXPLORER on comparison.( b × c ) a = 1 r r r b c a 3 1 r r b c Þ cos q = –1/3 3 8 2 2 16. (d): As a + 2 b is collinear with c r r r \ a + 2 b = Pc r r r r r r and b + 3 c is collinear with a \ b + 3 c = Qa . a1 a2 a 3 r r r 18. (d) : Let P is origin uur r uur r Let PA = a . b .x y x 1 + x - y 1 0 0 = x 1 1 y C 3 ® C 3 + C 1 = 1(1) = 1 x 1 + x which is independent of x and y.6 c = Pc . (d) : Let a = a1i + b1 j + c1 k r r r \ a 2 = a12 + b12 + c1 2 \ a ´ i = . l 2 = a × b are scalar quantities) r r Þ a || c r 19. (b) : We are given that points lies in the same plane. M. (a) : Using the condition of coplanarity of three vectors a a c r r Þ 1 0 1 = 0 Þ c = ab L × ( M ´ N ) = 0 c c b \ C is G. (a) : From given = sin q = 9 3 r r r r r r r 2 r r r r r l ( a + b ) × ( lb ´ lc ) = a × ( b + c ) ´ b r r v × u u × w r r r r 22.b1k + c1 j r r \ ( a ´ i ) 2 = b12 + c1 2 r Similarly ( a ´ ˆ j ) 2 = a12 + c1 2 r ( a ´ kˆ ) 2 = a12 + b1 2 r r r \ ( a ´ iˆ )2 + ( a ´ ˆ j )2 + ( a ´ kˆ )2 = 2( a12 + b12 + c1 2 ) r = 2 a 2 .3 ˆ j + kˆ ) C uur where CA = (2 .ˆ j + kˆ ) ( ai ˆ . 2 r r r 24. c2 c3 r r Þ -b × c = 1 2 \ 0 l 0 0 3 4 = 0 2l - 1 1 Þ l = 0.2v × w + 2u × w 17.2u × v . l = 2a . (d) : Þ ( ar ´ b ) ´ cr = 1 3 b r r r P(0) ® A(a ) ® ® C a + b 2 ® B( b ) r c a (As given) r r r ( a × c ) b .d1 = (5 – 1)i r + (4 r – 2)j + (k – 3k) \ W.v + w r r r r r r r r r Þ No value of l exist on real axis. (c) : [ a . v = 2. (b) : | 2uˆ ´ 3vˆ |= 1 Þ 6 | uˆ || vˆ || sin q |= 1 Þ sin q = 6 2uˆ ´ 3 vˆ is a unit vector for exactly one value of q. (a): Given = and v × w = 0 r r r r r u u Þ l 2 a × (lb + lc ) + l 2b × (lb ´ c ) = a × (c ´ b ) r r r Þ l 4 [ a b c ] = .a)i - 6k Þ a 2 – 3a + 2 = 0 Þ (a – 2)(a – 1) = 0 Þ a = 1. . w = 3 r r r 2 D < 0 Þ ( l 2 )2 + 1 = 0 Now u . (d) : Given ( a ´ b ) ´ c = a ´ ( b ´ c ) r r r r r r r r Þ Þ l1a = l 2 c ( b × c ) a = ( a × b ) c r r r r ( l1 = b × c . l = 2 and l = 2 b Thus a = 1 and b = 1 1 13. N are coplanar if 23. (c) : Total force F = F1 + F2 = 7i + 2j – 4k and displacement r r r d = d 2 . 2 r r r r r r 15. of a and b.M. 20. (a) : Now CA × CB = 0 A B ( i ˆ . uuur uuur 14..2 Qa .[ a b c ] Þ l 4 + 1 = 0 Also u = 1. è 2 ø r 21.3 ˆ j + kˆ ) (2 i ˆ .. We know = 1 + 4 + 9 + 0 = 14 that the vector L. c ] = a × ( b ´ c ) = b1 b2 b 3 c1 1 0 -1 = x 1 1 .(i) Now by (i) and (ii) we have r r r r a .a)iˆ - 2 ˆ j uur ˆ ˆ and CB = (1 . PB = b r r uuur \ PC = a + b 2 uur uur r r Now PA + PB = a + b r uuur æ ar + b ö = 2ç ÷ = 2 PC .D = F × d = 28 + 4 + 8 = 40 r r r 25. u .[(k . a éë ( a ´ b ) .OA = 6iˆ + 2 ˆ j . ( b ´ c ) . ( v ´ w ) + v . (v ´ w ) r r r r r r r r r .25 . n = |–3| = 3 r r r r r r r r r 29. (a ´ b ) ùû = é a .9 1 = = 2 b c 2 ´ 5´ 3 2 \ q = 60° v u r r r r r 36.u .0 = r r r r r r r r r 2 = éë a . (u ´ v ) . uuu r uuu r 1 uuur Þ AD = [ AB + AC ] 2 1 = [8i – 2j + 8k] 34. (u ´ w) + v . ( v ´ w ) r r r = u .a ´ a r r r r Þ a ´ br = c ´ a r r r b ´rc = c ´ a Similarly r r r r r \ a ´ b = b ´ c = c ´ a r r r i j k r r r (b) : Given c = a ´ b = 3 -5 0 6 3 0 r r c = 39 k \ r r Now a = 34. c + c . ( v ´ w ) + u . (b) : Using fact: ( ar ´ b ) 2 = a 2 b 2 – ( a .w . (u ´ w) + u .c 2 49 . (u ´ v ) r r r r r r r r r = u . (v ´ w ) r r r r r r r r + v .(k . ( v ´ w ) . (*) : AB = OB .w . a ) r r r r r r ( a 2 + b 2 + c 2 ) 2 32. (a) : a + b + c = 0 Consider (a + b + c) 2 r r r r r r = a 2 + b 2 + c 2 + 2 (a . (b) : Median through any vertex divide the opposite side into two equal parts uuur uuur uuur AB + AC = 2AD 33.u ´ w + v ´ w ] r r Q v ´ v = 0 r r r r r r r r r u . c + c . (c) : n r r u ´ v ˆ = r r now n u v | = 1 1 ˆ ) ´ ( - 2k = - kˆ 2 2 35. [k ´ (c ´ a )] where k = r r r r r r r r = a ´ b . (u ´ w) . c )a ] r r r r r r ar] r b´c r = (a ´ b ) . (b ´ c ) ù = 16 ë û ( ) r r r 37. . we need angle between r r r r r b and c so consider b + c = -a 2 2 2 Þ b + c + 2|b| |c| cos q = a Þ cos q = ) a 2 . ( - kˆ ) Now w . c ùû = (a ´ b ) [(b ´ c ) a ]c . n 28. [u ´ v . r ˆ = (i + 2 ˆ j + 3 kˆ ) . (a) : (u + v .v . b ) 2 = a 2 b 2 – a 2 b 2 cos 2q p = (4 × 2) 2 – (4 × 2) 2 cos 2 6 p 1 2 = 64 × sin = 64 × = 16 4 6 . a ) = - (12 + 2 2 + 32 ) = - 2 = –7 27. P = –6 Putting value either in (1) or in (ii) we get r these r r a + 2b + 6 c = 0 ( r r r 26.w. b = 45 and r r c = 39k = 39 r r r \ a : b : c = 34 : 45 : 39 r r r (b) : 3l c = 2m (b ´ a ) r r r Þ either 3l = 2m or c || b ´ a but 3l = 2m r r r r r r r (a) : We have a + b + c = 0 Þ ( a + b + c ) 2 = 0 r r r r r r r r r Þ | a |2 + | b |2 + | c |2 + 2( a × b + b × c + c × a ) = 0 r r r r r r Þ 25 + 16 + 9 + 2(a × b + b × c + c × a ) = 0 r r r r r r Þ 2( a × b + b × c + c × a ) = - 50 r r r r r r Þ (a × b + b × c + c × a ) = - 25 r r r r r r \ (a × b + b × c + c × a ) = 25 r r r (a) : Given a + b + c = 0. (a) : Consider [a ´ b b ´ c c ´ r r r r r = (a ´ b ) .2 ˆ j - kˆ | CD | = uuur uuur j - 2 kˆ | DA | = 17 DA = -2iˆ + 3 ˆ 41 r r r (c) : If rpossible say a + b + c = 0 r r Þ b + c = -a r r r r r a × b + c = . ( v ´ w ) (³ [a b c] = [b c a] = [c a b]) uuur uuur uuur 30. (v ´ w ) r r r r r r r r r = u .w ) .3 kˆ uuur \ | AB |= 49 = 7 uuur Similarly BC = 2iˆ . (u ´ v ) + w. a )c .95 Vector Algebra r r 31. 2 uuur \ AD = 33 r r r ˆ || ur ´ vr ³ u .3 ˆ j + 6 kˆ uuur \ | BC | = 49 = 7 uuur uuur CD = . Þ a (1 + 2Q) + c (–6 – P) = 0 Þ 1 + 2Q = 0 and – P – 6 = 0 Q = –1/2.[(b ´ c ) . (b ´ c ) ùû éë c . ( w ´ u ) . b + b .6iˆ .b 2 . u ´ v . c] a ù ë û r r r r r r r r r r r r . b + b . a ]c . n ˆ = 0 = v . Þ (a . é[(b ´ c) . .5. The teacher decided to give grace marks of 10 to each of the students. 5. 