PHYSICS Examination Papers 2008–2012 CONTENT n n CBSE Examination Paper–2008 (Delhi) CBSE Examination Paper–2008 (All India) CBSE Examination Paper–2009 (Delhi) CBSE Examination Paper–2009 (All India) CBSE Examination Paper–2009 (Foreign) CBSE Examination Paper–2010 (Delhi) CBSE Examination Paper–2010 (All India) CBSE Examination Paper–2010 (Foreign) CBSE Examination Paper–2011 (Delhi) CBSE Examination Paper–2011 (All India) CBSE Examination Paper–2011 (Foreign) CBSE Examination Paper–2012 (Delhi) CBSE Examination Paper–2012 (All India) CBSE Examination Paper–2012 (Foreign) 3 37 71 103 131 161 186 213 239 273 303 333 358 390 n n n n n n n n n n n n CBSE EXAMINATION PAPERS DELHI–2008 Time allowed : 3 hours General Instructions: (a) All questions are compulsory. (b) There are 30 questions in total. Questions 1 to 8 carry one mark each, questions 9 to 18 carry two marks each, questions 19 to 27 carry three marks each and questions 28 to 30 carry five marks each. (c) There is no overall choice. However, an internal choice has been provided in one question of two marks, one question of three marks and all three questions of five marks each. You have to attempt only one of the given choices in such questions. (d) Use of calculators is not permitted. (e) You may use the following values of physical constants wherever necessary: c = 3 ´ 108 ms - 1 h = 6 × 626 ´ 10 -34 Js e = 1× 602 ´ 10 -19 C 1 = 9 × 109 Nm2C– 2 4pe o m 0 = 4p ´ 10 -7 TmA -1 Maximum marks : 70 Boltzmann’s constant k = 1× 381 ´ 10 -23 J K -1 Avogadro’s number N A = 6 × 022 ´ 10 23 /mole Mass of neutron m n = 1× 2 ´ 10 -27 kg Mass of electron m e = 9 ×1´ 10 -31 kg Radius of earth = 6400 km CBSE (Delhi) SET–I ® 1. What is the direction of the force acting on a charge particle q, moving with a velocity v in a ® 2. 3. 4. 5. 6. 7. uniform magnetic field B ? Name the part of the electromagnetic spectrum of wavelength 10 - 2 m and mention its one application. An electron and alpha particle have the same de Broglie wavelength associated with them. How are their kinetic energies related to each other ? A glass lens of refractive index 1× 5 is placed in a through of liquid. What must be the refractive index of the liquid in order to make the lens disappear ? A 500 mC charge is at the centre of a square of side 10 cm. Find the work done in moving a charge of 10 mC between two diagonally opposite points on the square. State the reason, why heavy water is generally used as a moderator in a nuclear reactor. How does the fringe width of interference fringes change, when the whole apparatus of Young’s experiment is kept in a liquid of refractive index 1.3 ? 4 Xam idea Physics—XII 8. The plot of the variation of potential difference across a combination of three identical cells in series, versus current is as shown below. What is the emf of each cell ? V 6V 0 1A i 9. Derive the expression for the electric potential at any point along the axial line of an electric dipole ? 10. Define magnetic susceptibility of a material. Name two elements, one having positive susceptibility and the other having negative susceptibility. What does negative susceptibility signify ? 11. The oscillating magnetic field in a plane electromagnetic wave is given by B y = (8 ´ 10 - 6 ) sin [ 2 ´ 1011 t + 300 px ] T (i) Calculate the wavelength of the electromagnetic wave. (ii) Write down the expression for the oscillating electric field. Prove that an ideal capacitor, in an a.c. circuit does not dissipate power. OR Derive an expression for the impedance of an a.c. circuit consisting of an inductor and a resistor. 23 23 A nucleus 10 Na. Calculate the maximum kinetic energy of Ne undergoes b-decay and becomes 11 electrons emitted assuming that the daughter nucleus and anti-neutrino carry negligible kinetic energy. 23 ì mass of 10 Ne = 22 × 994466 u ü ï ï 23 í mass of 11 Na = 22 × 989770 u ý ï1 u = 931× 5 MeV / c 2 ï î þ Distinguish between an intrinsic semiconductor and P-type semiconductor. Give reason, why a P-type semiconductor crystal is electrically neutral, although n h >> n e ? Draw a ray diagram of a reflecting type telescope. State two advantages of this telescope over a refracting telescope. A ray of light passing through an equilateral triangular glass prism from air undergoes minimum deviation when angle of incidence is 3/4th of the angle of prism. Calculate the speed of light in the prism. The given inputs A, B are fed to a 2-input NAND gate. Draw the output wave form of the gate. A Input B Input t1 t2 t3 t4 t5 t6 12. 13. 14. 15. 16. 17. Examination Papers 5 18. A transmitting antenna at the top of a tower has a height of 36 m and the height of the receiving antenna is 49 m. What is the maximum distance between them, for satisfactory communication in the LOS mode ? (Radius of earth = 6400 km). 19. How is a wavefront defined ? Using Huygen’s construction draw a figure showing the propagation of a plane wave refracting at a plane surface separating two media. Hence verify Snell’s law of refraction. 20. A metallic rod of length l is rotated at a constant angular speed w, normal to a uniform magnetic field B. Derive an expression for the current induced in the rod, if the resistance of the rod is R. 21. The figure adjoining shows the V-I characteristics of a semiconductor diode. I (mA) 100 80 60 40 20 100 80 Vbr 60 40 20 O 0.2 0.4 0.6 0.8 10 20 30 V (Volt) I (mA) (i) Identify the semiconductor diode used. (ii) Draw the circuit diagram to obtain the given characteristic of this device. (iii) Briefly explain how this diode can be used as a voltage regulator. 22. An inductor 200 mH, capacitor 500 mF, resistor 10 W are connected in series with a 100 V, variable frequency a.c. source. Calculate the (i) frequency at which the power factor of the circuit is unity. (ii) current amplitude at this frequency. (iii) Q-factor. 23. Prove that the current density of a metallic conductor is directly proportional to the drift speed of electrons. OR A number of identical cells, n, each of emf E, internal resistance r connected in series are charged by a d.c. source of emf E ¢, using a resistor R. (i) Draw the circuit arrangement. (ii) Deduce the expressions for (a) the charging current and (b) the potential difference across the combination of the cells. 24. A potentiometer wire of length 1 m is connected to a driver cell of emf 3 V as shown in the figure. When a cell of 1× 5 V emf is used in the secondary circuit, the balance point is found to be 60 cm. On replacing this cell and using a cell of unknown emf, the balance point shifts to 80 cm. 6 Xam idea Physics—XII 3V A 1.5 V R B (i) Calculate unknown emf of the cell. (ii) Explain with reason, whether the circuit works, if the driver cell is replaced with a cell of emf 1 V. (iii) Does the high resistance R, used in the secondary circuit affect the balance point ? Justify our answer. 25. An electromagnetic wave of wavelength l is incident on a photosensitive surface of negligible work function. If the photo-electrons emitted from this surface have the de-Broglie wavelength l1 , æ 2mc ö 2 prove that l = ç ÷ l1 . è h ø 26. The energy level diagram of an element is given below. Identify, by doing necessary calculations, which transition corresponds to the emission of a spectral line of wavelength 102 × 7 nm. A B C - 3.4 eV D - 13.6 eV - 0.85 eV - 1.5 eV 27. Draw a plot of the variation of amplitude versus w for an amplitude modulated wave. Define modulation index. State its importance for effective amplitude modulation. 28. (a) Using Biot-Savart’s law, derive an expression for the magnetic field at the centre of a circular coil of radius R, l R x O I number of turns N, carrying current I. (b) Two small identical circular coils marked 1 and 2 carry equal currents and are placed with their geometric axes perpendicular to each other as shown in the figure. Derive x an expression for the resultant magnetic field at O. 2 O' R I OR Draw a schematic diagram of a cyclotron. Explain its underlying principle and working, starting clearly the function of the electric and magnetic fields applied on a charged particle. Deduce an expression for the period of revolution and show that it does not depend on the speed of the charged particle. Examination Papers 7 29. (a) For a ray of light travelling from a denser medium of refractive index n1 to a rarer medium of n refractive index n 2 , prove that 2 = sin i c , where i c is the critical angle of incidence for the media. n1 (b) Explain with the help of a diagram, how the above principle is used for transmission of video signals using optical fibres. OR (a) What is plane polarised light? Two polaroids are placed at 90° to each other and the transmitted intensity is zero. What happens when one more polaroid is placed between these two, bisecting the angle between them ? How will the intensity of transmitted light vary on further rotating the third polaroid? (b) If a light beam shows no intensity variation when transmitted through a polaroid which is rotated, does it mean that the light is unpolarised ? Explain briefly. 30. (a) Using Gauss law, derive an expression for the electric field intensity at any point outside a uniformly charged thin spherical shell of radius R and charge density s C/m 2 . Draw the field lines when the charge density of the sphere is (i) positive, (ii) negative. (b) A uniformly charged conducting sphere of 2 × 5 m in diameter has a surface charge density of q 100 mC/m 2 . Calculate the (i) charge on the sphere (ii) total electric flux passing through the sphere. OR 10 cm 10 cm (a) Derive an expression for the torque experienced by an electric dipole kept in a uniformly electric field. (b) Calculate the work done to dissociate the system of three charges –4q +2q 10 cm placed on the vertices of a triangle as shown. Here q = 1× 6 ´ 10 - 10 C. CBSE (Delhi) SET–II Questions different from Set – I Name the part of the electromagnetic spectrum of wavelength 10 2 m and mention its one application. 2. An electron and alpha particle have the same kinetic energy. How are the de-Broglie wavelengths associated with them related? 3. A converging lens of refractive index 1× 5 is kept in a liquid medium having same refractive index. What would be the focal length of the lens in this medium? 6. How does the angular separation of interference fringes change, in Young’s experiment, if the distance between the slits is increased? 11. Draw a ray diagram of an astronomical telescope in the normal adjustment position. State two draw backs of this type of telescope. 12. Calculate the distance of an object of height h from a concave mirror of focal length 10 cm, so as to obtain a real image of magnification 2. 1. 8 Xam idea Physics—XII 13. for all combinations of A. A X B C t1 t2 t3 t4 t5 t6 A Input B Input ® 15. How is a wavefront defined ? Using Huygen’s construction draw a figure showing the propagation of a plane wave reflecting at the interface of the two media. 18. why two independent sources of light cannot be considered as coherent sources. 19. and connected to a resistor R. . Draw the output wave form at X.c. Derive an expression for the potential energy of an electric dipole of dipole moment p in an ® electric field E. OR Derive an expression for the self-inductance of a long air-cored solenoid of length l and number of turns N. Prove that an ideal inductor does not dissipate power in an a. write down a truth table to find the final output. in a uniform magnetic field B. 5. is rotated at a constant angular speed w. 2. If the output of a 2 input NOR gate is fed as both inputs A and B to another NOR gate. Explain the variation of conductivity with temperature in (a) good conductors. How does the power of a convex lens vary. Draw a ray diagram of a compound microscope. Deduce expressions for : (i) Maximum emf induced in the coil (ii) Power dissipation in the coil. State the reason. (b) ionic conductors. An electron and a proton have the same de Broglie wavelength associated with them. 1. Also identify the gate. 12. B for the logic circuit shown below. CBSE (Delhi) SET–III Questions different from Set – I & II Name the absorbing material used to control the reaction rate of neutrons in a nuclear reactor. circuit. How are their kinetic energy related to each other? 7. Write the expression for its magnifying power. using the given inputs A. if the incident red light is replaced by violet light ? 9. area A. 25. 24. A coil of number of turns N. Show that the angle of incidence is equal to the angle of reflection. Define conductivity of a conductor. B. Examination Papers 9 15. 25. 50 Hz a. It is found that the power factor of the circuit is unity. which corresponds to the emission of a spectral line of wavelength 482 nm: 24. Obtain the expression for the mutual inductance of a pair of coaxial circular coils of radii r and R ( R > r ) placed with their centres coinciding. if the resistance of the rod is R. using the mathematical expression of the resistivity of a material.1.0. Plot a graph showing the variation of resistivity with temperature for a metallic conductor. Prove that the current density of a metallic conductor is directly proportional to the drift speed of electrons through the conductor. Calculate the inductance of the inductor and the current amplitude.4 eV B D .6 eV .5 eV Identify. OR Define resistivity of a conductor. An inductor of unknown value. normal to a uniform magnetic field B. using necessary calculations.13. 20. How does one explain such a behaviour. The energy levels of an element are given below: A C . A metallic rod of length l is rotated at an angular speed w. a capacitor of 100 mF and a resistor of 10 W are connected in series to a 200 V. Derive an expression for the (i) emf induced in the rod (ii) heat dissipation. .85 eV .3. (b) Write down the expression for the oscillating magnetic field. 16. the transition.c. 26.1 (a) Obtain the value of the wavelength of the electromagnetic wave. The oscillating electric field of an electromagnetic wave is given by: E y = 30 sin [ 2 ´ 1011 t + 300 px ] Vm . source. b = Þ b µ l for same D and d. fringe n 1× 3 1 width decreases to times. VA = VB . Fm = q v ´ B Obviously. Given l electron = l a de Broglie wavelength associated with a particle of mass m and energy E is h l= 2mE h h \ = 2m e E e 2m a E a 4.2 m belongs to microwaves. Force. Work done in moving charge q 0 = 10 mC from A to B is W = q 0 (VB . 3. It is used in RADAR. therefore. so heavy water molecules do not absorb fast neutorns. Heavy water has negligible absorption cross-section for neutrons and its mass is small.. Wavelength 10 .e. B q 500 mC O A \ 6. the wavelength decreases to l¢ = = × So..VA ) = 0 The basic principle of mechanics is that momentum transfer is maximum when the mass of colliding particle and target particle are equal. 7. the force on charged particle is perpendicular to both velocity v and magnetic field B ? 2. When the whole apparatus is immersed in a d l l transparent liquid of refractive index n = 1× 3. 1× 3 .10 Xam idea Physics—XII Solutions CBSE (Delhi) SET–I ® ® ® ® ® 1. refractive index of liquid = 1× 5. 5. Dl Fringe width. points A and B are at the same potential i. The glass lens will disappear in the liquid if the refractive index of liquid is equal to that of glass i. but simply slow them. The points A and B are equidistant from the centre of square where charge q = 500 mC is located. That is kinetic energy of electron and a-particle are in inverse ratio of these masses.e. l r + l û 4pe 0 ê ú ë r -l û q . having charges . then r > > l p 1 \ V= 4pe 0 r 2 10.l 2 = As q . Magnetic susceptibility: It is defined as the intensity of magnetisation per unit magnetising field.e.q ê ú ú= 2 2 4pe 0 ër . when i = 0.q) and V2 = V1 = 4pe 0 r . 2l 1 = 4pe 0 r 2 . 6 6 = 3e .q and + q at points A and B respectively. p = q . e = 2 V 9.e. V = 6 × 0 V \ From (1). Negative susceptibility of a substance signifies that the substance will be repelled by a strong magnet or opposite feeble magnetism induced in the substance. 8.q is AP = r + l Let V1 and V2 be the potentials at P due to charges + q and .. 2l = p (dipole moment) p 1 \ V= 2 4pe 0 r .(r . (r + l) (r – l) P Consider a point P on the axis of dipole at a distance r from A 2l B mid-point O of dipole.l 4pe 0 r + l \ Resultant potential at P due to dipole q q 1 1 V = V1 + V2 = 4 pe 0 (r .. c m = .l) 4pe 0 (r + l) é(r + l) .q respectively. From fig. M i.i r int where r int is effective (total) internal resistance.qê . The equation of terminal potential difference V = e eff .l 2 If point P is far away from the dipole.q to + q. –q +q O The distance of point P from charge + q is BP = r . ® Electric dipole moment. Consider an electric dipole AB. It has no unit.l r The distance of point P from charge . The separation between the charges is 2l.0 Þ e= =2V 3 i. . Then q 1 1 ( .Examination Papers 11 Let e be emf and r the internal resistance of each cell.i r int becomes …(1) V = 3 e . emf of each cell. Electric Potential due to an electric dipole at axial point. 2l. H Iron has positive susceptibility while copper has negative susceptibility.l) ù é 1 1 1 ù 1 . directed from . 6 T. K.1 12. According to right hand system of E. wavelength. circuit. we get B 0 = 8 ´ 10 .. VL = X L i = wLi where X L = wL is inductive reactance \ \ Þ Impedance. VR and VL are mutually perpendicular.6 ´ 3 ´ 108 = 2 × 4 ´ 10 3 Vm . The equation of b-decay of 23 10 Ne is Mass 23 ¾® 11 Na + e . l = 2p 1 = m 300p 150 ® ® ® w = 2 ´ 1011 rad s .. The applied voltage is V = V0 sin wt.me . while the voltage VL leads the current by an angle p × Thus.m N ( 11 Na) .2 × 4 ´ 10 3 sin ( 2 ´ 1011 t + 300px ) V m . Dm = m N ( 10 Ne) . The voltage VR and current I are in the same phase. so equation is E Z = . the electric field oscillates along negative Z-axis.12 Xam idea Physics—XII 11.1 . Power dissipated in a. P = Vrms I rms cos f where cos f = For an ideal capacitor R = 0 \ P = Vrms I rms cos f = Vrms R =0 Z I rms ´ 0 = 0 (zero). 2 2 V = VR + VL V = V0 sin wt V VL But VR = Ri.e. The resultant 2 of VR and VL is the applied voltage i. V = ( Ri) 2 + ( X L i) 2 V 2 Z = = R2 + XL i Z = R 2 + ( wL) 2 23 10 Ne I VR I 13.1 .e. Suppose the voltage across resistor is VR and that across inductor is VL . \ cos f = VR R VL L R Z i. k= 2p = 300p l (ii) E 0 = B 0 c = 8 ´ 10 .+ n 23 23 difference. (i) Standard equation of magnetic field is B y = B 0 sin ( wt + kx ) T Comparing this equation with the given equation.c. B. OR Let a circuit contain a resistor of resistance R and an inductor of inductance L connected in series. power dissipated in an ideal capacitor is zero. . (ii) Its resolving power is greater than refracting telescope due to larger aperture of mirror. 15.m e e 23 23 = m ( 10 Ne) . significant (of the order of mA).22 × 989770 = 0 × 004696 u \ Maximum K.E.Examination Papers 13 Changing nuclear masses into atomic masses Dm = m {( 23 10 Ne ) -10m }-{m ( e 23 11 Na ) -11m }. In) impurity. . Q = 0 × 004696 u ´ 931× 5 MeV/u = 4 × 37 MeV 14. MI M2 F Eye Advantages: (i) It is free from chromatic aberration.m ( 11 Na) = 22 × 994466 . minority charge carriers are electrons. (i) p-type semiconductor It is a semiconductor doped with p-type (like Al. Intrinsic semiconductor (i) (ii) It is a semiconductor in pure form. Intrinsic charge carriers are electrons and (ii) Majority charge carriers are holes and holes with equal concentration. (iii) Current due to charge carriers is feeble (iii) Current due to charge carriers is (of the order of mA). P-type semiconductor is electrically neutral because every atom. whether it is of pure semiconductor (Ge or Si) or of impurity (Al) is electrically neutral. . Clearly BB¢ = v1t. Given hT = 36 m. Given A = 60° . secondary spherical wavelets originate from these A' points. the frequency of wave remains unchanged but its speed and the wavelength both are changed. v= sin i sin r sin 45° 1 / 2 = = 2 = 1× 41 sin 30° 1 / 2 c 3 ´ 108 = = 2 × 13 ´ 10 8 ms . and R e = 6400 km = 6 × 4 ´ 10 6 m.e.e. Accordingly the output waveform Y = AB is shown in Output Y fig. Let v1 and v 2 be the speeds of waves in these media. which travel with speed v1 in the first medium and speed v 2 in the second medium. The output waveform is shown in fig. 90o i i B' As the wavefront AB advances. n 1× 41 17. output is obtained if either or both inputs are zero. . hR = 49 m. The output of NAND gate with inputs A and B is Y = AB i. 18. output is zero between intervals 0 to t 1 and t 4 to t 5 and in all other intervals it is ‘1’. Maximum LOS distance. In transversing from first medium to another medium. i= 3 3 A = ´ 60° = 45° 4 4 and r1 = r 2 = A = 30° 2 For minimum deviation i1 = i 2 = i \ Refractive index of prism. it strikes the points between X Y r A A and B¢ of boundary surface. n = = Speed of light in prism. This phenomenon is called refraction. d m = 2R e hT + 2R e hR Output waveform t1 t2 t3 t4 t5 t6 = 2R e ( hT + hR ) = 2 ´ 6 × 4 ´ 10 6 ( 36 + 49 ) = 3 × 578 ´ 10 3 ( 6 + 7) = 3 × 578 ´ 10 3 ´ 13 m = 46 × 5 ´ 10 3 m = 46 × 5 km 19. Proof of Snell’s law of Refraction using Huygen’s wave theory: When a wave starting from one homogeneous medium enters the another homogeneous medium.1 . i. Wavefront: A wavefront is a locus of all particles of medium vibrating in the same phase. Suppose a plane wavefront AB in first medium is incident obliquely on the boundary surface XY and its end A touches the surface at A at time t = 0 while the other end B reaches B the surface at point B¢ after time-interval t. Let XY be a surface separating the two media ‘1’ and ‘2’.14 Xam idea Physics—XII 16. it is deviated from its path. According to Huygen’s r 90o principle.. Consider a metallic rod OA of length l .. Assuming A as centre. A rod may be xvx x x x x x x supposed to be formed of a large number of small elements. and is called the Snell’s law.Examination Papers 15 First of all secondary wavelet starts from A. Clearly BB¢ = v1t and AA¢ = v 2 t. the point of wavefront traverses a distance BB¢ ( = v1t) in first medium and reaches B¢ . In the same time-interval t. Ð AA¢ B ¢ = 90° AA¢ v 2 t . the plane of x x x x x x x rotation being perpendicular to the magnetic field. In right-angled triangle AB¢ B. Hence A¢ B¢ will be the refracted wavefront.. one after the other and will touch A¢ B¢ simultaneously. we get v sin i . x x x 20.. which traverses a distance AA¢ ( = v 2 t) in second medium in time t. then area swept by the x x x x x x x element per second = v dx x x x x x x x x x x x The emf induced across the ends of element dA de = B = B v dx dt But v = xw \ de = B x w dx The emf induced across the rod \ e = ò B xw dx = Bw ò x dx 0 0 l l é x2 = Bw ê ê ë 2 Current induced in rod I = ù él2 -0 ú = Bw ê ú ê û0 ë 2 l ù 1 ú = B wl ú û 2 2 e 1 B wl 2 = × R 2 R . According to Huygen’s principle A¢ B¢ is the new position of wavefront AB in the second medium.(1) \ sin i = sin Ð BAB¢ = = 1 AB¢ AB¢ Similarly in right-angled triangle AA¢ B ¢ . we draw a spherical arc of radius AA¢ ( = v 2 t) and draw tangent B¢ A¢ on this arc from B¢ .(2) \ sin r = sin Ð AB¢ A¢ = = AB¢ AB¢ Dividing equation (1) by (2). where the secondary wavelet now starts. Ð ABB¢ = 90° vt BB¢ . x x x xO A dx Consider a small element of length dx at a distance x from centre. As the incident wavefront AB advances.(3) = 1 = constant sin r v2 The ratio of sine of angle of incidence and the sine of angle of refraction is a constant and is equal to the ratio of velocities of waves in the two media. x x x x l If v is the linear velocity of this element... Let the incident wavefront AB and refracted wavefront A¢ B¢ make angles i and r respectively with refracting surface XY. from. the secondary wavelets start from points between A and B¢ .. This is the second law of refraction. which is rotating with x x B x x x x angular velocity w in a uniform magnetic field B . (iii) Zener diode as a Voltage Regulator The Zener diode makes its use as a voltage regulator due to the Potential divider following property : – + When a Zener diode is operated in the breakdown region. The series resistance R absorbs the output voltage fluctuations so as to maintain constant voltage across the load..(1) E= × l Unregulated input Regulated output .3 H. the current through R and Zener diode also increases. If the input dc voltage increases. I 0 = 0 = Z R 100 2 = = 10 2A = 14 × 1 A 10 w L 100 ´ 200 ´ 10 . This causes an electric field at each point of the wire of strength V . Consider a uniform metallic wire XY of length l and cross-sectional area A. 22.1 3 6 wr C LC 200 ´ 10 ´ 500 ´ 10 Linear Resonant Frequency w 100 100 fr = r = = Hz = 15 × 9 Hz 2p 2 ´ 3 ×14 6 × 28 (ii) At resonant frequency f r impedance.. So. Z = R V V 2 \ Current amplitude. A potential difference V is applied across the ends X and Y of the wire.3 (iii) Q-factor = r = =2 R 10 23. voltage drop across R increases.16 Xam idea Physics—XII – + Semiconductor diode used is Zener diode (but the voltages quoted in fig. Vrms = 100 V (i) Angular (resonant) frequency wr at which power factor of the circuit is unity. 21. the K voltage across it remains practically constant for a large I IL change in the current. are much more than actual values Vbr is usually < 6 V). (i) R p n m A R = 10 W .6 F. is given by 1 1 1 wr L = Þ wr = = = 100 rad s . Given L = 200 mH = 200 ´ 10 . without any change in the voltage across zener diode. The Zener diode is connected across load such that it is Vin VZ RL V0 reverse biased. V (ii) The circuit diagram for reverse characteristics of Zener diode – + is shown in fig. C = 500 mF = 500 ´ 10 . R A simple circuit of a voltage regulator using a IZ Zener diode is shown in the Fig. e ( e = 1× 6 ´ 10 .ne e (a) Charging current.I ( nr ) ..ne . the electrons gain a drift velocity v d opposite to Current (I) direction of electric field. The volume of this cylinder = cross sectional area ´ height = A vd t If n is the number of free electrons in the wire per unit volume.r e.e) = ..r e. q .. Negative sign shows that the direction of current is opposite to the drift velocity.r e. then the number of free electrons in the cylinder = n ( A v d t) If charge on each electron is . e..neAv d .IR .e.(4) This is the relation between electric current and drift velocity.IR + e¢ = 0 Þ V = e¢ . then the total charge flowing in time t will be equal to the total charge on the electrons present within the cylinder PQ.neAv d t I= = t t i.19 C). then I I q Current in wire I = .ne Þ I= R + nr a d + – R e ¢ . If q be the charge passing through the vdt cross-section of wire in t seconds..neAv d t \ Current flowing in the wire. …(a) I= dc source R + nr (b) Potential difference across the combination V is given by . current I = ..IR + e¢ = 0 I e¢ .(2) V t + – The distance traversed by each electron in time t = average velocity ´ time = v d t If we consider two planes P and Q at a distance v d t in a conductor.. Numerically I = neAv d .Examination Papers E 17 P Q X Y Due to this electric field. OR (i) The circuit arrangement is shown in fig.(3) q = ( nA v d t) ( . then the total charge flowing through a cross-section of the wire .V .(5) I Current density.r c (ii) Applying Kirchhoff’s second law to the circuit b abcda V . \ J = = nev d A Þ J µ vd ... Kinetic energy of electrons. l2 = 80 cm 80 \ e2 = ´ 1× 5 V = 2 × 0 V 60 (ii) The circuit will not work if emf of driver cell is 1 V (less than that of cell in secondary circuit).e¢ + ne R + nr 24.1× 50) | = 12 ×1 eV Hence.ne) R + nr e¢ ( R + nr .e.19 Now. (i) Unknown emf e 2 is given by e 2 l2 = Þ e1 l1 e2 = l2 e1 l1 Given e1 = 1× 5 V. E k = energy of photon of e. or l1 = l2 1 = 2mE k 2mE k Using (1). since at balance point no current flows through galvanometer G i.13 × 6 . l1 = 60 cm. we get h2 æ hc ö 2m ç ÷ è lø . Plot of variation of amplitude versus w for amplitude modulated wave is shown in fig. DE = | . DE = hc = l = J eV = 66 ´ 3000 = 12 × 04 eV 1027 ´ 16 6 × 6 ´ 10 . cell remains in open circuit. wave hc …(1) = l h h2 de Broglie wavelength.34 ´ 3 ´ 108 ´ 1× 6 ´ 10 . 27. Ec Amplitude maEc 2 wc – wm wc wc+ w m w rad . (iii) No.34 6 × 6 ´ 10 ´ 3 ´ 108 l2 1 = 102 × 7 ´ 10 . 25.18 Xam idea Physics—XII Þ Þ V = e¢ - ( e¢ .m.1) + ne V= R + nr Þ V= e¢ ( R + nr ) . transition shown by arrow D corresponds to emission of l = 102 × 7 nm. because total voltage across wire AB is 1 V which cannot balance the voltage 1× 5 V.( ..9 102 × 7 ´ 10 -9 Þ æ 2mc ö 2 l =ç ÷ l1 è h ø 26. so its value is kept £ 1 for avoiding distortions. E ma = m Ec For effective amplitude modulation the modulation index determines the distortions. Here q = 90° . Consider a small current element ‘ab’ of length Dl. ® ® ® dl r I P where m 0 The magnitude of magnetic field is m Idl sin q dB = 0 4p r2 ® ® ® ® ® where q is the angle between current element I dl and position vector r . due to current element ab is m I Dl DB = 0 4p R 2 According to Maxwell’s right hand rule.. Its value is m 0 = 4p ´ 10 -7 Wb/A-m. Magnetic Field at the centre of circular loop: Consider a circular coil of radius R carrying current I in anticlockwise direction. The direction of magnetic field dB is perpendicular to the plane containing I dl and r . at which magnetic field is to be computed. (a) Biot Savart Law It states that the magnetic field strength ( dB) produced due to a current element (of current I and length dl) at a point having position ® vector r relative to current element is m I dl ´ r dB = 0 4p r 3 is permeability of free space. The coil may be supposed to be formed of a large number of current elements. the direction of magnetic field at O is upward. Therefore magnetic field produced at O. O is the centre of coil. because current element at each point of circular path is perpendicular to the radius. Therefore the resultant magnetic field at the centre will be the sum of magnetic fields due to all current elements. Say.Examination Papers 19 Modulation Index: The ratio of amplitude of modulating signal to the amplitude of carrier wave is called modulation index i.e. perpendicular to the plane of coil. The direction of magnetic field due to all current elements is the same. Thus m I Dl m 0 I B = S DB = S 0 = S Dl 4p R 2 4p R 2 But S Dl = total length of circular coil = 2pR (for one-turn) . According to Biot Savart law the magnetic field due to current element ‘ab’ at centre O is m I Dl sin q DB = 0 4p R2 dl a b R I where q is angle between current element ab and the line joining the element to the centre O. 28. but if the current in the coil is clockwise. devised by Lawrence and Livingston. then S Dl = N. 2p R or B = 0 2 2R 4p R \ B= Here current in the coil is anticlockwise and the direction of magnetic field is perpendicular to the plane of coil upward. is a device for accelerating ions to high speed by the repeated application of accelerating potentials. . 4 OR (a) Cyclotron: The cyclotron. m 0 I ( pR 2 ) × 4p 2. then the direction of magnetic field will be perpendicular to the plane of coil downward. B tan q = 2 B1 Þ Þ ® tan q = 1 p q= 4 (Q B 2 = B1 ) \ B is directed at an angle ® p with the direction of magnetic field B 1 . so the net magnetic field at O is 2 2 B = B1 + B2 = 2B1 (as B1 = B 2 ) C2 (2) I = 2 As R << x B= m 0 IR 2 2 (R 2 + x 2 ) 3/ 2 m 0 2 2 . x 3 x3 m 2 2 m 0 IA = 0 4p x3 = 2m 0 IR 2 where A = pR 2 is area of loop. 2pR m I m NI \ B = 0 × N. (1) B2 (b) Magnetic field due to coil 1 at point O B ® B1 = m 0 IR 2 ® R I C1 x 2 (R 2 + x 2 ) 3/ 2 m 0 IR 2 2 (R 2 + x 2 ) 3/ 2 along OC1 B1 O Magnetic field due to coil 2 at point O ® ® x B2 = ® along C 2 O ® Both B 1 and B 2 are mutually perpendicular.20 Xam idea Physics—XII m0 I × 2pR 4p R 2 m I or B= 0 2R If the coil contains N–turns. oscillator where m is the mass and r the radius of the path of ion in the dee and B is the strength of the magnetic field. The function of electric field is to accelerate the charged particle and so to impart energy to the charged particle. . = mv max = 2 2m Dee-2 When charged particle crosses the gap Beam between dees it gains KE = q V In one revolution.(1) R. = = 2nqV 2m Working: The principle of action of the apparatus is shown in fig. qvB = mv 2 r or r = mv qB .Examination Papers 21 Principle: The positive ions produced from a source are accelerated.. The positive ions produced from a source S at the centre are accelerated by a dee which is at negative potential at that moment. E. then. The phenomenon is continued till the ion reaches at the periphery of the dees where an auxiliary negative electrode (deflecting plate) deflects the accelerated ion on the target to be bombarded.. attained: If R be the radius of the path and v max the velocity of the ion when it leaves the periphery. Due to the presence of perpendicular magnetic field the ion will move in a circular path. The angular velocity w of the ion is given by.E. The magnetic field and the frequency of the applied voltages are so chosen that as the ion comes out of a dee. the kinetic energy gained = 2nqV q2B 2R 2 Thus K. S Magnetic Pole N S Dee Dee Magnetic Pole S Dee-1 q2B 2R 2 1 2 K. Expression for Period of Revolution and Frequency: Suppose the positive ion with charge q moves in a dee with a velocity v. The phenomenon is continued till the ion reaches at the periphery where an auxiliary negative electrode (deflecting plate) deflects the accelerated ion on the target to be bombarded. the dees change their polarity (positive becoming negative and vice-versa) and the ion is further accelerated and moves with higher velocity along a circular path of greater radius.F.E. it crosses the gap twice. Expression for K. Due to the presence of perpendicular magnetic field the ion will move in a circular path inside the dees. then qBR v max = m The kinetic energy of the ion when it leaves the apparatus is. The function of magnetic field is to provide circular path to charged particle and so to provide the location where charged particle is capable of gaining energy from electric field. therefore if it completes n-revolutions before emerging the does. .(5) T= qB w= This is the expression for period of revolution. r = 90° \ From (1) sin i c n n2 = 2 Þ = sin i c sin 90° n1 n1 ic n1 r = 90° n2 (b) Transmission of video signals using optical fibre.(4) =t = 2 qB 2p m or . sin i n 2 29.e. (a) Snell’s laws is …(1) = sin r n1 Critical angle is the angle of incidence in denser medium for which angle of refraction in rarer medium is 90° i. Clearly the period of revolution is independent of speed of particle.. Thus the ray within the fibre suffers multiple total internal reflections and finally strikes the other end at an angle less than critical angle for quartz-air interface and emerges in air. p pm .7 large number of such fibres held together form a light pipe and are used for communication of light signals. it suffers A B refraction from air to quartz and strikes the quartz-coating interface at an angle more than the critical angle and so suffers total internal reflection and strikes the opposite face again at an angle greater than critical angle and so again suffers total internal reflection..4 m coated all around with a material of refractive index 1× 5. It is a very long and thin pipe of quartz ( n = 1× 7) of thickness Coating n = 1..e. 1) ..(2) = r m The time taken by the ion in describing a semi-circle. i = i c ..22 Xam idea Physics—XII v qB (from eq.5 nearly » 10 . . When a light ray is incident on one end at a small angle of incidence... A n = 1. i.. in turning through an angle p is. An optical fibre is a device based on total internal reflection by which a light signal may be transmitted from one place to another with a negligible loss of energy.(3) t= = w Bq T pm . As there is no loss of energy in total internal reflection.. the light signal is transmitted by this device without any appreciable loss of energy. According to Malus law. Reason: If incident light is unpolarised. the intensity variation I = I 0 cos 2 q . so angle between ® ® E i and d S is zero at each point. the intensity decreases and if angle decreases. ® concentric with given shell. which is constant. The unpolarised and polarised light is represented as (a) Unpolarised light (b) Polarised light (c) Partially polarised light If ordinary unpolarised light of intensity I 0 ¢ is incident on first polaroid (A. electric flux through Gaussian surface . the intensity of transmitted light through a polaroid is always I 0 / 2. the incident light (of intensity I 0 ) is unpolarised. But if incident light is polarised. (b) Yes. we consider a spherical Gaussian surface of radiusr ( > R). (a) Electric field intensity at a point outside a uniformly charged thin spherical shell: Consider a uniformly charged thin spherical shell of radius R carrying charge Q. say) is I = I 0 cos 2 q Þ I = I 0 cos 2 90° = 0 When one more polaroid (C say) is placed between A and B making an angle of 45° with the transmission axis of either of polaroids. the intensity increases. To find the electric field outside the shell. say) I ¢ Intensity of light transmitted from first polaroid is I 0 = 0 2 Given angle between transmission axes of two polaroids A and B is initially 90°. Also the directions of normal at each point is radially outward.Examination Papers 23 OR (a) Plane Polarised Light: The light having vibrations of electric field vector in only one direction perpendicular to the direction of propagation of light is called plane polarised light. necessarily takes place. Hence. If E is electric field outside the shell. 30. then by symmetry electric field strength has same magnitude E 0 on the Gaussian surface and is directed radially outward. then intensity of light transmitted from A is I ¢ IA = 0 = I0 2 Intensity of light transmitted from C is I I C = I 0 cos 2 45° = 0 × 2 Intensity of light transmitted from polaroid B is I 1 I I B = I C cos 2 45° = 0 ´ = 0 2 2 4 This means that the intensity becomes one-fourth of intensity of light that is transmitted from first polaroid. On further rotating the polaroid C such that if angle between their transmission axes increases. intensity of light transmitted from second polaroid (B. 6 C / m 2 ) ´ 3 ×14 ´ ( 2 × 5 m) 2 = 19 × 625 ´ 10 . so charge enclosed by Gaussian surface is Q.4 C = 1× 96 ´ 10 . then f = 4pR 2 s C \ E0 = 1 4pR 2 s R 2 s = 4pe 0 r 2 e 0r 2 The electric field lines are shown in the fig. these are directed in radially inward direction. 4pR 2 = s . s = 100 mC/m = 100 ´ 10 2 C / m2 . Q = s . Hence. Gaussian surface is outside the given charged shell. the field lines are directed radially in outward direction and for negatively charged shell. by Gauss's theorem ® ® 1 E · d E= ´ charged enclosed 0 òS e0 1 Þ E 0 4pr 2 = ´Q e0 1 Q Þ E0 = 4pe 0 r 2 Thus. + + + + O + R + + + – – – – R O – – – – (a) Positively charged shell (a) Negatively charged shell -6 (b) Given.3 C = 1× 96 mC . p ( 2R) 2 = (100 ´ 10 . = ò E 0 dS cos 0 = E 0 . 4pr Now. electric field outside a charged thin spherical shell is the same as if the whole charge Q is concentrated at the centre. Diameter.24 Xam idea Physics—XII ® S ® Q 2 R r P EO dS = ò E · d S. D = 2R = 2 × 5 m (i) Charge on sphere. For a positively charged shell. If s is the surface charge density of the spherical shell. F1 = qE.qE = 0 (zero) As net force on electric dipole is zero. the magnitude of torque depends on orientation ( q) of the electric dipole relative to electric field.q and + q at separation 2l the dipole moment of electric dipole. .q is.3 ) 12 e0 8 × 86 ´ 10 = 2 × 21 ´ 108 Nm 2 C . \ torque t = magnitude of one force ´ perpendicular distance between lines of action of forces = qE ( BN) = qE ( 2l sin q) = ( q 2l) E sin q [using (1)] .. This couple is called the torque and is denoted by t.2l r ® Force: The force on charge + q is.(2) = pE sin q Clearly.. Maximum Torque: For maximum torque sin q should be the maximum. In vector form t =p ´E ® ® ® ® ® ® ® . it experiences no force but experiences a torque..F2 = qE . The charges of dipole are ...(3) Thus. t max = pE . opposite to the direction of field E ® ® Obviously forces F1 and F2 are equal in magnitude but opposite in direction. The torque tends to align the dipole moment along the direction of electric field. Torque ( t) is a vector quantity whose direction is perpendicular to both p and E . if an electric dipole is placed in an electric field in oblique orientation. along the direction of field E ® ® ® ® ® 2l q O +q B F1 = qE 2l sin q E F2 = – qE A q –q N The force on charge . hence net force on electric dipole in uniform electric field is F = F1 . As the maximum value of sin q = 1 when q = 90° \ Maximum Torque.. F2 = qE .(1) p = q.. so dipole does not undergo any translatory motion. Torque : The forces F1 and F2 form a couple (or torque) which tends to rotate and align the dipole along the direction of electric field.Examination Papers 25 (ii) Electric flux passing through the sphere 1 1 f= ( Q) = ´ (1× 96 ´ 10 .1 OR (a) Consider an electric dipole placed in a uniform electric field of strength E in such a way that its dipole moment p makes an angle q with the direction of E . 9 ´ 10 9 ´ 100 ´ (1× 6 ´ 10 .4q) . ( 2q) ( . focal length of converging lens is infinity i. glass lens behaves as a glass plate. p = Þ mv = l l h Þ l= mv 1. Þ l e = 86 × 5 ´ l a The focal length of lens in a liquid-medium is given by æ 1 1 1 ö ÷ = ( l n g .10 ) 2 = . ( . ( 2q) ù U= + + ú 4pe 0 ê 0 ×10 0 ×10 ë 0 ×10 û = 1 é .e. .8 J Work done to dissociate the system of charges W = .1÷ ç ÷ ç ÷ f l è nl ø è R1 R 2 ø Given n l = n g = 1× 5 1 or \ =0 fl = ¥ fl i.U = 2 × 3 ´ 10 .e..10q 2 ù 1 (100q 2 ) ê ú =4pe 0 ê 0 × 10 4 pe 0 ú ë û = .. h h 2..26 Xam idea Physics—XII (b) Potential energy of system i.e.2 × 3 ´ 10 .4q) q .1) ç ç ÷ fl è R1 R 2 ø ng öæ 1 1 æ 1 ö ç ÷ =ç . work done to assemble the system of charges 1 é q . Kinetic energy. \ Þ Since Ek = l= lµ ma > me 4m p ma le = = la me me Þ le = 1872 ´ 4 la ( mv) 2 2m h 1 mE k 2mE k 3.8 J CBSE (Delhi) SET–II Wavelength 10 2 m belongs to radio-waves. This is used to broadcast radio programmes to long distances. u = ? v magnification m=. Therefore. we have f v u 1 1 1 .=-2 Þ v = 2u u 1 1 1 From mirror formula = + . Given focal length f = . Angular separation of interference fringes in Young’s experiment. æ bö l bq ç = ÷ = è Dø d If distance between the slits ‘d’ is increased. (ii) The image formed is inverted and fainter. the angular separation decreases.Examination Papers 27 6. A C1 Fo A' F e' B' C2 Fe in At i fin ty Draw backs: (i) It is not free from chromatic aberration. the output is 1 when both inputs are 1. .10 cm. B fo fe 11. Output of first NOR gate C = AB B Output X = CC = C = AB C X = AB This is AND operation. 12.= + 10 2u u 3 1 Þ =2u 10 10 ´ 3 Þ u== . Accordingly the waveform output is shown in figure.15 cm 2 A 13. q1 = 90° and q 2 = 0° ).e. W1 = 0.pE cos q qE \ Total work done in bringing the electric dipole from infinity. (i) The work done (W1 ) in bringing the dipole perpendicular to electric field from infinity. therefore.pE cos q ® ® +q p qE –q In vector form U =..cos q 2 ) = pE (cos 90° .. As charges + q and . (ii) Work done (W2 ) in rotating the dipole such that it finally makes an angle q from the direction of electric field. i. along the field direction and opposite to field direction respectively. The potential energy of an electric dipole of an electric field is defined as the work done in bringing the dipole from infinity to its present position in the electric field. Suppose the dipole is brought from infinity and placed at orientation q with the direction of electric field.q are qE and qE. +q qE 2l qE –q From infinity (ii) Now the dipole is rotated and brought to orientation making an angle q with the field direction (i. e.. The work done in this process may be supposed to be done in two parts. i. therefore. (i) Let us suppose that the electric dipole is brought from infinity in the region of a uniform electric ® field such that its dipole moment p always remains perpendicular to electric field.p ×E …(1) .cos q) = .e.pE cos q = .q traverse equal distance under equal and opposite forces. Electric potential energy of electric dipole. The electric forces an charges + q and . net work done in bringing the dipole in the region of electric field perpendicular to field-direction will be zero.28 Xam idea Physics—XII t1 t2 t3 t4 t5 t6 A Input B Output X t1 t2 t3 t4 t5 t6 15. work done W2 = pE (cos q1 . U = W1 + W2 = 0 . 1 s= r S. the collision of electrons with fixed lattice ions/atoms increases so that relaxation time ( t) decreases. The reciprocal of resistivity (r) of a material is called its conductivity ( s).(1) where m 0 = 4p ´ 10 . the conductivity of metals decreases with rise of temperature. then N n= l 2 . (ii) Conductivity of ionic conductor increases with increase of temperature because with increase of temperature. OR Self Inductance of a long air-cored solenoid: Consider a long air solenoid having `n' number of turns per unit length. (i) Conductivity of a metallic conductor s = 1 ne 2 t ... If A is cross-sectional area of solenoid. Consequently.... the ionic bonds break releasing positive and negative ions which are charge carriers in ionic conductors. s T . then effective flux linked with solenoid of length ‘ l’ where N = nl is the number of turns in length ' l ' of solenoid. i.. unit of conductivity is mho m -1 (or siemen m -1 ). \ cos f = = 0 Z Z \ P = Vrms i rms cos f = 0 i. then magnetic field within the solenoid. Initially the variation of conductivity with temperature is linear and then it is non-linear. B = m 0 nI .. power dissipated by an ideal inductor in ac circuit is zero. Figure represents the variation of conductivity of metal with temperature.I..7 henry/metre is the permeability of free space.Examination Papers 29 18.(3) A æ N ö \ Self-inductance L = m 0 ç ÷ Al è l ø m N 2A .(2) F = nl (m 0 nI ) A = m 0 n 2 AlI \ Self-inductance of air solenoid F L = = m 0 n 2 Al I If N is total number of turns in length l . \ F = ( nl BA) Substituting the value of B from (1) ..(4) = 0 l 19. = r m l With rise of temperature. For ideal inductor R = 0. The power where cos f = P = Vrms i rms cos f R R . e. If current in solenoid is I .e. reaches point B¢ (of the surface). one after the other and will touch A¢ B¢ simultaneously.e. its points A1 . after travelling a distance BB¢. Now in right-angled triangles ABB¢ and AA¢ B¢ (both are equal to 90°) Ð ABB¢ = Ð AA¢ B¢ side BB¢ = side AA¢ (both are equal to vt) and side AB¢ is common i. angle of incidence i = angle of reflection r 25. i. 1 r i r i the point B of wavefront reaches the point B¢ of the X Y A B' surface.. As the wavefront AB advances further. the secondary wavelets starting from points between A and B¢. When coil rotates with angular speed w.e. C Law of Reflection: Let XY be a reflecting surface at which a wavefront is being incident obliquely.e. According to Huygen’s principle wavefront A¢ B¢ represents the new position of AB.. As the incident wavefront AB advances..30 Xam idea Physics—XII 24. A¢ B¢ is the reflected wavefront corresponding to incident wavefront AB. Now taking A as centre we draw a spherical arc of radius AA¢ ( = vt) and draw tangent A¢ B¢ on this arc from point B¢ . from where the secondary wavelet now starts. then emf induced in the coil df d e=-N = . A 3 K etc. the D i A wavefront touches the surface XY at A. then due to presence of reflecting surface. According to Huygen’s principle each point of wavefront acts as a source of secondary waves. it cannot advance further. (i) Suppose initially the plane of coil is perpendicular to the magnetic field B. therefore the incident and the reflected rays make equal angles with the normal drawn on the surface XY. strike the reflecting surface successively and send spherical secondary wavelets in the first medium. ® then after time t. incident wavefront AB and reflected wavefront A¢ B¢ make equal angles with the reflecting surface XY. both triangles are congruent. but the secondary wavelet originating from point A begins to spread in all directions in the first medium with speed v. N-turns. the angle between magnetic field B and normal to plane of coil is q = wt \ At this instant magnetic flux linked with the coil f = BA cos wt If coil constains. First of all the secondary wavelet starts from point A and traverses distance AA¢ ( = vt) in first medium in time t. \ Ð BAB¢ = Ð AB¢ A¢ i. the point B of wavefront.. In the same time t. Wavefront: A wavefront is a locus of particles of medium all vibrating in the same phase.e. As the rays are always normal to the wavefront. A 2 . i.N ( BA cos wt) dt dt = + NBA w sin wt q = wt B …(1) . Let v A' B A2 be the speed of the wavefront and at time t = 0. After time t. When the point A of wavefront strikes the reflecting surface. Cadmium is the absorbing material for neutrons produced in a nuclear reactor. the current induced. of proton) ö ÷. Power of a lens increases if red light is replaced by violet light because P = . e. P = eI = e ç ÷ = èRø R N 2 B 2 A 2 w2 sin 2 wt [using (1)] …(3) = R Average power dissipated in a complete cycle is obtained by taking average value of sin 2 wt 1 over a complete cycle which is × 2 1 i.1) ç ç f è R1 R 2 and refractive index is maximum for violet light in visible region of spectrum. 7. = h 2m p E p 5. ÷ ø æ 1 1 1 = ( a n g . of electron = 1840 ´ (K. 2. the initial phase difference cannot remain constant because light is emitted due to millions of atoms and their number goes on changing in a quite random manner. sin wt = 1 \ Maximum emf induced.E.Examination Papers 31 \ For maximum value of emf e.e. e max = NBAw (ii) If R is resistance of coil. l = 2mE k Given \ le = l p h 2m e E e Þ i. Coherent sources are defined as the sources in the which initial phase difference remains constant.E. In the case of two independent sources. h de Broglie wavelength. I = e R 2 æeö e …(2) \ Instantaneous power dissipated.. (sin 2 wt) av = 2 \ Average power dissipated Pav = N 2 B 2 A 2 w2 × 2R CBSE (Delhi) SET–III 1. Ee m p = » 1840 E p me K.. This is Boolean expression for OR gate. First gate is NOR gate.32 Xam idea Physics—XII 9. its output B" ve ö ÷ ÷ ø Y = C + C = C × C = C = A + B = A + B. uo Objective B A" A Fo O Fe' A' b vo Eye-piece ue D Eye E Fe B' v0 æ 1 1 Dç + ç u0 è f e ve v æD Dö ÷ or M =. its output C = A + B Magnifying power M =Second gate is also NOR gate. A B C Y Its truth table is A 0 1 0 1 B 0 0 1 1 Y 0 1 1 1 .0 ç + ÷ u0 ç è f e ve ø 12. transition shown by arrow B corresponds to emission of l = 482 nm. .9 ´ 1× 6 ´ 10 . K) the magnetic field must oscillate along Z-axis.( .Examination Papers 33 15.1 l E 0 = 30 Vm .1 Comparing with standard equation E y = E 0 sin ( wt + kx ) Vm .3 × 4 .9 = 6 × 6 ´ 10 . DE = | . E 30 \ B0 = 0 = = 10 . C 3 ´ 108 \ Equation of oscillating magnetic field is B Z = B 0 sin ( wt + kx ) T Þ Bz = 10 .1 .1 .7 T. The magnetic field produced by current carrying larger coil C1 in m I the vicinity of small coil C 2 is B1 = 0 1 × 2R The magnetic flux linked with shorter coil C 2 is m I f2 = B1 A 2 = 0 1 pr 2 2R Mutual Inductance M = 20. DE = f2 m 0 pr 2 henry. electric field is oscillating along Y-axis. we get 2p = 300 p m .7 sin ( 2 ´ 1011 t + 300 px ) T 16. l= 2p m 300p 1 = m = 6 × 67 ´ 10 . = I1 2R R I1 r C2 C1 hc 6 × 6 ´ 10 . w = 2 ´ 1011 rad s .0 × 85) | = 2 × 55 eV Hence.1 .34 ´ 3 ´ 108 = J l 482 ´ 10 . (a) Given equation is E y = 30 sin ( 2 ´ 1011 t + 300 px ) Vm .3 m 150 (b) The wave is propagating along X-axis.19 66 ´ 3000 = 1027 ´ 16 eV = 2.57 eV Now. k = \ Wavelength. B.34 ´ 3 ´ 108 482 ´ 10 . so according to ® ® ® right hand system of ( E. The volume of this cylinder = cross sectional area ´ height = A vd t If n is the number of free electrons in the wire per unit volume.neAv d t .19 C). the electrons gain a drift velocity v d opposite to direction of electric field. Relation between electric current and drift velocity: Consider a uniform metallic wire XY E P Q X Y of length l and cross-sectional area A..4 V Current amplitude.. Þ L= 1 w C 2 X L = XC 1 2 Þ wL = 1 wC = = 1 4p f C 2 2 ( 2pf ) C Given f = 50 Hz. If q be the charge passing through the cross-section of wire in t seconds. then q Current in wire . For power factor unity. A potential difference V is Current (I) applied across the ends X and Y of the wire. then the total charge flowing in time t will be equal to the total charge on the electrons present within the cylinder PQ..(3) q = ( nA v d t) ( .34 Xam idea Physics—XII 24.4 F 1 \ L= H = 0 × 10 H 2 4 ´ (3 ×14) ´ (50) 2 ´ 10 . I0 = 0 Z At resonance.. then the number of free electrons in the cylinder = n ( A v d t) If charge on each electron is .(1) E= V l + – Due to this electric field.. Z =R V0 200 2 \ I0 = = = 20 2 A R 10 = 20 ´ 1× 414 A = 28 × 3 A 25. This causes an electric field vdt at each point of the wire of strength I I V .(2) I= t The distance traversed by each electron in time t = average velocity ´ time = v d t If we consider two planes P and Q at a distance v d t in a conductor.e) = . C = 100 mF = 100 ´ 10 .e ( e = 1× 6 ´ 10 .6 F = 10 .. then the total charge flowing through a cross-section of the wire . At temperatures much lower than 0°C. a is called the temperature co-efficient of resistivity.. A rod may be supposed to be formed of a large number of small elements.2 0 50 100 150 Temperature T(K) Resistivity rT of copper as a function of temperature T 26. that is not too large.Examination Papers 35 \ Current flowing in the wire.. Numerically I = neAv d .T0 )] where r T is the resistivity at a temperature T and r 0 is the same at a reference temperature T0 .4 I 0. OR l R =r A J= If l = 1. Resistivity r (10–8 Wm) 0.neAv d t I= = t t i.(5) Current density. Negative sign shows that the direction of current is opposite to the drift velocity. \ Þ We know that. current I = .neAv d . deviates considerably from a straight line (Figure). …(1) r T = r 0 [1 + a (T . the resistivity of a metallic conductor is approximately given by. however. the graph. resistivity of a material is numerically equal to the resistance of the conductor having unit length and unit cross-sectional area. (1) implies that a graph of r T plotted against T would be a straight line.. then area swept by the element per second = v dx .(4) This is the relation between electric current and drift velocity. The relation of Eq. which is rotating with angular velocity w in a uniform magnetic field B . Consider a small element of length dx at a distance x from centre. If v is the linear velocity of this element. The resistivity of a material is found to be dependent on the temperature. q . Over a limited range of temperatures. A = 1 Þ r = R Thus. I = nev d A J µ vd ..e. Consider a metallic rod OA of length l .. the plane of rotation being perpendicular to the magnetic field. Different materials do not exhibit the same dependance on temperatures. Current induced in rod I = = e 2 B 2 w2 l 4 = R 4R . power dissipated.36 Xam idea Physics—XII The emf induced across the ends of element dA de = B = B v dx dt But v = xw \ de = B x w dx \ The emf induced across the rod e = ò B xw dx = Bw ò x dx 0 0 l l éx = Bw ê ê ë 2 2 ù ú ú û0 ù 1 ú = B wl ú û 2 2 l él2 = Bw ê -0 ê ë 2 e 1 B wl 2 = × R 2 R It circuit is closed. (b) There are 30 questions in total. (e) You may use the following values of physical constants wherever necessary: c = 3 ´ 108 ms . (d) Use of calculators is not permitted. 7. (c) There is no overall choice. Questions 1 to 8 carry one mark each. Identify the part of the electromagnetic spectrum to which the following wavelengths belong: (i) 10 . 3. 2. if the distance between the light source and the cathode of the cell is doubled? Draw an equipotential surface for a system. 4. one question of three marks and all three questions of five marks each. 5. an internal choice has been provided in one question of two marks. questions 9 to 18 carry two marks each. . However. . questions 19 to 27 carry three marks each and questions 28 to 30 carry five marks each.12 m How does the width of the depletion layer of a p-n junction diode change with decrease in reverse bias ? What is the nuclear radius of 125 Fe. consisting of two charges Q. 8.Q separated by a distance ‘ r ’ in air. how is the back emf induced in the coil related to it? An object is held at the principal focus of a concave lens of focal length f.1 m (ii) 10 .CBSE EXAMINATION PAPERS ALL INDIA–2008 Time allowed : 3 hours General Instructions: (a) All questions are compulsory. You have to attempt only one of the given choices in such questions. 6.1 h = 6 × 626 ´ 10 -34 Js e = 1× 602 ´ 10 -19 C 1 9 2 – = 9 × 10 Nm C 4pe o m 0 = 4p ´ 10 -7 TmA -1 2 Maximum marks : 70 Boltzmann’s constant k = 1× 381 ´ 10 -23 J K -1 Avogadro’s number N A = 6 × 022 ´ 10 23 /mole Mass of neutron m n = 1× 2 ´ 10 -27 kg Mass of electron m e = 9 ×1´ 10 -31 kg Radius of earth = 6400 km CBSE (All India) SET–I 1. if that of 27 Al is 3 × 6 fermi? When current in a coil changes with time. Where is the image formed? What is the geometrical shape of the wavefront when a plane wave passes through a convex lens? How does the stopping potential applied to a photocell change. 11. and (ii) width of the source slit is increased? A jet plane is travelling west at 450 ms .4 tesla and the angle of dip is 30°. OR Draw a plot of the binding energy per nucleon as a function of mass number for a large number of nuclei. Draw a plot of N as a function of time.0 1. Why do we need carrier waves of very high frequency in the modulation of signals ? A carrier wave of peak voltage 20 V is used to transmit a message signal. t 0 . Obtain the expression for the potential energy of an electric dipole of dipole moment p placed in an electric field E. i: 6. indicating clearly the directions of the oscillating electric and magnetic fields associated with it. Explain the energy release in the process of nuclear fission from the above plot. green.38 Xam idea Physics—XII 9. 17. State the law of radioactive decay. red wavelengths are incident on an isosceles right-angled prism. What should be the peak voltage of the modulating signal. 10. 1× 424. blue light are 1× 39. Write the expression for the magnetic moment when an electron revolves at a speed ‘ v’. light rays of blue. Explain with reason.1 . . in order to have a modulation index of 80%? ® ® 14. The following graph shows the variation of terminal potential difference V. If the horizontal component of earth’s magnetic field at that place is 4 ´ 10 .0 2.0 V (volt) 3. In the figure given below. 13.0 i (ampere) (i) Calculate the emf of each cell. find the emf induced between the ends of wings having a span of 30 m. across a combination of three cells in series to a resistor. The refractive index of the prism for red. What is meant by the transverse nature of electromagnetic waves ? Draw a diagram showing the propagation of an electromagnetic wave along the x-direction. 15. Define the term : magnetic dipole moment of a current loop. which ray of light will be transmitted through the face AC. How will the angular separation and visibility of fringes in Young’s double slit experiment change when (i) screen is moved away from the plane of the slits. (ii) For what current i will the power dissipation of the circuit be maximum ? 16. 12. versus the current. around an orbit of radius ‘ r ’ in hydrogen atom. green. If N 0 is the number of radioactive nuclei in the sample at some initial time. 1× 476 respectively. find out the relation to determine the number N present at a subsequent time. Write a typical nuclear reaction in which a large amount of energy is released in the process of nuclear fission. V Input A 0 t1 t2 t3 t4 t5 t6 t7 t8 t Input B 0 t1 t2 t3 t4 t5 t6 t7 t8 t 22. B as given below. A resistance R = 2 W is connected to one of the gaps in a R metre bridge. Distinguish between isotopes and isobars. with the help of a suitable figure. Find out the ratio of the drift speeds of the electrons in the two wires for the two cases. but their cross-sectional areas are in the ratio 2 : 3 and lengths in the ratio 1 : 2. 19. Draw the output wave-form in each case.c. of communication systems that use space wave mode propagation. Give one example for each of the species. are applied as input to (i) AND (ii) NOR and (iii) NAND gates. Two signals A. Give typical examples. it is found that the balance point further shifts by 20 cm (away from end A). X 23. An unknown resistance X > 2 W is connected in the other G gap as shown in the figure. How long will it take the activity to reduce to 3 ×125%? 21. which uses a wire of length 1 m. Two wires X. 20. Neglecting the end correction. The balance point is noticed J at ‘ l’ from the positive end of the battery. What does the term ‘LOS communication’ mean ? Name the types of waves that are used for this communication. source. calculate the value of unknown resistance X used.Examination Papers A 39 Red Green Blue 90° B C 18. What are permanent magnets ? What is an efficient way of preparing a permanent magnet ? Write two characteristic properties of materials which are required to select them for permanent magnets. A radioactive isotope has a half-life of 5 years. They are first connected in series and then in parallel to a d. . Y have the same resistivity. On B A (100–l ) cm l cm interchanging R and X . mutually perpendicular to each other and perpendicular to the direction of the beam. find the focal length of the lens in the liquid.c. State the condition for resonance to occur in a series LCR a. with the wavelength of the incident light. A beam of proton passes undeflected with a horizontal velocity v. Explain its working principle. 26. Q of the circuit. C 40 V R 30 V 2A Vrms OR Draw a labelled circuit arrangement showing the windings of primary and secondary coil in a transformer. A converging lens of refractive index 1× 5 and of focal length 15 cm in air.40 Xam idea Physics—XII 24. Calculate the (i) impedance. use the expression F = q ( v ´ B) to find the direction of the force F acting on it. Draw the input/output wave-forms indicating clearly the functions of the two diodes used. Draw a plot showing the variation of power of a lens. If the magnitudes of the electric and magnetic fields are 100 kV/m. How much current is drawn by the primary coil of a transformer which steps down 220 V to 22 V to operate device with an impedance of 220 ohm ? 29. 27. 50 mT respectively. if the proton beam carries a current of 0 × 80 mA. If a particle of charge q is moving with velocity v along the y-axis and the magnetic field B is ® ® ® acting along the z-axis. Write any two major sources of energy loss in this device. OR Draw a circuit diagram of a full-wave rectifier. (ii) wattless current of the given a. With a circuit diagram. 25. (b) Obtain the equivalent capacitance of the network given below. Define the quality factor. source used. Explain the underlying principle and working of a step-up transformer. ‘the electron revolves around the nucleus only in certain fixed orbits without radiating energy' can be explained on the basis of de-Broglie hypothesis of wave nature of electron. has the same radii of curvature for both sides. determine the charge and voltage across C 4 . (ii) force exerted by the beam on a target on the screen. circuit. If it is immersed in a liquid of refractive index 1× 7. 28. through a region of electric and magnetic fields. explain how a zener diode can be used as a voltage regulator. . (a) Derive an expression for the energy stored in a parallel plate capacitor C.c. charged to a potential difference V. calculate (i) velocity of the beam v.c. circuit and derive an expression for the resonant frequency. Show that Bohr’s second postulate. Draw a plot showing the variation of the peak current (i m ) with frequency of the a. For a supply of 300 V. will the object the be kept. Draw a labelled schematic sketch and write briefly its working. . The distance between the lenses is 15 cm. Draw a labelled ray diagram of a compound microscope and write an expression for its magnifying power. (i) How far from the objective lens. What is the minimum radius of the spherical shell required. A Van de Graaff type generator is capable of building up potential difference of 15 ´ 10 6 V. in which electrons with a maximum kinetic energy of 6 eV are emitted ? 4. 6 × 25 cm respectively. 2. The focal length of the objective and eye-lens of a compound microscope are 2 cm. The dielectric strength of the gas surrounding the electrode is 5 ´ 107 Vm . Using Ampere’s circuital law. 13. CBSE (All India) SET–II Questions different from Set–I What is the stopping potential of a photocell. what is the diameter of the image of moon formed by the objective lens? (Diameter of moon = 3 × 5 ´ 10 6 m and radius of lunar orbit = 3 × 8 ´ 108 m). in the normal adjustment position and write the expression for its magnifying power. derive an expression for the magnetic field along the axis of a toroidal solenoid. Identify the part of the electromagnetic spectrum to which the following wavelengths belong (i) 1 mm (ii) 10 .Examination Papers 100 pF C1 200 pF C2 200 pF C3 41 200 pF C4 + – 300 V OR Explain the principle on which Van de Graaff generator operates.1 . State the reason. 30. so as to obtain the final image at the near point of the eye? (ii) Also calculate its magnifying power. 5. why a photodiode is usually operated at a reverse bias. OR Draw a labelled ray diagram of an astronomical telescope. What is the angular magnification of the telescope? If this telescope is used to view moon. An astronomical telescope uses an objective lens of focal length 15 m and eye-lens of focal length 1 cm.11 m. X 26. Draw a plot showing the variation of power of a lens with the wavelength of the incident light. 19. The balance point is noticed J at ‘l’ cm from the positive end of the battery. calculate the focal length of the lens in the liquid. A diverging lens of focal length ‘F’ is cut into two identical parts each forming a plano-concave lens. On B A (100–l ) cm l cm interchanging R and X . calculate (i) velocity v of the beam. when the axis of the dipole makes an angle q with the electric field. State the factor. how the input/output signals differ in phase by 180°. An unknown resistance X > 5 W is connected in the other G gap as shown in the figure. mutually perpendicular to each other and normal to the direction of the beam. which controls : (i) wavelength of light. Neglecting the end correction. calculate the value of unknown resistance X used. If it is immersed in a liquid of refractive index 1× 7. OR Draw a labelled circuit diagram of a transistor amplifier in the common-emitter configuration. 4. A resistance R = 5 W is connected to one of the gaps in a R metre bridge. If the magnitudes of the electric and magnetic fields are 50 kV/m and 50 mT respectively. Draw a labelled circuit diagram of a full-wave rectifier and briefly explain its working principle. Derive an expression for the torque acting on an electric dipole. 20. 5. and (ii) intensity of light emitted by an LED. which uses a wire of length 1 m. if the maximum kinetic energy of electrons emitted is 5 eV ? Draw an equipotential surface for a uniform electric field. 27. What is the focal length of each part ? What is the stopping potential applied to a photocell.42 Xam idea Physics—XII 18. A diverging lens of refractive index 1× 5 and of focal length 20 cm in air has the same radii of curvature for both sides. A beam of proton passes undeflected with a horizontal velocity v. If a particle of charge q is moving with velocity v along the z-axis and the magnetic field B is acting along the x-axis. which is held in a uniform electric field. through a region of electric and magnetic fields. if the proton beam current is equal to 0 × 80 mA. 6. use the expression F = q ( v ´ B) to find the direction of the force F acting on it. (ii) force with which it strikes a target on a screen. Briefly explain. ® ® ® CBSE (All India) SET–III Questions different from Set–I & Set–II 2. it was found that the balance point further shifts by 20 cm away from end A. . The balance point is noticed J at ‘l’ from the positive end of the battery. Distinguish between paramagnetic and diamagnetic substances. mutually perpendicular to each other and normal to the direction of the beam. the depletion region of p-n junction decreases.12 m = 0 × 01 Å belongs to gamma rays. for Fe A = 125 \ Þ R Fe æ A Fe =ç RAl ç è AAl 5 R Fe = RAl 3 ö ÷ ÷ ø 1/ 3 æ 125 ö =ç ÷ è 27 ø 1/ 3 5 = ´ 3 × 6 fermi = 6 × 0 fermi 3 . Neglecting the end correction. How does this affect the size and intensity of the central diffraction band ? Draw a plot of the intensity distribution. Calculate the permeability and susceptibility of the iron-bar used. In a single slit diffraction experiment. 26. A resistance R = 4 W is connected to one of the gaps in a R metre bridge. if the proton beam current is equal to 0 × 80 mA. 2. 100 mT respectively. A beam of proton passes undeflected with a horizontal velocity v. 24. 3. calculate the value of unknown resistance X used. An unknown resistance X > 4 W is connected in the other G gap as shown in the figure. A magnetising field of 1500 A/m produces a flux of 2 × 4 ´ 10 .Examination Papers 43 11. Using Gauss’s law derive an expression for the electric field intensity at any point near a uniformly charged thin wire of charge/length l C/m.5 weber in a bar of iron of cross-sectional area 0 × 5 cm 2 . R Al = 3 × 6 fermi. (ii) force with which it strikes a target on the screen. calculate (i) velocity v of the beam. A = 27. (i) 10 . Nuclear radius. X 22. which uses a wire of length 1 m. (ii) 10 . through a region of electric and magnetic fields.1 m = 10 cm belongs to short radiowaves. If a particle of charge q is moving with velocity v along the x-axis and the magnetic field B is ® ® ® acting along the y-axis. it is found that the balance point further shifts by 20 cm (away from end A). If the magnitudes of the electric and magnetic fields are 50 kV/m. the width of the slit is made double the original width. use the expression F = q ( v ´ B) to find the direction of the force F acting on it. On B A (100–l ) cm l cm interchanging R and X. If the reverse bias across a p-n junction is decreased. 17. R = R 0 A 1/ 3 Þ R µ A 1/ 3 For Al. Solutions CBSE (All India) SET–I 1. Magnetic moment of a current loop: The torque on current loop is t = MB sin q.Þv=v f f 2 That is image will be formed between optical centre and focus of lens. The back emf induced in the coil opposes the change in current. As stopping potential does not depend on intensity.e. where q is angle between magnetic moment and magnetic field. Its direction is perpendicular to the plane of the loop. That is the magnetic moment of a current loop is defined as the torque acting on the loop when placed in a magnetic field of 1 T such that the loop is oriented with its plane normal to the magnetic field. evr M= 2 . Stopping potential remains unchanged.44 Xam idea Physics—XII 4. 1 1 1 1 1 1 = .f f 1 1 1 \ =. magnetic moment of a current loop is the product of number of turns. M = NIA i. u = .. the intensity of light incident on the photocell becomes one-fourth. Magnetic moment of Revolving Electron.. The wavefront is spherical of decreasing radius. 5. sin q = 1 or q = 90° then M = t. +Q –Q 9. towards the side of the object. current flowing in the loop and area of loop. t Þ M= B sin q If B = 1 T.f and for a concave lens f = . Also. Reason: On doubling the distance between the light source and the cathode of the cell. F 7. the stopping potential remains unchanged.Þ = + f v u v f u Here. 6. 8. 4 T) tan 30° 1 4 = 4 ´ 10 . \ ® ® ® In an electromagnetic wave. then the angular fringe width remains unchanged but s l fringes becomes less and less sharp. The wings of jet plane will cut the vertical component of earth’s magnetic field. so we need high frequency carrier waves.4 T 3 3 Induced emf across the wing e = Vvl Given v = 450 ms . so emf is induced across the wing.1 . This is called transverse nature of electromagnetic wave. the electric field vector oscillates along Y-axis and magnetic field vector oscillates along Z-axis. the interference pattern disappears. (i) Angular separation b q = \ V = ( 4 × 0 ´ 10 . . V = H tan q Given H = 4 × 0 ´ 10 .Examination Papers 45 b l = D d It is independent of D. The vertical component of earth’s magnetic field. so visibility of d fringes increases.4 ÷ ç ÷ ´ ( 450) ´ 30 V = 3 × 12 V è 3 ø 12.4 ´ = ´ 10 . If the condition < is S d not satisfied. B and K form a right handed system. so visibility of fringes decreases. Y Envelope of E E B X B Z Envelope of B 13. But the actual separation between fringes b = increases. Accordingly if a wave is propagating along X-axis. 11. perpendicular to the direction of propagation of wave. angular separation remains unchanged if screen is moved lD away from the slits.4 T. l = 30 m æ 4 ö e =ç ´ 10 . If E m is the peak value of modulating signal and EC that of carrier wave. Diagram is shown in fig. the three vectors E. the electric and magnetic field vectors oscillate. (ii) When width of source slit is increased. therefore. q = 30° 10. Transverse Nature of Electromagnetic Waves: In an electromagnetic wave. High frequency waves require antenna of reasonable length and can travel long distances without any appreciable power loss. W1 = 0. therefore.q traverse equal distance under equal and opposite forces. e. (i) Let e be emf and r the internal resistance of each cell.. (ii) Now the dipole is rotated and brought to orientation making an angle q with the field direction (i. i.q qE –q are qE and qE. The potential energy of an electric dipole of an electric field is defined as the work done in bringing the dipole from infinity to its present position in the electric field.p ×E 15. q1 = 90° and q 2 = 0° ). ma = Em Ec Given m a = 80% = 0 × 80. The electric forces an charges + q and . Suppose the dipole is brought from infinity and placed at orientation q with the direction of electric field.pE cos q = . Electric potential energy of electric dipole. (i) The work done (W1 ) in bringing the dipole perpendicular to electric field from infinity.cos q) = . The work done in this process may be supposed to be done in two parts. i.. work done W2 = pE (cos q1 .cos q 2 ) = pE (cos 90° .e. +q qE 2l qE –q From infinity (i) Let us suppose that the electric dipole is brought from infinity in the region of a uniform electric field such that ® p +q qE its dipole moment p always remains perpendicular to electric field. (ii) Work done (W2 ) in rotating the dipole such that it finally makes an angle q from the direction of electric field.. E c = 20 V \ E m = m a ´ E c = 0 × 80 ´ 20 V = 16 V 14.pE cos q \ Total work done in bringing the electric dipole from infinity.e.46 Xam idea Physics—XII Modulation index. The equation of terminal potential difference V = e eff . net work done in bringing the dipole in the region of electric field perpendicular to field-direction will be zero. therefore. As charges + q and .i r int becomes V = 3 e .pE cos q ® ® In vector form U =. along the field direction and opposite to field direction respectively. U = W1 + W2 = 0 .i r int …(1) . 