AIIMS 2017 Solved Paper.pdf

1AIIMS 2017 Solved Paper This Paper ''AIIMS 2017 Solved Paper'' is taken from our Book: ISBN : 9789386629746 2 AIIMS 2 0 17 PHYSICS 55. An organ pipe open at one end is vibrating in 107. The potential difference that must be applied to first overtone and is in resonance with another stop the fastest photoelectrons emitted by a pipe open at both ends and vibrating in third nickel surface, having work function 5.01 eV, harmonic. The ratio of length of two pipes is when ultraviolet light of 200 nm falls on it, must (a) 1 : 2 (b) 4 : 1 be (c) 8 : 3 (d) 3 : 8 (a) 2.4 V (b) – 1.2 V 5. The normal density of gold is r and its bulk (c) – 2.4 V (d) 1.2 V modulus is K. The increase in density of a lump 53. Hail storms are observed to strike the surface of of gold when a pressure P is applied uniformly the frozen lake at 300 with the vertical and on all sides is rebound at 600 with the vertical. Assume contact (a) K/r P (b) P/r K to be smooth, the coefficient of restitution is (c) r P/K (d) r K/P 1 1 102. Magnetic flux f in weber in a closed circuit of (a) e = (b) e = resistance 10W varies with time f (sec) as f = 6t2 3 3 – 5t + 1. The magnitude of induced current at t = (c) e = 3 (d) e = 3 0.25s is 146. In a npn transistor 1010 electrons enter the emitter (a) 0.2 A (b) 0.6 A in 10–6 s. 4% of the electrons are lost in the base. (c) 1.2 A (d) 0.8 A The current transfer ratio will be 4. Which of the following displacement (X) time (a) 0.98 (b) 0.97 graphs is not possible? (c) 0.96 (d) 0.94 x x 122. Each of the resistance in the network shown in fig. is equal to R. The resistance between the terminals A and B is (a) (b) O t O t L x X R R R A (c) (d) O t R R B t K M O (a) R (b) 5 R 2 (c) 3 R (d) 6 R 39. The binding energy per nucleon for 1H and 62. The wheel of a car is rotating at the rate of 1200 4 revolutions per minute. On pressing the 2 He respectively are 1.1 MeV and 7.1 MeV. The accelerator for 10 seconds it starts rotating at energy released in MeV when two 12 H nuclei 4500 revolutions per minute. The angular acceleration of the wheel is fuse to form 42 He is (a) 30 radian / second2 (b) 1880 degrees/ second2 (a) 4.4 (b) 8.2 (c) 40 radian / second2 (c) 24 (d) 28.4 (d) 1980 degree/second2  3 119. If a charge q is placed at the centre of the line (b) 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 35. The potential energy of a particle varies with (c) A x distance x from a fixed origin as V = where + + x+B A and B are constants. The dimensions of AB are (a) [M1 L5/ 2 T -2 ] (b) [M1 L2 T -2 ] (d) (c) 3/ 2 5/ 2 -2 [M L T ] + + 1 7/2 (d) [M L T ] -2 83. A light ray falls on a rectangular glass slab as shown. The index of refraction of the glass, if total internal reflection is to occur at the vertical 58. A point particle of mass 0.1 kg is executing S.H.M. face, is of amplitude of 0.1 m. When the particle passes 45º (a) 3/ 2 through the mean position, its kinetic energy is 8 × 10–3 Joule. Obtain the equation of motion of this (b) ( 3 + 1) particle if this initial phase of oscillation is 45º. 2 Glass æ pö y = 0.1sin ç ±4t + (c) ( 2 + 1) (a) è ÷ 4ø 2 æ pö (b) y = 0.2sin ç ±4t + ÷ (d) 5 /2 è 4ø 77. A bucket tied at the end of a 1.6 m long string is æ pö whirled in a vertical circle with constant speed. (c) y = 0.1sin ç ±2t + ÷ è 4ø What should be the minimum speed so that the water from the bucket does not spill when the pö æ bucket is at the highest position? (d) y = 0.2sin ç ±2t + ÷ è 4ø (a) 4 m/sec (b) 6.25 m/sec 99. The following configuration of gate is (c) 16 m/sec equivalent to (d) None of the above A OR 74. Which of the following figure shows the correct B equipotential surfaces of a system of two positive Y charges? AND (a) NAND + + (a) NAND gate (b) XOR gate (c) OR gate (d) NOR gate 18. Two long parallel wires P and Q are held perpendicular to the plane of paper with distance of 5 m between them. If P and Q carry current of 2.5 amp. and 5 amp. respectively in the same 4 direction, then the magnetic field at a point half- 114. In the equation X = 3YZ2 , X and Z are way between the wires is dimensions of capacitance and magnetic induction respectively. In MKSQ system, the (a) m 0 / 17 (b) 3 m0 / 2 p dimensional formula for Y is (c) m 0 / 2 p (d) 3 m 0 / 2 p (a) [M–3 L–2 T–2 Q–4] 70. A small block of mass m is kept on a rough (b) [M L–2] inclined surface of inclination q fixed in an (c) [M–3 L–2 Q4 T8] elevator. The elevator goes up with a uniform (d) [M–3 L–2 Q4 T4] velocity v and the block does not slide on the 66. Half lives for a and b emission of a radioactive wedge. The work done by the force of friction material are 16 years and 48 years respectively. on the block in time t as seen by the observer on When material decays giving a and b emission the inclined p lane will be simultaneously, time in which 3/4th material (a) zero (b) mgvt cos2q decays is (c) mgvt sin q 2 (d) mgvt sin 2q (a) 29 years (b) 24 years 123. A beam of light of wavelength 600 nm from a (c) 64 years (d) 12 years distance source falls on a single slit 1 mm wide 41. If a magnet is suspended at angle 30º to the and a resulting diffraction pattern is observed magnetic meridian, the dip needle makes an angle on a screen 2m away. The distance between the of 45º with the horizontal. The real dip is first dark fringes on either side of central bright fringe is (a) tan -1 ( 3 / 2 ) (a) 1.2 cm (b) 1.2 mm (c) 2.4 cm (d) 2.4 mm (b) tan -1 ( 3 ) 51. Figure here shows the vertical cross section of (c) tan -1 ( 3 / 2) a vessel filled with a liquid of density r. The normal thrust per unit area on the walls of the (d)tan -1 (2 / 3 ) vessel at the point P, as shown, will be 50. Gauss’s law states that (a) the total electric flux through a closed surface 1 P is times the total charge placed near the e0 H h closed surface. (b) the total electric flux through a closed surface 1 q is times the total charge enclosed by the e0 closed surface. (a) hrg (c) the total electric flux through an open surface (b) Hrg (c) (H – h) rg is 1 times the total charge placed near the (d) (H – h) rg cosq e0 47. If in the experiment of Wheatstone’s bridge, the open surface. positions of cells and galvanometer are (d) the line integral of electric field around the interchanged, then balance point will 1 (a) change boundary of an open surface is times the e0 (b) remain unchanged (c) depend on the internal resistance of cell and total charge placed near the open surface. resistance of galvanometer 18. A current carrying coil is subjected to a uniform (d) None of these magnetic field. The coil will orient so that its plane becomes (a) inclined at 45º to the magneic field  5 (b) inclined at any arbitrary angle to the 65. The temperature of the two outer surfaces of a magnetic field composite slab consisting of two materials (c) parallel to the magnetic field having coefficient of thermal conductivity K and (d) perpendicular to magnetic field 2K and thickness x and 4x respectively are T2 33. Radio waves and visible light in vacuum have and T1 (T2 > T 1). The rate of heat transfer (a) same velocity but different wavelength through the slab, in a steady state is (b) continuous emission spectrum æ A(T2 - T1 )K ö (c) band absorption spectrum çè ÷ø f with f equal to x (d) line emission spectrum x 4x 28. The work done in which of the following processes is equal to the internal energy of the K T2 2K T1 system? (a) Adiabatic process (b) Isothermal process (a) 1 (b) 1/2 (c) Isochoric process (c) 2/3 (d) 1/3 (d) None of these 37. The ratio of the longest to shortest wavelengths 90. Block A of weight 100 kg rests on a block B and in Brackett series of hydrogen spectra is is tied with horizontal string to the wall at C. Block B is of 200 kg. The coefficient of friction 25 17 (a) (b) between A and B is 0.25 and that between B and 9 6 1 9 4 surface is . The horizontal force F necessary (c) (d) 3 5 3 to move the block B should be (g = 10 m/s2) 41. Curie temperature is the temperature above which (a) a ferromagnetic material becomes (a) 1050 N A paramagenetic (b) 1450 N C (b) a paramagnetic material becomes (c) 1050 N B F diamagnetic (d) 1250 N (c) a ferromagnetic material becomes diamagnetic 129. In an A.C. circuit, the current flowing in (d) a paramagnetic meterial becomes inductance is I = 5 sin (100 t – p/2) amperes and ferromagnetic the potential difference is V = 200 sin (100 t) volts. 54. Five masses are placed in a plane as shown in The power consumption is equal to figure. The coordinates of the centre of mass are (a) 1000 watt (b) 40 watt nearest to y 2 3 kg (c) 20 watt (d) Zero 4 kg 1. Figure shows the electric lines of force emerging from a charged body. If the electric field at A and 5 kg B are E A and E B respectively and if the 1 displacement between A and B is r, then 0 1 kg 2 kg x 0 1 2 A r B (a) 1.2, 1.4 (b) 1.3, 1.1 (c) 1.1, 1.3 (d) 1.0, 1.0 29. Two spherical conductors A and B of radii a and (a) EA > EB (b) EA < EB b (b>a) are placed concentrically in air. The two (c) EA = EB/r (d) EA = EB/r2 are connected by a copper wire as shown in figure. Then the equivalent capacitance of the system is 6 proportionately. ab (a) 4pe 0 117. Assertion : Resonance is special case of forced b-a b B vibration in which the natural frequency of A (b) 4pe 0 (a + b) a vibration of the body is the same as the impressed frequency of external periodic force (c) 4pe 0 b and the amplitude of forced vibration is maximum (d) 4pe 0a Reason : The amplitude of forced vibrations of a 30. The magnitude of the de-Broglie wavelength (l) body increases with an increase in the frequency of electron (e), proton (p), neutron (n) and a- of the externally impressed periodic force. particle (a) all having the same energy of 1 MeV, 103. Assertion : Kirchoff’s juction rule can be in the increasing order will follow the sequence applied to a junction of several lines or a point (a) le, lp, ln, la (b) le, ln, lp, la in a line. (c) la, ln, lp, le (d) lp, le, la, ln Reason : When steady current is flowing, 54. The molar heat capacities of a mixture of two gases there is no accumulation of charges at any at constant volume is 13R/6. The ratio of number junction or at any point in a line. of moles of the first gas to the second is 1 : 2. The 102. Assertion : When a sphere is rolls on a horizontal respective gases may be table it slows down and eventually stops. (a) O2 and N2 (b) He and Ne Reason : When the sphere rolls on the table, (c) He and N2 (d) N2 and He both the sphere and the surface deform near the 39. By properly combining two prisms made of contact. As a result, the normal force does not different materials, it is not possible to have pass through the centre and provide an angular (a) dispersion without average deviation declaration. (b) deviation without dispersion 114. Assertion : When two semi conductor of p and (c) both dispersion and average deviation n type are brought in contact, they form p-n (d) neither dispersion nor average deviation junction which act like a rectifier. Reason : A rectifier is used to convent alternating ASSERTION - REASON TYPE QUESTIONS current into direct current. 