2x n is 4. x n be n observations. Statement 2 is false. If V A and V B represent the variances of the two populations. and another population B has 100 observations 151. Statement 2 is true; Statement 2 is 8. (c) (d) (2004) a correct explanation for Statement 1. (c) variance (d) mean (2013) Let x 1 . . The average marks of boys and girls combined is 50. If the standard deviation of the (a) Statement 1 is true. If the mean deviation about the median of the numbers a.. (d) Statement 1 is true. In a series of 2n observations. 2a. (a) Statement 1 is true. 2x n is 4s 2 . 1 + 100d from their mean is 255. then V A /V B is (a) 1 (b) 9/4 (c) 4/9 (d) 2/3... Then which one of the following gives possible values of a and b? (a) a = 3. (b) Statement 1 is true.. and let be their arithmetic mean and s 2 be their variance. 2x 2 .0 (d) 25. x n be n observations such that 2 5 11 13 å x1 = 400 and å xi = 80. (2007) Suppose a population A has 100 observations 101. STATISTICS All the students of a class performed poorly in Mathematics.0 (b) 10... 1 + d. Statement 2 is true; Statement 2 is observations is 2..1 (c) 20. 3. .. half of them equal a and 6 remaining half equal – a. Let x 1 . (d) Statement 1 is true.2 (d) 10.. Statement 2 is true. then the d is equal to (a) 20. (a) 2 (b) 2 (c) Statement 1 is false. (2005) n 2 - 1 is 11. .. In a frequency distribution.0 (2009) The mean of the numbers a. then its mode is approximately Statement2 : The sum of first n natural numbers is (a) 20. (2006) For two data sets.80. (c) Statement 1 is false. b... respectively. then |a| equals not a correct explanation for Statement 1.. Then a possible value of n among (b) (c) 6 (d) (a) 2 2 2 the following is (2010) (a) 18 (b) 15 Statement1 : The variance of first n even natural numbers (c) 12 (d) 9. the mean and median are 21 and 4 22 respectively.. (b) Statement 1 is true. (2009) n n .5 (b) 22. (2012) 9.. x 2 . 250.. 2.... The variance of the combined data set is 10. Statement 1 : Variance of 2x 1 .. . 102... . 6... the variances are given to be 4 and 5 and the corresponding means are given to be 2 and 4. respectively.. Statement 2 is true. 50a is 50. 8. Statement 2 : Arithmetic mean of 2x 1. then |a| equals (a) 4 (b) 5 (c) 2 (d) 3 (2011) If the mean deviation of number 1. 4. b = 6 (2008) The average marks of boys in class is 52 and that of girls is 42. x 2 . . 5.0 n (n + 1) and the sum of squares of first n natural numbers is (c) 24.. . Statement 2 is false. 10 is 6 and the variance is 6. 1 + 2d... . The percentage of boys in the class is (a) 80 (b) 60 (c) 40 (d) 20.. b = 7 (c) a = 5. 2x 2 . 200. b = 2 (d) a = 1. 152.96 JEE MAIN CHAPTERWISE EXPLORER CHAPTER 15 1. Which of the following statistical measures will not change even after the grace marks were given? (a) median (b) mode 7. b = 4 (b) a = 0.. not a correct explanation for Statement 1... Statement 2 is true; Statement 2 is a correct explanation for Statement 1. (2005) 2 n( n + 1)(2n + 1) 12. each of size 5. Statement 2 is true; Statement 2 is 1 2 . 00. (a) 16.66 (b) 177. (a) 2. The median of a set of 9 distinct observations is 20. Consider the following statements : (1) Mode can be computed from histogram (2) Median is not independent of change of scale (3) Variance is independent of change of origin and scale. (b) 3. If each of the largest 4 observations of the set is increased by 2. (b) 8. (2002) (d) is increased by 2. (2003) Answer Key 1. If the average marks of the complete (a) is decreased by 2 class is 72. Then the corrected variance is (a) 188. 2 å x = 2830. In an experiment with 15 observations on x. (b) 12. (2) and (3).97 Statistics 13. (a) 9. (a) . (b) 10. (a) 13. (c) 7. then 16. (c) 6. In a class of 100 students there are 70 boys whose average the median of the new set marks in a subject are 75. (c) 11. (a) 15. å x = 170 One observation that was 20 was found to be wrong and was replaced by the correct value 30. the following results were available. (b) 5. (d) 4.33 (c) 8.5. (a) 14.33 (d) 78. then what is the average of the girls? (b) is two times the original median (a) 73 (b) 65 (c) remains the same as that of the original set (c) 68 (d) 74. (2003) 14. Which of these is/are correct? (a) only (1) and (2) (b) only (2) (c) only (1) (d) (1). (2004) 15. 1 + d. .4 + 5...( x + 10)} Hence s 1 2 = s 2 2 n 2..( n + 1) 2 n = 4 n( n + 1)(2n + 1) × . + (2n )2 ) .x12 ) = -1. M= 25 a + 26 a = 25.5 a 2 Þ 50 = s= = å | ri . = x . 1 145 ...x 2 Þ 4 = å xi2 .x ) n 1 2 2 s2 = å {( xi + 10) .. JEE MAIN CHAPTERWISE EXPLORER (c) : 1 st solution : Variance doesn’t change with the change of origin.5 + 5. + n) Þ 2500 = 2| a| × 4. 3.1 + 5.(n + 1) 2 n 6 = 2 × (n + 1)(2n + 1) . (b) : The numbers are 1..M | MD( M) = n1x1 + n2 x2 n1 + n2 = 5 ´ 2 + 5 ´ 4 = 3 10 d1 = ( x1 ...P.. of 2x 1 .98 1. d2 = (x 2 .... n2 = 5. n1 = 5. x n .. x 2 = 4 x12 = (a) : Median is the mean of 25 th and 26 th observation.. x 2 . + xn ) = 2 x n = 2 nd solution : s12 = 4.M. = 4x \ Statement 2 is false.. 1 + 2d. 25 × 25. + 2n = 2 (1 + 2 + . \ 5 å xi2 = 40.(n + 1) 2 3 = (n + 1) [2(2n + 1) .M.1) = 3 3 6. ...5 + 1.3( n + 1)] 3 = (n + 1) n 2 . x 3 .1 = 10 \ s 2 = 1 {2| a| ´ (0.M. 2x n = 2( x1 + x 2 + . x 1 ..( x ) 2 n = 1 2 (2 + 4 2 + . 1 + 100d.90 55 11 (40 + 105) .5) } 50 55 11 = 10 2 11 2 5.. + 24. 10 10 10 2 N 2 n1s 12 + n2s 22 + n1d12 + n2 d2 n1 + n2 5.. 2 The numbers are in A. Variance = s 2 (b) : Statement 2 : A. 2x 2 . \ 5 s2 = 1 10 å yi2 = 105 (å xi2 + å yi2 ) . å yi = 20 ö 1 2 æ1 s1 = ç å xi2 ÷ . \ | a| = 4 2 = 2× (b) : 1 st solution: 2 s1 = 2 s 2 = Mean ( x ) = 4 üï x = 2 ý 5ïþ y = 4 We have å xi = 2 5 Variance = Þ å xi = 10 Similarly.9 = = = .. x1 = 2 s 2 2 = 5.5 + . A. (c) : Sum of first n even natural numbers = 2 + 4 + 6 + ..4 è5 ø 5 Þ 1 å xi2 = 8. 