2 × 0 (r int ) 3e 3 ´ 2 Þ r int = = =3 W 2×0 2×0 \ For maximum power. N = N 0 . e = 2 V (ii) For maximum power dissipation.(1) log e N = . emf of each cell. Suppose initially the number of atoms in radioactive element is N 0 and N the number of atoms after time t. If N 0 is initial number of radioactive nuclei. then at t = 0.lN dt dN Þ = . From fig.. i = 2 × 0 A.lt N Integrating . when V = 0. when i = 0. According to Rutherford and Soddy law dN where l is disintegration constant = . current in circuit = 1× 0 A 16.Examination Papers 47 where r int is effective (total) internal resistance.. dt where l is the decay constant.e. so log e N 0 = 0 + C Þ C = log e N 0 .lt + C where C is a constant of integration. e = 2 V and for maximum power dissipation. emf of each cell.0 6 Þ e= =2V 3 i.lN. i= = = 1× 0 A R + r int 3 + 3 Thus. µN dt dN or = . the effective internal resistance of cells must be equal to external resistance. From fig. dN i..e.. V = 6 × 0 V \ From (1). R = r int = 3 W 3´ 2 3e Current in circuit. \ Equation (1) gives 0 = 3e . 6 = 3e . Radioactive decay Law: The rate of decay of radioactive nuclei is directly proportional to the number of undecayed nuclei at that time.. external resistance.. 8.lt N0 The graph is shown in figure.6 8 Fe Bn MeV 20 56 A 180 The binding energy curve indicates that binding energy for nucleon of heavy nuclei is less than that of middle nuclei. we get log e N = .lt + log e N 0 Þ log e N .48 Xam idea Physics—XII Substituting this equation in (1).1 ( 0 × 6775) = 42 × 6° As angle of incidence at face AC is 45°.1 ( 0 × 7194) = 46° . Clearly a heavy nucleus breaks into two lighter nuclei then binding energy per nucleon will increase and energy will be released in the process.lt N0 N Þ = e .lt Þ N = N 0 e .log e N 0 = . Nuclear fission reaction is 235 1 1 ¾® 141 Ba + 92 92 U + 0n 36 Kr + 3 ( 0 n ) + 200 MeV 56 (slow neutron) 17. This process is called nuclear fission. The critical angle for green light C g is sin C g = 1 1 = 0 × 7022 = n g 1× 424 C g = sin .1 ( 0 × 7022) = 44 × 6° \ The critical angle for red light 1 1 sin C r = = Þ n r 1× 39 The critical angle for blue light 1 1 sin C b = = = 0 × 6775 n b 1× 476 Þ C b = sin .lt N Þ log e = . C r = sin . N0 N t OR The variation of binding energy per nucleon versus mass number is shown in figure. the current remains the same.Examination Papers 49 which is smaller than critical angle for red ray but greater than critical angles. ær tö æ e tö V vd = ç Þ vd = ç ÷E ÷ è m ø è mø l 1 For same temperature t is the same. 18. The magnetic field of the solenoid magnetises the rod. for green and blue rays. An efficient way to make a permanent magnet is to place a ferromagnetic rod in a solenoid and pass a current. The loss due to hysteresis is immaterial because the magnet in this case is never put to cyclic changes. 3 2He We know R æ 1 ö =ç ÷ R0 è 2 ø n . In this case we apply the formula for drift velocity. v d µ \ (v d ) X A 3 = Y = (v d ) Y AX 2 1 A (ii) When wires are connected in parallel : In parallel. Permanent Magnets:The magnets prepared from ferromagnetic materials which retain their magnetic properties for a long time are called permanent magnets. so v d µ l (v d ) X lY 2 \ = = (v d ) Y lX 1 19. r = Þ n= 2 2 ne t e tr \ Þ æ m ö ÷ eAv d i =ç ç e 2 tr ÷ è ø ie tr vd = mA or i= m Av d e tr For same temperature t is same. red-way will be transmitted through the face AC. so we use the relation i = neAv d . 20. The materials used for permanent magnet must have the following characteristic properties : (i) High retentivity so that the magnet may cause strong magnetic field. 2 3 Examples : 1 1 H. therefore. different atomic numbers ( Z ) are called isobars. m m Resistivity. mechanical ill-treatment and temperature changes. (i) When wires are connected in series : In series. 1 H 3 Examples : 1 H. 1 H. the potential difference is the same. (ii) High coercivity so that the magnetisation is not wiped out by strong external fields. Isotopes Isobars The nuclides having the same atomic number Z but The nuclides having the same atomic mass ( A ) but different atomic masses ( A ) are called isotopes. Space waves are used for LOS communication. Time interval t = 5 ´ 5 = 25 years Inputs A B 1 1 0 0 0 1 0 1 t4 t5 t6 t7 t8 AND Y=A.50 Xam idea Physics—XII Given \ or R 3 ×125 = 3 ×125% = R0 100 3 ×125 æ 1 ö =ç ÷ 100 è 2 ø 5 n or n 1 æ 1 ö =ç ÷ 32 è 2 ø n æ 1 ö æ 1 ö ç ÷ =ç ÷ è 2 ø è 2 ø Þ n =5 Given T = 5 years t As n= T t or \ =5 T 21. .B 0 1 0 0 0 1 0 0 NOR Y=A+B 0 0 0 1 1 0 1 0 NAND Y=A. LOS Communication: It means “Line of sight communication”.B 1 0 1 1 1 0 1 1 0 < t < t1 t1 < t < t 2 t2 <t <t3 t3 <t <t4 t 4 < t < t5 t5 < t < t 6 t 6 < t < t7 t7 < t < t8 0 A B t1 t2 t3 0 1 1 0 0 1 0 0 Output waveforms of the three gates: AND NOR NAND 22. If transmitting antenna and receiving antenna have heights hT and hR respectively. d R = 2R e hR \ Maximum line of sight distance. dm dT dR hT Earth hR Communication System using Space wave mode propagation are (i) LOS communication and Fig.(l + 20) 80 .l Equating (1) and (2) . microwave links and satellite communication The satellite communication is shown in fig.Examination Papers 51 In this communication the space waves (radio or microwaves) travel directly from transmitting antenna to receiving antenna. broadcast. The space wave used is microwave. Communication satellight spa ce w ave Transmitting antenna Earth Receiving antenna 23.l 100 . From ‘metre bridge’ formula R l = X 100 .l) Given R = 2 W \ …(1) X= ´ 2W l On interchanging R and X. so l + 20 l + 20 X …(2) = Þ X= ´ 2W R 100 .l Þ X= R l (100 . the balance point is obtained at a distance (l + 20) cm from end A. d T = 2R e hT where R e is radius of earth and radio horizon of receiving antenna. shows LOS communication system. d M = d T + d R = 2R e hT + 2R e hR (ii) Television. then Radio horizon of transmitting antenna. the current through R and Zener diode also increases.l Solving we get l = 40 cm 100 . Zener diode as a Voltage Regulator \ Unknown resistance. A simple circuit of a voltage regulator using a Zener diode is shown in the Fig.l) l + 20 ´2= ´2 l 80 . The series resistance R absorbs the output voltage fluctuations so as to maintain constant voltage across the load. X = The Zener diode makes its use as a voltage regulator due to the following property: When a Zener diode is operated in the breakdown region. So.40 X= ´ 2W 40 Þ X =3 W 24.l ´ 2W l 100 .52 Xam idea Physics—XII (100 . The Zener diode is connected across load such that it is reverse biased. the voltage across it remains practically constant for a large change in the current. voltage drop across R increases. R Unregulated input I IZ IL Regulated output Vin VZ RL V0 If the input dc voltage increases. I-V Characteristics I (mA) VR Zener voltage O VF VZ I(mA) . without any change in the voltage across zener diode. the terminal S1 is negative relative to S and S 2 is positive relative to S. ng -1 1× 5 .63 × 75 cm 1× 5 . n l = 1× 7 Focal length of lens in liquid. The direction of current i 2 due to diode II in load resistance is again from A to B. Thus for input a. signal the output current is a continuous series of unidirectional pulses. This output current may be converted in fairly steady current by the use of suitable filters. Refractive index n = A + Power of a lens P= B l2 . The circuit diagram for full wave rectifier using two junction diodes is shown in figure.1) ç ç ÷ f è R1 R 2 ø Clearly. where l is the wavelength. power of a lens µ ( n g . B =B $ j ® ® \ $ F = q (v $ i) ´ ( B $ j) = qvB k . The direction of current i1 due to diode I in load resistance R L is directed from A to B.Examination Papers 53 Input A. F =q v ´ B Given v = v $ i.C. In next half cycle. P1 S1 P B S I N RL A Output Suppose during first half cycle of input ac signal the terminal S1 is P2 S2 P II N positive relative to S and S 2 is negative relative to S. l è ø Given f a = 15 cm.1). Input wave form T 2T Time Output wave form of full wave rectifier T 2T Time 25.c. Therefore current flows in diode II and there is no current in diode I. æ 1 1 1 ö ÷ = ( n g . n g = 1× 5. Then diode I is reverse biased and diode II is forward biased. The plot is shown in figure. signal OR Full Wave Rectifier : For full wave rectifier we use two junction diodes. then diode I is forward biased and diode II is reverse biased. This implies that the power of a lens decreases with increase æ ö 1 of wavelength ç ç P µ 2 nearly ÷ ÷ .1× 7 l 26.1 fl = ´ fa = ´ 15 cm ng 1× 5 1 P -1 1× 7 nl = ® ® ® ® 0 × 5 ´ 1× 7 ´ 15 cm = . Therefore current flows in diode I and not in diode II. then X C = f1 fr f2 wr C and \ X L = wr L 1 = wr L wr C f Þ wr = 1 LC .54 Xam idea Physics—XII That is. equal to rate of change of linear momentum of the beam i. so force exerted on the target is zero.3 T \ v= 100 ´ 10 3 50 ´ 10 -3 = 2 ´ 10 6 ms . Condition for resonance to occur in series LCR ac circuit: For resonance the current produced in the circuit and emf applied must always be in the same phase.X L = 0 2 or XC = X L 1 If wr is resonant frequency. we get l 2pr = n mv l Þ mvr = n 2p This is Bohr’s second postulate.e. E = 100 kV/m = 100 ´ 10 3 V/m. As complete the Broglie wavelength may be in certain fixed orbits. The de Broglie wavelength …(1) l= mv Now for electron in orbit (for nth orbit) 2pr = n l Using (1). the condition is that electric and magnetic forces on the beam must be equal and opposite i.X L tan f= lm R im For resonance f= 0 Þ XC . so non-radiating electron can be only in certain fixed orbits. it exerts a force on the target. B = 50 mT = 50 ´ 10 . (i) For a beam of charged particles to pass undeflected crossed electric and magnetic fields. force is acting along Z-axis. eE = evB E Þ v= B Given. Phase difference ( f) in series LCR circuit is given by im XC . 28.1 E Fe (ii) The beam strikes the target with a constant velocity. Dp mv mv mvi mvi F= = = = = Dt Dt q / i q ne v B Fm Target where n is the number of protons striking the target per second.. if proton beam comes to rest. h 27.. However.e. wr fr w L i. Quality Factor ( Q ) : The quality factor is defined as the ratio of resonant frequency to the width of half power frequencies. Half power frequencies are the frequencies on either side of resonant frequency for which current reduces to half of its maximum value. In fig. R= R = = 15 W I 2 Impedance. f 1 and f 2 are half power frequencies.Examination Papers 55 Linear resonant frequency.e. Primary Secondary Primary Np Ns Np Ns (a) Two coils on top of each other (b)Two coils on separate limbs of the core . I wattless = I sin f = I .f 1 R (i) Potential difference across capacitance. Soft iron-core Secondary Transformer: Transformer is a device by which an alternating voltage may be decreased or increased. VC = X C I V \ Capacitive reactance.. This is based on the principle of mutual-induction. (a) and (b). ç C ÷ Z è ø 20 =2´ A = 1× 6 A 25 OR Arrangements of winding of primary and secondary coil in a transformer are shown in fig.w1 f 2 . 2 Z = R 2 + XC = (15) 2 + ( 20) 2 C R 30 V 2A 40 V Vrms = 225 + 400 = 625 W = 25 W (ii) The phase lead ( f) of current over applied voltage is Z X tan f = C Xc R æX ö R Wattless Current. X C = C I 40 = = 20W 2 V 30 Resistance. f r = wr 1 = 2p 2p LC The graph of variation of peak current i m with frequency is shown in fig. Q= = = r w2 . N S > N p ).... Similarly if the secondary circuit is open. then Power in primary = Power in secondary V p i p = VS i S iS V p N p 1 = = = i p VS N S r \ . e. We assume that Secondary there is no leakage of flux so that the Transformer flux linked with each turn of primary coil and secondary coil is the same.(5) .(1) ep =-Np Dt and emf induced in the secondary coil Df .(4) = = = r (say) Vp e p N p where r = NS is called the transformation ratio.56 Xam idea Physics—XII Step up Transformer: It transforms the alternating low voltage to alternating high voltage and in this the number of turns in secondary coil is more than that in primary coil...NS Dt From (1) and (2) eS NS = ep Np . the emf ( e p ) induced in the primary coil.(2) eS = .. Primary therefore the alternating emf of same Core laminated frequency is developed across the iron core secondary. Secondary Let N p be the number of turns in primary coil. then the potential difference VS across its ends will be equal to the emf ( e S ) induced in it.C. (i.. According to Faraday’s laws the emf induced in the primary coil Df .. If i p and i s are the instantaneous currents in Np primary and secondary coils and there is no loss of energy. mains) source is connected to the ends of primary coil. due to which the magnetic flux linked with the secondary coil changes continuously. therefore VS e S N S . will be equal to the applied potential difference (V p ) across its ends. NS the number of turns in secondary coil and f the magnetic flux (a) Step up linked with each turn... the current changes Primary continuously in the primary coil. Working: When alternating current (A.(3) If the resistance of primary coil is negligible.. This work is stored in the capacitor in the form of + – electrostatic potential energy.Examination Papers 57 In step up transformer. So VS > V p and i S < i p i. (a) When a capacitor is charged by a battery. which may be obtained by integration. the potential difference between its plates increases. When charge is given to capacitor. VS I S = V p I p V I Ip = S S Vp = 22 ´ 0 ×1 = 0 × 01A 220 A B 29.e.e.0 ç ç 2 è ö Q2 ÷= ÷ 2C ø If V is the final potential difference between capacitor plates. + – VB Consider a capacitor of capacitance C. (ii) Flux Leakage: Energy is lost due to coupling of primary and secondary coils not being perfect. i. N s > N p ® r > 1. the potential difference q between its plates V = C Now work done in giving an additional infinitesimal charge dq to capacitor q dW = V dq = dq C The total work done in giving charge from 0 to Q will be equal to the sum of all such infinitesimal works. work is done –Q – + +Q by the charging battery at the expense of its chemical + – energy. Initial charge on VA + – capacitor is zero. then Q = CV (CV ) 2 1 1 \ W= = CV 2 = QV 2C 2 2 . Let a charge Q be given to it in + – + – small steps. whole of magnetic flux generated in primary coil is not linked with the secondary coil. Reasons for energy losses in a transformer (i) Joule Heating: Energy is lost in resistance of primary and secondary windings as heat ( I 2 Rt). V 22 Current is secondary coil. Therefore total work Q Q q W = ò V dq = ò dq 0 0 C 1 é q2 ù = ê ú Cê ë 2 ú û Q = 0 1 C æ Q2 . Initial potential difference between + – capacitor plates = zero. step up transformer increases the voltage but decreases the current. Let at any instant VAB = V when charge on capacitor be q. I S = S = A = 0×1A Z 220 For an ideal transformer \ Current in primary coil.. 8 200 ´ 10 .58 Xam idea Physics—XII This work is stored as electrostatic potential energy of capacitor i. Q = C eq ´ V = 100 ´ 10 .8 coulomb Q Potential difference across C 4 = C4 = 3 ´ 10 . ) and the upper comb C 2 is connected to the inner surface of metallic sphere S. The lower comb C1 is connected to the positive terminal of a very high voltage source (HTS) ( » 104 volts. . Principle: It is based on the following two electrostatic phenomena: (i) The charge always resides on the outer surface of a hollow conductor. Construction.e. It consists of a large hollow metallic sphere S mounted on two insulating columns A and B and an endless belt of rubber or silk is made to run on two pulleys P1 and P2 by the means of an electric motor. C1 and C 2 are two sharp metallic spikes in the form of combs. (ii) The electric discharge in air or a gas takes place readily at the pointed ends of the conductors. Electrostatic potential energy.12 = 1× 5 ´ 10 2 = 150 V OR This is a machine that can build up high voltages of the order of a few million volts.12 ´ 300 = 3 ´ 10 . U = (b) 100pF C1 200pF C2 200pF C4 200pF + – 200pF C4 + – 200pF C3 + – 100pF + – Q2 1 = CV 2C 2 2 = 1 QV 2 100pF 200pF C4 100pF \ C eq = 100 pF Now.. the maximum electric field is at the surface of the shell. As the belt goes on revolving. . Thus the outer surface of metallic sphere S gains A B positive charge continuously and its potential rises to a very high value. it continues to take (+ ) charge upward.1 V \ R min = E max = 15 ´ 10 6 5 ´ 107 = 3 ´ 10 . to the C1 P1 surrounding air.Examination Papers 59 Working: When comb C1 is given very high potential.. E max = 5 ´ 107 Vm . then Q 1 . Van de Graaff generator is used to accelerate stream of charged particles to very high velocities. the generator is completely enclosed in an earthed connected steel tank which is filled with air under high pressure. When the potential of a metallic sphere gains very HTS high value.. These positive ions are carried upward by the moving belt..(1) E max = 4p e 0 R 2 min Also the potential From (1) and (2) Q 1 V= 4p e 0 R min V = R min E max . If maximum charge is Q. then it produces ions in its vicinity.(2) Here V = 15 ´ 10 6 V.. so produced. To prevent leakage of charge from the sphere. The comb C 2 collects positive charge from the belt which immediately moves to the outer surface of sphere S. Due to a charged shell (radius R).1 m = 30 cm. due to S action of sharp points. get sprayed on the belt due to the C2 P2 repulsion between positive ions and comb C1 . Such a generator is installed at IIT Kanpur which accelerates charged particles upto 2 MeV energy. The maximum potential is reached when the rate of leakage of charge becomes equal to the rate of charge transferred to the sphere. The pointed end of C 2 just touches the belt. The positive ions. which is collected by comb C 2 and transferred to outer surface of sphere S. the dielectric strength of surrounding air breaks down and its charge begins to leak. u 0 = ? (i) When final image is formed at least distance of distinct vision (D = 25 cm) : For eye lens : Here v e = .4 1 1 1 1 1 = == ue ve f e 25 6 × 25 25 or u e = .| u e | = 15 .5 = 10 cm For objective lens: 1 1 1 = f 0 v0 u0 . uo Objective B A" A Fo O Fe' A' vo Eye piece ue D Eye E Fe B' B" ve Magnifying power of compound microscope is v æ Döü ÷ï M =.60 Xam idea Physics—XII 30.5 cm As L = | v 0 | + | u e | Þ | v 0 | = L .25 cm 1 1 1 \ = f e ve ue Þ -1 . L = 15 cm.0 ç 1+ ç u0 è be ÷ ø ï for final image at distance of distinct vision ý L æ Dö ï ç1 + ÷ »f0 ç fe ÷ è ø ï þ v0 D L D for final image at infinity M =»u0 f e f0 fe Given f 0 = 2 × 0 cm. f e = 6 × 25 cm. B fo fe A C1 Fo A' F e' B' C2 Fe in At i fin ty Drawbacks: (i) It is not free from chromatic aberration.= . (ii) The image formed is inverted and fainter.Examination Papers 61 Þ 1 1 1 1 1 2 = = . f e = 1× 0 cm = 1× 0 ´ 10 .2 .2 × 59 cm.= u 0 v 0 f 0 8 × 75 2 2 ´ 8 × 75 u0 = - 2 ´ 8 × 75 6 × 75 \ u0 = .0 == .8 × 75 = = .f e = 15 .20 ç ÷ 2 ×5 è 6 × 25 ø fe ø è (ii) When final image is formed at infinity : In this case L = v 0 + f e Þ v 0 = L . |u0 | = 2 × 59 cm v D 8 × 75 æ 25 ö Magnification. fe because focal length of a lens has no concern with the aperture of lens. Þ 1 1 1 1 1 2 .=u 0 v 0 f 0 10 2 5 5 u 0 = .0 × =×ç ÷ = .2 m Angular magnification of telescope. (a) Given f 0 = 15 m. æ 10 æ 25 ö D ö ç 1+ ÷ =ç 1+ ÷ = .2 × 5 cm 2 M =v0 u0 That is distance of object from objective is 2 × 5 cm.1500 fe 1× 0 ´ 10 . M = . f 15 m=.6 × 25 = 8 × 75 cm For objective lens : 1 1 1 = f 0 v0 u0 Magnification.4 ´ 5 = .13 × 5 u0 f e 2 × 59 è 6 × 25 ø OR f0 Magnifying power m = × It does not change with increase of aperature of objective lens. . E k = eV0 Þ 6 eV = eV0 Þ V0 = 6V The stopping potential V0 = 6 volt (negative). 4. e.(1) ò B•dl Length of toroid = 2pr ® Number of turns in toroid = n ( 2pr ) current in one-turn = I \ \ Current enclosed by circular path = ( n 2pr ) • I Equation (1) gives . So. D d = r f0 Þ 3 × 48 ´ 10 D d = × f0 = ´ 15 m r 3 × 8 ´ 108 = 0 × 137 m = 13 × 7 cm 6 Moon Objective lens D a r a Image of moon d f0 CBSE (All India) SET–II 2. Now we apply Ampere’s circuital law to this circular path. Magnetic field due to a toroidal solenoid: A long solenoid shaped in P the form of closed ring is called a toroidal solenoid (or endless solenoid).. I I For points inside the core of toroid Consider a circle of radius r in the region enclosed by turns of toroid. The magnetic lines of force inside the toroid are in the form of concentric circles. The current causes the magnetic field inside the turns O of the solenoid. r Let n be the number of turns per unit length of toroid and I the current flowing through it. 5. ® ò B•dl ® ® =m I = ò Bdl cos 0 = B • 2pr .11 m = 0 ×1 Å belongs to gamma rays. The fractional change due to incident light on minority charge carriers in reverse bias is much more than that over the majority charge carriers in forward bias. Let D be diameter of moon. then from Fig. d diameter of image of moon formed by objective and r the distance of moon from objective lens.. By symmetry the magnetic field has the same magnitude at each point of circle and is along the tangent at every point on the circle. photodiodes are used to measure the intensity in reverse bias condition. 13.62 Xam idea Physics—XII Negative sign shows that the final image is inverted. (i) wavelength 1 mm belongs to the microwaves. (ii) wavelength 10 . i. . it experiences no force but experiences a torque.. Maximum Torque: For maximum torque sin q should be the maximum.(2) = pE sin q Clearly. The charges of dipole are . if an electric dipole is placed in an electric field in oblique orientation. Torque: The forces F1 and F2 form a couple (or torque) which tends to rotate and align the dipole along the direction of electric field..(3) Thus..2l r Force: The force on charge + q is.. This couple is called the torque and is denoted by t. In vector form t =p ´E ® ® ® ® ® ® ® . Torque ( t) is a vector quantity whose direction is perpendicular to both p and E . As the maximum value of sin q = 1 when q = 90° \ Maximum Torque. so dipole does not undergo any translatory motion. F2 = qE .q and + q at separation 2l the dipole moment of electric dipole. Consider an electric dipole placed in a uniform electric field of strength E in such a way that its dipole moment p makes an angle q with the direction of E . \ torque t = magnitude of one force ´ perpendicular distance between lines of action of forces = qE ( BN) = qE ( 2l sin q) = ( q 2l) E sin q [using (1)] . F1 = qE. hence net force on electric dipole in uniform electric field is F = F1 .F2 = qE . along the ® ® ® 2l q O +q B F1 = qE 2l sin q E F2 = – qE A q –q N direction of field E ® ® ® The force on charge .q is. opposite to the direction of field E ® ® Obviously forces F1 and F2 are equal in magnitude but opposite in direction. The torque tends to align the dipole moment along the direction of electric field...(1) p = q.qE = 0 (zero) As net force on electric dipole is zero.Examination Papers 63 B 2pr = m 0 ( n 2prI ) Þ B = m 0 nI 18. t max = pE .. the magnitude of torque depends on orientation ( q) of the electric dipole relative to electric field. 64 Xam idea Physics—XII 19. Refractive index n = A + Power of a lens P= B l2 , where l is the wavelength. æ 1 1 1 ö ÷ = ( n g - 1) ç çR -R ÷ f 2 ø è 1 P Clearly, power of a lens µ ( n g - 1). This implies that the power of æ ö 1 a lens decreases with increase of wavelength ç ç P µ 2 nearly ÷ ÷. l è ø The plot is shown in fig. Given f a = - 20 cm, n g = 1× 5, n l = 1× 7 Focal length of lens in liquid, ng -1 1× 5 - 1 fl = ´ fa = ´ ( -20) cm ng 1× 5 -1 -1 1× 7 nl = ® ® ® ® l 0 × 5 ´ 1× 7 ´ ( -20) cm = + 85 cm 1× 5 - 1× 7 (Converging) 20. Given, F =q v ´ B ® ® $, v = vk $ B = Bi $) ´ ( B i $ $) = qvBj F = q (vk E Fe That is, force is acting along y-axis. (i) For a beam of charged particles to pass undeflected crossed electric and magnetic fields, the condition is that electric and magnetic forces on the beam must be equal and opposite i.e., eE = evB E Þ v= B Given, E = 50 kV/m = 50 ´ 10 3 V/m, B = 50 mT = 50 ´ 10 - 3 T \ v= 50 ´ 10 3 50 ´ 10 -3 v B Fm Target = 1 ´ 10 6 ms - 1 (ii) The beam strikes the target with a constant velocity, so force exerted on the target is zero. However, if proton beam comes to rest, it exerts a force on the target, equal to rate of change of linear momentum of the beam i.e., Dp mv mv mvi mvi F= = = = = Dt Dt q / i q ne where n is the number of protons striking the target per second. Examination Papers 65 26. From ‘metre bridge’ formula R l = X 100 - l 100 - l Þ X= R l Given R = 5 W (100 - l) …(1) \ X= ´5W l On interchanging R and X, the balance point is obtained at a distance (l + 20) cm from end A, so l + 20 X = R 100 - (l + 20) l + 20 …(2) Þ X= ´5W 80 - l Equating (1) and (2) (100 - l) l + 20 ´5= ´5 l 80 - l Solving we get l = 40 cm 100 - l 100 - 40 ´5W = ´5W l 40 15 Þ X = = 7.5 W 2 27. Full Wave Rectifier: For full wave rectifier we use two junction diodes. The circuit diagram for full wave P1 S1 I N rectifier using two junction diodes is shown in figure. P Suppose during first half cycle of input ac signal the B RL A terminal S1 is positive relative to S and S 2 is negative S Output relative to S, then diode I is forward biased and diode II is reverse biased. Therefore current flows in diode I and P2 S2 P II N not in diode II. The direction of current i1 due to diode I in load resistance R L is directed from A to B. In next half cycle, the terminal S1 is negative relative to S and S 2 is positive relative to S. Then diode I is reverse biased and diode II is forward biased. Therefore current flows in diode II and there is no current in diode I. The direction of current i 2 due to diode II in load resistance is again from A to B. Thus for input a.c. signal the output current is a continuous series of unidirectional pulses. This output current may be converted in fairly steady current by the use of suitable filters. OR Common-Emitter Transistor Amplifier: Common-emitter transistor amplifier gives the highest gain and hence it is the most commonly employed circuit. Fig. depicts the circuit for a p-n-p transistor. In this circuit, the emitter is common to both the input (emitter-base) and output (collector-emitter) circuits and is grounded. The emitter-base circuit is forward biased and the base-collector circuit is reverse biased. \ Unknown resistance, X= Input A.C. signal Output Input signal 66 Xam idea Physics—XII C IB B Vi Ri – E VEE + IE + VCC – IC RL Output voltage V0 In a common-emitter circuit, the collector-current is controlled by the base-current rather than the emitter-current. Since in a transistor, a large collector-current corresponds to a very small base-current, therefore, when input signal is applied to base, a very small change in base-current provides a much larger change in collector-current and thus extremely large current gains are possible. Referring to fig., when positive half cycle is fed to the input circuit, it opposes the forward bias of the circuit which causes the collector current to decrease. It decreases the voltage drop across load R L and thus makes collector voltage more negative. Thus when input cycle varies through a positive half cycle, the output voltage developed at the collector varies through a negative half cycle and vice versa. Thus the output voltage in common-emitter amplifier is in antiphase with the input signal or the output and input voltages are 180° out of phase. CBSE (All India) SET–III 2. For a complete diverging lens. 1 æ 1 1ö = ( n g - 1) ç - - ÷ F è R Rø R Þ F =2 ( n g - 1) For each planoconcave lens 1 æ 1 1ö = ( n g - 1) ç - - ÷ F¢ è R ¥ø R Þ F¢ = = 2F ( n g - 1) 4. 5. E i.e., focal length of each half part will be twice the focal length of initial diverging lens. E k = eV0 Þ 5 eV = eV0 Þ V0 = 5V The stopping potential V0 = 5 volt (negative). Examination Papers 67 6. (i) Wavelength of light emitted depends on the nature of semiconductor. (ii) Intensity of light emitted depends on the forward current. 1 times the e0 11. Gauss Theorem : The net outward electric flux through a closed surface is equal to net charge enclosed within the surface i.e., ® ® 1 ò S E · dS = e 0 Sq dS1 Electric field due to infinitely long, thin E E E E E and uniformly charged straight wire: Consider an infinitely long line charge having linear charge density l coulomb S1 90 90 r metre - 1 (linear charge density means dS2 dS3 charge per unit length). To find the electric + + + + + + + + + + + + + + + + + + + field strength at a distance r , we consider a S2 S3 cylindrical Gaussian surface of radius r and length l coaxial with line charge. The cylindrical Gaussian surface may be divided into three parts: (i) Curved surface S1 (ii) Flat surface S 2 and (iii) Flat surface S 3 . By symmetry the electric field has the same magnitude E at each point of curved surface S1 and is directed radially outward. o o ® We consider small elements of surfaces S1 , S 2 and S 3 . The surface element vector d S 1 is directed ® ® ® ® ® ® ® ® ® along the direction of electric field (i. e., angle between E and d S 1 is zero); the elements d S 2 and d S 3 are directed perpendicular to field vector E (i. e., angle between d S 2 and E is 90° and so also angle between d S 3 and E). Electric Flux through the cylindrical surface ® ® S1 1 ® ® S2 ® ® S3 ® ® òS E ·d S =ò = E · d S1 + ò E ·d S2 + ò 2 E ·d S3 òS E dS1 cos 0° + òS = ò E dS1 + 0 + 0 S = E ò dS1 E dS 2 cos 90° + òS 3 E dS 3 cos 90° (since electric field E is the same at each point of curved surface) (since area of curved surface = 2p rl) = E 2p rl As l is charge per unit length and length of cylinder is l, therefore, charge enclosed by assumed surface = ( ll) \ By Gauss’s theorem ® ® 1 ò E · d S = e 0 ´ charge enclosed 68 Xam idea Physics—XII Þ Þ E . 2p rl = E= l 1 ( ll) e0 2pe 0 r Thus, the electric field strength due to a line charge is inversely proportional to r . 17. The angular size of central diffraction band, 2q = 2l 1 µ × When width of slit ‘a’ is doubled, the a a size of central band becomes half and the intensity is doubled. Intencity I0 –3l/a –2l/a l/a 0 l/a 2l/a 3l/a 22. From ‘metre bridge’ formula 100 - l R l = Þ X= R X 100 - l l (100 - l) Given R = 4 W \ …(1) X= ´ 4W l On interchanging R and X, the balance point is obtained at a distance (l + 20) cm from end A, so l + 20 X = R 100 - (l + 20) l + 20 …(2) Þ X= ´ 4W 80 - l Equating (1) and (2) (100 - l) l + 20 ´4= ´4 l 80 - l Solving we get l = 40 cm \ Unknown resistance, Þ ® ® ® ® 100 - l 100 - 40 ´ 4W = ´ 4W l 40 X =6W X= 24. Given, F =q v ´ B ® ® $, v = vi $ B = Bj $) = qvBk $ $) ´ ( Bj F = q (vi Examination Papers 69 That is, force is acting along z-axis. (i) For a beam of charged particles to pass undeflected crossed electric and magnetic fields, the condition is that electric and magnetic forces on the beam must be equal and opposite i.e., eE = evB E Þ v= B Given, E = 50 kV/m = 50 ´ 10 3 V/m, B = 100 mT = 100 ´ 10 - 3 T \ v= 50 ´ 10 3 100 ´ 10 - 3 v = 5 ´ 105 ms - 1 (ii) The beam strikes the target with a constant velocity, so force exerted on the target is zero. However, if proton beam comes to rest, it exerts a force on the target, equal to rate of change of linear momentum of the beam i.e., F= Dp mv mv mvi mvi = = = = Dt Dt q / i q ne E Fe v B Fm Target where n is the number of protons striking the target per second. 26. Distinction between diamagnetic, paramagnetic and ferromagnetic substances Property (i) Susceptibility ( c ) (ii) Permeability (m r ) (iii) Coercivity Example Diamagnetic - 1 £ c < 0 (negative and small) 0 £ m r < 1 (less than 1) High Gold Paramagnetic 0 < c < e (positive and small) 1 < m r < 1 + e (slightly greater than 1) Low Platinum The magnetic field lines near a diamagnetic substance and a paramagnetic substance are shown below: N S N S (i) Field lines near a Diamagnetic (ii) Field lines near a Paramagnetic 70 Xam idea Physics—XII Here, H = 1500 A/m, f = 2 × 4 ´ 10 - 5 Wb, A = 0 × 5 ´ 10 - 4 m 2 Q Þ B =m H f 2 × 4 ´ 10 - 5 m= = AH 0 × 5 ´ 10 - 4 ´ 1500 = 3 × 2 ´ 10 - 4 Wb/Am Now, Magnetic susceptibility, mr = m 3 × 2 ´ 10 - 4 = = 255 m0 4p ´ 10 - 7 c = m r -1 = 255 – 1 = 254 CBSE EXAMINATION PAPERS DELHI–2009 Time allowed : 3 hours General Instructions: (a) All questions are compulsory. (b) There are 30 questions in total. Questions 1 to 8 carry one mark each, questions 9 to 18 carry two marks each, questions 19 to 27 carry three marks each and questions 28 to 30 carry five marks each. (c) There is no overall choice. However, an internal choice has been provided in one question of two marks, one question of three marks and all three questions of five marks each. You have to attempt only one of the given choices in such questions. (d) Use of calculators is not permitted. (e) You may use the following values of physical constants wherever necessary: c = 3 ´ 10 8 ms 1 Maximum marks : 70 h = 6 × 626 ´ 10 -34 Js m 0 = 4p ´ 10 -7 TmA -1 e = 1 × 602 ´ 10 -19 C 1 = 9 × 109 Nm2C– 2 4pe o Boltzmann’s constant k = 1 × 381 ´ 10 -23 J K-1 Avogadro’s number N A = 6 × 022 ´ 10 23 /mole Mass of neutron mn = 1 × 2 ´ 10 -27 kg Mass of electron me = 9 × 1 ´ 10 -31 kg Radius of earth = 6400 km CBSE (Delhi) SET–I 1. 2. 3. 4. What is sky wave propagation? Write the following radiations in ascending order in respect of their frequencies : 1 X-rays, microwaves, UV-rays and radio waves 1 Magnetic field lines can be entirely confined within the core of a toroid, but not within a straight solenoid. Why? 1 You are given following three lenses. Which two lenses will you use as an eyepiece and as an objective to construct an astronomical telescope? 1 Lenses L1 L2 L3 Power (P) 3D 6D 10 D Aperture (A) 8 cm 1 cm 1 cm write the ratio of the intensities of original light and the transmitted light after passing through the analyser. ® B O q X Z-axis 14. .q and + q are located at points A ( 0. (ii) Does the charge gain kinetic energy as it enters the magnetic field? Justify your anwer. Xam idea Physics—XII If the angle between the pass axis of polarizer and the analyser is 45°. 1 Photoelectric The figure shows a plot of three curves a. 6. write its truth table and draw its logic symbol. How much work is done in moving a test charge from point P (7 . Predict from the graph the condition under which ‘V’ becomes equal to ‘E‘. tower be affected when the height of the tower is increased by 21 %? 2 Derive an expression for drift velocity of free electrons in a conductor in terms of relaxation time. What type of wavefront will emerge from a (i) point source. 12. and (ii) distant light source? 1 Two nuclei have mass numbers in the ratio 1 : 2. 0. A charge ‘q’ moving along the X-axis with a velocity v is subjected to a uniform magnetic field B acting along the Z-axis as it crosses the origin O. A B Y 0 1 2 3 4 5 6 7 .72 5. 0)? 10. + a) respectively. Y 15. 0. Identify the gate. B) and the output wavefrom (Y) of a gate. The following figure shows the input waveforms ( A . 11.V. 0. 2 (i) Trace its trajectory. 0. 8. b. What is the ratio of their nuclei densities? A cell of emf ‘E’ and internal resistance ‘r’ is connected across a variable resistor ‘R’. Plot a graph showing the variation of terminal potential ‘V’ with resistance R. c showing the current variation of photocurrent vs. (ii) Two charges . I 2 and I 3 having frequencies I3 n 1 . 9.a) and B ( 0. 0) to Q ( -3. By what percentage will the transmission range of a T. collector plate potential for I1 I2 three different intensities I 1 . a Point out the two curves for which the incident radiations have same frequency but different intensities. n 2 and n 3 respectively incident on a photosenitive c b surface. (i) Can two equi-potential surfaces intersect each other? Give reasons. 1 Collector plate potential 7. 13. 2 How does a charge q oscillating at certain frequency produce electromagnetic waves? 2 Sketch a schematic diagram depicting electric and magnetic fields for an electromagnetic wave propagating along the Z-direction. Indicate the direction of the magnetic field ® Z I dl O r P Y due to a small element d l at point P situated at a distance r from the element as shown in the figure. Why are high frequency carrier waves used for transmission? OR ® X What is meant by term ‘modulation’? Draw a block diagram of a simple modulator for obtaining an AM signal. . Which of them will result in the transition of a photon of wavelength 275 nm? A B C D 0 eV – 2 eV – 4. A proton and an alpha particle are accelerated through the same potential. Which one of the two has (i) greater value of de-Broglie wavelength associated with it. Find: (i) charge on each capacitor (ii) equivalent capacitance of the network (iii) energy stored in the network of capacitors 21. A thin conducting spherical shell of radius R has charge Q spread uniformly over its surface. A current I flows in a conductor placed perpendicular to the plane of the paper. a bright spot is seen at the centre of the shadow of the obstacle. 20. A radioactive nucleus ‘A’ undergoes a series of decays according to the following scheme : A ¾¾® A1 ¾¾® A2 ¾¾® A 3 ¾¾® A4 The mass number and atomic number of A are 180 and 72 respectively. when a tiny circular obstacle is placed in the path of light from a distant source. C2 C1 + 12 V – C3 a b a g (a) The energy levels of an atom are as shown below. What are these numbers for A4 ? 19. C 2 and C 3 of capacitance 6 mF each are connected to a 12 V battery as shown. Three identical capacitors C 1 . 17. Using Gauss’s law. derive an expression for an electric field at a point outside the shell. State Biot-Savart law. and (ii) less kinetic enrgy? Justify your answers. 18. Explain why? 23.5 eV – 10 eV (b) Which transition corresponds to emission of radiation of maximum wavelength? 22.Examination Papers 73 16. In a single slit diffraction experiment. Draw a graph of electric field E(r) with distance r from the centre of the shell or 0 £ r £ ¥. Determine the values of the resistances X and Y. 29. 1 × 44 and 1 × 47 respectively. the null point D is obtained 40 cm from the end A. a B G R 45° b c 28. (a) Derive an expression for the average power consumed in a series LCR circuit connected to a. with the help of a circuit diagram. green and blue wavelengths are 1 × 39. cross-sectional area A having N number of turns. (b) Define the quality factor in an a. Write its S. with the help of a labelled diagram. Three light rays red (R). When the two unknown resistances X and Y are inserted. 25. Trace the path of these rays after passing through face ‘ab’. OR Explain the principle and working of a cyclotron with the help of a schematic diagram. When a resistance of 10 W is connected in series with X. OR (a) Derive the relationship between the peak and the rms value of current in an a.c.I. the working of n-p-n transistor as a common emitter amplifier. working of a step-up transformer. Derive the expression for force per unit length between two long straight parallel current carrying conductors. define one ampere. green (G) and blue ( B) are incident on a right angled prism ‘abc’ at face ‘ab’. the null point shifts by 10 cm. (b) Derive an expression for self inductance of a long solenoid of length l. (ii) Explain. The refractive indices of the material of the prism for red. circuit. Out of the three which colour ray will emerge out of face ‘ac’? Justify your answer. Write the expression for cyclotron frequency. Hence. 27. units. The figure shows experimental set up of a meter bridge.c. . (i) Draw a circuit diagram to study the input and output characteristics of an n-p-n transistor in its common emitter configuration. Find the position of the null point when the 10 W resistance is instead connected in series with resistance ‘Y’. source in which the phase difference between the voltage and the current in the circuit is f. A step-up transformer converts a low voltage into high voltage.c. X B G A D C Y 26. Draw the typical input and output characteristics. Does it not violate the principle of conservation of energy? Explain.74 Xam idea Physics—XII State two points of difference between the interference pattern obtained in Young’s double slit experiment and the diffraction pattern due to a single slit. (b) Describe briefly. (a) Define self inductance. Why should the quality factor have high value in receiving circuits? Name the factors on which it depends. 24. circuit. CBSE (DELHI) SET–II Questions different from Set–I. 30. The equivalent capacitance of the combination between A and B in the given figure is 4 mF. 1 The nuclei have mass numbers in the ratio 1 : 3. Name the electromagnetic radiation to which waves of wavelength in the range of 10 -2 m belong. 1 What is ground wave propagation? 1 Unpolarized light is incident on a plane surface of refractive index m at angle i. 3 A 20mF C B a b a g (i) Calculate capacitance of the capacitor C. 8. Give one use of this part of EM spectrum. OR Draw the labelled ray diagram for the formation of image by a compound microscope. What are these numbers for A 4 ? 19. What is the ratio of their nuclear densities?1 The output of a 2-input AND gate is fed to a NOT gate. Write down its truth table. Explain briefly. 2 A radioactive nucleus ‘A’ undergoes a series of decays according to the following scheme: 2 A ¾® A 1 ¾® A 2 ¾® A 3 ¾® A 4 The mass number and atomic number of A 4 are 172 and 69 respectively. the image and the radius of curvature from the central point of the spherical surface. derive the expression of the lens maker’s formula. 1. 11. 6. how a p-n junction diode works as a half wave rectifier. 16. Explain why both the objective and the eye piece of a compound microscope must have short focal lengths. with the help of a circuit diagram. 2. 5. Hence. write the relation between the angle i and refractive index m. Derive the expression for the total magnification of a compound microscope. Give the name of the combination and its logic symbol.Examination Papers 75 OR How is a zener diode fabricated so as to make it a special purpose diode? Draw I-V characteristics of zener diode and explain the significance of breakdown voltage. Trace the rays of light showing the formation of an image due to a point object placed on the axis of a spherical surface separating the two media of refractive indices n1 and n2 . Draw a diagram to show refraction of a plane wavefront incident on a convex lens and hence draw the refracted wave front. . Establish the relation between the distances of the object. If the reflected light gets totally polarized. 22. What are these numbers for A 4 ? 19. 6. 2 A 1 1 B 0 1 2 3 4 5 6 7 8 (ii) If the output of the above AND gate is fed to a NOT gate. (iii) What will be the potential drop across each capacitor? 20. 3 CBSE (DELHI) SET–III Questions different from Set–I and Set–II. At what angle of incidence should a light beam strike a glass slab of refractive index 3.76 Xam idea Physics—XII (ii) Calculate charge on each capacitor if a 12 V battery is connected across terminals A and B. 3 An electron and a proton are accelerated through the same potential. Two nuclei have mass numbers in the ratio 2 : 5. (i) Sketch the output wavefrom from an AND gate for the inputs A and B shown in the figure. State Gauss’s law in electrostatic. name the gate of the combination so formed. Which one and the two has (i) greater value of de-Broglie wavelength associated with it and (ii) less momentum? Justify your answer. 7. 12. State Guass’s law in electrostatics. 3. 5. A radioactive nucleus ‘A’ undergoes a series of decays according to the following scheme : 2 A ¾® A 1 ¾® A 2 ¾® A 3 ¾® A 4 The mass number and atomic number of A are 190 and 75 respectively. 18. Using this law derive an expression for the electric field due to a uniformly charged infinite plane sheet. Use this law to derive an expression for the electric field due to an infinitely long straight wire of linear charge density l Cm -1 . such that the reflected and the refracted rays are perpendicular to each other? 1 What is space wave propagation? Name the part of electromagnetic spectrum which is suitable for : (i) radar systems used in aircraft navigation (ii) treatment of cancer tumours. What is the ratio of their nuclear densities? 1 Differentiate between a ray and a wavefront. 3 a b a g . 2. Polarizer Analyzer Here q = 45° \ 6. 7. so curves ‘a’ and ‘b’ have same frequency but different intensities. Magnetic field lines can be entirely confined within the core of a toroid because toroid has no ends. . Therefore. Radiowaves. UV rays. Wavefront from a point source – spherical Wavefront from a distant light source – plane. I transmitted = 0 cos 2 q 2 I0 I0/2 q I0/2 cos2 q 4. I Transmitted intensity. 8. 5. have the same area of plates and same separation between them. 2. we will use L 3 as eyepiece and L 1 as objective. X has air between the plates while Y contains a dielectric medium of Îr = 4. I Original I transmitted = I0 I transmitted = 2 cos 45° 2 = 4 1 Curves a and b have different intensities but same stopping potential. A solenoid is open ended and the field lines inside it which is parallel to the length of the solenoid.Examination Papers 77 23. (iii) What is the ratio of electrostatic energy stored in X and Y? X Y 12 V 3 Solutions CBSE (Delhi) SET–I 1. An astronomical telescope has an eyepiece of shorter aperture and shorter focal length while an objective of longer aperture and longer focal length. microwaves. (ii) Calculate the potential difference between the plates of X and Y. cannot form closed curved inside the solenoid. 3. Two parallel plate condition X and Y. X-rays. Skywave propagation is a mode of propagation in which communication of radiowaves (in the frequency range 30 MHz–40 MHz) takes place due to reflection from the ionosphere. Nuclear density is independent of mass number. (i) Calculate capacitance of each capacitor if equivalent capacitance of the combination is 4 mF. so ratio 1 : 1. 0.0) = 0. Transmission range of a TV tower d = 2hR If height is increased by 21 %. h ¢ = h + If d ¢ is the new range. a) +q Z Y Q (–3. V = IR = ç ÷R= r èR + rø 1+ R When R ® 0. – a) –q B (0. V = 0 E When R = r . ´ 100% = ´ 100% d d æ d¢ ö = ç . new height. V = 2 When R = ¥ . Reason: At the point of intersection. X P (7. Xam idea Physics—XII æ E ö E Terminal potential difference. 0. there will be two different directions of electric field.VP ) = q0 ( 0 . so work done. 11.1) ´ 100% = 10% èd ø . (i) No E V E 2 O r R 10. V = E The graph is shown in fig. 0) A (0.1 . W = q0 (VQ . As potential at every point on equatorial line is zero. (ii) Work done in moving test charge atom P to Q is zero.1÷ ´ 100% = (1.78 9. 0) Reason: Test charge is moved along the equatorial line of an electric dipole. which is not possible. 0.d Dd % increase in range. then d¢ = d 21 h = 1 × 21 h 100 h¢ = 1 × 21 = 1 × 1 h d¢ . 0. An oscillating electric charge produces oscillating electric field. the relaxation time t remains constant. (ii) direct collisions of electrons with atoms of metallic crystal lattice. due to which electrons are accelerated. then its initial velocity is zero. which in turn produces oscillating electric field and so on. This electric field exerts a force on free electrons. A potential difference V is applied across the conductor XY. The electric force on electron F = . If m is the mass of electron. the motion of electrons become random. 13. . In any way after a short duration t called relaxation time. so the velocity of electron after time t (i. The V magnitude of electric field strength E = and l its direction is from Y to X. Different ions vibrate in different directions and may be displaced by different amounts. Due to this potential difference an electric field E is produced in the conductor.e E ® ® ® ® ® l E X Y vd + V – (where e = + 1 × 6 ´ 10 -19 coulomb). Thus. thereby producing an electromagnetic wave propagating in free space. a ® ® ® = - eE m eE m ® ) t …(2) vd =0vd =- Þ ® et ® E m At given temperature. ® ® ® ® ® ® ® v = v d . t = t. . if we consider the average motion of an electron. so drift velocity remains constant. Consider a metallic conductor XY of length l and cross-sectional area A. These deflections may arise due to (i) ions of metallic crystal vibrate simple harmonically around their mean positions. we can imagine that the electrons are accelerated only for a short duration. e. then its acceleration ® a = F eE =m m …(1) This acceleration remains constant only for a very short duration. which produces oscillating magnetic field. since there are random forces which deflect the electron in random manner. As average velocity of random motion is zero. drift velocity v d ) is given by the relation v = u + a t (here u = 0.Examination Papers 79 12. so charge does not gain kinetic energy on entering the magnetic field. so work done by the magnetic force on charge is zero. The logic gate is NAND gate. . 15. Biot-Savart Law: It states that the magnetic field strength ( dB) produced due to a current element (of current I and length dl) at a point having position vector r relative to current element is dB = ® ® ® ® q r I m 0 I dl ´ r 4p r3 dl P where m 0 is permeability of free space. A Y B r O v Y q X Truth Table Inputs A 0 0 1 1 B 0 1 0 1 Output Y 1 1 1 0 16. (ii) No Reason: Magnetic force F m = q v ´ B ® ® ® ® ® B Z X' Force ( F m ) is perpendicular to velocity v . Its value is m 0 = 4p ´ 10 -7 Wb/A-m. (i) The trajectory is shown in fig.80 Xam idea Physics—XII Y Envelope of E E B X B Z Envelope of B 14. then by symmetry electric field strength has same magnitude E0 on the Gaussian surface and is directed radially ® .Examination Papers 81 The magnitude of magnetic field is m Idl sin q dB = 0 4p r2 where q is the angle between current element I dl and position vector r . The direction of current element is along Z-direction and that of r along Y-direction. To find the electric field outside the shell.176 g 172 a a 172 180 176 A ¾® A 1 ¾® A 2 ¾® A 3 ¾® A 72 70 71 69 69 4 Thus. OR Meaning of Modulation: The original low frequency message/information signal cannot be transmitted over long distances. mass number of A4 is 172 and atomic number is 69. at the transmitter end. ® ® ® The direction of magnetic field dB is perpendicular to the plane ® Z I B dl Y X' containing I dl and r . we consider a spherical Gaussian surface of radius r ( > R). (iii) High frequency carrier wave avoids mixing up of message signals. m(t) Am sin wmt (modilating signal) + c(t) = Ac sin wct (Carrier) Block diagram of simple modulator Square law device Band pass filter centered at wc AM wave Bx(t) + C x2(t) 18. Use of high frequency carrier wave in transmission of signals: (i) High frequencey carrier wave reduces the size of antenna as h = ® ® X l l or × 2 4 (ii) High frequency carrier wave radiates more power in space as P µ n 2 . 17. Electric field intensity at a point outside a uniformly charged thin spherical shell: Consider a uniformly charged thin spherical shell of radius R carrying charge Q. 19. is susperimposed on a high frequency carrier signal by a process known as modulation. concentric with given shell. Therefore. = 0 ( -i B= 0 3 4p r 2 4p r That is magnetic field is directed along negative X-direction. b . information contained in the low frequency message signal. so magnetic field $) ´ (r j $) m Idl ® m ( I dl k $) . If E is electric field outside the shell. 4pr 2 ® S ® ® ® R r Q P EO dS E·d S Now.d. Hence. Gaussian surface is outside the given charged shell. of 12 V across capacitor C 3 . Effective Capacitance of C 1 and C 2 is C C 6´6 C 12 = 1 2 = = 3 mF C 1+ C 2 6 + 6 Charge on each of capacitors C 1 and C 2 is same: q1 = q2 = C 12 V = ( 3 mF) ´ (12 V) = 36 mC Charge on capacitor C 3 . (a) The energy (E) of a photon of wavelength ( l) is given by . Hence. q 3 = C 3 V = ( 6 mF ´ 12 V) = 72 mC (ii) Equivalent capacitance of network C eq = C 12 + C 3 = 3 mF + 6 mF = 9 mF (iii) Energy stored in the network 1 1 U = C eq V 2 = ´ ( 9 ´ 10 -6 ) ´ (12) 2 = 6 × 48 ´ 10 -4 J 2 2 21.82 Xam idea Physics—XII outward. 20. (i) Capacitors C 1 and C 2 are in series across a 12 V supply while there exists p. so angle between E i and d S is zero at each point. Also the directions of normal at each point is radially outward. by Gauss's theorem ® ® 1 òS E 0 · d E = e 0 ´ charged enclosed Þ E0 4pr 2 = 1 1 Q ´ Q Þ E0 = e0 4pe 0 r 2 E Thus. electric field outside a charged thin spherical shell is the same as if the whole charge Q is concentrated at the centre. so charge enclosed by Gaussian surface is Q. electric flux through Gaussian surface = ò = ò E0 dS cos 0 = E0 . then f = 4 pR 2 s C \ E0 = 1 4 pR 2 s R 2 s = 4pe 0 r2 e 0r 2 R r The graph is shown in fig. If s is the surface charge density of the spherical shell. the energy is minimum.e. The waves diffracted at the edge of circular obstacle produce constructive interference at the centre of the shadow. . E = µ l l For minimum wavelength of emission.. All bright fringes are of equal intensity.4 × 5 eV) = 4 × 5 eV hc 1 Energy of Photon Emitted. Proton has greater de-Broglie wavelengths. (b) E = 0 .( . this transition corresponds to B since for transition B. Difference between Interference and Diffraction Interference 1. Fringe width of all fringes are equal. 2mp qp V lp la = ma qa . K = qV Kp q p e 1 = = = Ka q a 2 e 2 Kp < Ka i. proton has less kinetic energy.. The width of central maximum is maximum and goes on decreasing with increase of order of secondary maxima. producing a bright spot. This transition A corresponds to emission of radiation of maximum wavelength. 22. 2. 23. l = 2mqV lp = \ h . Diffraction The central maximum has maximum intensity and the intensity of secondary maxima goes on decreasing with increase of order.Examination Papers -34 ´ 3 ´ 10 8 hc 6 × 626 ´ 10 = J l 275 ´ 10 -9 83 E= = 6 × 626 ´ 10 -34 ´ 3 ´ 10 8 275 ´ 10 -9 ´ 1 × 6 ´ 10 -19 eV = 4 × 5 eV From fig. mp qp la = h 2ma qaV As ma = 4mp and qa = 2qp lp \ = 4´2 = 8 la (ii) Kinetic energy. \ lp > la i. h (i ) de-Broglie wavelength.e. then magnetic field within the solenoid.40) r Y 3 When a resistance of 10 W is connected in series with X. \ F = (nl BA) Substituting the value of B from (1) . F = NBA where N = nl is the number of turns in length ' l ' of solenoid. Xam idea Physics—XII (a) The self inductance is defined on the magnetic flux linked with the coil when unit current flows through it. l2 X \ = Y + 10 (100 . If A is cross-sectional area of solenoid. If current in solenoid is I .7 henry/metre is the permeability of free space.. Self Inductance of a long air-cored solenoid: Consider a long air solenoid having `n' number of turns per unit length.. X + 10 50r X + 10 \ = Þ =1 Y 50r Y Þ From (1). B = m 0 nI .. \ 3 Y= X 2 Þ Y = X + 10 3 X = X + 10 Þ X = 20 W 2 …(2) Y = 20 + 10 = 30 W When a resistance 10 W is inserted in series with Y. then effective flux linked with solenoid of length `l '. Let r = resistance per cm length of bridge wire X 40r X 2 = Þ = Y (100 .84 24. when the rate of change of current in the coil is 1 ampere/second. \ From (2).(2) F = nl (m 0 nI ) A = m 0 n 2 AlI \ Self-inductance of air solenoid F L = = m 0 n 2 Al I If N is total number of turns in length l .l2 ) . Or The self inductance is defined as the emf induced in the coil.. then N n= l N ö2 \ Self-inductance L = m 0 æ ç ÷ Al è l ø = m0 N 2A l …(1) (b) A l 25.(1) where m 0 = 4p ´ 10 . let the balancing length be l2 . the null point is obtained at ( 40 + 10) = 50 cm. The SI unit of self-inductance is henry (H). (1) B1 = 0 1 2p r According to Maxwell’s right hand rule or right hand palm rule no. The magnitude of force in each case remains the same. 1. It has been observed experimentally that when the currents in the wire are in the same direction. l2 = 33 × 3 cm.l2 = l 2 This gives null point length. b).Examination Papers 85 Þ l2 20 = 30 + 10 (100 . therefore the magnetic force on element ab of length DL \ \ DF = B1 I 2 DL sin 90° m I = 0 1 I 2 DL 2p r The total force on conductor of length L will be m I I m I I F = 0 1 2 S DL = 0 1 2 L 2p r 2p r Force acting on per unit length of conductor F m I I f = = 0 1 2 N/ m L 2p r \ . the direction of magnetic force will be towards PQ i. they experience a repulsive force (fig. P R P R b DL DF I1 Q r a B I2 S I1 Q B I2 a DL b DF S Consider a current–element ‘ab’ of length DL of wire RS.l2 ) Þ l2 100 . 26. (fig. .. Due to this magnetic field. On the other hand if the currents I 1 and I 2 in wires are in opposite directions.). Let the conductors PQ and RS carry currents I 1 and I 2 in same direction and placed at separation r.. a) and when they carry currents in opposite directions. the direction of B1 will be perpendicular to the plane of paper and directed downward. they experience an attractive force (fig.. each element of other wire experiences a force.e. The magnetic field produced by current-carrying conductor PQ at the location of other wire RS m I .(2) According to Fleming’s left hand rule. The direction of current element is perpendicular to the magnetic field.. the force will be attractive. Force per unit length between two long straight parallel conductors: Suppose two long thin straight conductors (or wires) PQ and RS are placed parallel to each other in vacuum (or air) carrying currents I 1 and I 2 respectively. the force will be repulsive. E. E.I. it crosses the gap twice.7 N/m 2p Thus 1 ampere is the current which when flowing in each of parallel conductors placed at separation 1 m in vacuum exert a force of 2 ´ 10 -7 on 1 m length of either wire. Due to the presence of perpendicular magnetic field the ion will move in a circular . r = 1 m. Due to the presence of perpendicular magnetic field the ion will move in a circular path. Expression for K. oscillator Dee-2 Beam When charged particle crosses the gap between dees it gains KE = q V In one revolution.E.86 Xam idea Physics—XII Definition of Ampere: In S. = q2 B2 R 2 2m = 2nqV Working: The principle of action of the apparatus is shown in fig. then m f = 0 = 2 ´ 10 . The positive ions produced from a source S at the centre are accelerated by a dee which is at negative potential at that moment. = q2 B2 R 2 1 2 mv max = 2 2m S Magnetic Pole N S Dee Dee Magnetic Pole S Dee-1 R. system of fundamental unit of current ‘ampere' has been defined assuming the force between the two current carrying wires as standard. Principle: The positive ions produced from a source are accelerated. The phenomenon is continued till the ion reaches at the periphery where an auxiliary negative electrode (deflecting plate) deflects the accelerated ion on the target to be bombarded. is a device for accelerating ions to high speed by the repeated application of accelerating potentials. OR The cyclotron. the kinetic energy gained = 2nqV Thus K. then qBR v max = m The kinetic energy of the ion when it leaves the apparatus is.F. therefore if it completes n-revolutions before emerging the does. attained: If R be the radius of the path and v max the velocity of the ion when it leaves the periphery. devised by Lawrence and Livingston. The force between two parallel current carrying conductors of separation r is F m I I f = = 0 1 2 N/ m L 2 pr If I 1 = I 2 = 1 A . K. 1) when it moves from one dee to the other. in turning through an angle p is.(2) The time taken by the ion in describing a semi-circle. qvB = mv 2 r or r = mv qB . Blue and Green rays will be totally reflected. The angular velocity w of the ion is given by. The function of magnetic field is to provide circular path to charged particle and so to provide the location where charged particle is capable of gaining energy from electric field. yet it takes the same time in each dee.. v qB (from equation 1) w= = r m . 27. So only red ray will be transmitted. Expression for Period of Revolution and Frequency: Suppose the positive ion with charge q moves in a dee with a velocity v. This condition is satisfied for red colour (m = 1 × 39). although the speed of the ion goes on increasing with increase in the radius (from eq. being known... i.e. Angle of incidence at face ac for all three colours. B can be calculated for q producing resonance with the high frequency alternating potential. then. (3) it is clear that for a particular ion. m From eq.. i = 45° Refractive index corresponding to critical angle 45° is 1 m= = 2 = 1 × 414 sin 45° The ray will be transmitted through face ‘ ac’ if i < ic .... p pm .e.(1) where m is the mass and r the radius of the path of ion in the dee and B is the strength of the magnetic field. The function of electric field is to accelerate the charged particle and so to impart energy to the charged particle.(3) t= = w Bq Thus the time is independent of the speed of the ion i..Examination Papers 87 path inside the dees. B G R R a 45° b B G c . The magnetic field and the frequency of the applied voltages are so chosen that as the ion comes out of a dee. the dees change their polarity (positive becoming negative and vice-versa) and the ion is further accelerated and moves with higher velocity along a circular path of greater radius. The phenomenon is continued till the ion reaches at the periphery of the dees where an auxiliary negative electrode (deflecting plate) deflects the accelerated ion on the target to be bombarded. w LI L 1 L Q= r = wr = .w1 Q= 1 R where w2 . Voltage. L C wr Q= w2 . 2 = 1 2 i. cos ( 2wt + f) = 0.cos ( 2wt + f)] 2 2 Average value of cos ( 2wt + f) over a complete cycle is zero i.e.88 28. RI R LC R Þ Also.. Xam idea Physics—XII (a) In series LCR circuit. quality factor must be very high. \ Average power over a complete cycle V I 1 Pav = V0 I 0 cos f = 0 0 cos f 2 2 2 Pav = Vrms I rms cos f (b) Quality Factor (Q): In series LCR circuit the ratio of the voltage drop across inductor (or capacitor) to the voltage drop across resistor under resonance condition is called the quality factor. V = V0 sin wt Current in circuit.w1 = band width of resonant curve. inductance and capacitance of the circuit.. The quality factor depends on the values of resistance. P = VI = V0 I 0 sin wt sin ( wt + f) 1 1 = V0 I 0 2 sin wt sin ( wt + f) = V0 I 0 [cos f . larger is the quality factor and selectivity (or sharpness of resonance) of the circuit. OR (a) Relationship between Peak and RMS Value of Current: Current I = I 0 sin wt 2 I 2 = I0 sin 2 wt Mean square current over full cycle 2 ( I 2 ) mean = I 0 (sin 2 wt) mean Mean value of sin 2 wt over full cycle is 2 1 . I = I 0 sin ( wt + f) Instantaneous Power. Smaller is the band width.e. That is why in receiving circuits. (sin wt) mean ò = 0 T sin 2 wt dt ò0 T dt . We assume that there is no leakage of flux so that the flux linked with each turn of primary coil and secondary coil is the same.Np Dt and emf induced in the secondary coil Df e S = .. (b) Transformer: Transformer is a device by which an alternating voltage may be decreased or increased.Examination Papers 89 \ 2 ( I 2 ) mean = I 0 . This is based on the principle of mutual induction. According to Faraday’s laws the emf induced in the primary coil Df ..NS Dt From (1) and (2) eS ep = NS Np . therefore .(2) If the resistance of primary coil is negligible.(1) e p = . Primary laminated iron core Core Secondary Working: When alternating current Step up source is connected to the ends of Secondary primary coil. then the potential difference VS across its ends will be equal to the emf ( e S ) induced in it. NS the number of turns in secondary coil and f the magnetic flux linked with each turn. will be equal to the applied potential difference (Vp ) across its ends. therefore the alternating emf of same frequency is developed across the secondary. 1 2 I0 2 \ Root mean square current I rms = ( I 2 ) mean = or I rms = 0 × 707 I 0 (A.C. mains) Primary This is required relation. . the current changes Transformer continuously in the primary coil. the emf ( e p ) induced in the primary coil.. due to which the magnetic flux linked with the secondary coil changes continuously. (i. Let N p be the number of turns in primary coil. Similarly if the secondary circuit is open.. e.(3) . Step up Transformer: It transforms the alternating low voltage to alternating high voltage and in this the number of turns in secondary coil is more than that in primary coil. N S > N p )... 90 Xam idea Physics—XII VS Vp where r = NS Np = eS ep = NS Np = r (say) ...(4) is called the transformation ratio. If ip and is are the instantaneous currents in primary and secondary coils and there is no loss of energy; then For about 100% efficiency, Power in primary = Power in secondary Vp ip = VS iS iS Vp N p 1 = = = ip VS N S r VS > Vp and iS < ip \ ...(5) In step up transformer, Ns > N p ® r > 1 ; So i.e. step up transformer increases the voltage. When output voltage increases, the output current automatically decreases to keep the power same. Thus, there is no violation of conservation of energy in a step up transformer. 29. (i) Characteristic Curves: The circuit diagram for determining the static characteristic curves of an n-p-n transistor in common-emitter configuration is shown in fig. + mA VBB + – Rh1 + VBE – B E IE – IB C + VCE – mA – IC + – VCC + Common Emitter Characteristics: (a) Input characteristics: These characteristic curves are obtained by plotting base current ( I B ) versus base-emitter voltage VBE for fixed collector-emitter voltage VCE . Fig. represents these characteristics. (b) Output characteristics: These characteristics are obtained by plotting collector current I C versus collector-emitter voltage VCE at a fixed value of base current I B . The base current is changed to some other fixed value and the observations of I C versus VCE are repeated. Fig. represents the output characteristics of a common-emitter circuit. V VBE CE = 10 IB VCE = 5V V Examination Papers 91 (ii) The circuit of common emitter amplifier using n-p-n transistor is shown below: IC C1 RB C IB B Vi + – E VBB IE + VCC – V0 Output waveform RL C2 Working: If a small sinusoidal voltage, with amplitude VS , is superposed on dc basic bias (by connecting the sinusoidal voltage in series with base supply VBB ), the base current will have sinusoidal variations superposed on the base current I B . As a consequence the collector current is also sinusoidal variations superimposed on the value of collector current I C , this will produce corresponding amplified changes in the value of output voltage V0 . The a.c. variations across input and output terminals may be measured by blocking the d.c. voltage by large capacitors. The phase difference between input signal and output voltage is 180°. The input and output waveforms are shown in Input waveform Output waveform figure. R Voltage gain Av = b L × Ri OR Fabrication: A zener diode is fabricated by heavily doping both p and n sides of the junction, due to which the depletion layer formed is extremely thin ( < 10 -6 m) and the electric field of the junction is extremely high (~ 5 ´ 10 6 V / m) even for small reverse bias voltage of about 5 V. I-V Characteristics of Zener Diode are shown in figure. I (mA) Vr VZ Reverse bias I (mA) Forward bias VF Significance of Breakdown Voltage: After breakdown voltage VZ , any variation in current through Zener diode, does not cause any chaneg in Zener voltage. This property of Zener diode is used for regulating supply voltages so that they remain constant. 92 Xam idea Physics—XII Half Wave Rectifier: The circuit diagram for junction diode as half wave rectifier is shown in Fig. A Input A.C. signal p n + Output voltage VDC Input p1 s1 RL p2 s2 B (a) Output – (b) Let during first half the cycle the secondary terminal S1 of trasformer be positive relative to S2 , then the junction diode is forward biased. Therefore the current flows and its direction in load resistance R L is from A to B. In next half cycle the terminal S1 is negative relative to S2 then the diode is in reverse bias, therefore no current flows in diode and hence there is no potential difference across load R L . Therefore the output current in load flows only when S1 is positive relative to S2 . That is during first half cycles of input a.c. signal there is a current in circuit and hence a potential difference across load resistance R L while no current flows for next half cycle. The direction of current in load is always from A to B. Thus a single p-n junction diode acts as a half wave rectifier. The input and output waveforms of half wave rectifier are shown in fig. (b). 30. Relation between u, v, n 1 and n 2 for a spherical surface: Let SPS¢ be a spherical refracting surface, which separates media ‘1’ and ‘2’. Medium ‘1’ is rarer and medium ‘2’ is denser. The refractive indices of media ‘1’ and ‘2’ are n1 and n2 respectively (n1 < n2 ). Let P be the pole and C the centre of curvature and PC the principal axis of spherical refracting surface. O is a point-object on the principal axis. An incident ray OA, after refraction at A on the spherical surface bends towards the normal CAN and moves along AB. Another incident ray OP falls on the surface normally and hence passes undeviated after refraction. These two rays, when produced backward meet at point I on principal axis. Thus I is the virtual image of O. n1 Rarer medium ‘1’ S i h R M P N A r O I u S' C v r B n2 Denser medium `2' Let angle of incidence of ray OA be i and angle of refraction be r i.e. Ð OAC = i Let and Ð NAB = r ...(1) ...(2) Ð AOP = a , Ð AIP = b and Ð ACP = g In triangle OAC , or i = g - a g =a +i In triangle AIC , g = b + r or r = g -b Examination Papers 93 From Snell’s law sin i sin r = n2 n1 ...(3) If point A is very near to P, then angles i, r , a , b , g will be very small, therefore sin i = i and sin r = r i n2 From equation (3) = r n1 Substituting values of i and r from (1) and (2) we get g - a n2 or = n1 ( g - a ) = n2 ( g - b) g - b n1 ...(4) The length of perpendicular AM dropped from A on the principal axis is h i.e. AM = h. As angles a , b and g are very small, therefore tan a = a , tan b = b , tan g = g Substituting these values in equation (4) n1 (tan g - tan a ) = n2 (tan g - tan b) As point A is very close to P , point M is coincident with P perpendicular AM h \ tan a = = = base MO PO AM h AM h tan b = = , tan g = = MI PI MC PC Substituting this value in (5), we get h ö æ h n1 ç ÷ = n2 è PC PO ø n1 n n or - 1 = 2 PC PO PC hö æ h ç ÷ è PC PI ø n - 2 PI ...(5) ...(6) Let u, v and R be the distances of object O, image I and centre of curvature C from pole P. By sign convention PO, PI and PC are negative i.e. u = - PO , v = - PI and R = - PC Substituting these values in (6), we get n1 n n n - 1 = 2 - 2 ( - R) ( - u) ( - R) ( - v) or or n1 R n2 v n1 u n1 u = = n2 R v n2 - n1 R n2 Lens Maker’s Formula: Suppose L is a thin lens. The refractive index of the material of lens is n2 and it is placed in a medium of refractive index n1 . The optical centre of lens is C and X ¢ X is principal axis. The radii of curvature of the surfaces of the lens are R 1 and R 2 and their 94 Xam idea Physics—XII poles are P1 and P2 . The thickness of lens is t, which is very small. O is a point object on the principal axis of the lens. The distance of O from pole X' P1 is u. The first refracting surface forms the image of O at I ¢ at a distance v¢ from P1 . From the refraction formula at spherical surface n2 n1 n2 - n1 = v¢ u R1 L n1 n2 t O u P1 C P2 v v' I I' X n1 ...(1) The image I ¢ acts as a virtual object for second surface and after refraction at second surface, the final image is formed at I. The distance of I from pole P2 of second surface is v. The distance of virtual object ( I ¢ ) from pole P2 is (v¢ - t). For refraction at second surface, the ray is going from second medium (refractive index n2 ) to first medium (refractive index n1 ), therefore from refraction formula at spherical surface n1 n2 n - n2 ...(2) = 1 v (v¢ - t) R2 For a thin lens t is negligible as compared to v¢ , therefore from (2) n1 n2 n - n1 =- 2 v (v¢ ) R2 Adding equations (1) and (3), we get æ 1 n1 n1 1 ö = (n2 - n1 ) ç ÷ v u è R1 R2 ø or i.e. where 1 n2 ...