86. Assertion: Ampere’s circuital law is independent Directions : Each of these questions contains two of Biot-Savart’s law. statements, Assertion and Reason. Each of these Reason: Ampere’s circuital law can be derived questions also has four alternative choices, only one from the Biot-savart’s law. of which is the correct answer. You have to select one 111. STATEMENT-1 : Mechanical energy is the sum of the codes (a), (b), (c) and (d) given below. of macroscopic kinetic & potential energies. (a) Assertion is correct, reason is correct; reason is STATEMENT-2 : Mechanical energy is that part a correct explanation for assertion. of total energy which always remain conserved. (b) Assertion is correct, reason is correct; reason 95. Assertion : A laminated core is used in is not a correct explanation for assertion transformers to increase eddy currents. (c) Assertion is correct, reason is incorrect Reason: The efficiency of a transformer increases (d) Assertion is incorrect, reason is correct. with increase in eddy currents. 96. Assertion : When a convex lens (µg= 3/2) of 77. Assertion: In the measurement of physical focal length f is dipped in water, its focal length quantities direct and indirect methods are used. 4 Reason : The accuracy and precision of becomes f . 3 measuring instruments along with errors in Reason : The focal length of convex lens in water measurements should be taken into account, becomes 4f. while expressing the result. 90. Assertion: The pressure of water reduces when it flows from a narrow pipe to a wider pipe. Reason: Since for wider pipe area is large, so flow of speed is small and pressure also reduces  7 116. Assertion : The Carnot cycle is useful in CHEMISTRY understandsing the performance of heat engines. Reason : The Carnot cycle provides a way of 5. Sodium metal crystallizes in a body centred cubic determining the maximum possible efficiency lattice with a unit cell edge of 4.29Å. The radius achievable with reservoirs of given temperatures. of sodium atom is approximately : [2017] 82. Assertion : Lenz's law violates the principle of (a) 5.72Å (b) 0.93Å conservation of energy. (c) 1.86Å (d) 3.22Å Reason : Induced emf always opposes the 8. Which of the following compounds is not an change in magnetic flux responsible for its antacid ? [2017] production. (a) Phenelzine 90. Assertion : A bullet is fired from a rifle. If the (b) Ranitidine rifle recoils freely, the kinetic energy of rifle is (c) Aluminium hydroxide more than that of the bullet. (d) Cimetidine Reason : In case of rifle bullet system, the law 21. The synthesis of alkyl fluorides is best accomplished of conservation of momentum violates. by : [2017] 104. Assertion : Orbital velocity of a satellite is (a) Finkelstein reaction greater than its escape velocity. (b) Swarts reaction Reason : Orbit of a satellite is within the (c) Free radical fluorination gravitational field of earth whereas escaping is (d) Sandmeyer's reaction beyond the gravitational field of earth. 20. In Bohr series of lines of hydrogen spectrum, 94. Assertion : In the absence of an external electric the third line from the red end corresponds to field, the dipole moment per unit volume of a which one of the following inter-orbit jumps of polar dielectric is zero. the electron for Bohr orbits in an atom of Reason : The dipoles of a polar dielectric are randomly oriented. hydrogen [2017] 70. Assertion : Identical springs of steel and copper (a) 5®2 (b) 4 ®1 are equally stretched. More work will be done on (c) 2 ® 5 (d) 3 ® 2 the steel spring 109. The ether that undergoes electrophilic Reason : Steel is more elastic than copper. substitution reactions is [2017] 107. Assertion : Electromagnets are made of soft iron. (a) CH3OC2H5 (b) C6H5OCH3 Reason : Coercivity of soft iron is small. 86. Assertion : The internal energy of a real gas is (c) CH3OCH3 (d) C2H5OC2H5 function of both, temperature and volume. 62. Aldol condensation will not be observed in Reason : Internal kinetic energy depends on (a) chloral [2017] temperature and internal potential energy (b) phenylacetaldehyde depends on volume. (c) hexanal 69. Statement-1 : The de-Broglie wavelength of a (d) nitromethane molecule (in a sample of ideal gas) varies inversely 84. The end product (C) in the following sequence as the square root of absolute temperature. of reactions is [2017] Statement-2 : The rms velocity of a molecule (in 1% HgSO CH MgX [O] HC º CH ¾¾ ¾ ¾¾ 4 ® A ¾¾ ¾ 3 ¾ ¾® B ¾¾® ¾ (C) a sample of ideal gas) depends on temperature. 20% H 2SO 4 H 2O 111. Assertion : Two longitudinal waves given by (a) acetic acid (b) isopropyl alcohol equations – yl (x, t) = 2a sin(wt – kx) and y2 (x, t) = a sin(2wt – 2kx) will have equal intensity. (c) acetone (d) ethanol Reason : Intensity of waves of given frequency 37. The reaction in same medium is proportional to square of Red P RCH 2 CH 2 COOH ¾¾¾® R - CH 2 - CH - COOH amplitude only. Br2 | Br is called as [2017] 8 (a) Reimer- Tiemann reaction 39. Which of the following relation represents (b) Hell-volhard Zelinsky reaction correct relation between standard electrode (c) Cannizzaro reaction potential and equilibrium constant? [2017] (d) Sandmeyer reaction nFE° I. log K = 24. A triglyceride can have how many different acyl 2.303 RT groups? [2017] nFE ° (a) 3 (b) 2 II. K = e RT (c) 1 (d) 4 -nFE° III. log K = 23. a - D-(+)-glucose and b-D-(+)-glucose are [2017] 2.303 RT -nFE° (a) conformers (b) epimers IV. log K = 0.4342 RT (c) anomers (d) enatiomers Choose the correct statement(s). 6. Which one of the following is not a (a) I, II and III are correct condensation polymer ? [2017] (b) II and III are correct (a) Melamine (b) Glyptal (c) I, II and IV are correct (c) Dacron (d) Neoprene (d) I and IV are correct 17. Teflon and neoprene are [2017] 115. At 25°C, the solubility product of Mg(OH)2 is 1.0 × 10–11. At which pH, will Mg2+ ions start (a) copolymers precipitating in the form of Mg(OH)2 from a (b) condensation polymers solution of 0.001 M Mg2+ ions? [2017] (c) homopolymers (a) 9 (b) 10 (d) monomers (c) 11 (d) 8 60. In the reaction 67. In the given reaction NaOH CO2 + HCl X Phenol ¾¾ ¾® (A) ¾¾¾¾® ¾ 140° (B), here B CH 3CH 2 CH = CHCH 3 ¾¾® is [2017] CH3CH 2COOH +CH3COOH (a) benzaldehyde (b) chlorobenzene The X is [2017] (c) benzoic acid (d) salicylic acid 39. The molar heat capacity of water at constant (a) C2 H5ONa pressure is 75 JK–1 mol–1. When 1kJ of heat is (b) Conc. HCl +Anhy.ZnCl2 supplied to 100 g of water, which is free to (c) Anh. AlCl3 expand, the increase in temperature of water is (d) KMnO4/OH– [2017] 128. The strongest ortho - para and strongest meta - (a) 6.6 K (b) 1.2 K directing groups respectively are [2017] (c) 2.4 K (d) 4.8 K (a) –NO2 and –NH2 51. The D f H° for CO2(g) , CO(g) and H2O(g) are – (b) –CONH2 and –NH2 (c) –NH2 and –CONH2 393.5, –110.5 and –241.8 kJ/mol respectively, the (d) –NH2 and –NO2 standard enthalpy change (in kJ) for the reaction 99. Volume of water needed to mix with 10 mL 10N CO2(g) + H2(g) ® CO(g) + H2O(g) is : [2017] HNO3 to get 0.1 N HNO3 is : [2017] (a) 524.1 (b) 41.2 (a) 1000 mL (b) 990 mL (c) – 262.5 (d) – 41.2 (c) 1010 mL (d) 10 mL 21. For the following reaction in gaseous phase 183. Hybridisation states of C in CH3+ and CH4 are 1 CO( g ) + O 2 ( g ) ® CO 2 ( g ), K p / K c is [2017] 2 (a) sp2 & sp3 (b) sp3 & sp2 [2017] (c) sp2 & sp2 (d) sp3 & sp3 (a) (RT)1/2 (b) (RT)–1/2 (c) (RT) (d) (RT)–1  9 192. Which of the following substances has the least 19. Which of the following fluorides does not exist? covalent character ? [2017] [2017] (a) Cl2O (b) NCl3 (a) NF5 (b) PF5 (c) PbCl2 (d) BaCl2 (c) AsF 5 (d) SbF5 4. The law of triads is applicable to a group of 125. Which of the following are peroxoacids of [2017] sulphur? [2017] (a) Cl, Br, I (b) C, N, O (a) H2SO5 and H2S2O8 (c) Na, K, Rb (d) H, O, N (b) H2SO5 and H2S2O7 57. Consider the following reaction occuring in basic (c) H2S2O7 and H2S2O8 medium [2017] (d) H2S2O6 and H2S2O7 35. For d block elements the first ionization potential 2MnO –4 (aq)+Br – (aq) ¾¾ ® is of the order [2017] 2MnO2 (s) + BrO3– (aq) (a) Zn > Fe > Cu > Cr How the above reaction can be balanced (b) Sc = Ti < V = Cr further? (c) Zn < Cu < Ni < Co (a) By adding 2 OH– ions on right side (d) V > Cr > Mn > Fe (b) By adding one H2O molecule to left side 57. Which of the following coordination compounds (c) By adding 2H+ ions on right side would exhibit optical isomerism? [2017] (d) Both (a) and (b) (a) pentamminenitrocobalt(III) iodide 19. On the basis of the following E° values, the (b) diamminedichloroplatinum(II) strongest oxidizing agent is : [2017] (c) trans-dicyanobis (ethylenediamine) [Fe(CN)6]4– ®[Fe(CN)6]3– + e– ; E° = – 0.35 V chromium (III) chloride Fe2+ ® Fe3+ + e–; E° = – 0.77 V (d) tris-(ethylendiamine) cobalt (III) bromide 4– 2+ 107. A solution of urea (mol. mass 56 g mol-1) boils (a) [Fe(CN)6] (b) Fe at 100.18°C at the atmospheric pressure. If Kf (c) Fe3+ (d) [Fe(CN)6]3– and Kb for water are 1.86 and 0.512 K kg mol-1 35. Consider the following cell reaction: [2017] respectively, the above solution will freeze at 2Fe( s) + O2 ( g ) + 4H + ( aq) ® [2017] (a) 0.654°C (b) - 0.654°C 2Fe2+ (aq) + 2H 2 O(l ); E ° = 1.67V (c) 6.54°C (d) - 6.54°C At [Fe2+] = 10–3 M, p(O2) = 0.1 atm and pH = 3, 44. Pure hydrogen sulphide is stored in a tank of the cell potential at 25ºC is 100 litre capacity at 20°C and 2 atm pressure. (a) 1.47 V (b) 1.77 V The mass of the gas will be [2017] (c) 1.87 V (d) 1.57 V (a) 34 g (b) 340 g 101. Which one of the following impurities present in (c) 282.68 g (d) 28.24 g colloidal solution cannot be removed by 85. The increasing order of stability of the following electrodialysis? [2017] free radicals is [2017] (a) Sodium chloride • • • (a) (C6H5)2 C H < (C6H5)3 C < (CH3)3 C < (b) Potassium sulphate • (CH3)2 C H (c) Urea • • • (d) Calcium chloride (b) (CH3)2 C H < (CH3)3 C < (C6H5)2 C H < • 27. In the Victor-Meyer’s test, the colour given by (C6H5)3C 1°, 2° and 3° alcohols are respectively. [2017] • • • (c) (CH3)2 C H < (CH3)3 C < (C6H5)2 C H < (a) red, colourless, blue • (C6H5)3 C (b) red, blue, colourless • • • (d) (C6H5)3C < (C6H5)2 C H < (CH3)3 C < (c) blue, red, violet • (d) red, blue, violet (CH3)2 C H 10 48. 