1 2 2 2 nd solution : s1 = å( xi . .x12 ) = 1 Given A.. + n 2 ) ..y 2 Þ 5 = å yi2 .æçè x +2 y ö÷ø n (n + 1) = n( n + 1) 2 n(n + 1) = n + 1 n 1 (å xi ) 2 ..(n + 1) 2 n = 1 2 2 × 2 (1 + 22 + .. æ1 ö 1 s 22 = ç å yi2 ÷ .16 è5 ø 5 Þ 1 å yi2 = 21.1 × (n . Then mean = 51 st term = 1 + 50d = x (say) . .( X ) 2 N 2 na 2 - 0 ; 2 = a 2 ; 2 = |a| 2 n 13. The new median remain unchanged. = 255 (given) 50 × 51 Þ d = 255 101 101 ´ 255 101 ´ 255 Þ d= = = 10.. (a) : According to problem X x 1 x 2 – – – – x n x n + 1 x n + 2 – – – x n + n (a) : The mean of a. + d + 0 + d + 2d + . 15.. 5.(2) using (1) we have a 2 + (7 – a) 2 = 25 Þ a 2 + 49 – 14a + a 2 = 25 Þ a 2 – 7a + 12 = 0 \ a = 3..8 = 34 Þ a 2 + b 2 – 12 × 7 + 72 + 21 = 34 \ a 2 + b 2 = 25 .D.. .x | n i =1 i 1 [50 d + 49 d + 48d + . + 50 d ] 101 + d + 0 + d + 2d + . x + y 5 a 2 a 2 – – – – – a 2 a 2 – – – a 2 2 å ( X . (a) : Mode can be computed by histogram Median will be changed if data’s are changed so (2) is correct.... + x 70 = 75 × 70 7200 - 5250 Þ 400 ³ 80 Þ 20 ³ 80 Þ n ³ 4 Þ n ³ 16 = 65 \ Average of 30 girls = n n n 30 n 9. 250 è 15 ø 15 Series B is obtained by adding a fixed quantity to each item ( x1 + x2 + .99 Statistics Mean deviation (M.D.... b. we know that variance is independent of change 16.... 10 is 6 a + b + 8 + 5 + 10 = 6 5 Þ a + b + 23 = 30 Þ a + b = 7 Þ n = 17 but not given in choice.. + x70 Again 1 = 75 10. ... Þ å ( xi .8 n ( a .6) 2 + ( b .6) 2 + 4 + 1 + 16 Þ = 6.... 102. 152. 2 = 14.) = = 1 101 å | x . 4 also b = 3.. + x 100 = 7200 variances is 1.1 50 ´ 51 2550 7.æ ´ 180 ö = 222 – (12) 2 = 78 Series B = 151.... (c) : Using fact.. of the numbers x 1 + . + 50d] 1 = × 2d (1 + 2 + ...(i) .M... 4 (a) : Let x and y are number of boys and girls in a class respectively. (a) : Using well known fact that root mean square of number 70 ³ A. . (b) : Total number of observations are 9 which is odd which means median is 5 th item now we are increasing 2 in the last four items which does not effect its value. . + 50) 101 1 50 × 51 50 × 51 = × 2 d × = d 101 2 101 But M. mode = 3 median – 2 mean = 3 × 22 – 2 × 21 = 3(22 – 14) = 3 × 8 = 24.... 11. 103. 52 x + 42 y = 50 x+ y x 4 x 4 = Þ x = 4 y Þ = and y 1 x + y 5 Again variance = 8.(1) Required percentage = x 4 ´ 100 = ´ 100 = 80%.A ) 2 = 6. (d) : åx = 170 and åx 2 = 2830 Increase in åx = 10 and åx¢ = 170 + 10 = 180 Increase in åx 2 = 900 – 400 = 500 then å x¢ 2 = 2830 + 500 = 3330 2 (a) : Series A = 101.. (b) : Using x = 100 of origin both series have the same variance so ratio of their \ x 1 + . 12..8 5 2 2 Þ a + b – 12(a + b) + 36 + 21+72=5×6. Variance depends on change of scale so (3) is not correct... 8..X ) N = å X 2 . \ n = 18 is correct number. 200 1 1 = ´ 3330 ....X ) S X = 0 \ X = ( X - X ) 2 d = value of X– X a a – – – – a –a –a – – – –a Value of X a a – – – – a –a –a – – – –a = 2 na 2 SX 0 = = 0 N 2 n 2 Now SD = å ( X .(ii) .. x + x2 + . + x 100 ) = 72 of series A. 153... A pair of fair dice is thrown independently three times. (2007) 1 arranged in some order will form an A... (a) Statement1 is true. four are 10.. equals 1 5 1 1 (b) (c) (d) (a) 7 14 50 14 (2009) 9. ú ë 2 û æ 11 ù (b) çè 12 . 3. 0. given that their maximum is 6. respectively. Statement2 is true; Statement2 is a correct explanation of Statement1. The from the set {1. ±4. from {1.P... Then P(B) is 4 2 3 have different colours is (a) 1/3 (b) 2/7 (c) 1/21 (d) 2/23 1 1 1 2 (a) (b) (c) (d) (2008) (2010) 2 6 3 3 Four numbers are chosen at random (without replacement) 11. 4 ú (d) çè 4 . ±5}. (b) Statement1 is true.. A die is thrown. It is given that the events A and B are such that blue and two are green. Each question has three alternative answers of which exactly one is correct. (c) Statement1 is true. given that the product of these digits is zero.. The Statement2 : If the four chosen numbers form an A. 8}. Let A be the event that the number obtained is greater than 3. If the probability of at least one failure is greater than or equal to ) 1 . then probabilities of I and II scoring a hit correctly are 0. The probability that their minimum is 3. probability of getting a score of exactly 9 twice is Statement1 : The probability that the chosen numbers when (a) 8/729 (b) 8/243 (c) 1/729 (d) 8/9. 3. 1 ú û æ 1 3ù æ 3 11 ù 9 (b) log 4 - log 3 10 10 1 (d) log 4 - log 3 10 10 (2009) 8. One ticket is selected at random from 50 tickets numbered 00. (2010) A multiple choice examination has 5 questions. 01. 2. 2. Statement2 is not a correct explanation for Statement1. . then the correct statement among the following is (a) P(C|D) < P(C) P(D) (b) P(C | D) = P(C ) (c) P(C|D) = P(C) (d) P(C|D) ³ P(C) 5. P ( A | B ) = and P ( B | A ) = . Statement2 is true. 49. ( In a binomial distribution B n. If C and D are two events such that C Ì D and P(D) ¹ 0.100 JEE MAIN CHAPTERWISE EXPLORER CHAPTER PROBABILITY 16 1. is . The second plane will bomb only if the ±2. Then P(A È B) is 2 3 (a) (b) (c) 0 (d) 1 (2008) 5 5 (2011) Consider 5 independent Bernoulli’s trials each with probability of success p. Two aeroplanes I and II bomb a target in succession. The probability that a student will get 4 or more correct answers just by guessing is 10 3 5 (b) 17 3 5 (c) 13 3 5 (d) 11 3 5 (2013) Three numbers are chosen at random without replacement 7. Then the probability that the sum of the digits on the selected ticket is 8. The probability that the three balls P ( A ) = . 12 úû (2011) û An urn contains nine balls of which three are red.P. ±3. Three balls are drawn at random without 1 1 2 replacement from the urn. Statement2 is true.2. 