(3) öæ 1 1 1 æ n2 1 ö - =ç - 1÷ ç ÷ v u è n1 ø è R1 R2 ø æ 1 1 1 1 ö - = ( 1 n2 - 1) ç ÷ v u è R1 R2 ø = n2 n1 ...(4) is refractive index of second medium (i.e. medium of lens) with respect to first medium. If the object O is at infinity, the image will be formed at second focus i.e. if u = ¥ , v = f 2 = f Therefore from equation (4) æ 1 1 1 1 ö - = ( 1 n2 - 1) ç ÷ f ¥ è R1 R2 ø i.e. æ 1 1 1 ö = ( 1 n2 - 1) ç ÷ f R R è 1 2ø ...(5) Examination Papers 95 This is the formula of refraction for a thin lens. This formula is called Lens-Maker’s formula. If first medium is air and refractive index of material of lens be n, then 1 n2 = n, therefore equation (5) may be written as æ 1 1 1 ö ...(6) = (n - 1) ç ÷ f è R1 R2 ø OR Compound Microscope: It consists of a long cylindrical tube, containing at one end a convex lens of small aperture and small focal length. This is called the objective lens (O). At the other end of the tube another co-axial smaller and wide tube is fitted, which carries a convex lens (E) at its Objective Eyepiece outer end. This lens is towards the eye and is called the eye-piece. The focal length and aperture of eyepiece are somewhat larger than those of objective lens. Cross-wires are mounted at a O E definite distance before the eyepiece. The entire tube can be moved forward and backward by the rack and pinion arrangement. Adjustment: First of all the eyepiece is displaced backward and forward to focus it on cross-wires. Now the object is placed just in front of the objective lens and the entire tube is moved by rack and pinion arrangement until there is no parallax between image of object and cross wire. In this position the image of the object appears quite distinct. Eyepiece uo Objective B A" A Fo O Fe' A' E Fe vo ue D Eye B' B" ve Compound microscope 96 Xam idea Physics—XII Magnifying power of a microscope is defined as the ratio of angle (b) subtended by final image on the eye to the angle (a ) subtended by the object on eye, when the object is placed at the least distance of distinct vision, i.e., b Magnifying power ...(1) M= . a B A D E Eye As object is very small, angles a and b are very small and so tan a = a and tan b = b. By definition the object AB is placed at the least distance of distinct vision. AB \ a = tan a = EA AB By sign convention EA = - D , \ a= -D and from figure b = tan b = A ¢ B¢ EA ¢ If ue is distance of image A ¢ B¢ from eye-piece E, then by sign convention, EA ¢ = - ue A ¢ B¢ and so, b= ( - ue ) Hence magnifying power, M = b A ¢ B¢ /( - ue ) A ¢ B¢ D = = × a AB / ( - D) AB ue By sign conventions, magnification of objective lens v0 A ¢ B¢ = AB ( - u0 ) \ M=v0 u0 × D ue ... (2) Using lens formula we get 1 1 1 = - for eye-lens, (i.e. using f = f e , v = - v e , u = - ue ), f v u or 1 1 1 = + ue f e v e 1 1 1 = f e - v e ( - ue ) M=v0 Magnifying power or M=- æ1 1ö Dç + ÷ u0 è f e v e ø v0 æ D D ö ç + ÷ u0 è f e v e ø When final image is formed at the distance of distinct vision, v e = D v æ Dö Magnification, M = - 0 ç1 + ÷ \ u0 è fe ø For greater magnification of a compound microscope, f e should be small. As f 0 < f e , so f q is small. Hence, for greater magnification both f 0 and f e should be small. Examination Papers 97 CBSE (Delhi) SET–II 1. 2. 5. 6. Microwave Use: Miscrowave oven or Radar. Ground Wave Propagation: Ground wave propagation is one in which electromagnetic waves glide on the surface of earth between two antennas on the ground. m = tan i The behaviour of a thin convex lens is shown in figure. Transmitted spherical wavefront Incident plane wavefront F f 8. 11. Nuclear density is independent of mass number, so ratio of nuclear densities is 1 : 1. Name of combination: NAND gate logic symbol. A Y B Truth Table of NAND gate is Inputs A 0 1 0 1 b- Output B 0 0 1 1 Y 1 1 1 0 16. g a a 180 176 176 172 172 A ¾® A 1 ¾® A 2 ¾® A 3 ¾® A 72 70 71 69 69 4 The mass number of A is 180 and atomic number is 72. 19. A 20mF C B (i) Given C AB = 4 mF Capacitance 20 mF and C (mF) are in series C ´ 20 \ C AB = C + 20 20 C or Þ 4 mF = C + 20 4C + 80 = 20 C . Q = C AB V = ( 4 mF) ´ (12 V) = 48 mC (iii) Potential drop across 20 mF capacitor 48 mC Q V1 = = = 2×4 V 20 mF 20 mF Q 48 mC Potential drop across C. charge per unit surface area) be s. q=90o dS3 o S1 E 90 S3 1 times e0 S2 dS2 E q=0o dS1 E q=0o B Sheet A To calculate the electric field strength near the sheet. Gauss Theorem: The net outward electric flux through a closed surface is equal to the net charge enclosed within the surface i. Let the surface charge density (i. we now consider a cylindrical Gaussian surface bounded by two plane faces A and B lying on the opposite sides and parallel to the charged sheet and the cylindrical surface perpendicular to the sheet (fig). Total electric flux òSE or ® •dS = •dS ® ® òSE 1 S1 ® • d S1 + ® òSE 2 ® • d S2 + ® òSE 3 ® ® • d S3 òSE ® =ò E dS1 cos 0° + òS2 E dS2 cos 0° + òS3 E dS 3 cos 90° = E ò dS1 + E ò dS2 = E a + E a = 2E a \ Total electric flux = 2E a . V2 = = = 9×6 V C 5 mF 20. ® ® 1 òS E · dS = e 0 Sq Let electric charge be uniformly distributed over the surface of a thin.98 Xam idea Physics—XII or Þ 16C = 80 C = 5 mF (ii) Charge on each capacitor.e. By symmetry the electric field strength at every point on the flat surface is the same and its direction is normal outwards at the points on the two plane surfaces and parallel to the curved surface. We have to calculate the electric field strength at any point distance r from the sheet of charge. non-conducting infinite sheet. the charge enclosed by Gaussian surface = s a . As s is charge per unit area of sheet and a is the intersecting area..e. Thus electric field strength due to an infinite flat sheet of charge is independent of the distance of the point and is directed normally away from the charge. Alternatively. CBSE (Delhi) SET–III 2. If the surface charge density s is negative the electric field is directed towards the surface charge. A ray is a perpendicular line drawn at any point on wavefront and represents the direction of propagation of the wave. A wavefront is a surface of constant phase.e. For proton. p e < p p so electron has less momentum. The reflected and refracted rays are mutually perpendicular at polarising angle. le = lp = h 2me . \ 2Ea = E= 1 (sa) e0 s × 2e 0 22. Momentum. eV h 2mp eV \ mp le = lp me (i) As me > mp . Space Wave Propagation: It is the straight line propagation of electromagnetic wave from transmitting antenna to receiving antenna both installed on the ground. p = 2meV µ m (ii) As me < mp .e. 1 Total electric flux = ´ (total charge enclosed by the surface) e0 i. (i) Microwave (ii) g -rays.. q = e . l= 2mqV For electron q = e .. 7. l e > l p i. 3. so from Brewster’s law ip = tan -1 (n) = tan -1 ( 3 ) = 60° .Examination Papers 99 According to Gauss’s theorem. electron has greater de-Broglie wavelength. . 6. h de-Broglie wavelength. Nuclear density is independent of mass number so ratio is 1 : 1. m = mp . space wave propagation is the line of sight (LOS) communication. 5. m = me . (output is 1 when both inputs are 1).100 12. We consider small elements of surfaces S1 .1 (linear charge density means charge per unit length). 0 1 2 3 4 5 6 7 8 (ii) NAND gate. the elements d S 2 and d S ® ® ® 3 ® ® ® ® ® are directed perpendicular to field vector E (i. e. Xam idea Physics—XII (i) Output waveform of AND gate is shown in fig.186 g a a 190 186 182 182 A ¾® A 1 ¾® A 2 ¾® A 3 ¾® A 75 73 74 72 72 4 The mass number of A 4 is 182 and atomic number is 72. ® ® 1 òS E · dS = e 0 Sq 1 times e0 Electric field due to infinitely long. b . Gauss Theorem: The net outward electric flux through a closed surface is equal to the net charge enclosed within the surface i. angle between d S 2 ® ® and E is 90° and so also angle between d S 3 and E). . . E E dS1 E E E 90o dS2 S2 S1 r 90o dS3 S3 + + + + + + + + + + + + + + + + + + + By symmetry the electric field has the same magnitude E at each point of curved surface S1 and is directed radially outward. S2 and S 3 . we consider a cylindrical Gaussian surface of radius r and length l coaxial with line charge. To find the electric field strength at a distance r . The cylindrical Gaussian surface may be divided into three parts : (i) Curved surface S1 (ii) Flat surface S2 and (iii) Flat surface S 3 . thin and uniformly charged straight wire: Consider an infinitely long line charge having linear charge density l coulomb metre . 18.. 19. . e. angle between E and d S 1 is zero). The surface element vector d S 1 is directed along the direction of electric field (i.e. charge enclosed by assumed surface = ( ll) By Gauss’s theorem \ ® ® 1 ò E · d S = e 0 ´ charge enclosed Þ Þ E . therefore.Examination Papers 101 Electric Flux through the cylindrical surface òS ® E ·dS =ò = = ® ® S1 E ·d S1 +ò ® ® S2 E ·d S2 +ò ® ® S3 E ·dS3 ® òS1 E dS1 cos 0° + òS2 E dS2 cos 90° + òS3 E dS 3 cos 90° òS E dS1 + 0 + 0 (since electric field E is the same = E ò dS1 at each point of curved surface) (since area of curved surface = 2p rl) = E 2p rl As l is charge per unit length and length of cylinder is l. e A (i) Capacitance of X. V = µ C C VX CY \ = = 4 Þ VX = 4VY VY CX Also VX + VY = 12 …(2) …(3) . 2p rl = E= l 2pe 0 r 1 ( ll) e0 23. CX = 0 d e e A e A Capacitance of Y. so Q 1 P. Thus. CY = r 0 = 4 0 d d CY \ =4 Þ CY = 4CX CX As X and Y are in series. the electric field strength due to a line charge is inversely proportional to r. so CX CY C eq = CX + CY Þ 4mF = CX. 4CX CX + 4CX Þ CX = 5 mF and CY = 4CX = 20 mF …(1) (ii) In series charge on each capacitor is same.d. d. across Y . VY = 2 × 4 V (ii) Energy stored in X Energy stored in Y Þ UX UY = Q 2 / 2CX Q / 2CY =4 2 = CY CX =4 . 4VY + VY = 12 Þ VY = 2 × 4V \ VX = 4 ´ 2 × 4 = 9 × 6 V Thus potential difference across X .102 Xam idea Physics—XII From (1) and (2). V X = 9 × 6 V. P. (b) There are 30 questions in total. What is the maximum kinetic energy of the photoelectrons emitted? 1 Two nuclei have mass numbers iin the ratio 1 : 8. However. What is the focal length of the combination? 1 The stopping potential in an experiment on photoelectric effect is 1 × 5 V. 7. questions 9 to 18 carry two marks each. What is its frequency range? 1 An electron does not suffer any deflection while passing through a region of uniform magnetic field. What is the ratio of their nuclear radii? 1 Give the logic symbol of NOR gate. (d) Use of calculators is not permitted. 6. 5. What is the elecrostatic potential due to an electric dipole at an equatorial point? 1 Name the EM waves used for studying crystal structure of solids. one question of three marks and all three questions of five marks each. . (e) You may use the following values of physical constants wherever necessary: c = 3 ´ 10 8 ms 1 Maximum marks : 70 h = 6 × 626 ´ 10 -34 Js m 0 = 4p ´ 10 -7 TmA -1 e = 1 × 602 ´ 10 -19 C 1 = 9 × 109 Nm2C– 2 4pe o Boltzmann’s constant k = 1 × 381 ´ 10 -23 J K-1 Avogadro’s number N A = 6 × 022 ´ 10 23 /mole Mass of neutron mn = 1 × 2 ´ 10 -27 kg Mass of electron me = 9 × 1 ´ 10 -31 kg Radius of earth = 6400 km CBSE (All India) SET–I 1. You have to attempt only one of the given choices in such questions. an internal choice has been provided in one question of two marks. 8. 3. 2. Questions 1 to 8 carry one mark each. What is the direction of the magnetic field? 1 How would the angular separation of interference fringes in Young’s double slit experiment change when the distance between the slits and screen is doubled? 1 Two thin lenses of power +6 D and – 2 D are in contact.CBSE EXAMINATION PAPERS ALL INDIA–2009 Time allowed : 3 hours General Instructions: (a) All questions are compulsory. questions 19 to 27 carry three marks each and questions 28 to 30 carry five marks each. 4. (c) There is no overall choice. (b) Plot a graph showing the variation of potential energy of a pair of nucleons as a function of their separtion. Why? (b) The small ozone layer on top of the stratosphere is crucial for human survival. 2 (a) The mass of a nucleus in its ground state is always less than the total mass of its constituents – neutrons and protons. Explain. If the radius if reduced to half. 15. A charge q is enclosed by a spherical surface of radius R. . Plot a graph showing the variation of the angle of deviation with the angle of incidence.I. R1 = 1 W R 2 = 5 W R3 = 4 W R4 = 2 W + – 4V 13. 2 Define the term ‘linearly polarised light. 2 R5 = 2 W 11. Xam idea Physics—XII 10. Increasing the current sensitivity may not necessarily increase the voltage sensitivity of a galvanometer. 12. A ray of light passes through a triangular prism. Justify.104 9. How are these surfaces different from that of a constant electric field along Z-direction? 2 Define electric flux. 2 Write the function of (i) Transducer and (ii) Repeater in the context of communication system. Draw 3 equipotential surfaces corresponding to a field that uniformly increases in magnitude but remains constant along Z-direction. units. These parts are then connected in parallel across a 3 × 0 volt battery. 18. It is then cut into two equal parts. Find the current drawn from the battery. OR Write two factors justifying the need of modulation for transmission of a signal. Answer the following questions : 1 (a) Optical and radio telescopes are built on the ground while X-ray astronomy is possible only from satellites orbiting the Earth. Write its S. Why? Define current sensitivity and voltage sensitivity of a galvanometer. 2 Calculate the current drawn from the battery in the given network. 16. 17.’ When does the intensity of transmitted light become maximum. how would the electric flux through the surface change 2 Define refractive index of a transparent medium. 2 14. when a polaroid sheet is rotated between two crossed polaroids? 2 A wire of 15 W resistance is gradually stretched to double its original length. Explain briefly how it works and how its is used to accelerate the charged particles. explain the formation of depletion region in a p-n junction. Give a brief description with the help of suitable diagrams indicating how these waves are propagated. Derive the expression for the electric field at the surface of a charged conductor. 3 Distinguish between sky wave and space wave propagation. (ii) In a meter bridge balance point is found at a distance l1 with resistance R and S as shown in the figure. After some time the battery is disconnected and a dielectric slab of dielectric constant K is inserted between the plates. How did the scattering of a -particles of a thin foil of gold provide an important way to determine an upper limit on the size of the nucleus? Explain briefly. Find the wavelength of light from the second source. A positive point charge ( + q) is kept in the vicinity of an uncharged conducting plate. (ii) A jet plane is travelling towards west at a speed of 1800 km/h. How does its width change when the junction is (i) forward biased. 3 Draw a schematic sketch of a cyclotron. monochromatic light of wavelength 630 nm illuminates the pair of slits and produces an interference pattern in which two consecutive bright fringes are separated by 8 × 1 mm. When an unknown resistance X is connected in parallel with the resistance S. 2 X (i) State the principle of working of a meter bridge. Find the expression for X in terms of l1 . 3 Draw a plot showing the variation of binding energy per nucleon versus the mass number A. OR 3 20. Draw the input and output waveforms of the signal. 4V (i) State Faraday’s law of electromagnetic induction. Explain with the help of this plot the release of energy in the processes of nuclear fission and fusion. 24. With the help of a suitable diagram. A parallel plate capacitor is charged by a battery. (i) Show that time period of ions in a cyclotron is independent of both the speed and radius of circular path. In Young’s double slit experiment. 2 l1 A B R S G 21. What is the voltage difference developed between the ends of the wing having a span of 25 m.Examination Papers 105 19. 28. 25. Sketch electric field lines originating from the point on to the surface of the plate. Write the expression for its voltage gain. be affected? Justify your answer. the balance point shifts to a distance 3 l . 26. l2 and S. and (ii) reverse biased? 3 Give a circuit diagram of a common emitter amplifier using an n-p-n transistor. if the Earth’s magnetic field at the location has a magnitude of 5 ´ 10 -4 T and the dip angle of 30°? + – 22. . How would (i) the capacitance. What is the effect on the interference fringes if the monochromatic source is replaced by a source of white light? 3 Draw a schematic arrangement of the Geiger-Marsden experiment. 27. 23. (ii) the electric field between the plates and (iii) the energy stored in the capacitor. Another source of monochromatic light produces the interference pattern in which the two consecutive bright fringes are separated by 7 × 2 mm. (ii) the external force required to move the arm. Depict the pattern of magnetic field lines around them. (ii) Write three distinct advantages of a reflecting type telescope over a refracting type telescope. Give one example to illustrate this law. What is the power of the lens required to enable the person to read clearly a book held at 25 cm from the eye? 5 . and (iii) the power dissipated as heat. If an object is placed 30 cm in front of the convex lens. The loop is kept in a uniform magnetic field ‘B’ directed downward perpendicular to the plane of the loop. derive the mirror formula for a concave mirror.” Justify this statement. F (i) What is the direction of the magnetic moment of E the current loop? S (ii) When its the torque acting on the loop (a) N maximum.106 Xam idea Physics—XII (ii) What is resonace condition? How is it used to accelerate the charged particles? 5 OR (a) Two straight long parallel conductors carry currents I 1 and I 2 in the same direction. (b) A convex lens of focal length 10 cm is placed coaxially 5 cm away form a concave lens of focal length 10 cm. The arm RS is moved with a uniform speed ‘v’. (b) A rectangular current carrying loop EFGH is kept in a uniform magnetic field as shown in the fig. 5 30. Deduce an expression for : (i) the emf induced across the arm ‘RS’. (b) The near point of a hypermetropic person is 50 cm from the eye. “The Lenz’s law is a consequence of the principle of conservation of energy. (a) What are eddy currents? Write their two applications. Deduce the expression for the force per unit length between them. (a) (i) Draw a labelled ray diagram to show the formation of image in an astronomical telescope for a distant object. (b) Deduce an expression for the mutual inductance of two long co-axial solenoids but having different radii and different number of turns. find the position of the final image formed by the combined system. (b) Figure shows a rectangular conducting loop PQRS in which arm RS of length ' l' is movable. 5 OR (a) With the help of a suitable ray diagram. (b) zero? 5 H G 29. P R l v Q S 5 OR (a) State Lenz’s law. What is the ratio of their nuclear radii? 1 The maximum kinetic energy of a photoelectron is 3 eV. 1. 22. (ii) What are the necessary conditions for this phenomenon to occur? (i) State the law that gives the polarity of the induced emf. 21. What is its stopping potential? 1 (i) State the principle on which the working of an optical fiber is based. (a) In a single slit diffraction experiment.Examination Papers 107 CBSE (All India) SET–II Questions different from Set–I. 7. Find the potential difference between two points. For what value of ‘d’ will: (i) the first minimum fall at an angle of diffraction of 30°. (ii) Draw V-I characteristics of a p-n junction diode in (a) forward bias. a slit of which ‘d’ is illuminated by red light of wavelength 650 nm. (b) reverse bias. What is the focal length of the combination? 1 Give the logic symbol of NAND gate. Find the capacitive reactance and the rms current. 1 . 8. one lying on the sphere and the other on the shell. 6. 3 25. (i) With the help of circuit diagrams distinguish between forward biasing and reverse biasing of a p-n junction diode. 5. 9.s respectively. Define the term ‘potential energy’ of charge ‘q’ at a distance ‘r’ in an external electric field. (b) How would the charge between the two flow if they are connected by a conducting wire? Name the device which works on this fact. 1. What is the work done in moving a test charge q through a distance of 1 cm along the equatorial axis of an electric dipole? 1 Two thin lenses of power + 4 D and –2 D are in contact. Another small conducting sphere of radius r carrying charge ‘q’ is introdcued inside the large shell and is placed at its centre. 3 CBSE (All India) Set–III Questions different from Set–I and Set–II. 3 OR (a) A charge +Q is placed on a large spherical conducting shell of radius R. and (ii) the first maximum fall at an angle of diffraction of 30°? (b) Why does the intensity of the secondary maximum become less as compared to the central maximum? 3 23. 50 Hz source. 2 (ii) A 15 × 0 mF capacitor is connected to 220 V. Use Gauss’s law to derive the expression for the electric field between two uniformly charged large parallel sheets with surface charge densities s and . 1 Two nuclei have mass numbers in the ratio 8 : 125. take for its activity to reduce to 1/8th of its initial value? 3 22. It is then cut into two equal parts. Zero. 11. What is the maximum kinetic energy of the photoelectrons emitted? 1 Two thin lenses of power +5 D and –2 × 5 D are in contact. 7. 2 A wire of 20 W resistance is gradually stretched to double its original length. 3. Xam idea Physics—XII The stopping potential in an experiment on photoelectric effect is 2 V. 27. so angular separation (b q ) will remain unchanged. 2..e. How long will a radioactive isotope. X-Rays Frequency range : 3 ´ 10 16 Hz – 3 ´ 10 19 Hz. 8. Solutions CBSE (All India) SET–I 1.I. b q is independent of distance between the slits and screen. d = separation between coherent sources. Draw its I-V characteristics. What is the ratio of their nuclear radii? 1 (i) What is the relation between critical angle and refractive index of a material? (ii) Does critical angle depend on the colour of light? Explain. angle between v and B is 0° or 180°. 3 Define the activity of a radionuclide. units. Another source of monochromatic light produces the interference pattern in which the two consecutive bright fringes are separated by 8 mm. Find the current drawn from the battery. 1 Two nuclei have mass numbers in the ratio 27 : 125. 16. Angular separation between fringes bq = l d ® ® where l = wavelength. Magnetic field is parallel or antiparallel to velocity of electron i. Write its S. Find the wavelength of light from the second source. 4. monochromatic light of wavelength 600 nm illuminates the pair of slits and produces an interference pattern in which two consecutive bright fringes are separated by 10 mm. These parts are then connected in parallel across a 4 × 0 volt battery. 5. . What is the effect on the interference fringes if the monochromatic source is replaced by a source of white light? 3 Explain with the help of a circuit diagram how a zener diode works as a DC voltage regulator. 25. What is the focal length of the combination? 1 Give the logic symbol of AND gate. whose half life is T years.108 4. 2 In Young’s double slit experiment. Give a plot of the activity of a radioactive species versus time. for the same potential difference between them. Symbol of NOR gate. v . where E is electric field strength. Refractive Index : Refractive index of a medium is the ratio of speed of light in vacuum to the c speed of light in medium i. F = = m = 25 cm \ P 4 Kmax = eVs = e (1 × 5 V) = 1 × 5 eV = 1 × 5 ´ 1 × 6 ´ 10 -19 J = 2 × 4 ´ 10 19 6. Nuclear radius. Net power of lens combination P = + 6 D . while for increasing electric field. R = R 0 A 1/ 3 \ æA ö =ç 1÷ R 2 è A2 ø R1 A Y B ® 1/ 3 1 ö 1/ 3 1 =æ = ç ÷ è 8ø 2 8. On decreasing the radius of the spherical surface to half there will be no effect on the electric flux.I. unit of electric flux is volt metre. Electric flux through entire closed surface is f = ò E . n = . 10. 9. in the direction of increasing field. For constant electric field E d1 d2 For increasing electric field V E d1 d2 2V 3V E V 2V d 1 = d2 3V d1 > d2 Difference: For constant electric field.. Electric flux through surface element D S ® ® ® ® is Df = E. the separation between these surfaces decreases.dS S ® ® As electric flux through the surface = 1 ´q e0 11. J 7. S. the equipotential surfaces are equidistant for same potential difference between these surfaces. D S DS cos q q DS E = EDS cos q.e.Examination Papers 109 5. Electric Flux: The total number of electric lines of force diverging normally from a surface is called the electric flux through that surface.2 D = + 4 D 1 1 Focal length. 110 Xam idea Physics—XII Alternatively : It is defined as the ratio of sine of angle of incidence to the sine of angle of refraction in medium i. The five resistors form a balanced Wheatstone’s bridge. R ¢ = R 1 + R5 = 1 + 2 = 3 W The effective resistance of R 4 and R 3 in series is R ¢¢ = R 4 + R 3 = 2 + 4 = 6 W \Equivalent resistance of network between A and B R ¢ R ¢¢ 3´6 R AB = = =2W R ¢ + R ¢¢ 3 + 6 Current drawn from battery. I = E 4 = = 2A R AB 2 A R1 1W 5W 2W R4 R2 4W R5 2W B R3 + – 4V 13. 14. R 2 is ineffective. (a) The visible radiations and radiowaves can penetrate the earth's atmosphere but X-rays are absorbed by the atmosphere.i x 12. Sq = = I C . The effective resistance of R 1 and R5 in series.e.. Since R1 R4 = R5 R 3 So. Current sensitivity : It is defined as the deflection of coil per unit current flowing in it. The equivalent circuit is shown in fig. sin i n= sin r Y Angle of deviation. q NAB Current Sensitivity.d d dm i1 i i2 Angle of incidence. hence ozone layer is crucial for human survival. (b) The ozone layer absorbs ultraviolet and other low wavelength radiations which are harmful to living cells of human bodies and plants. This is maximum when sin 2q = 1 or q = 45° .. final resistance = Current drawn from battery. R ¢ = n 2 R = ( 2) 2 ´ 15 = 60 W Resistance of each half part = 60 = 30 W 2 30 = 15 W 2 When both parts are connected in parallel.Examination Papers 111 Voltage sensitivity : It is defined on the deflection of coil per unit potential difference across its ends.e. (a) The mass of a nucleons in ground state is always less than the total mass of its constituents neutrons and protons. SV = = V GC where G is resistance of galvanometer. When length of a given wire is made n-times by strecting it. because this mass difference appears in the form of binding energy to hold the nucleons inside the nucleus. . but at the same time. hence inreasing current sensitivity does not è Gø necessarily increase the voltage sensitivity. The unpolarised and polarised light is represented as (a) Unpolarised light (b) Polarised light (c) Partially polarised light Intensity of transmitted light is maximum when the polaroid sheet makes an angle of 45° with the pass axis. then the current sensitivity ( µ N) is doubled. its resistance becomes n 2 times i. Justification: When number of turns N is doubled. 15. V I= R 3×0 = = 0×2A 15 17. so voltage Nö sensitivity æ ç SV µ ÷ will remain unchanged. Linearly Polarised Light: The light having vibrations of electric field vector in only one direction perpendicular to the direction of propagation of light is called plane (or linearly) polarised light. q NAB Voltage Sensitivty. 16. the resistance of galvanometer coil (G) will also be doubled. Electric field on the Surface of a Charged Conductor: Let electric charge be uniformly distributed over the surface of a thin. 19. A +100 Repulsive B D Attractive MeV 0 –100 C 1 2 r (fm) 3 4 18. amplifies and retransmits it to the receiver sometimes with a change in a carrier frequency. +q – – – – – To calculate the electric field strength near the sheet. OR Need of Modulation: Modulation is needed for (i) Practicable size of antenna. charge per unit surface area) be s. o . By B A symmetry the electric field strength at every point on the flat Sheet surface is the same and its direction is normal outwards at the points on the two plane surfaces and parallel to the curved surface. (i) Transducer: A device which convert one form of energy into the other. Let the surface charge density (i.e. The electric field lines are shown in fig.112 Xam idea Physics—XII (b) Part AB represents repulsive force and part BCD represents attractive force.. (ii) Repeater: A repeater picks up the signal from the transmitter. non-conducting infinite sheet. We have to calculate the electric field strength at any point distance r from the sheet of charge. we now dS3 consider a cylindrical Gaussian surface bounded by two plane S1 E 90 S3 S2 faces A and B lying on the opposite sides and parallel to the charged dS1 dS2 sheet and the cylindrical surface E E perpendicular to the sheet (fig). (ii) More effective power radiation by an antenna. (i) Metre Bridge: Meter bridge is based on the principle of Wheatstone’s bridge. lr P R R At balance = Þ = Q S (100 . C increases to K times C V V¢ = K K e0 A d µ K) 1 times. \ As s is charge per unit area of sheet and a is the intersecting area. If the surface charge density s is negative the electric field is directed towards the surface charge. The resistance of wire is divided into two resistances P and Q. OR (i) The capacitance of capacitor increases to K times (since C = (ii) The potential difference between the plates becomes Reason: V = \ As E = Q . the charge enclosed by Gaussian surface = s a According to Gauss’s theorem. U = As Q = constant. K (iii) Energy stored by the capacitor. K V 1 and V is decreased.. 1 Total electric flux = ´ (total charge enclosed by the surface) e0 i. d K Q2 .l) r S .e. C is increased. 2C 1 times. \ 2Ea = E= 1 (sa) e0 s × 2e 0 Thus electric field strength due to an infinite flat sheet of charge is independent of the distance of the point and is directed normally away from the charge. Q same. and so energy stored by capacitor decreases to 20. therefore.Examination Papers 113 Total electric flux òSE or ® •dS •dS ® = =ò òSE 1 S1 ® • d S1 + ® òSE 2 ® •d S2 + ® òSE 3 ® •dS 3 ® òSE ® ® E dS1 cos 0° + òS2 E dS2 cos 0° + òS3 E dS 3 cos 90° = E ò dS1 + E ò dS2 = E a + E a = 2E a Total electric flux = 2E a . electric field decreases to times. R is known resistance and S is unknown resistance. l1 ö = ç ÷ X l2 è 100 . Resistance box Resistance wire (S) D G A (P) l cm B (Q) (100–l ) cm C .l2 ç ÷ è X + Sø Dividing (2) by (1). Df Df e µÞe =-k Dt Dt where k is a constant of proportionality.l2 ø S Þ X= l2 æ 100 . The induced emf is proportional to the rate of change of magnetic flux linked with the coil. t in second.l1 ö ç ÷ -1 l1 è 100 . l = 25 m. Df In S. so emf is induced across the wing.l1 …(1) When X and S are in parallel.114 Xam idea Physics—XII Þ æ 100 . (2) (R) (Rh) + – Cell Rheostat K (i) Faraday’s Laws of Electromagnetic Induction: (a) Whenever there is a change in magnetic flux linked with of a coil... system f is in weber. . q = 30°. e in volt. i.. e µ Dt (b) emf induced in the coil opposes the change in flux. then k = 1.N Dt (ii) The wing of horizontal travelling plane will cut the vertical component of earth’s magnetic field.l2 ø 21. an emf is induced in the coil.l ö unknown resistance.e. where Be is earth’s magnetic field and q is angle of dip Induced emf of wing e = V v l = ( Be sin q) v l Given Be = 5 × 0 ´ 10 4 T.I. let resistance XS S¢ = X+S In second case l2 R = æ XS ö 100 . S = ç ÷R è l ø (ii) In first case l1 R = S 100 . so e = Dt Df If the coil contains N-turns. we get X + S l1 æ 100 . The vertical component of earth's field is given by V = Be sin q.e. then e = .. Df i. b 2 = 7 × 2 mm. The Schematic arrangement of Geiger-Marsdon Experiment (also known as Rutherford Scattering Experiment) is shown in fig. The conclusion is that the entire positive charge of atom is concentrated in a small volume called the nucleus. b1 = b2 = \ Þ b2 b1 = l 1D d l 2D d l2 l1 b2 b1 l1 4 4 5 m/s = 500 m/s 18 ´ sin 30° ) ´ 500 ´ 25 ´ 0 × 5) ´ 500 ´ 25 = 3 × 1 V …(1) …(2) l2 = Given b 1 = 8 × 1 mm. 23. . we obtain an interference pattern consisting of a central white fringe having on both sides symmetrically a few coloured fringes and then uniform illumination. This shows that the number of a -particles undergoing head on collision is very small. l 1 = 630 mm \ æ 7 × 2 mm ö l2 = ç ÷ ´ 630 mm è 8 × 1 mm ø = 560 mm Use of white light: When white light is used to illuminate the slit. ZnS screen Lead bricks Beam of a-particles f Gold foil Source of a-particles Most of a-particles (~ 10– 4 m thick) Detector Observations: (i) Only a small fraction of number of a -particles rebound back.Examination Papers 115 v = 1800 km/h = 1800 ´ \ e = (5 × 0 ´ 10 = (5 × 0 ´ 10 22. Sky Wave Propagation: In sky wave propagation. 24.116 Xam idea Physics—XII At the distance of head on approach. the entire kinetic energy of a -particle is converted into electrostatic potential energy. Space Wave Propagation: In space wave propagation the radiowaves transmitted from antenna reach the receiving antenna through a line of sight (straight) propagation. dM dT hT Earth hR Space wave propagation . while space wave propagation is direct. This distance of head on approach gives an upper limit of the size of Nucleus nucleus (denoted by r0 ) and is given by 1 ( Ze) ( 2e) Ek = r0 4pe 0 r0 Þ r0 = 1 2Ze 2 4pe 0 Ek This is about 10 -14 m. Distinction between sky wave and Space wave Propagation: Ionospheric layers Earth Sky wave propagation Sky wave propagation is achieved by ionospheric reflection of radiowaves. the radiowaves transmitted from antenna get reflected from the ionosphre and thereby reach the receiving antenna. line of sight propagation from the transmitted to the receiver. The range of such a tranmission is limited by the curvature of the earth. The field Ei is directed from n-region to p-region.6 cm) on either side of the junction becomes free from n p n mobile charge carriers and hence is called the depletion layer. R – Ei E p n 26. This sets up a potential difference and hence an internal electric field Ei across the junctions. Some charge carriers combine + + + with opposite charges to neutralise each other. and hence the width of depletion layer decreases. The symbol of p-n junction diode is shown in Fig. Thus Ei near the junction there is an excess of positively p Depletion charged ions in n-region and an excess of negatively layer charged ions in p-region. Ei E R + p – n (b) Forward current (ii) Under reverse biasing the applied potential difference causes a field which is in the same direction as the field due to internal potential barrier. so that some of electrons of n-region diffuse to p-region while some of holes of p-region + + + diffuse into n-region. This field stops the further diffusion of charge carriers. Thus the layers ( » 10 . VB Formatium of Depletion Layer: At the junction there – + is diffusion of charge carriers due to thermal + + + agitation.4 cm to 10 . Common emitter amplifier using n-p-n transistor + K (c) Reverse current The circuit of common emitter amplifier using n-p-n transistor is shown below: IC C1 RB C IB B Vi + – E VBB IE + VCC – V0 Output waveform RL C2 . Effect of Forward and Reverse Bias: (i) Under forward biasing the applied potential difference causes a field which acts opposite to the potential barrier. This results in an increase in barrier voltage and hence the width of depletion layer increases.Examination Papers 117 25. This results in reducing the potential barrier. nucleons get more tightly bound. e. The nuclei having mass number 56 and about 56 have maximum binding energy – 5·8 MeV and so they are most stable. the binding is energy per nucleon of fused heavier nucleus more than the binding energy per nucleon of lighter nuclei. 8 O16 .g. .0 3. Input waveform Output waveform The variation of binding energy per nucleon versus mass number is shown in figure.0 2. The nuclei having mass number below 20 and above 180 have relatively small binding energy and hence they are unstable. 2 He 4 . so again energy would be released in nuclear fusion. R Voltage gain Av = b L × Ri 27. When two very light nuclei ( A £ 10) join to form a heavy nucleus.0 5. Explanation: When a heavy nucleus ( A ³ 235 say) breaks into two lighter nuclei (nuclear fission).0 O16 8.0 4. The input and output waveforms are shown in figure. 9.0C12 F18 He4 N14 7.0 0 H2 Li7 Fe56 U238 Binding Energy per Nucleon (in MeV) 20 40 60 80 100 120 140 160 180 200 220 240 Mass Number Inferences from graph 1.0 0.. this indicates that these nuclei are relatively more stable than their neighbours. the binding energy per nucleon increases i.e.0 6. 6 C12 . 3.0 1. Some nuclei have peaks. This implies that energy would be released in nuclear fission.118 Xam idea Physics—XII The phase difference between input signal and output voltage is 180°. 2. Due to the presence of perpendicular magnetic field the ion will move in a circular path inside the dees.Examination Papers 119 28. the dees change their polarity (positive becoming negative and vice-versa) and the ion is further accelerated and moves with higher velocity along a circular path of greater radius.(4) =t = 2 qB . The positive ions produced from a source S at the centre are accelerated by a dee which is at negative potential at that moment... v qB (from equation 1) .e. in turning through an angle p is. (i) Expression for Period of Revolution Suppose the positive ion with charge q moves in a dee with a velocity v. i.. the applied alternating potential should also have the same semi-periodic time (T / 2) as that taken by the ion to cross either dee. then.(1) where m is the mass and r the radius of the path of ion in the dee and B is the strength of the magnetic field..(3) t= = w Bq Thus the time is independent of the speed of the ion.. Working: The principle of action of the apparatus is shown in fig. The function of magnetic field is to provide circular path to charged particle and so to provide the location where charged particle is capable of gaining energy from electric field.. The magnetic field and the frequency of the applied voltages are so chosen that as the ion comes out of a dee. The angular velocity w of the ion is given by..(2) w= = r m The time taken by the ion in describing a semi-circle.. qvB = mv 2 r or r = mv qB .. (ii) Resonance Condition: The condition of working of cyclotron is that the frequency of radio frequency alternating potential must be equal to the frequency of revolution of charged particles within the dees. p pm .e. Magnetic Pole N S Dee Dee Magnetic Pole S Dee-1 S R.F.. i. The phenomenon is continued till the ion reaches at the periphery of the dees where an auxiliary negative electrode (deflecting plate) deflects the accelerated ion on the target to be Beam bombarded. Now for the cyclotron to work. oscillator Dee-2 The function of electric field is to accelerate the charged particle and so to impart energy to the charged particle. This is called resonance condition. T pm . As wire of loop carrying r opposite current is nearer.. B I1 I2 . b)..120 Xam idea Physics—XII or T= 2p m qB . Let the conductors PQ and RS carry currents I 1 and I 2 in same direction and placed at separation r. Obviously. each element of other wire experiences a force..(5) This is the expression for period of revolution. the direction of B1 will be perpendicular to the plane of paper and directed downward. so the net froce acting on the loop is repulsive.e. (i) The magnetic field lies due to two current carrying parallel wires are shown in figure. 1. It has been observed experimentally that when the currents in the wire are in the same direction. Consider a current–element ‘ab’ of length DL of wire RS. The direction of current element is perpendicular to the magnetic field.). they experience an attractive force (fig. Due to this magnetic field. F m I I The force between parallel wires = 0 1 2 N / m l 2 pr (ii) We know that parallel currents attract and opposite 1 currents repel and F µ . a) and when they carry currents in opposite directions. therefore the magnetic force on element ab of length DL m I DF = B1 I 2 DL sin 90° = 0 1 I 2 DL 2p r \ The total force on conductor of length L will be m I I m I I F = 0 1 2 S DL = 0 1 2 L 2p r 2p r \ Force acting on per unit length of conductor F m I I f = = 0 1 2 N/ m L 2p r P R b DL DF I1 Q r a B I2 S .(2) According to Fleming’s left hand rule.. the direction of magnetic force will be towards PQ i. the force will be attractive.. (fig.(1) B1 = 0 1 2p r According to Maxwell’s right hand rule or right hand palm rule no. The magnetic field produced by current-carrying conductor PQ at the location of other wire RS m I . they experience a repulsive force (fig. OR (a) Suppose two long thin straight conductors (or wires) PQ and RS are placed parallel to each other in vacuum (or air) carrying currents I 1 and I 2 respectively.. period of revolution is independent of speed of charged particle and radius of circular path. df d (b) (i) Induced emf | e| = = ( BA) Q R dt dt d æ dA ö = B ç ÷ = B ( lx) x v l è dt ø dt dx = Bl = Blv dt P S (ii) Force on arm RS = BIl e ö B2 l 2v æ Blv ö =Bæ l= ç ÷. The cylinder. (a)] . 29. (a) when seen from the magnet side [fig. According to Lenz’s law. This process is used in extracting a metal from its ore.Examination Papers 121 (b) (i) Direction of magnetic moment m of the current loop is perpendicular to the plane of paper and directed downward. These currents are sometimes so strong. (iii) Power dissipated on heat P = Fv = B2 l 2v B2 l 2v 2 v= R¢ R¢ ® ® ® ® OR (a) Lenz's law: According to this law “the direction of induced current in a closed circuit is always such as to oppose the cause that produces it. Application of Eddy Currents: 1. Induction Motor : The eddy currents may be used to rotate the rotor. begins to rotate in the direction of the field. these currents tend to reduce to relative motion between the cylinder and the field. eddy currents are produced in it. (ii) Torque acting on the current loop is (a) maximum when m is perpendicular to B. the metal to be heated is placed in a rapidly varying magnetic field produced by high frequency alternating current. the direction of current induced in the coil is such as to oppose the S N N approach of north pole. that the metallic plate becomes red hot. these currents are called eddy currents. the induced currents are set up in the plate. (b) Minimum when M is parallel to B.” Example: When the north pole of a coil is brought near a closed coil.l= Bç ÷ èR¢ø è R¢ ø R¢ where R ¢ = resistance. 2. Its principle is : When a metallic cylinder (or rotor) is placed in a rotating magnetic field. therefore. (a) Eddy currents: When a metallic plate is placed in a time varying magnetic field. This necessitates an anticlockwise current in the coil. Strong eddy currents are set up in the metal produce so much heat that the metal melts. the magnetic flux linked with the plate changes. For this the nearer face of coil behaves as north pole. The arrangement of heating the metal by means of strong induced currents is called the induction furnace. This is the principle of induction motion. Induction Furnace: In induction furnance. For this the nearer face of coil behaves as south pole. when seen from the (b) magnet side (fig. Magnetic flux linked with inner solenoid ( S 2 ). Thus Lenz’s law is based on conservation of energy. This necessitates a clockwise current in the coil. in each case whenever there is a relative motion between a coil and the magnet. number of turns N 1 and N 2 each of length ‘l’ Suppose I 1 is the current in outer solenoid. b). This work appears in the form of electric energy of coil. a force begins to act which opposes the relative motion. Conservation of Energy in Lenz’s Law: Thus. magnetic field at the axis. (b) Suppose two co-axial solenoids S1 and S2 of S1 radii r1 and r2 . Therefore to maintain the relative motion. a mechanical work must be done.122 Xam idea Physics—XII Similarly when north pole of the magnet is moved away from the coil. 2 f 2 = ( N 2 ) B1 A1 = N 2 (m 0 n1 I 1 ) pr1 2 = m 0 n1 N 2 pr1 I1 \ Mutual inductance of two solenoid system f 2 M = 2 = m 0 n1 N 2 pr1 I1 But 30. the direction of current in the coil will be such as to attract the S S N magnet. B1 = m 0 n1 I 1 where n1 = number of turns/meter of outer S2 solenoid. (a) (i) n1 = B N1 l \ M= 2 m 0 N 1 N 2 pr1 l fo fe A C1 Fo A' F e' B' C2 Fe in At i fin ty Astronomical telescope (ii) Advantages of Reflecting Telescope over Refracting Telescope: (a) Less chromatic aberration (c) High resolving power (b) Less spherical aberration (d) High intense image . f 2 = . focus F and centre of curvature C.Examination Papers 123 (b) The position of image formed by convex lens is L1 L2 u = 30 cm d 1 1 1 = f 1 v 1 u1 Þ 1 1 1 1 1 = + = v 1 f 1 u1 10 30 Þ v 1 = 15 cm For concave lens u2 = 15 . AB BC \ = A ¢ B¢ B¢ C Now in DDNF and A ¢ B¢ F (each equal to 90°) ÐDNF = ÐA ¢B¢ F (opposite angles) ÐDFN = ÐA ¢ FB¢ \ Both triangles are similar DN FN AB FN or (Q AB = DN) = = A ¢ B¢ B¢ F A ¢ B¢ B¢ F Comparing (1) and (2).5 = + 10 cm.10 cm 1 1 1 1 1 1 1 1 \ = Þ = + =+ f 2 v 2 u2 v 2 f 2 u2 10 10 Þ v2 = µ That is final image is formed at infinity. The image formed by mirror is A ¢ B¢ . In DABC and DA ¢ B¢ C (each equal ÐABC = ÐA ¢ B¢ C to 90°) (opposite ÐACB = ÐA ¢ CB¢ angles) Both triangles are similar. OR (a) Mirror Formula: M 1 M 2 is a concave mirror having pole P . A An object AB is placed in front of mirror with point B O on the principal axis. we get BC FN = B¢ C B¢ F M1 D C B' F i r I A' N P M2 u …(1) …(2) …(3) . The perpendicular dropped from point of B incidence D on principal axis is DN. That is u = .2 fv f v + uf = uv Dividing both sides by uvf . so work done W = q DV = 0 (zero).u Distance of image from mirror PB¢ = .50 cm 1 1 1 \ =+ f 50 25 Þ f = 50 cm = 0 × 50 m 1 1 Power.PB¢ PB¢ .v .PF By sign convention Distance of object from mirror PB = .PC BC FP FP or …(4) \ = = B¢ C B¢ F PC .124 Xam idea Physics—XII If aperture of mirror is very small.( .(. At every point on equatorial axis. so FN = FP PB .uf + uv + 2 f 2 . P = = D= 2D f 0 × 50 CBSE (All India) SET–II 1.2 f ) -f = Þ . we get .2 f Substituting these values in (4).u .2f + v or = -f -v+ f .( . 5.vf = . Net power.u + 2f .25 cm.v Focal length of mirror PF = .2 f . we get 1 1 1 + = u v f The corresponding formula for thin lens is 1 1 1 = f v u 1 1 1 (b) = f v u The corrective lens must form the image of letters of book placed at 25 cm (near point) of hypermetropic eye.R = . P = P1 + P2 = + 4 D .f ) Þ 2f 2 . 6.f Radius of curvature of mirror PC = . the point N will be very near to P . the potential is zero.v) . .2 D = + 2 D 1 1 \ F = = = 0 × 50 m = 50 cm P 2 Logic symbol of NAND gate. v = . the ray is totally reflected in first (denser) medium. This phenomenon is called total internal reflection. i > C). (i) The working of optical fiber is based on total internal reflection. (ii) Conditions: (a) Ray of light must go from denser medium to rarer medium. (b) Angle of incdience must be greater than critical angle (i. Vs = (Ek ) max e = =3V 9. Nuclear radius R = R 0 A 1/ 3 \ æA ö =ç 1÷ R 2 è A2 ø R1 1/ 3 8 ö =æ ç ÷ è 125 ø 3 eV e 1/ 3 = 2 5 8. d sin q = ( 2n + 1) n = 1. e.Examination Papers A Y 125 B 7. 21. . q = 30°. I = = 22. (i) Lenz’s Law: The polarity of induced emf is such that it tends to produce a current which opposes the change in magnetic flux that produces it. (Ek ) max = eVs Stopping potential. (a) (i) For nth minima. n = 1. Statement: When a light ray goes from denser to rarer medium at an angle greater than critical angle. d sin q = nl Given l = 650 nm = 650 ´ 10 -9 m. XC = 2pfC = 1 2 ´ 3 × 14 ´ 50 ´ 15 ´ 10 -6 E XC 220 = 1 × 03 A 212 × 1 W = 212 × 1 W RMS Current. 1 (ii) Capacitive reactance. q = 30° \ d= 1 ´ 650 ´ 10 nl = sin q sin 30° -9 = 650 ´ 10 -9 0 ×5 = 1300 ´ 10 -9 m = 1 × 3 ´ 10 -6 m = 1 × 3 mm (ii) For nth maxima. l = 650 ´ 10 -9 m l 2 . charge per unit surface area) be s. the charge enclosed by Gaussian surface = s a + – According to Gauss’s theorem.e. E2 = i 2e 0 2e 0 ® – – – – – – – –s ® E = E1 + E2 = ® s $ i e0 . the wavelets from lesser and lesser part of slit produce constructive interference. To calculate the electric field strength near the sheet.126 Xam idea Physics—XII \ d= ( 2n + 1) l 2 sin q = 3 ´ 650 ´ 10 -9 2 ´ 0 ×5 = 1 × 95 ´ 10 -6 m = 1 × 95 mm (b) To produce secondary maxima. Total electric flux o ò E •dS S ® ® = ò E • d S1 + S1 S1 ® ® ò E •d S2 + S2 ® ® ò E •dS 3 S3 ® ® or òSE ® •dS ® =ò E dS1 cos 0° + òS2 E dS2 cos 0° + òS3 E dS 3 cos 90° = E ò dS1 + E ò dS2 = Ea + Ea = 2Ea \ Total electric flux = 2Ea. As s is charge per unit area of sheet and a is the intersecting area. non-conducting infinite sheet.. \ 2Ea = 1 (sa) e0 + + + E1 + + + + +s s E= × 2e 0 ® E2 E1 = s $ ® s $ i. By B A symmetry the electric field Sheet strength at every point on the flat surface is the same and its direction is normal outwards at the points on the two plane surfaces and parallel to the curved surface. we now dS3 consider a cylindrical Gaussian S1 E 90 S3 S2 surface bounded by two plane faces A and B lying on the opposite sides and parallel to the charged dS dS2 1 sheet and the cylindrical surface E E perpendicular to the sheet (fig). Let electric charge be uniformly distributed over the surface of a thin. + – 1 Total electric flux = ´ (total charge enclosed by the surface) + – e0 i. We have to calculate the electric field strength at any point distance r from the sheet of charge.. 23. Let the surface charge density (i.e. (b) When both shells are connected by a conducting wire. The external field E is much stronger than internal field Ei .Examination Papers 127 OR (a) For external points. so that the external field is established to help the internal field Ei as shown in Fig. the charge may be supposed to be concentrated at the centre. so that an external electric field E is established directed from p to n-end to oppose the internal field Ei as shown in Fig. Ei E p n R – p + n – + Reverse biasing K (c) Reverse current . therefore the current flow stops. the potential is same as on the surface 1 æq Qö \ V (r) = ç + ÷ 4pe 0 è r R ø \ Potential difference. V (r) . whole charge of inner shell will flow to outer shell. (i) Forward Bias: In this arrangement the positive terminal of battery is connected to p-end and negative terminal to n-end of the crystal. Device Working on this principle is Van de Graaff Generator. Under the biasing the holes in p-region and the electrons in n-region are pushed away from the junction to widen the depletion layer and hence increases the potential barrier.ù 4pe 0 ê ër R ú û R +q V(R) r O 25. so 1 æQ + qö V ( R) = ç ÷ 4pe 0 è R ø +Q V(r) For internal points. Ei Ei E p n p n R + p – n + K – (a) No current (b) Forward current Reverse Bias: In this arrangement the positive terminal of battery is connected to n-end and negative terminal to p-end of the crystal.V ( R) = = 1 é q Q Q + qù + 4pe 0 ê R ú ër R û 1 1 1 qé . I = = A = 0×2A Rnet 20 . 19 J 7. resistance of each part = 40 W 40 ´ 40 When there parts are connected in parallel. 5. net resistance Rnet = = 20 W 40 + 40 V 4×0 Current. Maximum kinetic energy. P = P1 + P2 = + 5 . 16. æA ö =ç 1÷ R 2 è A2 ø R1 1/ 3 27 ö 1/ 3 3 =æ = ç ÷ è 125 ø 5 (i) Relation between refractive index (n) and critical angle (C) is 1 n= sin C (ii) Yes. 11. Potential energy of a charge q in an external electric field is defined as the work-done in bringing the charge q from infinity to a distance r in an external electric field produced by the other charge(s). B 8. Resistance of stretched wire R ¢ = n 2 R = ( 2) 2 ´ 20 = 80 W When wire is cut into two equal parts. (Ek ) max = eVs = e ( 2 V) = 2 eV = 2 ´ 1 × 6 ´ 10 -19 J = 3 × 2 ´ 10 Net power. F = = m = 0 × 4 m = 40 cm P 2 ×5 Symbol of AND gate. it increases with increase of wavelength being maximum for red and minimum for violet.2 × 5 = + 2 × 5 D 1 1 \: Focal length. critical angle depends on wavelength or colour of light.128 Xam idea Physics—XII (ii) V-I Characteristics of (a) forward bias and (b) reverse bias: I (mA) Avalanche breakdown (–) V O (+) (a) Forward bias F V R (b) Reverse bias I (µA) CBSE (All India) SET–III 1. A Y 4. So. If the input dc voltage increases. b= \ Þ Dl d b1 b2 = l1 l2 b2 b1 l1 ´ 600 nm = 480 nm l2 = = 8 mm 10 mm On replacing monochromatic light by white light. Zener diode as a Voltage Regulator Unregulated input IZ Vin VZ RL V0 When a Zener diode is operated in the breakdown region. The Zener diode is connected across load such that it is reverse biased. 25. The series resistance R absorbs the output voltage fluctuations so as to maintain constant voltage across the load. voltage drop across R increases. without any change in the voltage across zener diode. the voltage across it remains practically constant for a large change in the current. A simple circuit of a voltage regulator using a Zener diode is shown in the figure. I-V Characteristics I (mA) VR Zener voltage O VF VZ I(mA) Regulated output The Zener diode makes its use as a voltage regulator due to the following property : R I IL . the interference pattern will contain central white fringe surrounded on either side by few coloured fringes.Examination Papers 129 22. the current through R and Zener diode also increases. S.I. unit of activity is becquerel (= 1 disintegration/second). æ dN ö Activity R ç = ÷ = lN è dt ø Decay constant l = \ Activity log e 2 T (log e 2) N T . R2 = N1 (log e 2) N 2 T2 T2 A A0 2 A0 4 O T 2T t A0 R= \ R1 = (log e 2) N 1 T1 R1 æ N ö æT ö For two elements = ´ =ç 1÷ç 2÷ R2 T1 N 2 è N 2 ø è T1 ø Numerical: Given \ Þ R 1 ön =æ ç ÷ R0 è 2ø R 1 t Time taken = . Xam idea Physics—XII The activity of a radioactive element at any instant is equal to its rate of decay at that instant. The plot is shown in figure.130 27.n= = R0 8 T Half life Þ Þ æ1 ö ç ÷ è 2ø 3 n æ1 ö = æ1 ö ç ÷ ç ÷ è 8ø è 2ø 1ö =æ ç ÷ è 2ø n Þ n= 3 t =3 T t = 3T years . . 1 Unpolarised light of intensity I is passed through a polaroid. 4. What is the intensity of the light transmitted by the polaroid? 1 Why are coherent soruces required to create interference of light? 1 3. electric field? 1 Name the electromagnetic radiations which are produced when high energy electrons are bombarded on a metal target. questions 19 to 27 carry three marks each and questions 28 to 30 carry five marks each. (e) You may use the following values of physical constants wherever necessary: c = 3 ´ 10 8 ms 1 Maximum marks: 70 h = 6 × 626 ´ 10 -34 Js m 0 = 4p ´ 10 -7 TmA -1 e = 1 × 602 ´ 10 -19 C 1 = 9 × 109 Nm2C– 2 4pe o Boltzmann’s constant k = 1 × 381 ´ 10 -23 J K-1 Avogadro’s number N A = 6 × 022 ´ 10 23 /mole Mass of neutron mn = 1 × 2 ´ 10 -27 kg Mass of electron me = 9 × 1 ´ 10 -31 kg Radius of earth = 6400 km CBSE (Foreign) SET–I 1. 1 Draw the wavefront coming out of a convex lens when a point source of light is placed at its focus. 2. (c) There is no overall choice. 5.CBSE EXAMINATION PAPERS FOREIGN–2009 Time allowed: 3 hours General Instructions: (a) All questions are compulsory. one question of three marks and all three questions of five marks each. You have to attempt only one of the given choices in such questions. an internal choice has been provided in one question of two marks. Questions 1 to 8 carry one mark each. current density. (d) Use of calculators is not permitted. questions 9 to 18 carry two marks each. drift speed. Why is it necessary that the field lines from a point charge placed in the vicinity of a conductor must be normal to the surface of the conductor at every point? 1 A steady current flows in a metallic conductor of non-uniform cross-section. 6. (b) There are 30 questions in total. However. Which of these quantities is constant along the conductor: Current. 9. Xam idea Physics—XII 8.132 7. 2 A wire of length L is bent round in the form of a coil having N turns of same radius. n (i) Why is the slope same for both lines? (ii) For which material will the emitted electrons have greater kinetic energy for the incident radiations of the same frequency ? Justify your answer. If a steady current I flows through it in a clockwise direction. Give its two important applications. If a -particle is replaced by a proton. 2 15. 12.13 × 6 eV . Draw a schematic diagram of a localised-wave describing the wave nature of the moving electron. V0 M1 M2 11. 14. Find the potential difference between A and C. A 5 cm B 3 cm E C 10. The energy of the electron in the ground state of hydrogen atom is . 2 OR The sum of two point charges is 7 m C. 2 Name the electromagnetic radiations having the wavelength range from 1 mm to 700 nm. 2 Derive an expression for the de-Broglie wavelength associated with an electron accelerated through a potential V. 2 Figure shows variation of stopping potential (V0 ) with the frequency ( n) for two photosensitive materials M1 and M2 . find the magnitude and direction of the magnetic field produced at its centre. 13. In the Rutherford scattering experiment the distance of closest approach for an a -particle is d 0 . (i) What does the negative sign signify? (ii) How much energy is required to take an electron in this atom from the ground state to the first excited state? 2 . 1 Figure shows a sheet of aluminium foil of negligible thickness placed between the plates of a capacitor. How will its capacitance be affected if: (i) the foil is electrically insulated? (ii) the foil is connected to the upper plate with a conducting wire? 2 Three points A. Calculate the value of each charge. They repel each other with a force of 1 N when kept 30 cm apart in free space. how much kinetic energy in comparison to a -particle will it require to have the same distance of closest approach d 0 ? 1 State the Faraday’s law of electromagnetic induction. B and C lie in a uniform electric field (E) of 5 ´ 10 3 NC -1 as shown in the figure. If this logic gate is connected to NOT gate. B = 0 and (ii) A = 1. A J B 22.Examination Papers 133 16. Draw the logic symbol of the gate whose truth table is given below: Input A 0 0 1 1 B 0 1 0 1 Output Y 1 0 0 0 17. what will be output when (i) A = 0. B = 1? Draw the logic symbol of the combination. 2 (i) What is the line of sight communication? (ii) Why is it not possible to use sky wave propagation for transmission of TV signals? (i) How are eddy currents reduced in a metallic core? (ii) Give two uses of eddy currents. ® . Deduce an expression for the electric field at a point on the equatorial plane of an electric dipole of length 2a. 3 State Kirchhoff’s rules. and (2) when key (K) the key ( K)i s closed? 3 Deduce the expression for the torque experienced by a rectangular loop carrying a steady current ‘I’ and placed in a uniform magnetic field B. Use Kirchhoff’s rules to show that no current flows in the given circuit.’ Is it scalar or vector? 2 2 20. e1 = 2 V r1 e2 = 2 V r2 2V R1 21. 19. (i) What is the purpose of using high resistance R 2 ? (ii) How does the position of balance point ( J ) change e G when the resistance R 1 is decreased? R2 (iii) Why cannot the balance point be obtained (1) when the emf e is greater than 2 V. (b) Figure shows the circuit diagram of a potentiometer for determining the emf ‘e’ of a cell of negligible internal resistance. (a) State the Principle of working of a potentiometer. 18. Define the term ‘electric dipole moment. Indicate the direction of the torque acting on the loop. Why is the emf maximum when the plane of the armature is parallel to the magetic field? 5 OR Draw a labelled diagram of a step-up transformer and explain briefly its working. how will the width of central maximum be affected if (i) the width of the slit is doubled. Derive the expression for its magnifying power in normal adjustment. 3 Depict the field-line pattern due to a current carrying solenoid of finite length. A ray incident on another face at an angle of 45° is refracted and reflected from the silver coated face and retraces its path. Deduce the expressions for the secondary voltage and secondary current in terms of the number of turns of primary and secondary windings. what is the power of the combination? 5 OR Draw a labelled ray diagram to show the image formation by an astronomical telescope. 3 State the working principle of an AC generator with the help of a labelled diagram. (ii) the wavelength of the light used is increased? 26.134 Xam idea Physics—XII OR Deduce the expression for magnetic dipole moment of an electron revolving around the nucleus in a circular orbit of radius ‘r’. If a convex lens of focal length 50 cm is placed in contact coaxially with a concave lens of focal length 20 cm. 28. (b) Define power of a lens and give its S. 3 23.I. Indicate the direction of the magnetic dipole moment. Derive an expression for the instantaneous value of the emf induced in coil. (a) What is meant by half life of a radioactive element? 3 27. (b) The half life of a radioactive substance is 30 s. 3 What is meant by detection of a signal in a communication system? With the help of a block diagram explain the detection of AM signal. Find the refractive index of the material of the prism. (i) In what way do these lines differ from those due to an electric dipole? 24. How is the power transmission and distribution over long distances done with the use of transformers? 5 29. Write two basic features which can distinguish between a telescope and a compound microscope. 5 . 3 25. One face of a prism with a refracting angle of 30° is coated with silver. units. Justify your answer in each case. Calculate (i) the decay constant. (a) Why do we not encounter diffraction effects of light in everyday observations? (b) In the observed diffraction pattern due to a single slit. (a) Draw a ray diagram for formation of image of a point object by a thin double convex lens having radii of curvatures R 1 and R 2 and hence derive lens maker’s formula. and (ii) time taken for the sample to decay by 3/4th of the initial value. (ii) Why can’t two magnetic field lines intersect each other? State the conditions under which total internal reflection occurs. 1 A logic gate is obtained by applying output of OR gate to a NOT gate. Write the symbol and truth table of this gate. Define self-inductance. What should be the magnitude and direction of the electric field so that the particle moves undeviated along the same path? 2 n +q X 13. 2 A 10 cm B 6 cm C E OR The sum of two point charges is 9 µC. 12. 2 . (b) With the help of a labelled circuit diagram explain the use of a p-n junction diode as a full wave rectifier. Three points A. Explain how a transistor in active state exhibits a low resistance at its emitter base junction and high resistance at its base collector junction. units. B and C lie in a uniform electric field (E) of 5 ´ 10 3 NC -1 as shown in the figure. (b) If the potential difference used to accelerate electrons is doubled. Find the potential difference between A and C. 1 Sketch the shape of wavefront emerging from a point source of light and also make the rays. Draw the input and output waveforms. 3. Name the characteristic of the incident radiation that is kept constant in this experiment. 5 OR Draw a circuit diagram of an n-p-n transistor with its emitter base junction forward biased and base collector junction reverse biased. Draw a circuit diagram and explain the operation of a transistor as a switch. 7. by what factor does the de-Broglie wavelength associated with the electrons change? 2 14. Give its S.Examination Papers 135 30.I. (a) Explain the formation of ‘depletion layer' and ‘barrier potential’ in a p-n junction. 5 CBSE (Foreign) SET–II Questions different from Set–I. (a) Draw a graph showing variation of photo-electric current ( I ) with anode potential (V ) for different intensities of incident radiation. Calculate the value of each charge. Describe briefly its working. Name the gate so formed. 11. They repel each other is force of 2 N when kept 30 cm apart in free space. 2 Y A point charge is moving with a constant velocity B perpendicular to a uniform magnetic field as shown in the figure. A 800 pF capacitor is charged by a 100 V battery. 8. What is the electrostatic energy stored? 2 OR The sum of two point charges is 7 m C. 2 -3 Name the elecromagnetic radiations having the wavelength range from 1 nm to 10 nm. (i) What will happen to the plates when they are released? (ii) Draw the pattern of the electric field lines for the system. Calculate the value of each charge. (b) If a low frequency signal in the audio frequency range is to be transmitted over long distances. After some time the battery is disconnected. 2 10. 3 (a) Define the terms (i) ‘amplitude modulation’ and (ii) ‘modulation index’. Xam idea Physics—XII (i) What happens when a diamagnetic substance is placed in a varying magnetic field? (ii) Name the properties of a magnetic material that make it suitable for making (a) a permanent magnet and (b) a core of an electromagnet. 1 Define a wavefront. Give its S. 2 Q + Q – 13. 1 Name the electromagnetic radiation which can be produced by klystron or a magnetron valve. What is its effect on the capacitance of the capacitor? 1 Define mutual inductance. 27. 3 CBSE (Foreign) SET–III Questions different from Set–I and Set–II.136 23. The capacitor is then connected to another 800 pF capacitor. 26. 1 P1 P2 Figure shows two large metal plates P1 and P2. units. 4. They repel each other with a force of 1 N when kept 30 cm apart in free space.I. Give its two important applications. 9. 3 (a) What is meant by half life of a radioactive element? (b) The half life of a radioactive substance is 20 s. Calculate: (i) the decay constant and (ii) time taken for the sample to decay by 7/8th of the initial value. 6. . A metal plate is introdcued between the plates of a charged parallel plate capacitor. tightly held against each other and placed between two equal and unlike point charges perpendicular to the line joining them. 1. explain briefly the need of translating this signal to high frequencies before transmission. n 2 or n 3 ) is the highest? 2 (i) How does angle of dip change as one goes from magnetic pole to magnetic equator of the Earth? (ii) A uniform magnetic field gets modified as shown below when two specimens X and Y are placed in it. 16.Examination Papers 137 14. paramagnetic or ferromagnetic. Calculate: (i) the decay constant. 2 The graph below shows variation of photo-electric current with collector plate potential for different frequencies of incident radiations. and (ii) time taken for the sample to decay by 3/4th of the initial value. Y X (iii) How is the magnetic permeability of specimen X different from that of specimen Y? 26. Name the gate so formed. Write the symbol and truth table of this gate. (ii) Which frequency ( n 1 . 3 3 . A logic gate is obtained by applying output of AND gate to a NOT gate. Photo-electric current n1 n2 n3 Collector plate potential (i) Which physical parameter is kept constant for the three curves? 24. (a) What is meant by half life of a radioactive element? (b) The half life of a radioactive substance is 30 s. Identify whether specimens X and Y are diamagnetic. Mathematically. Intensity of light transmitted through the polaroid = I × 2 Coherent sources are required for sustained interference. so capacitance is doubled. X-rays. Surface of a conductor is an equipotential surface and field lines are always directed from higher to lower potential. Wavefront is plane as shown in fig. Distance AB = 4 cm . 7.138 Xam idea Physics—XII Solutions CBSE (Foreign) SET–I 1. so potential of B = potential of C i. 8. Current is constant along a conductor. 4. (i) No effect on capacitance if foil is electrically neutral. The line joining B to C is perpendicular to electric field. q = 2e) Ek = 4pe 0 d0 Ek ¢ = Ek ¢ Ek = 1 ( Ze) ( e) (for proton.e. the intensity at a point will go on changing with time. q = e) 4pe 0 d0 1 2 Þ Ek ¢ = Ek 2 That is KE of proton must be half on comparison with KE of a -particle. The magnitude of the induced emf in a circuit is equal to the time rate of change of magnetic flux through the circuit. VB = VC . 3. 1 ( Ze) ( 2e) (for a -particle. the effective separation between plates becomes half. If sources are incoherent. F Plane wavefront 5. (ii) If foil is connected to upper plate with a conducting wire.. 2. 6. 10. so field lines in the vicinity of a conductor must be normal to the surface of conductor. the induced emf is given by Df e=Dt 9. L L = N ´ 2 pr Þ r = 2 pN B= m 0 NI 2r = m 0 NI 2 ( L / 2pN) Þ B= m 0 pN 2 L I The direction of magnetic field is normal to plane of coil in downward direction. If Ek is the kinetic energy of electron. Þ Infrared radiations q1 = 5 mC.q2 ) 2 = ( q1 + q2 ) 2 .(i) l= = p mv where m is mass and v is velocity of electron. Expression for de Broglie Wavelength associated with Accelerated Electrons The de Broglie wavelength associated with electrons of momentum p is given by h h .q2 = 3 ´ 10 -6 C Solving (1) and (3)... 13.Examination Papers 139 Potential difference between A and C = E ´ ( AB) = 5 ´ 10 3 ´ ( 4 ´ 10 -2 ) = 200 volt OR q1 + q2 = 7 ´ 10 -6 C q1 q2 1 =1 4pe 0 ( 0 × 30) 2 or q1 q2 = 1 9 ´ 10 9 … (1) Þ q1 q2 = ( 4pe 0 ) ( 0 × 30) 2 …(2) ´ 9 ´ 10 -2 = 10 -11 ( q1 . (ii) for therapeutic purposes. q2 = 2 mC …(3) Applications: (i) Taking photograph during fog and smoke etc..4 ´ 10 -11 = 49 ´ 10 -12 .(ii) \ Equation (i) gives l = .4 q1 q2 = (7 ´ 10 -6 ) 2 . we get q1 = 5 ´ 10 -6 C.40 ´ 10 -12 = 9 ´ 10 -12 q1 .. then EK = Þ 2 p2 1 1 æ p ö mv 2 = m ç = ÷ 2 2 èmø 2m pö æ çsince p = mv Þ v = ÷ è mø p = 2mEK h 2mEK . q2 = 2 ´ 10 -6 C 11. 12. 140 Xam idea Physics—XII If V volt is accelerating potential of electron. the gate becomes OR gate. B = 1 gives output 1.3 × 4 eV 4 Energy required to take electron from ground state to first excited state DE = E2 . (i) A = 0. = hn . 16. The diagram of wave packet describing the motion of a moving electron is shown. then Kinetic energy.. æ hö (i) The slope of stopping potential (V0 ) versus frequency ( n) is equal to ç ÷ which is universal èeø constant. 14. . (ii) A = 1. so slope is same for both lines. A Y B When it is connected to a NOT gate.E.. (ii) Energy of electron in H-atom in nth orbit is Rhc 13 × 6 En = =2 n n2 For first excited state n = 2 13 × 6 E2 = eV = .E1 = . EK = eV \ Equation (ii) gives h l= 2meV .hn 0 As threshold frequency n 0 is lesser for M 1 . (i) Negative sign shows that electron in ground state is bound in H-atom due to attractive force between electron and nucleus.13 × 6 eV .( .(iii) This is the required expression for de Broglie wavelength associated with electron accelerated to potential of V volt. The given truth table is of NOR gate.3 × 4 eV) = 10 × 2 eV 15. The logic symbol is shown in fig. (ii) K. B = 0 gives output 0. so K. will be greater for M 1 for same frequency n.E. (ii) Eddy currents are used in (a) induction furnace (b) induction motor. this reducing the strength of eddy currents. A B A+B A+B 17.e. Ionospheric layers do not reflect back such high frequency signals. The electric dipole moment is a vector quantity whose magnitude is equal to the product of charge on one dipole and distance between them..q to + q. AP = BP = r 2 + l \ E1 = ® E2 ® 2 E = E1cosq + E2cosq E E1 E1sinq q P q 1 2 4p e 0 r + l q 1 2 4p e 0 r + l 2 along B to P E2 E2sinq r = 2 along P to A A –q l 2l O l q +q B \ Resultant electric field at P is E = E1 cos q + E2 cos q q 1 But E1 = E2 = 2 4p e 0 (r + l 2 ) and cos q = OB l = PB r2 + l 2 = l (r + l 2 ) 1/ 2 2 \ E = 2E1 cos q = 2 ´ = q 1 l × 2 2 2 4p e 0 (r + l ) (r + l 2 ) 1/ 2 2ql 1 4p e 0 (r 2 + l 2 ) 3 / 2 But q.2l = p = electric dipole moment . 19. (i) A metallic core cuts the path of eddy currents. 18. From figure. Derivation of electric field at a point of equatorial plane: Consider a point P on broad side on the position of dipole formed of charges + q and . (b) TV signals have high frequency range 100 to 200 MHz. (a) LOS Communication: The propagation of a radio wave in a straight line from transmitting to receiving antenna on the ground is called line of sight communication. Hence. Its direction is from . sky waves cannot be used for transmission of TV signals. ® 2l –q p +q p = q2 l ® Electric dipole moment is a vector quantity.q at separation 2l. i.Examination Papers 141 The combination is shown in fig. e. SV = 0 This law is based on conservation of energy.. if terminals of one of the batteries are interchanged.Ir2 . 20. i. Numerical: Applying Kirchhoff’s second law SV = 0 to given closed circuit along the path abcda.e. Kirchhoff’s Rules: (i) First law (or junction law): The algebraic sum of currents meeting at any junction is zero. so l compared to r and so equation (3) gives p 1 E= 2 4p e 0 (r ) 3 / 2 E= p 1 4p e 0 r 3 ® p 1 parallel to BA 4p e 0 r 3 2 may be neglected as i.2 . so given question is wrong.142 Xam idea Physics—XII \ E= p 1 2 4p e 0 (r + l 2 ) 3 / 2 . However.Ir1 + 2 = 0 Þ I=0 b 2V r2 c . i. SI = 0 This law is based on conservation of charge.Ir2 .. + 2 .e.. we have l < < r . electric field strength due to a short dipole at broadside on position E= . I a 2V r1 d ..