2CuFeS2 + O 2 ¾¾ ® Cu 2 S + 2FeS + SO 2 Reason : The phenomenon in which ore is mixed Which process of metallurgy of copper is with suitable flux and coke is heated to fusion is represented by above equation? [2017] known as smelting. (a) Concentration (b) Roasting 247. Assertion : Both rhombic and monoclinic (c) Reduction (d) Purification sulphur exist as S8 but oxygen exists as O2. 85. Which of the following are intermediates in [2017] Sandmeyer reaction ? [2017] Reason : Oxygen forms pp – pp multiple bond due to small size and small bond length but (i) C 6 H 5 N + º NCl - (ii) C6 H5 N+ º N pp – pp bonding is not possible in sulphur. g 107. Assertoin : Aniline does not undergo Friedel- (iii) C 6 H 5 (iv) C6H5Cl (a) (ii) and (iii) (b) (i) and (iv) Crafts reaction. [2017] (c) (ii) and (iv) (d) (i) and (ii) Reason : –NH2 group of aniline reacts with AlCl3 (Lewis acid) to give acid-base reaction. 53. When zeolite (hydrated sodium aluminium silicate) is treated with hard water the sodium 123. Assertion : Equal moles of different substances ions are exchanged with [2017] contain same number of constituent particles. (a) H+ ions (b) Ca2+ ions [2017] Reason : Equal weights of different substances (c) SO 4 2- ions (d) OH– ions contain the same number of constituent 16. A laboratory reagent imparts green colour to the particles. flame. On heating with solid K2Cr2O7 and conc. 89. Assertion : HClO4 is a stronger acid than HClO3. H2SO4 it evolves a red gas. Identify the reagent [2017] [2017] Reason : Oxidation state of Cl in HClO4 is +VII (a) CaCl2 (b) BaCl2 and in HClO3 +V. (c) CuCl2 (d) None of these 108. Assertion : Lithium carbonate is not so stable to heat. [2017] ASSERTION / REASON Reason : Lithium being very small in size Directions : Each of these questions contain two statements, Assertion and Reason. Each of these polarizes large CO32 - ion leading to the questions also has four alternative choices, only formation of more stable Li2O and CO2 one of which is the correct answer. You have to 154. Assertion : [Fe(CN)6]3– is weakly paramagnetic select one of the codes (a), (b), (c) and (d) given while [Fe(CN)6]4– is diamagnetic. [2017] below. Reason : [Fe(CN)6]3– has +3 oxidation state (a) Assertion is correct, reason is correct; reason is while [Fe(CN)6]4– has +2 oxidation state. a correct explanation for assertion. 150. Assertion : If one component of a solution obeys (b) Assertion is correct, reason is correct; reason is Raoult’s law over a certain range of composition, not a correct explanation for assertion the other component will not obey Henry’s law in that range. [2017] (c) Assertion is correct, reason is incorrect Reason : Raoult’s law is a special case of (d) Assertion is incorrect, reason is correct. Henry’s law. 133. Assertion : The enthalpy of physisorption is 115. Assertion : Gases do not liquefy above their greater than chemisorption. [2017] critical temperature, even on applying high Reason : Molecules of adsorbate and adsorbent pressure. [2017] are held by van der Waal’s forces in Reason : Above critical temperature, the physisorption and by chemical bonds in molecular speed is high and intermolecular chemisorption. attractions cannot hold the molecules together 101. Assertion : Coke and flux are used in smelting. because they escape because of high speed. [2017]  11 172. Assertion : Aniline is better nucleophile than BIOLOGY anilium ion. [2017] 67. Match column I with column II and choose the Reason : Anilium ion have +ve charge. correct option. 171. Assertion : Benzene exhibit two different bond Column-I Column-II lengths, due to C – C single and C = C double A. Family I. tuberosum bonds. [2017] B. Kingdom II. Polymoniales Reason : Actual structure of benzene is a C. Order III. Solanum hybrid of following two structures. D. Species IV. Plantae E. Genus V. Solanaceae ¬¾® (a) A – IV; B – III; C – V; D – II; E – I (b) A – V; B – IV; C – II; D – I; E – III 125. Assertion : Galvanised iron does not rust. (c) A – IV; B – V; C – II; D – I; E – III [2017] (d) A – V; B – III; C – II; D – I; E – IV Reason : Zinc has a more negative electrode 34. Consider the following statements regarding the potential than iron. major pigments and stored food in the different 118. Assertion : Atomic radius of gallium is higher groups of algae and choose the correct option than that of aluminium [2017] (i) In chlorophyceae, the stored food material Reason : The presence of additional d-electron is starch and the major pigments are chlorophyll- offer poor screening effect for the outer electrons a and d. from increased nuclear charge. [2017] (ii) In phaeophyceae, laminarian is the stored 140. Assertion : The radius of the first orbit of food and major pigments are chlorophyll-a and b. hydrogen atom is 0.529Å. [2017] (iii) In rhodophyceae, floridean starch is the Reason : Radius of each circular orbit (rn) - 0.529Å stored food and the major pigments are (n2/Z), where n = 1, 2, 3 and Z = atomic number. chlorophyll-a, d and phycoerythrin. 153. Assertion : NF 3 is a weaker ligand than (a) (i) is correct, but (ii) and (iii) are incorrect N(CH3)3. [2017] (b) (i) and (ii) are correct, but (iii) is incorrect Reason : NF3 ionizes to give F– ions in aqueous (c) (i) and (iii) are correct, but (ii) is incorrect solution. (d) (iii) is correct, but (i) and (ii) are incorrect 114. Assertion : SN2 reaction of an optically active 58. Column-I contains organisms and column-II aryl halide with an aqueous solution of KOH contains their excretory structures. Choose the always gives an alcohol with opposite sign of correct match form the options given below. rotation. [2017] Column- I Column -II Reason : SN2 reactions always proceed with (Organism) (Excretory inversion of configuration. structures) A. Cockroach I. Nephridia 152. Assertion : Magnetic moment values of B. Cat fish II. Malpighian actinides are lesser than the theoretically tubules predicted values. [2017] C. Earthworm III. Kidneys Reason : Actinide elements are strongly D. Balanoglossus IV. Flame cells paramagnetic. E. Flatworm V. Proboscis 107. Assertion : Sedatives are given to patients who gland are mentally agitated and violent. [2017] (a) A – I; B – III; C – II; D – IV; E – V Reason : Sedatives are used to suppress the (b) A – III; B – I; C – II; D – V; E – IV activities of central nervous system. [2017] (c) A – II; B – I; C – III; D – V; E – IV 107. Assertion : In vulcanisation of rubber, sulphur (d) A – II; B – III; C – I; D – V; E – IV cross links are introduced. [2017] 65. In which one of the following the genus name, Reason : Vulcanisation is a free radical initiated its two characters and phylum are not correctly matched ? chain reaction. 12 Genus Two characters Phylum 2(C51H 98 O6 ) + 145O 2 ¾¾ ® 102CO 2 name +98H 2O + Energy (i) Body segmented (a) Pila Mollusca The R.Q of above reaction is Mouth with radula (a) 1 (b) 0.7 (ii) Spiny skinned (c) 1.45 (d) 1.62 (b) Asterias Echinodermata 74. Assertion : Water and electrolytes are almost Water vascular system fully absorbed in the large intestine. (iii) Pore bearing Reason : In large intestine, haustral contractions (c) Sycon Porifera Canal system (slow segmenting movements) roll the forming (iv) Jointed appendages faeces over and over, causing absorption of (d) Periplaneta Arthropoda water and electrolytes. Chitinous exoskeleton 71. Assertion : A cerebellum is related with skillful 47. Assertion : In a DNA molecule, A–T rich parts voluntary movement and involuntary activity melt before G–C rich parts. like body balance, equilibrium etc. Reason: In between A and T there are three Reason : It is part of hind brain and it is situated H–bond, whereas in between G and C there are behind the pons. two H-bonds. 74. In a practical test, a student has to identify the 83. Nucleotides are building blocks of nucleic acids. Each organisms in which syngamy does not occur. In nucleotide is a composite molecule formed by those organisms the female gamete undergoes (a) base-sugar-phosphate. development to form new organisms without (b) base-sugar-OH. fertilization. This phenomenon is called "X". (c) (base-sugar-phosphate)n. Identify the organisms and the phenomenon "X". (d) sugar-phosphate. (a) Frog, Parthenogenesis 44. Match the description (given in column I) with (b) Lizards, Gametogenesis correct stage of prophase I (given column II) (c) Rotifers, Embryogenesis and choose the correct option. (d) Honeybee, Parthenogenesis Column I Column II 47. Assertion : Endosperm is a nutritive tissue and it is triploid. A. Chromosomes are I. Pachytene Reason: Endosperm is formed by fusion of moved to spindle secondary nucleus to second male gamete. It is equator used by developing embryo. B. Centromere splits and II. 74. The given figure represents the cross bridge cycle Zygotene in skeletal muscle. What does the step B in the figure chromatids move apart represents? C. Pairing between III. Anaphase Myosin head ADP homologous (high-energy configuration) P1 chromosomes takes A place Thin filament D. Crossing between IV. Metaphase homologous ATP ADP Thick filament ADP hydrolysis P1 chromosomes (a) A – I; B – II; C – III; D – IV D B (b) A – II; B – III; C – IV; D – I ATP (c) A – IV; B – III; C – II; D – I ATP Myosin head (low-energy configuration) (d) A – III; B – I; C – IV; D – II 34. Refer the given equation and answer the (a) Attachment ofC myosin head to actin forming question. cross bridge. (b) Release of phosphate. Myosin changes shape to pull actin.  