31 . . Statement2 is false.3 and the set of all possible values of common difference is {±1. p = (c) èç 2 . Let B be the event that the number obtained is less than 5.. .. 20}. if the probability of 4 9 at least one success is greater than or equal to . 85 12. then n is 10 greater than 1 (a) log 4 + log 3 10 10 4 (c) log 4 - log 3 10 10 (2012) 3. is (a) 1 4 (b) 2 5 (c) 3 8 (d) 1 5 4. 02. (d) Statement1 is false.. (a) 2. …. then p lies in the interval 32 é 1 ù (a) ê 0. 6. 25. where A stands for P ( A Ç B ) = 4 4 complement of event A. (d) (d) (d) (d) 3. Then the variance of distribution of success is (a) 8/3 (b) 3/8 (c) 4/5 (d) 5/4. 17. The probability that they contradict each other when asked to speak on a fact is 13. 10.05 A and B are events such that P (A È B) = 3/4.87 (d) 0.2 (b) 0. 18.7 19. B. 15. 12. 19. (2004) average of 5 phone calls during 10minute time intervals. At a telephone enquiry system the number of phone calls (a) 7/20 (b) 1/5 regarding relevant enquiry follow Poisson distribution with a (c) 3/20 (d) 4/5. (2005) 1 16. The probability that the target is hit (c) 0. Then events A and B are (a) equally likely but not independent (b) equally likely and mutually exclusive (c) mutually exclusive and independent (d) independent but not equally likely. 17. P ( A ) = 2/3 then P ( A Ç B ) is (a) 5/12 (b) 3/8 (c) 5/8 (d) 1/4. (d) êë 3 .20 0.06 (d) 0. 20. the probability P(E È F) is: Answer Key 1. 5 6 55 e (2006) 20.07 0. Mr. (2002) For the events E = {X is a prime number} and F = {X < 4}. (a) (c) (d) (d) 5. 7. Getting an odd number is considered a success. B. 21. (2003) 22. (b) (b) (d) (d) 6. (2007) for B is 3/4. 13.08 0. C are mutually exclusive events such that P( A) = 3x + 1 . 22. (2002) A problem in mathematics is given to three students A. (2005) e 15. A die is tossed 5 times. then P (X = 1) is (a) 1/16 (b) 1/8 (c) 1/4 (d) 1/32.15 0. The probability that all the three apply for the same house is (a) 1/9 (b) 2/9 (c) 7/9 (d) 8/9. P( B ) = 1 . (2004) by the second plane is (a) 0. P (A Ç B) = 1/4. while this probability (c) 0. Three persons apply for the houses.14. The mean and variance of a random variable X having a binomial distribution are 4 and 2 respectively. The probability that A speaks truth is 4/5. Five horses are in a race. 3 úû (c) [0. The probability that there is at the most one phone call during a 10minute time period is 6 5 6 6 (a) e (b) (c) (d) 5 .101 Probability (a) 0. 16. Let A and B be two events such that P ( A È B ) = .x and P(C ) = 1 . Events A.50. C and their respective probability of solving the problem is 1/2. A selects two of the horses at random and bets on them. 2 û ú 8 P(X) : 0. (2003) 14. 24.35 (b) 0. 18. Then set of 3 4 2 possible values of x are in the interval 1 2 3 4 5 6 7 é1 13 ù (b) êë3 . A selected the winning horse is (a) 3/5 (b) 1/5 (c) 2/5 (d) 4/5. Probability that the problem is solved is (a) 3/4 (b) 1/2 (c) 2/3 (d) 1/3. (c) (d) (b) (a) . A random variable X has the probability distribution: X : é1 2 ù (a) êë3 .2 x .23 0. Three houses are available in a locality.10 0. 6 1 and P ( A ) = 1 . (d) (d) (d) (a) (a) 2. 11. 8. The probability that Mr. Then the probability of 2 successes is: (a) 128/256 (b) 219/256 (c) 37/256 (d) 28/256. The mean and the variance of a binomial distribution are 4 and 2 respectively. 9. 3 úû 4. 1] é1 1 ù . 23. 14. Each applies for one house without consulting others. (d) (d) (a) (c) (2003) 23. 1/3 and 1/4.5) equals (a) 0 (b) 2/e 2 3 (c) 3/e 2 (d) 1 - 2 . A random variable X has Poisson distribution with mean 2.77 first misses the target. (2005) 21.12 0. The P (X > 1. (2004) 25. (2002) 24. . 3. 2. 4} We have A È B = {1. . P (C ) Ì P(D).. the number must be of the form either x0 which are 5 in numbers. 9.1 / 2] 32 2 5... 5.. 2. i. (c) : From the definition of independence of events P ( A / B ) = P ( A Ç B ) P ( B ) . 9} for the product of digits to be zero.. Þ n(log104 – log103) ³ 1 P(C Ç D) = P (C ) Þ n³ We have. 2. 84 7 6. 02. 1. 4} and b Î {0. 9} The only number common to both = 00 Thus the number of numbers in S. 5. 4 1∙1∙2 Þ 8 P( B / A) = C3 2 1 P( A Ç B) = = = 5 P( B) C2 10 5 Taking logarithm on base 10 we have 8 C3 P(C Ç D ) 3. (d) : P(C | D ) = P(D ) \ n log10(3/4) £ log1010 –1 Þ n(log103 – log104) £ –1 as C Ì D. (d) : A = {4. 1. 4. 49} is of the form ab where a Î {0. 8} without replacement. 6} = S Where S is the sample space of the experiment of throwing a die.. . = 5´2 + 1 = 11 7. 2. 2. (d) : P(correct answer) = 1/3 The total number of ways n(S) = 20C 4 4 æ 1ö æ 2ö æ 1 ö 5 The required probability = 5C4 ç ÷ ç ÷ + 5 C5 ç ÷ è 3ø è 3ø è 3 ø The desired probability = 57 20 C 4 = 57 ´ 24 1 = 20 ´ 19 ´ 18 ´ 17 85 Now statement2 is false and statement1 is true..P. 1.’s with common difference 6 = 57 1 log10 4 - log10 3 8.P. 3.’s with common difference 3 = 11 Number of A. 2.’s with common difference 5 = 5 2 Number of A. (d) : Probability of at least one success 35 35 35 = 1 – No success = 1 – n C n q n where q = 1 – p = 3/4 2. (b) : n(S) = 9 C3 = 9 ´ 8 ´ 7 = 84 6 n(E ) = 3C 1 ∙ 4C 1 ∙ 2C 1 = 3 × 4 × 2 = 24. (d) :3 numbers are chosen from {1. Hence P(A È B) = 1 10.æç ö÷ ³ è 4 ø 10 Let B be the event that the minimum of chosen numbers is 3. .P. The desired probability = 4 1 æ3ö 1 æ3ö ³ç ÷ Þ ç ÷ £ è 4 ø 10 10 è 4 ø 24 2 = . 3.. 01. the product of whose digits is zero = 10 + 5 – 1 = 14 Of these the number whose sum of digits is 8 is just one.’s with common difference 4 = 8 Number of A...P. 1. we want 1 . 3...’s with common difference 1 = 17 Number of A. 3. for it is a sure event.102 JEE MAIN CHAPTERWISE EXPLORER 1.P.’s with common difference 2 = 14 Number of A. 6} Also B = {1. P(S) = 1. Let A be the event that the maximum of chosen 4 3 9 numbers is 6. 