(iv) Its direction is parallel to the axis of dipole from positive to negative charge.. (ii) Second law (or loop law): The algebraic sum of potential differences of different circuit elements of a closed circuit (or mesh) is zero.. then current is zero.(iii) 2 If dipole is infinitesimal and point P is far away.Ir1 + 2 = 0 Þ I (r1 + r2 ) = 4 I= 4 r1 + r2 b 2V c I a 2V r1 d r2 This current is non-zero. Let V be the potential difference across the portion of the wire of length l whose resistance is R.e. the terminal potential difference of cell is zero. When jockey is far from balance point. the potential gradient of potentiometer wire increases. potential difference across any portion of potentiometer wire is directly proportional to length of the wire of that portion. (2) When key ( K) is closed. (Since V = kl Þ l = 0 for V = 0) . (b) (i) The purpose of high resistance R 2 is to reduce the current through the galvanometer. (a) Principle of Potentiometer: Potentiometer works on the fact that the fall of potential across any portion of the wire is directly proportional to the length of that portion provided the wire is of uniform area of cross-section and a constant current is flowing through it.. (ii) When resistance R 1 is decreased. V Here.e. the fall of potential per unit length of wire.Examination Papers 143 21. so balance point cannot be between Aand B. this saves the galvanometer and the cell (of emf e) from being damaged. V=IR rl R= A V=Ir V µI l A æ where K = Ir ö ç ÷ è Aø (if I and A are constant) As \ or i. (iii) (1) The balance point is not obtained because maximum emf across potentiometer wire is 2 V. then from Ohm’s law. If I is the current flowing through the wire. = K = is called potential gradient.. B1 + – K Rh + A + – C R () K1 e HR P2 P1 J G – B Suppose A and r are resepectively the area of cross-section and specific resistance of the material of the wire. l i. so balance point ( J ) shifts to longer length of wire. RS and SP are F1 . the perpendicular distance between F1 and F3 is b sin q. The net force and torque acting on the loop will be determined by the forces acting on all sides of the loop. therefore the magnitude of each of forces F1 and F3 is F = IlB sin 90° = IlB. Therefore. Let at any instant the normal to the plane of loop make an angle q with the direction of magnetic field B and I be the current in the loop. QR. each side of the loop will experience a force. then t = NI AB sin q ® ® ® ® ® ® ® ® . The length of loop = PQ = RS = l and breadth = QR = SP = b. The sides QR and SP make angle ( 90° . therefore the resultant force of F1 and F3 is zero. The sides PQ and RS of current loop are perpendicular to the magnetic field. the resultant of F2 and F4 is zero. According to Fleming’s left hand rule the forces F2 and F4 are equal and opposite but their line of ® ® ® ® N' ® ® ® ® ® ® Q F1 I N F2 q R Axis of loop or normal to loop B F3 P S F4 F1=IlB I (Upward) loo p q b q ax is N of B I (Downward) b sin q F3=IlB action is same. We know that a force acts on a current carrying wire placed in a magnetic field. \ t = IAB sin q If the loop contains N-turns. but their lines of action are different. but they form a couple called the deflecting couple. breadth b suspended in a uniform magnetic field B .e. Suppose that the forces on sides PQ. Xam idea Physics—XII Torque on a current carrying loop: Consider a rectangular loop PQRS of length l. F2 . According to Fleming’s left hand rule the forces F1 and F3 acting on sides PQ and RS are equal and opposite.q) = Blb cos q. Therefore these forces cancel each other i.144 22. \ Moment of couple or Torque. F3 and F4 respectively. t = (Magnitude of one force F ) ´ perpendicular distance = ( BIl ) × (b sin q) = I ( lb) B sin q But lb = area of loop = A (say) Torque. When the normal to plane of loop makes an angle q with the direction of magnetic field B . Therefore each of the forces F2 and F4 acting on these sides has same magnitude F ¢ = Blb sin ( 90° .q) with the direction of magnetic field. In vector form Ml = ® e ® L 2me .. A = pr 2 Magnetic moment due to orbital motion. Direction of torque is perpendicular to direction of area of loop as well as the direction of magnetic field i. M l = e L 2m e …(iv) . Relation between magnetic moment and angular momentum Orbital angular momentum of electron L = me v r where me is mass of electron.e...(v) .. Dividing (ii) by (iii)..(iii) This is expression of magnetic moment of revolving electron in terms of angular momentum of electron. ev evr M l = IA = ( pr 2 ) = 2 pr 2 ® ® ® ® ® .. The revolving electron is equivalent to electric current e I= L T e– 2p r where T is period of revolution = me v r N e ev . The direction of magnetic moment is along the axis.Examination Papers 145 In vector form t = NI A ´ B. we get M l ev r / 2 e = = L me v r 2me Magnetic moment..(i) \ I= = 2 pr / v 2 pr Area of current loop (electron orbit).. OR Magnetic moment of an electron moving in a circle: Consider an electron revolving around a nucleus ( N) in circular path of radius r with speed v.(ii) This equation gives the magnetic dipole moment of a revolving electron. along I A ´ B.. the width of central fringes increases. n= r1 = A . è aø (ii) When wavelength of light used is increased (b q µ l). these will be two directions of magnetic field which is impossible. while in the case of an electric dipole the field lines begin from a positive charge and end on a negative charge or escape to infinity. (b) The angle of incidence i > critical angle C. l = 0 × 6931 0 × 6931 -1 = s = 0 × 0231s -1 T 30 26.r2 = 30° . the (angular) width of central fringe æ ç µ ÷ is halved.146 23. (a) The half-life of a radioactive sample is defined as the time in which the mass of sample is left one half of the original mass. (ii) Numerical: Given A = 30° When ray is incident normally on the other face. (i) The conditions for total internal reflection are (a) The ray must travel from a denser into a rarer medium. (ii) Two magnetic field lines cannot intersect because at the point of intersection. Xam idea Physics—XII Axis I Field fines of a current carrying solenoid Field fines of an electric dipole Axis (i) Difference: Field lines of a solenoid form continuous current loops.0 = 30° sin i1 sin r1 = sin 45° sin 30° 30° 45° r1 1/ 2 = 2 = 1 × 414 1/2 (a) We do not encounter diffraction effects of light in everyday observations. so r2 = 0. size of obstacle/aperture must be comparable with wavelength of light but in daily observations size of obstacle/aperture is much larger than the wavelength of light. 2l Angular width of central fringe b q = a 1ö (b) (i) If the width of slit is doubled. 24. . To observe diffraction. n = Þ 25. (b) Give T = 30 s (i) Decay constant. As r1 + r2 = A Þ i1 = 45° Refractive index. it retraces its path. the magnetic field is generated by a permanent magnet. whose output is the required message signal m(t). The method is shown in the form of a block diagram.' the envelope of which is the message signal. (iii) Slip Rings: The slip rings R 1 and R 2 are the two metal rings to which the ends of armature coil are connected. These rings are fixed to the shaft which rotates the armature coil so that the rings also rotate along with the armature. It is based on the principle of electromagnetic induction. AC generator: A dynamo or generator is a device which converts mechanical energy into electrical energy. the magnetic field is produced by an electromagnet. The rectified wave is passed through an envelope detector. 28. Explanation of Detection with the help of a block diagram: AM wave Rectifier Envelope Detector (b) m(t) Out put (a) (c) time time Rectified wave time Output AM input wave The modulated carrier wave contains frequencies wc ± wm . It can revolve round an axle between the two poles of the field magnet. The modulated signal is passed through a rectifier. . while in the case of large power dynamo. Detection: Detection is the process of recovering the modulating signal from the modulated carrier wave. In the case of a low power dynamo. Construction: It consists of the four main parts: (i) Field Magnet: It produces the magnetic field. The detection means to obtain message signal m(t) of frequency wm . (ii) Armature: It consists of a large number of turns of insulated wire in the soft iron drum or ring. The drum or ring serves the two purposes : (i) It serves as a support to coils and (ii) It increases the magnetic field due to air core being replaced by an iron core.Examination Papers 147 (ii) N æ 1 ön =ç ÷ N0 è 2 ø Þ n= 1t =2 T 3 æ 1 ön 1 ö 2 æ 1 ön = ç ÷ or æ ç ÷ =ç ÷ è 2ø è 2ø 4 è 2ø or t = 2T = 2 ´ 30 = 60 s This gives 27. The output current in external load R L is taken through these brushes. (b)]. (iv) Brushes: These are two flexible metal plates or carbon rods ( B1 and B2 ) which are fixed and constantly touch the revolving rings. It produces rectified wave [fig. the wire ab moves upward and cd a downward. In the external circuit. the current d B1 flows along B1 R L B2 . These coils are kept insulated from each other and from the iron-core. so that the direction of induced current is shown in fig. the maximum value is e 0 = NBAw. so the R2 direction of current is reversed and in external B2 circuit it flows along B2 R L B1 . Thus the direction of induced emf and current changes in the external circuit after each half revolution. The terminals of primary coils are connected to AC mains and the terminals of the secondary coil are connected to external circuit in which alternating current of desired voltage is required. Considering the armature to be in vertical position and as it rotates in anticlockwise N B S Field magnet direction. then sin wt = 1. Out of these coils one coil is called primary coil and other is called the secondary coil. because the average value of ac over full cycle is zero. A area of coil and B the magnetic induction.148 Xam idea Physics—XII w Working: When the armature coil is rotated in the Armature coil strong magnetic field. The number of turns in these coils are different. Construction: It consists of laminated core of soft iron. If N is the number of turns in coil. then induced emf df d e== {NBA (cos 2p f t )} = 2p NBA f sin 2p f t dt dt Obviously. OR Transformer: Transformer is a device by which an alternating voltage may be decreased or increased. the magnetic flux linked b with the coil changes and the current is induced in the coil. the emf produced is alternating and hence the current is also alternating.C. the Slip rings Brushes Load RL wire ab moves downward and cd upward. Current produced by an ac generator cannot be measured by moving coil ammeter. The direction of current R1 remains unchanged during the first half turn of armature. so emf is maximum. its direction being given by Fleming’s right c hand rule. f the frequency of rotation. but are coupled through mutual induction. mains) Primary Primary laminated iron core Core Secondary Step up Secondary Transformer . The source of energy generation is the mechanical energy of rotation of armature coil. During the second half revolution. When plane of armature coil is parallel to magnetic field. on which two coils of insulated copper wire are separately wound. Step up Transformer: It transforms the alternating low voltage to alternating high (A. This is based on the principle of mutual-induction. will be equal to the applied potential difference (Vp ) across its ends.. due to which the magnetic flux linked with the secondary coil changes continuously. (i.. then the potential difference VS across its ends will be equal to the emf ( e S ) induced in it. We assume that there is no leakage of flux so that the flux linked with each turn of primary coil and secondary coil is the same. So VS > Vp and iS < ip i. \ .. therefore VS e S N S .. Let N p be the number of turns in primary coil. e.(v) In step up transformer... this steps down the current. Power Transmission Over Long Distances The power (electrical energy) is transmitted to long distances by the use of transformers. Working: When alternating current source is connected to the ends of primary coil.. The voltage output of the generator is stepped up by means of step up transformer. Similarly if the secondary circuit is open.(iv) = = = r (say) Vp e p N p where r = NS Np is called the transformation ratio.NS Dt From (1) and (ii) eS ep = NS Np . Power in primary = Power in secondary Vp ip = VS iS iS Vp N p 1 = = = ip VS N S r Ns > N p ® r > 1 . therefore the alternating emf of same frequency is developed across the secondary. the current changes continuously in the primary coil..(ii) If the resistance of primary coil is negligible. step up transformer increases the voltage. .e.. the emf ( e p ) induced in the primary coil. . so power loss I 2 R is significantly reduced.(i) e p = . then For about 100% efficiency.. If ip and is are the instantaneous currents in primary and secondary coils and there is no loss of energy.Np Dt and emf induced in the secondary coil Df e S = . According to Faraday’s laws the emf induced in the primary coil Df . At the receiving station the voltage is stepped down to 220 V for domestic supply.Examination Papers 149 voltage and in this the number of turns in secondary coil is more than that in primary coil. NS the number of turns in secondary coil and f the magnetic flux linked with each turn.(iii) . N S > N p ).. therefore from (ii) n1 n2 n . The thickness of lens is t. The refractive n1 n1 n2 index of the material of lens is n2 t and it is placed in a medium of X X' refractive index n1 .=ç .(ii) = 1 v (v¢ .1÷ ç ÷ v u è n1 ø è R1 R2 ø æ 1 1 1 1 ö .. therefore from refraction formula at spherical surface n1 n2 n .e.n2 .n1 ) ç ÷ u è R1 R2 ø öæ 1 1 1 æ n2 1 ö ... From the refraction formula at spherical surface n2 n1 n2 .t) R2 For a thin lens t is negligible as compared to v ¢ . the ray is going from second medium (refractive index n2 ) to first medium (refractive index n1 ). O is a point object on the principal axis of the lens. The first refracting surface forms the image of O at I ¢ at a distance v¢ from P1 .1) ç ÷ v u è R1 R2 ø = n2 n1 . If the object O is at infinity. The distance of I from pole P2 of second surface is v. medium of lens) with respect to first medium.(i) = v¢ u R1 The image I ¢ acts as a virtual object for second surface and after refraction at second surface. The radii of curvature of the surfaces of the lens are R 1 and R 2 and their poles are P1 and P2 .. (a) Lens Maker’s Formula: Suppose L is a thin lens..e. The optical P1 C P2 I I' O v centre of lens is C and X ¢ X is u v' principal axis.. which is very small.(iv) is refractive index of second medium (i. the image will be formed at second focus i. if u = ¥ . the final image is formed at I.2 v (v¢ ) R2 Adding equations (i) and (iii).n1 .n1 .. The distance of O from pole P1 is u.(iii) =.t).e.. we get n1 v or i. where 1 n2 - n1 æ 1 1 ö = (n2 . The distance of virtual object ( I ¢ ) from pole P2 is (v¢ . For refraction at second surface.150 Xam idea Physics—XII L 29. v = f 2 = f .= ( 1 n2 . Examination Papers 151 Therefore from equation (iv) æ 1 1 1 1 ö . when the final image and the object both lie at infinite distance from the eye. then 1 n2 = n.1) ç ÷ f ¥ è R1 R2 ø i.(v) This is the formula of refraction for a thin lens. 1 100 Power of a lens.5 × 0 D = . It is defined as the reciprocal of focal length in metres... P2 = \ 1 1 D= = -5×0 D F2 . P1 = 1 1 D= = 2×0 D F1 0 × 50 Power of concave lens. This formula is called Lens-Maker’s formula..3 × 0 D OR Astronomical Telescope: Magnifying power of astronomical telescope in normal adjustment is defined as the ratio of the angle subtended at the eye by the final image to the angle subtended at the eye. Numerical: Power of convex lens.0 × 20 Power of combination of lenses in contact P = P1 + P2 = 2 × 0 D . If first medium is air and refractive index of material of lens be n. F0 a b .e. by the object directly. æ 1 1 1 ö = ( 1 n2 .(vi) (b) Power of a Lens: The power of a lens is its ability to deviate the rays towards its principal axis. P = diopters = diopters f ( in metres) f ( in cm) The SI unit for power of a lens is dioptre (D).1) ç ÷ f è R1 R2 ø ..= ( 1 n2 . therefore equation (v) may be written as æ 1 1 1 ö = (n .1) ç ÷ f è R1 R2 ø . 6 cm) on either side of the junction becomes free n p n . Put in equation (2).f e = focal length of eye lens. (a) Formation of Depletion Layer and Potential Barrier VB At the junction there is diffusion of charge carriers – + due to thermal agitation. The field Ei is directed from n-region to p-region.4 cm to 10 . m= tan a In DA ¢B ¢ C 2 . therefore. Compound Microscopes Objective lens is of large focal length and Both objective and eye lenses are of small focal eye lens is of small focal length. The diameter of objective is kept large to increase (i) intensity of image. Objective is of small aperture. tan b From equation (1). C 2 B ¢ = . Distinction Telescope 1. m = b a (1) As angles a and b are small. Some charge + + + carriers combine with opposite charges to + + + neutralise each other. (ii) resolving power of telescope. a » tan a and b » tan b. In DA ¢B ¢ C 1 .fe where C 1 B ¢ = f 0 = focal length of objective lens. or tan b = tan a = m= m= A ¢B ¢ C2 B¢ A ¢B ¢ C 1B¢ (2) A ¢B ¢ C 1 B ¢ C 1 B ¢ ´ = C 2 B ¢ A ¢B ¢ C 2 B ¢ f0 .152 Xam idea Physics—XII Magnifying power. Thus the layers ( » 10 . Negative sign of m indicates that final image is inverted. Objective is of very large aperature. Thus near the junction there Ei is an excess of positively charged ions in n-region p Depletion and an excess of negatively charged ions in layer p-region. 30. so that some of electrons + + + of n-region diffuse to p-region while some of holes of p-region diffuse into n-region. lengths but focal length of eye lens is larger than that of objective lens. 2. This field stops the further diffusion of charge carriers. This sets up a potential difference called potential barrier and hence an internal electric field Ei across the junctions. In next half cycle. Therefore current flows in diode II and there is no current in diode I. P1 S1 I n p RL B A Suppose during first half cycle of input ac S Output signal the terminal S1 is positive relative to S and S2 is negative relative to S. As soon as a hole (in P-region) combines with an electron. Thus for input a. This output current may be converted in fairly steady current by the use of suitable filters. then diode I is S2 P2 p II n forward biased and diode II is reverse biased. The circuit diagram for full wave rectifier using two junction diodes is shown in figure. The . the terminal S1 is negative relative to S and S2 is positive relative to S. signal the output current is a continuous series of unidirectional pulses. most electrons (about 98%) starting from emitter region cross the base region and reach the collector while only a few of them (about 2%) combine with an equal number of holes of base-region and get neutralised.C. The direction of current i2 due to diode II in load resistance is again from A to B. signal (b) Full Wave Rectifier: For full wave rectifier we use two junction diodes. As the base region is very thin. a covalent bond of crystal atom of base region breaks releasing an electron-hole pair. Then diode I is Output wave form reverse biased and diode II is of full wave 2T Time rectifier T forward biased. Input A. The direction of current i1 due to Input wave form diode I in load resistance R L is 2T Time T directed from A to B. thus the depletion layer of emitter-base junction is eliminated. The symbol of p-n junction diode is shown in Figure. OR E n-p-n C Input R1 B R2 Output – + VEE – + VCC n-p-n Comman base configuration Base Current and Collector Current: Under forward bias of emitter-base junction.c. Therefore current flows in diode I and not in diode II. the electrons in emitter and holes in base are compelled to move towards the junction.Examination Papers 153 from mobile charge carriers and hence is called the depletion layer. I c = I nc + I leakage . This causes a current in collector circuit. causing the emitter current. Base and Collector Currents: Applying Kirchhoff’s I law at terminal O . . Its direction in external circuit is from emitter to base. .154 Xam idea Physics—XII electron released is attracted by positive terminal of emitter battery VEE . Thus. This is also input dc voltage (VC ). This current is called the leakage current. i. This is the fundamental relation between currents in the bipolar transistor circuit. The hole released in the base region compensates the loss of hole neutralised by electrons. R L is the load resistance in output circuit. due to reverse – – + + O VEE VCC biasing of collector-base junction. the emitter current I E is the sum of base current I B and the collector current I C . The dc output voltage is taken across collector-emitter terminals. called the collector current. VBB and VCC are two dc supplies which bias base-emitter and emitter collecter junctions respectively. (ii) leakage current ( I leakage ) due to minority charge carriers. In addition to this the collector current is also due to flow of minority charge carriers under reverse bias of base-collector junction. e. Emitter Current: When electrons enter the emitter battery VEE from the base causing base current or electrons enter the collector battery VCC from the collector causing collector current. Relation between Emitter. we get I E = I B + IC That is. Depletion layer n E IE e– – IE + B IB I e– – + IC p n C IC B The electrons crossing the e– base and entering the collector. Operation: The circuit diagram of n-p-n transistor in CE configuration working as a switch is shown in fig. A transistor can be used to turn current ON or OFF rapidly in electrical circuits. collector current is formed of two components: (i) Current ( I nc ) due to flow of electrons (majority charge carriers) moving from emitter to collector. are attracted towards the positive terminal of collector battery VCC . an equal number of electrons enter from emitter battery VEE to emitter. giving rise to a feeble base current ( I B ). Let VBB be the input supply voltage. In the process an equal number of electrons leave the negative terminal of battery VCC and enter the positive terminal of battery VEE . Transistor as a Switch A switch is a device which can turn ON and OFF current is an electrical circuit. The process continues. Examination Papers 155 C RB IB B E + Vi – VBB 1 VBE 2 VCE IC + RL VO – VCC Applying Kirchhoff’s second law to input and output meshes (1) and (2). With further increase in Vi . the increase in I C is linear and so decrease in output voltage V0 is linear. Beyond Vi = 1 V. Hence. (i) between cut off state and active state and (ii) Vi 0. so above equations take the form and …(iv)Let us see the change in V0 ( = VCE ) = VCC . Thus. from (iv). Therefore. In case of Si transistor.6 V 1. there is no collector current ( I C = 0). so transistor will be in cut off state. the change in collector current and hence in output voltage V0 is non-linear and the transistor goes into saturation. I C begins to flow and increase with increase of Vi . we get the graph as shown in fig. V0 = VCC . when Vi is less than 0 × 6 V. V0 decreases upto Vi = 1 V. thus showing that the transitions (i) from cut off to active state and from active to saturation state are not sharply defined. Cut off Active region region V0 Saturation region AV .0 V between active state and saturation state. we get VBB = I B R B + VBE and VCC = I C R L + VCE Vi = VBE + I B R B …(iii) …(i) …(ii) We have VBB = Vi and VCE = V0 . the barrier voltage across base-emitter junction is 0 × 6 V.] The curve shows that there are non-linear regions. the output voltage further decrease towards zero (though it never becomes zero). When Vi becomes greater than 0 × 6 V.I C R L V0 due to a change in Vi . If we plot V0 versus Vi . [This characteristics curve is also called transfer characteristic curve of base biased transistor. from (iv) with I C = 0. For particle to move undeviated. Parallel rays (b) Plane wavefront 11. it is said to be ‘switched ON’. The switching circuits are designed in such a way that the transistor does not remain in active state. Name of gate formed is NOR gate symbol. it is said to be ‘switched off’ but when it is conducting ( I C is not zero). When transistor is non-conducting ( I C = 0). so wavefront is plane (figure). so the transistor is ‘switched OFF’ and if it is high enough to derive the transistor into saturation ( I C is high and so V0 ( = VCC . qE +q v´ B =0 Þ ® ® ® ® E=.156 Xam idea Physics—XII Now we are in the position to explain the action of transistor as a switch. The rays coming out of the convex lens. CBSE (Foreign) SET–II 3.I C R L ) is low.v ´ B ® ® . When the current in a coil is changed. Thus we can say low input switches the transistor is OFF state and high input switches it ON. V0 is high ( I C = 0 and V0 = VCC ). Truth Table Inputs A Y B Output B 0 0 1 1 Y 1 0 0 0 A 0 1 0 1 12. a back emf is induced in the same coil. very near to zero. This phenomenon is called self-induction. when point source is at focus. 7. so the transistor is ‘switched ON’. are parallel. As long as input voltage Vi is low and unable to overcome the barrier voltage of the emitter base junction. j $) ´ ( .. (a) (i) Amplitude Modulation In amplitude modulation. E ma = m Ec . When amplitude of information increases.80 ´ 10 -12 = 1 ´ 10 -12 Solving (1) and (2) 16.e. h 1 l= µ 2mqV V If potential difference V is doubled. (ii) Modulation Index: The modulation index of an amplitude modulated wave is defined as the ratio of the amplitude of modulating signal (Em ) to the amplitude of carrier wave (Ec ) i. ® ® ® $ B =-Bk $) = . (b) de-Broglie wavelength. (b) Electromagnets are made of soft iron which is characterised by high retentivity and low coercivity. the amplitude of modulated wave increases and vice versa.4q1 q2 = 81 ´ 10 -12 .q2 ) = ( q1 + q2 ) 2 . 23.Examination Papers 157 $.( -vi i. the amplitude of modulated (carrier) wave varies in accordance with amplitude of information (signal) wave.. q1 = 5 mC. In microwave ovens to heat the food. magnitude of electric field is vB and its direction is along positive Y-axis.e. Given v = . (a) The frequency of incident radiation was kept constant. amplitude modulation index. Microwaves: Uses: 1. In this case the amplitude of modulated wave is not constant. the de-Broglie 1 wavelength is decreased to times. q2 = 4 mC (ii) (a) Permanent magnets are made of steel which is characterised by high retentivity and high coercivity. 2.Bk $) = vBj $ E = .vB ( . After OR q1 + q2 = 9 ´ 10 -6 C 1 q1 q2 =2 4pe 0 r 2 Þ q1 q2 = 2 ´ ( 0 × 30) 2 9 ´ 10 9 n0 I I3 I2 I1 I3 > I2 > I1 …(1) = 20 ´ 10 -12 …(2) ( q1 . (i) A diamagnetic substance is attracted towards a region of weaker magnetic field. 2 14. In Radar for aircraft navigation. 26.v i \ 13. 6931 T = 0 . (b) (i) Given T = 20 s Decay constant. By introducing the metal plate between the plates of charged capacitor.158 Xam idea Physics—XII For modulated wave. then e0 A C= 1ö d-t æ ç1 . C = 0 ( d .t). Þ Þ æ1 ö ç ÷ è 2ø 3 0 . effective separation between plates is decreased from d to ( d . . (iii) The various information signals transmitted at low frequency get mixed and hence can not be distinguished. l = (ii) Fraction decayed = Fraction remained \ 7 8 N 7 1 =1. 27.t) Obviously.= N0 8 8 N æ 1 ön = ç ÷ gives N0 è 2 ø 1 æ 1 ön =ç ÷ 8 è 2ø Number of half lives. E . 6931 20 s -1 = 0 × 0346 s -1 = 0.0346 s–1 1 ön =æ ç ÷ è 2ø t = 3T = 3 ´ 20 s = 60 s CBSE (Foreign) SET–III 1. the capacitance of capacitor increases. (ii) The power radiated at audio frequency is quite small. Reason: It t is thickness of metal plate.Emin ma = max Emax + Emin (b) The modulation is needed due to (i) Transmission of audiofrequency electrical signals need long impracticable antenna. n = 3.÷ è Kø e A For metal plate K = ¥ . hence transmission is quite lossy. (a) Half-life period: The half-life period of a radioactive substance is defined as the time in which one-half of the radioactive substance is disintegrated. they will tend to move away from one another.i Mutual Inductance: The mutual inductance of two coils is defined as the magnetic flux linked with the secondary coil when the current in primary coil is 1 ampere. (ii) to detect fracture in bones. the wavefront is spherical. Wavefront: The locus of particles of a medium vibrating in the same phase is called a wavefront. . C = C 1 + C 2 = 1600 pF 1 Energy stored = (C 1 + C 2 ) V 2 2 1 = ´ 1600 ´ 10 -12 ´ (50) 2 J 2 = 2 ´ 10 -6 J 13. The distant wavefront is plane. Electromagnetic radiation produced by a Klystron or a Magnetron valve is microwave. P1 – – +Q – – – P2 + + + + + –Q –Q +Q 10. P1 P2 – – – – – – – – – + + + + + + + + + 9. When plates are released. From a point source. 6. Common potential. +Q –Q (ii) The field pattern is shown in fig. V = = C 1V1 + 0 C1 + C2 ( 800 pF) ´ 100 800 + 800 V = 50 V Net capacitance. stone in gall bladder and kidney etc.Examination Papers 159 4. (i) Charges induced on outer surfaces of P1 and P2 are -Q and +Q respectively. X-rays Uses: (i) To study crystal structure. plate P1 moving towards +Q and P2 towards -Q due to attraction. while for a line source the wavefront is cylindrical. 8. Frequency n 1 is highest. Intensity of incident radiations was kept constant.160 14. Xam idea Physics—XII NAND gate. (ii) X is diamagnetic and Y is ferromagnetic. (i) Angle of dip decreases from 90° to 0° as one goes from magnetic pole to magnetic equator of earth. 24. 0 × 6931 -1 26. Truth Table A A Y B 0 0 1 1 Y 1 1 1 0 0 1 0 1 B 16. (b) (i) l= s » 0 × 014 s -1 50 (ii) N æ 1 ön =ç ÷ N0 è 2 ø n= 2 Þ Þ n æ1 ö = æ1 ö ç ÷ ç ÷ è 4ø è 2ø t = 2T = 2 ´ 50 s = 100 s . 1 h = 6 × 626 ´ 10 -34 Js e = 1× 602 ´ 10 -19 C m 0 = 4p ´ 10 -7 TmA -1 1 = 9 × 109 Nm2C– 2 4pe o Boltzmann’s constant k = 1× 381 ´ 10 -23 J K -1 Avogadro’s number N A = 6 × 022 ´ 10 23 /mole Mass of neutron m n = 1× 2 ´ 10 -27 kg Mass of electron m e = 9 ×1´ 10 -31 kg Radius of earth = 6400 km Maximum marks: 70 CBSE (Delhi) SET–I 1. In which orientation. . questions 9 to 18 carry two marks each. (d) Use of calculators is not permitted. Questions 1 to 8 carry one mark each.CBSE EXAMINATION PAPERS DELHI–2010 Time allowed: 3 hours General Instructions: (a) All questions are compulsory.q and +3q. a dipole placed in a uniform electric fields is in (i) stable. (c) There is no overall choice. You have to attempt only one of the given choices in such questions. (b) There are 30 questions in total. A plot of magnetic flux ( f) versus current ( I ) is shown in the figure for A two inductors A and B. Which part of electromagnetic spectrum has largest penetrating power? 3. (e) You may use the following values of physical constants wherever necessary: c = 3 ´ 108 ms . Which of the two has larger value of self inductance? f B 4. an internal choice has been provided in one question of two marks. questions 19 to 27 carry three marks each and questions 28 to 30 carry five marks each. (ii) unstable equilibrium? 2. What is the electric flux due to this configuration through the surface ‘S’? I +2q +3q –q S . However. Two charges +2q and -q are enclosed within a surface ‘S’. +2q. Figure shows three point charges. one question of three marks and all three questions of five marks each. Arrange the following electromagnetic radiations in ascending order of their frequencies: (i) Microwave (ii) Radio wave (iii) X-rays (iv) Gamma rays Write two uses of any one of these. How is photodiode used to measure light intensity? Lamp C 12. How does the brightness of the lamp change on reducing the (i) capacitance. Draw the circuit diagram of an illuminated photodiode in reverse bias. A glass lens of refractive index 1× 45 disappears when immersed in a liquid. (a) The bluish colour predominates in clear sky. Source 13. 10. Calculate the energy Q released per fission in MeV. 9. State reason to explain these observations.6 MeV is split into two fragments Y and Z of mass numbers 110 and 130.5 MeV per nucleon. find the refractive index of the material of the lens. . (ii) paramagnetic substance is placed in an external magnetic field. The binding energy of nucleons in Y and Z is 8. A wire of resistance 8 R is bent in the form of a circle. An electron is accelerated through a potential difference of 100 volts. What is the A B effective resistance between the ends of a diameter AB ? 8. What is the value of refractive index of the liquid? 6. (b) Violet colour is seen at the bottom of the spectrum when white light is dispersed by a prism. and (ii) the frequency? Justify your answer.162 Xam idea Physics—XII 5. An electric lamp having coil of negligible inductance connected in series with a capacitor and an AC source is glowing with certain brightness. State the conditions for the phenomenon of total internal reflection to occur. If focal length of the lens is 12 cm. What is the ratio of radii of the orbits corresponding to first excited state and ground state in a hydrogen atom? 7. What is the de-Broglie wavelength associated with it? To which part of the electromagnetic spectrum does this value of wavelength correspond? 16. Explain the function of a repeater in a communication system. A heavy nucleus X of mass number 240 and binding energy per nucleon 7. (ii) Why is core of an electromagnet made of ferromagnetic materials? OR Draw magnetic field lines when a (i) diamagnetic. Which magnetic property distinguishes this behaviour of the field lines due to the two substances? 11. (i) Write two characteristics of a material used for making permanent magnets. 15. The radii of curvature of the faces of a double convex lens are 10 cm and 15 cm. 17. 14. What is space wave propagation? Give two examples of communication system which use space wave mode. . I l Give reasons to explain that the loop will experience a net force but no torque. 20. how a potentiometer is used to determine the internal resistance of a given cell. Describe briefly. Plot a graph showing the variation of stopping potential with the frequency of incident radiation for two different photosensitive materials having work functions W1 and W2 (W1 > W2 ). 23. After sometime the battery is disconnected and a dielectric slab with its thickness equal to the plate separation is inserted between the plates. Write two reasons why a galvanometer can not be used as such to measure current in a given circuit. (a) Depict the equipotential surfaces for a system of two identical positive point charges placed a distance ‘d’ apart. 26. (ii) Plot a graph showing variation of activity of a given radioactive sample with time.Examination Papers 163 18.I. A parallel plate capacitor is charged by a battery. what are their corresponding values for D ? 25. Apply Ampere’s circuital law to calculate the magnetic field at a point ‘r’ in the region for (i) r < a and (ii) r > a. with the help of a circuit diagram. (b) Deduce the expression for the potential energy of a system of two point charges q1 and q 2 ® ® ® brought from infinity to the points r1 and r 2 respectively in the presence of external electric ® field E . 24. The current is uniformly distributed across the cross-section. Write the expression for the magnetic moment ( m) due to a planar square loop of side ‘l’ carrying a steady current I in a vector form. Calculate the maximum distance upto which the signal transmitted from the tower can be received. 22. (iii) The sequence of stepwise decay of a radioactive nucleus is a b- D ¾¾® D1 ¾¾® D2 If the atomic number and mass number of D2 are 71 and 176 respectively. A long straight wire of a circular cross-section of radius ‘a’ carries a steady current ‘I’. I1 21. Write the expression for Brewster angle in terms of the refractive index of denser medium. Write the principle of working of a potentiometer. What is an unpolarized light? Explain with the help of suitable ray diagram how an unpolarized light can be polarized by reflection from a transparent medium. A TV tower is 80 tall. How will (i) the capacitance of the capacitor. unit. (i) Define ‘activity’ of a radioactive material and write its S. Name any two factors on which the current sensitivity of a galvanometer depends. l In the given figure this loop is placed in a horizontal plane near a long straight conductor carrying a steady current I 1 at a distance l as shown. Write the expression for this force acting on the loop. (ii) potential difference between the plates and (iii) the energy stored in the capacitor be affected? Justify your answer in each case. OR State the underlying principle of working of a moving coil galvanometer. On what factors does the (i) slope and (ii) intercept of the lines depend? 19. OR Draw a ray diagram to show the working of a compound microscope. with the help of a suitable diagram. (a) Explain the formation of depletion layer and potential barrier in a p-n junction. State the conditions under which it is (i) maximum and (ii) minimum.164 Xam idea Physics—XII 27. Draw a plot of intensity distribution and explain clearly why the secondary maxima become weaker with increasing order ( n) of the secondary maxima. In a compound microscope. If a resistance of 12 W is connected in parallel with S. with the help of a labelled diagram. Deduce an expression for the total magnification when the final image is formed at the near point. If the eye piece has a focal length of 5 cm and the final image is formed at the near point. Device ‘X’ Input A Y B Output (c) Identify the logic gate represented by the circuit as shown and write its truth table. Show diagrammatically how an alternating emf is generated by a loop of wire rotating in a magnetic field. Derive the expression for the instantaneous current J and its phase relationship to the applied voltage. the null point occurs at 50 × 0 cm from A. Define ‘power factor’. how this principle is used to obtain the diffraction pattern by a single slit. Show. State Huygen’s principle. Write the expression for the instantaneous value of the emf induced in the rotating loop. Determine the values of R and S. . In a meter bridge. 29. the null point is found at a distance of 40 cm from A. Obtain the condition for resonance to occur. OR A series LCR circuit is connected to an ac source having voltage v = v m sin wt. Describe briefly. Name the device and draw its circuit diagram. 12W R G A S B 28. 30. the basic elements of an AC generator. an object is placed at a distance of 1× 5 cm from the objective of focal length 1× 25 cm. the input waveform is converted into the output wave from a device ‘X’. State its underlying principle. (b) In the figure given below. estimate the magnifying power of the microscope. What is can be done to have larger deflection in the galvanometer with the same battery? (ii) State the related law. will take place in: (i) charge on plates? (ii) electric field intensity between the plates? (iii) capacitance of the capacitor? Justify your answer in each case. A parallel plate capacitor. (i) Why does the Sun appear reddish at sunset or sunrise? (ii) For which colour the refractive index of prism material is maximum and minimum? 17. The radius of innermost electron orbit of a hydrogen atom is 5 × 3 ´ 10 -11 m. transmission? What is common between these waves and light waves? 2 11. What is the range of frequencies used for T.Examination Papers 165 OR (a) With the help of the circuit diagram explain the working principle of a transistor amplifier as an oscillator. The battery used to charge it remains connected. a semiconductor and an insulator on the basis of energy band diagrams. A dielectric slab of thickness d and dielectric constant k is now placed between the plates. CBSE (Delhi) SET–II Questions uncommon to Set–I 3. What is the maximum distance between them for satisfactory communication in line of sight mode? . What change. What is the de-Broglie wavelength associated with it? To which part of the electromagnetic spectrum does this wavelength correspond? 19.V. (i) When primary coil P is moved towards secondary coil S (as shown in the figure below) the galvanometer shows momentary deflection. is charged to a potential difference V. each with plate area A and separation d. Calculate the 3 refractive index of lens material. What is the radius of orbit in the second excited state? 6. (b) Distinguish between a conductor. if any. An electron is accelerated through a potential difference of 144 volts. 20. 14. S P – G V + 10. (i) Why is communication using line of sight mode limited to a frequencies above 40 MHz? (ii) A transmitting antenna at the top of a tower has a height 32 m and the height of the receiving antenna is 50 m. A biconvex lens has a focal length times the radius of curvature of either surface. Which part of electromagnetic spectrum is absorbed from sunlight by ozone layer? 9. B A.0 cm from A.166 Xam idea Physics—XII 22. An electron is accelerated through a potential difference of 64 volts. l2 and S. (b) Bulb gets dimmer if the coil Q is moved towards left. the null point is found at a distance of 60. In a meter bridge. (a) The bulb ‘B’ lights. whose focal length is 0 × 3 m and the refractive index of the material of the lens is 1× 5. If now a resistance of X is connected in parallel with S. In a metre bridge. 7. the null point occurs at l2 cm. 15. Write the expression for Bohr’s radius in hydrogen atom. Find the radius of curvature of the convex surface of a plano-convex lens. 20. 14.C Source Q P 13. . (ii) Give the formula that can be used to determine refractive index of material of a prism in minimum deviation condition. the null point is found at a distance of l1 cm from A. Obtain a formula for X in terms of l1 . 11. 5. R G A B S CBSE (Delhi) SET–III Questions uncommon to Set–I Which part of electromagnetic spectrum is used in radar systems? Calculate the speed of light in a medium whose critical angle is 30°. Give reason to explain the following observations: 4. Determine the values of R and S. What is the range of frequencies used in satellite communication? What is common between these waves and light waves? 12. What is the de-Broglie wavelength associated with it? To which part of the electromagnetic spectrum does this value of wavelength correspond? (i) Out of blue and red light which is deviated more by a prism? Give reason. If now a resistance of 5 W is connected in series with S. the null point occurs at 50 cm. A coil Q is connected to low voltage bulb B and placed near another coil P is shown in the figure. If the distance between the plates is doubled. Electric flux. r n = e0 h2 n2 2 µ n2 pme For I excited state. (ii) In unstable equilibrium P. The capacitor is then disconnected from the source. is maximum. q = 0).. Inductor A has larger value of self-inductance. so L A > L B i. f = ´ (net charge enclosed by surface S) e0 q 1 = ´ ( 2q . state with reason how the following will change: (i) electric field between the plates. (i) In stable equilibrium the dipole moment is parallel to the direction of electric field (i. refractive index of liquid = refractive index of lens = 1× 45 6. so q = p so dipole moment is antiparallel to electric field. (ii) capacitance. 3. A parallel plate capacitor is charged to a potential difference V by a d..Examination Papers X 167 R G A S B 27. and (iii) energy stored in the capacitor. For disappearance of glass lens in liquid. f = L I For same current fA > fB . source. Solutions CBSE (Delhi) Set-I 1.e. g-rays have largest penetrating power. n = 2 For ground state. 1 4.q) = e0 e0 5.e.E. n = 1 r2 4 = r1 1 \ .c. 2. 9. (ii) Angle of incidence must be greater than critical angle (C). electron-hole pairs are created at the junction. It is a reversed biased p-n junction. 10. amplifiers and retransmits it to the receiver. When p-n junction is reversed biased with no current. illuminated by radiation. A repeater is a combination of a receiver and a transmitter. 11. so when current is passed in ferromagnetic material it gains sufficient magnesium immediately on passing a current through it.168 Xam idea Physics—XII 7. Repeaters are used to increase the range of communication of signals. A typical example of repeater station is a communication satellite. (i) For permanent magnet the material must have high retentivity and high coercivity (e. so effective resistance across ends of diameter AB is 4R ´ 4R = = 2R 4R + 4R 8. Light p n m A p Symbol – + (a) (b) n Photodiode . Two parts each of resistance 4R are connected in parallel. When the junction is illuminated with light. due to which additional current begins to flow across the junction. sometimes with a change in carrier frequency.. (ii) Ferromagnetic material has high retentivity. Conditions for total internal reflection are: (i) Light must travel from denser to rarer medium.g. steel). the current is solely due to minority charge carriers. A repeater picks up the signal from the transmitter. OR X Y (b) Paramagentic subsistence (a) Diamagnetic subsistence The magnetic susceptibility of diamagnetic substance is small and negative but that of paramagnetic substance is small and positive. a very small reverse saturated current flows across the junction called the dark current. Their frequency range is 3 GHz to 300 Ghz. ultraviolet rays.1) ç + ÷ 12 è 10 15 ø 5 = ( n . capacitive reactance X C = 1 wC C V decreases. they can penetrate thick iron blocks. microwaves. magnetrons and gunn diodes. Uses of Electromagnetic Spectrum (i) g-rays are highly penetrating. (ii) X-rays are used in medical diagnostics to detect fractures in bones.1) ´ 30 30 1 30 or n = 1 + Þ n -1 = ´ Þ n = 1 + 0 × 5 = 1× 5 5 12 60 15. R 2 = . g-rays are produced in nuclear reactions. Given R1 = 10 cm. 13.1) ç çR -R ÷ f 2 ø è 1 1 1ö æ1 Þ = ( n . (iii) Radiowaves are used for broadcasting programmes to distant places. X-rays and gamma rays.15 cm. They are used in radar systems used in air craft navigation and microwave users in houses. capacitive reactance X C = increases. hence impedance of circuit. they are used to destroy cancer cells. f = 12 cm Refractive index n = ? Lens-maker’s formula is æ 1 1 1 ö ÷ = ( n . (i) When capacitance is reduced. As a result brightness of bulb is reduced. (ii) When frequency in decreases. Due to high energy. They are used in engineering to check flaws in bridges. namely. presence of stone in gallbladder and kidney. they are used to produce nuclear reactions.Examination Papers 169 Lamp 12. so current decreases. In physics X-rays are used to study crystal structure. As a result the brightness Z Source 1 increases and hence 2pnC impedance of circuit increases. klystrons. According to frequency range. Z = R 2 + X C2 increases and so current I = of the bulb is reduced. l = h 2meV = 6 × 63 ´ 10 -34 2 ´ 9 ×1 ´ 10 -31 ´ 1× 6 ´ 10 -19 ´ 100 m . In medicine. tuberculosis of lungs. 14. In ascending power of frequencies: radiowaves. they are divided into following groups (1) Medium frequency band or medium waves 0·3 to 3 MHz (2) Short waves or short frequency band 3 MHz — 30 MHz (3) Very high frequency (VHF) band 30 MHz to 300 MHz (4) Ultrahigh frequency (UHF) band 300 MHz to 3000 MHz (iv) Microwaves are produced by special vacuum tubes. The refractive index of proton is maximum for violet and minimum for red. so it is scattered most. 16. Energy stored by d K Q2 the capacitor. The graph of stopping potential Vs and frequency (n) for two metals 1 and 2 is shown in fig. Q V Reason: V = .M ´ c 2 = 8 × 5 ´ 240 MeV – 7 × 6 ´ 240 MeV = (8 × 5 . (b) While light consists of infinite wavelengths starting from 400 nm (violet) to 750 nm (red). the potential drop across. . C increases to K times... Now rheostat ( Rh) is so adjusted that on touching the jockey J at ends A and B of 17. As Q = constant.any portion of wire is directly proportional to the length of that portion i. so prism separates constituent colours of white light and causes maximum deviation for violet colour. 2 (w2/e) 19. V µl Method: (i) Initially key K is closed and a potential difference is applied across the wire AB. Principle: If constant current is flowing through a wire of uniform area of cross-section at constant temperature. e. V ¢ = C K V 1 (iii) As E = and V is decreased. electric field decreases to times. l ø è In visible light blue colour has minimum wavelength. Vs h (i) Slope of graph tan q = and depends on h and e.e. C is increased. 2 1 18. Q = ( M y + M z ) c 2 .170 Xam idea Physics—XII = 1× 227 ´ 10 -10 m = 1× 227 Å This wavelength corresponds to X-ray region of em spectrum. q q e n (ii) Intersect of lines depend on the work function. therefore. that is why bluish colour predominates in a clear sky.7 × 6) ´ 240 MeV = 0 × 9 ´ 240 MeV = 216 MeV æ 1 ö (a) The intensity of scattered light varies inversely as fourth power of wavelength ç ç i. K 20. U = . and so energy stored by capacitor 2C 1 decreases to times. (i) The capacitance of capacitor increases to K times Ke 0 A (since C = µ K) d 1 (w1/e) 1 (ii) The potential difference between the plates becomes K times. That is why violet colour is seen at the bottom of spectrum when white light is dispersed through a prism. I µ 4 ÷ ÷. Q same. ® ® 2 ® m =I A For square loop A = l \ l l m $ m = I l2 n $ is unit vector normal to loop. Let this position on wire be P2 and AP2 = l2 ..(i) (iii) Now a suitable resistance R is taken in the resistance box and key K1 is closed. where n Magnetic field due to current carrying wire at the location of loop is directed downward perpendicular to plane of loop. In this situation the cell is in closed circuit. emf e = kl1 . Again. Suppose that in this position the + – Rh potential gradient on the wire is k.. C In this situation the cell is in R () open circuit.(ii) V = kl2 e l1 Dividing (i) by (ii).. the deflection in B1 K the galvanometer is on both sides. the position of null point is obtained on the wire by using jockey J. we get = V l2 æ l1 ö æ e ö r =ç -1 ÷ R = ç -1 ÷ ç ÷R è V ø è l2 ø From this formula r may be calculated. therefore the terminal potential difference (V ) of cell will be equal to the potential difference across external resistance R. i. Let this e + – HR position be P1 and AP1 = l1 . (ii) Now key K1 is kept open and the P2 P1 position of null deflection is + – J B A obtained by sliding and pressing G the jockey on the wire.. I1 l P I l z S R Q x y . therefore the K1 terminal potential difference will be equal to the emf of cell. . 21. Magnetic moment due to a planar square loop of side l carrying current I is \ Internal resistance of cell.e..e.Examination Papers 171 potentiometer wire. i.. l $ j´ 0 1 $ j. 2p $) $) + m 0 I 1 ( -k = m 0 l ( -i 2p ( 2l) ® ® ® ® ® From on side RS.172 Xam idea Physics—XII Force on QR and SP are equal and opposite. $ FPQ = I l ´ B1 li $) $ ´ u 0 I 1 ( -k = Ili 2pl m II $ = 0 1 j . dr = r . ® ® ® ® Work done in bringing charge q1 from µ at location r1 is W1 = q1V (r1 ) Work done in bringing q 2 against the electric field at location r 2 is W2 = q 2 V (r 2 ) Work done on q 2 against the electric field due to q1 is ® ® ® r r12 q1 q 2 1 $ W3 = ò 12 F12 . 22. so net force on these sides is zero. ® ® Suppose q1 and q 2 are two charges brought from infinity at locations r1 and r 2 respectively in an external electric field. FRS ® ® m 0 II 1 $ j. 4p ® ® ® m II Torque t = r ´ F = . 2p æ -m II ö + ( -21$ j) ´ ç 0 1 $ j ÷ = zero è 4p ø Net force F = FPQ ´ FRS = That is loop experiences a repulsive force but no torque. (b) Potential energy of a system of two charges in an external electric field. ® ® ® ® ® Let V (r1 ) and V (r 2 ) be the potentials at positions r1 and r 2 due to external electric field E . In this case work is done in bringing charges q 1 and q 2 against their own electric fields and external electric fields. (a) Equipotential surfaces due to two identical charges is shown in fig. ( dr ) ¥ 4pe 0 ò¥ r2 r 1 1 =q1 q 2 ò 12 dr ¥ 4pe 0 r2 . Force no side PQ. 2D . called polarising angle. . 24. A = 180 (ii) Z . Brewster’s law: n = tan i p .1 = 71 Þ Atomic number of D. then reflected light is plane polarised with its electric vector perpendicular to the plane of incidence when the refracted and reflected rays make a right angle with each other. (i) The activity of a radioactive substance is the rate of decay or the number of disintegrations per second of the substance. (ii) A0 A A0 2 A0 4 O T 2T t (iii) The sequence is represented as A Z a D ¾® A-4 Z . Unpolarised light: The light having vibrations of electric field vector in all possible directions perpendicular to the direction of wave propagation is called the ordinary (or unpolarised) light.Examination Papers 173 ér -1 ù 1 =q1 q 2 ê ú 4pe 0 ê ú¥ ë -1 û ® ® r /2 = 1 q1 q 2 4pe 0 r12 where r12 = | r 2 .1 D2 (i) Given A . b- ¾® A-4 Z .4 = 176 Þ Mass number of D. Unpolarised light ip r' Polarised light Partially polarised light n r If unpolarised light falls on a transparent surface of refractive index n at a certain angle i p . Z = 72.r1 | 23.r1 | \ Potential energy of system = Work done in assembling the configuration ® ® q1 q 2 1 U = W1 + W2 + W3 = q1V (r1 ) + q 2 V (r 2 ) + 4pe 0 ® ® | r 2 . Suppose that the current is uniformly distributed over whole cross-section of the wire. The cross-section of wire is circular. Current per unit cross-sectional area. passing through internal point Q.174 Xam idea Physics—XII 25. I …(i) i= pa 2 Magnetic Field at External Points: We consider a circular path of radius r ( > a) passing through external point P concentric with circular cross-section of wire. This shows that for external points the current flowing in wire may be supposed to be concerned at the axis of cylinder. In this case the assumed circular path encloses only a path of current carrying circular cross-section of the wire. So ® line integral of magnetic field B around the circular path ® ® ò B × dl = ò B dl cos 0° = B 2pr Current enclosed by path = Total current on circular cross-section of cylinder = I By Ampere’s circuital law ® ® ò B × dl = m ´ current enclosed by path Þ Þ B 2pr = m 0 ´ I m I B= 0 2pr This expression is same as the magnetic field due to a long current carrying straight wire. By symmetry the strength of magnetic field at every point of circular path is same and the direction of magnetic field is tangential to path at every point. concentric with circular cross-section of the wire. Magnetic Field due to a straight thick wire of uniform cross-section: Consider an infinitely long cylinderical wire of radius a. carrying current I. \ Current enclosed by path = current per unit cross-section ´ cross section of assumed circular path æ I = i ´ pr 2 = ç ç 2 è pa \ By Ampere’s circuital law ® ® ö Ir 2 2 ÷ ´ p r = ÷ a2 ø ò B × dl = m 0 ´ current closed by path . Magnetic Field at Internal Points: Consider a circular path of radius r ( < a). torque t acts on the coil. given by t = NIAB sin q where q is the angle between the normal to plane of coil and the magnetic field of strength B. then in every position of coil the plane of the coil. so magnetic field at surface of wire m I (maximum value) Bs = 0 2pa The variation of magnetic field strength ( B) with distance (r ) from the axis of wire for internal and external points is shown in figure. At surface of cylinder r = a. m I m 0I m I B outside = 0 = = 0 a ö 3pa 2pr æ 2p ç a + ÷ 2ø è m 0 Ir m 0 I ( a / 2) m 0 I Binside = = = 4pa 2pa 2 2pa 2 \ B outside 4 = Binside 3 m 0I × 2pa Maximum value of magnetic field is at the surface given by B outside = OR Principle: When current ( I ) is passed in the coil. magnetic field strength inside the current carrying wire is directly proportional to distance of the point from the axis of wire. t = NIAB A galvanometer cannot be used as such to measure current due to following two reasons. (ii) A galvanometer is a very sensitive device. as in the case of cylindrical pole pieces and soft iron core. When the magnetic field is radial. hence if connected as such it will not measure current of the order of ampere. so it will increase the resistance of circuit and hence change the value of current in the circuit. . so that q = 90° and sin 90° = 1 Deflecting torque. (iii) Strength of magnetic poles (B): It increases with increase of strength of poles. N is the number of turns in a coil. it gives a full scale deflection for the current of the order of microampere. is parallel to the magnetic field lines. (i) A galvanometer has a finite large resistance and is connected in series in the circuit. (ii) Area of coil A: It increases with increase of area of coil.Examination Papers 175 Þ B × 2pr = m 0 ´ B= m 0 Ir 2pa 2 Ir 2 a2 Clearly. Current sensitivity of galvanometer depends on (i) Number of turns N: It increases with increase of number of turns. In first case l1 = 40 cm l1 R = S 100 . The drum or ring serves the two purposes: (i) It serves as a support to coils and (ii) It increases the magnetic field due to air core being replaced by an iron core.50 3 From (i) S= R 2 Substituting this value in (ii). while in the case of large power dynamo. so 12S …(ii) S¢ = 12 + S R 50 …(iii) \ = = 1 Þ S¢ = R S ¢ 100 . Maximum coverage distance d = 2R e h = 2 ´ 6400 ´ 10 3 ´ 80 = 32 ´ 10 3 m = 32 km 27. (ii) Armature: It consists of a large number of turns of insulated wire in the soft iron drum or ring. S =6W 28. 26. Space wave propagation is a straight line propagation of electromagnetic wave from transmitting antenna to recieving antenna both installed in the ground. S ¢ = R 18R 3 \ =R Þ 18 = 12 + R 3 2 12 + R 2 3 or Þ R =6 R =4W 2 3 \ S = R =6W 2 R =4W . It can revolve round an axle between the two poles of the field magnet.176 Xam idea Physics—XII (iv) Torsional rigidity of suspension: It increases with decrease of torsional rigidity of suspension. the magnetic field is generated by a permanent magnet. .l1 Þ R 40 2 = = S 60 3 …(i) In second case when S and 12 W are in parallel balancing length l2 = 50 cm. the magnetic field is produced by an electromagnet. we get æ3 ö 12 ´ ç R ÷ è 2 ø = 18R S¢ = 3 æ3 ö 12 + ç R ÷ 12 + R 2 è2 ø Also from equation (iii). AC generator consists of the four main parts: (i) Field Magnet: It produces the magnetic field. In the case of a low power dynamo. voltage across inductance L is VL and voltage across capacitance C is VC . a) On account of being in series. Principle: When the armature coil is rotated in the strong magnetic field.VL ) are mutually perpendicular and the phase difference between them is 90°. the voltage VL will lead the current by angle 90° while the voltage VC will lag behind the current by angle 90° (fig. The voltage VR and current i are in the same phase. the current (i ) flowing through all of them is the same.VL ) will also be V. therefore their resultant potential difference = VC .. From fig. the Armature coil direction of induced emf and current changes b in the external circuit after each half c revolution.VL ) 2 . (fig.VL ) 2 Þ V = VR 2 + (VC . b). inductance L and capacitance C are connected in series and an alternating source of voltage V = V0 sin wt is applied across it.VL (if VC > VC ). V 2 = VR 2 + (VC . These rings are fixed to the shaft which rotates the armature coil so that the rings also rotate along with the armature. Clearly VC and VL are in opposite directions. If N is the number of turns in coil.Examination Papers 177 (iii) Slip Rings: The slip rings R1 and R 2 are the two metal rings to which the ends of armature coil are connected. its direction being given by Fleming’s right hand rule. R VL L VR C VC V=V0 sin wt (a) Suppose the voltage across resistance R is VR . then induced emf df d e== {NBA (cos 2p f t )} dt dt = 2p NBA f sin 2p f t R1 Slip rings R2 d B1 Brushes Load RL B2 OR Suppose resistance R. w the direction of current is reversed. The direction of current remains unchanged during the first half turn of armature.. the resultant of VR and (VC . the magnetic flux linked with the coil changes and the current is induced in the coil. During the second half revolution. (iv) Brushes: These are two flexible metal plates or carbon rods ( B1 and B 2 ) which are fixed and constantly touch the revolving rings. As applied voltage across the circuit is V. A area of coil and B the a magnetic induction.(i) . Thus. Thus VR and (VC . f the N B S Field magnet frequency of rotation. The output current in external load R L is taken through these brushes. cos f = .e. For minimum power cos f = 0 or R = 0 i.e.XL tan f= C R For resonance f= 0 Þ XC . (ii) The new disturbances from each point spread out in all directions with the velocity of light and are called the secondary wavelets.X L ) 2 i æ 1 ö Z = R 2 + (X C .178 Xam idea Physics—XII But where X C \ \ i.. Z For maximum power cos f = 1 or Z =R i... circuit should be free from ohmic resistance. Principle: (i) Every point on a given wavefront may be regarded as a source of new disturbance. Z = = R 2 + ( X C .X L ) 2 = R 2 + ç . Phase difference ( f) in series LCR circuit is given by X . circuit is purely resistive.e.wL ÷ è wC ø 2 2 Instantaneous current I = Condition for resonance to occur in series LCR ac circuit: For resonance the current produced in the circuit and emf applied must always be in the same phase. then X C = wr C and \ X L = wr L 1 = wr L wr C Þ wr = 1 LC R Power factor is the cosine of phase angle f.X L = 0 or XC = X L 1 If wr is resonant frequency. VC = X C i and VL = X L i 1 = = capacitance reactance and X L = wL = inductive reactance wC . i.(ii) V = ( R i) 2 + ( X C i ..e.wL ÷ w C è ø V0 sin ( wt + f) æ 1 ö R2 + ç .X L i) 2 V Impedance of circuit. (iii) The surface of tangency to the secondary wavelets in forward direction at any instant gives the new position of the wavefront at that time. VR = R i .. 29. Let us illustrate this principle by the following example: . each with a different phase.AP = BN In D ANB . starting from different points of illuminated slit. 3. (c) Draw a tangent A1 B1 common to all these circles in the forward direction. ÐANB = 90° \ and ÐBAN = q BN or BN = AB sin q \ sin q = AB As AB = width of slit = a \ Path difference.. These points are the sources of secondary wavelets.. then path difference l a sin q = l Þ sin q = a . This gives maximum intensity at the central point C. This was done by Fresnel by rigorous calculations. … on the wavefront AB. If point P on screen is such that the path difference between rays starting from edges A and B is l . D = BP . be the section of a wavefront in a homogeneous isotropic medium at t = 0. we have to sum up their contribution. Hence the waves starting from all points of slit arrive in the same phase. the intensity of central band is 3 3 maximum and goes on decreasing on both S 4 sides. Propagation of wavefront from a point A1 A source: A A2 A1 When monochromatic light is made A2 1 1 incident on a single slit. (b) At time t the radius of these secondary wavelets is vt. the angle q is zero.(i) D = a sin q To find the effect of all coherent waves at P.. According to Fresnel the diffraction B2 B (a) (b) pattern is the result of superposition of a 1 large number of waves. but the main features may be explained by simple arguments given below: At the central point C of the screen. draw circles of radius vt. We have to find the position of the wavefront at time t using Huygens’ principle. 4 Let AB be a slit of width ‘a’ and a parallel 5 5 B2 beam of monochromatic light is incident on B B1 B it. we get diffraction pattern on a screen placed behind the slit. The path difference between rays diffracted at points A and B. 2 2 The diffraction pattern contains bright and dark bands. Let v be the velocity of light in the given medium. Let q be the angle of diffraction for waves reaching at point P of screen and AN the perpendicular dropped from A on wave diffracted from B. Taking each point as centre. 2. . (a) Take the number of points 1.Examination Papers 179 Let AB shown in the fig. For n = 2.. sin q = q = I0 l a .180 Xam idea Physics—XII If angle q is small. v æ Dö M =. it is one-seventh and so on.. OR Eyepiece uo Objective B A" A Fo O Fe' A' E Fe vo ue L1 D Eye B' B" ve Magnifying power of microscope. for n = 3. the contribution of slit decreases. it is one-fifth.0 ç 1+ ÷ ÷ u0 ç f e ø è .(ii) –4l a –3l a –2l a –l a 0 l a 2l a 3l a 4l a The intensity of secondary maxima decreases with increase of order n because with increasing n. 1× 5 cm.4 cm to 10 . The field Ei is directed from n-region to p-region. signal P2 S2 p II n (c) Logic gate is AND gate.6 cm) on either side of the junction becomes free from mobile charge carriers and hence is called the depletion layer. D = 25 cm 1 1 1 Formula gives = f 0 v0 u 0 1 1 1 1 1 1 = + Þ = 1× 25 v 0 1× 5 v 0 1× 25 1× 5 Þ v 0 = 7 × 5 cm 7 ×5 æ 25 ö M =ç1 + ÷ = .30 1× 5 è 5 ø 30.5 ´ 6 = . (a) Formation of depletion layer and potential barrier: VB At the junction there is diffusion of charge – + carriers due to thermal agitation. This sets up a potential difference called potential barrier and hence an internal electric field Ei across the p n junctions. f e = 5 cm. Thus near the p n Depletion junction there is an excess of positively charged layer ions in n-region and an excess of negatively charged ions in p-region. (b) The box contains the circuit of full wave rectifier. so that some of + + + electrons of n-region diffuse to p-region while + + + some of holes of p-region diffuse into n-region. The symbol of p-n junction diode is shown in Fig. P1 S1 p B S I n RL A Output Input A. This field stops the further diffusion of charge carriers. Thus the layers ( » 10 . f 0 = 125 cm.C. Its truth table is A 0 1 0 1 B 0 0 1 1 Y 0 0 0 1 Output Input signal . Some charge carriers combine with opposite + + + Ei charges to neutralise each other.Examination Papers 181 Given u 0 = . 182 Xam idea Physics—XII OR (a) Principle: An oscillator converts dc into ac.1) = Þ n = 1 + = 2R R 4 4 4 è ø rn = e0 h2 n2 . (ii) Faraday law: The induced emf is directly proportional to rate of change of magnetic flux linked with the circuit. (i) For larger deflection to coil P should be moved at a faster rate. L 1 + – npn L C + – CBSE (Delhi) SET–II 3. Ozone layers absorbs ultraviolet rays. (b) If the valence and conduction bands overlap. 76–890 MHz. the substance is a semiconductor.1) ç ç ÷ f è R1 R 2 ø For biconvex lens R1 = + R. the substance is referred as a conductor. æ 1 1 1 ö ÷ 11. 10. Radius of nth orbit of hydrogen atom pme 2 For inner most orbit n = 1 e h 2 (1) 2 \ h= 0 pme 2 For second excited state n = 3 e h 2 (3) 2 r3 = 0 pme 2 r Þ 3 = 9 Þ r 3 = 9r1 = 9 ´ 5 × 3 ´ 10 -11 m = 3 × 77 ´ 10 -10 m r1 6.1) ç ÷ Þ ( n . 9. Speed of waves is same for TV waves and light waves. If the valence and conduction bands have a forbidden gap more than 3 eV. R 2 = . = ( n . A fraction of output voltage or current is fed back to the input circuit in the same phase as the input signal and the oscillations produced in LC circuit are amplified. the substance is an insulator. If the valence and condition bands have a small forbidden gap ( = 1 eV).R 2 Given f = R 3 3 3 3 7 æ2ö \ = ( n . electric field strength d remains the same. (ii) Refractive index of prism material is maximum for violet and minimum for red colour. and V = constant. At the time of sunrise and sunset. blue colour is scattered the most and red colour enters our eyes. l = h 2meV = 6 × 63 ´ 10 -34 2 ´ 9 ×1 ´ 10 -31 = 1 ´ 10 -10 m = 1 Å -19 ´ 1× 6 ´ 10 ´ 144 This corresponds to X-rays. (i) The charge Q = CV . because at these frequencies antennas are relatively smaller and can be placed at heights of many wavelengths above the ground Because of LOS mode direct waves get blocked by the curvature of earth. According to Lord Rayleigh the intensity of scattered light 1 1 I µ Þ Iµ 4 (wavelength) l4 As l blue < l red . (i) The light is scattered by air molecules. the receiving antenna must be very high to intercept the LOS wave. .Examination Papers 183 14. there. (iii) The capacitance of capacitor increases as K > 1. d = constant. 20. V (ii) A electric field E = . 19. l¢ = 50 cm R 50 R \ = Þ =1 S + 5 100 .50 S +5 …(i) 3 3 6 × 4 + 10 ´ 10 3 3 6×4 6 × 4 = 18 ´ 10 ´ 2 × 53 m = 45 × 5 ´ 10 m = 45 × 5 km 22. For receiving signals beyond horizon. so sky appears blue.l) Given balancing length l = 60 × 0 cm. C = increases. (ii) Maximum LOS distance d m = 2R e hT + 2R e hR Given hT = 32 m. therefore. 17. so sunrise and sunset appear red. R e = 6 × 4 ´ 10 6 m d m = 2 ´ 6 × 4 ´ 10 6 ´ 32 + 2 ´ 6 × 4 ´ 10 6 ´ 50 = 8 ´ 10 3 = 18 ´ 10 R l = S (100 . R 60 R 3 = Þ = S 40 S 2 When a resistance of 5 W is connected in series with S. (i) The line of sight (LOS) mode is limited to frequencies above 40 MHz. accordingly blue colour is scattered the most and red the least. hR = 50 m . charge on plates increases. V = same. 12.1) ç ç ÷ f è R1 R 2 ø 1 æ1 1ö = ( n . r1 = \ æ 1 1 1 ö ÷ gives = ( n . (b) When coil Q moves towards left the rate of change of magnetic flux linked with Q decreases and so lesser current is induced in Q. 1 1 5. de-Broglie wavelength l= h 2meV = = 12 × 27 V Here V = 64 V 6 × 28 ´ 10 -34 2 ´ 9 ×1 ´ 10 -31 ´ 1× 6 ´ 10 -19 ´ V ´ 10 -10 m = 12 × 27 V Å .R 7.1) ´ 0 × 3 m = 0 ×15 m = 15 cm 14. For a plano-convex lens R1 = ¥. n = = =2 sin C sin 30° Speed of light in medium v = 1× 5 ´ 108 m/s = 0 × 529 ´ 10 -10 m pme 2 11. 13. Microwaves are used in operating a RADAR. n = 1× 5 \ R = (1× 5 . S = 10 W …(ii) CBSE (Delhi) SET-III 4.184 Xam idea Physics—XII or R =S + 5 Solving (i) and (ii) R = 15 W. (a) The bulb B lights due to induced current in coil Q because of change in magnetic flux linked with it on a consequence of continuous variation of magnitude of alternating current flowing in P. Range of frequencies used for satellite communication 5 × 925 . Bohr’s radius.1) ç + ÷ f èµ Rø 1 n -1 = Þ R = ( n .4 × 2 GHz (Downlink) Speed of wave is same for these waves and light waves. R 2 = .1) f f R e0 h2 c 3 ´ 108 = = 1× 5 ´ 10 8 m / s n 2 or Given f = 0 × 3 m.6 × 425 GHz (Uplink) 3 × 7 . (ii) Refractive index æ A + dm ö sin ç ÷ è 2 ø n= æAö sin ç ÷ è2ø \ l= where A = angle of prism d m = angle of minimum deviation. because deviation d = ( n . we get X + S l1 æ 100 .l2 ø l1 27. (i) No change.l2 ç çX +S÷ ÷ è ø Dividing (ii) by (i).Examination Papers 185 12 × 27 Å= Å = 1× 53 Å 8 64 This corresponds to X-ray region of em spectrum. 12 × 27 …(i) …(ii) S ö ÷ -1 ÷ ø æ 100 .l1 ö ÷ Þ X= = ç ÷ X l2 ç l2 è 100 . (i) By a prism blue light is deviated more than red light.l 2 è q s Q E= or e0 e0 A e A 1 Q C = 0 or C µ d d Q W= Q2 1 or W µ 2C C .l1 When X and S are in parallel. 20. In first case l1 R = S 100 . (ii) Halved. 15.1) A refractive index n is more for blue than red light.l1 ç ç 100 . (iii) Doubled. let resistance XS S¢ = X +S l2 R In second case = æ XS ö 100 . Is it a scalar or a vector quantity? 2. an internal choice has been provided in one question of two marks. Define self-inductance of a coil. unit is JC -1 . (c) There is no overall choice. Questions 1 to 8 carry one mark each. What is its value for a hydrogen atom? . You have to attempt only one of the given choices in such questions. experiences a force due to a magnetic field along the +y-axis. questions 9 to 18 carry two marks each. However. A beam of a particles projected along +x-axis. (e) You may use the following values of physical constants wherever necessary: c = 3 ´ 108 ms .CBSE EXAMINATION PAPERS ALL INDIA–2010 Time allowed: 3 hours General Instructions: (a) All questions are compulsory. (b) There are 30 questions in total. questions 19 to 27 carry three marks each and questions 28 to 30 carry five marks each. Define ionisation energy.1 h = 6 × 626 ´ 10 -34 Js e = 1× 602 ´ 10 -19 C 1 = 9 × 109 Nm2C– 4pe o m 0 = 4p ´ 10 -7 TmA -1 2 Maximum marks: 70 Boltzmann’s constant k = 1× 381 ´ 10 -23 J K -1 Avogadro’s number N A = 6 × 022 ´ 10 23 /mole Mass of neutron m n = 1× 2 ´ 10 -27 kg Mass of electron m e = 9 ×1´ 10 -31 kg Radius of earth = 6400 km CBSE (All India) SET–I 1. (d) Use of calculators is not permitted. 4. one question of three marks and all three questions of five marks each.I. What is the focal length of the combination? 5. What is the direction of the magnetic field? x e particle z y 3. Name the physical quantity whose S. A converging lens is kept co-axially in contact with a diverging lens – both the lenses being of equal focal lengths. Write its SI units. Give its one use.] 18. Write two important conclusions which you can draw regarding the nature of nuclear forces. one of copper and the other of aluminium. Draw a sketch of a plane electromagnetic wave propagating along the z-direction. Write two factors justifying the need of modulating a signal. Write Einstein’s photoelectric equation. Find the ratio of their de-Broglie wavelengths. find the ratio of drift velocity of electrons in the two wires. the surface charge density and n 13. 17. OR Draw a plot of the binding energy per nucleon as a function of mass number for a large number of nuclei. If the number density of electrons in X is twice that in Y. Show that the electric field at the surface of a charged conductor is given by E = n e0 $ is a unit vector normal to the surface in the outward direction. A charge ‘q’ is placed at the centre of the shell. A carrier wave of peak voltage 12 V is used to transmit a message signal. 14. 10. 8. 2 £ A £ 240. 9. Two conducting wires X and Y of same diameter but different materials are joined in series across a battery. Compare (i) the induced emf and (ii) the current produced in the two coils. Deduce the expression for the magnetic dipole moment of an electron orbiting around the central nucleus. 