13 (c) Attachment of new ATP to myosin head. (a) A – V; B – IV; C – I; D – III; E – IV The cross bridge detaches. (d) Splitting of ATP into ADP and Pi. Myosin (b) A – I; B – II; C – III; D – IV; E – V cocks into its high energy conformation. (c) A – III; B – V; C – II; D – IV; E – I 91. The figure given below shows the sectional (d) A – III; B – I; C – V; D – II; E – IV view of ovary. Select the option which gives correct identification of marked structure (A to 50. Assertion : In humans, the gamete contributed by the male determines whether the child D) and its feature. produced will be male or female. A B Reason : Sex in humans is a polygenic trait depending upon a cumulative effect of some C genes on X-chromosome and some on Y- chromosome. 52. Assertion : Replication and transcription occur in the nucleus but translation takes place in the cytoplasm. Reason : mRNA is transferred from the nucleus into cytoplasm where ribosomes and amino acids are available for protein synthesis. 64. The given figure shows the structure of D nucleosome with their parts labelled as A, B & C. (a) A: Primary follicle, it is also called gamete Identify A, B and C. mother cell. A B (b) B: Corpus luteum, it cannot be formed and added after birth. (c) C: Graafian follicle, mature follicle which C ruptures to release secondary oocyte. (d) D: Tertiary follicle, a large number of this follicle degenerates during the phase from Core of histone molecules birth to puberty. (a) A – DNA; B – H1 histone; 53. Select the correct match of the techniques C – Histone octamer given in column I with its feature given in (b) A – H1 histone; B – DNA; column II. C – Histone octamer Column I Column II (c) A – Histone octamer; B – RNA; A. ICSI I Artificially introduction of C – H1 histone semen into the vagina or uterus. (d) A – RNA; B – H1 histone; C – Histone octamer B. IUI II Transfer of ovum collected 59. Match the codons given incolumn I with their from a donor into the fallopian respective amino acids given in column II and tube where fertilization occur choose the correct answer. C. IUT III Formation of embryo by Column -I Column -II directly injecting sperm into the ovum of the zygote or early (Codons) (Amino acids) D. GIFT IV Transfer embryo (with upto 8 A UUU I. Serine blastomeres) into a fallopian B GGG II. Methionine tube. C UCU III. Phenylalanine E. ZIFT V Transfer of embryo with more D CCC IV. Glycine than 8 blastomeres into the uterus E AUG V. Proline 14 (a) A – III; B – IV; C – I; D – V; E – II 41. Biodiversity loss occurs due to the reasons given (b) A – III; B – I; C – IV; D – V; E – II below. (c) A – III; B – IV; C – V; D – I; E – II (i) Habitat loss and fragmentation (d) A – II; B – IV; C – I; D – V; E – III (ii) Co-extinction 89. According to Hardy-Weinberg principle, allele and (iii) Over-exploitation genotype frequencies in a population will remain (iv) Alien species invasion constant from generation to generation in the Identify the correct reasons. absence of other evolutionary influences. It makes (a) (i) and (ii) (b) (i), (ii) and (iii) several assumptions which were given below. (c) (ii), (iii) and (iv) (d) (i), (ii), (iii) and (iv) i. Random Mating 43. Assertion: Communities that comprise of more ii. Sexual Reproduction species tend to be more stable. iii. Non-overlapping Generations Reason: A higher number of species results in iv. Occurrence of Natural Selection less animal variation in total biomoss. v. Small size of population 8. Euro II norms stipulate that sulphur be controlled Identify two assumptions which do not meet for at _____ ppm in diesel and _____ ppm in petrol. a population to reach Hardy-Weinberg (a) 350; 150 (b) 150; 350 Equilibrium? (c) 350; 250 (d) 150; 250 (a) iv and v (b) ii and iv 43. Assertion : Eutrophication shows increase in (c) iii, iv and v (d) i, ii and iii productivity in water. 47. Assertion : Somatic embryos can be induced Reason : With increasing eutrophication, the from any cell in plant tissue culture. diversity of the phytoplankton increases. Reason : Any living plant cell is capable of dif- 37. Assertion : In a food chain, members of ferentiating into somatic embryos. successive higher levels are fewer in number. 51. Assertion : A major advantage of tissue culture Reason : Number of organisms at any trophic is protoplast fusion. level depends upon the availability of organisms Reason : A hybrid is formed by the fusion of which serve as food at the lower level. naked protoplasts of two plants. 53. Assertion : Species are groups of potentially 42. Which one of the following statement regarding interbreeding natural populations which are BOD is true? isolated from other such groups. (a) The greater the BOD of waste water, more Reason : Distinctive morphological characters is its polluting potential. are displayed due to reproductive isolation. (b) The greater the BOD of waste water, less is its polluting potential. 46. Assertion : Insertion of recombinant DNA within (c) The lesser the BOD of waste water, more is the coding sequence of b-galactosidase results its polluting potential. in colourless colonies. (d) The lesser the BOD of waste water, less is Reason : Presence of insert results in inactivation its polluting potential. of enzyme b-galactosidase known as insertional 33. Which of the following statement is not correct inactivation. about cloning vector ? 87. Assertion: Artificially acquired passive immunity (a) ‘Ori’ is a sequen ce r esponsible for results when antibodies or lymphocytes produced controlling the copy number of the linked DNA. outside the host are introduced into a host. (b) Selectable marker selectively permitting the Reason: A bone marrow transplant given to a growth of the non-transformants. patient with genetic immunodeficiency is an (c) In order to link the alien DNA, the vector example of artificially acquired passive immunity. needs to have single recognition site for the 71. Assertion : Interstitial cell is present in the region commonly used restriction enzymes. outside the seminiferous tubule called interstitial (d) The ligation of alien DNA is carried out at a spaces. restriction site present in one of the two antibiotic Reason : Interstitial cells provide nutrition to the resistance genes. sertoli cells.  15 53. Assertion : Inflammation of a skeletal joint may 56. Arrange the following ecosystems in increasing immobilize the movements of the joint. Reason order of mean NPP (Tonnes / ha / year) :Uric acid crystals in the joint cavity and A. Tropical deciduous forest ossification of articular cartilage lead to this. 48. Assertion : Auxins help to prevent fruit and leaf B. Temperate coniferous forest drop at early stages. C. Tropical rain forest Reason : Auxins promote the abscission of older D. Temperate deciduous forest mature leaves and fruits. (a) B <A< D < C 56. Assertion : The squamous epithelium is made (b) D < B <A< C of a single thin layer of flattened cells with irregular boundaries. (c) A<C<D<B Reason : They are found in walls of blood vessels (d) B < D < A< C and air sacs of wings. 22. Fungi are filamentous with the exception of "X" 56. Assertion : Ambulacral system plays a major which is unicellular. Identify X. role in locomotion of echinoderm. (a) Yeast (b) Albugo Reason : Hydraulic pressure of fluid and (c) Mucor (d) Lichen contraction of muscle of tube feet make possible 24. Which of the following statements is not correct movement of echinoderm. for viruses? 49. Assertion : TMV is a virus which causes mosaic (a) Viruses are obligate parasites. disease. (b) Viruses can multiply only when they are Reason : TMV has RNA as genetic material. inside the living cells. 17. Which of the following is a modified stem for the (c) Viruses cannot pass through bacterial filters. protection of plants from browsing animals? (d) Viruses are made up of protein and DNA or (a) Tendrils (b) Thorns RNA (never both DNA and RNA). (c) Rhizome (d) Tuber 46. Which of the following statements regarding 74. Which of the following was most similar to mod- cyanobacteria is incorrect? ern man? (a) It is also called blue green algae. (a) Java man (b) Neanderthal man (b) They are chemosynthetic autotrophs. (c) Homo habilis (d) Cro-Magnon man (c) It forms blooms in polluted water bodies. 76. Explant is required to be disinfected before plac- ing in culture. This is done by (d) It is unicellular, colonial or filamentous, (a) autoclaving marine or terrestrial bacteria. (b) ultra-violet rays 21. Leaves of dicotyledonous plants possess (c) clorax or hypochlorite _________ venation, while _________ (d) X-rays venation is the characteristic of most 2. Which of the following is a viral disease of poul- monocotyledons. try birds? (a) reticulate and parallel (a) Anthrax (b) Ranikhet (b) parallel and reticulate (c) Coccidiosis (d) None of these (c) reticulate and perpendicular 26. The free-living fungus Trichoderma can be used for (d) obliquely and parallel 21. (a) Leaves of dicotyledonous plants possess (a) killing insects reticulate venation while parallel venation is the (b) biological control of plant diseases characteristics of most monocotyledonous. In (c) controlling butterfly caterpillars reticulate venation, the main veins of leaf form (d) producing antibiotics numerous irregular branches and as a result a 34. In Urn shaped age pyramid of the population net like arrangements is formed. Reticulated the trend of growth is venation is the most common vein formation in (a) Rapid (b) Stable leaves. It can be found in the leaves of maple (c) Declining (d) Stationary trees, oak trees and rose bushes. In parallel venation veins are arranged parallel to each other. 