08 The required probability = 1/14. (c) : Number of A. 4} or of the form 0x which are 10 in numbers because x Î {0. P(C | D) = P(C ) P ( D) As 0 < P(D) £ 1 we have P(C|D) ³ P (C ) 4.. because x Î{0.p 5 ³ Þ p 5 £ 31 32 1 1 thus p £ \ p Î [0. 2. (a) : Probability of at least one failure = 1 – P (no failure) = 1 – p 5 Now 1 .P. (d) : Any number in the set S = {00..e. . P ( B ) + P ( A Ç B ) 1 3 1 = 1 .77 1 5 3 1 P(B) = \ P ( B ) = 4 4 Now we needed P(A) P ( B ) + P(B) P ( A ) 19. 3..62 P(F) = P(1 or 2 or 3) (events 1.0.2 . B.(ii) .) = P(1) + P(2) + P(3) = ..(1) .... (6. (d) : We know that poission distribution is given by -l r P ( x = r ) = e l where l = 5 r ! Now P(x = r £ 1) = P(x = 0) + P(x = 1) = -l e -l l e 6 + = e -5 (1 + 5) = 5 .3. (c) : No.. (d) : P ( A È B ) = 1 . 0! 1! e e . 4). 9 33 16. we have from (1) and (2) P(A)P(B/A) = P(B)P(A/B) Þ ..7. (b) : Possibility of getting 9 are (5.P ( A È B ) = 1 .P ( B ) + 6 4 4 P ( B ) = 1 - 1 Þ P ( B ) = 4 = 1 2 6 12 3 1 = 1 ´ 3 = P ( A ) P ( B ) now P ( A Ç B ) = 4 3 4 so even are independent but not equally likely as P(A) ¹ P(B).. 5).103 Probability Then P(B) ∙ P(A/B) = P(A Ç B) Interchanging the role of A and B in (1) P(A)P(B/A) = P(B Ç A) As A Ç B = B Ç A. è 9 ø è 9 ø 243 12. C are mutually exclusive \ 0 £ P(A) + P(B) + P(C) £ 1 0 £ P(A).35 P(E È F) = P(E) + P(F) – P(E Ç F) = . npq = 2 2 1 npq = \ q = p = and n = 8 Þ 4 2 np 8 æ 1 ö Now P(X = r) = 8 C r ç ÷ è 2 ø (Use p(X = r) = n C r p r q n–r 8 15. of horses = 5 4 3 ´ 5 4 Probability that ‘A’ must win the race = 1 - P ( A ) P ( B ) \ Probability that A can’t win the race = = 1 - 12 2 = 20 5 22.P ( A ) . (d) : A. P(B).3 = 0.2. (d) : Given np = 4 and npq = 2 npq 2 1 1 = = so p = 1 – 1/2 = q = 2 np 4 2 Now npq = 2 \ n = 8 \ BD is given by P(X = r) = 8 C r p r q n – r 8 28 æ 1 ö P(X = r = 2) = 8 C 2 ç ÷ = 256 è 2 ø 18. (d) : P (I) = 0. of favourable cases No.l lr ( l = mean ) 14. 5. 7 \ ‘E’ denote prime ‘F’ denote the number < 4 \ P(E) = P(2 or 3 or 5 or 7) (Events 2.5) = P(2) + P(3) + . 3.2 = 0.. ¥ = 1 – [P(0) + P(1)] -2 2 ù é e ´ 2 3 = 1 .. (a) : P(A) = = 4 5 \ P( A ) = 4 1 3 1 7 ´ + ´ = 5 4 4 5 20 20. 13.ê e -2 + = 1 .7)(0. P (II ) = 1 .E) = P(2) + P(3) + P(5) + P(7) = . (3. P(C) £ 1 Now on solving (i) and (ii) we get 1 1 £ x £ 3 2 . 3). (a) : No. (d) : P ( X = r ) = r ! \ P(X = r > 1.50 P(E Ç F) = P(2 or 3) = P(2) + P(3) = . (b) : From the given table prime numbers are 2. 5.35 = . 8 1 æ 1 ö = ÷ = 2 16 ´ 16 32 è ø \ p(X = 1) = 8 C 1 ç 21. P (II) = 0.(2) 1 2 1 1 2 1 × = P ( B ) × Þ P ( B ) = × × 2 = 4 3 2 4 3 3 11. 6) 4 1 Probability of getting score 9 in a single throw = p = 36 = 9 Required probability = probability of getting score 9 exactly 2 twice = 3 C2 æç 1 ö÷ ´ æç 8 ö÷ = 8 ..14. P ( I ) = 1 .8 Required probability = P ( I Ç II) = P ( I ) P (II) = (0. 3 are m.0.2) = 0. 2 úû ë e 17. of houses = 3 = No. 2.50 – . of applicants = 3.(i) . 7 are M. (d) : Given mean np = 4.E. \ Total number of events = 3 3 (because each candidate can apply by 3 ways) Required probability = 3 = 1 . (4.62 + . or P(A È B È C) = 1 – P ( A ) P ( B ) P (C ) 1 2 3 = 1 – .P ( A Ç B ) 3 By using P(B) = P(A È B) + P(A Ç B) – P(A) 2 1 P ( A Ç B ) = - = 5/12 \ 3 4 25. = 3/4 2 3 4 = 2 ´ 10 32 5 æ 1 ö ÷ è 2 ø = . (d) : n = 5 p = q = 1/2 P(X = r) = 5 C r ç ÷ è 2 ø x i f i f i x i f i x i 2 5 (0) 0 5 æ 1 ö ÷ è 2 ø (1) 5 C 1 ç (2) 5 C 1 ç (3) 5 C 3 ç (4) 5 C 4 ç (5) 5 C 5 ç 5 æ 1 ö ÷ è 2 ø 5 æ 1 ö ÷ è 2 ø 5 æ 1 ö ÷ è 2 ø æ 1 ö ç ÷ è 2 ø 0 1 ´ 5 32 5 32 2 2 3 ´ 10 32 3 2 4 ´ 5 32 4 2 5 × 1 32 5 2 0 10 32 10 32 5 32 1 32 240 80 å f i x i 2 = 32 32 5 x = mean = 2 å f i = 1 å f i x i = å f i xi2 . (a) : Given P (A È B) = 3/4 P(A Ç B) = 1/4 P ( A ) = 2/3 2 \ P(B) = P ( A Ç B ) = P ( B ) . (a) : Given P(A) = 1/2 \ P ( A ) = 1/2 2 P(B) = 1/3 \ P ( B ) = 3 1 3 \ P (C ) = P(C) = 4 4 Now problem will be solved if any one of them will solve the problem. \ P(at least one of them solve the problem) = 1 – probability none of them can solve the problem.104 JEE MAIN CHAPTERWISE EXPLORER n æ 1 ö 23.çæ å fi x i ÷ö Now variance = å f i çè å fi ÷ø 2 240 25 - 32 4 40 5 = 32 4 A B 24. . 1 æç 5 ö÷ = p . The height of the tower is (a) a / 3 (b) a 3 (c) 2 a / 3 (d) 2 a 3. In a DPQR. A tower stands at the centre of a circular park. then (2013) (a) 7 3 2 1 m 3 + 1 (c) 7 3 ( 3 + 1) m 2 (b) 7 3 2 1 m 3 - 1 (d) 7 3 ( 3 - 1) m 2 4. and the angle of elevation of the top of the tower from A or B is 30°. p ÷ö for which the function. If x. è 2 2 ø . è ø (2007) 13. then (a) 2x = 3y = 6z (b) 6x = 3y = 2z (c) 6x = 4y = 3z (d) x = y = z (2013) 2.x 2 -1 æ x ö f ( x) = 4 + cos . ABCD is a trapezium such that AB and CD are parallel and BC ^ CD. 2 ÷ ë ø é pö (b) ê0.1 æç x ö÷ + cosec . 5 13 4 Then tan2a = (a) 25 16 (b) 56 33 (c) 19 12 (d) 20 (2010) 17 7. BC = p and CD = q. A and B are two points on the boundary of the park such that AB (= a) subtends an angle of 60° at the foot of the tower. (2007) . if 3 sin P + 4 cos Q = 6 and 4 sin Q + 3cos P = 1. 