15. so binding energy per nucleon remains constant. (i) Identify the logic gates marked P and Q in the given logic circuit. B = 1. What should be the peak voltage of the modulating signal in order to have a modulation index of 75%? 16. State clearly the three salient features observed in photoelectric effect. . 7. Does this decrease in speed imply a decrease in the energy carried by the light wave? Justify your answer. (ii) outer surface of the shell? (b) Write the expression for the electric field at a point x > r 2 from the centre of the shell. When light travels from a rarer to a denser medium. 11. (a) What is the surface charge density on the (i) inner surface. which can be explained on the basis of the above equation.Examination Papers 187 6. A spherical conducting shell of inner radius r1 and outer radius r 2 has a charge ‘Q’. B = 0 and A = 1. Justify your answer. An a-particle and a proton are accelerated from rest by the same potential. the speed decreases. ® s $. Depict clearly the directions of electric and magnetic fields varying sinusoidally with z. How do you explain the constancy of binding energy per nucleon in the range 30 < A < 170 using the property that nuclear force is short-ranged? Nuclear forces are short ranged. so every nucleon interacts with their neighbours only. Name the part of electromagnetic spectrum whose wavelength lies in the range of 10 -10 m. where s is 12. Draw a plot of potential energy of a pair of nucleons as a function of their separation. Two identical loops. A P B B C Q X (ii) Write down the output at X for the inputs A = 0. are rotated with the same angular speed in the same magnetic field. Determine (a) equivalent capacitance of the network and (b) charge on each capacitor. Why is there an upper limit to frequency of waves used in this mode? + – 20. (ii) Why must both the objective and the eye-piece of a compound microscope have short focal lengths? 23. I 2 and I 3 in the circuit diagram shown. is 2 × 2 V. In Young’s double slit experiment. The screen is 1.0 m away from the slits. Find the internal resistance of the cell. Explain briefly its working. Use these rules to write the expressions for the currents I 1 . State Kirchhoff’s rules. C1 C3 C4 A 500 V B 22. when a cell is connected across it. the two slits 0 ×15 mm apart are illuminated by monochromatic light of wavelength 450 nm. When the terminals of the cell are also connected to a + – resistance of 5 W as shown in the circuit. (b) How will the fringe pattern change if the screen is moved away from the slits? 24. (i) Draw a neat labelled ray diagram of an astronomical telescope in normal adjustment. Write any two factors on which internal resistance of a cell depends. V The reading on a high resistance voltmeter. (ii) An astronomical telescope uses two lenses of powers 10 D and 1 D. Which mode of propagation is used by short wave broadcast services having frequencies range from a few MHz upto 30 MHz? Explain diagrammatically how long distance communication can be achieved by this mode. (ii) dark fringe from the central maximum. Explain briefly its working. A network of four capacitors each of 12 mF capacitance is connected to a 500 V supply as shown in the figure. What is its magnifying power in normal adjustment? OR (i) Draw a neat labelled ray diagram of a compound microscope.188 Xam idea Physics—XII 19. R = 5W C2 K 21. the voltmeter reading drops to 1× 8 V. A I1 E1 = 2V r1 = 4W B E2 = 1V r2 = 3W D E3 = 4V r3 = 2W F C I2 I3 E . (a) Find the distance of the second (i) bright fringe. unit of current. (b) What is the importance of a radial magnetic field and how is it produced (c) Why is it that while using a moving coil galvanometer as a voltmeter a high resistance in series is required whereas in an ammeter a shunt is used? OR (a) Derive an expression for the force between two long parallel current carrying conductors. A third polaroid is placed between the two making an angle q with the pass axis of the first polaroid. S P Q R x=0 x=b x = 2b OR Draw a schematic diagram of a step-up transformer. (b) Use this expression to define S. how is this ratio related to the currents in the two coils? How is the transformer used in large scale transmission and distribution of electrical energy over long distances? . 28. twice the size of the object. B (c) A long straight wire AB carries a current I. Deduce the expression for the secondary to primary voltage in terms of the number of turns in the two coils. What is P Proton d the force experienced by the proton and what is its direction? A 29. obtain the expressions for the flux and the induced emf. The field extends from x = 0 to x = b and is zero for x > b. In what orientations will the transmitted intensity be (i) minimum and (ii) maximum? 27. A proton P travels with a speed v. Explain its working principle. When the arm PQ is pulled outward from x = 0 to x = 2b and is then moved backward to x = 0 with constant speed v. (a) With the help of a diagram. In an ideal transformer. Give its relationship with the half-life. Write the expression of the intensity of light transmitted from the second polaroid. (a) Write symbolically the b . explain the principle and working of a moving coil galvanometer.Examination Papers 189 25.decay process of 32 P. An illuminated object and a screen are placed 90 cm apart. I. at a distance d from it in a I direction opposite to the current as shown in the figure. State Faraday’s law of electromagnetic induction. Sketch the variations of these quantities with distance 0 £ x £ 2b. How does an unpolarised light get polarised when passed through polaroid? Two polaroids are set in crossed positions. Determine the focal length and nature of the lens required to produce a clear image on the screen. Figure shows a rectangular conductor PQRS in which the conductor PQ is free to move in a uniform magnetic field B perpendicular to the plane of the paper. 26. parallel to the wire. Assume that only the arm PQ possesses resistance r. 15 (b) Derive an expression for the average life of a radionuclide. A beam of electrons projected along +X axis. v v B . (ii) current amplification factor. Which of the following has the shortest wavelength: Microwaves. A rectangular loop and a circular loop are moving out of a uniform magnetic field to a field-free region with a constant velocity ‘v’ as shown in the figure.190 Xam idea Physics—XII 30. CBSE (All India) SET–II Questions uncommon to Set–I 1. OR (a) Draw the circuit arrangement for studying the input and output characteristics of an n-p-n transistor in CE configuration. (b) Describe briefly with the help of a circuit diagram how an n-p-n transistor is used to produce self-sustained oscillations. experiences a force due to a magnetic field along the + y-axis. With the help of these characteristics define (i) input resistance. and (ii) the highest permitted energy level to the first permitted level. (ii) reverse bias. (a) Draw the circuit diagrams of a p-n junction diode in (i) forward bias. Explain in which loop do you expect the induced emf to be constant during the passage out of the field region. What is the direction of the magnetic field? e z y 5. The magnetic field is normal to the loops. x 4. X-rays 12. Ultraviolet rays. How are these circuits used to study the V-I characteristics of a silicon diode? Draw the typical V-I characteristics? (b) What is a light emitting diode (LED)? Mention two important advantages of LEDs over conventional lamps. Find the ratio of energies of photons produced due to transition of an electron of hydrogen atom from its: (i) second permitted energy level to the first level. Arrange the following in descending order of wavelength: X-rays. when a cell is connected across it. I 2 and I 3 in the network. In Young’s double slit experiment. The ground state energy of hydrogen atom is – 13 × 6 eV. The screen is 1× 0 m away from this slits.Examination Papers 191 19. the voltmeter reading drops to 1× 5 V. calculate the object and image distances. C2 C1 C3 C4 500 V 20. is 2 × 0 V. (ii) dark fringe from the central maximum. Write any two factors on which internal resistance of a cell depends. Determine (a) equivalent capacitance of the network and (b) charge on each capacitor. Radio waves. Apply Kirchhoff’s rules to the loops ACBPA and ACBQA to write the expressions for the currents I 1 . (b) How will the fringe pattern change if the screen is moved away from ths slits? 23. E1 = 6V P I1 0. (a) Find the distance of the second (i) bright fringe.5W A 1W I2 I3 Q E2 = 10V C R = 12W B 27. the two slits 0 ×12 mm apart are illuminated by monochromatic light of wavelength 420 nm. Infrared light 8. A network of four capacitors each of 15 mF capacitance is connected to a 500 V supply as shown in the figure. CBSE (All India) SET–III Questions uncommon to Set–I and Set–II 5. The image obtained with a convex lens is erect and its length is four times the length of the object. Find the internal resistance of the cell. When the terminals of the cell are also connected to a resistance of 3 W as shown in the circuit. What are the kinetic and potential energies of electron in this state? . Blue light. + V – + – R = 3W K 22. State Kirchhoff’s rules. If the focal length of the lens is 20 cm. The reading on a high resistance voltmeter. 2. The screen is 1× 0 m away from the slits. By Fleming’s left hand rule magnetic field must be along negative Z-axis 3. State Kirchhoff’s rules. is 2 × 5 V. It is a scalar quantity. Electric potential. The self inductance is defined on the magnetic flux linked with the coil when unit current flows through it.f respectively. (b) How will the fringe pattern change if the screen is moved away from the slits? 26. 4. when a cell is connected across it. Find the internal resistance of the cell. Let focal length of converging and diverging lenses be + f and . the two slits 0 × 20 mm apart are illuminated by monochromatic light of wavelength 600 nm. (a) Find the distance of the second (i) bright fringe. A convex lens is used to obtain a magnified image of an object on a screen 10 m from the lens. I 2 and I 3 in the given circuit. Apply these rules to the loops PRSP and PRQP to write the expressions for the currents I 1 . In Young’s double slit experiment. The unit of self-inductance is henry (H). find the focal length of the lens. Or The self inductance is defined as the emf induced in the coil. When the terminals of the cell are also connected to a resistance + – of 5 W are shown in the circuit. 200 W S 5V 60 W I2 I3 I1 20W (milliammeter) Q R P 4V Solutions CBSE (AI) Set–I 1. Power of diverging lens P2 = f f 1 1 Power of combination P = P1 + P2 = . If the magnification is 19. + – 24. The V reading on a high resistance voltmeter. Write any two factors on which internal resistance of a cell depends.= 0 \ f f .192 Xam idea Physics—XII 21. (ii) dark fringe from the central maximum. when the rate of change of current in the coil is 1 ampere/second. 1 1 Power of converging lens P1 = . the voltmeter reading drops to 2 × 0 V. R = 5W K 25. its frequency remains unchanged. The ionisation energy for hydrogen atom is 13. \ n x eAv x = n y eAv y ny vx n y 1 Þ \ = = = v y n x 2n y 2 vx : v y =1 : 2 7. total charge on inner surface -q and on outer surface it is Q + q. According to quantities theory. The minimum energy required to remove an electron from atom to infinitely for away is called the ionisation energy.Examination Papers 193 \ Focal length of combination F = 1 1 = = ¥ (infinite) P 0 F = ¥. 4pe 0 x 2 . whole charge acts at centre. The revolving electron is equivalent to electric e– current me e r I= N T 2p r where T is period of revolution = v e ev \ I= = 2pr / v 2pr Area of current loop (electron orbit).6 eV. For same diameter. so electric field at distance x > r 2 . used to study crystal structure 8. X-ray. ev evr M l = IA = ( pr 2 ) = 2pr 2 10. A = pr 2 Magnetic moment due to orbital motion. charge induced on inner surface is –q and on outer surface it is + q. (a) Charge Q resides on outer surface of spherical conducting shell. when light travels from a rarer to denser medium. No. 1 Q+q E( x ) = . 5. 6. the energy of a light beam depends on frequency and not on speed 9. In series current is same i X = iY . So. cross-sectional area is same. Consider an electron revolving around a nucleus ( N) in circular path of L radius r with speed v. q Surface charge density on inner surface = 4pr12 Q+ q Surface charge density on outer surface = 2 4pr 2 r2 q -q O r1 Q +q P (b) For external points. Due to charge q placed at centre. just outside the surface of A dS3 the conductor. The flux of electric field through the curved surface of the box is zero. Let us consider a point P at which electric field strength is to be calculated. . just outside the surface of the conductor is perpendicular to the surface of the conductor in the neighbourhood of P. then ® ® S1 ® ® S2 ® ® S3 ® ® Total electric flux òS E× d S = ò =ò E× d S 1 + ò E× d S 2 + ò ® E× d S 3 ® S1 S E dS1 cos 0 + òS 0× d S 2 + 2 òS 3 E dS 3 cos 90° = ò E dS1 = Ea As the charge enclosed by the cylinder is ( sa) coulomb. e.(i) Þ Ea = ( sa) E= e0 e0 Thus the electric field strength at any point close to the surface of a charged conductor of any shape is equal to 1 / e 0 times the surface charge density s. the dS S1 other base D lying inside the conductor and the curved surface being perpendicular to the surface of the B conductor. as electric field strength inside the conductor is zero.. As the surface of the conductor is equipotential surface... since there is no component of electric field E normal to curved surface. the electric field strength E at P. This is known as Coulomb’s law. this charge under electrostatic equilibrium will redistribute and the electric field inside the conductor is zero (i. Let a charge Q be given to a conductor. Now consider a small cylindrical C dS1 E S2 box CD having one base C passing through P . Let the area of each flat base be a. This may be analytically seen as: If S1 and S 2 are flat surfaces at C and D and S 3 is curved surface. Let the surface charge density on the Conductor S3 surface of the conductor in the neighbourhood of P be D dS2 P s coulomb / metre 2 . Electric field is along x-axis and magnetic field is along y-axis x E B z B y 12.194 Xam idea Physics—XII 11. Also the flux of electric field through the base D is zero. The electric field strength is directed radially away from the conductor if s is positive and towards the conductor if s is negative. Ein = 0). we have. 1 Total electric flux = ´ charge enclosed e0 1 s or . using Gauss’s theorem. Therefore the resultant flux of electric field through the entire surface of the box is same as the flux through the face C. increases. hence transmission is quite lossy. the kinetic energy of emitted electron increases. Einstein's photoelectric equation is E k = hn . . w are same for both loops so induced emf is same for both loops. E ma = m Ec Peak voltage of modulating signal. A. df d 13. e (ii) Current induced I = R As resistance R is less for copper loop. de Broglie wave length l = = 2mE 2mqV For a-particle. \ (i) Explanation of frequency law: When frequency of incident photon (n). la = lp = h 2m a q a V h 2m p q p V \ But \ ma q = 4. (iii) The various information signals transmitted at low frequency get mixed and hence can not be distinguished. Intensity has no effect on kinetic energy of photoelectrons. For proton. (ii) The power radiated at audio frequency is quite small. then If n ® s $ E= n e0 Obviously electric field strength near a plane conductor is twice of the electric field strength near a non-conducting thin sheet of charge.( BA cos wt) dt dt = BAw sin wt As B. (i) Induced emf e = = .W for a single photon ejecting a single electron. Modulation index. 75 em = m a ´ E c = ´ 12 = 9 V 100 16. The modulation is needed due to (i) Transmission of audio frequency electrical signals need long impracticable antenna. a = 2 mp qp mpqp la = lp ma qa la 1 1 1 = × = lp 4 2 2 2 15.Examination Papers 195 $ is unit vector normal to surface in outward direction. h h 14. so current induced is larger in copper loop. Thus there is no time lag between incidence of light and emission of photoelectrons. the increase in intensity will increase the number of ejected electrons.0 0. In other words. as one photon ejects one electron. Part AB represents repulsive force and Part BCD represents attractive force. A +100 Repulsive B MeV 0 Attractive –100 C 1 2 r (fm) 3 4 Conclusions: (i) Nuclear forces are attractive and stronger.0 5. 17. 9. (iii) Explanation of no time lag law: When the energy of incident photon is greater than work function. then electrostatic force. Frequency has no effect on photocurrent.0C12 F18 He4 N14 7. photocurrent will increase with increase of intensity.196 Xam idea Physics—XII (ii) Explanation of intensity law: When intensity of incident light increases. OR The variation of binding energy per nucleon versus mass number is shown in figure.0 0 H2 Li7 Fe56 U238 Binding Energy per Nucleon (in MeV) 20 40 60 80 100 120 140 160 180 200 220 240 Mass Number .0 2.0 3. the number of incident photons increases. (ii) Nuclear forces are charge-independent.0 6.0 O16 8.0 1.0 4. the photoelectron is immediately ejected. There is an upper limit of frequency because for frequency higher than 30 MHz the radiowaves penetrate through the ionosphere and escape. B Ionospheric layers 20. Sky wave communication (a) C1 .1÷ R èV ø 10 æ 2×2 ö \ r =ç .Examination Papers 197 Since nuclear forces are short-ranged. The internal resistance of a cell depends on (i) distance (l) between electrodes. C = A × B = 0 × 0 = 1 \ X =1 + 0 =1 When A = 1. V = 1× 8 V. B = 0. The mode of propagation used by short wave broadcast services having frequency range from a few MHz upto 30 MHz is sky wave propagation. 18. their equivalent capacitance C ¢ is given by 1 1 1 1 = + + C ¢ C1 C 2 C 3 1 1 1 = + + 12 12 12 Þ C¢ = 4 m F C 4 is in parallel with C ¢. (i) P is ‘NAND’ gate and Q is ‘OR’ gate. Given. R = 5 W æE ö r = ç . so equivalent capacitance of network C eq = C ¢ + C 4 = 4 + 12 = 16 mF (b) Charge on capacitor C 4 is q4 = C 4V = (16 mF) ´ 500 V = 8000 mC = 8 mC . The diagram is shown in fig. E = 2 × 3 V. and (iii) nature and concentration of electrolyte. C 2 and C 3 are in series. (ii) area ( A) of immersed part of electrode. binding energy per nucleon remains constant. Therefore. A C P (ii) C = A × B \ X = C + B = A × B + B B Q X When A = 0. B = 1 C = A × B =1 ×1 = 1 = 0 \ X = 0 + 1 =1 19. every nucleon interacts with their neighbours only.1÷ ´ 5 W = W = 1× 1 W 9 è 1× 8 ø 21. For normal adjustment position this image also lies at first focus ( Fe ) of eye lens ( L 2 ). is magnified and inverted.e == . inverted and diminished. B fo fe A C1 Fo A' F e' B' C2 Fe in At i fin ty (ii) M = - f0 P 10 D =. The eyepiece is so adjusted that the image A¢ B¢ lies between the first focus Fe ¢ and the eyepiece E. The objective lens forms the real.10 fe P0 1D OR (i) Eyepiece uo Objective B A" A Fo O Fe' A' E Fe vo ue D Eye B' B" ve Working: Suppose a small object AB is placed slightly away from the first focus F0 ¢ of the objective lens. which acts as an object for eyepiece. Magnifying power of compound microscope is . Thus the final image A¢ ¢ B¢ ¢ formed by the microscope is inverted and magnified and its position is outside the objective and eyepiece towards objective lens. This image acts as an object for eye lens and final image is formed at infinity. The image A ¢B ¢ is real. The eyepiece forms its image A¢ ¢ B¢ ¢ which is virtual. (i) Working: Suppose AB is the point object whose end A is on the axis of telescope.198 Xam idea Physics—XII 22. The final image A ¢¢B ¢¢ (say). erect and magnified. inverted and magnified image A¢ B¢ . The objective lies ( L1 ) forms image A ¢B ¢ of object AB at the second principal focus F0 . .1 Applying Kirchoff’s II law to mesh ABCEA E3 = 4V r3 = 2W I3 -2 .÷ 2 =çn . I 3 = I1 + I 2 = A 13 .4 I 1 + 3 I 2 + 1 = 0 …(ii) Þ 4I 1 . Given d = 0 × 25 mm = 0 ×15 ´ 10 -3 m.. 23.3I 2 = . i. so fringe width b = Dl increases.Examination Papers 199 M =»M =- v0 u0 L f0 v0 u0 æ Döü ç1 + ÷ï ç be ÷ è ø ï for final image at distance of distinct vision ý æ Dö ç1 + ÷ ï ç fe ÷ è ø ï þ D L D for final image at infinity »fe f0 fe (ii) For large magnifying power. SV = 0 E1 = 2V r1 = 4W From Kirchhoff’s first law I1 …(i) I 3 = I1 + I 2 For applying Kirchhoff’s second law to mesh ABDC E2 = 2V r2 = 3W I2 -2 .e. Kirchhoff’s Rules: ( i) The algebraic sum of currents meeting at any junction is zero. we get 2 7 I 1 = A.4 I 1 .÷ 2ø 2 è 2 ø 0 ×15 ´ 10 -3 è = 4 × 5 ´ 10 -3 m = 4 × 5 mm (b) If screen is moved away from the slits. I 2 = 1 .2 I 3 + 4 = 0 Þ 4I 1 + 2I 3 = 2 or 2I 1 + I 3 = 1 Using (i) we get Þ 2I 1 + ( I 1 + I 2 ) = 1 or …(iii) 3I 1 + I 2 = 1 Solving (ii) and (iii). d 24. f 0 and f e both have to be small. D = 1× 0 m (a) Distance of second bright maximum from central maximum ( n = 2) y2 = nDl 2 ´ 1× 0 ´ 450 ´ 10 -9 = m = 6 ´ 10 -3 m = 6 mm -3 d 0 ×15 ´ 10 Distance of second dark fringe from central maximum ( n = 2) 1 ö Dl æ 1 ö 1× 0 ´ 450 ´ 10 -9 æ y¢ =ç2 .e.3I 1 = A 3 13 9 so. SI = 0 (ii) The algebraic sum of potential differences across circuit elements of a closed circuit is zero. l = 450 nm = 450 ´ 10 -9 m. D increases. i. we get T = 0 × 6931 t 26.lt ü ¥ ù ¥ æ e ..lt \ t= S tl ( N 0 e .[ 0 . . the mean life time of a radioactive element is reciprocal of its decay constant.e.ò0 1 ç ç -l ÷ ú êï .1e 0 +n (b) If N 0 is the total number of nuclei at t = 0.. dN t= = Total number of nuclei N0 St lNdt = N0 Also we have N = N 0 e . then mean life time Total life time of all the nuclei St . Thus. t = .(i) T= l 1 Mean life . we get éì . l i.lt ¥ 1 1 =e = ...lt ï ü ù 1ï ê =l 0+ í ý ú ê ú lï l ï î þ0 û ë 1 . Integrating by parts. (a) 32 P 15 ¾® 32 16 S + .e. t=lò ¥ 0 t e . The electric vectors (associated with the propagating light wave) along the direction of the aligned molecules get absorbed..lt ö ï te ï ÷ dt ú t = l êí ý .l ï þ0 è ø û ëî ¥ é ì e .lt dt N0 As nuclei decay indefinitely. if an unpolarised light wave is incident on such a polaroid then the light wave will get linearly polarised with the electric vector oscillating along a direction perpendicular to the aligned molecules. Polaroid: A polaroid consists of long chain molecules aligned in a particular direction.lt ) dt = l S t e . Relation Between Mean Life and Half Life 0 × 6931 Half life .200 Xam idea Physics—XII 25.(ii) t= l [ ] Substituting value of l from (ii) in (i).lt dt. we may replace the summation into integation with limits from t = 0 to t = ¥ i..1] = 0 l l l 1 Thus. Examination Papers P1 I0 I1 = I0/2 P3 q I2 I3 P2 201 Let intensity of incident unpolarised light on first polaroid be I 0 .q) = ç 0 cos 2 q ÷ sin 2 q è 2 ø I0 I = ( 2 cos q sin q) 2 = 0 sin 2q 8 8 (i) Intensity I 3 will be minimum. given by t = NIAB sin q where q is the angle between the normal to plane of coil and the magnetic field of strength B.30 cm. I Intensity of light transmitted through 1st polaroid P1 is I 1 = 0 . N is the number of turns in a coil. . v = 60 cm 1 1 1 gives = f v u 1+ 2 1 1 = + = 60 30 60 (convex lens) Þ f = 20 cm 28. 2 Intensity of light transmitted through polaroid P3 is I I 2 = 0 cos 2 q 2 Angle between pass-axis of P3 and P2 is ( 90 .q) \ Intensity of light transmitted through polaroid P2 is æI ö I 3 = I 2 cos 2 ( 90 . (a) Priciple and working: When current ( I ) is passed in the coil. | v | = 60 cm By sign convention u = . when sin 2q = 0 Þ q = 0° (ii) Intensity I 3 will be maximum when sin 2q = 1 Þ q = 45° 27. Given …(i) u + v = 90 cm O I v = gives O u |v| 2= u |u | F Screen I = 20 v or (numerically) …(ii) | v| =2|u| From (i) and (ii) | u | = 30 cm. torque t acts on the coil. If a voltmeter has very high resistance.. across a resistance in an electrical circuit. NIAB = C q NAB . (c) A voltmeter is used to measure p.. hence reading will be true. the radial magnetic field makes the scale linear. then in every position of coil the plane of the coil. it will not affect the resistance of circuit. It is connected in parallel across the resistance. That is why while using a moving coil galvanometer on a voltmeter. so torque is t = NIAB. . as in the case of cylindrical pole pieces and soft iron core. That is why while using a moving coil galvanometer as an ammeter.resistance and so reading will be true. t = NIAB If C is the torsional rigidity of the wire and q is the twist of suspension strip. To produce radial magnetic field pole pieces of permanent magnet are made cylindrical and a soft iron core is placed between them. The soft iron core helps in making the field radial and reduce energy losses produced due to eddy currents. An ammeter is used to measure current in circuit. (b) Importance (or function) of uniform radial magnetic field: In radial magnetic field sin q = 1.e.202 Xam idea Physics—XII H Suspension wire M Coil NIBl b T1 T2 N S N S N S NIBl Coiled strip (b) Magnetic lines of force of radial magnetic field (a) (c) When the magnetic field is radial.d. then restoring torque = C q For equilibrium. a shunt (small resistance in parallel) is used. qµI deflection of coil is directly proportional to current flowing in the coil and hence we can construct a linear scale. If an ammeter has very low resistance it will not affect the circuit . so that q = 90° and sin 90° = 1 Deflecting torque. is parallel to the magnetic field lines. This makes the deflection ( q) proportional to current.e. In other words. hence it is connected in series with the circuit.(i) \ q= I C i. deflecting torque = restoring torque i. a high resistance in series is required. each element of other wire experiences a force.I.. 1. The force between two parallel current carrying conductors of separation r is F m I I f = = 0 1 2 N/ m L 2pr If I 1 = I 2 = 1 A. (fig.I.(ii) (b) Definition S. therefore the magnetic force on element ab of length DL DF = B1 I 2 DL sin 90° = P R P R b DL DF I1 Q r a B I2 S I1 Q B I2 a DL b DF S m 0 I1 I 2 DL 2p r \ The total force on conductor of length L will be m I I m I I F = 0 1 2 S DL = 0 1 2 L 2p r 2p r \ Force acting on per unit length of conductor F m I I f = = 0 1 2 N/ m L 2p r . the direction of B1 will be perpendicular to the plane of paper and directed downward. b). It has been observed experimentally that when the currents in the wire are in the same direction...). a) and when they carry currents in opposite directions. Let the conductors PQ and RS carry currents I 1 and I 2 in same direction and placed at separation r. unit of Current (Ampere): In S. r = 1 m. The magnetic field produced by current-carrying conductor PQ at the location of other wire RS m I .7 N/m 2p .Examination Papers 203 OR (a) Suppose two long thin straight conductors (or wires) PQ and RS are placed parallel to each other in vacuum (or air) carrying currents I 1 and I 2 respectively. The direction of current element is perpendicular to the magnetic field. they experience an attractive force (fig. Due to this magnetic field.. Consider a current–element ‘ab’ of length DL of wire RS. system of fundamental unit of current ‘ampere' has been defined assuming the force between the two current carrying wires as standard.(i) B1 = 0 1 2p r According to Maxwell’s right hand rule or right hand palm rule no. then m f = 0 = 2 ´ 10 . they experience a repulsive force (fig. (c) Magnetic field due to current carrying wire is perpendicular to plane of paper – downward.( Bldx ) dt dt . e µ Dt Df If the coil contains N-turns. 2 The magnetic flux from x = b to x = 2b is zero. When PQ moves a small distance from x to x + dx . B =..e. then magnetic flux linked = BdA = Bldx The magnetic field is from x = 0 to x = b. Faraday’s law of electromagnetic induction states that whenever there is a change in magnetic flux linked with of a coil.204 Xam idea Physics—XII Thus 1 ampere is the current which when flowing in each of parallel conductors placed at separation 1 m in vacuum exert a force of 2 ´ 10 -7 on 1 m length of either wire.N Dt Let length of conductor PQ = l As x = 0. ® m I $ i. then e = .. in 2pd ® the plane of paper perpendicular to direction of v and to the right. S P Q R x=0 dx x=b x = 2b Induced emf. an emf is induced in the coil. 29. Df i.0 k i ÷= 2pd è 2pd ø m evI That is the magnetic force has magnitude 0 and is directed along positive x-axis ie. e = - df d = . The induced emf is proportional to the rate of change of magnetic flux linked with the coil.. so final magnetic flux = SBldx = Bl S dx (increasing) = Blb 1 Mean magnetic flux from x = 0 to x = b is Blb.e. magnetic flux f = 0.0 k 2pd ® ® ® Force F = q v ´ B æ m I $ ö m 0 evI $ = e ( -v$ j) ´ ç . Examination Papers 205 = . the current changes continuously in the primary coil..C. mains) Primary Primary laminated iron core Core Secondary Step up Secondary Transformer .Bl where v = dx = .Blv dt dx = velocity of arm PQ from x = 0 to x = b.(iii) If the resistance of primary coil is negligible.. then the potential difference VS across its ends will be equal to the emf ( e S ) induced in it. NS the number of turns in secondary coil and f the magnetic flux linked with each turn. will be equal to the applied potential difference (V p ) across its ends.. Similarly if the secondary circuit is open. and induced emf will be in opposite direction.. due to which the magnetic flux linked with the secondary coil changes continuously. We assume that there is no leakage of flux so that the flux linked with each turn of primary coil and secondary coil is the same. the emf ( e p ) induced in the primary coil. Let N p be the number of turns in primary coil. but now area is decreasing so magnetic flux is decreasing. e = Blv Graph is shown in figure. According to Faraday’s laws the emf induced in the primary coil Df ep =-Np Dt and emf induced in the secondary coil Df eS = .NS Dt From (i) and (ii) eS NS = ep Np (A.(ii) . therefore the alternating emf of same frequency is developed across the secondary. dt During return from x = 2b to x = b.. From x = b to x = 0 f x=0 x=b x = 2b e x=0 x=b x = 2b From x = 0 to x = b OR Principle: When alternating current source is connected to the ends of primary coil. therefore ..(i) . the induced emf is zero. ..206 Xam idea Physics—XII VS e S N S = = = r (say) Vp e p N p where r = . 30.(iv) NS is called the transformation ratio.e. So VS > V p and i S < i p i. then For about 100% efficiency. When output voltage increases. there is no violation of conservation of energy in a step up transformer. so power loss I 2 R is negligible. The current in wires at this voltage is quite small. If i p and i s are the instantaneous currents in Np primary and secondary coils and there is no loss of energy... Step up transformer is used at power house to transmit power at high voltage 11000 V or 33000 V. Power in primary = Power in secondary V p i p = VS i S iS V p N p 1 = = = i p VS N S r \ . At town. (a) (i) Forward Bias: Ei E p n R + p – n K Forward current + – (ii) Reverse Bias: Ei E p n R – p + n K – + Reverse biasing Reverse current . N s > N p ® r > 1.(v) In step up transformer. This saves enormous electrical energy. Thus. step up transformer increases the voltage. the output current automatically decreases to keep the power same.. the step down transformer is used to supply power at 220 V. Examination Papers 207 The battery is connected to the diode through a potentiometer (or rheostat) so that the applied voltage to the diode can be changed. For different values of voltages, the value of the current is noted. A graph between V and I is obtained as in figure. Note that in forward bias measurement, we use a milliammeter since the expected current is large while a micrometer is used in reverse bias to measure current. I (mA) 100 80 60 40 20 100 80 Vbr 60 40 20 O 10 20 30 0.2 0.4 0.6 0.8 V (volt) I (mA) V-I characteristics of a silicon diode (b) Light Emitting Diode (LED): Light p n R + – LED symbol A light emitting diode is simply a forward biased p-n junction which emits spontaneous light radiation. When forward bias is applied, the electron and holes at the junction recombine and energy released is emitted in the form of light. For visible radiation phosphorus doped GaAs is commonly used. The advantages of LEDs are: (i) Low operational voltage and less power. (ii) Fast action with no warm up time. (iii) Emitted light is nearly monochromatic radiation. (iv) They have long life. 208 Xam idea Physics—XII OR (a) Characteristic Curves: The circuit diagram for determining the static characteristic curves of an n-p-n transistor in common-emitter configuration is shown in figure. + mA VBB + – Rh1 + VBE – B E IE – IB C + VCE – mA – IC + + – VCC Common Emitter Characteristics: (i)Input characteristics: These characteristic curves are obtained by plotting base current ( I B ) versus base-emitter voltage VBE for fixed collector-emitter voltage VCE . Fig. represents these characteristics. (ii) Output characteristics: These characteristics are VCE = 5V obtained by plotting collector current I C versus V collector-emitter voltage VCE at a fixed value of base current I B . The base current is changed to some other fixed value and the observations of I C versus VCE are VBE repeated. Fig. represents the output characteristics of a common-emitter circuit. Input Resistance. It is the ratio of change in base-emitter voltage ( DVBE ) to the corresponding change in base current ( DI B ) at constant collector-emitter voltage VCE , i. e., æ DVBE ö ÷ ri = ç ç DI ÷ B øVCE = constant è The input resistance of a common emitter circuit is of the order of a few hundred ohms. Current amplification factors of a transistor (a and b): The current gain a is defined as the ratio of change in collector current to the change in emitter current for constant value of collector voltage in common base configuration i.e., æ DI C ö ÷ …(i) a =ç ç DI ÷ E øVC = constant è Practical value of a ranges from 0 × 9 to 0 × 99 for junction transistor. The current gain b is defined as the ratio of change in collector current to the change in base current for constant value of collector voltage in common emitter configuration i. e., æ DI C ö ÷ …(ii) b =ç ç DI ÷ B øV = constant è C CE The value of b ranges from 20 to 200. The current gains a and b are related as b a or b = a= 1+b 1-a …(iii) = 10 IB V Examination Papers 209 (b) A transistor as an Oscillator: Circuit Operation. When the collector supply voltage is switched on by closing switch S, collector current starts increasing and the capacitor C is charged. When the capacitor attains maximum charge, it discharges through coil L , setting up oscillations of natural frequency. 1 f = 2p ( LC) R1 L' L C + VCC – C1 R2 n-p-n S RE CE These oscillations induce a small voltage in coil L¢ by mutual induction. This induced voltage is the feed back voltage; its frequency is same as that of resonant LC circuit but its magnitude depends on the number of turns in L¢ and coupling between L and L¢ . The feedback voltage is applied between the base and emitter and appears in the amplified form in the collector circuit. A part of this amplifier energy is used to meet losses taking place in oscillatory circuit to maintain oscillations in tank circuit and the balance is radiated out in the form of electromagnetic waves. Positive Feed back. The feed back applied in tuned collector oscillator circuit is positive. This may be seen as follows: A phase shift of 180° is created between the voltages of L and L¢ due to transformer action. A further phase shift of 180° arises between base-emitter and collector circuit due to transistor action in CE configuration. Thus the net phase becomes 360° (or zero); which is the required condition for a positive feed back. Due to positive feed back the energy fed back to the tank circuit is in phase with the generated oscillations, thus maintaining oscillations. CBSE (All India) SET–II æ 1 1 ö 3 1. E I = Rhc ç ç 2 - 2÷ ÷ = Rhc 2 ø 4 è1 æ 1 1 ö E II = Rhc ç ç 2 - 2÷ ÷ = Rhc ¥ ø è1 E 3 Ratio I = E II 4 $´ B 4. Fm = q v ´ B Þ Fm $ j = - evi ® ® ® ® For validity of this equation, the direction of magnetic field must be along z-axis, since $) = - $ $´ k (i j 5. X-rays has shortest wavelength. 210 Xam idea Physics—XII x x x x x x x x x x x x x x x x x x x x x x (b) x x x x x 12. In rectangular coil the induced x x x x x x x emf will remain constant x x x x x x x because in this the case rate of x x x x x x x v change of area in the magnetic x x x x x x x field region remains constant, (a) while in circular coil the rate of change of area in the magnetic field region is not constant. 19. (i) Equivalent capacitance of C1 , C 2 and C 3 in series is C ¢ 1 1 1 1 = + + C ¢ C1 C 2 C 3 C ¢ = 5 mC C eq = C ¢ + C 4 = 5 mF + 15 mF = 20 mF (ii) Charge on C 4 q 4 = C 4 V = 500 ´ 15 mC = 7 × 5 mC (iii) Charge on C1 , C 2 and C 3 is q1 = q 2 = q 3 = C ¢ V = 5 mF ´ 500 V = 2500 mC = 2 × 5 mC é æE ö ù æ 2×0 ö 20. êr = ç - 1÷ R = ç - 1÷ ´ 3 = 1 W ú ø è 1× 5 ø ë èV û Þ 22. Given d = 0 ×12 mm = 0 ×12 ´ 10 -3 m l = 420 nm = 420 ´ 10 -9 m, D = 1× 0 m (a) (i) æ nDl ö 2 ´ 1× 0 ´ 420 ´ 10 y2 = ç ÷= è d ø 0 ×12 ´ 10 -3 -9 C1 v C2 C3 C4 500 V = 7 ´ 10 -3 m = 7 mm hö l æ 1 ö 1× 0 ´ 420 ´ 10 -9 æ y¢ = n = 2 ç ÷ ç ÷ 2 2ø d è 2 ø 0 ×12 ´ 10 -3 è = 3 1× 0 ´ 4 × 2 ´ 10 -7 ´ = 5 × 25 ´ 10 -3 m = 5 × 25 mm -4 2 1× 2 ´ 10 E1 = 6V P I1 0.5W A 1W I2 I3 Q E2 = 10V C R = 12W B 23. From Kirchhoff’s law …(i) I 3 = I1 + I 2 Applying Kirchhoff’s II law to loop ACBPA -12I 3 - 0 × 5I 1 + 6 = 0 …(ii) 0 × 5I 1 + 12I 3 = 6 Applying Kirchhoff’s II law to loop ACBQA - 12I 3 - 1I 2 + 10 = 0 …(iii) I 2 + 12I 3 = 10 Examination Papers 211 27. Given I v = = 4 Þ v = 4u 0 u From lens formula 1 1 1 = - , f v u 1 1 1 = 20 4u u Þ u = - 15 cm thus, v = 4 ´ ( -15) = - 60 cm Object distance = 15 cm Image distance from lens = 60 cm. I O F' u v F CBSE (AI) Set–III 5. Radiowaves, Infrared light, Blue light, X-rays. 8. Kinetic energy, K= 1 1 e2 mv 2 = × 2 4pe 0 2r 1 e2 4pe 0 r 1 e2 4pe 0 2r …(i) …(ii) …(iii) Potential energy, U = Total energy E =K +U =- Comparing equations (i), (ii), (iii), we have K = - E and U = 2E Given (in ground state) E = - 13 × 6 eV \ Kinetic energy, K = 13 × 6 eV Potential energy U = 2 ´ ( - 13 × 6 eV) = - 27 × 2 eV 21. v = 10 m Real image is formed on screen, so image is inversed. v \ = - 19, u v 10 u=- =- m 19 19 1 1 1 Lens formula = - gives f v u 1 1 19 20 = + = =2 f 10 10 10 1 f = m = 50 cm 2 212 Xam idea Physics—XII æE ö æ 2 ×5 ö 24. r = ç - 1÷ R = ç - 1÷ ´ 5 W 4 è ø è 2×0 ø 5 = W = 1× 25 W 4 25. (a) (i) For II bright fringe, æ mDl ö 2 ´ 1× 0 ´ 600 ´ 10 y2 ç = ÷= d ø è 0 × 20 ´ 10 -3 For II dark fringe, 1 ö Dl æ 1 ö 1× 0 ´ 600 ´ 10 -9 æ y¢ =ç2 - ÷ 2 =çn - ÷ 2ø d è 2 ø 0 × 20 ´ 10 -3 è = 4 × 5 ´ 10 -3 m = 4 × 5 mm 26. From Kirchhoff’s I law I 3 = I1 + I 2 …(i) Applying Kirchhoff’s II law to loop PRSP - 20I 3 - 200I 2 + 5 = 0 Þ 40I 2 + 4I 3 = 1 Applying Kirchhoff’s II law to loop PRQP - 20I 3 - 60I 1 + 4 = 0 Þ 15I 1 + 5I 3 = 1 -9 = 6 ´ 10 -3 m = 6 mm …(ii) …(iii) CBSE EXAMINATION PAPERS FOREIGN–2010 Time allowed: 3 hours General Instructions: (a) All questions are compulsory. (b) There are 30 questions in total. Questions 1 to 8 carry one mark each, questions 9 to 18 carry two marks each, questions 19 to 27 carry three marks each and questions 28 to 30 carry five marks each. (c) There is no overall choice. However, an internal choice has been provided in one question of two marks, one question of three marks and all three questions of five marks each. You have to attempt only one of the given choices in such questions. (d) Use of calculators is not permitted. (e) You may use the following values of physical constants wherever necessary: c = 3 ´ 108 ms - 1 h = 6 × 626 ´ 10 -34 Js e = 1× 602 ´ 10 -19 C m 0 = 4p ´ 10 -7 TmA -1 1 = 9 × 109 Nm2C– 2 4pe o Boltzmann’s constant k = 1× 381 ´ 10 -23 J K -1 Avogadro’s number N A = 6 × 022 ´ 10 23 /mole Mass of neutron m n = 1× 2 ´ 10 -27 kg Mass of electron m e = 9 ×1´ 10 -31 kg Radius of earth = 6400 km Maximum marks: 70 CBSE (Foreign) SET–I 1. A charge Q mC is placed at the centre of a cube. What is the electric flux coming out from any one surface? 2. What is the characteristic property of a diamagnetic material? 3. Which part of the electromagnetic spectrum is used in satellite communciation? 4. A metallic sphere is placed in a uniform electric field a as shown in the figure. Which path is followed by b electric field lines and why? c d 5. Why does the sky appear blue? 6. Name an experiment which shows wave nature of electrons. Which phenomenon was observed in this experiment using an electron beam? 214 Xam idea Physics—XII × × × × × 7. Two loops of different shapes are moved in a region of × c × × × g × uniform magnetic field in the directions marked by arrows as shown in the figure. What is the direction of the induced a × × × × × d f b current in each loop? × × × × e × 8. State Bohr’s quantisation condition for defining stationary B orbits. 9. In standard AM broadcast, what mode of propagation is used for transmitting a signal? Why is this mode of propagation limited to frequencies upto a few MHz? 10. Write the truth table for the following circuit. Name the equivalent gate that this circuit represents. A Y B ® 11. Define drift velocity. Write its relationship with relaxation time in terms of the electric field E applied to a conductor. A potential difference V is applied to a conductor of length L. How is the drift velocity affected when V is doubled and L is halved? OR Define ionic mobility. Write its relationship with relaxation time. How does one understand the temperature dependence of resistivity of a semiconductor? 12. If both the number of protons and the number of neutrons are conserved in a nuclear reaction like 12 6 C + 12 6C ¾® 20 10 Ne + 4 2 He, in what way is mass converted into energy? Explain. 13. Draw a ray diagram to show the formation of the image in a myopic eye. Show with the help of a ray diagram how this defect is corrected. 14. An a-particle and a proton moving with the same speed enter the same magnetic field region at right angles to the direction of the field. Show the trajectories followed by the two particles in the region of the p magnetic field. Find the ratio of the radii of the circular paths which a the two particles may describe. 15. State the principle of working of a potentiometer. Define potential B gradient and write its S.I. unit. 16. Define the resolving power of a microscope. How is this affected when (i) the wavelength of illuminating radiations is decreased, and (ii) the diameter of the objective lens is decreased? Justify your answer. 17. Two long co-axial solenoids of the same length but different radii and different number of turns are wound one over the other. Deduce the expression for the mutual inductance of this arrangement. 18. How are X-rays produced? Write their two important uses. (a) How is the electric field due to a charged parallel plate capacitor affected when a dielectric slab is inserted between the plates fully occupying the intervening region? (b) A slab of material of dielectric constant K has the same area as the plates of a parallel plate 1 capacitor but has thickness d. 20. 4W A B 1W C 12 W D 4W 6W 16 V 1W (a) Compute the equivalent resistance of the network.5 mm from the centre of the screen. refractive index of air-glass = 1× 5). (c) the width of the central maximum. is used to obtain interference fringes in a Young’s double slit experiment. 23. 25. (a) Plot a graph comparing the variation of potential ‘V’ and electric field ‘E’ due to a point charge ‘Q’ as a function of distance ‘R’ from the point charge. OR A beam of light consisting of two wavelengths. Find the 2 expression for the capacitance when the slab is inserted between the plates. What is the least distance from the central maximum where the bright fringes due to the both the wavelengths coincide? The distance between the slits is 2 mm and the distance between the plane of the slits and screen is 120 cm. becomes the same. Write two important advantages that the reflecting telescope has over a refracting type. by drawing trajectories of the scattered a-particles. (b) Obtain the voltage drops VAB and VCD . (b) the distance of the second maximum from the centre of the screen. . in the two cases. It is observed that the first minimum is at a distance of 2. 22. A network of resistors is connected to a 16 V battery of internal resistance of 1 W as shown in the figure. (b) Find the ratio of the potential differences that must be applied across the parallel and the series combination of two identical capacitors so that the energy stored. Draw a schematic diagram of a reflecting telescope (Cassegrain). where d is the separation between the plates. 650 nm and 520 nm. Draw a schematic arrangement of the Geiger – Marsden experiment for studying a-particle scattering by a thin foil of gold. (a) How is the focal length of a spherical mirror affected when the wavelength of the light used is increased? (b) A convex lens has 20 cm focal length in air. Find (a) the width of the slit. Dsecribe briefly.Examination Papers 215 19. What is its focal length in water? (Refractive index of air-water = 1× 33. how this study can be used to estimate the size of the nucleus. A parallel beam of monochromatic light of wavelength 500 nm falls normally on a narrow slit and the resulting diffraction pattern is obatined on a screen 1 m away. 21. 24. (c) Sketch the magnetic field lines for a finite solenoid. OR (a) What are eddy currents? How are these currents reduced in the metallic cores of transformers? (b) A step down transformer operates on a 2 × 5 KV line. How are these field lines different from the electric field lines from an electric dipole? OR (a) Using Biot-Savart Law. the use of a Zener diode as a voltage-regulator. Explain briefly the reasons why wave theory of light is not able to explain the observed features in photoelectric effect. (b) Draw a labelled block diagram of a simple modulator for obtaining an AM signal. (b) Use it to derive an expression for magnetic field insdie. (c) A photodiode is operated under reverse bias although in the forward bias the current is known to be more than the current in the reverse bias. (a) State briefly any two reasons explaining the need for modulating a signal. . calculate: (i) the power output. Find the voltage (rms) across the resistor. an inductor of H and a capacitor of mF are connecetd in series across p p a source of alternating voltage of 140 sin 100 pt volts. 28. OR (a) Draw the circuit diagram of a base-biased n-p-n transistor in C-E configuration. deduce an expression for the magnetic field on the axis of a circular current loop. It supplies a load with 20 A. How do we explain the working of a transistor as a switch using the characteristic? 30.Vi characteristics). What is the ® O 29. The ratio of the primary winding to the secondary is 10: 1. (ii) the voltage. Define the terms ‘threshold frequency’ and ‘stopping potential’ in the study of photoelectric emission. magnetic field B at O due to (i) straight segments (ii) the semi-circular arc? 5 50 (a) A resistor of 400 W. On closing the switch. (b) Explain with the help of a circuit diagram. (a) State Ampere’s circuital law. (a) Draw I-V characteristics of a Zener diode. show that the sum of the energies in the capacitor and inductor is constant in time in the free oscillations of the LC circuits. (Given 2 = 1× 4) (b) An ideal capacitor having a charge q = q 0 cos wt is connected across an ideal inductor ‘L’ through a switch ‘S’. resolve the paradox. 27. (c) A straight wire carrying a current of 12 A is bent into a semi-circular arc of radius 2 × 0 cm as shown. along the axis of an air cored solenoid.216 Xam idea Physics—XII 26. Explain giving reason. Explain how this circuit is used to obtain the transfer characteristic (Vo . (b) Draw the magnetic field lines due to a current carrying loop. the inductor and the capacitor. and (iii) the current in the secondary. Is the algebraic sum of these voltages more than the source voltage? If yes. If the transformer is 90% efficient. Show p the trajectory followed by the two particles in the magnetic field. How are infra-red rays produced? Write their two important uses. and (ii) out of the field. Collector current (IC) in mA 10 8.VCE ) of an n-p-n transistor in C-E configuration is shown in the figure. A Y B . Under what condition is the error in determining the unknown resistance minimized? 12. A circular loop is moved through the region of uniform magnetic field.5 Base current (IB) 60 mA 50 mA 40 mA 30 mA 20 mA 8 6 4 2 0 2 10 mA 4 6 8 10 12 14 16 Collector to emitter voltage (VCE) in volts CBSE (Foreign) SET–II Questions uncommon to Set–I 4. Find the direction of induced current (clockwise or anticlockwise) when the loop moves: (i) into the field. An electron and a proton moving with the same speed enter the same magnetic field region at right angles to the direction of the field. Calculate (i) the output resistance r 0 and (ii) the current amplification factor b ac . Wrtie the truth table for the following circuit. Name the gate that this circuit represents. B 8. 10. Find the ratio of the radii of the circular paths which the particles may e describe. 11.Examination Papers 217 (b) The typical output characteristics ( I C . Name the electromagnetic radiation used to destroy cancer cells and write its frequency range. B 13. State the principle on which the working of a meter bridge is based. What is the direction of induced current in the loop? I 9. A rectangular loop of wire is pulled to the right.218 Xam idea Physics—XII 18. . Draw a ray diagram to show the formation of the image in a far-sighted (hypermetropic) eye.2 W of 0. (a) Plot a graph comparing the variation of potential ‘V’ and electric field ‘E’ due to a point charge ‘Q’ as a function of distance ‘R’ from the point charge. 19. in the two cases. (b) A convex lens has 10 cm focal length in air. To which part of the electromagnetic spectrum do the waves emitted by radioactive nuclei belong? What is its frequency range? 6. Calculate the value of current flowing through the 3 W resistance. 23. refractive index of air-glass = 1× 5). (a) Plot a graph comparing the variation of potential ‘V’ and electric field ‘E’ due to a point charge ‘Q’ as a function of distance ‘R’ from the point charge. Also. 15. Two cells E1 and E 2 of EMF’s 5 V and 9 V and internal resistances 0. name the gate. E1 E2 21.3 W and 1. Draw the output wavefrom for the following gate.5 W 3W 22. How are microwaves produced? Write their two important uses. What is its focal length in water? (Refractive index of air-water = 1× 33. 6W 4.3 W 1. becomes the same. Show with the help of a ray diagram how this defect is corrected. (a) How is the focal length of a spherical mirror affected when it is immersed in water.2 W respectively are connected to a network of 5V 9V resistances as shown in the figure. (b) Find the ratio of the poetntial differences that must be applied acros the parallel and the series combination of two capacitors C1 and C 2 with their capacitances in the ratio 1: 2 so that the energy stored. Draw a ray diagram to show the formation of the image in a myopic eye. 1 A 0 1 B 0 t1 t2 t3 t4 t5 t1 t2 t3 t4 t5 A B Y 13. CBSE (Foreign) SET–III Questions uncommon to Set–I and Set–II 5. Show with the help of a ray diagarm how this defect is corrected. away from the long straight wire through which a steady current I flows upwards. . According to Lord Rayleigh the intensity of scattered light 1 1 I µ Þ Iµ 4 (wavelength) l4 As l blue < l red . 1 £ c £ 0. Davission-Germer experiment shows wave nature of electrons. Diamagnetic substances: These are the substances in which feeble magnetism is produced in a direction opposite to the applied magnetic field. Reason: There are no electric field lines within a metallic sphere and field lines are normal at each point of the surface. in the two cases. becomes the same. The light is scattered by air molecules. ® These substances have small negative values of magnetism M and susceptibility c and positive low value of relative permeability m r . Short radiowaves l < 10 m or v > 30 MHz are used in satellite communication. The phenomenon of diffraction of electron beam was observed in this experiment to produce magnetic field upward. Loop abc is entering the magnetic field. lead.e. 5. 6.8 W Solutions CBSE (Foreign) SET–I 1. 20. copper. Path (d) is followed by electric field line.Examination Papers 219 (b) Find the ratio of the potential differences that must be applied across the parallel and the series combination of two capacitors C1 and C 2 with their capacitances in the ratio 1: 3 so that the energy stored. 7. These substances are repelled by a strong magnet. nitrogen (at STP) and sodium chloride. water. 0 < m r < 1 The examples of diamagnetic substances are bismuth. Calculate the steady current through the 2 W resistor in the circuit shown below. The induced current always opposes the change in magnetic flux. Loop defg is .m 6e 0 2. Q mV . 4. 3. so sky appears blue. so magnetic flux linked with loop tends to increase. accordingly blue colour is scattered the most and red the least. 2W A B 3W 2 mF 4W 6V 2. so current induced in loop abc is anticlockwise to produce magnetic field upward to oppose the increase in flux. i. antimony. .220 Xam idea Physics—XII leaving the magnetic field. 2. so flux linked with it tends to decrease. 9. Drift velocity is defined as the average velocity with which the free electrons get drifted towards the positive end of the conductor under the influence of an external electric field applied. This mass defect appears as energy of reaction. 2p h …(ii) mvr = n . In fact the number of protons and number of neutrons are same before and after a nuclear reaction. the electromagnetic wave penetrate and escape. Sky wave propagation. ® vd ® . 8. When temperature increases.e. 12. e = charge of electron E = electric field applied Becomes 4 times. This difference is called the mass defect. K 2p Integer n is called the principal quantum number. In this sense a nuclear reaction is an example of mass-energy interconversion. Above a frequency of few MHz. n = 1. Y = A + B = A + B The equivalent gate is OR gate. 3. Quantum Condition: The stationary orbits are those in which angular momentum of electron is an h integral multiple of i. but the binding energies of nuclei present before and after a nuclear reaction are different. covalent bonds of neighbouring atoms break and charge carrier become free to cause conductive.e. the induced current will be clockwise to produce magnetic field downward to oppose the decrease in magnetic flux. The Truth Table.. so resistivity of semi-conductor decreases with rise of temperature. Input A 0 1 0 1 B 0 0 1 1 Output Y 0 1 1 1 11. It is given by eE =t m where m = mass of electron. This equation is called Bohr’s quantum condition. OR Mobility of an ion is defined as the drift velocity per unit electric field i. v et m= d = E m 2 Its unit is m /Vs. 10. unit of potential gradient is volt/metre. there is a fall of potential along the wire from A to B. 14. Due to these reasons the image is formed in front of the retina. G When a steady current is passed through a + – E potentiometer wire AB. The circuit containing battery Primary circuit B1 is the main circuit and the circuit P P P2 B 1 containing battery B 2 is the secondary A – + J circuit. 16. but fails to see the far away objects distinctly. This defect is due to Eye lens Parallel rays from object F I Eye lens (a) Image formation by myopic eye (b) Corrected myopia Corrective lens Retina I (a) decrease in focal length of the eye lens. It is a device to measure the potential K – difference across a circuit element + Rh Driver battery accurately.I.Examination Papers 221 13. Myopia or shortsightedness: Myopia is the defect of eye in which a person can see only nearby objects. q r p ( m / q) p ( m p / e) 1 = = = r a ( m / q) a (( 4m p ) / 2e) 2 B1 a × p a p × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × 15. (b) Spreading of the eye-sphere. Unit of resolution of microscope 1× 22l dq = 2n sin q 1 Resolving power µ unit of resolution n sin q µ l . The fall of potential per unit length along potentiometer wire is called the potential gradient. Radius of charged particle in magnetic field mv r= qB m µ for same v and B. If L is length of wire AB and V is the potential difference across it then V Potential gradient k = L The S. For the working of potentiometer Secondary E circuit emf of battery B1 > emf of battery B 2 . so resolving power decreases. M 21 = 2 = 0 1 2 2 I1 l If n1 is number of turns per unit length of outer solenoid and r 2 is radius of inner solenoid. Production of X-rays: When high energetic electrons strike a metallic target of high atomic weight and high melting point. E (b) Let C be capacitance of each capacitor. M = m 0 n1 N 2 pr 2 18. the magnetic field produced within this solenoid. then given Us =U p 1 1 2 2 Þ C S VS = C pV p 2 2 Vp Cs C/2 1 Þ = = = Vs Cp 2C 2 .. then 2 . The total flux linkage with inner coil of N 2 -turns.222 Xam idea Physics—XII (i) When wavelength l decreases. 19. They are used in engineering to check flaws in bridges. (a) The graph of variation of potential and electric field due to a point charge Q with distance R from the point charge is shown in fig. m N N æ m N I ö F2 = N 2 f2 = N 2 B1 A 2 = N 2 ç 0 1 1 ÷ A 2 = 0 1 2 A 2 I 1 l l è ø F m N N A By definition Mutual Inductance.. presence of stone in gallbladder and kidney. E V or C In series arrangement net capacitance C s = . C p = 2C 1 Energy stored U = CV 2 2 If VS and VP are potential differences applied across series and parallel arrangements. m N I . tuberculosis of lungs. V V 2 E R In parallel arrangement. (ii) When diameter of obejctive lens decreases. where A 2 is the cross-sectional area of inner solenoid. r1 17. In physics X-rays are used to study crystal structure. resolving power increases. Uses: X-rays are used in medical diagnostics to detect fractures in bones. Mutual Inductance of Two Co-axial Solenoids : N1 Consider two long co-axial solenoid each of length l with number of turns N1 and N 2 wound r2 N2 one over the other. In production of X-rays mechanical energy of electrons is converted with electromagneitic energy of X-rays. net capacitance.(ii) B1 = 0 1 1 l The flux linked with each turn of inner solenoid coil is f2 = B1 A 2 . Number of turns per unit N length in order (primary) solenoid. n = 1 × If I 1 is l l the current flowing in primary solenoid. q decreases. X-rays are produced. D = 1 m If a is width of slit.9 = d 1 ´ 10 . n1 < n 2 If n1 = n. there must be a difference of 1 in order of l1 and l 2 . n = 2 1 ö 1 ´ 5 ´ 10 -7 æ \ ( y 2 ) max = ç 2 + ÷ = 5 ´ 10 -4 m = 0 × 5 mm 2 ø 2 × 5 ´ 10 -3 è æ 2Dl ö (c) Width of central maximuim. y min = nDl1 4 ´ 1 ´ 650 ´ 10 .Examination Papers 223 20.520) nm 130 \ Least distance.3 = 2 × 6 ´ 10 . sin q1 = q1 = 1 D y1 l \ = D a y1 = 2 × 5 mm = 2 × 5 ´ 10 -3 m \ a= lD 5 ´ 10 -7 ´ 1 = = 2 ´ 10 -4 m = 0 × 2 mm -3 y1 2 × 5 ´ 10 1 ö Dl æ (b) Position of nth maximum. (a) Given l = 500 nm = 5 ´ 10 -7 m. y n = ç n + ÷ 2ø a è For second maximum.l 2 ( 650 . As l1 > l 2 . n 2 = n + 1 nDl1 ( n + 1) Dl 2 \ ( y n ) l1 = ( y n + 1 ) l2 Þ = d d l2 520 nm 520 or n = Þ nl1 = ( n + 1) l 2 Þ n= = =4 l1 .3 m = 2 × 6 mm Here D = 120 cm = 1× 20 m and d = 2 mm = 2 ´ 10 -3 m y min = nDl1 4 ´ 1× 2 ´ 650 ´ 10 -9 = m d 2 ´ 10 -3 = 1× 56 ´ 10 -3 m = 1× 56 mm . then for first minimum l sin q1 = a y For small q1 . ç = ÷ a ø è = Separation between first minima on either side of centre of screen = 2 × 5 + 2 × 5 = 5 mm OR For least distance of coincidence of fringes. suffer scattering through angles more than 90°. then a few of them experience such a strong repulsive force that they turn back. Therefore. A 2 . in 1911. This implies that “most part of the atom is hollow. The entire apparatus is placed in a vacuum chamber to prevent any energy loss of a-particles due to their collisions with air molecules. K etc. The number of a-particles scattered in any direction is counted by scintillation counter. This arrangement provides a fine beam of a-particles. This implies that when fast moving positively charged a-particles come near gold-atom. H. Marsden performed an important experiment called Geiger-Marsden experiment (or Rutherford’s scattering experiment). Observations and Conclusions (i) Most of a-particles pass through the gold foil undeflected. e. a-particles are scattered by this foil. 3. This scattering takes place in all possible directions. 2. Geiger. called the nucleus. Source of a-particles : The radioactive source polonium emits high energetic alpha ( a –) particles. polonium is used as a source of a-particles. ZnS Screen Incident beam of a-particles Nucleus Detector The distance of closest approach of a-particle gives the estimate of nuclear size. The foil taken is thin to avoid multiple scattering of a-particles. while a still smaller number (nearly 1 in 8000) retrace their path. If Ze is charge of nucleus E k kinetic energy of a particle 2e charge on a-particle the size of nucleus r 0 is given by . This source is placed in an enclosure containing a hole and a few slits A1 . the a-particles are scattered due to collision with gold atoms. At the suggestion of Rutherford. to ensure that a-particle be deflected by a single collision with a gold atom. On this basis Rutherford concluded that whole of positive charge of atom is concentrated in a small central core. Method: When a-particle beam falls on gold foil.6 m. are placed in front of the hole.” (ii) a-particles are scattered through all angles.224 Xam idea Physics—XII 21. Some a-particles (nearly 1 in 2000). Thin gold foil : It is a gold foil* of thickness nearly 10 .. It consists of 1. Scintillation counter : By this the number of a-particles scattered in a given direction may be counted. and E. i. (a) Wavelength of light has no effect on focal length of a spherical mirror. (iii) The negative charges (electrons) do not influence the scattering process.1 ´ 20 cm » 80 cm fl = ´ fa = ng æ 1× 5 ö . the permittivity of medium becomes Ke 0 . I = 4 ´ 2 = 8 V 24. Equivalent resistance of network R AD = R AB + R BC + RCD = 2 + 1 + 4 = 7 W. E q so final electric field between the plates.14 1 the order of 10 . fa = 20 cm. E 0 = q s q/ A = = e0 e0 A e0 After introduction of dielectric.10 m. K .5. E = = 0 AK e 0 K 1 i.e. This implies that nearly whole mass f of atom is concentrated in nucleus. 000 10 atom.. n l = 3 ng -1 1× 5 . electric field reduces to times. while size of atom is of 10 . therefore the size of nucleus is about times the size of = . ng = 1.10 10.Examination Papers 225 Ek = r0 = 1 ( Ze) ( 2e) 4pe 0 r0 1 2Ze 2 4pe 0 E k Calculations show that the size of nucleus is of the order of 10 . 22. (a) Initial electric field between the plates. 4 (b) Given. E 16 (b) Current in circuit I = = =2A R + r 7 +1 VAB = R AB I = 2 ´ 2 = 4 V VCD = RCD .1÷ ç -1 è 1× 33 ø nl (a) R AB = R BC 4´ 4 = 2 W. RCD = =4W 12 + 6 23.14 m. 4+4 12 ´ 6 = 1 W. (ii) Its resolving power is greater than refracting telescope due to larger aperture of mirror. If charges on plates are + Q and . C= e0 A t d -t + K = e0 A 1 ö æ d -t ç 1 . area of each plate being A. C = or.Q. Secondary mirror Eyepiece Advantages : (i) It is free from chromatic aberration. Let a dielectric slab of dielectric constant K and thickness t < d be placed between the plates. t = \ d 2 e0 A e0 A = dæ 1ö dæ 1ö d .t ). the separation between the plates being d. . Q Q = E1 ( d .226 Xam idea Physics—XII (b) Consider a parallel plate capacitor. E 2 = = Ke 0 K e 0 A \ The potential difference between the plates VAB = work done in carrying unit positive charge from one plate to another = SEx (as field between the plates is not constant).t ) + t e0 A K e0 A \ VAB = Q é t ù d -t + ú e0 A ê K ë û Q Q = Q t ö VAB æ ç d -t + ÷ e0 A è K ø \ Capacitance of capacitor.÷ ç1 + ÷ 2è Kø 2è Kø Objective mirror C= 25.ç1 . then surface charge density Q s= A Q s The electric field between the plates in air. E = = e0 e0 A Q s The electric field between the plates in slab.t ) + E 2 t = (d . The thickness of air between the plates is ( d .÷ K ø è Here. the light of frequency less than threshold frequency can not emit electrons. the energy transferred by light waves will not go to a particular electron. but according to experimental observations. but according to experimental observations. These waves will then. It is denoted by V0 (or VS ). The observed characteristics of photoelectric effect could not be explained on the basis of wave theory of light due to the following reasons. wc . whatever the intensity of incident light may be. the light of any frequency can emit electrons from metallic surface provided the intensity of light be sufficient to provide necessary energy for emission of electrons. Stopping Potential : The minimum retarding potential applied to anode of a photoelectric tube which is just capable of stopping photoelectric current is called the stopping potential. It is denoted by v 0 . (iii) According to wave theory. but it will be distributed uniformly to all electrons present in the illuminated surface. according to wave theory. (a) The modulation is needed due to (i) Transmission of audiofrequency electrical signals need long impracticable antenna. electrons will take some time to collect the necessary energy for their emission. 2wm . 27.Examination Papers 227 26. 2wc and allows wave of frequency wc . The output of Bandpass filter is AM wave. (iii) The various information signals transmitted at low frequency get mixed and hence can not be distinguished. the light propagates in the form of wavefronts and the energy is distributed uniformly over the wavefronts. (ii) The power radiated at audio frequency is quite small. (i) According to wave theory. provide more energy to electrons of metal. consequently the energy of electrons will increase. (ii) According to wave theory. whatever the intensity of light may be. Thus. Therefore. hence transmission is quite lossy. . With increase of intensity of light.wm and wc + wm . the amplitude of waves and the energy stored by waves will increase. m(t) + Am sin wmt c(t) (Modulating signal ) AC sin wct (carrier wave) x(t) Square law device y(t) Bandpass Filter centred at wc AM wave B x(t) + cx2(t) The modulating signal is superposed on carrier wave of high frequency ( » MHz). The time for emission will be more for light of less intensity and vice versa. Threshold Frequency : The minimum frequency of incident light which is just capable of ejecting electrons from a metal is called the threshold frequency. the kinetic energy of photoelectrons must depend on the intensity of incident light. the kinetic energy of photoelectrons does not depend on the intensity of incident light. (b) Amplitude Modulation: The block diagram is shown in fig. The resultant wave so obtained is sent to square law device which producses wave y(t) = Bx (t) + Cx 2 (t) This is finally sent to Bandpass filter which rejects dc and sinusoids of frequencies wm . But experimental observations show that the emission of electrons take place instantaneously after the light is incident on the metal. dl = m 0 I (b) Magnetic Field Due to a Current Carrying Long Solenoid: S N A solenoid is a long wire wound in the form of a close-packed helix. To construct a solenoid a large number of closely packed turns of – + insulated copper wire are wound on a cylindrical tube of card-board or china clay. Let I be the current flowing in the solenoid. then N n= l m 0 NI or B= l Derivation: Consider a symmetrical long solenoid having number of turns per unit length equal to n.. then by right hand rule. then magnetic field at axis of long solenoid B = m 0 nI If there are N turns in length l of wire. carrying current.228 Xam idea Physics—XII 28. then the magnetic field within the B solenoid is uniform. Applying Ampere’s law to this path ® ® ò B · dl = m ´ 0 (since net current enclosed by path is zero) As dl ¹ 0 \ B = 0 This means that the magnetic field outside the solenoid is zero. The a b magnetic field due to a circular loop is along its axis and the current in upper and lower straight parts of solenoid is equal and opposite. Due to this the magnetic field in a direction perpendicular to the axis of solenoid is zero and so the resultant magnetic field is along the axis of the solenoid. ® ® ò B . If there are ‘n’ number of turns per metre length of solenoid and I amperes is the current flowing. (a) It states that the line integral of magnetic field induction along a closed path is equal to m 0 -times the current enclosed by the path i. The reason is that the solenoid may be supposed to be formed of a large number of circular current elements. When an d c s r electric current is passed through the solenoid. with practically no magnetic field outside it. If the solenoid is long and the B p q successive insulated copper turns have no l gaps. Field outside the solenoid: Consider a closed path abcd. a magnetic field is produced within the solenoid. The line integral of magnetic field B along path pqrs is ® . the magnetic field is parallel to the axis of the solenoid. Field Inside the solenoid: Consider a closed path pqrs.e. equation (i) gives òpqrs B · dl = ò B · dl = Bl . ® òpq B · dl = ò B dl = Bl ® ® ( pq = l say) For paths qr and sp... with its plane perpendicular to the plane of paper. Let P be a point of observation on the axis of this circular loop at a distance x from its centre O. B and d l are mutually perpendicular.(i) ® ® For path pq.(ii) By Ampere’s law ò B · dl = m 0 ´ net current enclosed by path \ Bl = m 0 ( nl I ) \ B = m 0 nI This is the well known result. B = 0 (since field is zero outside a solenoid) \ B · dl = 0 ® ® pq ® ® ® ® In view of these. (c) I Axis I The magnetic field lines of magnet (or current carrying solenoid) form continuous closed loops and are directed from N to S pole outside the magnet and S to N pole inside the magnet and forms closed loops while in the case of an electric dipole the field lines begin from positive charge and end on negative charge or escape to infinity.. The ® magnitude of the magnetic induction dB at point P due to this element is given by .Examination Papers ® ® pq ® ® qr ® ® rs ® ® sp ® ® 229 òpqrs \ B · dl = ò ® B · dl + ò B · dl + ò B • dl + ò B · dl . ® ® sp ® \ òqr òrs B · dl = ò ® ® B · d l = ò B dl cos 90° = 0 ® For path rs. Consider a small element of length dl of the coil at point A. B and dl are along the same direction.. OR (a) Magnetic field at the axis of a circular loop: Consider a circular loop of radius R carrying current I. ..(ii) ® In figure dB has been represented by P Q. the magnitude of the magnetic ® induction dB is given by. PM and PN ¢ along the axis and perpendicular to the axis respectively.. Thus the resultant ® ® magnetic induction B at axial point P is along the axis and may be evaluated as follows: ® The component of dB along the axis.. B = = tesla. As the angle between I dl and r is 90°.. while those parallel to the axis will be added up. m I dl sin 90° m 0 I dl dB = 0 = × 4p r2 4p r 2 ® ® ® .. Now consider another small current element `of length dI ¢ at A ¢ . The magnetic induction dB due ® m I dl ¢ to this element has been represented by P Q¢ whose magnitude is 0 and which can also 4p r 2 be resolved into two components. then by symmetry the components of magnetic induction perpendicular to the axis will be cancelled out. namely PM and PN along and perpendicular to the axis respectively.(iii) ..(v) Therefore. The vector dB or ( P Q) can be resolved into two components.(iv) Therefore the magnitude of resultant magnetic induction at axial point P due to the whole circular coil is given by m 0 IR m 0 IR B= ò dl = ò dl 2 2 3/ 2 4p ( R + x ) 4p ( R 2 + x 2 ) 3 / 2 But ò dl = length of the loop = 2pR m 0 IR 4p ( R 2 + x 2 ) 3 / 2 m 0 IR 2 2 (R 2 + x 2 ) 3/ 2 ( 2 p R) ... then . If the coil contains N turns. m I dI dB x = 0 sin a 4p r 2 R But sin a = and r = ( R 2 + x 2 ) 1/ 2 r \ dB x = m 0 I dl R m 0 IR m 0 IR × = dl = dl 2 3 r 4p r 4p r 4p ( R 2 + x 2 ) 3 / 2 ..230 Xam idea Physics—XII dB = m 0 I dl sin a 4p r2 ® ® ® . Thus if we consider the magnetic induction produced by the whole of the circular coil.(i) ® ® The direction of dB is perpendicular to the plane containing dl and r and is given by right hand screw rule. . q = ® B2 p ® ® I S dl sin m 0 SI d l ´ r m 0 m Ipr 2 n $= 0 $ \ = = n 3 2 4p 4p 4p r 2 r r m I = 0 . m I dl ´ r dB= 0 4p r 3 (i) For straight segments q = 0 or p ® ® ® ® Þ \ $ =0 dl ´ r = dl r sin 0 n p 2 ® B1 = 0 (ii) For semicircular arc S dl = pr . L = H. 5 50 50 29. we get V0 = 140 V.. w = 100p Þ 2pn = 100p Þ n = 50 Hz V 140 Vrms = 0 = = 100 V 2 1× 4 5 Inductive reactance X L = wL = 100 p ´ = 500 W p .. C = mF = ´ 10 -6 F p p p E = 140 sin 140pt Comparing with V = V0 sin wt.(vi) (b) Magnetic field lines due to a circular current loop: (c) Magnetic field due to a current carrying element. 