16 20. In stems, the protoxylem lies towards the (a) (ii), (iii) and (v) (b) (ii), (iii) and (iv) ___________and the metaxylem lies towards (c) (iv) and (v) (d) (ii) and (v) the ____________ of the organ. 76. In alcoholic fermentation, NAD+ is produced (a) centre; periphery during the (b) periphery; centre (a) reduction of acetyldehyde to ethanol. (c) periphery; periphery (b) oxidation of glucose. (d) centre; centre 25. Male cockroach can be identified from the female (c) oxidation of pyruvate to acetyl coA. by the presence of (d) hydrolysis of ATP to ADP. (a) long antennae 64. Which of the following statement is true ? (b) wingless body (a) Pepsin cannot digest casein. (c) elongated abdomen (b) Trypsin can digest collagen. (d) anal styles (c) Pepsin cannot digest collagen. 32. The sensory papillae in frogs are associated (d) Chymotrypsin can digest casein. with 43. Human immuno deficiency virus (HIV) has a (a) smell (b) hearing protein coat and a genetic material which is (c) respiration (d) touch 22. In earthworms setae are present in all segments (a) Single stranded DNA. except (b) Single stranded RNA. (a) first and the last segments (c) Double stranded RNA. (b) first segment and the clitellum (d) Double stranded DNA. (c) first segment 122. Which one of the following pairs of diseases is (d) clitellum and last segments viral as well as transmitted by mosquitoes? 35. Which of the following statements is/are not (a) Elephantiasis and dengue incorrect? (b) Yellow fever and sleeping sickness (i) Water and minerals, and food are generally moved by a mass or bulk flow system. (c) Encephalitis and sleeping sickness (ii) Bulk flow can be achieved either through a (d) Yellow fever and dengue positive hydrostatic pressure gradient or a 24. Which variety of rice was patented by a U.S. negative hydrostatic pressure gradient. company even though the highest number of (iii) The bulk movement of substances through varieties of this rice is found in India ? the conducting tissues of plants is called (a) Sharbati Sonara (b) Co-667 translocation. (c) Basmati (d) Lerma Roja (iv) Xylem translocates organic and inorganic solutes, mainly from roots to the aerial parts of 16. Which of the following hormone acts upon the the plants. renal tubule and blood capillaries ? (v) Phloem translocates water, mineral salts, (a) Glucagon (b) Aldosterone some organic nitrogen and hormones, from the (c) Vasopressin (d) Glucocorticoids leaves to other parts of the plants.  1 SOLUTION – AIIMS 2 0 17 equivalent resistance between A and B is PHYSICS given by hc hc K 107. (d) Kmax = -W = - 5.01 = l l R R 12375 - 5.01 l (in Å) A R M B 12375 = –5.01 = 6.1875 – 5.01 = 2000 R R 1.17775 ; 1.2 V 53. (b) Components of velocity before and after L collision parallel to the plane are equal, So v sin 60° = u sin 30° .......(1) 1 1 1 2 1 = + = = Components of velocity normal to the plane R¢ 2 R 2 R 2 R R are related to each other i.e., R ¢ = R v cos 60° = e u (cos 30°) ........(2) cos 60° w - w1 62. (d) a= 2 Þ cot 60° = e cot 30° Þ e = t 2 - t1 cot 30° 1 2 p ´ 1200 1 w1 = = 40p ; Þ e= 3 Þ e= . 60 3 3 2 p ´ 4500 146. (c) No. of electrons reaching the collector, w2 = = 150p 60 96 nC = ´ 1010 = 0.96 ´ 1010 100 110p a= rad / sec2 n ´e 10 Emitter current, IE = E t Now, p radian = 180° nC ´ e Collector current, IC = 180 t \ 1 rad = degree \ Current transfer ratio, p I nC 0.96 ´ 1010 11p ´180 a= C = = = 0.96 \ a= degree /sec 2 IE nE 1010 p = 1980 degree/sec2 122. (a) The equivalent circuit is shown in fig. Since 55. (a) For 3rd harmonic/2nd over tone of organ the Wheatstone’s bridge is balanced, pipe open at ends therefore no current will flow through the l2 arm KL. Equivalent resistance between AKM = R + R = 2 R Equivalent resistance between ALM = R + R= 2R The two resistances are in parallel. Hence l/ 4 l l/ 4 2 3ν A C B Þ n2 = Q Q 2l 2 x q x 2 2 For 1st overtone of organ pipe open at one x end l1 x x Let AC = , BC = 2 2 The force on A due to charge q at C, ® 1 Q.q F CA = . along AC l/ 2 l/ 4 4pe 0 ( x / 2) 2 3ν The force on A due to charge Q at B Þ n1 = 4l1 ® 1 Q2 ® F AB = . along BA 3n 3n l 1 4pe0 x 2 Given n1 = n 2 Þ = or 1 = 2l 2 4l1 l2 2 The system is in equilibrium, then two oppositely directed force must be equal, i.e., p DV p total force on A is equal to zero. 5. (a) K= or = ; ® ® DV / V V K ® ® F CA + F AB = 0 Þ F CA = - F AB M ' M Also r = and r = ; 1 4Q.q –1 Q 2 V V - DV . 2 = . -1 4 pe 0 x 4 pe 0 x 2 r' V 1 æ DV ö \ = = = ç1 - ÷ Q r (V - DV ) (1 - DV / V) è V ø Þq=- 4 æ DV ö p r¢ p 35. (d) B = x = [L]; A x = Vx; A = V x » ç1 + ÷ = 1+ or -1 = è V ø K r K = M L2 T -2 L1 / 2 = M L5 / 2 T -2 pr or r¢ - r = (Q D V << V) AB = (ML5 / 2 T -2 ) (L) = [M1 L7 / 2 T -2 ] K sin 45° 102. (a) e= - df -d 2 dt = dt ( ) 6t - 5t + 1 = -12t + 5 83. (a) For point A, a mg = sin r e = – 12 (0.25) + 5 = 2 volt 1 e 2 Þ sin r = i = = = 0.2A. 2 a mg R 10 4. (d) is not possible, because at a particular time for point B, sin (90 – r) = gma (90 – r) is critical angle. t, displacement cannot have two values. 39. (c) The chemical reaction of process is 45º Air 212 H ® 42 He 90 – r r A Energy released = 4 ´ (7.1) - 4(1.1) = 24 eV 119. (d) Let q charge is situated at the mid position 90 – r B of the line AB. The distance between AB is x. A and B be the positions of charges Q Glass and Q respectively.  3 1 2 \ cos r = g m a = 1 æ dy ö 1 a mg m çè ÷ø = mw2 a2 = 8 × 10–3 joule 2 dt max 2 1 Þ a mg = 1 cos r or × (0.1) w2 × (0.1)2 = 8 × 10–3 2 1 1 = = Solving we get w = ± 4 1 - sin r 2 1 1- Substituting the values of a, w and f in the 2 amg2 equation of S.H.M., we get y = 0.1 sin (± 4t + p /4) metre. 2 1 2 a m 2g 99. (b) Þ a mg = = 1 2 a m 2g -1 Y1 1- A 2 a m g2 B Y 2 3 Þ 2 a mg - 1 = 2 Þ a mg = 2 77. (a) Since water does not fall down, therefore Y2 the velocity of revolution should be just sufficient to provide centripetal acceleration Y1 = A + B, Y2 = A . B at the top of vertical circle. So, Y = (A + B)gAB = AgA + AgB + BgA + BgB v = (g r ) = {10 ´ (1.6)} = (16) = 4 m/sec. = 0 + AgB + BgA + 0 = AgB + BgA 74. (c) Equipotential surfaces are normal to the This expression is for XOR electric field lines. The following figure shows the equipotential surfaces along m0 2i2 m 2 i1 m 4 18. (c) B = - 0 = 0 (i 2 - i1 ) with electric field lines for a system of two 4 p (r / 2) 4 m (r / 2) 4 p r positive charges. m0 4 m = (5 - 2.5) = 0 . 4p 5 2p 70. (a) Since block does not slide on wedge so displacement is zero & hence work done by force is zero. u f 58. (a) The displacement of a particle in S.H.M. is given by q in y = a sin (wt + f) gs mg m q dy velocity = = wa cos (wt + f) 123. (d) The distance between the first dark fringe dt on either side of the central bright fringe The velocity is maximum when the particle = width of central maximum passes through the mean position i.e., 2Dl 2 ´ 2 ´ 600 ´ 10-9 æ dy ö = = çè ÷ø dt max = w a a 10-3 –3 = 2.4 × 10 m = 2.4 mm The kinetic energy at this instant is given by 4 51. (c) Pressure is proportional to depth from the = m2 m 2 g = m2 (mass of A and B) g free surface and is same in all directions. 47. (b) 1 300 = (100 + 200)g = g = 100g newton -1 - 2 4 2 3 3 [X] M L T A 114. (d) [Y] = = 2 [Z ] M 2 T -4 A - 2 \ F = F1 + F2 = 25 g + 100 g = 25g = 125 × 10 N æ Qö \ F = 1250 N = M -3L-2Q4T4 çèQ A = ÷ø T 129. (d) Power, P = Ι r.m.s ´ Vr.m.s ´ cos f 66. (b) Effective half life is calculated as In the given problem, the phase difference 1 1 1 = + between voltage and current is p/2. Hence T T1 T2 P = Ι r.m.s ´ Vr.m.s ´ cos(p / 2) = 0. 1 1 1 = + Þ T = 12 years 1. (a) Figure indicates the presence of some T 16 48 positive charge to the left of A. 3 Time in which will decay is 2 half lives = \ EA > EB (Q rA < rB) 4 65. (d) For slab in series, we have 24 years 41. (d) Angle of dip, d = 45° x 4x 3x Req = R1 + R2 = + = tan 45 1 2 KA 2KA KA \ tan d ¢ = tan d = = = cos q cos 30 º 3/2 3 Now, in a steady state rate of heat transfer through the slab is given by \ Real dip d ¢ = tan -1 (2 / 3) dQ T2 - T1 (T2 - T1 ) …(i) 50. (b) Gauss’s law is applicable only for closed = = KA surface and for the charge placed inside it dt R eq 3x not near it. 1 dQ æ A(T2 - T1 )K ö Total electric flux, fÎ = Q Given =ç ÷ø f …(ii) e0 dt è x r r r r Comparing (i) and (ii), we get f = 1/3 18. (d) t = (M ´ B) , where | M | = i.A 37. (a) For Bracket series = MB sin q where q is angle between Magnetic moment 1 é1 1ù 9 r =Rê 2 - 2ú= R r l max ë4 5 û 25 ´ 16 & B . For q = 0 t = 0 & coil is in stable equilibrium. Hence plane of coil must be 1 é1 1 ù R l 25 perpendicular to magnetic field. and = R ê 2 - 2 ú = Þ max = l min ë4 ¥ û 16 l min 9 M B 41. (a) i 1´ 0 + 2 ´ 2 + 3 ´ 0 + 4 ´ 2 + 5 ´ 1 54. (c) X C.M. = 1+ 2 + 3 + 4 + 5 33. (a) In vacuum velocity of all EM waves are 4 + 8 + 5 17 same but their wavelengths are different. = = = 1.1 28. (a) In adiabatic process 15 15 DQ = 0 1´ 0 + 2 ´ 0 + 3 ´ 2 + 4 ´ 2 + 5 ´ 1 \ DW = – DU YC.M = 1+ 2 + 3 + 4 + 5 90. (d) F1 = Force of friction between B and A = m1m1g 6+8+5 = = 1.3 = 0.25 × 100 × g = 25 g newton 15 F2 = Force of friction between (A + B) and surface  5 29. (c) All the charge given to inner sphere will 117. (c) Amplitude of oscillation for a forced damped pass on to the outer one. So capacitance F0 / m that of outer one is 4p Î0 b . oscillatory is A = , (w - w0 2 ) + (bw / m) 2 2 h 1 where b is constant related to the strength 30. (c) l = so h µ 2m E m of the resistive force, w0 = k / m is Since m a > m n > m p > m e natural frequency of undamped oscillator so de-Broglie wavelength in increasing order (b = 0) will be When the frequency of driving force la , ln , lp , le (w) » w0 , then amplitude A is very larger.. For w < w0 or w > w0 , the amplitude n1Cv1 + n 2 Cv2 Cv mix = decreases. 