2 ÷ . where 0 £ a . AB is a vertical pole with B at the ground level and A at the top. He moves away from the pole along the line BC to a point D such that CD = 7 m.2 . If A = sin 2 x + cos 4 x.105 Trigonometry CHAPTER TRIGONOMETRY 17 1. Let cos(a + b) = (d) 3 13 £ A £ 4 16 13 £ A £ 1 16 (2011) 4 5 p and let sin(a – b) = . z are in A. (2013) Then the height of the pole is tan A cot A + can be written as 1 . From D the angle of elevation of the point A is 45°. then the angle R is equal to 5 2 10. A man finds that the angle of elevation of the point A from a certain point C on the ground is 60°. The expression p2 + q2 (b) p2 cosq + q2 sin q (d) ( p2 + q2 )sin q p cos q + qsin q (b) tanA + cotA sinA cosA + 1 (a) (b) (c) (d) A is false and B is true both A and B are true both A and B are false A is true and B is false (2009) 9. A false statement among the following is (a) there is a regular polygon with r/R = 1/2 (b) there is a regular polygon with r/R = 1 / 2 (c) there is a regular polygon with r/R = 2 / 3 (d) there is a regular polygon with r/R = 3 / 2 8. let r and R be the radii of the inscribed and the circumscribed circles.1 + log (cos x ) is defined. b £ . 2 ÷ ë ø (c) [0.. The largest interval lying in çæ -p . tan –1 y and tan –1 z are also in A.P.cot A 1 - tan A (a) secA cosecA + 1 (c) secA + cosecA(d) 3 2 If cos(b – g) + cos(g – a) + cos(a – b) = - . and tan –1 x. is è 2 ø é p pö (a) ê . If ÐADB = q. then for all real x (a) 1 £ A £ 2 (c) (b) 3 £ A £ 1 4 6. then the values of x is è5ø è 4 ø 2 (a) 4 (b) 5 (c) 1 (d) 3. If sin . (2007) 12. y.P. For a regular polygon. Let A and B denote the statements A : cos a + cos b + cos g = 0 B : sin a + sin b + sin g = 0 (2010) (d) (2008) 4 17 (2008) 11. p] æ p pö (d) ç . The value of is cot æ cosec -1 + tan -1 ö è (a) p/4 (b) 3p/4 (c) 5p/6 (d) p/6 3 3 ø (2012) 5 6 3 (a) (b) (c) 17 17 17 5. then AB is equal to (a) (c) p2 + q2 cos q p cos q + qsin q ( p2 + q2 )sin q 2 ( pcos q + qsin q ) 3.4 . If u = a 2 cos 2 q + b 2 sin 2 q + a 2 sin 2 q + b 2 cos 2 q (c) tana (d) cot(a/2) (2002) then the difference between the maximum and minimum values 31. the altitudes from the vertices A. In a triangle PQR.b 4 3 is of cos 2 (4 + 7) (1 + 7) 3 3 6 - 6 (c) - (d) . (b) (b) (a) (d) (a) . The upper 3/4 2 (a) 4 (b) 2sin2a tan –1 (3/5) at a point in the horizontal plane through its foot and (c) –4sin 2a (d) 4sin 2a. In a triangle ABC. (2003) 17. If sina + sinb = –21/65.P. medians AD and BE are drawn. A possible height of the vertical pole is p 18. a cos 2 2 2 14. a ¹ 0 then (a) b = a + c (b) b = c –1 –1 (c) c = a + b (d) a = b + c (2005) 28. then 4x 2 – 4xy cosa + y 2 is equal to th portion of a vertical pole subtends an angle 26.. If AD = 4. sides are in H. If r is the inradius and R is (a) 40 m (b) 60 m 2 (c) 80 m (d) 20 m. (2003) the roots of ax 2 + bx + c = 0. b and c 16. (2003) river is 60º and when he retires 40 meters away from the tree 2 2 the angle of elevation becomes 30º. sinB. b be such that p < a – b < 3p. Then the greatest angle of the triangle is 1 1 2 -1 é 1 30. (d) (b) (c) (d) (*) 5. then the area of the DABC is (d) G. (a) (b) (c) - (d) 3 4 65 65 130 130 (2004) 15.P. (2006) 24. 23. The sides of a triangle are sina. (2005) at a distance 40 m from the foot. (c) A. 30. If in a triangle ABC. if ÐR = . In a triangle with sides a. 15. 29. 13. In a triangle ABC. has a solution for ( ) ( ) ( ) ( ) ( ) 1 (a) all real values (b) | a | < 20. 20. 25. let ÐC = . b. 24. and cosa + cosb = –27/65. is a p p p (b) a cot 2 n (a) 2 cot 2 n æQö P and tan ç ÷ are 19. 21. 17. 18. (2005) (a) 16/3 (b) 32/3 (c) 64/3 (d) 8/3. (b) ArithmeticGeometric progression 25. cosa and 1 + sin a cos a for (c) a > b and b < c (d) a < b and b > c. 11.P. (2006) . 3p] satisfying the equation 2sin 2 x + 5sin x – 3 = 0 is 2 C + c cos 2 A = 3 b . B. (2002) p some 0 < a < . Let a. cot ë (cos a ) 2 ùû + tan éë (cos a ) 2 ùû = x (a) 120º (b) 90º then sin x = (c) 60º (d) 150º. (d) (c) (d) (c) (*) (b) 2. then the value (1 - 7 ) (4 - 7) (b) (a) a . 19. (2004) (a) a > b > c (b) a < b < c 21.P. 2 2 (c) 2(a + b ) (d) (a – b) 2 . The trigonometric equation sin x = 2sin a. A person standing on the bank of a river observes that the angle 2 of elevation of the top of a tree on the opposite bank of the 1 1 1 (c) | a | ³ (d) 2 < | a | < . (c) (a) (d) (c) (a) 6. r > r > r (which are the ex 1 2 3 (a) 40 m (b) 30 m radii) then (c) 20 m (d) 60 m. (d) (b) (c) (c) (a) 3. The sum of the radii of inscribed and circumscribed circles for (c) c + a (d) a + b + c (2005) an n sides regular polygon of side a.106 JEE MAIN CHAPTERWISE EXPLORER 23.P. C on opposite (a) are in G. 28. (2004) (a) 1 (b) cot 2 (a/2) 22. 8. 10. then 2(r + R) equals (a) a + b (b) b + c 27. 27. 16. 14. (2003) the circumradius of the triangle ABC. 2p) of u 2 is given by is (a) (a + b) 2 (b) 2 a 2 + b 2 (a) 2 (b) 3 (c) 0 (d) 1.cos -1 y = a .P. 26. If tan 2 2 è 2 ø a p p (c) 4 cot 2 n (d) a cot n . 22. 31. 7. The number of value of x in the interval [0. (a) (c) (a) (a) (a) 4. then the sides (a) 4 (b) 6 (c) 1 (d) 2. If cos -1 x . sinC are in (c) satisfy a + b = c (d) are in A.P. 12. c. then sinA. ÐDAB = p/6 and ÐABE = p/3. then tanx is () ( ) a. (b) are in H. (2004) (2002) Answer Key 1. The number of solutions of tan x + sec x = 2cos x in [0. 9. (2003) (a) H. The breadth of the river is 29. If in a DABC. If 0 < x < p and cosx + sinx = 1/2. n = 4. we get p2 + q2 sin q AB BD = Þ AB = sinq sin(q + b) sin(q + b ) As tan b = sin(q + b) = sinq cos b + cos q sinb q 2 p +q 2 + cos q × p + q 2 = pcosq + q sin q p2 + q2 2 2 We then get AB = ( p + q )sin q p cos q + qsin q 3. Remark : y ¹ 0 is implicit to make any of the choice correct. (d) : Using sine rule in triangle ABD. (a) : sin 2 A cos 2 A + cos A(sin A . we have q = sin q × 5. n 2 n 6 p 2 But cos = will produce no value of n. we get x = y = z. 2 6 3 5 + 4 12 = 36 + 20 = 56 3 5 48 - 15 33 1 .cos 3 A (sin A - cos A)cos A sin A (sin A . Again A = t + (1 – t) 2 = t 2 – t + 1.b) = = 1 . (b) : cos(a + b) = 4/5 giving tan(a + b) = 3/4 Also sin(a – b) = 5/13 given tan(a – b) = 5/12 r R p 1 Let cos = . (d) : As x. t ³ 0..107 Trigonometry 1.cot A 1 - tan A = = 3 £ A £ 1 .P. R = .cos A)(sin 2 A + sin A cos A + cos 2 A) = (sin A - cos A)sin A cos A 1 + sin A cos A = 1 + sec Acosec A sin A cos A 4. \ n = 6. acceptable.xz ÷ø æ 2 y ö æ x + z ö Þ tan -1 ç = tan -1 ç 2 ÷ è 1 . (c) : We have = cos p 2 Thus tan(a + b) + tan(a .P. n 3 cos 1 p p p < cos < cos 3 n 4 As 1 2 < < 2 3 Þ p p p > > Þ 3 < n < 4 (impossible) 3 n 4 2 Þ cos 8. where minimum is 3/4 p 1 p p = Þ = . (c) : A = sin 2 x + cos 2 x We have cos 4 x £ cos 2 x sin 2 x = sin 2 x Adding sin 2 x + cos 4 x £ sin 2 x + cos 2 x \ A £ 1. 