4r directed perpendicular to plane of paper downward.Examination Papers 231 B= m 0 NIR 2 2 (R 2 + x 2 ) 3/ 2 tesla. R = 400 W. VC ) 2 = (80) 2 + ( 60) 2 = 100 V p × 2 This resolves the paradox. X C = 1 = wC 1 = 200 W æ 50 ö -6 100p ´ ç ÷ ´ 10 è pø Impedance Z = R 2 + ( X L . the free electrons of conductor experience Lorentz force and traverse closed paths.VC ) leads VR by an angle 2 V = VR + (VL . which causes heating of conductor. so induced currents are induced in the body of x x x x field conductor. (b) Given V p = 2 × 5 kV = 2 × 5 ´ 10 -3 V. I = This is because VR and (VL .200) 2 = 500 W Vrms 100 = = 0×2 A Z 500 Voltage across resistor. rms current in circuit. r = (i) h = Pout Pin Vs 1 = . which are equivalent to small current loops. OR x x x x x x x (a) Eddy Currents: When a conductor is placed in a varying Conductor x x x x magnetic field the magnetic flux linked with the conductor x x x x Varying magnetic changes. VL = X L I = 500 ´ 0 × 2 = 100 V Voltage across capacitor.VC ) are not in phase but (VL . they cause heating effect and sometimes the conductor becomes red-hot. etc. I p = 20 A Input power. These currents are the eddy currents. Eddy current losses may be reduced by using laminated soft iron cores in galvanometers. VR = RI = 400 ´ 0 × 2 = 80 V Voltage across inductor. Pout = hPin = 0 × 9 ´ 50 kW = 45 kW 1 (ii) Voltage across secondary Vs = rV p = ´ 2 × 5 ´ 10 3 V = 250 V 10 P 45 ´ 10 3 (iii) Current in secondary I s = out = = 180 A Vs 250 .40 = 160 V Yes algebraic sum of voltages (160 V) is more than the source voltage (100 V). Efficiency h = 90% = 0 × 9 V p 10 Power output. In varying magnetic field. and making holes in conductor. transformers.X C ) 2 = ( 400) 2 + (500 . Pin = V p I p = 2 × 5 ´ 10 3 ´ 20 = 50 ´ 10 3 W = 50 kW Turn ratio. The x x x x currents induced in the conductor are called the eddy x x x x x x x currents. VC = X C I = 200 ´ 0 × 2 = 40 V Algebraic sum of voltages = 80 + 100 .232 Xam idea Physics—XII Capacitive reactance. then V0 = VZ = I Z × R Z = I L R L Applying Kirchhoff’s law to the mesh containing resistance R.4 0. The series resistance R absorbs the output voltage fluctuations so as to maintain constant voltage across the load. The value Vin VZ RL V0 of the series resistance is so chosen that the diode operates in the breakdown region under the Zener voltage VZ across it. I Z the current through Zener diode and I L the current through load. Zener diode and supply voltage Vin . the voltage across it remains practically constant for a large change in the current. The operation of the circuit may be explained as follows : I IL R Let Vin be the unregulated input dc voltage and V0 be the output voltage IZ across R L to be regulated and VZ be the Zener voltage of the diode.I L If R Z is Zener diode resistance. Let I be the current drawn from supply.2 0.8 V (volt) I (mA) (b) Zener diode as a Voltage Regulator The Zener diode makes its use as a voltage regulator due to the following property : When a Zener diode is operated in the breakdown region. (a) I-V Characteristics of Zener diode: I (mA) 100 80 60 40 20 100 80 Vbr 60 40 20 O 10 20 30 0. Then obviously I = I Z + I L or I Z = I .6 0.Examination Papers 233 30. we have Unregulated input Regulated output . The Zener diode is connected across load such that it is reverse biased. A simple circuit of a voltage regulator using a Zener diode is shown in the Fig. RI . there is no current conduction i. OR æ DVCE ö ÷ (b) (i) Output resistance r 0 = ç ç DI ÷ C ø I = constant è B Taking I B = 60 mA = constant For DVCE = (16 .. This implies V0 = Vin .RI ) becomes constant. DI C = (5 × 8 . A further increase of input voltage Vin does not result in the corresponding increase in V0 or VZ but merely increases the voltage drop across R. the breakdown point is reached and the voltage across the diode VZ = (Vin . represents the plot of output voltage V0 versus Regulated output voltage input voltage Vin .RI Fig. (c) A photodiode is used in reverse bias. although in forward bias current in more. When the input voltage Vin is lower than the Zener voltage VZ of diode. I Z = 0. It is clear from the graph that the VZ output voltage remains constant when the diode is in Zener region. the series resistance R for a given range of input voltage be so chosen that (i) the diode operates in Zener region and VZ (ii) current should not exceed a certain value to cause Input voltage burn out of diode.3 × 8) mA = 2 mA DI B = ( 40 . then current in reverse bias because in reverse bias it is easier to observe change in current with change in light intensity.30) mA = 10 mA 2 mA 2 ´ 10 -3 \ b ac = = = 200 10 mA 10 ´ 10 -6 Output voltage V0 .. we have …(ii) V0 = VZ = Vin .e. It may be pointed out that for maintaining constant regulated output. Thus in breakdown region.234 Xam idea Physics—XII RI + VZ = Vin i.8) mA = 0 × 5 mA 12 \ r0 = W = 24 ´ 10 3 W = 24 kW -3 0 × 5 ´ 10 (ii) Current amplification factor æ DI c ö ÷ b ac = ç ç DI ÷ è B øVCE = cosntant Taking VCE = 8 V. …(i) VZ = Vin . As input voltage Vin is increased so that it becomes equal to VZ .4) V= 12 V Corresponding DI C = (8 × 5 .e. These waves are used for long distance photography and for therapeutic purposes. but fails to see the far away objects distinctly. Working of meter bridge is based on Wheatstone bridge. Infrared rays produced by hot bodies and molecules. The symbol and truth table are shown in fig. Truth Table A A B Y B 0 0 1 1 Y 0 0 0 1 0 1 0 1 18. Myopia or shortsightedness: Myopia is the defect of eye in which a person can see only nearby objects. × × × × × 12.Examination Papers 235 CBSE (Foreign) SET–II 4. 10. Frequency range 1019 Hz—10 22 Hz. This defect is due to Corrective lens F I Eye lens (a) Image formation by myopic eye (b) Corrected myopia Eye lens Parallel rays from object Retina I . (i) anticlockwise (ii) clockwise 8. g-rays. re m e = rp mp As mass of proton m p » 1840 ´ mass of electron ( m e ) me 1 \ » m p 1840 \ re 1 = r p 1840 e B × × × × × × × × × × 13. 11. Trajectories are shown in fig. × × × × × mv As r = ®r µ m × × × × × p qB × × × × × Ratio of radii of electron path and proton path. The output Y = A + B = A × B = AB That is equivalent gate is ‘AND’ gate. The error may be minimised by taking balancing length near the middle of the bridge. If a myopic eye has a far point at F. Remedy: To eliminate this defect a concave lens of suitable focal length is used. equivalent capacitance. C 2 = 2C In series combination of C1 and C 2 . (b) spreading of the eye-sphere. equivalent capacitance CC C ´ 2C 2 CS = 1 2 = = C C1 + C 2 C + 2C 3 In parallel combination of C1 and C 2 . This image will act as an object for eye lens and the final image ( I ) will be formed at the retina [Fig (b)].236 Xam idea Physics—XII (a) decrease in focal length of the eye lens. Clearly. . then the parallel rays from infinity will be incident on concave lens and form its vertical image at F. 21. Due to these reasons the image is formed in front of the retina. I = 6´3 + 0 × 3 + 1× 2 = 8 W 6+3 E 4 = = 0 ×5 A R 8 Potential difference across parallel combination of 3 W and 6 W æ 6´3ö V = IR = 0 × 5 ´ ç ç6 +3÷ ÷ =1 V è ø \ Current in R1 = 3 W resistance V 1 I1 = = A R1 3 22. Net emf of circuit E = E 2 . The equivalent focal length of concave lens and eye lens should be increased to a value such that the distinct image of far away objects is formed at the retina.5 = 4V Net resistance of circuit. (a) No change. The focal length of a concave mirror does not depend on the nature of the medium. for elimination of myopia the focal length of corrective concave lens will be equal to the distance of far point of myopic eye from the eye lens. U series = C sVs2 2 1 2 U parallel = C p V p 2 Given U series = U parallel 1 1 2 \ C sVs2 = C p V p 2 2 Vp C s æ 2C / 3 ö 2 Þ = =ç ÷= Vs C p è 3C ø 9 23.E1 = 9 . (b) C1 = C. C p = C1 + C 2 = C + 2C = 3C 1 Energy stored. R eq = 4 × 5 + Curernt in circuit. nl = 1. magnetrons and gunn diodes.5 1 -1 1. Suppose the near point of a normal eye is at N and that of a hypermetropic eye is at O. Direction of induced current in loop is clockwise. Farsightedness or Hypermetropia: Hypermetropia is the defect of eye in which a person can see only farther objects but fails to see nearer objects distinctly. in loop current in clockwise. 15. namely. Output waveform is shown in fig.33 ´ 10 = cm = 39. mv Radius. The paths of proton ( p) and deuteron ( d) are shown in figure.5 ´ 1. The equivalent focal length of corrective convex lens and eye lens should be decreased to a value such that the distinct image of nearby objects is formed at the retina. To eliminate this defect a convex lens of suitable focal length is used. so in nearer side of loop the current will be upward i. (b) Contraction of eye-sphere. Due to these reasons the image of a nearby object is formed behind the retina.33 n g -1 1. Remedy: The near point of hypermetropic eye is displaced from D = 25 cm to some distant point..5 -1 fe = ´ fa = ´ 10 cm ng 1. ng = 1. r = qB m pv rp = eB md v rd = eB rp mp 1 = = rd m d 2 t1 t2 t3 t4 t5 × × × p d × × × × × × × × × × × × × × × × × × × × × × × × × × × 13. g-rays Frequency range 1019 Hz—10 22 Hz. fa = 10 cm. Their frequency range is 3 GHz to 300 Ghz.5.e. then this image will act as the . 6. They are used in radar systems used in air craft navigation and microwave users in houses.12 cm 0. Microwaves are produced by special vacuum tubes. Y = AB The gate is NAND gate. This defect is due to (a) Increase in focal length of eye lens. 14. Reason: Induced current opposes the motion of loop away from wire.Examination Papers 237 (b) Given. 9. as similar currents attracts. klystrons.33 nl 0.17 CBSE (Foreign) SET–III 5. The corrective convex lens forms the image of near point ( N) at point O. 2´3 R eq = + 2 × 8 = 1× 2 + 2 × 8 = 4 W 2+3 Net emf.238 Xam idea Physics—XII object for eye lens and the final image ( I ) will be formed at the retina. V ¢ = IR ¢ = 1× 5 ´ 1× 2 = 1× 8 V Current in R1 = 2 W resistance V ¢ 1× 8 I1 = = = 0×9 A R1 2 . E = 6 V Current in circuit. In steady state there is no current in capacitor branch.8 W I1 B Potential difference across parallel combination of 2 W and 3 W resistances. so equivalent circuit is shown in fig. C p = C1 + C 2 = C + 3C = 4C Given Us =U p 1 1 2 Þ C sVs2 = C p V p 2 2 Vp C s 3C / 4 3 Þ = = = Vs Cp 4C 16 20. (b) Let C1 = C. I = E 6 = = 1× 5 A R eq 4 6V 2W A I 3W I 2. Retina Normal near point N I O Normal near point N I D = 25 cm (a) Image formation by hypermetropic eye Near point of defective eye Corrective convex lens (b) Corrected hypermetropia 19. Therefore the corrective lens enables to form the distinct image of near point ( N) at retina. C 2 = 3C In series combination C s = C1C 2 3 ´ 3C 3 = = C C1 + C 2 C + 3C 4 In parallel combination. Net resistance of circuit. (e) You may use the following values of physical constants wherever necessary: c = 3 ´ 108 ms . 5. (b) There are 30 questions in total.1 h = 6 × 626 ´ 10 -34 Js e = 1× 602 ´ 10 -19 C 1 = 9 × 109 Nm2C– 4pe o m 0 = 4p ´ 10 -7 TmA -1 2 Maximum marks: 70 Boltzmann’s constant k = 1× 381 ´ 10 -23 J K -1 Avogadro’s number N A = 6 × 022 ´ 10 23 /mole Mass of neutron m n = 1× 2 ´ 10 -27 kg Mass of electron m e = 9 ×1´ 10 -31 kg Radius of earth = 6400 km CBSE (Delhi) SET–I 1. What can you say about the direction of electric and magnetic field vectors ? 3. (d) Use of calculators is not permitted. questions 19 to 27 carry three marks each and questions 28 to 30 carry five marks each. Show graphically. Name the type of magnetic materials it represents. if Q is (i) positive (ii) negative? Q O A B 2. A point charge Q is placed at point O as shown in the figure. Questions 1 to 8 carry one mark each. (c) There is no overall choice. However. A potentiometer now measures the potential difference between the terminals of the cell as V. You have to attempt only one of the given choices in such questions. A resistance R is connected across a cell of emf e and internal resistance r. Write the expression for ‘r’ in terms of e.9983. questions 9 to 18 carry two marks each. The permeability of a magnetic material is 0. negative. . or zero.CBSE EXAMINATION PAPERS DELHI–2011 Time allowed: 3 hours General Instructions: (a) All questions are compulsory. A plane electromagnetic wave travels in vacuum along z-direction. 4. the variation of the de-Broglie wavelength (l) with the potential (V) through which an electron is accelerated from rest. one question of three marks and all three questions of five marks each. an internal choice has been provided in one question of two marks. Is the potential difference VA – VB positive. V and R. Two uniformly large parallel thin plates having charge densities +s and –s are kept in the X-Z plane at a distance ‘d’ apart. each of 1 mF capacitance connected to a battery of 6 V. After sometimes ‘S’ is left open and dielectric slabs of dielectric constant K = 3 are inserted to fill completely the space between the plates of the two capacitors.4 G. Figure shows two identical capacitors. each of dipole moment ‘p’ are kept at an angle of 120° as shown in the figure. What is the ratio of their resolving power? Which telescope will you prefer and why? Give reason. Initially switch ‘S’ is closed. 11. doping level in base is increased slightly. what is the magnitude and direction of this field? OR Two small identical electrical dipoles AB and CD. Determine the magnitude of the earth’s magnetic field at the place. its wavelength changes but frequency remains the same. How will the (i) charge and (ii) potential difference between the plates of the capacitors be affected after the slabs are inserted? S 6V 1 mF C1 1 mF C2 12. Explain. are used as the objective lenses in two astronomical telescopes having identical eyepieces.240 Xam idea Physics—XII 6. If a particle of mass m and charge ‘–q’ remains stationary between the plates. When monochromatic light travels from one medium to another. . A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip down at 60° with the horizontal. Sketch an equipotential surface due to electric field between the plates. Whatr is the resultant dipole moment of this combination? If this system is subjected to electric field ( E) directed along + X direction. Two convex lenses of same focal length but of aperture A1 and A2 (A2 < A1). In a transistor. 9. The horizontal component of the earth’s magnetic field at the place is known at to be 0. 8. Define the term ‘wattless current’. C1 and C2. How will it affect (i) collector current and (ii) base current? 7. what will be the magnitude and direction of the torque acting on this? D –q Y A +q 120° X’ X +q B –q C Y’ 10. A A B Y X B t1 t2 t3 t4 t5 t6 t7 14. Define the terms (i) drift velocity. Find (i) their momenta. 16. Draw the output waveform at X. 15. C1 C2 G K 19. (ii) relaxation time. A conductor of length L is connected to a dc source of emf e. Name the semiconductor device that can be used to regulate an unregulated dc power supply. for r > R and r < R.00 nm. . 21. A current is induced in coil C1 due to the motion of current carrying coil C2. Also. explain its working principle. (a) Write any two ways by which a large deflection can be obtained in the galvanometer G. An electron and a photon each have a wavelength 1. (i) Define modulation index. how will the drift velocity change? 20. 17. using the given inputs A and B for the logic circuit shown below. Draw the transfer characteristic curve of a base biased transistor in CE configuration. How are infrared waves produced? Why are these referred to as ‘heat waves’? Write their one important use. If this conductor is replaced by another conductor of same material and same area of cross-section but of length 3L. With the help of I-V characteristics of this device. (ii) Why is the amplitude of modulating signal kept less than the amplitude of carrier wave? 18.Examination Papers 241 13. (b) Suggest an alternative device to demonstrate the induced current in place of a galvanometer. Explain clearly how the active region of the Vo versus Vi curve in a transistor is used as an amplifier. Draw a graph showing the variation of electric field with r. identify the logic operation performed by this circuit. Using Gauss’s law obtain the expression for the electric field due to a uniformly charged thin spherical shell of radius R at a point outside the shell. Depict in the plot the number of undecayed nuclei at (i) t = 3T1/ 2 and (ii) t = 5T1/ 2 . 25.5 eV D – 10 eV 26.242 Xam idea Physics—XII 22. Deduce the expression for the fringe width. (b) When unpolarized light passes from air to a transparent medium. Which of the shown transitions will result in the emission of a photon of wavelength 275 nm? Which of these transitions correspond to emission of radiation of (i) maximum and (ii) minimum wavelength? 0 eV A B C – 2 eV – 4. R2 = R3 = 15 W. with the help of suitable diagram. under what condition does the reflected light get polarized? The energy levels of a hypothetical atom are shown below. State the law of radioactive decay. Write the frequency range for each of the following: (i) Standard AM broadcast (ii) Television (iii) Satellite communication Describe Young’s double slit experiment to produce interference pattern due to a monochromatic source of light. 23. Calculate the equivalent resistance of the circuit and the current in each resistor. (ii) the energy of the photon and (iii) the kinetic energy of electron Draw a schematic diagram showing the (i) ground wave (ii) sky wave and (iii) space wave propagation modes for em waves. R1 I1 E A I2 R2 I4 R4 I3 R3 I1 B . R4 = 30 W and E = 10 V. R1 = 4 W. how the transverse nature of light can be demonstrated by the phenomenon of polarization. OR Use Huygen’s principle to verify the laws of refraction. 24. In the circuit shown. Plot a graph showing the number (N) of undebased nuclei as a function of time (t) for a given radioactive sample having half life T1/ 2 . 27. (a) Describe briefly. Examination Papers 243 28. Name the type of magnetic materials it represents. (a) Draw a ray diagram to show refraction of a ray of monochromatic light passing through a glass prism. 30. Use this law to derive the expression for the magnetic field due to a circular coil carrying current at a point along its axis. 29. (ii) Write any two sources of energy loss in a transformer. State Biot-Savart law.9 × 10–5. . Deduce the expression for the refractive index of glass in terms of angle of prism and angle of minimum deviation. state the underlying principle of a cyclotron. is incident on the convex side of spherical refracting surface of radius of curvature R. Is there an upper limit on the energy acquired by the particle? Give reason. Using this diagram. OR Derive an expression for the impedance of a series LCR circuit connected to an AC supply of variable frequency. How does a circular loop carrying current behave as a magnet? OR With the help of a labelled diagram. (b) Draw a ray diagram to show the image formation by a concave mirror when the object is kept between its focus and the pole. The susceptibility of a magnetic material is 1. Explain clearly how it works to accelerate the charged particles. giving the mathematical expression for it. Explain briefly how the phenomenon of resonance in the circuit can be used in the tuning mechanism of a radio or a TV set. describe briefly the underlying principle and working of a step up transformer. Does it violate law of conservation of energy? Explain. CBSE (Delhi) SET–II Questions uncommon to Set–I 2. OR (a) Obtain lens makers formula using the expression n 2 n1 ( n 2 . (b) Explain briefly how the phenomenon of total internal reflection is used in fibre optics. Plot a graph showing variation of current with the frequency of the applied voltage. (i) With the help of a labelled diagram. Show that cyclotron frequency is independent of energy of the particle. derive the magnification formula for the image formed.n1 ) = v u R Here the ray of light propagating from a rarer medium of refractive index (n1) to a denser medium of refractive index (n2). (iii) A step up transformer converts a low input voltage into a high output voltage. C1 C2 N S A A 11. Figure shows two identical capacitors C1 and C2 each of 1 mF capacitance. identify the logic operation performed by this circuit. After sometimes ‘S’ is left open and dielectric slabs of dielectric constant K = 5 are inserted to fill completely the space between the plates of the two capacitors. connected to a battery of 5 V. How is forward biasing different from reverse biasing in a p-n junction diode? .244 Xam idea Physics—XII 4. What will be the direction of induced current in each coil as seen from the magnet? Justify your answer. A plane electromagnetic wave travels in vacuum along x-direction. A magnet is quickly moved in the direction indicated by an arrow between two coils C1 and C2 as shown in the figure. How will the (i) charge and (ii) potential difference between the plates of the capacitors be affected after the slabs are inserted? S 5V 2m F C1 2m F C2 13. Also. using the given inputs A and B for the logic circuit shown below. Initially switch ‘S’ is closed. Draw the output wave form at X. What can you say about the direction of electric and magnetic field vectors? 10. A B Y X 16. Plot a graph showing the number (N) of undecayed nuclei as a function of time (t) for a given radioactive sample having half life T1 . 9. Explain how a depletion region is formed in a junction diode. Work out the equivalent resistance of the circuit and the current in each resistor. 2 2 CBSE (Delhi) SET–III Questions uncommon to Set–I & II 2. State the law of radioactive decay.50 nm. 24. Initially switch ‘S’ is closed. 2 Depict in the plot the number of undecayed nuclei at (i) t = 2 T1 and (ii) t = 4T1 .5 mF capacitance. 14. Find (i) their momenta. An electron and a photon each have a wave length of 1. The susceptibility of a magnetic materials is –4. Name the type of magnetic materials it represents. R4 = 10 W and E = 6 V. connected to a battery of 2 V.5 mF C . R1 = 4 W. A plane electromagnetic wave travels in vacuum along y-direction.Examination Papers 245 20. (ii) the energy of the photon and (iii) kinetic energy of the electron. After sometimes ‘S’ is left open and dielectric slabs of dielectric constant K = 2 are inserted to fill completely the space between the plates of the two capacitors.5 mF C1 1.2 × 10 -6 . R1 I1 E A I2 R2 I4 R4 I3 R3 I1 B 23. Figure shows two identical capacitors C1 and C2 each of 1. What can you say about the direction of electric and magnetic field vectors? 7. R2 = R3 = 5 W. In the circuit shown. How will the (i) charge and (ii) potential difference between the plates of the capacitors be affected after the slabs are inserted? S 2V 1. An electron and a photon each have a wavelength of 2 nm. R 2 = R 3 = 10 W.246 Xam idea Physics—XII 16. A C Y B D 17. Work out the equivalent resistance of the circuit and the current in each resistor. Write the truth table for the logic circuit shown below and identify the logic operation performed by this circuit. R1 I1 E A I2 R2 I4 R4 I3 R3 I1 B 25. A B S N S N 23. Find (i) their momenta (ii) the energy of the photon (iii) the kinetic energy of the electron. R 4 = 20 W and E = 6 V. In the circuit shown. R1 = 2 W. . Predict the polarity of the capacitor when the two magnets are quickly moved in the directions market by arrows. (i) collector current decreases slightly and (ii) base current increases slightly. When doping level in base is increased slightly. along (–) Y-axis. The equipotential surface is at a distance d/2 +s from either plate in X-Z plane. q + + + + + + + d z y x V=0 + + + + + + + . 8. vertically downward. so magnetic material is diamagnetic.VB is negative. So. 2. l µ . Electric field vector along X-axis Magnetic field vector along Y-axis. then surface (i) weight mg acts. Current flowing in an ac circuit without any net dissipation of power is called wattless current.. d/2 –s so mg = qE mg E= .VB is positive. The potential due to a point charge decreases with increase of distance. 1 5. 7.Examination Papers 247 Solutions CBSE (Delhi) SET–I 1. For a particle Equipotential of charge (–q) at rest between the plates.e. so graph is shown in figure below. vertically downward (ii) electric force qE acts vertically upward. When monochromatic light travels from one medium to another its wavelength and speed both change such that v1 v 2 = = frequency l1 l 2 so frequency remains unchanged. 9. r = ç -1÷ R èV ø 4. i. in case (i) VA . For case (ii) VA . æe ö 3. mr < 1. V l V 6. 4 × 10–4 T If Be is earth’s magnetic field. 6 V while the charge on capacitor C2 remains 6 mC. p. Potential difference across C2 becomes q 6 mC ¢ = 2 = =2V V2 C¢ 6 mF 2 ® p 60° p 120° S 1 mF C1 1 mF C2 . ® ® ® ® ® ® +q B –q C Y’ Torque. 10. the p. across C1 remains 6 V.d. making an angle 60° with Y-axis or 30° with X-axis.÷ è 2ø 2 2 = 2 p 2 . the new capacitance of each capacitor becomes ¢ = C¢ C1 2 = 3 ´ 1 mF = 3 mF ¢ = C1 ¢ V1 = (3 mF) ´ 6 V = 18 mC (i) Charge on capacitor C1.8 G 11.4 G = 0.4 ´ 10 -4 T Þ Be = = = cos q 0 ×5 cos 60 o = 0.8 × 10–4 T = 0. t = P ´ E ( t is perpendicular to both P and E ) 1 = p E sin 30° = pE. then H = Be cosq.p 2 = p.248 Xam idea Physics—XII OR Resultant dipole moment ® pr = p 2 + p 2 + 2 pp cos120 o = 2 p + 2 p 2 cos120 o 2 D –q Y A +q 120° X’ X æ 1ö = 2 p + (2 p ) ´ ç . Angle of dip.4 ´ 10 -4 T 0.d. q1 Charge on capacitor C2 remains 6 mC (ii) Potential difference across C1 remains 6 V. H 0. After insertion of dielectric between the plates of each capacitor. across each capacitor is 6V V1 = V2 = 6 V C1 = C 2 = 1 m C \ Charge on each capacitor q1 = q 2 ( = CV ) = (1 mF) ´ (6 V) = 6 mC When switch S is opened. When switch S is closed. q = 60° H = 0. 2 Direction of torque is along negative Z-direction. In the active region. Zener diode can be used to regulate an unregulated dc power supply. We prefer telescope of higher resolving fe power to view the fine details of the object. A ×B 13. This results in an increase in the voltage drop across RC . i. where A is aperture 122l fo = same for both. telescope having convex lens of aperture A1.e. m = That is logic operation is OR. Resolving power R = \ R1 A1 = R2 A2 A .Examination Papers 249 12. Forward biased Vz VR (volt) Reverse biased I (mA) VF (volt) I (m A) V-I characteristics of zener diode Principle: In reverse breakdown (zener) region. They are referred as heat waves because they are readily absorbed by water molecules in most materials. 16. Y = A + B = = A ×B = A + B = A + B (A × B ) + A × B Magnification of telescope. which increases their thermal motion. A B Output Waveform 14. 15. a small increase of Vi results in a large (almost linear) increase in I C .. a very small change in voltage across the zener diode produces a very large change in current through the circuit but the voltage across the zener remains constant. Use: For therapeutic purpose and long distance photography. Infrared waves are produced by hot bodies and molecules. . so they heat up the material. If E is electric field outside the shell. Electric field intensity at a point outside a uniformly charged thin spherical shell: Consider a uniformly charged thin spherical shell of radius R carrying charge Q.. Due to induced current the bulb begins to glow.E min m a = max E max + E min E m < E c . et e 1 Drift velocity. 19. 18.250 Xam idea Physics—XII V0 Cut off region Active region Saturation region q AV O Vi 17. E ma = m Ec For modulated wave. (ii) Relaxation Time: The time of free travel of a free electron between two successive collisions of electron with lattice ions/atoms is called the relaxation time.e. (i) Modulation Index: The modulation index of an amplitude modulated wave is defined as the ratio of the amplitude of modulating signal ( E m ) to the amplitude of carrier wave ( E c ) i. (i) Drift Velocity: When a potential difference is applied across a conductor. amplitude modulation index. the free electrons drift towards the direction of positive potential. concentric with ® given shell. Vd = a m L L When length L is made 3L. then by symmetry electric field strength has . (b) The induced current can be demonstrated by connecting a torch bulb (in place of galvanometer) in Coil C1. E . To find the electric field outside the shell. we consider a spherical Gaussian surface of radius r ( > R). to avoid distortion (ii) The amplitude of modulating signal is kept less than the amplitude of carrier wave to avoid distortion. drift velocity becomes one-third. 20. (ii) by placing a soft iron laminated core at the centre of coil C1. The small average velocity of free electrons along the direction of positive potential is called the drift velocity. (a) The deflection in galvanometer may be made large by ( i) moving coil C2 towards C1 with high speed. 4pr 2 Now. by Gauss’s theorem ® ® R r òS Ei . Q P EO dS = ò E 0 dS cos 0 = E 0 . we consider a spherical Gaussian surface of radius ® r ( < R). Also the directions of normal at each point is radially outward. Gaussian surface is inside the given charged shell. dS Þ Ei 4pr 2 = 1 ´0 e0 = 1 ´ charge enclosed e0 Þ Ei = 0 . then by symmetry electric field strength has the same magnitude Ei on the Gaussian surface and is directed radially ® outward. Hence. Also the directions of normal at each point is radially outward. by Gauss's theorem ® ® 1 ò S E 0 · d E = e 0 ´ charged enclosed 1 1 Q Þ E 0 4pr 2 = ´ Q Þ E0 = e0 4pe 0 r 2 Thus. so angle between ® ® ® E0 S ® R r E i and d S is zero at each point. If s is the surface charge density of the spherical shell. electric field outside a charged thin spherical shell is the same as if the whole charge Q is concentrated at the centre. then f = 4pR 2 s C \ E0 = 1 4pR 2 s R 2 s = 4pe 0 r 2 e 0r 2 Electric field inside the shell (hollow charged conducting sphere): The charge resides on the surface of a conductor. To find the electric field inside the shell. Hence. electric flux through Gaussian surface = ò · d S. concentric with the given shell. Gaussian surface is outside the given charged shell. d S = ò Ei dS cos 0 = Ei . If E is the electric field inside the shell. electric flux through Gaussian surface ® S ® E = ò Ei . 4pr 2 Now. so charge enclosed by Gaussian surface is Q. so angle between E i and ® d S is zero at each point. so charge enclosed by Gaussian surface is zero. Hence. Hence. Thus a hollow charged conductor is equivalent to a charged spherical shell.Examination Papers 251 same magnitude E 0 on the Gaussian surface and is directed radially outward. 1 l 1× 00 ´ 10 .25 ´ 3 ´ 108 J = 19 × 89 ´ 10 . E = hn = h × = pc = 6 × 63 ´ 10 .17 1× 6 ´ 10 .34 = = 6 × 63 ´ 10 .25 kg ms . Given l = 1× 00 nm = 1× 00 ´ 10 .19 eV = 1× 24 ´ 10 3 eV = 1 × 24 keV p2 1 me v 2 = 2 2m e J 2 × 42 ´ 10 .19 J = 22.19 eV = 1 × 51 eV (iii) Kinetic energy of electron E k = = ( 6 × 63 ´ 10 .9 m (i) Momenta of electron and photon are equal.17 J = 19 × 89 ´ 10 .31 = 2 × 42 ´ 10 . LOS Three propagation modes of em waves Frequency Range of Different Services (i) Standard AM broadcast 540–1600 kHz .25 ) 2 2 ´ 9 ×1 ´ 10 . given by p= h 6 × 63 ´ 10 .9 c h = c l l (ii) Energy of photon.19 1× 6 ´ 10 .252 Xam idea Physics—XII 21. 2 GHz downlink.e. Obviously point M is equidistant from S1 and S 2 .S1 P » S 2 N As D > > d. sin q = 2 S1S 2 MP In DMOP.425 GHz uplink 3.S1 P = S 2 N = . OM D yd Path difference D = S 2 P . Consider a point P on the screen at a distance y from M.Examination Papers 253 (ii) Television 54–890 MHz (iii) Satellite Communication 5. n = 0. 1.7–4. Therefore the path difference between the two waves at point M is zero. D = nl yd \ = nl . tan q = OM As q is very small \ sin q = q = tan q S 2 N MP \ = S1S 2 OM y MP \ S 2 N = S1S 2 =d. 2. Young’s Double slit experiment: Coherent sources are those which have exactly the same frequency and are in this same phase or have a constant difference in phase. therefore Ð S 2 S1 N = q is very small \ Ð S 2 S1 N = Ð MOP = q S N In D S1S 2 N .925–6. Let the distance of the screen from the coherent sources be D. 3. the midpoint of S1 and S 2 on the screen. Thus the point M has the maximum intensity.. P S1 S d O S2 y q q M N D The path difference between two waves reaching at P from S1 and S 2 is D = S 2 P . Expression for Fringe Width: Let S1 and S 2 be two coherent sources separated by a distance d. Let M be the foot of the perpendicular drawn from O. Draw S1 N perpendicular from S1 on S 2 P. Conditions: (i) The sources should be monochromatic and originating from common single source. 23. i. K D . (ii) The amplitudes of the waves should be equal..(i) \ D (i) Positions of bright fringes (or maxima): For bright fringe or maximum intensity at P. the path difference must be an integral multiple of wavelength ( l) of light used. . i. the path l difference must be an odd number multiple of half wavelength...1) 2 y...d l where n = 1..(ii) yn = ..y n = ( n + 1) = . d (ii) Positions of dark fringes (or minima): For dark fringe or minimum intensity at P. b = y n + 1 . 2d 2 d This equation gives the distance of nth dark fringe from point M.(n . For Bright Fringes: If y n + 1 and y n denote the distances of two consecutive bright fringes \ y= from M.y n 1 Dl 1 Dl = (n + ) .(v) .(iv) \ Fringe width.) 2 d 2 d Dl 1 1 Dl = (n + .e. we get nDl . b = y n + 1 . we get 1 Dl . D = ( 2n . yn = ( n ..) 2 d 2 d \ Fringe width. fringe width is the same for bright and dark fringes equal to Dl b= d The condition for the interference fringes to be seen is s l < b d . Therefore writing y n for y. It is denoted by w.) 2 d (iii) Fringe Width b : The distance between any two consecutive bright fringes or any two consecutive dark fringes is called the fringe width.254 Xam idea Physics—XII nDl .n + ) = d 2 2 d Thus.. \ = ( 2n . Therefore writing y n for y. d This equation gives the distance of nth bright fringe from the point M..1) lD 1 Dl or y= = (n . 2.. .1) D 2 ( 2n . then we have Dl nDl and y n = d d Dl nDl Dl ..) . then we have 1 Dl 1 Dl yn + 1 = ( n + ) .(iii) yn = ( n . 3. d d d For Dark Fringes: If y n + 1 and y n are the distances of two consecutive dark fringes from y n + 1 = ( n + 1) M. Ð ABB¢ = 90° vt BB¢ . which traverses a distance AA¢ ( = v 2 t) in second medium in time t. we draw a spherical arc of radius AA¢ ( = v 2 t) and draw tangent B¢ A¢ on this arc from B¢ . Suppose a plane wavefront AB in first medium is incident obliquely on the boundary surface XY and its end A touches the surface at A at time t = 0 while the other end B reaches the surface at point B¢ after time-interval t. refracted ray and the normal at the point of incidence all lie in the same plane. This phenomenon is called refraction. According to Huygen’s principle.(i) \ sin i = sin Ð BAB¢ = = 1 AB¢ AB¢ Similarly in right-angled triangle AA¢ B ¢ . In the same time-interval t. From a point source. the point of wavefront traverses a distance BB¢ ( = v1t) in first medium and reaches B¢ . As the lines drawn normal to wavefront denote the rays. A¢ B ¢ and surface XY are in the plane of paper. Let v1 and v 2 be the speeds of waves in these media. Assuming A as centre.. therefore we may say that the incident ray. secondary spherical wavelets originate from these points. Rectilinear Propagation of Light: According to Newton’s corpuscular theory. the frequency of wave remains unchanged but its speed and the wavelength both are changed. it strikes the points between A and B¢ of boundary surface. Proof of Snell’s law of Refraction using Huygen’s wave theory: When a wave starting from one homogeneous medium enters the another homogeneous medium. Clearly BB¢ = v1t and AA¢ = v 2 t. The envelope of all wavelets at a given instant gives the position of a new wavefront. Hence A¢ B¢ will be the refracted wavefront. According to Huygen’s principle A¢ B¢ is the new position of wavefront AB in the second medium.. Clearly BB¢ = v1t. from. As the wavefront AB advances. The locus of particles of a medium vibrating in the same phase is called a wavefront. which travel with speed v1 in the first medium and speed v 2 in the second medium. it is deviated from its path. the path of light is a straight line. while for a line source the wavefront is cylindrical. As the incident wavefront AB advances. Let XY be a surface separating the two media ‘1’ and ‘2’. In right-angled triangle AB¢ B. the wavefront is spherical. (ii) Each point of a wavefront acts as a source of secondary wavelets. one after the other and will touch A¢ B¢ simultaneously. In transversing from first medium to another medium. The distant wavefront is plane. This is the first law of refraction. the secondary wavelets start from points between A and B¢ . Second law: Let the incident wavefront AB and refracted wavefront A¢ B¢ make angles i and r respectively with refracting surface XY. Ð AA¢ B ¢ = 90° . First law: As AB.Examination Papers 255 OR The assumptions of Huygen’s theory are: (i) A source sends waves in all possible directions. First of all secondary wavelet starts from A. where the secondary wavelet now starts. therefore the perpendicular drawn on them will be in the same plane. but according to wave theory the rectilinear propagation of light is only approximate. the light is almost completely transmitted through B and no change is observed in the light coming out of B. the intensity of light emerging out of B decreases and becomes zero when the axis of B is perpendicular to that of A. This is the second law of refraction. If B is further rotated. (a) Light from a source S is allowed to fall normally on the flat surface of a thin plate of a tourmaline crystal. Unpolarised light S A Unpolarised light S (b) (a) Polarised light No light B Polarised light Polarised light If now the crystal A is kept fixed and B is gradually rotated in its own plane. and is called the Snell’s law. we see that the intensity of light transmitted through B is maximum when axes of A and B are parallel and minimum when they are at right angles. when reflected and refracted rays are perpendicular to each other. Thus.. therefore B the incident and refracted rays make angles i and r with the o 90 i normal drawn on the surface XY i.. we get v sin i = 1 = constant sin r v2 .(ii) Dividing equation (i) by (ii). the character of transmitted light remains unchanged.256 Xam idea Physics—XII \ sin r = sin Ð AB¢ A¢ = AA¢ v 2 t = AB¢ AB¢ . If the light transmitted through A is passed through B. Now another similar plate B is placed at some distance from A such that the axis of B is parallel to that of A.(iii) As the rays are always normal to the wavefront. From this experiment. . i and r are the angle of i B' Y incidence and angle of refraction respectively. the rotation of crystal B would not produce any change in the intensity of light. According X r A o r 90 to equation (iii): The ratio of sine of angle of incidence and the sine of A' angle of refraction is a constant and is equal to the ratio of velocities of waves in the two media..e. it is obvious that light waves are transverse and not longitudinal. because. 24. if they were longitudinal. Only a part of this light is transmitted through A. cut parallel to its axis. (b) The reflected ray is totally plane polarised. the intensity begins to increase and becomes maximum when the axes of A and B are again parallel. If now the plate A is rotated. so their effective resistance (R) is given by 1 1 1 1 1 1 1 = + + = + + R R 2 R 3 R 4 15 15 30 Þ R =6 W Number of undecayed nuclei at t = 3T1/2 is F I1 R1 A I2 I4 C I3 E R2 R4 R3 G I1 B D .Examination Papers 257 25. hc hc (i) DE = Þ l = l DE For maximum wavelength DE should be minimum.5 eV. The number of nuclei undergoing the decay per unit time. E = 10 V. No N = No e–lt N No 2 No No 8 32 T1/2 2T1/2 3T1/2 t 4T1/2 5T1/2 No N and at t = 5T1/2. is proportional to the total number of nuclei present in the sample at that instant. (ii) For minimum wavelength DE should be maximum. This corresponds to transition D. 26. This corresponds to transition A. = l 275 ´ 10 -9 ´ 1× 6 ´ 10 -19 This corresponds to transition ‘B’. Given R1 = 4W. 8 32 27. Equivalent Resistance: R2. Energy of photon wavelength 275 nm E= 6 × 63 ´ 10 -34 ´ 3 ´ 108 hc eV = 4. at any instant. R3 and R4 are in parallel. R4 = 30W. R2 = R3 = 15W. it is o . .4 A Thus..(ii) Also. with its plane perpendicular to the plane of paper. I2 = I3 = 0.2 A \ I2 = I3 = 2 × 0. ® ® ® where m 0 The magnitude of magnetic field is m Idl sin q dB = 0 4p r2 ® ® ® ® ® where q is the angle between current element I dl and position vector r . Consider a small element of length dl of ® the coil at point A.(iii) Þ Substituting values of I2. we get 2I4 + 2I4 + I4 = 1 A Þ I4 = 0. so equivalent resistance Req = R + R1 = 6 + 4 = 10 W. The direction of magnetic field dB is perpendicular to the plane containing I dl and r . I3 and I4. As the angle between I dl and r is 90°. I1 = 1A.(i) This current is divided at A into three parts I2. The magnitude of the magnetic induction dB at point P due to this element is given by m I dl sin a . . Its value is m 0 = 4p ´ 10 -7 Wb/A-m. Biot-Savart Law It states that the magnetic field strength ( dB) produced due to a current element (of current I and ® length dl) at a point having position vector r relative to current element is m I dl ´ r dB = 0 4p r 3 is permeability of free space.(i) dB = 0 4p r2 ® ® ® ® ® ® The direction of dB is perpendicular to the plane containing dl and r and is given by right hand screw rule. I3 in (ii).. Let P be a point of observation on the axis of this circular loop at a distance x from its centre O. Currents: E 10 I1 = = =1 A R eq 10 .4A and I4 = 0.. \ I2 + I3 + I4 = 1 A .. I2 R2 = I3R3 = I4R4 I2 × 15 = I3 × 15 = I4 ×30 Þ I2 = I3 = 2I4 .. Magnetic field at the axis of a circular loop: Consider a circular loop of radius R carrying current I.258 Xam idea Physics—XII R1 is in series with R.2A 28. the magnitude of the magnetic induction dB is given by.2 = 0.... .. namely PM and PN along and perpendicular to the axis respectively. then by symmetry the components of magnetic induction perpendicular to the axis will be cancelled out.. The magnetic induction ® ® dl q r P dB due to this element has been represented by P Q¢ whose I m I dl ¢ magnitude is 0 and which can also be resolved into two 4p r 2 components.(ii) In figure dB has been represented by P Q.Examination Papers 259 dB = ® m 0 I dl sin 90° m 0 I dl = × 4p r2 4p r 2 ® ® ® .(iv) Therefore the magnitude of resultant magnetic induction at axial point P due to the whole circular coil is given by m 0 IR m 0 IR B= ò dl = ò dl 2 2 3/ 2 4p ( R + x ) 4p ( R 2 + x 2 ) 3 / 2 But ò dl = length of the loop = 2pR B= = m 0 IR 4p ( R 2 + x 2 ) 3 / 2 m 0 IR 2 2 (R 2 + x 2 ) 3/ 2 m 0 NIR 2 2 (R 2 + x 2 ) 3/ 2 ( 2 p R) . If the coil contains N turns... Thus the resultant magnetic induction B at axial point P is along the axis and may be evaluated as follows: ® The component of dB along the axis.. PM and PN ¢ along the axis and perpendicular to the axis respectively.. while those ® parallel to the axis will be added up..(vi) . m I dI dB x = 0 sin a 4p r 2 R But sin a = and r = ( R 2 + x 2 ) 1/ 2 r m I dl R m 0 IR \ dB x = 0 × = dl 4p r 2 r 4p r 3 dB x = m 0 IR 4p ( R 2 + x 2 ) 3 / 2 dl . then B= tesla. Now consider another small current element `of length dI ¢ at A ¢ . The vector dB or ( P Q) can be resolved into two components.(iii) . . Thus if we consider the magnetic induction produced by the whole of the circular coil.(v) Therefore.. tesla.. i.. Infact every current carrying coil is equivalent to a magnetic dipole (or magnet) of magnetic moment m = IA = I × p r 2 .e.. Due to the presence of perpendicular magnetic field the ion will move in a circular path inside the dees.. The phenomenon is continued till the ion reaches at Dee-1 the periphery of the dees where an auxiliary negative electrode (deflecting plate) deflects the accelerated ion on the target to be S R.. S Dee Dee Working: The principle of action of the apparatus is shown in fig. The positive ions produced from a source S at the centre Magnetic Pole are accelerated by a dee which is at negative potential at that S moment. p pm . . then.F..(iii) t= = w Bq Thus the time is independent of the speed of the ion i. The magnetic field and the frequency of the applied voltages are so chosen that as the ion comes out of a dee.(ii) w= = r m The time taken by the ion in describing a semi-circle. in turning through an angle p is. OR Principle: When a charged particle is kept in a magnetic field Magnetic Pole it experiences a force and the perpendicular magnetic field N causes the particle to spiral many times.260 Xam idea Physics—XII A magnetic needle placed at the center and axis of a circular coil shows deflection. i) when it moves from one dee to the other.e. The angular velocity w of the ion is given by. Expression for Period of Revolution and Frequency: Suppose the positive ion with charge q moves in a dee with a velocity v. the dees change their polarity (positive becoming negative and vice-versa) and the ion is further accelerated and moves with higher velocity along a circular path of greater radius. The function of electric field is to accelerate Dee-2 the charged particle and so to impart energy to the charged particle.(i) where m is the mass and r the radius of the path of ion in the dee and B is the strength of the magnetic field. Beam The function of magnetic field is to provide circular path to charged particle and so to provide the location where charged particle is capable of gaining energy from electric field. oscillator bombarded. i) .. although the speed of the ion goes on increasing with increase in the radius (from eq.. qvB = mv 2 r or r = mv qB . yet it takes the same time in each dee. This implies that a circular coil behaves as a magnet. v qB (from eq.. This is called resonance condition.. \ Frequency of revolution of particles qB 1 f = = T 2pm This frequency is called the cyclotron frequency. (iii) it is clear that for a particular ion.(iv) =t = 2 qB 2p m or ..(v) T= qB From eq. when produced meet at N.(i) = (i1 + i 2 ) . Resonance Condition: The condition of working of cyclotron is that the frequency of radio frequency alternating potential must be equal to the frequency of revolution of charged particles within the dees. This is the expression for period of revolution. period of revolution is independent of speed of charged particle and radius of circular path. Ð OFG = i1 . B can be calculated for producing q resonance with the high frequency alternating potential. The angle of Q R refraction for this face is r1 . Clearly the cyclotron frequency is independent of speed of particle.r 2 In D FOG .. Let Ð FNG = q . the applied alternating potential should also have the same semi-periodic time (T / 2) as that taken by the ion to cross either dee.. Obviously. i. The incident ray EF and emergent ray GH when produced meet at O.. T pm .r 2 ) i .r1 and Ð OGF = i 2 . (a) Let PQR be the principal section of the prism..(r1 + r 2 ) The normals FN1 and GN 2 on faces PQ and PR respectively. The refractive index of material of prism for this ray is n. The angle dm between these two rays is called angle of d deviation ‘d‘. The angle of incidence on this face is r 2 (into prism) and angle of refraction d (into air) is i 2 . A A ray of monochromatic light EF is incident on N1 N2 O d (i –r ) (i1–r1) 2 2 face PQ at angle of incidence i1 . Now for the cyclotron to work.Examination Papers 261 m being known. The refracting angle of the prism is A.e.r1 ) + (i 2 . P 29. This i1 i2 r1 r F q 2 G ray enters from rarer to denser medium and so is deviated towards the normal FN and gets N E H refracted along the direction FG.. d is exterior angle x i1 i i2 \ d = Ð OFG + Ð OGF = (i1 . The refracted ray FG becomes incident on face PR and is refracted away from the normal GN 2 and emerges in the direction GH. i.. r1 + r 2 + q = 180° In quadrilateral PFNG.(vii) Minimum Deviation: From equation (v).. Ð PFN = 90° .. the angle of deviation first decreases. there are two angles of incidences i1 and i 2 .(iv) . we get d = i1 + i 2 . we get r + r = A or r = A / 2 A + dm and i + i = A + d m or i = 2 æ A + dm ö sin ç ÷ 2 è ø sin i Hence from Snell’s law. Thus for two angles of incidence i1 and i 2 . the angle of deviation is the minimum.262 Xam idea Physics—XII In D FGN. therefore if angle of incidence be i 2 .. Ð PGN = 90° \ A + 90° + q + 90° = 360° or A + q = 180° Comparing (ii) and (iii). n= = sin r æ A ö sin ç ÷ è 2 ø (b) An optical fibre is a device based on total internal reflection by which a light signal may be transmitted from one place to another with a negligible loss of energy. Thus the ray within the fibre suffers multiple total internal reflections and finally strikes the other end at an angle less than critical angle for quartz-air interface and emerges in air.(vi) . from a small value. For minimum deviation i1 and i 2 become coincident. there will be one angle of deviation. .i1 = i 2 = i (say) So from (vii) r1 = r 2 = r (say) Hence from (iv) and (vi).A or i1 + i 2 = A + d sin i1 sin i 2 From Snell’s law n = = sin r1 sin r 2 .(iii) .. it suffers refraction from air to quartz and strikes the quartz-coating interface at an angle more than the critical angle and so suffers total internal reflection and strikes the opposite face again at an angle greater than critical angle and so again suffers total internal reflection. becomes minimum for a particular angle of incidence and then begins to increase. the angles of deviation corresponding to different angles of incidence and then plot i (on X-axis) and d (on Y-axis). When a light ray is incident on one end at a small angle of incidence. but for one and only one particular value of angle of incidence (i).(ii) . then angle of emergence will be i1 . As the path of light is reversible. we get a curve as shown in figure... r1 + r 2 = A Substituting this value in (i).e... If we determine experimentally... This minimum angle of deviation is represented by d m . Obviously for one angle of deviation ( d). it is clear that the angle of deviation depends upon the angle of incidence i1 . Clearly if angle of incidence is gradually increased..(v) .. .t) R2 For a thin lens t is negligible as compared to v¢ .(iii) =.n2 . therefore from refraction formula at spherical surface n1 n2 n .n1 .1÷ ç ÷ ç v u è n1 ø è R1 R 2 ö ÷ ÷ ø ö ÷ ÷ ø . O is a point object on the principal axis of the lens.. The thickness of lens is t.. The distance of virtual object ( I ¢ ) from pole P2 is (v¢ . The distance of I from pole P2 of second surface is v...2 v (v¢ ) R2 Adding equations (i) and (iii). the ray is going from second medium (refractive index n 2 ) to first medium (refractive index n1 ). The distance of O from pole P1 is u. The radii of curvature of the surfaces of the lens are R1 and R 2 and their poles are P1 and P2 . The n1 n1 n2 refractive index of the t material of lens is n 2 and it is X placed in a medium of X' P1 C P2 I I' O v refractive index n1 . we get æ 1 n1 n1 1 = ( n 2 .7 A B OR L (a) Lens Maker’s Formula: Suppose L is a thin lens. the final image is formed at I. The first refracting surface forms the image of O at I ¢ at a distance v¢ from P1 .(i) = v¢ u R1 The image I ¢ acts as a virtual object for second surface and after refraction at second surface.=ç ..n1 ) ç ç v u è R1 R 2 or öæ 1 1 1 æ n2 1 ç . which is very small. For refraction at second surface.t).5 263 n = 1.(ii) = 1 v (v¢ . From the refraction formula at spherical surface n 2 n1 n 2 . The u v' optical centre of lens is C and X ¢ X is principal axis. therefore from (ii) n1 n 2 n .n1 .Examination Papers Coating n = 1. v = f 2 = f Therefore from equation (iv) æ 1 1 1 1 ö ÷ .= ( 1 n 2 .= ( 1 n 2 ..(vi) = ( n .1) ç ç ÷ f è R1 R 2 ø (b) Ray Diagram: The ray diagram of image formation for an object between focus (F) and pole (P) of a concave mirror is shown in fig..1) ç ç ÷ f è R1 R 2 ø This is the formula of refraction for a thin lens. then 1 n 2 = n.e..e. This formula is called Lens-Maker’s formula..1) ç ç ÷ v u è R1 R 2 ø n where 1 n 2 = 2 is refractive index of second medium (i.. ÐAPB = ÐBPQ = i Also. medium of lens) with respect to n1 first medium. i. the image will be formed at second focus i.e.1) ç ç ÷ f ¥ è R1 R 2 ø i.(iv) . A' M A h C F Q B u i i i P v B' h' Magnification: m = Size of image (A ¢B ¢) Size of object (AB) From fig.(2) . tan i = BP A ¢B ¢ In D A ¢PB ¢. ÐBPQ = ÐA ¢PB ¢ = i AB In D APB.e. If first medium is air and refractive index of material of lens be n.(v) = ( 1 n 2 ... if u = ¥ ..(1) . æ 1 1 1 ö ÷ . therefore equation (v) may be written as æ 1 1 1 ö ÷ ...264 Xam idea Physics—XII æ 1 1 1 1 ö ÷ . tan i = B ¢P . If the object O is at infinity. . If i p and i s are the instantaneous currents in Np 30. We assume that there is no leakage of flux so that the flux linked with each turn of primary coil and secondary coil is the same. then the potential difference VS across its ends will be equal to the emf ( e S ) induced in it.(iv) = = = r (say) Vp e p N p where r = NS is called the transformation ratio.. N S > N p ). the emf ( e p ) induced in the primary coil.(v) \ = = = i p VS N S r .(i) ep =-Np Dt and emf induced in the secondary coil Df .(ii) eS = . primary and secondary coils and there is no loss of energy. NS the number of turns in secondary coil and f the magnetic flux linked with each turn... therefore the alternating emf of same frequency is developed across the secondary. will be equal to the applied potential difference (V p ) across its ends. due to which the magnetic flux linked with the secondary coil changes continuously... then For about 100% efficiency. m = = AB BP v v or m= or m = –u u (i) Principle: It is based on the principle of mutual inductance and transforms the alternating low voltage to alternating high voltage and in this the number of turns in secondary coil is more than that in primary coil...NS Dt From (i) and (ii) eS NS .. Power in primary = Power in secondary V p i p = VS i S iS V p N p 1 .. the current changes continuously in the primary coil.. According to Faraday’s laws the emf induced in the primary coil Df . Working: When alternating current source is connected to the ends of primary coil. therefore VS e S N S . (i. Similarly if the secondary circuit is open.Examination Papers 265 From (1) and (2) AB A ¢B ¢ = BP B ¢P A ¢B ¢ B ¢P Þ Magnification.(iii) = ep Np If the resistance of primary coil is negligible. Let N p be the number of turns in primary coil. e. Primary Np Ns VR f i V=V0 sin wt (a) VC (b) . N s > N p ® r > 1. So VS > V p and i S < i p i. The voltage VR and current i are in the same phase.e. therefore their resultant potential difference = VC .e.VL ) will also be V. From fig. there is no violation of conservation of energy in a step up transformer.VL ) are mutually perpendicular and the phase difference between them is 90°. a) On account of being in series. Thus VR and (VC .. (b) Flux Leakage: Energy is lost due to coupling of primary and secondary coils not being perfect.. whole of magnetic flux generated in primary coil is not linked with the secondary coil. the voltage VL will lead the current by angle 90° while the voltage VC will lag behind the current by angle 90° (fig.VL (if VC > VC ). the resultant of VR and (VC . Soft iron-core Secondary Two coils on separate limbs of the core (ii) Reasons for energy losses in a transformer: (a) Joule Heating: Energy is lost in resistance of primary and secondary windings as heat ( I 2 Rt). voltage across inductance L is VL and voltage across capacitance C is VC .266 Xam idea Physics—XII In step up transformer. Clearly VC and VL are in opposite directions. R VL L VR C VC VL 90o (VC–VL) 90 o Suppose the voltage across resistance R is VR . i. inductance L and capacitance C are connected in series and an alternating source of voltage V = V0 sin wt is applied across it. b). Thus. step up transformer increases the voltage. (fig. (iii) When output voltage increases. OR (a) Expression for Impedance in LCR series circuit: Suppose resistance R. the output current automatically decreases to keep the power same. As applied voltage across the circuit is V. the current (i ) flowing through all of them is the same. .(i) ..X L = 0 1 1 Þ = wL Þ w2 = wC LC 1 Resonant frequency wr = \ LC Graph of variation of current (i) with frequency (f): The resonance in series LCR circuit occurs when inductive reactance is equal to capacitance reactance i.X L ) 2 i 2 \ Impedance of circuit. so X C .c.X L ) 2 = R 2 + ç .VL ) 2 VR = R i . VC = X C i and VL = X L i 1 = = capacitance reactance and X L = wL = inductive reactance wC V = ( R i) 2 + ( X C i . source becomes 1 . fr 2p LC Graph is shown in fig. im = V/R i fr f .VL ) 2 Þ V = VR 2 + (VC .wL ÷ è wC ø 2 The phase difference ( f) between current and voltage f is given by X -XL tan f = C R Resonant Frequency: For resonance f = 0. Z = i. Alternatively when the frequency of applied a.(ii) V But where X C \ 2 = VR 2 + (VC .wL ÷ è wC ø V0 sin ( wt + f) æ 1 ö R2 + ç .e.e. X L = X C .X L i) 2 V = R 2 + (X C .. Instantaneous current I = æ 1 ö Z = R 2 + (X C ..Examination Papers 267 .. E. Y = AB X = AB × AB = AB = AB That is the logic operation is AND operation. CBSE (Delhi) SET–II 2. q¢ 10 mC ¢ = 2 = V2 =1 V C¢ 10 mF 2 13. C1 ¢ = C1 ¢ V1 = (10 mF) ´ 5 V = 50 mC Charge on C1 . r r r (so that K. Magnetic field along Z-direction. C1 = V´1 = V1 = 5V Potential difference across C2. Therefore output is 1 only. q¢ 2 = q 2 = 10 mC (ii) Potential difference across. so the current in coil C2 will also be clockwise. B form a right handed system) 10. Initial charge on C1 and C2 is q1 = q2 = C1V = (2 mF) × 5 V = 10 mC S 5V 2 mF C1 2 mF C2 (i) When switch S is opened the p. New capacitance of C1 and C2 after filling dielectric. The circuit remains connected with an areal coil through the phenomenon of mutual inductance. so material is paramagnetic. so the direction of current in coil C1 will be clockwise. The output waveform is shown in figure. then waves produce alternating voltage of frequency f. The near face of coil C2 will also become S-pole to oppose the approach of magnet. 11. 1 When capacitor C is in circuit is varied then for a particular value of capacitance C. so the radio or TV gets tunned. . According to Lenz’s law. across C1 remains 5V and charge on C2 remains 10 mC. Suppose a radio or TV station is transmitted a programme at frequency f. due to which an emf of same frequency is induced in LC circuit. when both inputs are 1.268 Xam idea Physics—XII Use of phenomenon of resonance in tuning a ratio or TV set: A radio or a TV set has a L-C circuit with capacitor of variable capacitance C. Susceptibility is small positive.d. The near face of coil C1 will become S-pole. the direction of induced current is such that it opposes the relative motion between coil and magnet. in areal. Electric field along Y-direction. 4. ¢ = C2 ¢ = 5 ´ 2 = 10 mF. 2p LC the resonance occurs and maximum current flows in the circuit. q1 Charge on C2. f = . . equation (iii) gives 2I 4 + 2I 4 + I 4 = 1 Þ I4 = 1 A = 0.2 = 0.4 A \ I 1 = 1A.2 A 23.(iii) . Effective resistance of R2.. 1. (ii) The current is due to leakage of minority charge carriers through the junction and is very small of the order of mA..4A. Forward Bias: (i) Within the junction diode the direction of applied voltage is opposite to that of built-in potential. (iii) The diode offers very large resistance in reverse bias. I 2 = I 3 = 0... Req = R1 + R = 4 + 2 = 6 W E 6 Current in R1 .(v) . Given l = 1.(iv) I 2 = I 3 = 2 ´ 0.42 ´ 10 -25 kg m s -1 -9 l 150 . ´ 10 –9 . I 2 + I 3 + I 4 = 1 A and I 2 R2 = I 3 R3 = I 4 R4 Þ 5I 2 = 5I 3 = 10I 4 Þ I 2 = I 3 = 2I 4 Using equation (iv). R3 and R4 in parallel.. 1 1 1 1 1 1 1 2 + 2 +1 20. (ii) The current is due to diffusion of majority charge carriers through the junction and is of the order of milliamperes.50 × 10 m. Reverse Bias: (i) The direction of applied voltage and barrier potential is same.Examination Papers 269 A B t1 Output X t2 t3 t4 t5 t6 t7 16... 2.. is = + + = + + = R R 2 R 3 R 4 5 5 10 10 R=2W Þ R1 is in series with R. I 4 = 0.(ii) . I 1 = = = 1A R eq 6 Also.50 nm = 1. so equivalent resistance. (iii) The diode offers very small resistance in the forward bias.. (i) As momentum of electron = momentum of photon h 6.(i) .2 A 5 .63 ´ 10 -34 = = = 4. Susceptibility of material is negative.7 ´ 10 -19 eV = 0. 9. so given material is diamagnetic. Thus near the junction there is an excess of positively charged ions in n-region and an excess of negatively . This is the law of radioactive decay.67 eV 1.42 ´ 10 -25 ) 2 = J 2m e 2 ´ 9 ×1´ 10 -31 = 1. 4 16 CBSE (Delhi) SET–III 2. = No N = No e–lt N No 2 No 4 No 16 T1/2 2T1/2 3T1/2 t 4T1/2 Number of undecayed nuclei at t = 2T1/2 is No N and at t = 4T1/2. at any instant. Some charge carriers combine with opposite charges to neutralise each other. Magnetic field along Z-direction. so that some of electrons of n-region diffuse to p-region while some of holes of p-region diffuse into n-region.42 × 10–25 × 3 × 108 J = 13.26 × 10–17 J = 13.6 ´ 10 -19 24.270 Xam idea Physics—XII (ii) Energy of photon.0. Formation of depletion layer: At the junction there is diffusion of charge carriers due to thermal agitation. it is o .26 ´ 10 -17 eV = 828. Electric field along X-direction. is proportional to the total number of nuclei present in the sample at that instant. 7. E = hc = pc l = 4. The number of nuclei undergoing the decay per unit time.6 ´ 10 -19 (iii) Kinetic energy of electron EK = P 2 ( 4.07 × 10–19 J 1.7 eV 1. A 0 1 0 1 B 0 0 1 1 Y 0 1 1 1 17.Examination Papers 271 charged ions in p-region. the plate A of capacitor will be positive and plate B will be negative.. R 4 = 20 W. ¢ = 6 mC.4 cm to 10 . C = A . left face of coil will act as N-pole and right face as S-pole. V1¢ = 2 V.D... Truth table.20 I2 = I3 = 2I4 I2 + I3 + I4 = 1 A L C . D = B Y = CD = A B = A + B = A + B The logic circuit performs OR-operation. V2 =1 V ¢ 3 mF C2 16. R 2 = R 3 = 10 W. Þ Þ and I2R2 = I3R3 = I4R4 I2. their effective resistance R is 1 1 1 1 1 1 1 2 + 2 +1 = + + = + + = R R 2 R 3 R 4 10 10 20 20 20 R= Þ =4 W 5 Req = R1 + R = 2 + 4 = 6 W E 6 Current in R1.6 cm) on either side of the junction becomes free from mobile charge carriers and hence is called the depletion layer.10 = I4.(ii) .D. Current induced in coil will oppose the approach of magnet. For this the current in coil will be anticlockwise as seen from left. therefore. (i) Charge on C1 . Charge on C 2 .. This field stops the further diffusion of charge carriers. R1 = 2 W.10 = I3. The field Ei is directed from n-region to p-region. Thus the layers ( » 10 . q¢ 14. I1 = = =1A R eq 6 Also. P. A S N B S N 23. This sets up a potential difference called potential barrier and hence an internal electric field Ei across the junctions. q1 2 = 3 mC q ¢ 3 mC ¢ = 2 = (ii) P. across C 2 . E = 6 V Equivalent Resistance: R 2 = R 3 and R 4 are in parallel. across C1 . therefore.(i) . 315 × 10–25 kg m s–1 = l 2 ´ 10 -9 hc = pc l 9.38 eV = 9.6 ´ 10 -19 = 6.945 ´ 10 -17 = 3. l = 2 nm = 2 × 10–9 m (i) Momentum of electron = Momentum of photon = p = (ii) Energy of photon.1´ 10 -31 = 0. E = h 6.945 ´ 10 -17 J = eV .4A.604 × 10–19 J = 0.315 ´ 10 -25 ´ 3 ´ 108 J 1.216 × 102 eV = 0.272 Xam idea Physics—XII Solving (i) and (ii) I2 = I3 = 0.63 ´ 10 -34 = 3. I4 = 0.62 keV (iii) Kinetic energy of electrons EK = p2 (3.315 ´ 10 -25 ) 2 1 m eV 2 = = J 2 2m e 2 ´ 9.2 A 25. 6. However. Write its S. one question of three marks and all three questions of five marks each. (b) There are 30 questions in total. Questions 1 to 8 carry one mark each. 2. Predict the polarity of the capacitor. You have to attempt only one of the given choices in such questions. unit. an internal choice has been provided in one question of two marks.I.1 h = 6 × 626 ´ 10 -34 Js e = 1× 602 ´ 10 -19 C 1 = 9 × 109 Nm2C– 4pe o m 0 = 4p ´ 10 -7 TmA -1 2 Maximum marks: 70 Boltzmann’s constant k = 1× 381 ´ 10 -23 J K -1 Avogadro’s number N A = 6 × 022 ´ 10 23 /mole Mass of neutron m n = 1× 2 ´ 10 -27 kg Mass of electron m e = 9 ×1´ 10 -31 kg Radius of earth = 6400 km CBSE (All India) SET–I 1.CBSE EXAMINATION PAPERS ALL INDIA–2011 Time allowed: 3 hours General Instructions: (a) All questions are compulsory. C S N S N . How are radio waves produced? 5. (d) Use of calculators is not permitted. Define electric dipole moment. questions 9 to 18 carry two marks each. Two bar magnets are quickly moved towards a metallic loop connected across a capacitor ‘C’ as shown in the figure. (c) There is no overall choice. Write any two characteristic properties of nuclear force. What is the potential at the centre of the sphere? 4. Where on the surface of Earth is the angle of dip 90°? 3. A hollow metal sphere of radius 5 cm is charged such that the potential on its surface is 10 V. questions 19 to 27 carry three marks each and questions 28 to 30 carry five marks each. (e) You may use the following values of physical constants wherever necessary: c = 3 ´ 108 ms . In the given circuit. Show that no work is done by this force on the charged particle. A thin straight infinitely long conducting wire having charge density l is enclosed by a cylindrical surface of radius r and length l. æ 1 ö 10. Write the expression for this force. state clearly how the release in energy in the processes of nuclear fission and nuclear fusion can be explained. 17. where r is the distance èr ø between the two charges of each pair of charges: (1 mC. Write the expression for Lorentz magnetic force on a particle of charge ‘q’ moving with velocity v in a magnetic field B . What are eddy currents? Write any two applications of eddy currents. use Kirchhoff’s rules to determine the potential at point B. . its axis coinciding with the length of the wire. OR A steady current (I1) flows through a long straight wire. balance point was observed at J with AJ = l. A parallel plate capacitor is being charged by a time varying current. 2 mC) and (2 mC – 3 mC). (ii) reverse biased? 8. 16. What happens to the width of depletion layer of a p-n junction when it is (i) forward biased. Explain briefly how Ampere’s circuital law is generalized to incorporate the effect due to the displacement current. 12. Define the term ‘stopping potential’ in relation to photoelectric effect. 1A D 2A 2V ® ® R 2W R R1 1 A 1V C 2A 15. Net capacitance of three identical capacitors in series is 1 mF. 9. Show with the help of a diagram how the magnetic field due to the current I1 exerts a magnetic force on the second wire. assuming point A to be at zero potential. 18. In the meter bridge experiment. Plot a graph showing the variation of coulomb force (F) versus ç ç 2÷ ÷. What is sky wave communication? Why is this mode of propagation restricted to the frequencies only up to few MHz? 14. What will be their net capacitance if connected in parallel? Find the ratio of energy stored in the two configurations if they are both connected to the same source. Another wire carrying steady current (I2) in the same direction is kept close and parallel to the first wire. 13. Find the expression for the electric flux through the surface of the cylinder. 11. Interpret the graphs obtained.274 Xam idea Physics—XII 7. Using the curve for the binding energy per nucleon as a function of mass number A. A X Z B Y . Draw a suitable diagram to show amplitude modulation using a sinusoidal signal as the modulating signal. in turn. In which case will the stopping potential be higher? Justify your answer. 21. Draw a labelled diagram of a full wave rectifier circuit. v1 > v 2 . in (i) a medium of refractive index 1. Write briefly any two factors which demonstrate the need for modulating a signal. how will the balance point get affected? R X G l A J B 19. A convex lens made up of glass of refractive index 1. 24. Hence. 23. (ii) a medium of refractive index 1. (c) an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image. Use the mirror equation to show that (a) an object placed between f and 2f of a concave mirror produces a real image beyond 2f. What are the kinetic and potential energies of the electron in this state? 25. (a) Using de Broglie’s hypothesis. State its working principle. 22.65. (a) Will it behave as a converging or a diverging lens in the two cases? (b) How will its focal length change in the two media? 20. (b) The ground state energy of hydrogen atom is –13. (b) a convex mirror always produces a virtual image independent of the location of the object.33. Bohr’s second postulate of quantization of energy levels in a hydrogen atom. What would be the new position of balance point? (ii) If the galvanometer and battery are interchanged at the balance position. Draw a plot showing the variation of photoelectric current with collector plate potential for two different frequencies. Show the input-output waveforms. identify the logic operation carried out by this circuit. of incident radiation having the same intensity.5 is dipped. Write its truth table. Draw the logic symbol of the gate it corresponds to.6 eV. You are given a circuit below. explain with the help of a suitable diagram.Examination Papers 275 (i) The values of R and X were doubled and then interchanged. 27.f. (i) Write the expression for the instantaneous value of the e. (c) Outline the necessary steps to convert a galvanometer of resistance RG into an ammeter of a given range. Also calculate the length of the microscope. OR A giant refracting telescope at an observatory has an objective lens of focal length 15 m. State the importance of coherent sources in the phenomenon of interference. (b) A horizontal straight wire of length L extending from east to west is falling with speed v at right angles to the horizontal component of Earth’s magnetic field B. consume powers P1 and P2 respectively. giving its labelled diagram.m.” Justify this statement.8 × 108 m. The coil of an a. circuit containing a pure inductor. A compound microscope uses an objective lens of focal length 4 cm and eyepiece lens of focal length 10 cm. find the angular magnification of the telescope. in turn. What is the source of energy generation in this device? OR (a) Show that in an a. Deduce the expression for the alternating e. (a) State the principle of the working of a moving coil galvanometer.42 × 106 m and the radius of the lunar orbit is 3. 28. obtain the expression for the magnetic field due to a long solenoid at a point inside the solenoid on its axis. Calculate the magnifying power of the compound microscope. An object is placed at 6 cm from the objective lens. generated in the coil. if the entire experimental apparatus of Young is immersed in water? . each of area A.276 Xam idea Physics—XII 26. State the working of a. induced in the wire. generator having N turns. what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is 3.f. generator with the help of a labelled diagram.c. (ii) What is the direction of the e. Two heating elements of resistance R1 and R2 when operated at a constant supply of voltage.c. V.c. How does the fringe width get affected.0 cm is used.? (iii) Which end of the wire is at the higher potential? 30. is rotated with a constant angular velocity w . Hence. (c) How is the magnetic field inside a given solenoid made strong? 29.m.m. Deduce the expressions for the power of their combination when they are. In Young’s double slit experiment to produce interference pattern. OR (a) Using Ampere’s circuital law. the voltage is ahead of current by p /2 in phase. If this telescope is used to view the moon.f. deduce the expression for the fringe width. (b) “Increasing the current sensitivity of a galvanometer may not necessarily increase its voltage sensitivity. (b) In what respect is a toroid different from a solenoid? Draw and compare the pattern of the magnetic field lines in the two cases. connected in (i) series and (ii) parallel across the same voltage supply. If an eyepiece lens of focal length 1. obtain the conditions for constructive and destructive interference. What is the potential at the centre of the sphere? 2. (c) If a monochromatic source of light is replaced by white light. (b) Show that the angular width of the first diffraction fringe is half of that of the central fringe. Draw the logic symbol of the gate it corresponds to. Hence. Using this principle explain how a diffraction pattern is obtained on a screen due to a narrow slit on which a narrow beam coming from a monochromatic source of light is incident normally. What is ground wave communication? On what factors does the maximum range of propagation in this mode depend? 21. 14. use Kirchhoff’s rules to determine the potential at point B. 1A D 3A 4V B R 2W R R1 1 A 2V C 3A 17. Can a transformer be used to step up or step down a d. Write its truth table. A hollow metal sphere of radius 10 cm is charged such that the potential on its surface is 5V.c. Where on the surface of Earth is the angle of dip zero? 9. What will be their net capacitance if connected in parallel? Find the ratio of energy stored in the two configurations if they are both connected to the same source.Examination Papers 277 OR (a) State Huygen’s principle. State the principle of working of a transformer. identify the logic operation carried out by this circuit. A X Z B Y . voltage? Justify your answer. assuming point A to be at zero potential. 12. what change would you observe in the diffraction pattern? CBSE (All India) SET–II Questions uncommon to Set–I 1. In the given circuit. You are given a circuit below. How are X-rays produced? 