54. (c) n1 + n 2 103. (a) 102. (b) 13R n1Cv1 + 2n1Cv2 é n1 1 ù 114. (b) Study of junction diode characteristics Þ = êQ n = 2 ú shows that the junction diode offers a low 6 n1 + 2n1 ë 2 û resistance path, when forward biased and 13R high resistance path when reverse biased. Þ = Cv1 + 2C v2 2 This feature of the junction diode enables Possible values are, it to be used as a rectifier. 86. (d) Ampere’s circuital law can be derived from 3R 5R Biot-Savart law and is not independent of Cv1 = , C v2 = 2 2 Biot-Savart law. \ Gases are monatomic (like He) and 111. (d) diatomic (likeN2) 95. (d) Large eddy currents are produced in non- 39. (d) We can combine two prisms in such a way laminated iron core of the transformer by (i) deviation is zero but dispersion not the induced emf, as the resistance of bulk (ii) dispersion is zero but deviation is not. iron core is very small. By using thin iron But in any situation both deviation & sheets as core the resistance is increased. dispersion can not be zero Laminating the core substantially reduces simultaneously. the eddy currents. Eddy current heats up the core of the transformer. More the eddy ASSERTION - REASON TYPE QUESTIONS currents greater is the loss of energy and the efficiency goes down. æ3 ö 77. (a) ç - 1÷ amg -1 116. (a) Carnot cycle has maximum efficiency. 96. (d) fw = f = f è2 ø =4f æ amg ö æ 3/ 2 ö 82. (a) Lenz's law (that the direction of induced emf ç - 1÷ ç - 1÷ è a mw ø è 4/3 ø is always such as to oppose the change that cause it) is direct consequence of the 90. (d) Pressure of water reduces when it comes law of conservation of energy. from wide pipe to narrow pipe. According 90. (d) Law of conservation of linear momentum to equation of continuity, av = constant. is correct when no external force acts. As the water flows from wider tube to When bullet is fired form a rifle then both narrow tube, its velocity increases. should possess equal momentum but According to Bernouli prinicple, where velocity is large pressure is less. p2 different kinetic energy. E = \ 2m Kinetic energy of the rifle is less than that 6 of bullet because E µ 1/m CHEMISTRY 104. (d) 94. (a) 5. (c) In bcc the atoms touch along body diagonal \ 2r + 2r = 3a 70. (a) Work done 1 1 3a 3 ´ 4.29 = ´ Stress ´ Strain = ´ Y ´ (Strain)2 . \ r= = = 1.857Å 2 2 4 4 Since, elasticity of steel is more than copper, 8. (a) Phenelzine is an antidepressant, while hence more work has to be done in order to others are antacids. stretch the steel. 21. (b) Alkyl fluorides are more conveniently 107. (b) Electromagnets are magnets, which can be prepared by heating suitable chloro – or turnd on and off by switching the current bromo-alkanes with organic fluorides such as AsF3, SbF3, CoF2, AgF, Hg2F2 etc. This on and off. reaction is called Swarts reaction. As the material in electromagnets is subjected to cyclic changes (magnification CH 3Br + AgF ¾¾ ® CH 3 F + AgBr and demangetisation), the hysteresis loss of the material must be small. The material 2CH3CH 2 Cl + Hg 2 F2 ¾¾ ® should attain high value of I and B with low 2CH3CH 2 F + Hg 2Cl2 value of magnetising field intensity H. As 20. (a) The lines falling in the visible region soft iron has small coercivity, so it is a best comprise Balmer series. Hence the third line choice for this purpose. 86. (a) In real gas, intermolecular force exist. Work would be n1 =2, n 2 = 5 i.e. 5 ® 2. has to be done in changing the distance 109. (b) between the molecules. Therefore, internal 62. (a) only those compounds which have a-H energy of real gas is sum of internal kinetic give Aldol condensation and internal potential energy which are function of temperature and volume O a || respectively. Also change in internal energy (a) Cl3C CH of a system depends only on initial and final sates of the system. a 69. (a) de-Broglie wavelength associated with gas (b) C 6 H 5 C H 2 CHO molecules varies as 1 (no a-H) (a-H) 1 a lµ (c) C 4H9 C H 2 CHO (d) CH3NO2 T (a-H) 1 2 2 84. (c) 111. (b) intensity, I = rw A v 2 CH MgX 1%HgSO \ Intensity depend upon amplitude, HC º CH ¾¾¾ ¾ ¾4 ® CH CHO ¾¾ ¾ 3 3 ¾ ¾® 20%H 2SO4 [A ] H 2O frequency as well as velocity of wave. Also I1 = I2 [O] CH 3CHOHCH 3 ¾¾® ¾ CH 3COCH 3 [ B] Acetone [C] 37. (b) 24. (a) Since glycerol has three –OH groups, it can have three acyl (similar or different) groups 23. (c) Anomers are those diastereomers that differ in configuration at C – 1 atom. Since a - D - (+) - glucose and b – D – (+)  7 glucose differ in configuration at C – 1 atom - 1 so they are anomers. \ Kp = Kc ( RT ) 2 6. (d) Neoprene is an addition polymer of Kp - 1 isoprene. or = ( RT ) 2 Kc Cl | 39. (c) DG = –2.303 RT log K O 2 or peroxides nCH 2 = CH - C = CH 2 ¾¾¾¾¾¾ ¾ ® –nFE° = –2.303 RT log K Chloroprene nFE°(I) log K = æ Cl ö 2.303 RT ç | ÷ nFE° = 0.4342 —çè CH 2 - C = CH - CH2 — ÷ ø RT ........ (i) Neoprene nFE° ln K = 17. (c) RT - nFE ° OH ONa K = e RT NaOH ....... (ii) 60. (d) ¾¾ ¾¾® 115. (b) ˆˆ† Mg ++ + 2 OH - Mg(OH) 2 ‡ˆˆ Phenol Ksp = [Mg++][OH–]2 OH 1.0 × 10–11 = 10–3 × [OH–]2 CO 2 + HCl COOH ¾¾ ¾ ¾¾¾® 10-11 140° C [OH- ] = = 10-4 -3 10 Salicylic acid \ pOH = 4 39. (c) Given Cp = 75 JK–1 mol–1 \ pH + pOH = 14 100 \ pH = 10 n= mole , Q = 1000 J DT = ? 18 67. (d) A doubly bonded carbon atom having an alkyl group is oxidised to aldehyde which Q = nCpDT Þ DT = 1000 ´ 18 = 2.4 K is further oxidised to carboxylic acid. 100 ´ 75 (i) KMnO ,OH - CH3CH 2CH = CH CH3 ¾¾¾¾¾¾¾ 4 ® 51. (b) DH = S é DH f° products ù + (ii) H êë úû CH3CHO + CH3CH 2CHO -S é DH f° reactants ù êë úû ¯ ¯ CH3COOH CH3CH 2COOH DH° = [ D H f° (CO)(g) + DH f° (H 2 O)(g)] – 128. (d) [DH f° (CO 2 )(g) + DH f° (H 2 )(g)] 99. (b) Given N1 = 10N, V1 = 10 ml, N2 = 0.1N, V2 = ? = [– 110.5 + ( – 241.8)]– [– 393.5 + 0] = 41.2 N1V1 = N 2V2 21. (b) For a gaseous phase reaction Kp and Kc or 10 × 10 = 0.1 × V2 are related as 10 ´ 10 Dn g or V2 = , V2 = 1000 ml K p = K c ( RT ) 0.1 For the given reaction, Volume of water to be added 1 = V2 – V1 = 1000 – CO(g) + O2 ( g ) ® CO2(g) 10 = 990 ml. 2 1 Dng = 1– (1 + 0.5) = – 0.5 or - 2 8 183. (a) Hybridisation of carbon in CH 3+ is sp2 and 101. (c) Electrodialysis involves movement of ions in CH4 its hybridisation is sp3 towards oppositely charged electrodes. 192. (d) According to Fajan's rule : Urea being a covalent compound does not 1 dissociate to give ions and hence it cannot Covalent character µ be removed by electrodialysis.However all size of cation the other given compounds are ionic which can undergo dissociation to give oppositely µ size of anion charged ions and thus can be separated. Among the given species order of size of 27. (b) cations 19. (a) NF5 does not exist because N does not form N3+ < O2+ < Pb2+ < Ba2+ pentahalides due to the absence of d-orbital order of size of anions O2– > Cl–. in its valence shell. While P, As and Sb form Hence the order of covalent character is pentahalides of the general formula MX5 NCl3 > Cl2O > PbCl2 > BaCl2 (where, M = P, As and Sb) due to the presence of vacant d-orbitals in their BaCl2 is least covalent in nature. respective valence shell. 4. (a) According to the law of triads the atomic 125. (a) Both have peroxy linkage wt of the middle element is arithmatic mean 35. (a) The ionisation energies increase with of I and III. increase in atomic number. However, the At.wt of Cl + At wt of I trend is irregular among some d-block At wt of Br = 2 elements. On the basis of electronic 57. (d) Since reaction is occuring in basic medium configuration, the therefore 2OH– are added on right side. Zn : 1s 2 2s 2 p 6 3s 2 p6 d 10 4s 2 2MnO -4 (aq) + Br– (aq) ¾¾ ® 2MnO2(s) + BrO–3 (aq) + 2OH–(aq) Fe : 1s 2 2s 2 p 6 3s 2 p 6 d 6 4s 2 Now, hydrogen atoms can be balanced by Cu : 1s 2 2s 2 p6 3s 2 p 6 d 10 4s1 adding one H2O molecule to the left side 2MnO 4- (aq) + Br - (aq) + H 2 O(l ) ¾¾ ® Cr : 1s 2 2s 2 p6 3s 2 p 6 d 5 4s1 2MnO2(s) + BrO 3- (aq) + 2 OH - (aq) IE1 follows the order : Zn > Fe > Cu > Cr 19. (c) From the given data we find Fe3+ is 57. (d) The optical isomers are pair of molecules strongest oxidising agent. More the which are non superimposable mirror positive value of E°, more is the tendency images of each other. to get oxidized. Thus correct option is (c). en en 35. (d) Here n = 4, and [H+] = 10– 3 (as pH = 3) Applying Nernst equation 0.059 [Fe 2 + ]2 en Co Co en E = Eº – log n [H + ]4 (p O2 ) 0.059 (10-3 )2 en en = 1.67 - log 4 (10-3 ) 4 ´ 0.1 [Co(en)3]3+ [Co(en)3]3+ Mirror (dextro) (laevo) 0.059 = 1.67 - log107 The two optically active isomers are 4 collectivity called enantiomers. = 1.67 – 0.103 = 1.567  9 107. (b) As DTf = Kf. m 101. (b) Both assertion and reason are true but DTb = Kb. m reason is not the correct explanation of assertion. Non fusible mass present in ore DT f DTb Hence, we have m = = in mixing with suitable flux are fused which Kf Kb are then reduced by coke to give free metal. Kf 247. (a) or DT f = DTb 107. (a) Kb 123. (c) Equal moles of different substances contain Þ [DTb = 100.18 - 100 = 0.18°C] same number of constituent particles but equal weights of different substances do 1.86 = 0.18 × = 0.654°C not contain the same number of consituent 0.512 particles. As the Freezing Point of pure water is 89. (b) Both Assertion and Reason are true but 0°C, reason is not the correct explanation of DTf = 0 –Tf assertion. Greater the number of negative 0.654 = 0 – Tf atoms present in the oxy-acid make the acid \ Tf = – 0.654 stronger. In general, the strengths of acids Thus the freezing point of solution will that have general formula (HO)m ZOn can be – 0.654°C. be related to the value of n. As the value of n increases, acidic character also increases. PV m 44. (c) n= = The negative atoms draw electrons away RT M from the Z-atom and make it more positive. MPV 34 ´ 2 ´ 100 The Z-atom, therefore, becomes more m= = = 282.68gm RT 0.082 ´ 293 effective in withdrawing electron density 85. (b) The order of stability of free radicals away from the oxygen atom that bonded to • • • • hydrogen. In turn, the electrons of H – O (C6 H 5 ) 3 C > (C 6 H 5 ) 2 CH > (CH 3 ) 3 C > (CH 3 ) 2 C H bond are drawn more strongly away from The stabilisation of first two is due to the H-atom. The net effect makes it easier resonance and last two is due to inductive from the proton release and increases the effect. acid a strength. 