2.cos A) sin A(cos A - sin A) = sin 3 A . (b) : cos(b – g) + cos(g – a) + cos(a – b) = – 3/2 Þ (cosb cosg + sinb sing) + (cosg cosa + sing sina) + (cosa cosb + sina sinb) = – 3/2 Þ 2(cosb cosg + cosg cosa + cosa cosb) + 2( sinb sing + sing sina + sina sinb) + 3 = 0 Þ {cos 2a + cos 2b + cos 2g + 2(cosa cosb + cosb cosg + cosg cosa)} 2 + {sin a + sin 2b + sin 2g + 2(sina sinb + sinb sing + sing sina)} = 0 Þ (cosa + cosb + cosg) 2 + (sina + sinb + sing) 2 = 0 Which yields simultaneously cosa + cosb + cosg = 0 and sina + sinb + sing = 0 . cos tan A cot A + 1 . n 2 n 3 \ n = 3. n i.xz ÷ø è 1 . y. z are in A. (d) :3 sin P + 4cos Q = 6 4 sin Q + 3cos P = 1 Þ 16 + 9 + 24 (sin (P + Q)) = 37 Þ 24 (sin (P + Q)) = 12 Þ sin ( P + Q) = But if R = 1 1 5 p p Þ sin R = Þ R = or 2 2 6 6 5 p p 1 then P < and then 3sin P < 6 6 2 and so 3sin P + 4 cos Q < 6. then 2tan –1 y = tan –1 x + tan –1 z æ x + z ö 2 tan -1 y = tan -1 ç è 1 . n 2 p p Thus we get = n 4 1 p + 4 ( ¹ 6) Thus.. p 3 p p = Þ = . acceptable. acceptable.b) p . 4 7.´ 4 12 p .tan(a + b)tan(a .e. tan –1 y and tan –1 z are in A.y ø … (i) Thus y 2 = xz … (ii) From (i) and (ii). Þ 2y = x + z tan –1 x. 7 of B from C i.1 £ 1 and cos x > 0 2 or 0 £ x £ 4 and . BD to BC is 3 :1 Q tan x < 0 Þ tan x = 8 BD BD 3 = 3. a b c 17.x 2 1 = cos a 2 4 a 30° B x 12.B 2 ) Þ Þ 18. (c) : 0 < x < p 60° 45° 7 C D Þ h- h = 7 Þ h ( 3 . JEE MAIN CHAPTERWISE EXPLORER (c) : 1 st Solution : Let height of the pole AB be h.sin -1 çæ ÷ö ÷ 5 è 5 ø ø è2 A x = cos æ sin -1 4 ö = cos æ cos -1 3 ö = 3 ç ç 5 5 ÷ø 5 ÷ø 5 Þ x = 3.. 510°.p < x < p \ 0 £ x < p .36 -4 ± 7 2 nd Solution : tan x = = 6 3 We use the fact that the ratio of distance of B from D and that -4 . b ..sin -1 çæ ÷ö è5ø 2 è 5 ø BC = h cot 60° = h / 3 BD = h cot 45° = h As BD – BC = CD æp x 4 ö = sin ç . 2 2 2 y = a 2 \ 60° a ) = cos -1 AB + (1 . Then p -1 æ x ö -1 æ 4 ö 13.cos -1 Þ xy y 2 + 1 . 2 D . So number of values are 4 7 3 3 7 3 like 30°. (d) : sin ç ÷ + sin ç ÷ = è5ø è 5 ø 2 x p 4 Þ sin -1 çæ ÷ö = .1) = 7 3 3 Þ h= 7 3 7 3( 3 + 1) 7 3 = = ( 3 + 1)m 2 2 3 - 1 1 Given cos x + sin x = 2 1 1 + sin 2 x = Þ 4 2 tan x -3 = 4 1 + tan 2 x Þ 3tan 2 x + 8tan x + 3 = 0 (By squaring both sides) -8 ± 64 .sin x ¹ - 3 2 1 there sin x = 2 .÷ = (1 . è è h B 14. (b) : cot æ cosec -1 + tan -1 ö è 3 3 ø \ Altitudes are in H. we know that each trigonometrical function 60° 45° assumes same value twice in 0 £ x £ 360°. CD = ×7 = ( 3 + 1) Then BD = 2 3 -1 3 - 1 16. B D C 3 2 \ 2 D . c Î A.A2 ) (1 . (a) : OP = Tower OAB is equilateral triangle \ OA = OB = AB = a In D AOP.cos -1 B ( cos -1 x . the height of the pole = 2 2 2 × Area of D \ = AD 5 2 a 10.e.(A)(Q C = 90º) and .P.108 9. tan 30° = OP OA Þ OP = a 3 A 30° a O Þ 1 1 1 .x ) ç 1 ÷ 2 ø 4 ø è è 4x 2 – 4xy cosa + y 2 = 4 (1 – cos 2a) = 4sin 2a.× ÷ø çè 4 3 11. 390°. Î H.P.P. (b) : f (x) is defined if -1 £ . 150°. B C D In our problem 0° £ x £ 540°. so that = 2 BC CD 15. (d) : Using cos -1 A . 2 D Î HP = cot æ tan -1 + tan -1 ö a a b c è ø 4 3 3 2 ö æ + ç -1 4 3 ÷ æ -1 17 ö 6 = cot ç tan 3 2 ÷ = cot è tan 6 ø = 17 1 . c = 2 R sin C \ c = 2R . (a) : 2 æ xy ö y 2 ö æ 2 ç cos a . Þ a . . (a) : 2sin x + 5 sinx –3 = 0 3 - 1 A 1 Þ sin x = . (c) : Altitude from A to BC is AD A 1 AD ´ BC 7 3 Area of D = ( 3 + 1)m As AB = BD. (B) Q Q tan P + tan = 1 .b ) = cos 2 = = 4 ´ 65 ´ 65 (130) ´ (130) 2 130 a . (c) : ÐR = 90° \ ÐP + ÐQ = 90° Q P 90 Q P \ = - . (a) : If a = sin a..109 Trigonometry tan C = r 2 s-c A 2(1 + cos a cos b + sin a sin b) = \ r = (s – c) ( tan 2[1 + cos (a – b)] = R ) B C a + b + c = - c ; 2r = a + b – c 2 adding (A) and (B) we get 2(r + R) = a + b. and r be the radius of inscribed circle then R = a a p p cosec and r = cot 2 2 2n n \ R + r = a æ p pö ç cosec + cot ÷ 2 è n n ø pö æ 1 + cos ÷ aç n = a cot p = ç ÷ 2 ç sin p ÷ 2 2 n ç ÷ n ø è A O R π π R n n B C L B OL = r OB = R ..tan a tan b = 1 . (a) : If R be the radius of circumcircle of regular polygon of n sides. (c) : sin a + sin b = - 65 27 and cos a + cos b = - 65 by squaring and adding we get E p/3 B O C D 1 8 3 16 3 ´ 4 = Area of triangle ADB = ´ 2 9 9 16 3 32 3 = Area of triangle ABC = 2 × 9 9 + b 4 sin 2 q cos 2 q + a 2 b 2 sin 4 q + cos 4 q = a 2 + b 2 + 2 A OB 25. (d) : 2a cos 2 Q æ ö P 2 ç Q tan .ç ÷ ç ÷ è 40 ø è 160 ø h 2 – 200h + 6400 = 0 (h – 40) (h – 160) = 0 h = 40 or h = 160 27. (*) : = tan 30° AO OA 8 3 uuur = Þ OB = 9 3 26. a 30° 20. c = 1 + sin a cos a then cos c = 60° O .b = 1 Þ c = a + b.tan A tan a 22. (c) : Breadth of river OC = AC cos 60° 30° = 40 cos 60° B C 2 2 2 2 2 21. . u 2 = a 2 + b 2 + 2ab = (a + b) 2 Now Maximum u 2 – Minimum u 2 = 2(a 2 + b 2 ) – (a 2 + b 2 + 2ab) = a 2 + b 2 – 2ab = (a – b) 2 = a 2 + b 2 + 2 + 21 23. = 45 2 2 2 2 2 1 tan Q / 2 tan P / 2 = Þ 1 1 + tan Q / 2 Þ (65)2 (65) 2 1170 130 ´ 9 9 (a . b = cos a.b 3 = \ cos 2 130 r C = tan 45° = 1 2 (21)2 + (27) 2 Þ Þ Þ 3 = 5 3/4 h b a A 1/4 h h æ h ö +ç÷ 40 è 160 ø æ h öæ h ö 1 . tan are roots of ax + bx + c = 0 ÷ 2 2 è ø Þ 1170 æa -bö As p < a – b < 3p then cos ç ÷ = negative è 2 ø 19.tan P × tan 2 2 2 2 Þ - Þ Þ Þ Þ b c = 1 - a a A c . (d) : u 2 = a 2 (cos 2q + sin 2q) + b 2 (sin 2q + cos 2q) + 2 C + 2c cos 2 A/2 =3b (from given) 2 a(1 + cos C) + c(1 + cos A) = 3b a + c + a cos C + c cos A = 3b (a cos C + c cos A = b projection formula) a + c + b = 3b a + c = 2b 24. u 2 = 2(a 2 + b 2 ) and Min. (a) : a = A + b \ b = A – a tan A .b 2 ö a 2 b 2 + ç ÷ sin 2 q ç 2 ÷ø è \ u 2 will be maximum or minimum according as q = p/4 or q = 0° \ Max.sin a cos a \ cos c = –1/2 2 sin a cos a (a 4 ) ( 2 2 ( 2 a b + a -b 2 2 ) ) sin Þ 2 2 q cos q æ a 2 . 