4. Net capacitance of three identical capacitors in series is 2 mF. A X Z B Y . A hollow metal sphere of radius 6 cm is charged such that the potential on its surface is 12 V. It is immersed in a liquid of refractive index 1. assuming point A to be at zero potential. In the given circuit. 22. What is the potential at the centre of the sphere? 3. Mention various energy losses in a transformer. Calculate its new focal length. Write its truth table.3. Where on the surface of Earth is the vertical component of Earth’s magnetic field zero? 12. A converging lens has a focal length of 20 cm in air. use Kirchhoff’s rules to determine the potential at point B. You are given a circuit below.278 Xam idea Physics—XII CBSE (All India) SET–III Questions uncommon to Set – I & II 1. It is made of a material of refractive index 1. Draw the logic symbol of the gate it corresponds to. 1A D 4A 6V R 2W R1 A 3V C 4A 16. 14. Hence.6. 20. What is space wave communication? Write the range of frequencies suitable for space wave communication. How are microwaves produced? 7. identify the logic operation carried out by this circuit. Examination Papers 279 Solutions CBSE (All India) SET-I 1. the width of depletion layer increases. (ii) Nuclear forces are charge . The minimum retarding (negative) potential of anode of a photoelectric tube for which photoelectric current stops or becomes zero is called the stopping potential. therefore. electric flux = × charge enclosed. Charge enclosed by the cylindrical surface = ll 1 By Gauss Theorem. Potential at centre of sphere = 10 V. 9. Electric dipole moment is defined as the numerical product of either charge and the distance between the charges. and is directed from negative to positive charge. (i) Nuclear forces are short range attractive forces. Current induced in coil will oppose the approach of magnet. . F = . Radio waves are produced by the accelerated motion of charges in conducting wires. For this the current in coil will be anticlockwise as seen from left. left face of coil will act as N-pole and right face as S-pole. The SI unit of electric dipole moment is coulomb metre (Cm). the plate A of capacitor will be positive and plate B will be negative. therefore. 2. 4pe o r2 1 1 The graph between F and is a straight line of slope q1 q 2 passing through origin. 5. the width of depletion layer decreases. eo 1 = ( l l) eo q1 q 2 1 10.independent. Angle of dip is 90° at the poles. A S N B S N (i) When forward biased. 8. 3. 6. The upper plate is positive and lower plate is negative. 2 4 pe r o 7. (ii) When reverse biased. 4. W = 0 ® ® ® ® ® ò q ( v ´ B ) . v dt ® ® ® OR Suppose two long thin straight conductors (or wires) PQ and RS are P R P R placed parallel to each other in vacuum (or air) carrying currents b I2 I 1 and I 2 respectively. Lorentz magnetic force. they experience a repulsive force. magnitude of the slope is more for attraction. 11. W = F m × S = ò Fm . It has been DL observed experimentally that when the a B DF DF currents in the wire are in the same a B DL direction. Fm = q v ´ B r ® ® ® Work done. they experience an attractive b I2 I1 force and when they carry currents in I1 opposite directions. therefore. Consider a current–element ‘ab’ of length DL of wire RS. attractive force is greater than repulsive force.280 Xam idea Physics—XII 1/r2 F Repulsive F 1/r2 Attractive Since. The magnetic field produced by current-carrying conductor PQ at the location of other wire RS m I B1 = 0 1 2p r m I DF = B1 I 2 DL sin 90° = 0 1 I 2 DL 2p r \ The total force on conductor of length L will be m I I m I I F = 0 1 2 S DL = 0 1 2 L 2p r 2p r \ Force acting on per unit length of conductor F m I I f = = 0 1 2 N/ m L 2p r . v = 0 Work. v dt = As \ ( v ´ B) . Q S Q S r Let the conductors PQ and RS carry currents I 1 and I 2 in same direction and placed at separation r. then we apply Ampere’s Circuital Law to two closed loops C1 and C2. VA + 1 + 2× 1 – 2 = VB. Induction Furnace: In induction furnance. eddy currents are produced in it. Its principle is: When a metallic cylinder (or rotor) is placed in a rotating magnetic field. these currents tend to reduce to relative motion between the cylinder and the field. The arrangement of heating the metal by means of strong induced currents is called the induction furnace. A time varying current is flowing through the capacitor.(ii) òC B × dl = 0 2 . Displacement current and generalised Ampere’s Circuital Law: Consider a parallel plate capacitor.. the magnetic flux linked with the plate changes. 2. As VA = 0 \ VB = 1 + 2 – 2 = 1 V. Strong eddy currents are set up in the metal produce so much heat that the metal melts. Sky wave propagation is a mode of propagation in which communication of radiowaves in frequency range 30 MHz–40 MHz takes place due to reflection from the ionosphere. In transformer frames there is a huge loss of energy due to production of eddy currents.(i) .. being charged by a battery. these currents are called eddy currents. For frequencies higher than few MHz. These currents are sometimes so strong. so these currents are undesirable in transformer. the metal to be heated is placed in a rapidly varying magnetic field produced by high frequency alternating current.. that the metallic plate becomes red hot. The cylinder. the induced currents are set up in the plate. This is the principle of induction motion.Examination Papers 281 12. 14. If we consider only the conduction current I. Induction Motor: The eddy currents may be used to rotate the rotor. This process is used in extracting a metal from its ore. Applying Kirchhoff’s II law along the path ACDB.. therefore. the sky waves penetrate the ionosphere and are not reflected back. Eddy currents: When a metallic plate is placed in a time varying magnetic field. According to Lenz’s law. then we get Ø Ø òC and 1 B × dl = m 0 I Ø Ø . begins to rotate in the direction of the field. Application of Eddy Currents: 1. Current in 2 W resistor is 1 A. 13. IA R R 1V A C C1 I C2 I 2W B 15. Applying Kirchhoff’s first law at power P. we get I = Id Hence. Maxwell postulated the existence of displacement current which is produced by time varying electric field. dl = m 0 ( I + m 0 e 0 dt ) 16.. dl = m 0 ( I + I d ) = m 0 ç dt ø è But EA = Electric flux fE Ø Ø df E \ B ò . then in series 1 1 1 1 3 = + + = Cs C C C C I ( ) or C = 3Cs = 3 × 1 mF = 3 mF When these capacitors are connected in parallel..282 Xam idea Physics—XII Since there cannot be any conduction current in region between the capacitor plates. If s (t) is the surface charge density on capacitor s (t) q (t) plates and q(t) is the charge.(iii) But this condition is violated by equations (i) and (ii). dE 1 dg(t) = dt Ae 0 dt dg(t) dE or = e0 A dt dt This is expression for displacement current (ld). equation (i) and (ii) take the forms Ø Ø Ø Ø ò C1 B × dl = m 0 I and ò C2 B × dl = m 0 I d = m 0 I Id P The total current is the sum of the conduction current and displacement current. we must expect Ø Ø Ø Ø 2 ò C1 B × dl = ò C B ×dl . Thus. Let C be the capacitance of each capacitor. Cp = 3 C = 3 ´ 3 = 9 mF When these two combinations are connected to same source the potential difference across each combination is same. Hence Ampere’s Circuital Law seems to be inconsistent in this case. 2 1 U s 2 C SV C 1 mF 1 = = = s = 2 9 mF 9 U p 1 C pV Cp 2 Þ Us: Up = 1: 9 . modified form of Ampere’s circuital law is Ø Ø dE ö æ I + e0 A ÷ ò B. Therefore. net capacitance. Ratio of energy stored. As C1 and C2 are very close. where A is area of = e0 Ae 0 each plate. then time varying electric field E (t) = . 100 .0C12 F18 He4 N14 7. This implies that energy would be released in nuclear fission.. the binding energy per nucleon increases i.0 6.l When both R and X are doubled and then interchanged. . (i) = X r (100 .l) Þ R l .l¢ From (i) and (ii).l¢ Þ l¢ = (100 .(ii) X l¢ = R 100 .l) Þ (ii) If galvanometer and battery are interchanged.0 3. so again energy would be released in nuclear fusion... the new balance length becomes l¢ given by 2X l¢ = 2R (100 .. When two very light nuclei ( A £ 10) join to form a heavy nucleus.0 0.0 2.0 O16 8.(i) = X 100 . there is no effect on the balance point. The variation of binding energy per nucleon versus mass number is shown in figure.0 1. R rl 18. nucleons get more tightly bound.l l¢ = l 100 .0 4.Examination Papers 283 17.0 5. 9.e.0 0 H2 Li7 Fe56 U238 Binding Energy per Nucleon (in MeV) 20 40 60 80 100 120 140 160 180 200 220 240 Mass Number Explanation: When a heavy nucleus ( A ³ 235 say) breaks into two lighter nuclei (nuclear fission).l¢) . the binding is energy per nucleon of fused heavier nucleus more than the binding energy per nucleon of lighter nuclei. 65 increases) 1. The stopping potential (Vs) is higher for radiations of frequency n1 .33 20.5. (iii) The various information signals transmitted at low frequency get mixed and hence can not be distinguished. The modulation is needed due to (i) Transmission of audio frequency electrical signals need long impracticable antenna.33 ng 15 . The plots are shown in fig. so f l and f a are of opposite sign. so convex lens in liquid (nl = 1. (ii) The power radiated at audio frequency is quite small.5 -1 1. . hence transmission is quite lossy.65 ng 15 .33) behaves as a convergent lens 1. nl = 1.5 -1 1.284 Xam idea Physics—XII 19.5. I n1 = n2 L = Constant n1 n2 (Vs)1 (Vs)2 Saturation photocurrent Collector potential 21. f 1 = f a = -5. so fl and fa are of same sign. n l = 1. \ = >1 n l 133 .5 -1 (ii) Focal length.65 a diverging lens (ii) ng = 1. f 2 = f a = 4f a (Focal length increases) 1.65 behaves as n l 1.5f a (Focal length becomes negative and its magnitude 1. = < 1. Stopping potential is directly proportional to the frequency of incident radiation. Focal length of lens in liquid (l) ng -1 a n g -1 fl = fa = fa ng l n g -1 -1 nl (a) ( i) ng = 1. so convex lens in liquid nl = 1.5 -1 (b) (i) Focal length. f > 0.e. f are negative) u f 2f 2f u f 1 1 1 or <. f < 0. > > (as u. . image lies v behind the mirror.<2f u f 1 1 1 1 or < .. (b) For a convex mirror. Hence. i.. Mirror equation is 1 1 1 = + f v u or 1 1 1 = v f u (a) For a concave mirror. 1 1 1 1 1 1 For u between f and 2f implies lies between and i.e. image is virtual whatever the value of u may be. is negative.e..Examination Papers 285 EC (a) Carrier wave ec = Ec sin wct Em (b) Modulating wave (em) Em Ec Envelope (c) Modulated wave e(t) 22.. This means that the image lies beyond 2f and it is real. f is negative. f is positive i. so v must be positive i. For a real object (on the left of mirror). e..e. For a real object on the left u is negative. v This implies that v is negative and greater than 2f . 1 1 1 1 1 1 = + implies = f v u v f u 1 As u is negative and f is positive.<0 f 2f f u 1 1 or < <0 2f v 1 i. must be positive. so f < u < 0 1 1 This implies... 1 1 1 = v f u For a concave mirror..(ii) 2pr = nl .(i) |f | >1 | f | -| u | Input RL Tapped secondary D2 diode Output Principle: It works on the principle that p-n junction conducts when it in forward bia and does not do so when it in reverse bias. the stationary orbits are those which contain complete de-Broglie wavelength. v i. Full Wave Rectifier D1 diode .(i) l= mv According to de Broglie’s condition of stationary orbits. (a) According to de Broglie’s hypothesis.e. Input wave form T 2T Time Output wave form of full wave rectifier T 2T Time 24. ..> 0 f u 1 Then from (1) > 0 or v is positive.. 23..e. magnification m = i. .= u u-f As u is negative and f is positive. f v Magnification.. image is on the right and hence virtual..286 Xam idea Physics—XII (c) For a mirror. f is negative f < 0 As u is also negative. h . m = . image is enlarged. The logic symbol and truth table are shown: Truth Table A 0 A B Y B 0 0 1 1 Y 0 0 0 1 1 0 1 . (iii).27 × 2 eV (in ground state) 25. The output Y = A + B = A × B = AB That is equivalent gate is ‘AND’ gate.Examination Papers 287 Substituting value of l from (ii) in (i).13 × 6 eV) = .(iii) Þ mvr = n 2p This is Bohr’s postulate of quantisation of energy levels.E and U = 2E Given \ Kinetic energy. (ii)..13 × 6 eV K = 13 × 6 eV U = 2 ´ ( . Potential energy E = . we have K = .. Potential energy. we get h 2pr = n mv h . n=3 (b) Kinetic energy. K= 1 1 e2 mv 2 = × 2 4pe 0 2r 1 e2 4pe 0 r 1 e2 4pe 0 2r …(i) …(ii) …(iii) U =- Total energy E = K + U = - Comparing equations (i). Length of microscope L = v o + u e 1 1 1 For eye lens = f e ve ue 1 1 1 1 1 = = .2 Let D be diameter of moon. u e = ?) m Angular magnification of telescope.ue ve f e 25 10 50 Þ u e = .25 cm. D d = r f0 Þ 27.288 Xam idea Physics—XII 26. Given fo = 4 cm. fe = 10 cm u o = – 6 cm Magnifying power of microscope 1v 1 æ Dö M=. R = R1 + R2 As heating elements are operated at same voltage V. then from Fig. R1 = P P1 P2 . f e = 1× 0 cm = 1× 0 ´ 10 -2 ( v e = D = .. d= 3 × 48 ´ 10 6 D × f0 = ´ 15 m = 0 × 137 m = 13 × 7 cm r 3 × 8 ´ 108 .5 = -7 6 è 10 ø Negative sign shows that the image is inverted.( cm) = -7 ×14 7 \ L = v o + u e = 12 + 7 ×14 = 19 ×14 cm OR Given f 0 = 15 m. f 15 m= 0 = = 1500 fe 1× 0 ´ 10 .= v o f o u o 4 6 12 Þ \ vo = 12 cm 12 æ 25 ö m = – ç1 + ÷ = -2 ´ 3.(i) (i) In series combinations Net resistance.o ç 1+ ÷ ç 1u o 1 è f e ÷ ø From lens formula 1 1 1 = f o vo uo 1 1 1 1 1 3 -2 = = + = .=.. we have V V2 V2 and R 2 = R = 2 . d diameter of image of moon formed by objective and r the distance of moon from objective lens. (i) \ q= I C i. is parallel to the magnetic field lines.. When the magnetic field is radial.. so that q = 90° and sin 90° = 1 Deflecting torque. H Suspension wire M Coil NIBl b T1 T2 N S N S N S NIBl Coiled strip (b) Magnetic lines of force of radial magnetic field (a) (c) . as in the case of cylindrical pole pieces and soft iron core. When current ( I ) is passed in the coil. qµI deflection of coil is directly proportional to current flowing in the coil and hence we can construct a linear scale.e. then in every position of coil the plane of the coil. deflecting torque = restoring torque i.e. NIAB = C q NAB . can experience a torque. N is the number of turns in a coil. (a) Principle: It works on the principle that a current carrying coil when kept inside a uniform magnetic field. torque t acts on the coil. then restoring torque = C q For equilibrium. given by t = NIAB sin q where q is the angle between the normal to plane of coil and the magnetic field of strength B.Examination Papers 289 \ From equation (i) V2 V2 V2 1 1 1 = + Þ = + P P1 P2 P P1 P2 (ii) In parallel combination 1 1 1 Net resistance = + R R1 R 2 P P P = P = P1 + P2 = 1 + 2 Þ 2 2 V V V2 28. t = NIAB If C is the torsional rigidity of the wire and q is the twist of suspension strip. e..e.Ig This is the working equation of conversion of galvanometer into ammeter. S = I g .I g S = I g G or IS = I g (S + G ) S or . G Substituting value of I S from (i). shunt required. (ii) Clearly the voltage sensitivity depends on current sensitivity and the resistance of galvanometer. then it is not certain that voltage sensitivity will be increased.. we get or (I . the increase of current sensitivity does not imply the increase of voltage sensitivity. The resistance of an ammeter is to be made as low as possible so that it may read current without any appreciable error. Let G be the resistance of galvanometer and I g the current required for full scale deflection. (c) Conversion of galvanometer into ammeter: Rg An ammeter is a low resistance galvanometer and i ig G is connected in series in a circuit to read current directly in ‘a’. then from fig. If the solenoid is long . Thus. Suppose this galvanometer is to converted into ammeter of range I ampere and the value of shunt required is S. .... Therefore to convert a (i – ig) S galvanometer into ammeter a shunt resistance. To construct a solenoid a large number of closely packed turns of insulated copper wire are wound on a cylindrical tube of card-board or china clay. SV = ç ÷ = è V ø GC Dividing (ii) by (i) SV 1 = SI G Þ SV = 1 SI G .(ii) Ig = I S +G GI g i. carrying current. If we increase current sensitivity and resistance G is larger.I g ) Also potential difference across A and B (VAB ) = I S . S = . a magnetic field is produced within the solenoid.(i) I = I g + I S Þ I S = (I .. (i) . If I S is current in shunt.(iii) I .. (i. OR (a) Magnetic Field Due to a Current Carrying Long Solenoid: A solenoid is a long wire wound in the form of a close-packed helix. When an electric current is passed through the solenoid. small resistance in parallel) is connected across the coil of galvanometer... S I = ç ÷ = C èIø æ q ö NAB Voltage sensitivity.I g ) S = I g G or IS ..290 Xam idea Physics—XII æ q ö NAB (b) Current sensitivity. ® ® sp ® \ òqr B · dl = ò B · d l = ò B dl cos 90° = 0 ® . B and dl are along the same direction. Field Inside the solenoid: Consider a closed path pqrs. the magnetic field is parallel to the axis of the solenoid. Applying Ampere’s law to this path ® ® ò B · dl = m ´ 0 (since net current enclosed by path is zero) As dl ¹ 0 \ B = 0 This means that the magnetic field outside the solenoid is zero. then N n= l m 0 NI or B= l Derivation: Consider a symmetrical long solenoid having number of turns per unit length equal to n. B and d l are mutually perpendicular.(i) For path pq. The line integral of magnetic field B along path pqrs is ® ® pq ® ® ® ® ® qr ® ® rs ® òpqrs \ B · dl = ò ® B · dl + ò B · dl + ò B • dl + ò ® ® sp ® ® B · dl . The reason is that the B p q solenoid may be supposed to be formed of a l large number of circular current elements. Due to this the magnetic field in a direction a b perpendicular to the axis of solenoid is zero and so the resultant magnetic field is along the axis of the solenoid.. òpq B · dl = ò B dl = Bl ® ® ( pq = l say) For paths qr and sp. with practically no magnetic field outside it.Examination Papers 291 and the successive insulated copper turns have d c s r no gaps. Field outside the solenoid: Consider a closed path abcd. then by right hand rule. then the magnetic field within the solenoid is uniform. The B magnetic field due to a circular loop is along its axis and the current in upper and lower straight parts of solenoid is equal and opposite. Let I be the current flowing in the solenoid. If there are ‘n’ number of turns per metre length of solenoid and I amperes is the current flowing.. then magnetic field at axis of long solenoid B = m 0 nI If there are N turns in length l of wire. N B S Field magnet the wire ab moves upward and cd downward. magnetic lines do not exist outside the body. the wire ab moves downward and cd RL upward. the magnetic field may be made strong by (i) passing large current and (ii) using laminated coil of soft iron./Toroid is closed whereas solenoid is open on both sides./Magnetic field is uniform inside a toroid whereas for a solenoid. Ä Ä . w 29. so that a the direction of induced current is shown in fig. the magnetic flux linked with b the coil changes and the current is induced in the c coil. Inside a given solenoid.(ii) ò B · dl = m 0 ´ net current enclosed by path \ Bl = m 0 ( nl I ) \ B = m 0 nI (b) In a toroid. so the direction of current is reversed and in R2 external circuit it flows along B 2 R L B1 . Thus the B2 direction of induced emf and current changes in the external circuit after each half revolution. equation (i) gives òpqrs By Ampere’s law B · dl = ò ® B · dl = Bl . B = 0 (since field is zero outside a solenoid) ® ® \ òrs B · dl = 0 ® ® pq ® ® ® In view of these. During the second half Slip rings Brushes Load revolution. Working: When the armature coil is rotated in the Armature coil strong magnetic field.292 Xam idea Physics—XII For path rs. it is different at the two ends and cenre. Considering the armature to be in vertical position and as it rotates in anticlockwise direction. its direction being given by Fleming’s right hand rule.. In d the external circuit. (Any one) Q S Ä Ä Ä Ä Ä Ä Ä P Ä O Ä Ä (c) The magnetic field lines of toroid are circular having common centre. the current flows along B1 R L B 2 .. B1 The direction of current remains unchanged during R1 the first half turn of armature. V V ì cos wt ü V0 i = 0 ò sin wt dt = 0 í cos wt ý =L L î w þ wL V æ p ö = . That is inductance acts like a resistance in ac circuit. then induced emf df d e== {NBA (cos 2p f t )} dt dt = 2p NBA f sin 2p f t Obviously. the emf produced is alternating and hence the current is also alternating. L di V + e=0 Þ V -L =0 dt di di V or or V =L = dt dt L V sin w t di V0 sin wt V=V0 sin wt or or = di = 0 dt (a) dt L L Integrating with respect to time ‘t’. The magnitude and direction of AC changes periodically. Current produced by an ac generator cannot be measured by moving coil ammeter. The source of energy generation is the mechanical energy of rotation of armature coil.Examination Papers 293 If N is the number of turns in coil. the rate of change of current in the circuit dt at that instant. due to which there is a continual change in magnetic flux linked with the coil.. f the frequency of rotation. As a result the current in the circuit is reduced.0 sin ç ..(iii) .. An alternating potential difference is applied across its ends. then instantaneous induced emf di e=-L dt According to Kirchhoff’s second law in closed circuit. OR (a) AC circuit containing pure inductance: Consider a coil of self-inductance L and negligible ohmic resistance.÷ wL 2 ø è current ö ÷ ø …(ii) This is required expression for p æ or i = i 0 sin ç wt 2 è . an induced emf is produced in the coil.(i) V = V0 sin wt di If i is the instantaneous current in the circuit and . because the average value of ac over full cycle is zero. A area of coil and B the magnetic induction.wt ÷ wL 2 è ø or i= V0 p ö æ sin ç wt . The instantaneous value of alternating voltage applied . which opposes the applied voltage. Therefore according to Faraday’s law.. If coherent sources are not taken... (iii) East-send will be at higher potential. we note that current lags behind the applied voltage by an angle . we get y = a1 sin wt + a 2 sin ( wt + f) Using trigonometric relation sin ( wt + f) = sin wt cos f + cos wt sin f .. we get y = a1 sin wt + a 2 (sin wt cos f + cos wt sin f) . then equation of waves may be expressed as . Clearly the amplitude of resultant disturbance is A and phase difference from first wave is q.(iv) = ( a1 + a 2 cos f) sin wt + ( a 2 sin f) cos wt Let . the frequency of w each wave is and amplitudes of electric field are a1 and a 2 respectively. we get ..(ii) y 2 = a 2 sin ( wt + f) According to Young’s principle of superposition.. 30..(v) a1 + a 2 cos f = A cos q And .. 2 (b) (i) Induced emf e = B H vL Where BH is horizontal component of earth’s magnetic field directed from S to N.294 Xam idea Physics—XII V0 wL is the peak value of alternating current where i0 = . the 2p electric fields of waves at a point are y1 and y 2 respectively and phase difference is f. we get ( a1 + a 2 cos f) 2 + ( a 2 sin f) 2 = A 2 cos 2 q + A 2 sin 2 q or a12 + a 22 cos 2 f + 2a1 a 2 cos f + a 22 sin 2 f = A 2 (cos 2 q + sin 2 q) A 2 = a12 + a 22 (cos 2 f + sin 2 f) + 2a1 a 2 cos f or A 2 = a12 + a 22 + 2a1 a 2 cos f As cos 2 q + sin 2 q = 1...(i) y1 = a1 sin wt . will change continuously and a sustained interference pattern will not be obtained.(vii) y = A cos q sin wt + A sin q cos wt = A sin ( wt + q) This is the equation of the resultant disturbance. If at any time t...(iii) y = y1 + y 2 Substituting values of y1 and y 2 from (i) and (ii) in (iii). The values of A and q are determined by (v) and (vi). Thus. Then equation (iv) gives . the phase difference between two interfering waves.. Squaring (v) and (vi) and then adding.. coherent sources provide sustained interference pattern.. Suppose two coherent waves travel in the same direction along a straight line. the resultant displacement at that point will be .(vi) a 2 sin f = A sin q where A and q are new constants... (ii) West to east.(iv) p Also comparing (i) and (iii). I µ A 2 or I = KA 2 watt / m 2 where K is a constant which depends on properties of medium and the frequency of wave.. I min = a12 + a 22 .1) p .. 1.(xv) = ´ ( 2n . 2p .. \ Intensity of resultant wave I = A 2 = a12 + a 22 + 2a1 a 2 cos f . the minimum intensity is obtained in the region of superposition at those points where waves meet in opposite phase or the phase difference between the waves is odd multiple of p or l path difference between the waves is odd multiple of and minimum intensity = ( a1 .a 2 ) 2 2 which is less than the sum of intensities of the individual waves by an amount 2a1 a 2 .1) p = ( 2n ..(viii) As the intensity of a wave is proportional to its amplitude i.(ix) Clearly the intensity of resultant wave at any point depends on the amplitudes of individual waves and the phase difference between the waves at the point.... 7p .. K = 1. n = 1. therefore for convenience. 2.(xii) ´ Phase difference = ´ 2 np = nl 2p 2p Clearly the maximum intensity is obtained in the region of superposition at those points where waves meet in the same phase or the phase difference between the waves is even multiple of p or path difference between them is the integral multiple of l and maximum intensity is ( a1 + a 2 ) 2 which is greater than the sum intensities of individual waves by an amount 2a1 a 2 .e.. = 2np ( n = 0.. we may take....(xiii) . Destructive Interference: For minimum intensity at any point cos f = .. 4p . I max = a12 + a 22 + 2a1 a 2 = ( a1 + a 2 ) 2 Path difference D= . .(xi) l l . f = p ... ..a 2 ) 2 Path difference. 2. 3 . 6p ...... 3p . D= .(x) .. 5p ....2a1 a 2 = ( a1 . Constructive Interference: For maximum intensity at any point cos f = + 1 or phase difference f = 0..Examination Papers 295 \ Amplitude. then the units of intensity I will not be watt / m 2 but arbitrary.1) 2p 2 Clearly.. = ( 2n ..) The maximum intensity. A = a12 + a 22 + 2a1 a 2 cos f ...1 or phase difference.(xiv) l ´ Phase difference 2p l l .. In interference the frequencies of two waves are same and medium is same. In this case the minimum intensity.. Thus the point M has the maximum intensity. D = nl .(i) \ D (i) Positions of bright fringes (or maxima): For bright fringe or maximum intensity at P. sin q = 2 S1S 2 MP In DMOP.S1 P = S 2 N = . Variation of Intensity of light with position x is shown in fig. the path difference must be an integral multiple of wavelength ( l) of light used. Expression for Fringe Width: Let S1 and S 2 be two coherent sources separated by a distance d.. This implies that the interference is based on conservation of energy. Let the distance of the screen from the coherent sources be D. the midpoint of S1 and S 2 on the screen. i.. OM D yd Path difference D = S 2 P . Let M be the foot of the perpendicular drawn from O. Therefore the path difference between the two waves at point M is zero.S1 P » S 2 N As D > > d. Draw S1 N perpendicular from S1 on S 2 P. tan q = OM As q is very small \ sin q = q = tan q S 2 N MP \ = S1S 2 OM y MP \ S 2 N = S1S 2 =d. therefore Ð S 2 S1 N = q is very small \ Ð S 2 S1 N = Ð MOP = q S N In D S1S 2 N . P S1 S d O S2 y q q M N D The path difference between two waves reaching at P from S1 and S 2 is D = S 2 P .296 Xam idea Physics—XII I Imax Imin –2l –l O x l 2l From equations (xii) and (xvi) it is clear that the intensity 2a1 a 2 is transferred from positions of minima to maxima.e. Obviously point M is equidistant from S1 and S 2 . Consider a point P on the screen at a distance y from M. 1) D 2 ( 2n . D = ( 2n . the fringe-width decreases ( l w < l a ). 2d 2 d This equation gives the distance of nth dark fringe from point M.(v) .(iv) = .y n 1 Dl 1 Dl = (n + ) . \ = ( 2n . For Bright Fringes: If y n + 1 and y n denote the distances of two consecutive bright fringes \ from M.) 2 d 2 d \ Fringe width... fringe width is the same for bright and dark fringes equal to Dl b= d When entire apparatus is immersed in water. d d d and y n are the distances of two consecutive dark fringes from For Dark Fringes: If y n + 1 M. Therefore writing y n for y... yn = ( n .y n = ( n + 1) Dl nDl Dl . 3. Therefore writing y n for y..) 2 d (iii) Fringe Width b : The distance between any two consecutive bright fringes or any two consecutive dark fringes is called the fringe width. we get 1 Dl . K D nDl \ y= .) .) 2 d 2 d Dl 1 1 Dl = (n + . d This equation gives the distance of nth bright fringe from the point M.. then we have 1 Dl 1 Dl yn + 1 = ( n + ) . we get nDl . the path l difference must be an odd number multiple of half wavelength.1) lD 1 Dl or y= = (n . i.. 1. 3..e. b = y n + 1 .Examination Papers 297 yd = nl .(n . . b = y n + 1 .d l where n = 1. n = 0.(ii) yn = .n + ) = d 2 2 d Thus.. It is denoted by w.1) 2 y. d (ii) Positions of dark fringes (or minima): For dark fringe or minimum intensity at P.. then we have Dl nDl y n + 1 = ( n + 1) and y n = d d \ Fringe width.(iii) yn = ( n .. 2. 2. .. Diffraction of light at a single slit: When monochromatic light is made incident on a single slit. The path difference between rays diffracted at points A and B.(i) D = a sin q To find the effect of all coherent waves at P. each with a different phase. The principle may be stated in three parts as follows: (i) Every point on a given wavefront may be regarded as a source of new disturbance.AP = BN In D ANB . The diffraction pattern contains bright and dark bands. Explanation: Let AB be a slit of width ‘a’ and a parallel beam of monochromatic light is incident on it. (iii) The surface of tangency to the secondary wavelets in forward direction at any instant gives the new position of the wavefront at that time. ÐANB = 90° and ÐBAN = q BN or BN = AB sin q \ sin q = AB As AB = width of slit = a \ Path difference. we have to sum up their contribution.. (ii) The new disturbances from each point spread out in all directions with the velocity of light and are called the secondary wavelets.298 Xam idea Physics—XII OR (a) Huygen’s principle is useful for determining the position of a given wavefront at any time in the future if we know its present position. starting from different points of illuminated slit.. P A Light from source M1 q C q O M2 90o N B q Let q be the angle of diffraction for waves reaching at point P of screen and AN the perpendicular dropped from A on wave diffracted from B. D = BP . but the main features may be explained by simple arguments given below: . . This was done by Fresnel by rigorous calculations. the intensity of central band is maximum and goes on decreasing on both sides. According to Fresnel the diffraction pattern is the result of superposition of a large number of waves. we get diffraction pattern on a screen placed behind the slit. each of width . Thus the general condition of minima is a . with n as integer. This means that the contributions from the two halves of slit AO and OB are 2 2 opposite in phase and so cancel each other. M 1 in AO. This gives maximum intensity at the central point C.Examination Papers 299 At the central point C of the screen. such that M 1 M 2 = . Then 2 path difference between waves arriving at P and starting from M 1 and M 2 will be a l sin q = . . the angle q is zero. Similarly it may be shown that the intensity is zero for nl sin q = ..(iii) a sin q = nl Secondary Maxima: Let us now consider angle q such that 3l sin q = q = 2a which is midway between two dark bands given by l 2l sin q = q = and sin q = q = a a I0 –4l a –3l a –2l a –l a 0 l a 2l a 3l a 4l a . there is a corresponding point M 2 in OB... Thus equation (2) gives the angle of diffraction at which intensity falls to zero. then path difference l a sin q = l Þ sin q = a l If angle q is small.(ii) sin q = q = a a Minima: Now we divide the slit into two equal halves AO and OB. If point P on screen is such that the path difference between rays starting from edges A and B is l . Now for 2 a every point.. Hence the waves starting from all points of slit arrive in the same phase. (c) If monochromatic light is replaced by white light. C S = 3 In parallel. the current in primary coil remains same. Þ 17. I = 2 A Applying Kirchhoff’s law along the path ACDB VA + 2 + 2 ´ 2 . X-rays are produced by sudden deceleration or acceleration of electrons. 4. the angular width of violet is least and that of red is maximum.4 = VB VB = VA + 2 = 0 + 2 = 2V The mode of wave propagation in which wave glides over the surface of the earth is called ground wave communication. Explanation: When dc voltage source is applied across a primary coil of a transformer.300 Xam idea Physics—XII (b) Angular width of central maximum (b q ) C = 2l a 2l l l Angular width of first maximum. . The maximum range of propagation in this mode depends on (i) transmitted power and (ii) frequency (less than a few MHz) 12. the fringe width of the first diffraction fringe is half that of the central fringe. 5 V. (b I ) = . Current in 2W resistor. CBSE (All India) SET–II 1. Angle of dip is zero at equator of earth’s surface./In an X-ray tube. each diffraction band gets splited into the number of coloured bands. 2.= a a a bI 1 = (b q ) C 2 Hence. C 9. Transformer cannot step up or step down a dc voltage. so there is no change in magnetic flux and hence no voltage is induced across the secondary coil. CP = 3C CP =9 CS Þ C P = 9 ´ 2 mF = 18 mF 1 C sV 2 Es C 2 1 2 Required ratio = = = s = = . Reason: No change in magnetic flux. In series. Ep 1 C p 18 9 C pV 2 2 Working of a transformer is based on the principle of mutual induction. 14. Current in 2 W resistor is 3 A VA + 3 + 3 × 2 – 6 = VB Þ VB = VA + 3 = 0 + 3 = 3 V 16. Truth table A A Y B B 0 1 0 1 X=A 1 1 0 0 Y=B 1 0 1 0 Z 0 1 1 1 0 0 1 1 CBSE (All India) SET–III 1. At magnetic equator. etc. When a wave propagates in a straight line. magnetrons. Frequency range: Above 40 MHz. a n g -1 20. Energy losses in a transformer are due to (i) Flux leakage (ii) Joule heating in resistance of winding (iii) Eddy currents (iv) Hysteresis 14. klystrons.Examination Papers 301 21. 12 V 3. from the transmitting antenna to the receiving antenna. Gunn diodes. 12. The circuit represents OR gate. namely. f l = ´ fa a ng -1 P a nl 1× 6 -1 = ´ 20 cm cm A q æ 1× 6 ö q Light from M1 -1÷ ç O C è 1× 3 ø M2 90 N source q 0 × 6 ´ 1× 3 = ´ 20 cm B 0 ×3 = 52 cm o . 7. Microwaves are produced by special vacuum tubes. its mode of propagation is called space wave communication. Logic symbol: A 0 A y B Truth table: B 0 0 1 1 Y 0 0 0 1 1 0 1 . B = AB The circuit represents AND gate.302 Xam idea Physics—XII 22. y = A + B = A . m. You have to attempt only one of the given choices in such questions. each having the same momentum. 2. (c) There is no overall choice. Write its (i) rms (ii) average value over a complete cycle. (e) You may use the following values of physical constants wherever necessary: c = 3 ´ 108 ms . (d) Use of calculators is not permitted. Two insulated charged copper spheres A and B if identical size have charges q A and q B respectively. . Write two uses of microwaves. Define the term ‘threshold frequency’ in relations to photoelectric effects. Questions 1 to 8 carry one mark each. Draw the logic circuit of NAND gate and write its truth table. an internal choice has been provided in one question of two marks. However. What is the function of 'Repeater' in communication system? 6. A narrow beam of protons and deuterons. questions 9 to 18 carry two marks each.1 h = 6 × 626 ´ 10 -34 Js e = 1× 602 ´ 10 -19 C 1 = 9 × 109 Nm2C– 4pe o m 0 = 4p ´ 10 -7 TmA -1 2 Maximum marks: 70 Boltzmann’s constant k = 1× 381 ´ 10 -23 J K -1 Avogadro’s number N A = 6 × 022 ´ 10 23 /mole Mass of neutron m n = 1× 2 ´ 10 -27 kg Mass of electron m e = 9 ×1´ 10 -31 kg Radius of earth = 6400 km CBSE (Foreign) SET–I 1. What are the new charges on A and B? 4.c. A third sphere C of the same size but uncharged is brought in contact with the first and then in contact with the second and finally removed from both.CBSE EXAMINATION PAPERS FOREIGN–2011 Time allowed: 3 hours General Instructions: (a) All questions are compulsory. questions 19 to 27 carry three marks each and questions 28 to 30 carry five marks each. one question of three marks and all three questions of five marks each. How is the mean life of a radioactive sample related to its half life? 8. 7. (b) There are 30 questions in total. What would be the ratio of the circular paths described by them? 5. enters a region of uniform magnetic field directed perpendicular to their direction of momentum. The peak value of e.f. 3. is E 0 . in a. If the two capacitors still have equal capacitance. Calculate the amount of work done in rotating a dipole. How is this behaviour explained? 11.002603 u. of dipole moment 3 ´ 10 -8 cm.f. Show how the equation for Ampere's circuital law. viz. Current in a circuit falls steadily from 5. (a) You are required to select a carbon resistor of resistance 47 kW ± 10% from a large collection. Why is modulation index generally kept less than one? 13. obtain the relation between dielectric constants K. If an average e. 1u = 931 MeV / c 2 12.007825 u (ii) mass of helium nucleus = 4. C1 l/2 K l K2 d d K1 C2 . For an amplitude modulated wave.V. calculate the self-inductance of the circuit. Two identical parallel plate (air) capacitors C1 and C 2 have capacitances C each.0 A to 0.304 Xam idea Physics—XII 9. The between their plates is now filled with dielectrics as shown.0 A in 100 ms. Calculate the modulation index. Plot a graph showing temperature dependence of resistivity for a typical semiconductor.m. in a uniform electric field of intensity 10 4 N/C. Draw a block diagram showing the important components in a communication system. 17. 10. estimate the amount of energy in MeV released in this process of fusion (Neglect the masses of electrons and neutrinos) Given: (i) mass of 1 1 H = 1. When four hydrogen nuclei combine to form a helium nucleus. What is the function of a transducer? 14. dl = m I 16. K1 and K 2 . ® ® ò B . the maximum amplitude is found to be 10 V while the minimum amplitude is 2. 18. from its position of stable equilibrium to the position of unstable equilibrium. (ii) Where on the surface of the Earth is the vertical component of the Earth's magnetic field zero? 15. Explain the following: (i) Why do magnetic lines of force form continuous closed loops? (ii) Why are the field lines repelled (expelled) when a diamagnetic material is placed in an external uniform magnetic field? OR (i) Name the three elements of the Earth's magnetic field. of 200 V is induced. What should be the sequence of colour bands used to code it? (b) Write the characteristics of manganin which make it suitable for making standard resistance. 2 eV. Define the term 'barrier potential'. For what value of q is the intensity of emergent light zero? . Explain why? 20. Draw transfer characteristics of common emitter n-p-n transistor. State the principal of the device that can build up high voltages of the order of a few million volts. what will be the energy of emitted electrons? 21. 22. show that (i) the intensity of diffraction fringes decreases as the order (n) increases. how a p-n junction is formed. Point out the region in which the transistor operates as an amplifier. The intensity at the central maxima (O) in a Young's double slit experiment is I 0. Light of wavelength 2000 Å falls on a metal surface of work functions 4. Name the important process that occur during the formation of a p-n junction. f = +10 –10 + 30 cm O 30 cm 5 cm 10 cm 24. show that the intensity at point P would be 0 .5 eV.Examination Papers 305 19. Draw its labelled diagram. Define the following terms used in transistor amplifiers: (i) Input resistance (ii) Output resistance (iii) Current amplification factor 25. Find the position of the image formed of the object 'O’ by the lens combination given in the figure. What is the kinetic energy (in eV) of the fastest electrons emitted from the surface? (i) What will be the change in the energy of the emitted electrons if the intensity of light with same wavelength is doubled? (ii) If the same light falls on another surface of work functions 6. 23. 4 P S1 d S2 D O y OR In the experiment on diffraction due to a single slit. If the distance OP I equals one-third of the fringe width of the pattern. Explain briefly. with the help of a suitable diagram. (i) Light passes through two polaroids P1 and P2 with axis of P2 making an angle q with the pass axis of P1 . (ii) angular width of the central maximum is twice that of the first order secondary maximum. A stage reaches in this device when the potential at the outer sphere cannot be increased further by piling up more charge on it. Explain. having N closely wound turns and area of ® ® ® cross-section A. 26. using Huygens’ principle. (ii) Hence verify Snell's law. (c) Determine the current quality factor at resonance for a series LCR circuit with L = 1. (b) Show that mutual inductance of solenoid 1 due to solenoid 2. 8 where I 0 is the intensity of light on the polaroid P1 . (a) Show that a planar loop carrying a current I. (c) A power transmission line feeds power at 2200 V with a current of 5 A to s step down transformer with its primary winding having 4000 turns. (i) A plane wavefront approaches a plane surface separating two media. find out the expression for the torque acting on it. (iii) When a light wave travels from rarer to denser medium. Does it imply reduction its energy? Explain. Using the postulates of Bohr's model of hydrogen atom.c.e. how a potentiometer is used to measure the internal resistance of a given cell. (c) A galvanometer coil of 50 W resistance shows full scale deflection for a corrent of 5 mA. How will you convert this galvanometer into a voltmeter of range 0 to 15 V? OR (a) Draw a schematic sketch of a cyclotron. State the underlying principal of potentiometer. Calculate the number of turns and the current in the secondary in order to get output power at 220 V. Describe briefly. is the same as that of 2 due to 1 i. explain its working principal and deduce the expression for the kinetic energy of the ions accelerated. 1. 28.00 mH. source having peak voltage of 100 V. them. (a) Derive the expression for the mutual inductance of two long coaxial solenoids of same length l having radii r1 and r 2 (r 2 > r1 and l >> r2 ). giving the necessary circuit diagram. 30. (b) Two long and parallel straight wires carrying currents of 2 A and 5 A in the opposite directions are separated by a distance of 1 cm. Find a value of b for which the intensity of light emerging from P2 is 0 . a low power factor implies large power loss in transmission. M 12 . 27.00 nF and R = 100 W connected to an a. .. explain and show how a refracted wavefront is constructed. M 21 . If medium 'one’ is optically denser and medium 'two’ is optically rarer. (b) When this loop is placed in a magnetic field B. obtain an expression for the frequency of radiation emitted when atom make a transition from the higher energy state with quantum number ni to the lower energy state with quantum number n f ( n f < ni ). 29. possesses a magnetic moment m = N I A . (b) For circuits used for transporting electric power. OR (a) An alternating voltage v = v m sin w t applied to a series LCR circuit drives a current given by i = i m sin ( w t + f).306 Xam idea Physics—XII (ii) A third polaroid is placed between P1 and P2 with its pass axis making an angle b with the I pass axis of P1 . the speed decreases. Find the nature and magnitude of the magnetic force between. Deduce an expression for the average power dissipated over a cycle. What is the function of a transmitter in a communication system? 5. derive an expression for the refractive index of the material of the prism in terms of the angle minimum deviation and its refracting angle. 12. 1. You are given an air filled parallel plate capacitor C1 . d/2 K1 C2 K2 d d . the maximum amplitude is found to be 12 V while the minimum the amplitude is 2V. Trace the paths of these rays reasoning out 45° the difference in their behaviour. Write two uses of infrared rays. What is (i) rms and (ii) average value of the current for a complete cycle? 2. When they are brought in contact with each other and finally separated.47.39. from its position of stable equilibrium to the position of unstable equilibrium. Calculate the amount of work done in rotating a dipole.44 R and 1. Calculate the modulation index. in electric field of intensity 10 4 N/C. Hence. The B refractive indices of the material of the prism for red. what are the new charges on them? 4. Draw the logic circuit of AND gate and write its truth table. Why is modulation index generally kept low? 13. For an amplitude wave. Trace the path of ray passing through the prism. 7. 1.Examination Papers 307 OR (i) A ray of monochromatic light is incident on one of the faces of an equilateral triangular prism of refracting angle A. a (ii) Three light rays red (R). G green and blue wavelengths are respectively 1. of dipole moment 5 ´ 10 -8 cm. Find the capacitances of the capacitor C 2 if area of the plates is A distance between the plates is d. 10. The space between its plates is now filled with slabs of dielectric constants K1 and K 2 as shown in C 2 . b c CBSE (Foreign) SET–II Questions uncommon to Set-I. Two insulated charged copper spheres A and B of identical size have charges q A and q B respectively. Show the variation of photocurrent with collector plate potential for different frequencies but same intensity of incident radiation. green (G) and blue (B) are incident on the right angled prism abc at face ab. 3. The current flowing through a pure inductor of inductance 4 mH is i = 12 cos 300 t ampere. 12. in uniform electric field of intensity 5 ´ 10 4 N/C. of 200V is induced. calculate the self-inductance of the circuit. When they are brought in contact with each other and then separated. What is the kinetic energy (in eV) of (i) the fastest and (ii) the slowest electrons emitted from the surface? If the same light falls on another surface of work function 5. 2. Calculate the modulation index.f. Find the capacitances of the capacitor C 2 if area of the plates is A and distance between the plates is d. of 200 V is induced. Two insulated charged copper spheres A and B of identical size have charges q A and -3q A respectively. what will be the energy of emitted electrons? .f. What is the kinetic energy (in eV) of (i) the fastest and (ii) the slowest electronic emitted from the surface? If the same light falls on another surface of work function 5. Light of wavelength 2500 Å falls on a metal surface of work function 3. What is the (i) rms and (ii) average value of current for a complete cycle? 6. The space between its plates is now filled with slabs of dielectric constants K1 and K 2 as shown in C 2 . 9. The current flowing through a pure inductor of inductance 2 mH is i = 15 cos 300 t ampere. Calculate the amount of work done in rotating a dipole. what will be the energy of emitted electrons? CBSE (Foreign) SET–III Questions uncommon to Set-I & II.5 V. What is the function of a Receiver in a communication system? 8. 3. of dipole moment 2 ´ 10 -8 cm. Draw the logic circuit of NOT gate and write its truth table.5 eV. C1 K1 C2 K2 d d 14.0 to 0. 27.0 A in 300 ms. For an amplitude modulated wave the maximum amplitude is found to be 15 V While the minimum amplitude is 3V. If an average e. from its position of stable equilibrium to the position of unstable equilibrium.5 eV. Current in a circuit falls steadily from 3.m. what are the new charges on them? 4. Write two uses of X-rays. 15. Show the variation of photocurrent with collector plate potential for different intensity but same frequency of incident radiation. If an average e. Light of wavelength 2400 Å falls on a metal surface of work function 3.0 A in 10 ms. You are given an air filled parallel plate capacitor C1 .6 eV.0 A to 0.m. 7. Why is modulation index generally kept less than one? 26. Current in a circuit steadily from 2.308 Xam idea Physics—XII 15. the self-inductance of the circuit. Examination Papers 309 Solutions CBSE (Foreign) SET–I 1. rp rd 1 q = p Bq 2 æ çQ qvB = mv ç r è ö ÷ ÷ ø [for constant momentum (P)] qd q p = =1 qp qp Hence. on amplifier and a receiver which picks up signal from the transmitter. It is different for different metal . Mean life ( t) and half life (T1/ 2 ) are related as: T t = 1/ 2 0. Logic circuit of NAND gate: A Y B Truth table Input A 0 0 1 1 B 0 1 0 1 Output Y 1 1 1 0 7. Threshold frequency is defined as the minimum frequency of incident radiation which can cause photoelectric emission. E 0 = peak value of emf E (i) rms value [ E rms ] = 0 2 (ii) average value [ E av ] = zero q q + 2q B 3. 6. New charge on A is A and New charge on B is A 2 4 4.6931 . A repeater which is a combination of a transmitter. amplifies and retransmits it to the receiver. Charge on deutron ( q d ) = charge on proton ( q p ) qd = q p Radius of circular path (r ) = rµ so. r p : r d = 1 :1 5. 2. m ( 2 He) 4 Energy released ( Q) = [ 4. (ii) In radar 9.4. resistivity decreases with increase in temperature.cos180° ] = 3 ´ 10 -4 [1 .A min 10 . But the effect of increase in n has higher impact than decrease of t.72 MeV. .2 8 = = = 0 × 67 A max + A min 10 + 2 12 Generally. Block diagram of communication system: Communication System Information Message source signal Transmitted Received User of Channel Receiver signal signal Message information signal Transmitter Noise Function of a transducer is to convert one form of energy into another form.cos q 2 ) = (3 ´ 10 -8 ) (10 4 ) [cos 0° . So. Variation of resistivity (r) with temperature (T ) is shown below: r T Explanation: In semiconductor the number density of free electrons ( n) increases with increase in temperature (T ) and consequently the relaxation period decreases. the modulation index is kept less then one to avoid distortion. 13. P = 3 ´ 10 -8 cm .m ( 2 He)] ´ 931 MeV = [ 4 ´ 1.310 Xam idea Physics—XII 8.( -1)] W = 6 ´ 10 -8 J 10. A max = 10V A min = 2 V Modulation index = A max . E = 10 4 N/C At stable equilibrium ( q1 ) = 0° At unstable equilibrium ( q 2 ) = 180° Work done in rotating dipole is given by: W = PE (cos q1 . 12.007825 . 11. Uses of microwaves: (i) In long distance communications. m ( 1 1 H) . Energy released = Dm ´ 931 MeV 4 Dm = 4 m ( 1 1 H) .002603] ´ 931 MeV = 26. Displacement current and generalised Ampere’s I Circuital Law: Consider a parallel plate capacitor. equation (i) and (ii) take the forms Ø Ø Ø Ø ò C1 B × dl = m 0 I and ò C2 B × dl = m 0 I d = m 0 I .(ii) Ø Ø and B × dl = 0 òC 2 Since there cannot be any conduction current in region between the capacitor plates. So. (ii) When a diamagnetic substance is placed in an external magnetic field. where A is area of = e0 Ae 0 each plate. Maxwell postulated the existence of displacement current which is produced by time varying electric field. magnetic lines of force are repelled.. N S OR (i) Elements of earth’s magnetic field: (a) Angle of declination ( q) (b) Angle of dip ( d) ( c) Horizontal component of earth’s magnetic field ( B H ) (ii) At equator.(i) .. a feeble magnetism is induced in opposite direction. we must expect Ø Ø Ø Ø 2 ò C1 B × dl = ò C B ×dl . (i) Magnetic lines of force form continuous closed loops because a magnet is always a dipole and as a result. 15. being charged by a battery. dg(t) dE 1 dg(t) dE or = = e0 A dt Ae 0 dt dt dt This is expression for displacement current (ld).(iii) But this condition is violated by equations (i) and (ii)... A time varying current is flowing through the capacitor. the net magnetic flux of a magnet is always zero..Examination Papers 311 14. then time varying electric field E (t) = . we get I = Id Hence. then we apply Ampere’s Circuital Law to two closed loops C1 and C2. then we get Ø Ø C1 C2 I òCB × dl = m 1 0 I . If we consider only the conduction current I. Hence Ampere’s Circuital Law seems to be inconsistent in this case. As C1 and C2 are very close.. Applying Kirchhoff’s first law at power P. If s (t) is the surface charge density on capacitor s (t) q (t) plates and q(t) is the charge. Therefore. 5 × 0) A = . modified form of Ampere’s circuital law is Ø Ø dE ö æ I + e0 A ÷ ò B.5 × 0) I P Id ( ) (f = LI ) L = 4×0 H 17.5 × 0 A Time taken ( Dt) = 100 ´ 10 -3 S Induced emf ( e) = 200 V Induced emf ( e) is given by Df e=Dt D ( LI ) =Dt DI e = -L Dt or L = -e. (100 ´ 10 -3 ) Dt =DI ( . Violet. dl = m 0 ( I + I d ) = m 0 ç dt ø è But EA = Electric flux fE Ø Ø df E \ ò B. Let A ® area of each plate. Thus. (a) Resistance = 47 kW ± 10% = 47 ´ 10 3 W ± 10% Sequence of colour should be: Yellow. dl = m 0 ( I + m 0 e 0 dt ) 16.312 Xam idea Physics—XII The total current is the sum of the conduction current and displacement current. Î A Let initially C1 = C = 0 = C 2 d After inserting respective dielectric slabs: ¢ = KC C1 Î0 ( A / 2) K 2 Î0 ( A / 2) and C¢ + 2 = K1 d d Î0 A = ( K1 + K 2 ) 2d C C¢ 2 = ( K1 + K 2 ) 2 From (i) and (ii) ¢ = C¢ C1 2 C KC = ( K1 + K 2 ) 2 …(i) …(ii) . Orange and Silver (b) (i) Very low temperature coefficient of resistance. ( 200) . Change in current ( DI ) = ( 0 × 0 . (ii) High resistivity 18. . Now suppose a small conducting sphere of radius ‘ r ’ is introduced inside the spherical shell and placed at its centre. V (r ) . Thus the outer surface of metallic sphere S gains positive charge continuously and its potential rises to a very high value. The maximum charge that may be given to outer shell which may cause discharge in air. so produced. The charge spreads uniformly over whole surface of the shell. The pointed end of C 2 just touches the belt. it continues to take (+ ) charge upward.U ( R) = .Examination Papers 313 1 ( K1 + K 2 ) 2 19. The comb C 2 collects HTS positive charge from the belt which immediately moves to the outer surface of sphere S. This forms the principle of Van de Graaff generator. so that both the sphere and shell have same centre O. Working: When comb C1 is given very high potential. As r < R . which is collected by comb C 2 and transferred to outer surface of sphere S. The positive ions. These positive ions are carried upward by the moving belt. As the belt C1 P1 goes on revolving. As charge flows from higher to lower potentials S therefore. if we connect the small sphere and large C2 shell by a conducting wire.d x q q q é x -1 ù 1 dx = ê ú = 2 4pe 0 x 4pe 0 ê ú R 4pe 0 ë -1 û æ1 1 ö ç . the charge flows from P2 sphere to outer shell whatsoever the charge on outer shell may be. carrying charge Q. A B get sprayed on the belt due to the repulsion between positive ions and comb C1 . This device is Van de Graaff generator. Principle: Suppose we have a large spherical conducting shell of radius R. directed radially outward 4pe 0 x 2 K= The potential difference between the sphere and the shell V (r ) . due to action of sharp points. so electric field strength at a distance x from the centre O is q 1 E= .U ( R) is positive. then it produces ions in its vicinity. The electric field in the region inside the small sphere and large shell is due to charge + q only.÷ èr R ø This is independent of charge Q on the large spherical shell.ò =-ò r R r R ® ® r Q R O P x E r +q E . 314 Xam idea Physics—XII When the potential of a metallic sphere gains very high value, the dielectric strength of surrounding air breaks down and its charge begins to leak, to the surrounding air. The maximum potential is reached when the rate of leakage of charge becomes equal to the rate of charge transferred to the sphere. To prevent leakage of charge from the sphere, the generator is completely enclosed in an earthed connected steel tank which is filled with air under high pressure. Van de Graaff generator is used to accelerate stream of charged particles to very high velocities. Such a generator is installed at IIT Kanpur which accelerates charged particles upto 2 MeV energy. 20. l = 2000 Å = 2000 ´ 10 -10 m W0 = 4 × 2 eV h = 6 × 63 ´ 10 -34 hc = W0 + KE l hc or K.E. = - W0 l = ( 6 × 63 ´ 10 -34 ) ´ (3 ´ 108 ) ( 2000 ´ 10 -10 ´ 1 1× 6 ´ 10 -19 eV - 4 × 2 eV ) = ( 6 × 2 - 4 × 2) eV = 2 × 0 eV (i) The energy of the emitted electrons does not depend upon intensity of incident light, hence the energy remains unchanged. (ii) For this surface, electrons will not be emitted as the energy of incident light ( 6 × 2 eV) is less than the work function (6 × 5 eV) of the surface. VB 21. At the junction there is diffusion of charge carriers due – + to thermal agitation; so that some of electrons of + + + n-region diffuse to p-region while some of holes of p-region diffuse into n-region. Some charge carriers + + + combine with opposite charges to neutralise each + + + other. Thus near the junction there is an excess of Ei positively charged ions in n-region and an excess of p n Depletion negatively charged ions in p-region. This sets up a layer potential difference called potential barrier and hence an internal electric field Ei across the junctions. Barrier potential: During the formation of a p-n junction the electrons diffuse from n region to p-region and holes diffuse from p-region to n-region. This forms recombination of charge carriers. In this process immobile positive ions are collected at a junction toward n region and negative ions at a junction toward p-region. This causes a p.d. across the unbiased junction. This is called potential barrier or barrier potential. lD 22. Fringe width (b) = d b lD y= = 3 3d Examination Papers 315 yd lD d l Þ Dp = × = D 3d D 3 2p 2p p 2p Df = . Dp = × = l l 3 3 Intensity at point P = I 0 cos 2 Df Path diff ( Dp) = 2p ù é = I 0 êcos ú 3 û ë æ1ö = I0 ç ÷ è2ø I0 = 4 2 2 OR (i) In diffraction due to a single slit the path difference is given by: where, a is the width of the slit Dx = a sin q l For maxima: Dx = ( 2x + 1) 2 l Dx = a sin q = ( 2x + 1) 2 3.l For n = 2, Dx = 2 Let us divide the slit into three equal parts. If we take first two parts of slit, the path difference between rays diffracted from the extreme ends of first two parts 2 2 3l a sin q = a ´ =l 3 3 2a l Then the first two parts will have a path difference of and cancel the effect of each other. 2 The remaining third part will contribute to the intensity at a point between two minima. This is called first secondary maxima. In similar manner we can show that the intensity of the other secondary maxima will go on decreasing. (ii) The general maxima has between first minima on either side of the central maxima. We know for first minima. a sin q = l Þ aq = l Q for small angle sin q ~ -q y1 tan q = D y1 Þ q= D l Þ D = y1 = y 2 a lD Hence, whole width on secondary maxima on one side is . d 316 Xam idea Physics—XII 2 lD . a So, angular width of the central maxima is twice that of the first order secondary maximum. The angular width of the central maxima = –Ist Minima y1 q D y2 Central max Ist Minima 23. For first lens, u1 = - 30 cm, f 1 = + 10 cm 1 1 1 \ From lens formula, = f 1 v1 u1 1 1 1 1 1 3 -1 Þ = + = = v1 f 1 u1 10 30 30 Þ v1 = 15 cm This means that the image formed by first lens is at a distance of 15 cm to the right of first lens. This image serves as a virtual object for second lens. For second lens, f 2 = - 10 cm, u 2 = 15 - 5 = + 10 cm 1 1 1 1 1 \ = + =- + Þ v2 = ¥ v2 f 2 u 2 10 10 This means that the real image is formed by second lens at infinite distance. This acts as an object for third lens. For third lens, f 3 = + 30 cm, u 3 = ¥ 1 1 1 1 1 From lens formulae, = + = + v 2 f 3 u 3 30 ¥ Þ v 3 = 30 cm i.e., final image is formed at a distance 30 cm to the right of third lens. The ray diagram of formation of image is shown in figure. f1= 10 cm I2 at ¥ O I1 I3 30 cm 5 cm 10 cm Examination Papers 317 24. V0 Cut off region Active region Saturation region Vi In active region the transistor is used as an amplifier. (i) Input Resistance: It is the ratio of change in emitter base voltage ( DVEB ) to the corresponding change in emitter current ( DI E ) at constant collector-base voltage (VCB ) i. e. æ DVEB ö ÷ Input resistance ri = ç ç DI ÷ E øVCB = constant è Physically input resistance is the hindrance offered to the signal current. The input resistance is very small, of the order of a few ohms, because a small change in VEB causes a large change in I E . (ii) Output Resistance: It is the ratio of change in collector-base voltage to the corresponding change in collector current at constant emitter current I E æ DVCB ö ÷ i.e., r0 = ç ç DI ÷ C ø I E = constant è The output resistance is very high, of the order of several-tens kilo ohm because a large change in collector-base voltage causes a very small change in collector current. (iii) Current amplification factors of a transistor (a and b): The current gain a is defined as the ratio of change in collector current to the change in emitter current for constant value of collector voltage in common base configuration, i.e., æ DI C ö ÷ …(i) a =ç ç DI ÷ E øVC = constant è Practical value of a ranges from 0.9 to 0.99 for junction transistor. The current gain b is defined as the ratio of change in collector current to the change in base current for constant value of collector voltage in common emitter configuration i. e., æ DI C ö ÷ …(ii) b =ç ç DI ÷ B øVC = constant è The value of b ranges from 20 to 200. The current gains a and b are related as b a or b = a= 1+b 1-a 25. (i) At q = 90°, the intensity of emergent light is zero. I (ii) Intensity of light coming out polariser P1 = 0 2 …(iii) 318 Xam idea Physics—XII æI ö Intensity of light coming out from P3 = ç 0 ÷ cos 2 b è 2 ø I Intensity of light coming out from P2 = 0 cos 2 b cos 2 ( 90 - b) 2 I0 2 2 = × cos b. sin b 2 I é( 2 cos b . sin b) 2 ù = 0 ê ú 2 ê 2 ú ë û I0 I= (sin 2 b) 2 8 But it is given that intensity transmitted from P2 is I I= 0 8 I0 I0 So, = (sin 2 b) 2 8 8 or, (sin 2 b) 2 = 1 p p sin 2 b = sin Þ b= 2 4 26. Suppose m be the mass of an electron and v be its speed in nth orbit of radius r. The centripetal force for revolution is produced by electrostatic attraction between electron and nucleus. mv 2 1 ( Z e) ( e) … (i) = r 4p e 0 r2 or, So, 1 Z e2 4p e 0 r 1 Kinetic energy [ K ] = mv 2 2 mv 2 = 1 Z e2 4p e 0 2 r 1 ( Z e) ( -e) Potential energy = 4p e 0 r K= =Total energy, E = KE + PE = 1 Z e2 æ 1 Z e2 + çç 4p e 0 r 4p e 0 2r è 1 Z e2 4p e 0 2r ö ÷ ÷ ø 1 Ze 2 4p e 0 r E =- Examination Papers 319 For nth orbit, E can be written as E n 1 Z e2 so, En = 4p e 0 2 rn Again from Bohr's postulate for quantization of angular momentum. nh mvr = 2p nh v= 2p mr Substituting this value of v in equation (i), we get m é nh ù 1 Ze 2 = ú r ê 4p e 0 r 2 ë 2pmr û or, or, r= rn = e0 h2 n2 p m Ze 2 e0 h2 n2 p m Ze 2 Ze 2 æ e h2n2 2ç 0 ç p mZ e 2 è , ö ÷ ÷ ø m Z 2 e4 8 e0 h2 n2 me 4 3 8 e2 0 ch 2 … (ii) Substituting value of r n in equation (ii), we get En = 1 4p e 0 =- or, En = - Z 2 Rhc n2 where R = R is called Rydberg constant. For hydrogen atom Z = 1, -Rch En = n2 B1 + – K Rh + A e + – C R () K1 HR P2 P1 J G – B If ni and n f are the quantum numbers of initial and final states and Ei & E f are energies of electron in H-atom in initial and final state, we have 320 Xam idea Physics—XII Ei = -Rhc ni2 and Ef = -Rhc n f2 If n is the frequency of emitted radiation. we get Ei - E f n= h - Rc ö -Rc æ ÷ n= -ç ç 2 2 ÷ ni è nf ø é1 1 ù ú n = Rc ê ê n f2 ni2 ú ë û 27. Principle of potentiometer: If constant current is flowing through a wire of uniform area of cross-section at constant temperature, the potential drop across- any portion of wire is directly proportional to the length of that portion V µl Determination of internal resistance of potentiometer. (i) Initially key K is closed and a potential difference is applied across the wire AB. Now rheostat ( Rh) is so adjusted that on touching the jockey J at ends A and B of potentiometer wire, the deflection in the galvanometer is on both sides. Suppose that in this position the potential gradient on the wire is k. (ii) Now key K1 is kept open and the position of null deflection is obtained by sliding and pressing the jockey on the wire. Let this position be P1 and AP1 = l1 . In this situation the cell is in open circuit, therefore the terminal potential difference will be equal to the emf of cell, i.e., emf e = kl1 ...(i) (iii) Now a suitable resistance R is taken in the resistance box and key K1 is closed. Again, the position of null point is obtained on the wire by using jockey J. Let this position on wire be P2 and AP2 = l2 . In this situation the cell is in closed circuit, therefore the terminal potential difference (V ) of cell will be equal to the potential difference across external resistance R, i.e., ...(ii) V = kl2 e l1 Dividing (i) by (ii), we get = V l2 æ l1 ö æ e ö \ Internal resistance of cell, r = ç -1 ÷ R = ç -1 ÷ ç ÷R V l è ø è 2 ø From this formula r may be calculated. (a) Torque ( t) on the loop is given by: ® ® ® 28. t = NI A ´ B which can be written as, Examination Papers ® ® ® ® 321 t =M ´ B where, M is the magnetic dipole moment given by ® ® N F2 Q F1 I F3 P S F4 F1=IlB I (Upward) lo o p M =N I A (b) Torque on a current carrying loop: Consider a rectangular loop PQRS of length l, breadth b suspended in a uniform magnetic field B . The length of loop = PQ = RS = l and breadth = QR = SP = b. Let at any instant the normal to the plane of loop make an angle q with the direction of magnetic field B and I be the current in the loop. We know that a force acts on a current carrying wire placed in a magnetic field. Therefore, each side of the loop will experience a force. The net force and torque acting on the loop will be determined by the forces acting on all sides of the loop. Suppose that the forces on sides PQ, QR, RS and SP are F1 , F2 , F3 and F4 respectively. The sides QR and SP make angle ( 90° - q) with the direction of magnetic field. Therefore each of the forces F2 and F4 acting on these sides has same magnitude F¢ = Blb sin ( 90° - q) = Blb cos q. According ® ® ® ® N' b sin q F3=IlB ® ® ® ® ® ® Axis of loop or normal to loop B R q b q ax q is N of B I (Downward) to Fleming’s left hand rule the forces F2 and F4 are equal and opposite but their line of action is same. Therefore these forces cancel each other i.e. the resultant of F2 and F4 is zero. The sides PQ and RS of current loop are perpendicular to the magnetic field, therefore the magnitude of each of forces F1 and F3 is F = IlB sin 90° = IlB. According to Fleming’s left hand rule the forces F1 and F3 acting on sides PQ and RS are equal and opposite, but their lines of action are different; therefore the resultant force of F1 and F3 is zero, but they form a couple called the deflecting couple. When the normal to plane of loop makes an angle q with the direction of magnetic field B , the perpendicular distance between F1 and F3 is b sin q. ® ® ® ® ® ® ® ® it crosses the gap twice. The phenomenon is continued till S R.95 kW. then qBR v max = m The kinetic energy of the ion when it leaves the apparatus is. therefore if it completes n-revolutions before emerging the does.322 Xam idea Physics—XII \ Moment of couple or Torque. Due to Dee-1 the presence of perpendicular magnetic field the ion will move in a circular path. the kinetic energy gained K. t = (Magnitude of one force F ) ´ perpendicular distance = ( BIl ) × ( b sin q) = I (lb) B sin q But lb = area of loop = A (say) \ Torque.e.50 = 29 50 W = 2. Expression for K. then t = NI AB sin q ® ® ® In vector form t = NI A ´ B .. oscillator the ion reaches at the periphery where an auxiliary negative electrode Dee-2 (deflecting plate) deflects the accelerated ion on the target to be Beam bombarded.E. q2B 2R 2 1 2 mv max = 2 2m When charged particle crosses the gap between dees it gains KE = q V In one revolution. attained: If R be the radius of the path and v max the velocity of the ion when it leaves the periphery. E.F.50 Ig 5 ´ 10 -3 = 3000 . (c) G = 50 W I g = 5 m A = 5 ´ 10 -3 A V = 15 V The galvanometer can be converted into a voltmeter when a high resistance R is connected in series with it. = . t = IAB sin q If the loop contains N-turns. along I A ´ B . Direction of torque is perpendicular to direction of area of loop as well as the direction of ® ® magnetic field i. (a) OR Principle: The positive ions produced from a source are accelerated. Value of R is given by: V 15 R= -G = . m N N æ m N I ö F2 = N 2 f2 = N 2 B1 A 2 = N 2 ç 0 1 1 ÷ A 2 = 0 1 2 A 2 I 1 l l è ø F m N N A By definition Mutual Inductance.Examination Papers 323 Thus = 2nqV q2B 2R 2 K. From (i) F M= 2 I1 If I 1 = 1 ampere. m N I ... when current flowing in primary coil is 1 ampere. a = 1 cm = 1 ´ 10 -2 m Force between two parallel wires per unit length is given by m I I F= 0 × 1 2 2p a 2´5 -7 = 2 ´ 10 ´ = 20 ´ 10 -5 N (Repulsive) -2 1 ´ 10 29. M = F2 i. M 21 = 2 = 0 1 2 2 I1 l If n1 is number of turns per unit length of outer solenoid and r 2 is radius of inner solenoid. 2 then M = m 0 n1 N 2 pr 2 . n = 1 × If I 1 is the l current flowing in primary solenoid.E.(ii) B1 = 0 1 1 l The flux linked with each turn of inner solenoid coil is f2 = B1 A 2 . (b) Due to current I 1 through solenoid of radius r1 .. flux linked with second solenoid N 2 f2 = M 21 I 1 …(i) . Mutual Inductance of Two Co-axial r1 N1 Solenoids: Consider two long co-axial solenoid each of length l with number of r2 N2 turns N1 and N 2 wound one over the other.e.(i) F2 µ I 1 or F2 = MI 1 where M is a constant. and is called the coefficient of mutual induction or mutual inductance. The current I 1 is flowing in primary coil C1 .. The total flux linkage with inner coil of N 2 -turns. I 2 = 5 A .. Number of turns per unit length in order N l (primary) solenoid. due to which an effective magnetic flux F2 is linked with secondary coil C 2 . = = 2nqV 2m (b) I 1 = 2 A . the magnetic field produced within this solenoid. where A 2 is the cross-sectional area of inner solenoid. the mutual inductance between two coils is numerically equal to the effective flux linkage with secondary coil. (a) Suppose there are two coils C1 and C 2 . By experiments . l Hence Using (1) as l >> r 2 and r 2 >> r1 m 0 N1 N 2 pr12 l which is same as expression of mutual inductance derived in part (a) above.cos ( 2wt + f)] 2 Average power for complete cycle .B) . P = Vi = Vm sin wt .324 Xam idea Physics—XII But flux due to current I 1 in first solenoid (using Ampere’s circuital law) will be = N æ ö N 2 f2 = N 2 ( pr12 ) ç m 0 1 I 1 ÷ l è ø M 21 = m 0 N1 I 1 .cos ( A + B) 1 \ Instantaneous power. N s = ? I s = ? Vs I p N s = = Vp I s N p N 220 5 = = s 2200 I s 4000 22 p 5 = 220 p I s 1 5 = p Is I s = 50 A N 5 = s I s 4000 N 5 = s 50 4000 N s = 400 OR (a) V = Vm sin wt i = i m ( wt + f) and instantaneous power. I 0 = 5A. N p = 4000 Vs = 220 V . P = Vmi m [cos ( wt + f .cos ( wt + f + wt)] 2 1 … (i) = Vm i m [cos f . \ M 21 = M 12 (c) V p = 2200V . sin ( wt + f) 2 From trigonometric formula 2 sin A sin B = cos ( A . i 0 sin ( wt + f) = Vm i m sin wt sin ( wt + f) 1 = Vmi m 2 sin wt .wt) . But for a complete cycle. Q = ? L = 1 × 00 mH = 1 ´ 10 -3 H. ® 1 P = Vmi m cos f 2 V0 i 0 = cos f 2 2 ® P = Vrms i rms cos f (b) The power is P = Vrms I rms cos f . Q = 10 (i) When a wave starting from one homogeneous medium enters the another homogeneous medium. This phenomenon is called refraction. Let v1 and v 2 be the speeds of waves in these media. Hence. we know is I 2 R. cos ( 2wt + f) = 0 P= \ Average power. E 0 = 100 V I0 = E0 2 = \ 1 ö æ R 2 + ç wL ÷ wC ø è V 100 I= = R 100 I 0 =1 A I 1 2 1. (c) I v = ? . C = 1. then current considerably increases when voltage is constant.cos ( 2wt + f) ] 2 where cos ( 2wt + f) is the mean value of cos ( 2wt + f) over complete cycle. If cos f is small.Examination Papers ® 325 1 Vm i m [cos f . In transversing from first medium to another medium. the frequency of wave remains unchanged but its speed and the wavelength both are changed. B i X A A' 90o i r 90 o B' r Y . Let XY be a surface separating the two media ‘1’ and ‘2’.00 nF = 1 ´ 10 -9 F R = 100 W .44 Iv = 0 = = = = 0 × 707 A 2 2 2 2 I v = 0 × 707 A Q= 1 R L 1 1 × 0 ´ 10 -3 1 = = ´ 10 3 = 10 9 C 100 1 × 0 ´ 10 100 E0 Z 1 ü ì ï at resonance wL = ï í wC ý ï ï î Hence Z = R þ 30. Power loss. it is deviated from its path. power loss increases. Ð AA¢ B ¢ = 90° AA¢ v 2 t . According to Huygen’s principle A¢ B¢ is the new position of wavefront AB in the second medium.. which travel with speed v1 in the first medium and speed v 2 in the second medium.. secondary spherical wavelets originate from these points.(ii) \ sin r = sin Ð AB¢ A¢ = = AB¢ AB¢ Dividing equation (i) by (ii). and is called the Snell’s law. Hence A¢ B¢ will be the refracted wavefront.(iii) = 1 = constant sin r v2 As the rays are always normal to the wavefront. In right-angled triangle AB¢ B. A¢ B ¢ and surface XY are in the plane of paper. Ð ABB¢ = 90° vt BB¢ . therefore the perpendicular drawn on them will be in the same plane. which traverses a distance AA¢ ( = v 2 t) in second medium in time t. Assuming A as centre. Clearly BB¢ = v1t. This is the first law of refraction. i and r are the angle of incidence and angle of refraction respectively. we get v sin i . the secondary wavelets start from points between A and B¢ .e. (ii) Proof of Snell’s law of Refraction using Huygen’s wave theory: When a wave starting from one homogeneous First law: As AB. therefore we may say that the incident ray. we draw a spherical arc of radius AA¢ ( = v 2 t) and draw tangent B¢ A¢ on this arc from B¢ . it strikes the points between A and B¢ of boundary surface. According to Huygen’s principle. As the incident wavefront AB advances. In the same time-interval t. (iii) No. from. where the secondary wavelet now starts. Second law: Let the incident wavefront AB and refracted wavefront A¢ B¢ make angles i and r respectively with refracting surface XY. the point of wavefront traverses a distance BB¢ ( = v1t) in first medium and reaches B¢ ..326 Xam idea Physics—XII Suppose a plane wavefront AB in first medium is incident obliquely on the boundary surface XY and its end A touches the surface at A at time t = 0 while the other end B reaches the surface at point B¢ after time-interval t. .. refracted ray and the normal at the point of incidence all lie in the same plane. This is the second law of refraction. According to equation (3): The ratio of sine of angle of incidence and the sine of angle of refraction is a constant and is equal to the ratio of velocities of waves in the two media..