48. (b) 108. (a) Lithium carbonate is unstable to heat; 85. (a) lithium being very small in size polarises 53. (b) Na zeolite + CaCl 2 ® Ca zeolite + 2NaCl a large CO32 - ion leading to the formation 16. (b) The reagent is BaCl2 which imparts green of more stable Li2O and CO2. colour to flame. BaCl 2 forms chromyl 154. (b) Both Assertion and Reason are true but chloride (which is red in colour), when Reason is not the correct explanation of treated with K2Cr2O7 and conc. H2SO4. statement-1. [Fe(CN) 6 ] 3– is weakly paramagnetic as it has unpaired electrons 2BaCl2 + K 2Cr2O7 + 3H 2SO4 ¾¾ ® while [Fe(CN)6]2– has no unpaired electron. K 2SO4 + 2BaSO4 + 2CrO2Cl 2 + 3H 2 O \ It is diamagnetic. Chromyl chloride 150. (b) (red gas) 115. (a) ASSERTION / REASON 172. (a) It is fact that aniline is better nucleophile than anilium ion. Anilium ion contain +ve 133. (d) Assertion is false but Reason is true. The charge, which reduces the tendency to enthalpyof chemisorption is of the order of 40 - 400 kJmol–1 while for physical adsorption donate lone pair of electron C6 H5 NH3+ . it is of the order of20 - 40 kJmol–1. Anilium ion 10 171. (c) Benzene has a uniform C – C bond BIOLOGY distance of 139 pm, a value intermediate between the C – C single. (154 pm) and C 67. (b) A - V; B - IV; C - II; D - I; E - III = C double (134 pm) bonds. 34. (d) In chlorophyceae, the stored food material 125. (a) Zinc metal which has a more negative is starch and the major pigments are electrode potential than iron will provide chlorophyll- a and b. In phaeophyceae, electrons in preference of the iron, and laminarian is the stored food and major therefore corrodes first. Only when all the pigments are chlorophyll a, c and zinc has been oxidised, the iron start to rust. fucoxanthin. 118. (c) Atomic radius of gallium is less than that 58. (d) of aluminium. 65. (a) Molluscans are soft bodied animals. Their 140. (a) Both assertion and reason are true and body is unsegmented with a distinct head, reason is the correct explanation of muscular foot and visceral hump. In Pila, assertion. the buccal cavity contains a rasping organ, n2 h2 n2 the radula with transverse rows of teeth. Radius, rn = = ´ 0.529Å.rn 4pe 2 mZ Z In the following questions, a statement of Assertion is For first orbit of H-atom followed by a statement of Reason. n=1 (a) If both Assertion and Reason are true and the Reason is the correct explanation of the (1)2 Assertion. r1 = ´ 0.529Å = 0.529Å (b) If both Assertion and Reason are true but the 1 Reason is not the correct explanation of the 153. (c) It is correct statement that NF3 is a weaker Assertion. ligand than N(CH3)3, the reason is that (c) If Assertion is true but Reason is false. fluorine is highly electronegative therefore, (d) If both Assertion and Reason are false. it with draw electrons from nitrogen atom. 47. (c) In a DNA molecule, A-T rich parts melt Hence, the lone pair of nitrogen atom cannot before G-C rich parts because there are two be ligated. While N(CH3)3 is a strong ligand H-bond between A and T whereas in because CH3 is electron releasing group. between G and C, there are three H-bond. 114. (d) Assertion is false, because aryl halides do 83. (a) Nucleotides are the building blocks of not undergo nucleophilic substitution nucleic acid. Each nucleotide consists of under ordinary conditions. This is due to three parts: a sugar (ribose for RNA and resonance, because of which the carbon– chlorine bond acquires partial double bond deoxyribose for DNA), a phosphate, and a character, hence it becomes shorter and nitrogenous base. stronger and thus cannot be replaced by 44. (c) A – IV, B – III, C – II, D – I nucleophiles. However Reason is true. Metaphase – 152. (b) The magnetic moments are lesser than the Chromosomes are moved to spindle fibre. fact that 5f electrons of actinides are less Anaphase – Centr omere effectively shielded which results in splits and chromatids moveapart. quenching of orbital contribution. Zygotene – Pairing between 107. (a) A small quantity of sedative produces a homologous chromosomes takes place. feeling of relaxation, calmness an d Pachytene – C rossing drowsiness. between homologous chromosomes 107. (b) Vulcanisation is a process of treating occurs. natural rubber with sulphur or some compounds of sulphur under heat so as to modify its properties. This cross-linking give mechanical strength to the rubber.  11 34. (b) The ratio of the volume of CO2 liberated to Reason is the correct explanation of the the volume of oxygen absorbed per Assertion. molecule during respiration is called (b) If both Assertion and Reason are true but the Respiratory Quotient (RQ). The value of RQ Reason is not the correct explanation of the Assertion. indicates the types of respiratory substrate. (c) If Assertion is true but Reason is false. Volume of CO2 evolved (d) If both Assertion and Reason are false. RQ = 47. (a) Male gamete (n) + secondary nucleus (2n) Volume of O 2 consumed = primary endosperm nucleus which develops 102 into endosperm (3n) RQ = = 0.7 Endosperm is the reserve food used by 145 developing embryo. In the following questions, a statement of Assertion is 74. (b) Step A: Attachment of myosin head to actin followed by a statement of Reason. forming cross bridge. (a) If both Assertion and Reason are true and the Step B: Release of phosphate. Myosin Reason is the correct explanation of the changes shape to pull actin. Assertion. (b) If both Assertion and Reason are true but the Step C: Attachment of new ATP to myosin Reason is not the correct explanation of the head. The cross bridge detaches. Assertion. Step D: Splitting of ATP into ADP and Pi. (c) If Assertion is true but Reason is false. Myosin cocks into its high energy (d) If both Assertion and Reason are false. conformation. 74. (a) 91. (c) Oogonia are called as gamete mother cell. In the following questions, a statement of Assertion is Corpus luteum is formed as a temporary followed by a statement of Reason. endocrine structure after the ovulation. It (a) If both Assertion and Reason are true and the is involved in the production of relatively Reason is the correct explanation of the high levels of progesterone and moderate Assertion. levels of estradiol and inhibin A to maintain pregnancy. A large number of primary (b) If both Assertion and Reason are true but the follicles degenerate during the phase from Reason is not the correct explanation of the birth to puberty. Assertion. 53. (d) ICSI (Intracytoplasmic sperm injection) - (c) If Assertion is true but Reason is false. Formation of embryo by directly injecting (d) If both Assertion and Reason are false. sperm into the ovum 71. (b) Hind brain consists of cerebellum located IUI (intrauterine insemination) - Artificial dorsally to medulla oblongate and pons varolii. It contains centres for maintenance introduction of semen into the vagina or of posture and equilibrium of the body and uterus for the muscle tone. All activities of the IUT (Intra uterine transfer) - Transfer of cerebellum are involuntary but may involve embryo with more than 8 blastomeres into learning in their early stages. the uterus 74. (d) Parthenogenesis is a form of reproduction GIFT (Gamete intra fallopian transfer) - in which an unfertilized egg develops into a Transfer of ovum collected from a donor new individual, occurring commonly among into the fallopian tube where fertilization insects and certain other arthropods. occurs In the following questions, a statement of Assertion is ZIFT (Zygote intra fallopian transfer) - followed by a statement of Reason. (a) If both Assertion and Reason are true and the Transfer of the zygote or early embryo (with 12 upto 8 blastomeres) into a fallopian tube. UCU– Serine In the following questions, a statement of Assertion is CCC – Proline followed by a statement of Reason. AUG – Methionine (a) If both Assertion and Reason are true and the 89. (a) Occurrence of natural selection and small Reason is the correct explanation of the size of population do not meet for a Assertion. population to reach Hardy-Weinberg (b) If both Assertion and Reason are true but the Equilibrium. For Hardy-Weinberg Reason is not the correct explanation of the equilibrium to be reached, natural selection Assertion. should not be occurring. If populations are (c) If Assertion is true but Reason is false. undergoing natural selection at the locus (d) If both Assertion and Reason are false. 50. (c) In human, the gamete contributed by the under consideration, allele frequencies will male determines whether the child produced be changing in a specific direction and will be male or female. Sex in humans is a continuously, Hardy-Weinberg Equilibrium polygenic trait depending upon cumulative predicts that allele frequencies will stay effect of some genes present on Y- constant. It assumes that population size is chromosome. Only sex in human is very large. amonogenic trait. 47. (a) Somatic embryos are non- zygotic embryo In the following questions, a statement of Assertion is like structures that develop into from any type of tissue in plant tissue culture. followed by a statement of Reason. 51. (b) An important technique of tissue culture, (a) If both Assertion and Reason are true and the somatic hybridization results in the Reason is the correct explanation of the production of somatic hybrid plants. Two Assertion. different plant varieties each with a desirable (b) If both Assertion and Reason are true but the character can be made to undergo Reason is not the correct explanation of the protoplast fusion, which further can be grown into a new plant. Assertion. (c) If Assertion is true but Reason is false. 