150°.p £ sin -1 x £ p ú 2 2 û ë D D D > > s . (*) : sin –1 x = 2 sin –1 a Þ Þ - p p £ 2 sin –1 a £ 2 2 p p Þ £ sin –1 a £ 4 4 æ pö æpö Þ sin ç .110 JEE MAIN CHAPTERWISE EXPLORER 28. sin x = –1 2 so there are three solution like x = 30°. (b) : tan x + sec x = 2 cos x 1 + sin x = 2 cos 2 x sin 150° = ½ sin 30° = ½ 1 + sin x = 2(1 – sin 2 x) Þ 2 sin 2 x + sin x – 1 = 0 Þ (2 sin x – 1) (1 + sin x) = 0 1 sin 270° = – 1 Þ sin x = .÷ £ a £ sin ç ÷ è 4ø è4ø Þ - 1 2 Þ |a| £ £ a £ 1 2 1 2 No choice is matched. (a) : As r 1 > r 2 > r 3 Þ éQ sin -1 x = 2sin -1 a ù ê ú ê and . (a) : Using tan –1q + cot –1q = \ sin x = sin p = x 2 p = 1 2 31.a s . 29. 270° .b s - c Þ s-a s-b s-c < < D D D a > b > c 30. (c) Statement1 is false. (c) 5. (a) (a) 2. (2013) (b) Statement 1 is true. R : Suman is honest The negation of the statement “Suman is brilliant and dishonest Statement1 : r is equivalent to either q or p. if and only if Suman is rich” can be expressed as Statement2 : r is equivalent to ~ ( p « ~ q). (c) Every rational number x Î S satisfies x £ 0. Statement2 is true. Let S be a nonempty subset of R . then I (c) Statement1 is false. Statement2 : ~ (p « ~ q) is a tautology. (2010) not a correct explanation for Statement1. Statement1 : ~ (p « ~ q) is equivalent to p « q. Statement2 is true; Statement2 is a 4. (b) There is no rational number x Î S such that x £ 0. Statement2 : (p ® q) « (~ q ® ~ p) is a tautology. correct explanation for Statement1 (2009) (c) I will become a teacher and I will not open a school. Statement2 is true; Statement2 is (d) Statement1 is true. Statement2 is false (c) ~ P Ù (Q « ~ R) (d) ~ (Q « (P Ù ~ R)) (b) Statemen1 is false. Statement2 is true; Statement2 is (b) I will not become a teacher or I will open a school. Statement2 is false 2. Statement2 is true. (a) Statement1 is true. and r be the statement Q : Suman is rich “x is a rational number iff y is a transcendental number”. (b) 6. Statement2 is true will open a school”. q be the P : Suman is brilliant statement “y is a transcendental number”. (d) Either I will not become a teacher or I will not open a 6. (a) ~ Q « ~ P Ù R (b) ~ (P Ù ~ R) « Q (a) Statement1 is true. (a) Statement1 is true. (d) Statement1 is true. (b) Statement1 is true. Statement2 is true; Statement2 is P : There is a rational number x Î S such that x > 0. 7. (d) 4. The negation of the statement “If I become a teacher. is (a) Neither I will become a teacher nor I will open a school. Consider the following correct explanation for Statement1 statement: (d) Statement1 is true. Consider : (a) There is a rational number x Î S such that x £ 0. Let p be the statement “x is an irrational number”. (c) 3. The statement p ® (q ® p) is equivalent to (a) p ® ( p « q) (b) p ® ( p ® q) school. Statement2 is a not a correct explanation for Statement1 correct explanation for Statement1. Statement2 is (d) x Î S and x £ 0 Þ x is not rational. Statement1 : (p Ù ~ q) Ù (~ p Ù q) is a fallacy. Statement2 is false. 5. not a correct explanation for Statement1 (2008) Which of the following statements is the negation of the statement P ? Answer Key 1. Statement2 is true (2011) (c) Statement1 is true. Consider the following statements 7. (c) .111 Mathematical Logic CHAPTER MATHEMATICAL LOGIC 18 1. Statement2 is true. (2012) (c) p ® ( p Ú q) (d) p ® ( p Ù q) (2008) 3. (a) : 1 st solution : Let's prepare the truth table for the statements. . (d) : The statement can be written as P Ù ~R Û Q p q ~p ~q p Ù ~q ~p Ù q (p Ù ~q) Ù (~p Ù q) T T F F F F F T F F T T F F F T T F F T F F F T T F F F Thus the negation is ~ (Q « P Ù ~ R) Then Statement1 is fallacy. 3. 2 nd solution : ~ (~ p Ú q) Ù ~ (~ q Ú p) ~(p « ~q) is NOT a tautology because it’s statement value is not º ~ ((~ p Ú q) Ú (~ q Ú p)) º ~ ((p ® q) Ú (q ® p)) º ~ T always true. . ((p ® q) ® (~ q ® ~ p) Ù ((~ q ® ~ p) ® (p ® q)) = ((~ p Ú q) ® (q Ú ~ p) Ù ((q Ú ~ p) ® (~ p Ú q)) º T Ù T º T. The negation would be : There is no rational number x Î S such that x > 0 which is equivalent to all rational numbers x Î S satisfy x £ 0. then I will open a school’’ Negation of the given statement is ‘‘ I will become a teacher and I will not open a school’’ ( . Thus Statement1 is true because its negation is false. (a) : The given statement r º ~ p « q The Statement1 is r 1 º (p Ù ~ q) Ú (~ p Ù q) The Statement2 is r 2 º ~ (p « ~ q) = (p Ù q) Ú (~ q Ù ~ p) we can establish that r = r 1 Thus Statement1 is true but Statement2 is false. 2. ~ (p ® q) = p Ù ~ q) 6. . 4. (c) : The given statement is ‘‘If I become a teacher. we conclude that ~(p « ~q) is equivalent to (p « q). (c) : Let’s simplify the statement p ® (q ® p) = ~ p Ú (q ® p) = ~ p Ú (~ q Ú p) = – p Ú p Ú ~ q = p ® (p Ú q) 7. (c) : The given statement is P : at least one rational x Î S such that x > 0.112 JEE MAIN CHAPTERWISE EXPLORER 1. Then Statement2 is true. (b) : Let’s prepare the truth table T T F F T T T T F F T F F T F T T F T T T F F T T T T T Then Statement2 is tautology. p T T F q T F T ~q F T F p « q T F F p « ~q F T T F F T T F ~(p « ~q) T F F T As the column for ~(p « ~q) and (p « q) is the same. p q ~ p ~ q p ® q ~ q ® p (p ® q) ® (~ q ® p) 5.