42. (a) BOD is the method of determining the (d) If both Assertion and Reason are false. amount of oxygen required by 52. (a) In eukaryotes, the replication and microorganisms to decompose the transcription takes place in the nucleus. waste present in the water supply. It is mRNA came out from the nucleus through a measure of organic matter present in the nuclear pore. In cytoplasm, translation the water. If the quantity of organic occurs. In prokaryote, there is no nuclear wastes in the water supply is high then membrane, so replication, transcription and translation all occur in the cytoplasm. the number of decomposing bacteria 64. (a) Nucleosome is a structural unit of a present in the water will also be high. eukaryotic chromosome which consists of As a result, BOD value will increase. a length of DNA coiled around a core of 33. (b) Selectable marker selectively permitting the histones and are thought to be present only growth of the transformants. during interphase of cell cycle. In the given 41. (d) Biodiversity refers to the variety found in figure of nucleosome structure, the parts biota from genetic make-up of plants and marked as A, B and C are respectively DNA, animals to cultural diversity. The main cause H1 histones and histone octamer. of the loss of biodiversity can be attributed 59. (a) A-III, B-IV, C-I, D-V, E-II to the influence of human beings on the UUU – world's ecosystem. The important factors Phenylalanine causing loss of biodiversity are - habitat GGG – Glycine loss, habitat fragmentation, disturbances,  13 over exploitation of resources, pollution, chromogenic substrate. The plasmid in the exotic species, co-extinction, alien species bacteria, lacking an insert produces blue invasion, intensive agriculture and forestry. coloured colonies, while those plasmids with 43. (a) Communities with higher number of species an insert do not produce any colour due to are more stable as it can resist occasional insertional inactivation of the enzyme, b- disturbances. A stable community should galactosidase. show less variation in productivity from 87. (b) Artificially acquired passive immunity year to year and resistant towards by alien results when antibodies or lymphocytes species. that have been produced outside the host 8. (a) The Govt. of India through a new auto fuel are introduced into a host. This type of policy has laid out a roadmap to cut down immunity is immediate short lived, lasting the vehicular pollution in Indian cities. For only a few weeks to a few months. An example, Euro II norms stipulate that sulphur example is bone marrow transplant given to be controlled at 350 ppm in diesel and 150 a patient with genetic immunodeficiency. ppm in petrol. 71. (c) Leydig cells, also known as interstitial cells, 43. (b) Eutrophication is a natural process which are found adjacent to the seminiferous literally means well nourished or enriched. tubules in the testicle. They produce It is a natural state in many lakes and ponds testosterone in the presence of luteinizing which have a rich supply of nutrients. hormone (LH). Eutrophication becomes excessive, when 53. (a) Painful inflammation of the synovial abnormally high amount of nutrient from membrane of the joints results in stiffening sewage, fertilizers, animal wastage and of joints and painful movement Uric acid detergent, enter streams and lakes causing accumulation in the joints can lead to painful excessive growth or blooms of movement of joint. microorganisms. With increasing 48. (b) Auxin delays abscission of young leaves eutrophication, the diversity of th e and fruits. Its effect is through non- phytoplankton community of a lake formation of abscission zone below a leaf increases and the lake finally becomes or fruit. Abscission zone cuts off nutrients dominated by blue - green algae. and water supply. However, auxin promotes 37. (d) When food is made available automatically the abscission of mature or older leaves and the next higher level of organism in the fruits. hierarchy should increase. This is because 56. (b) when the forest cover got depleted it led to 56. (a) The water vascular system is a unique organ the increase in the number of endangered system that functions in locomotion, species. If the deer population is more, it feeding, respiration and excretion. automatically leads to an increase in the tiger Ambulacral canal is connected to outside population. through external tube feet. Hydraulic 53. (b) A group of individuals resembling each pressure of fluid and contraction of muscle other in morphological, physiological, of tube feet make possible movement of biochemical and behavioural characters Echinoderm. constitute a species. Such individuals can 49. (a) Nostoc is a cyanobacterium. Cyanobacteria breed among themselves but cannot breed are gram (–) ve prokaryotes which perform with members other than their own to oxygenic photosynthesis like plants. produce fertile offsprings. New species are Cyanobacteria can be unicellular (e.g. formed mainly due to reproductive isolation. spirulina), colonial. (e.g. Nostoc) or 46. (a) Alternative markers have been developed filamentous (e.g. Oscillatoria) that can differentiate recombinants from non-recombinants based upon their ability to produce colour in presence of a 14 17. (b) Thorn is a stiff, sharp-pointed woody 21. (a) Leaves of dicotyledonous plants possess projection on the stem or other part of a reticulate venation while parallel venation plant. Thorns are found in many plants such is the characteristics of most as Citrus, Bougainvillea. They protect monocotyledonous. In reticulate venation, plants from grazing animals. the main veins of leaf form numerous 74. (d) The skeleton of Cro-Magnon was almost irregular branches and as a result a net like identical to the modern man. arrangements is formed. Reticulated 76. (c) Before transferring on the culture medium, venation is the most common vein formation the explant is first of all disinfected by sur- in leaves. It can be found in the leaves of face sterilization using clorx water, sodium maple trees, oak trees and rose bushes. In or calcium hypochlorite solution or parallel venation veins are arranged parallel methiolate. Too much care must be taken in to each other. this operation so that the cells do not die. 20. (a) The first formed primary xylem elements are 2. (b) Coccidiosis is a protozoan disease. called protoxylem and the later formed 26. (b) Trichoderma is a free-living saprophytic primary xylem is called metaxylem. In stems, fungi that most commonly lives on dead the protoxylem lies towards the centre (pith) organic matter in the soil and rhizosphere and the metaxylem lies towards the (root ecosystem). It inhibits pathogens periphery of the organ. This type of primary through release of gliotoxin, viridin, xylem is called endarch. gliovirin and trichodermin like substances. 25. (d) Both the sexes of cockroach have anal cerci 34. (c) In Urn Shaped pyramid the individuals be- which are jointed structures. But in the male, low the reproductive age are fewer in num- in addition, there is a paired unjointed ber than the individuals of reproductive age. needle-like anal style, which serve to 56. (d) Net primary productivity (NPP) is the biom- distinguish between the male and the ass or storage of energy by green plants. It female. is equal to the gross primary productivity 32. (d) Frog has different types of sense organs minus loss due to respiration. The produc- like organs of touch (sensory papillae), taste tivity generally increases from polar regions (taste buds), smell (nasal epithelium), vision toward the tropics, because of the increas- (eyes) and hearing (tympanum with internal ing sunlight and temperature. ears). 22. (d) Except the first, the last and clitellar segment 22. (a) Yeast being a unicellular fungus does not in each segment bear a ring of tiny curved, show filamentous nature. It is a microscopic chitinous structure known as setae. Setae fungus consisting of a single oval cell that helps in locomotion and copulation. reproduces by budding. 35. (c) Statements (iv) and (v) are not correct. 24. (c) Virus is a small infectious agent that replicates only inside the living cells of other (iv) Xylem is associated with the organisms. Viruses can infect all types of translocation of mainly water, mineral salts, life forms, from animals and plants to some organic nitrogen and hormones from microorganisms, including bacteria and roots to the aerial parts of the plants. archaea. Viruses can pass through bacterial (v) Phloem translocates a variety of proof filters as they are smaller than organic and inorganic solutes mainly from bacteria. the leaves to other parts of the plants. 46. (b) Cyanobacteria are photosynthetic (containing a blue photosynthetic pigment) 76. (a) Alcoholic fermentation is a process in which autotrophs. They are prokaryotic and molecules such as glucose etc. are represent the earliest known form of life on converted into cellular energy and thereby the Earth.  15 produce ethanol and carbon dioxide as metabolic waste products. During alcoholic fermentation, NAD+ is produced when acetaldehyde is reduced to ethanol. 64. (d) Milk protein can be digested by pepsin and chymotrypsin. 43. (b) The human immunodeficiency virus is a lentivirus that causes the acquired immunodeficiency syndrome, a condition in humans in which progressive failure of the immune system allows life-threatening opportunistic infections and cancers to thrive. HIV has a protein coat and a genetic material which is single stranded RNA. 122. (d) Yellow fever and dengue are viral diseases, and they are transmitted by mosquitoes. 24. (c) Basmati rice was patented by a US company even though the highest number of varieties of this rice is found in India. 16. (c) ADH (or vasopressin) is secreted by posterior pituitary gland. It acts on kidney tubule and blood capillaries and concentrates the urine by promoting the reabsorption of water and salts into the cortical collecting ducts.