(i) \ sin i = sin Ð BAB¢ = = 1 AB¢ AB¢ Similarly in right-angled triangle AA¢ B ¢ . As the wavefront AB advances. Because energy of wave depends on its frequency and not on its speed.. As the lines drawn normal to wavefront denote the rays. therefore the incident and refracted rays make angles i and r with the normal drawn on the surface XY i. Clearly BB¢ = v1t and AA¢ = v 2 t. First of all secondary wavelet starts from A. one after the other and will touch A¢ B¢ simultaneously. .(vii) = sin r1 sin r 2 For minimum deviation i1 and i 2 become coincident.. The incident ray EF and emergent ray GH when produced meet at O.i1 = i 2 = i (say) So from (vii) r1 = r 2 = r (say) Hence from (iv) and (vi)..(iv) r1 + r 2 = A Substituting this value in (i).Examination Papers 327 OR (i) Let PQR be the principal section of the prism. Ð OFG = i1 . i. we get ..r1 ) + (i 2 . we get r + r = A or r = A / 2 A + dm and i + i = A + d m or i = 2 æ A + dm ö sin ç ÷ 2 è ø sin i Hence from Snell’s law. The refracted ray FG N H becomes incident on face PR and is refracted E Q R away from the normal GN 2 and emerges in the direction GH... The angle of q 2 G refraction for this face is r1 .(ii) r1 + r 2 + q = 180° In quadrilateral PFNG.. n= = sin r æ A ö sin ç ÷ è 2 ø . Let Ð FNG = q In D FGN. The refracting angle of the prism is A. This N1 N2 O d (i –r ) (i1–r1) ray enters from rarer to denser medium and so is 2 2 deviated towards the normal FN and gets i1 i2 r1 r F refracted along the direction FG. when produced meet at N. Ð PFN = 90° . The refractive A index of material of prism for this ray is n. .r 2 ) .(r1 + r 2 ) The normals FN1 and GN 2 on faces PQ and PR respectively.r1 and Ð OGF = i 2 . Ð PGN = 90° . P A ray of monochromatic light EF is incident on face PQ at angle of incidence i1 .(v) d = i1 + i 2 .. d is exterior angle \ d = Ð OFG + Ð OGF = (i1 ...(vi) i1 + i 2 = A + d sin i1 sin i 2 From Snell’s law n = .(i) = (i1 + i 2 ) ...(iii) \ A + 90° + q + 90° = 360° or A + q = 180° Comparing (ii) and (iii). .e.. The angle between these two rays is called angle of deviation ‘d‘.A or ... The angle of incidence on this face is r 2 (into prism) and angle of refraction (into air) is i 2 .r 2 In D FOG . i = 45° Refractive index corresponding to critical angle 45° is 1 m= = 2 = 1× 414 sin 45° The ray will be transmitted through face ‘ ac’ if i < i c . Blue and Green rays will be totally reflected. The function of transmitter is to convert the message signal produced by the source of information into a form suitable for transmission through the channel. CBSE (Foreign) SET–II 1.328 Xam idea Physics—XII (ii) a a B Blue Green Red 90° b 45° c b B G 45° c R G R Angle of incidence at face ac for all three colours. The charge on both spheres would be = . qA + qB . So only red ray will be transmitted. 2 4. This condition is satisfied for red colour (m = 1× 39). Photo current i0 = 12 2 = 12 2 ´ 2 2 =6 2A 2 =0A n03 n 02 n01 n03 >n02 >n01 Collector potential V03 V02 V01 3. Peak value (i 0 ) = 12 A (i) i rms = (ii) i av 2. Logic circuit of AND gate: Truth Table Inputs A Y B Output B 0 1 0 1 Y 0 0 0 1 A 0 0 1 1 10.2 10 = = = 0 × 714 12 + 2 4 To minimize distortion of signal due to noise signal from atmosphere and electrical disturbances modulation index is kept less than one. 7. e A 13.cos q1 ) = . A max = 12 V A min = 2 V Modulation index = A max . Dipole moment P = 5 ´ 10 -4 Cm. (i) Doctors use infrared lamps to treat skin diseases and releave the pain of sore muscles (ii) In electronic devices for example semiconductor light emitting diodes.A min A max + A min 12 .cos 0) = 10 ´ 10 -4 J 12. Electric field ( E) = 10 4 N/C At stable equilibrium q1 = 0° At unstable equilibrium q 2 = 180° Work done = .(5 ´ 10 -8 ) (10 4 ) (cos 180 .Examination Papers 329 5. C1 = 0 d 1 1 1 d d = + = + e A e A 2 × K1 e 0 A 2 × K 2 e 0 A C2 K1 0 K2 0 d/2 d/2 2 × e 0 A é K1 K 2 ù 1 d é1 1 ù Þ = + C2 = ê ú ê ú C 2 2 e 0 A ë K1 K 2 û d ë K1 + K 2 û é K K ù C 2 = 2C1 ê 1 2 ú ë K1 + K 2 û é 2 K1 K 2 ù C 2 = C1 ê ú ë K1 + K 2 û .PE (cos q 2 . 3 × 5ú eV 2500 ´ 10 -10 1 × 6 ´ 10 -19 ê ú ë û = ( 4 × 97 .330 Xam idea Physics—XII 15.3 × 5) eV = 1 × 47 eV (i) KE of fastest electron = 1 × 47 eV (ii) KE of slowest electron = 0 eV If the same light (having energy 4.5 eV).3q A = . Change on each = q A . Photo current I3 I2 I1 I3 >I2 >I1 V0 Collector plate potential 3.qA 2 . Wavelength of incident radiation ( l) = 2500 Å Work function (W0 ) = 3 × 5 eV hc = W0 + KE max l hc KE max = . 97 eV) falls on the surface (of work function 5. CBSE (Foreign) SET–III 2.2 × 0 A Time taken ( D t) = 10 ms = 10 ´ 10 -3 s From law of induction Induced emf is given by: Df D( LI ) e==Dt Dt DI e = -L Dt L =e Dt ( 200) (10 ´ 10 -3 ) =DI ( -2) L =1 × 0 H 27.2 × 0) A = . Change in current ( DI ) = ( 0 × 0 . then no photoelectrons will emit.W0 l é( 6 × 63 ´ 10 -34 ) (3 ´ 108 ) ù 1 =ê ´ . ( -1)) = 20 ´ 10 -4 J. Logic circuit of NOT gate A Input Output Truth Table Input A 0 1 Output Y 1 0 7.cos 180] = (10 ´ 10 -4 ) [1 . (i) In the detection of fracture.D ( LI ) = Dt . 9. Peak value of current (i 0 ) = 15 A i 15 15 (i) i rms = 0 = = ´ 2 =7 ×5 2 A 2 2 2 2 (ii) i av = 0 6. amplifies and reproduces the original message signal from it. deformity of the bones/skeletal system.Examination Papers 331 4.3 × 0 A 12. (ii) In study of crystal structure. q 2 = 180° Work done = PE (cos q1 . Charge in current ( D I ) = ( 0 × 0 . The function of receiver is to receive the modulated wave and it demodulates. C1 = Time taken ( D t) = 300 ms = 300 ´ 10 -3 s Df Induced emf ( e) = Dt .cos q 2 ) = ( 2 ´ 10 -8 ) (5 ´ 10 4 ) [cos 0 . Dipole moment ( P) = 2 ´ 10 -8 Cm Electronic field ( E) = 5 ´ 10 4 N/C q1 = 0° . e0 A d K1 e 0 ( A / 2) K 2 e 0 ( A / 2) C2 = + d d e0 A = [ K1 + K 2 ] 2d [K + K 2 ] C 2 = C1 1 z 14. 8.3 × 0) A = . E.W0 l é( 6 × 63 ´ 10 -34 ) (3 ´ 108 ) ù 1 =ê ´ .E.[ 200] [300 ´ 10 -3 ] [ -3 × 0 ] L = 20. l = 2400 Å = 2400 ´ 10 -10 m. 18 eV) falls on the surface of work function 5.3 × 6) eV = 1 × 58 eV (i) K. 0 H 15. 26. A min = 3 V A . of fastest electron = 1 × 58 eV (ii) K.E.A min Modulation index ( m) = max A max + A min 15 . of slowest electron = 0 eV If the same light (having energy 5. .332 Xam idea Physics—XII DI DI Dt L = -e. A max = 15 V.3 × 6ú eV -10 -19 2400 ´ 10 1 × 6 ´ 10 ê ú ë û = (5 × 18 . 5 eV. DI e=-L = . = .3 12 = = = 0 × 67 15 + 3 18 The modulation index is kept low to reduce distortion. then no photoelectron will emit. work function (W0 ) = 3 × 6 eV hc K. 6. 4. Predict the directions of induced currents in metal rings 1 and 2 lying in the same plane where current I in the wire is increasing steadily.42 A. Find the angle of incidence for this ray. Show on a graph. Name the physical quantity which remains same for microwaves of wavelength 1 mm and UV radiations of 1600 Å in vacuum.CBSE EXAMINATION PAPERS DELHI–2012 Time allowed: 3 hours General Instructions: As given in CBSE Examination Papers Delhi–2011. . does it mean that all the free electrons of the metal are moving in the same direction? 2. Maximum marks: 70 CBSE (Delhi) Set–I 1. 9. When electrons drift in a metal from lower to higher potential. (ii) with resistance R1 only (iii) with R1 and R2 in series combination (iv) with R1 and R2 in parallel combination. Under what condition does a biconvex lens of glass having a certain refractive index act as a plane glass sheet when immersed in a liquid? 7. but not necessarily in that order. State de-Broglie hypothesis. A ray of light. OR Mention the function of any two of the following used in communication system: (i) Transducer (ii) Repeater (iii) Transmitter (iv) Bandpass Filter 11.05 A. 1. The currents measured in the four cases are 0. What is the value of vertical component of earth’s magnetic field at equator? 3. Why should electrostatic field be zero inside a conductor? 5. 10.4 A and 4. 1 I 2 8. The current in the circuit is measured in four different situations: (i) without any external resistance in the circuit. incident on an equilateral glass prism (m g = 3 ) moves parallel to the base line of the prism inside it. The horizontal component of the earth’s magnetic field at a place is B and angle of dip is 60°.2 A. 1. Distinguish between ‘Analog and Digital signals’. the variation of resistivity with temperature for a typical semiconductor. A cell of emf E and internal resistance r is connected to two external resistances R1 and R2 and a perfect ammeter. Identify the currents corresponding to the four cases mentioned above. (ii) The current at the resonating frequency. Q I P I 14. Define self-inductance of a coil. R = 40 W connected to a variable frequency 240V source. Show that magnetic energy required to build up the current I in a coil of 1 self inductance L is given by LI2. The figure shows a series LCR circuit with L = 5.0 H. However. A constant and uniform magnetic field B parallel to the axis is present everywhere. 2 17. based on the concept of displacement current? 15. R C L (i) The angular frequency of the source which drives the circuit at resonance. A metallic rod of ‘L’ length is rotated with angular frequency of ‘w’ with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius L.334 Xam idea Physics—XII 12. The susceptibility of a magnetic material is –2. Draw a plot showing the variation of (i) electric field (E) and (ii) electric potential (V) with distance r due to a point charge Q. Deduce the expression for the emf between the centre and the metallic ring. when an ac source is used.6 × 10–5. . Identify the type of magnetic material and state its two properties. C = 80 mF. 13. then. 19. about an axis passing through the centre and perpendicular to the plane of the ring. 16. How does one explain this. (iii) The rms potential drop across the capacitor at resonance. the current flows continuously. Calculate. Two identical circular wires P and Q each of radius R and carrying current ‘I’ are kept in perpendicular planes such that they have a common centre as shown in the figure. Find the magnitude and direction of the net magnetic field at the common centre of the two coils. no current flows. What is the reason. The current in the forward bias is known to be more (~mA) than the current in the reverse bias (~mA). to operate the photodiode in reverse bias? 18. When an ideal capacitor is charged by a dc battery. In the figure a long uniform potentiometer wire AB is having a constant potential gradient along its length. I = 5A 4 cm 2A 10 cm 2A (a) Using Bohr’s second postulate of quantization of orbital angular momentum show that the circumference of the electron in the nth orbital state in hydrogen atom is n times the de Broglie wavelength 1cm associated with it. What is the maximum number of spectral lines which can be emitted when it finally moves to the ground state? 22. The null points for the two primary cells of emfs e1 and e 2 connected in the manner shown are obtained at a distance of 120 cm and 300 cm from the end A. Also find the potential difference between A and D. How is the sensitivity of a potentiometer increased? 21.Examination Papers 335 20. If the loop and the wire are coplanar. 300 cm 120 cm A e1 e2 e1 e2 B OR Using Kirchoff’s rules determine the value of unknown resistance R in the circuit so that no current flows through 4 W resistance. A straight long wire carrying 5A current is kept near the loop as shown. find (i) the torque acting on the loop and (ii) the magnitude and direction of the force on the loop due to the current carrying wire. F 1W E 4W 1W 6V I A 9V B 3V C R D . A rectangular loop of wire of size 4 cm × 10 cm carries a steady current of 2A. Find (i) e1 / e 2 and (ii) position of null point for the cell e1 . (b) The electron in hydrogen atom is initially in the third excited state. Explain its working.I. Draw a diagram showing an amplitude modulated wave by superposing a modulating signal over a sinusoidal carrier wave. how this affect the size and intensity of the central diffraction band? Explain the principle of a device that can build up high voltages of the order of a few million volts. find the height of the final image when it is formed 25 cm away from the eye piece.336 Xam idea Physics—XII 23. the intensity of light at a point on the screen where path difference is l. If this telescope is used to view a 100 m high tower 3 km away. 24. Write Einstein’s photoelectric equation. OR (a) Define electric flux. (i) What characteristic property of nuclear force explains the constancy of binding energy per nucleon (BE/A) in the range of mass number ‘A’ lying 30 < A < 170? (ii) Show that the density of nucleus over a wide range of nuclei is constant independent of mass number A. (a) Why are coherent sources necessary to produce a sustained interference pattern? (b) In Young’s double slit experiment using monochromatic light of wavelength l. What is the significance of the negative sign in the expression for the voltage gain? . RL is the load ri resistance and ri is the input resistance of the transistor. units. A small telescope has an objective lens of focal length 150 cm and an eye piece of focal length 5 cm. State clearly how this equation is obtained using the photon picture of electromagnetic radiation. (ii) negatively charged? Define magnifying power of a telescope. of the amplifier is given by A V = . Use Huygen’s principle to explain the formation of diffraction pattern due to a single slit illuminated by a monochromatic source of light. 25. 27.where b ac is the current gain. Show that the voltage gain b R AV. When the width of slit is made double the original width. prove that the electric field at a point due to a uniformly charged infinite plane sheet is independent of the distance from it. Draw a simple circuit of a CE transistor amplifier.ac L . 29. Find out the intensity of light at a point where path difference is l / 3. is K units. (b) Using Gauss’s law. Is there any restriction on the upper limit of the high voltages set up in this machine? Explain. Find the position of the object relative to the objective in order to obtain an angular magnification of 30 in normal adjustment. 30.25 cm and 5 cm respectively. Write its S. Write the three salient features observed in photoelectric effect which can be explained using this equation. Write any two factor which justify the need for modulating a signal. Draw a schematic diagram and explain the working of this device. OR How is the working of a telescope different from that of a microscope? The focal lengths of the objective and eyepiece of a microscope are 1. (c) How is the field directed if (i) the sheet is positively charged. 26. 28. Write its expression. The susceptibility of a magnetic material is 2. I v 10. Explain its working and show the output. The conductor is carrying current I in the direction shown in the figure. The direction of current in both the loops is clockwise as seen from O which is equidistant from the both loops. Find the magnitude of the net magnetic field at point O. 14. Derive the expression for the self inductance of a long solenoid of cross sectional area A and length l. 2r P O Q I 2I . having n turns per unit length. Identify the type of magnetic material and state its two properties.Examination Papers 337 OR (a) Draw the circuit diagram of a full wave rectifier using p-n junction diode. Two identical circular loops. P and Q. Why must electrostatic field be normal to the surface at every point of a charged conductor? 6. (b) Show the output waveforms (Y) for the following inputs A and B of (i) OR gate (ii) NAND gate t1 t2 t3 t4 t5 t6 t7 t8 A B CBSE (Delhi) Set–II Questions uncommon to Set–I 1. 16. each of radius r and carrying currents I and 2I respectively are lying in parallel planes such that they have a common axis. Predict the direction of induced current in a metal ring when the ring is moved towards a straight conductor with constant speed v.6 × 10–5. input waveforms. A rectangular loop of wire of size 2 cm × 5 cm carries a steady current of 1 A. Predict the direction of induced current in metal rings 1 and 2 when current I in the wire is steadily decreasing? 1 I 2 . Explain. A straight long wire carrying 4 A current is kept near the loop as shown in the figure.0 H. (iii) the rms potential drop across the inductor at resonance. CBSE (Delhi) Set–III Questions uncommon to Set–I & II 6.338 Xam idea Physics—XII 20. (a) Why are coherent sources necessary to produce a sustained interference pattern? I = 4A (b) In Young’s double slit experiment using mono-chromatic light of wavelength l. using a proper diagram the mode of propagation used in the frequency range above 40 MHz. 2 cm 5 cm 1A 1cm 27. A series LCR circuit with L = 4. is K units. 21. the intensity of light at a point on the screen where path difference is l. 3 22. C = 100 mF and R = 60 W is connected to a variable frequency 240 V source as shown in R C L Calculate: (i) the angular frequency of the source which derives the circuit at resonance. find (i) the torque acting on the loop and (ii) the magnitude and direction of the force on the loop due to current carrying wire. If the loop and the wire are coplanar. Why is electrostatic potential constant throughout the volume of the conductor and has the same value (as inside) on its surface? 8. Find out the intensity of light at a 2l point where path difference is . (ii) the current at the resonating frequency. Name the three different modes of propagation of electromagnetic waves. Identify the nature of magnetic material and state its two properties. (iii) the rms potential drop across the inductor at resonance. each of radius r and carrying equal currents are kept in the parallel planes having a common axis passing through O. Define mutual inductance between two long coaxial solenoids. The relative magnetic permeability of a magnetic material is 800. find the (i) torque acting on the loop and (ii) the magnitude and direction of the force on the loop due to the current carrying wire.5 cm × 4 cm carries a steady current of 1 A. 23. calculate (i) the angular frequency of the source which drives the circuit at resonance. The figure shows a series LCR circuit with L = 10. R C L . Explain. C = 40 mF.Examination Papers 339 9. (ii) the current at the resonating frequency. 2r P O Q I I 21. using a proper diagram the mode of propagation used in the frequency range from a few MHz to 40 MHz. Find out the expression for the mutual inductance of inner solenoid of length l having the radius r1 and the number of turns n1 per unit length due to the second outer solenoid of same length and n 2 number of turns per unit length. 16. Two identical circular loops. R = 60 W connected to a variable frequency 240 V source. 2A 2. P and Q. Find the magnitude of the net magnetic field at O. Name the three different modes of propagation of electromagnetic waves. If the loop and the wire are coplanar. The direction of current in P is clockwise and in Q is anti-clockwise as seen from O which is equidistant from the loops P and Q. A straight wire carrying 2 A current is kept near the loop as shown.0 H.5 cm 1A 2cm 4 cm 25. A rectangular loop of wire of size 2. 12. . Zero 3.340 Xam idea Physics—XII Solutions CBSE (Delhi) Set–I 1. can display the wave like properties. r t 4. 1 Clockwise 2 Anti clockwise 8. 2." h i. 60° i r 30° 60° From the figure. so electric field is zero. 6. Velocity (c = 3×108 m/s) This is because both are electromagnetic waves. (mathematically) l= mv 9. No. 5. The charge inside the conductor is zero. According to hypothesis of de Broglie "The atomic particles of matter moving with a given velocity. When m L = m g where m L = Refractive index of liquid and mg = Refractive index of glass 7. we see r = 30° .e. i = 1. the effective resistance is more than (i) but less than (ii) and (iii). (iii) Transmitter: A device which processes the incoming message signal so as to make it suitable for transmission through a channel and for its subsequent reception. effective resistance is more than (i) and (iv) but less than (iii) So. e 11. .2 A e (ii) i = R1 + r Here. effective resistance of circuit is minimum so current is maximum. Analog signals: They are the continuous variations of voltage or current. (ii) Repeater: It is a combination of a receiver and a transmitter. i = 1.Examination Papers 341 We know Þ Þ n 21 = sin i sin r Þ 1 2 3= sin i sin 30° sin i = 3 sin 30° = 3 ´ Þ i = 60° 10. (iv) Bandpass filter: A bandpass filter blocks lower and higher frequencies and allows only a band of frequencies to pass through. Two properties: (i) This material expels the magnetic field lines. So. Hence.05 A e (iii) i = r + R1 + R 2 In this situation effective resistance is maximum so current is minimum.42 A e (iv) i = R R r+ 1 2 R1 + R 2 In this situation. i = 4. So.4 A 12. Digital signals: They are the signals which can take only discrete values OR (i) Transducer: A device which converts energy from one form to another form. (i) i = r where e = emf r = Internal resistance In this situation. (ii) They have the tendency to move from stronger to weaker part of the external magnetic field. The magnetic material having negative susceptibility is diamagnetic in nature. i = 0. displacement current df and id = ic at all instants. This causes change in electric field between the plates of the capacitor which causes the electric flux to change and gives rise to a displacement current in the region between the plates of the capacitor. When an ideal capacitor is charged by dc battery.0 e or V 1. charge flows (momentarily) till the capacitor gets fully charged.5 1.5 2.5 r V E 2. 14. id = e 0 E dt 1 1 15. Here e µ and V µ 2 r r Y 3.5 0.5 3. Due to changing current. dq When an ac source is connected then conduction current i c = keep on flowing in the dt connecting wire. Q I BP B BQ q = 45° I P Bp ® directed vertically upward BQ ® horizontally directed \ We have 2 B = B2 p + BQ 2R The net magnetic field is directed at angle of 45° with either of the fields.0 2.0 X V E B P = BQ = m 0I Þ 2R B = 2 Bp = 2 m 0I 2R Þ B= m 0I .342 Xam idea Physics—XII 13. charge deposited on the plates of the capacitor changes with time. As we know.0 0.5 1.0 1.0 2. The induced emf = B dt d × × × × × e = ( BA) \ fB = BA cos f Q dt × × ×R × × dA Q f = 0° =B w dt q = wt fB = BA × × × P × dA O where = Rate of change of area of loop formed by the dt × × × × × sector OPQ. the opposing di induced emf is given by e = -L dt If the source of emf sends a current i through the inductor for a small time dt.e. p' = p + Dp But Dn = Dp and n >> p Dp Dn Hence. the total amount of work done by source of emf when the current increases from its initial values (i = 0) to its final value (I) is given by éi 2 ù W = ò Li di = L ò i di = L ê ú ê ë2ú û0 0 0 1 2 W = LI 2 This work done gets stored in the inductor in the form of energy. As the current starts growing.Examination Papers 343 16. 1 \ U = LI2 2 17. n >> p. Hence photodiodes are used in reverse bias condition for measuring light intensity. × × × × × I I I . Energy stored in an inductor: Consider a source of emf connected to an inductor L. of both types of carriers would equally increase in number as n' = n + Dn. i. df 18. On illumination.e. the no. Self inductance of a coil is numerically equal to the magnetic flux linked with the coil when the current through coil is 1A.. then the amount of work done by the source. << (fractional change in minority n p carrier) Fractional change due to photo-effects on minority carrier dominated reverse bias current is more easily measurable than the fractional change in majority carrier dominated forward bias current. Consider the case of n-type semiconductor. The majority carrier (electron) density is larger than the minority hole density.. is given by di dw = e i dt = L i dt = Li di dt Hence. Let q be the angle between the rod and the radius of the circle at P at time t. i. the fractional change in majority carrier. 344 Xam idea Physics—XII The area of the sector OPQ = pR 2 ´ where Hence 19. t = 0 (ii) Force acting on the loop mI I æ 1 1 ö F = 1 2 lç . e = B´ q 1 = R2q 2p 2 2 d æ1 2 ö 1 2 dq B w R = ç R q ÷ = BR dt è 2 dt 2 ø 2 20. (i) We know wr = Angular frequency at resonance 1 1 = 50 rad/s wr = = LC 5 ´ 80 ´ 10 -6 (ii) Current at resonance V 240 =6A I rms = rms = R 40 (iii) Vrms across capacitor Vrms = I rms X C 1 6 ´ 10 6 = 1500 V = 6´ = 50 ´ 80 ´ 10 -6 4 ´ 10 3 (i) Torque 't' = MB sin q where q = 0° Therefore. -2 5 5 è ø 10 Direction: Towards the conductor or attractive (i) According to Bohr's second postulate h mvr n = n 2p nh 2pr n = mv h h But By de Broglie hypothesis = =l mv p \ 2pr n = nl (ii) For third excited state n = 4 For ground state n = 1 Hence possible transitions are ni = 4 to nf = 3. R = Radius of the circle. 20 ´ 10 -8 æ 1 ö -6 4 –5 ç1 . 1 ni = 3 to nf = 2. 2.÷ N = 20 ´ 10 ´ N = 1.6 × 10 N. 1 .÷ ÷ 2p ç è r1 r 2 ø [Q As M and B are parallel] æ 1 1 = 2 × 10–7 × 2 × 5 × 10–1 ç ç -2 5 ´ 10 -2 è 10 = ö ÷ ÷ ø 21. .. Let m = mass of nucleon .. (i) Let k = potential gradient in V/cm e1 + e 2 = 300k e1 – e 2 = 120k e 7 We can solve.(ii) . –9 + 6 + 4 × 0 + 2I = 0 I = 1.(i) 23.Examination Papers 345 ni = 2 to nf = 1 Total no. (iii) By decreasing potential gradient OR Applying Kirchhoff's loop rule for loop ABEFA...(i) .5 A Þ For loop BCDEB 3 + IR + 4 × 0 – 6 = 0 IR = 3 \ Putting the value of I from (i) we have 3 ´ R =3 Þ R = 2W 2 Potential difference between A and D through path ABCD 9 – 3 – IR = VAD 3 or 9 – 3 – ´ 2 = VAD Þ VAD = 3V 2 (i) Saturation or short range nature of nuclear forces. of transitions = 6 n=4 n=3 n=2 n=1 22.. balancing length for cell e1 is 210 cm. 1 = e2 3 (ii) From equation (i) e1 + e 2 = 300k 3 \ e1 + e1 = 300k Þ e1 = 210k 7 Therefore. (ii) We have R = R0 A 1 3 . ) \ hn = f0 + K max Three salient features are: (i) Cut-off potential of the emitted electrons is proportional to n.5 (amplitude modulated wave) 25. (b) We know I = 4 I 0 cos 2 f 2 for path difference l. f (ii) Photo electric emission of electrons is possible only when v > v0 = 0 h (iii) Max. (ii) impart maximum K.346 Xam idea Physics—XII mA mA m = = 3 4 4 1 3 3 4 æ p R0 A p R0 pç R 0 A 3 ö ÷ 3 3 ø 3 è Hence r is independent of A. K. to the emitted electron (Kmax.5 0. 26.E.5 Em Ec 1 Envelope 1. Einstein photo electric equation is hn = f0 + K max .5 1 1. Two factors justifying the need for modulation: (i) Practical size of antenna (ii) To avoid mixing up of signals from different transmitters. Phase difference f= 3 3 f p Intensity of light I ' = 4I 0 cos 2 = 4I 0 cos 2 = I 0 2 3 K Þ I¢ = 4 . 24. The energy (hn) carried by a photon of frequency n is absorbed by electrons on the surface to (i) overcome the work function of metal f0 . \ Density (r) = 0. phase difference f = 2p Intensity of light = K Hence. is independent of intensity of incident radiations.E. (a) This is because coherent sources are needed to ensure that the positions maxima and minima do not change with time. K = 4I 0 cos 2 p = 4I 0 l 2p For path difference . then the size of central maxima will be reduced to half and intensity will be four times. Now suppose a small conducting sphere of radius ‘ r ’ is introduced inside the spherical shell and placed at its centre. i. S1 The point O is maxima because contribution from each half of the slit S1S2 is in phase. C1 and C 2 are two sharp metallic spikes in the form of combs. the path a S O difference is zero. Principle: Suppose we have a large spherical conducting shell of radius R. if we connect the small sphere and large shell by a conducting wire. so that both the sphere and shell have same centre O. According to Huygen's principle "The net effect at any point due to a number of wavelets is equal to P sum total of contribution of all wavelets with proper phase difference. S2 At point P (i) If S2P – S1P = nl Þ the point P would be D minima.U ( R) = . V (r ) . Slit Screen l (ii) If S 2 P – S1 P = ( 2n + 1) Þ the point would be 2 maxima but with decreasing intensity. As charge flows from higher to lower potentials therefore.e..÷ èr R ø This is independent of charge Q on the large spherical shell. carrying charge Q. The maximum charge that may be given to outer shell which may cause discharge in air. 28.ò =-ò r R r R ® ® E . Van de Graaff. As r < R . directed radially outward 4pe 0 x 2 The potential difference between the sphere and the shell V (r ) . The lower comb C1 is connected to the positive terminal of a very high voltage source (HTS) ( » 104 volts. The charge spreads uniformly over whole surface of the shell. in 1931 designed an electrostatic generator which is capable of producing very high potential of the order of 10 MV. 2 lD The width of central maxima = a When the width of the slit is made double the original width. ) and the upper comb C 2 is connected to the inner surface of metallic sphere S. It consists of a large hollow metallic sphere S mounted on two insulating columns A and B and an endless belt of rubber or silk is made to run on two pulleys P1 and P2 by the means of an electric motor. This forms the principle of Van de Graaff generator. The electric field in the region inside the small sphere and large shell is due to charge + q only. . Construction. the charge flows from sphere to outer shell whatsoever the charge on outer shell may be.U ( R) is positive.d x q q 1 dx = 4pe 0 x 2 4pe 0 é x -1 ù q ê ú = 4 pe 1 ê ú 0 ë ûR r æ1 1 ö ç . so electric field strength at a distance x from the centre O is q 1 E= .Examination Papers 347 27. the high voltages can be built up to the breakdown field of the surrounding medium. non-conducting infinite sheet. Thus the outer surface of metallic sphere S gains positive A B charge continuously and its potential rises to a very high value. 1 times the net charge e0 òE s ® . The maximum potential is reached when the rate of leakage of charge becomes equal to the rate of charge transferred to the sphere. Van de Graaff generator is used to accelerate stream of charged particles to very high velocities. charge per unit surface area) be s. C2 get sprayed on the belt due to the repulsion between P2 positive ions and comb C1 .. We have to calculate the electric field strength at any point distance r from the sheet of charge.e. due to S action of sharp points. . Yes. then it produces ions in its vicinity. The pointed end of C 2 just touches the belt. Such a generator is installed at IIT Kanpur which accelerates charged particles upto 2 MeV energy. the generator is completely enclosed in an earthed connected steel tank which is filled with air under high pressure. The comb C 2 collects positive charge from the belt which immediately moves to the outer surface of sphere S. which is collected by comb C 2 and transferred to outer surface of sphere S. These positive ions are carried upward by the moving belt. the dielectric strength of surrounding air breaks down and its charge begins to leak. To prevent leakage of charge from the sphere. it continues to take (+ ) charge upward.e.. When the potential of a metallic sphere gains very HTS high value. (b) Gauss’s Theorem It states that the total electric flux through a closed surface is equal to enclosed by the surface i. As the belt goes on revolving. f= ò E .dS = ® 1 Sq e0 Let electric charge be uniformly distributed over the surface of a thin. The positive ions. OR (a) Electric flux: It is defined as the total number of electric field lines passing through an area normal to them: Also. so produced. d s ® ® The SI unit is Nm2/C or volt-metre. Let the surface charge density (i. to the C1 P1 surrounding air.348 Xam idea Physics—XII Working: When comb C1 is given very high potential. the charge enclosed by Gaussian surface = s a According to Gauss’s theorem. Total electric flux ò SE or ® •dS •dS ® = ò SE 1 ® • d S1 + ® ò SE 2 ® • d S2 + ® ò SE 3 ® • d S3 ® ò SE ® ® =ò S1 E dS1 cos 0° + òS 2 E dS 2 cos 0° + òS 3 E dS 3 cos 90° = E ò dS1 + E ò dS 2 = E a + E a = 2E a \ Total electric flux = 2E a . (c) (i) Away from the charged sheet.. we now consider a cylindrical Gaussian surface bounded by two plane faces A and B lying on the opposite sides and parallel to the charged sheet and the cylindrical surface perpendicular to the sheet (fig). If the surface charge density s is negative the electric field is directed towards the surface charge.e. (ii) Towards the plane sheet. . f æ f ö b f or m= = 0 m = 0 ç1 + e ÷ a fe fe è Dø 29. (a) Magnifying power of telescope is the ratio of the angle subtended at the eye by the image to the angle subtended at the unaided eye by the object. 2Ea = ( sa) e0 s \ E= × 2e 0 Thus electric field strength due to an infinite flat sheet of charge is independent of the distance of the point and is directed normally away from the charge. 1 Total electric flux = ´ (total charge enclosed by the surface) e0 1 i. As s is charge per unit area of sheet and a is the intersecting area. By symmetry the electric field strength at every point on the flat surface is the same and its direction is normal outwards at the points on the two plane surfaces and parallel to the curved surface.Examination Papers q=90o dS3 o S1 E 90 S3 349 S2 dS2 E o q=0 dS1 E q=0o B Sheet A To calculate the electric field strength near the sheet. 350 Xam idea Physics—XII (b) Using. size of final image = 30 × 10–4 × 100 m = 30 cm OR (a) Difference in working of telescope and microscope (i) Objective of telescope forms the image of a very far off object at or within the focus of its eyepiece. fe = 5 cm Angular magnification m = 30 Now. The microscope does the same for a small object kept just beyond the focus of its objective. 1 1 1 = f 0 v0 u0 1 1 1 = 150 v 0 -3 ´ 105 1 1 1 Þ = 5 150 3 ´ 10 v0 Þ Þ v0 = 3 ´ 105 cm ~ – 150 cm 1999 v 0 150 ´ 10 -2 m 10 -2 ~ = – u0 3000 m 20 Hence.25 cm. magnification due to the objective lens m0 = m 0 = 0. the lens equation for objective lens. total magnification m = m e ´ m 0 m = 6 × 5 × 10–4 = 30 × 10–4 Hence. (iii) The objective of a telescope has large focal length and large aperture while the corresponding for a microscope have very small values. m = me × mo . (b) Given fo = 1.05 ´ 10 -2 Using lens formula for eye piece 1 1 1 = f e ve ue Þ 1 1 1 = 5 -25 u e Þ ue = -25 cm 6 \ Magnification due to eyepiece m e = -25 =6 25 6 Hence. (ii) The final image formed by a telescope is magnified relative to its size as seen by the unaided eye while the final image formed by a microscope is magnified relative to its absolute size. Examination Papers 351 In normal adjustment. the output.25 . Then applying Kirchhoff's law to the output loop. 46 cm = – 1. we have VBE + Vi = VBE + I B R B + DI B ( R B + Ri ) Þ Vi = DI B ( R B + Ri ) Þ Vi = rDI B Change in IB causes a change in IC DI I Hence. Circuit diagram of CE transistor Amplifier. We first assume that Vi = 0. increases. which is measured between collector and ground. IC C1 RB C C2 RL IB B E Vi + – + VCC – VBB V0 Output waveform IE When an ac input signal Vi (to be amplified) is superimposed on the bias VBB. b ac = C = C DI B I B As DVCC = DVCE + R L DI C = 0 Þ DVCE = -R L DI C The change in VCE is the output voltage V0 Þ V0 = -R L DI C = b ac DI B R L The voltage gain of the amplifier is . angular magnification of eyepiece d 25 =5 me = = + fe 5 Hence m 0 = 6 v But m 0 = 0 u0 v Þ -6 = 0 u0 Þ v 0 = . VCC = VCE + I C R L Similarly the input loop gives VBB = VBE + I B R B When Vi is not zero.6u 0 Applying lens equation to the objective lens 1 1 1 1 1 1 = = Þ f 0 v0 u0 1.6u 0 u 0 Þ u 0 = -1.5 cm 30. Thus for input AC signal the output current is a continuous series of unidirectional pulses.352 Xam idea Physics—XII V0 DVCE -b ac DI B R L R = = -b ac L = Vi rDI B rDI B r Negative sign in the expression shows that output voltage and input voltage have phase difference of p. AV = OR (a) p1 Input AC signal to be rectified s1 D1 Centre tap s2 P2 N2 P1 N1 i1 + A i2 RL Output B – p2 D2 Working: The AC input voltage across secondary s1 and s 2 changes polarity after each half cycle. then output frequency will be 2f hertz because for each cycle of input. Therefore. Waveform at P1 2T O T 2 T 3 T 2 t Waveform at P2 O T 2 Due to D1 Due to D2 T 2 T Due to D1 T 3 T 2 3 T 2 2T t Due to D2 2T t Output waveform O (across RL) In a full wave rectifier. The direction of current (i1 ) due to diode D1 in load resistance R L is directed from A to B. the terminal s1 is negative and s 2 is positive relative to centre tap O. the terminal s1 is positive relative to centre tap O and s 2 is negative relative to O. if input frequency is f hertz. diode D1 conducts while diode D2 does not. Thus the current in load resistance R L is in the same direction for both half cycles of input AC voltage. In next half cycle. . Then diode D1 is forward biased and diode D2 is reverse biased. two positive half cycles of output are obtained. Suppose during the first half cycle of input AC signal. The diode D1 is reverse biased and diode D2 is forward biased. The direction of current (i 2 ) due to diode D2 in load resistance R L is still from A to B. diode D2 conducts while D1 does not. Therefore. Clockwise 10. Magnetic field inside the solenoid B = m 0 nI n ® number of turns per unit length. i.e. 14.Examination Papers 353 (b) Output waveforms for the following inputs A and B of OR gate and NAND gate. t1 t2 t3 t4 t5 t6 t7 t8 A B OR output NAND output CBSE (Delhi) Set–II ® 1. f¢ = Magnetic field through each turn = BA = m 0 nIA Total magnetic flux linked with solenoid f = Nf¢ = nl ´ m 0 nIA = m 0 n 2 AlI = LI Þ L = m 0 n 2 Al L ® self inductance of solenoid and l ® length of solenoid.. 6. The material having positive susceptibility is paramagnetic material. (ii) When a paramagnetic material is placed in an external field the field lines get concentrated inside the material. and the field inside is enhanced. 2r Clockwise Clockwise BP O BQ Q X-axis P I Current Q 2I Current . Properties (i) They have tendency to move from a region of weak magnetic field to strong magnetic field. I ® current flowing. 16. they get weakly attracted to a magnet. In the static situation. E has to ensure that the free charges on the surface do not experience any force. K = 4I 0 cos 2 p = 4I 0 2l When path difference = 3 2p Phase difference. . (ii) Force acting on the loop m ii æ1 1 ö F = 0 1 2 lç . I ' = 4I 0 cos 2 22.÷ ÷ 2p ç è r1 r 2 ø æ 1 ö 1 ÷ = 2 × 10–7 × 4 × 1 × 5 × 10–2ç ç -2 -2 ÷ 3 ´ 10 ø è 10 2 80 = 40 ´ 10 -7 ´ N = ´ 10 -7 N = 26.66 × 10–7N = 2. (i) Torque = MB sin q ® ® 4p K æ1ö = 4I 0 ç ÷ = I 0 = 3 4 è2ø 2 As M and B are parallel. intensity. f = 3 Hence. (i) w = (ii) I = 1 LC = 1 4 ´ 100 ´ 10 -6 = 50 rad/s 21. e 240 = = 4A R 60 (iii) VL = I X L = Iw L = 4 × 50 × 4 = 800V (a) Coherent sources are needed to ensure that the positions of maxima and minima do not change with time.354 Xam idea Physics—XII ® | BP| = ® m 0r 2 I 2(r + r ) m 0 ( 2I )r 2 2(r 2 + r 2 ) ® 2 2 3 2 = = 2 u0 I 4 2r m 0I Pointing towards P Pointing towards Q | BQ| = ® 3 ® | B| = | B Q | – | B p | = 4 2r m 0I 4 2r m 0I 4 2r So. phase difference f = 2p Thus.67 mN 3 3 Direction: Towards the conductor or attractive. magnetic field at point O has a magnitude 20. (b) We know that I = I 1 + I 2 + 2 I 1 I 2 cos f Þ I = 4I 0 cos 2 f 2 (I1 = I2 = I0) For path difference l. So t = 0. (ii) When a paramagnetic material is placed in an external field the field lines get concentrated inside the material. (iii) Space waves (a) LOS (Line of Sight) Communication dM dT dR hT Earth hR (b) Satellite communication Communication satellite ave ce w a p S Transmitting antenna Earth Receiving antenna Above 40 MHz. Electric field intensity is zero throughout the volume and the potential just inside. has to be equal to potential on the surface. . For properties The material having positive susceptibility is paramagnetic material. anticlockwise 9. the mode of propagation used is via space waves a space wave travels in a straight line from the transmitting antenna to the receiving antenna. 12. Mutual Inductance: It is defined as the magnetic flux linked to the second coil when unit current is flowing in the first coil. i. Magnetic field along X-direction. 8.. they get weakly attracted to a magnet.Examination Papers 355 27. Three mode of propagation of electromagnetic waves are: (i) Ground waves. The magnetic material is paramagnetic in nature.e. In coil 1. CBSE (Delhi) Set–III 6. (ii) Sky waves. Space waves are used for the line of sight (LOS) communication as well as satellite communication. and the field inside is enhanced. clockwise In coil 2. Properties (i) They have tendency to move from a region of weak magnetic field to strong magnetic field. 356 Xam idea Physics—XII N2 r2 N1 r1 r1 < r2 l N1 f1 = M 12 I 2 N1 f1 = ( n. the net magnetic field at the point O has magnitude 21. Clockwise 2r O BP P I BQ Q 2I X-axis Anticlockwise ® | BP | = m 0r 2 I 2(r 2 + r 2 ) ® 3 2 = m 0I 4 2r = ® | BQ | Pointing towards P ® | B | = | B P | + | BQ | = 2 m 0I 4 2r = m 0I 2 2r m 0I 2 2r . Ionospheric layers Sky wave communication . So. Three mode of propagation of electromagnetic wave (i) Ground wave propagation (ii) Sky wave propagation (iii) Space wave propagation. l) ( pr12 )(m 0 n 2 I 2 ) M 12 =m 0 n1 n 2 pr12 l 16. ÷ ÷ 2p ç è r1 r 2 ø æ ö 1 1 ÷ = 2 × 10–7 × 2 × 1 × 4 × 10–2 ç ç -2 -2 ÷ 4.Examination Papers 357 The Ionospheric layers act as a reflector for a certain range of frequencies (from few MHz to 40MHz).5 .5 ´ 10 ø è 2 ´ 10 é 4. 5 9 ë û Direction of force is towards conductor (attractive). (i) w = = 50 rad/s = LC 10 ´ 40 ´ 10 -6 (ii) Current at resonance V 240 = 4A I rms = rms = R 60 (iii) Vrms across inductor Vrms = I rms ´ L = I rms ( wL) = 4 × 50 × 10 = 2000V . (i) Torque on the loop 't' = MB sin q = M ´ B [Q M and B are parallel] t=0 (ii) Magnitude of force m I I læ 1 1 ö F = 0 1 2 ç .44 × 10–7N ú= 2 ´ 4 .2 ù 8 ´ 5 ´ 10 -7 = 16 × 10–7 × ê = 4. ® ® ® ® 23. 1 1 25. Hence these waves reach from the transmitting antenna to the receiving antenna via their reflection from the ionosphere. the angles of refraction in three media A. one of copper and the other of manganin have the same resistance. 0) C (6. Which wire is thicker? 2. 25° and 35° respectively. What is the electric flux passing through each face of the cube? 9. Mention the two characteristic properties of the material suitable for making core of a transformer. Two wires of equal length. What are the directions of electric and magnetic field vectors relative to each other and relative to the direction of propagation of electromagnetic waves? 3. (i) Calculate the potential difference between A and C. 3) B E (2. A bar magnet is moved in the direction indicated by the arrow between two coils PQ and CD. Which one has greater de-Broglie wavelength and why? 7. (ii) At which point (of the two) is the electric potential more and why? (2. 0) A . Maximum marks: 70 CBSE (AIl India) Set–I 1. A proton and an electron have same kinetic energy. 8. Predict the directions of induced current in each coil P Q C D N A S A 5. A charge ‘q’ is placed at the centre of a cube of side l. A test charge ‘q’ is moved without acceleration from A to C along the path from A to B and then from B to C in electric field E as shown in the figure. In which medium would the velocity of light be minimum? 6.CBSE EXAMINATION PAPERS ALL INDIA–2012 Time allowed: 3 hours General Instructions: As given in CBSE Examination Papers Delhi–2011. For the same value of angle incidence. How does the angular separation between fringes in single-slit diffraction experiment change when the distance of separation between the slit and screen is doubled? 4. B and C are 15°. An object AB is kept in front of a concave mirror as shown in the figure. Find the work done in rotating it through the angle of 180° 11. State the underlying principle of a transformer. In the given block diagram of a receiver. (i) Show that the net force acting on it is zero. (ii) How will the position and intensity of the image be affected if the lower half of the mirror’s reflecting surface is painted black? 14. A B C F (i) Complete the ray diagram showing the image formation of the object. 15.Examination Papers 359 10. identify the boxes labelled as X and Y and write their functions. Will the ammeter show a momentary deflection during the process of charging? If so. A capacitor of capacitance of ‘C’ is being charged by connecting it across a dc source along with an ammeter. 16. A light bulb is rated 100 W for 220 V ac supply of 50 Hz. Receiving antenna Received signal Amplifier X Detector Y Output 17. An electric dipole is held in a uniform electric field. How is the large scale transmission of electric energy over long distances done with the use of transformers? 12. (ii) The dipole is aligned parallel to the field. Describe briefly with the help of a circuit diagram. Mention its two advantages over the refracting telescope. Calculate (i) the resistance of the bulb. (ii) the rms current through the bulb . Draw a labelled ray diagram of a reflecting telescope. how the flow of current carries in a p-n-p transistor is regulated with emitter-base junction forward biased and base-collector junction reverse biased. 13. how would you explain this momentary deflection and the resulting continuity of current in the circuit? Write the expression for the current inside the capacitor. Plot a graph to show the variation of current with frequency of the source. current I remaining the same. Find (i) the frequency of the source. A series LCR circuit is connected to an ac source. Mention three ‘different modes of propagation used in communication system. An object is kept at 40 cm in front of L1. Find the separations between L1. . L2 and L3 each of focal length 20 cm. Calculate the value of the resistance R in the circuit shown in the figure so that the current in the circuit is 0.2 A B 5W C 15W 30W 3V R 0. A circular coil of N turns and radius R carries a current I.2 A. How will the (i) energy stored and (ii) the electric field inside the capacitor be affected when it is completely filled with a dielectric material of dielectric constant ‘K’? 20. 19. What would be the potential difference between points B and E? 8V 0. derive the expression for the impedance of the circuit. Define the terms (i) ‘cut-off voltage’ and (ii) ‘threshold frequency’ in relation to the phenomenon of photoelectric effect. 24.360 Xam idea Physics—XII OR An alternating voltage given by V = 140 sin 314 t is connected across a pure resistor of 50 W. Using Einstein’s photoelectric equation show how the cut-off voltage and threshold frequency for a given photosensitive material can be determined with the help of a suitable plot/graph. Explain with the help of a diagram how long distance communication can be achieved by ionospheric reflection of radio waves.2 A E 10W 10W D 21. as shown. Deduce the expression for the electrostatic energy stored in a capacitor of capacitance ‘C’ and having charge ‘Q’. L1 L2 L3 I 40cm 20cm 22. Calculate the ratio of the magnetic moments of the new coil and the original coil. The final real image is formed at the focus ‘I’ of L3. 18. L2 and L3. Using the phasor diagram. (ii) the rms current through the resistor. It is unwound and rewound to make another coil of radius R/2. 23. You are given three lenses L1. explaining the nature of its variation. Draw the necessary circuit diagram and explain its working. calculate the distance of closest approach to the nucleus of Z = 80. derive the condition for (i) constructive interference and (ii) destructive interference at a point on the screen. If an electron makes a transition from an energy level – 0. (b) Two polaroids ‘A’ and ‘B’ are kept in crossed position.28 mm.6 eV. (a) In Young’s double slit experiment. the role of the two important processes involved in the formation of a p-n junction. Write any two characteristic features of nuclear forces.4 eV. (b) Identify the equivalent gate for the following circuit and write its truth table. 26.4 m away. OR (a) How does an unpolarised light incident on a polaroid get polarised? Describe briefly. If the two slits are separated by 0. with the help of a necessary diagram. In a Geiger–Marsden experiment. 28. calculate the wavelength of the spectral line emitted. Define relaxation time of the free electrons drifting in a conductor. 800 nm and 600 nm is used to obtain the interference fringes in a Young’s double slit experiment on a screen placed 1. To which series of hydrogen spectrum does this wavelength belong? 27. (b) Name the device which is used as a voltage-regulator. How should a third polaroid ‘C’ be placed between them so that the intensity of polarised light transmitted by polaroid B reduces to 1/8th of the intensity of unpolarised light incident on A? 29. How is it related to the drift velocity of free electrons? Use this relation to deduce the expression for the electrical resistivity of the material. Draw the necessary circuit diagram and explain its working. Draw a plot of potential energy of a pair of nucleons as a function of their separations. (a) Describe briefly. A A' Y B' B . with the help of a diagram. (b) A beam of light consisting of two wavelengths. OR (a) Explain briefly the principle on which a transistor-amplifier works as an oscillator. Mark the regions where the nuclear force is (i) attractive and (ii) repulsive. the polarisation of light by reflection from a transparent medium. How will the distance of closest approach be affected when the kinetic energy of the a-particle is doubled? OR The ground state energy of hydrogen atom is – 13. calculate the least distance from the central bright maximum where the bright fringes of the two wavelengths coincide.Examination Papers 361 25.85 eV to –3. when an a-particle of 8 MeV energy impinges on it before it comes momentarily to rest and reverses its direction. (ii) the rms current through the resistor. (a) Write the expression for the force. (ii) At which point (of the two) is the electric potential more and why? (1.362 Xam idea Physics—XII ® 30. . how the magnetic field due to one conductor acts on the other. current ‘I’ remaining the same. Calculate (i) the resistance of the bulb. giving reasons. 0) 16. Find (i) the frequency of the source. OR (a) Explain. with the help of a suitable diagram. CBSE (All India) Set–II Questions uncommon to Set–I 2. How does this affect the size and intensity of the central diffraction band? 11. Hence deduce the expression for the force acting between the conductors. OR An alternating voltage given by V = 280 sin 50 pt is connected across a pure resistor of 40 W. 0) (5. Calculate the ratio of the magnetic moments of the new coil and the original coil. Prove that the torque t acting on the loop is given by t = m × B. moving with a velocity v in the presence of both electric field E and magnetic field B . (i) Calculate the potential difference between A and C. A circular coil of ‘N’ turns and diameter ‘d’ carries a current ‘I’. A charge q is placed at the centre of a cube of side l. In a single-slit diffraction experiment. (ii) the rms current through the bulb. where m is the magnetic moment of the loop. Obtain the condition under which the particle moves undeflected through the fields. A proton and an electron have same velocity. Mention the nature of this force. Which one has greater de-Broglie wavelength and why? 8. F . A light bulb is rated 200 W for 220 V ac supply of C A 50 Hz. What is the electric flux passing through two opposite faces of the cube? 5. (b) Two long straight parallel conductors carrying steady current I1 and I2 are separated by a distance ‘d’. It is unwound and rewound to make another coil of diameter ‘2d’. (b) A rectangular loop of size l × b carrying a steady current I is placed in a uniform magnetic ® ® ® ® ® ® ® ® ® field B. acting on a charged particle of charge ‘q’. the basic difference in converting a galvanometer into (i) a voltmeter and (ii) an ammeter. 17. Explain briefly. 3) B from A to C along the path from A to B and then from B to C in electric field E as shown in the E figure. A test charge ‘q’ is moved without acceleration (1. the width of the slit is made double the original width. L2 and L3. What would be the potential difference between points A and B? 6V 0. How does its frequency change? 9.51 eV. You are given three lenses L1. Find the separations between L1. An object is kept at 20 cm in front of L1.2 A B 5W C 10W 15W 2V 0. To which series of hydrogen spectrum does this wavelength belong? CBSE (All India) Set–III Questions uncommon to Set – I & II 1. m being the me permeability of the medium and e its permittivity. If an electron makes a transition from an energy level – 0. A charge ‘q’ is placed at the centre of a cube. in Young’s double-slit experiment. A circular coil of closely wound N turns and radius r carries a current I. What is the electric flux passing through the cube? 1 4.Examination Papers 363 19. calculate the distance of closest approach to the nucleus of Z = 75.2 A R A 30W 5W D 27. How does the fringe width. when an a-particle of 5 MeV energy impinges on it before it comes momentarily to rest and reverses its direction.2 A. as shown. The final real image is formed at the focus ‘I’ of L3. Calculate the value of the resistance R in the circuit shown in the figure so that the current in the circuit is 0. The speed of an electromagnetic wave in a material medium is given by v = . L1 L2 L3 I 20cm 15cm 23. Write the expressions for the following: (i) the magnetic field at its centre (ii) the magnetic moment of this coil .85 eV to – 1. In a Geiger– Marsden experiment. calculate the wavelength of the spectral line emitted.6 eV. change when the distance of separation between the slits and screen is doubled? 3. L2 and L3 each of focal length 15 cm. How will the distance of closest approach be affected when the kinetic energy of the a-particle is doubled? OR The ground state energy of hydrogen atom is –13. 2 A D 5W C 10W . (ii) At which point (of the two) is the electric potential more and why? 20.2 A A B 30W 15W 10W 5V R 0. L2 and L3. (–2. Describe briefly with the help of a circuit diagram. 2) OR B An alternating voltage given by V = 70 sin 100 p t is E connected across a pure resistor of 25 W. You are given three lenses L1. –1) (2. (ii) the rms current through the bulb. Calculate the value of the resistance R in the circuit shown in the figure so that the current in the circuit is 0. Find (i) the frequency of the source. Explain briefly the following terms used in communication system: (i) Transducer (ii) Repeater (iii) Amplification 22. (–2. as shown. (i) Calculate the potential difference between A and C. L2 and L3 each of focal length 10 cm. What would be the potential difference between points A and D? 10V 0.2 A. (ii) the rms current through the resistor. L1 L2 L3 I 15cm 10cm 26. –1) C A 16. A test charge ‘q’ is moved without acceleration from A to C along the path from A to B and them from B to C in electric field E as shown in the figure. A light bulb is rated 150 W for 220 V ac supply of 60 Hz. 12. The final real image is formed at the focus ‘I’ of L3. the paths of current carriers in an n-p-n transistor with emitter-base junction forward biased and base-collector junction reverse biased.364 Xam idea Physics—XII 10. An object is kept at 15 cm in front of L1. Calculate (i) the resistance of the bulb. Find the separations between L1. There waves are perpendicular to each other and perpendicular to the direction of propagation. and lm = lCu . Angular separation would remain same. P 1 A Q C D 2 A N S In the figure..(i) l . v µ sin r .Examination Papers 365 Solutions CBSE (All India) SET–I 1. R Cu = r Cu For manganin. R Cu = R m Equation (i) and equation (ii) rm A = m >1 r Cu A Cu Hence A m > A Cu Manganin wire is thicker.. From Snell's law n = = sin r v For given i. N pole is receding away coil (1). sin i c 5. r is minimum in medium A. The direction of current will be from P to Q in coil (1) and from C to D in coil (2).. so in coil (1).. We know R =r For copper. Þ 2. b Dl / d l 3.(ii) lCu A Cu . A But r m > r Cu . Rm = r m lm Am . Both electric field and magnetic fields are electromagnetic waves. . so current in coils will flow clockwise as seen from the side of magnet. Angular separation is q = = = D D d Since q is independent of D. 4. so velocity of light is minimum in medium A. the nearer faces will act as S-pole and in coil (2) the nearer face will also act as S-pole to oppose the approach of magnet towards coil (2). 366 Xam idea Physics—XII 6. de Broglie wavelength h l= 2m E k For same K. We know.E. By Gauss's Theorem in electrostatics ® ® q f = ò E . Hence. ds = e0 Total flux through all six faces would be f' = 6f where f = Flux through one face. potential difference between A and C C® ® VC – VA = -ò E . . (i) From the given diagram. The principle of transformer is based upon the principle of mutual induction which states that due to continuous change in the current in the primary coil an emf gets induced across the secondary coil. of electron and proton. 7. VC – VA = 4E (ii) VC > VA Because direction of electric field is in decreasing potential. (i) The dipole moment of dipole is ® B +q ® +qE | P | = q ´ ( 2a) ® Force on –q at A = –qE ® 2a ® –qE ® ® –q Force on +q at B = + q E N Net force on the dipole = qE – qE = 0 A (ii) Work done on dipole W = DU = PE(cos q1 . q q Hence. Two characteristic properties: (i) Low hysteresis loss (ii) Low coercivity 8. at the consumer end. As m p > me Þ le > l p Hence wavelength of electron is greater than that of proton. a step down transformer step down the voltage. the step up transformers step up the output voltage which reduces the current through the cables and hence reduce resistive power loss. Then.cos q 2 ) = PE(cos 0°.cos180° ) W = 2 PE 11. 6f = Þ f= e0 6e 0 9. At the power generating station. dl cos180° = E × 4 = 4E A 10. Under the forward biasing of emitter-base region. As a hole reaches terminal C. The holes entering the collector move under the aiding reverse bias towards terminal C. (i) Image formed will be inverted diminished between C and F. As soon as a hole combines with the electron. Objective mirror Secondary mirror Eyepiece Advantages: (i) It is free from chromatic aberration. e. 12. Yes.region move towards the base. Current inside the capacitor d fE I D = e0 dt 13. The emitter-base junction is given a small forward bias. (ii) Its resolving power is greater than refracting telescope due to larger aperture of mirror. This causes the collector current I C . A B' F B C A' D P (ii) No change in position of image and its intensity will get reduced. 14. because of the production of displacement current between the plate of capacitor on account of changing electric field.. Due to thin base E C most of holes (about 98%) entering it pass onto collector while a very few of them (nearly 2%) combine with the electrons of base. Both these currents I B and I C combine to form the emitter current I e i. the large scale transmission of electric energy over long distances done with the use of transformers is taken place. the positive p n p holes of P . while base collector junction is given a large reverse bias. I E = I B + IC .Examination Papers 367 Hence. a fresh B IC IE electron leaves the negative terminal of battery VEE and enters IB the base. This causes a very small base current I B . an electron + – + – VEE VCC leaves the negative terminal of battery VCC and neutralises the hole. 15. 9 V 2 V 98 . 9 = 1. X ® IF stage (Intermediate Frequency stage) Its function is to change the carrier frequency to lower frequency. V 2 rms 220 ´ 220 = = 48 W P 100 P 100 (b) I rms = = 0. 16. Receiving antenna Received signal Amplifier X Detector Y Output Here. As length of wire remains the same R N1 ´ 2pR = N 2 ´ 2p 2 N2 = 2N \ Magnetic moment of a coil. Y ® Amplifier Its function is to amplify the signal because the detected signal may not be strong enough to use by the user. 17. m = NAI For the coil of radius R.45 A Vrms 220 (a) R = OR (i) V = 140 sin 314t Q V = V0 sin wt Hence w = 314 (ii) Vrms = V0 = 140 Þ Þ 2pn = 314 n = 50 Hz = 98.97 A I = rms = R 50 18. magnetic moment 2N1 I pR 2 N IpR 2 = 1 4 2 m 2 = N 2 IA 2 = . magnetic moment m1 = N1 IA1 = N1 I pR 2 2 For the coil of radius R 2.368 Xam idea Physics—XII Obviously the holes are the charge carriers within the p-n-p transistor while the electrons are charge carriers in external circuit. the potential difference between its plates V = . Therefore total work Q Q q W = ò V dq = ò dq 0 0 C 1 é q2 ù = ê ú Cê ú ë 2 û Q = 0 1 C æ Q2 0 ö Q2 ç . Let at any instant A + +Q + + + + + + + VAB = V B –Q – – – – – – – – VA VB q when charge on capacitor be q. K E (ii) In the presence of dielectric the electric field is reduced to E = 0 . Here.Examination Papers 369 Now. m2 = 1:2 m1 19. K 20.2 + 15 × 0.2 + R × 0. U= = CV 2 = QV 2C 2 2 When battery is disconnected 2 Q2 Q0 U (i) Energy stored will be decreased. C Now work done in giving an additional infinitesimal charge dq to capacitor q dW = V dq = dq C The total work done in giving charge from 0 to Q will be equal to the sum of all such infinitesimal works. work is done by the charging battery at the expense of its chemical energy. This work is stored in the capacitor in the form of electrostatic potential energy.. The energy becomes. then Q = CV (CV ) 2 1 1 = CV 2 = QV 2C 2 2 This work is stored as electrostatic potential energy of capacitor i.2 = 8 – 3 \ W= . the potential difference between its plates increases. Q2 1 1 Electrostatic potential energy. When a capacitor is charged by a battery.e. Consider a capacitor of capacitance C. RBCD = 5 W + 10 W = 15 W Effective resistance between B and E 1 1 1 1 = + + Þ R BE = 5W R BE 30 10 15 Applying Kirchhoff’s Law 5 × 0. Initial potential difference between capacitor plates = zero. Initial charge on capacitor is zero. energy is reduced to times the initial energy. U = 0 = = 0 2C 2KC 0 K 1 Thus. Let a charge Q be given to it in small steps.÷= ç 2 2 ÷ 2C è ø If V is the final potential difference between capacitor plates. When charge is given to capacitor. which may be obtained by integration. the distance between lens L2 and L3 does not matter. Given f 1 = f 2 = f 3 = 20 cm For lens L1 u1 = – 40 cm 1 1 1 By lens formula . Hence.hn 0 l eV0 = hn .. there is a minimum frequency h Slope = e of light below which no photo electric emission will take place.2 × 5 = 1 volt 21.f = hn . this frequency is called as threshold frequency. = v3 u 3 f 3 1 1 1 Þ = 20 u 3 20 1 = 0 Þ u3 = ¥ Þ u3 Thus lens L2 should produce image at infinity. As the image formed by lens L2 lies at infinity. v3 = 20 cm. then.n 0 e e Clearly.e.370 Xam idea Physics—XII Þ Hence R = 5W VBE = IR BE = 0.hn 0 h h V0 = n . its objective should be at focus. (ii) For a given material. The image formed by lens L1 is at 40 cm on the right side of lens L1 which lies at 20 cm left of lens L2 i. for L2. V0 X By Einstein's photo electric equation n0 n hC KE max = . (i) Cut off or stopping potential is that minimum value Y of negative potential at anode which just stops the photo electric current. Intercept = – f/e . focus of lens L2. the distance between L1 and L2 = 40 + 20 = 60 cm.= v1 u1 f 1 1 1 1 Þ = Þ v1 = 40 cm v1 20 -40 For lens L3 f3 = 20 cm. 22. Hence. V0 – n graph is a straight line. Hence. u3 = ? 1 1 1 By lens formula. the distance between L2 and L3 can have any value. inductance L and capacitance C are connected in series and an alternating source of voltage V = V0 sin wt is applied across it. b).. therefore their resultant potential difference = VC . R VL L VR C VC VL 90o (VC–VL) 90 o VR f i V=V0 sin wt (a) VC (b) Suppose the voltage across resistance R is VR . i. im lm im 2 f1 f fr f2 .wL ÷ w C è ø 2 The phase difference ( f) between current and voltage f is given by X -XL tan f = C R The graph of variation of peak current i m with frequency is shown in fig. VC = X C i and VL = X L i 1 where X C = = capacitance reactance and X L = wL = inductive reactance wC \ \ Impedance of circuit.Examination Papers 371 23. the current amplitude is maximum. At resonant frequency. V But 2 = VR 2 + (VC . Thus VR and (VC .VL ) 2 .X L ) 2 i 2 æ 1 ö Z = R 2 + (X C . the resultant of VR and (VC .(ii) VR = R i .(i) .X L ) 2 = R 2 + ç . the current (i ) flowing through all of them is the same.VL ) are mutually perpendicular and the phase difference between them is 90°. From fig. The voltage VR and current i are in the same phase. Instantaneous current I = V = ( R i) 2 + ( X C i ..e. With increase in frequency. As applied voltage across the circuit is V.X L i) 2 Z= V = R 2 + (X C . (fig.. Expression for Impedance in LCR series circuit: Suppose resistance R.VL ) will also be V.VL ) 2 Þ V = VR 2 + (VC . voltage across inductance L is VL and voltage across capacitance C is VC . current first increases and then decreases..wL ÷ è wC ø V0 sin ( wt + f) æ 1 ö R2 + ç . the voltage VL will lead the current by angle 90° while the voltage VC will lag behind the current by angle 90° (fig.VL (if VC > VC ). Clearly VC and VL are in opposite directions. a) On account of being in series. 25. Part AB represents repulsive force and Part BCD represents attractive force. A +100 Repulsive B MeV 0 D Attractive –100 C 1 2 r (fm) 3 4 . reach the receiving antenna after reflection from the ionosphere which acts as a reflector for radio waves. Three mode of propagation of electromagnetic waves: (i) Ground waves (ii) Sky waves (iii) Space waves Sky wave propagation is used for long distance communication by ionospheric reflection of radio wave Ionospheric layers Sky wave communication When radio waves (frequency range 3MHz – 30 MHz).372 Xam idea Physics—XII 24. emitted from the transmitting antenna. E1 = -13. Ek = 8MeV 1 ( Ze) ( 2e) K= 4pe 0 r0 Þ r0 = r0 = = As 1 ( Ze)( 2e) 2Ze 2 = 4pe 0 K 4pe 0 K 9 ´ 10 9 ´ 2 ´ 80 ´ (1. The relation between t and v d is .6 ´ 10 -19 ) 2 8 ´ 10 6 ´ (1.6 .. Then. n B = 2.6 Similarly. æ 1 1 1 ö ÷ We know. R = 1.097 ´ 107 ç l = 4862Å ç ÷ Þ 2 l 42 ø è2 27. Z = 80.Examination Papers 373 Conclusions: (1) Nuclear forces are attractive and stronger. Relaxation time of free electrons drifting in a conductor is the average time elapsed between two successive collisions.88 × 10–14 m 1 K If kinetic energy (K) of a-particle is doubled.6 – 0. It corresponds to Balmer series. 26. – 3.85eV to E B = -3. then electrostatic force.6 eV When electron undergoes transitions from E A = -0. (2) Nuclear forces are charge-independent. from equation (i) -13.(i) En = eV n2 For n = 1. the distance of closest approach will become half. = Rç 2 2 ÷ ç l n n A ø è B Here. Let r 0 be the distance of closest approach where the K.E. n A = 4. Given.4 = Þ nB = 2 2 nB r0 µ Hence electron transits from n = 4 to n = 2.6 ´ 10 -19 ) 18 ´ 80 ´ 1.097 ´ 107 m–1 æ 1 1 1 ö ÷ = 1.6 ´ 10 -10 8 ´ 10 6 m = 2. of a-particle is converted into its potential energy.4eV Then.85 = Þ nA = 4 2 nA -13. OR As we know: -13.. the resultant displacement at that point will be y = y1 + y 2 Substituting values of y1 and y 2 from (i) and (ii) in (iii).(i) ... = I ne 2 tA By Ohm's Law R = V I V E ® vd ® V l A l æ m öl ÷ =r l R =ç ç ne 2 t ÷ A A è ø m \ r= ne 2 t 28.(iii) . If at any time 2p t.. we get y = a1 sin wt + a 2 sin ( wt + f) Using trigonometric relation sin ( wt + f) = sin wt cos f + cos wt sin f ... the frequency w of each wave is and amplitudes of electric field are a1 and a 2 respectively... The current flowing through the conductor is æ e E ö ne 2 EA I = – neA v d = n e A ç t÷ = t m è m ø If V is the potential difference applied across the two ends.374 Xam idea Physics—XII eE =t m Let a conductor of length 'l' and area of cross-section ‘A’ with electron density n. then electric field ( E) = Then current becomes n A e 2 Vt I= ml V ml Then.(ii) According to Young’s principle of superposition.(v) . then equation of waves may be expressed as y1 = a1 sin wt y 2 = a 2 sin ( wt + f) . we get Let y = a1 sin wt + a 2 (sin wt cos f + cos wt sin f) = ( a1 + a 2 cos f) sin wt + ( a 2 sin f) cos wt a1 + a 2 cos f = A cos q .. (a) Conditions of Constructive and Destructive Interference: Suppose two coherent waves travel in the same direction along a straight line.(iv) . the electric fields of waves at a point are y1 and y 2 respectively and phase difference is f... .. 2. In this case the minimum intensity. Squaring (v) and (vi) and then adding.. The values of A and q are determined by (v) and (vi).. 2p .. f = p . 3 ......(xii) .. 6p ..... 1..(vii) and a 2 sin f = A sin q where A and q are new constants.. 7p . Then equation (iv) gives y = A cos q sin wt + A sin q cos wt = A sin ( wt + q) This is the equation of the resultant disturbance..... f = 0.. . 4p .(vi) .Examination Papers 375 ..(xiv) = ( 2n . 2... = 2np ( n = 0... 5p .a 2 ) 2 .(viii) As cos 2 q + sin 2 q = 1. .. I max = a12 + a 22 + 2a1 a 2 = ( a1 + a 2 ) 2 Path difference D= l l ´ Phase difference = ´ 2 np = nl 2p 2p .) The maximum intensity. n = 1. we get As the intensity of a wave is proportional to its amplitude in arbitrary units I = A 2 \ Intensity of resultant wave I = A 2 = a12 + a 22 + 2a1 a 2 cos f .(xi) . A = a12 + a 22 + 2a1 a 2 cos f . Clearly the amplitude of resultant disturbance is A and phase difference from first wave is q.1 or phase difference... Constructive Interference : For maximum intensity at any point cos f = + 1 or phase difference.. 3p .(ix) Clearly the intensity of resultant wave at any point depends on the amplitudes of individual waves and the phase difference between the waves at the point.2a1 a 2 = ( a1 . Destructive Interference : For minimum intensity at any point cos f = ..1) p . I min = a12 + a 22 . we get ( a1 + a 2 cos f) 2 + ( a 2 sin f) 2 = A 2 cos 2 q + A 2 sin 2 q or a12 + a 22 cos 2 f + 2a1 a 2 cos f + a 22 sin 2 f = A 2 (cos 2 q + sin 2 q) A 2 = a12 + a 22 (cos 2 f + sin 2 f) + 2a1 a 2 cos f or \ A 2 = a12 + a 22 + 2a1 a 2 cos f Amplitude.(x) Clearly the maximum intensity is obtained in the region of superposition at those points where waves meet in the same phase or the phase difference between the waves is even multiple of p or path difference between them is the integral multiple of l and maximum intensity is ( a1 + a 2 ) 2 which is greater than the sum intensities of individual waves by an amount 2a1 a 2 ..(xiii) ... D= l ´ Phase difference 2p l l = ´ ( 2n . Imax I Imin –2l –l O x l 2l From equations (xii) and (xvi) it is clear that the intensity 2a1 a 2 is transferred from positions of minima to maxima. the minimum intensity is obtained in the region of superposition at those points where waves meet in opposite phase or the phase difference between the waves is odd l multiple of p or path difference between the waves is odd multiple of and minimum 2 2 intensity = ( a1 .376 Xam idea Physics—XII Path difference.4 ´ 800 ´ 10 -9 0. n2 = n +1 n D l1 ( n + 1) Dl 2 = \ (yn ) l1 = ( yn + 1 ) l2 Þ d d Þ n l1 = (n + 1) l 2 l2 600 = =3 Þ n= l1 . there must be a difference of 1 in order of l1 and l 2 .1) p = ( 2n .a 2 ) which is less than the sum of intensities of the individual waves by an amount 2a1 a 2 ..1) 2p 2 . n1 < n 2 If n1 = n. This implies that the interference is based on conservation of energy.28 × 10–3m For least distance of coincidence of fringes. Variation of Intensity of light with position x is shown in fig.600 nDl1 y min = d = 3 ´ 1. (b) Given l1 = 800 nm = 800 ´ 10 -9 m l 2 = 600nm = 600 ´ 10 -9 m D = 1.l 2 800 .(xv) Clearly.28 ´ 10 -3 = 12000 × 10–6 = 12 × 10–3m .28 mm = 0. As l1 > l 2 .4 m d = 0.. sin q) 2 2 2 According to question I0 I (cos q sin q ) 2 = 0 2 8 I0 2 or. (b) Let the angle between the pass axis of A and C = q I Intensity of light passing through A = 0 2 I0 Intensity of light passing through C = cos2q 2 I Intensity of light passing through B = 0 cos 2 q . 2 . D A Incident ray ip r' O r Refracted ray B Reflected ray 90° E C Whenever unpolarised light is incident on the boundary between two transparent media. I æ 2 sin q cos q ö ç ÷ = 0 2 8 è ø sin 2q = 1 2q = 90° Þ q = 45° The third polaroid is placed at q = 45°. sin 2 q = 0 (cos q . The incident light thus gets linearly polarised. it lets only those of its electric vectors that are oscillating along a direction perpendicular to its aligned molecules to pass through it. the reflected light gets partially or completely polarised. when an unpolarised light falls on a polaroid. cos 2 ( 90 .Examination Papers 377 OR (a) A polaroid consists of long chain molecules aligned in a particular direction. The electric vectors along the direction of the aligned molecules get absorbed.q) 2 I0 I = cos 2 q . When reflected light is perpendicular to the refracted light. the reflected light is a completely polarised light. So. The circuit contains tuned circuit made of variable capacitor C and an inductor L in the collector circuit and hence is named as tuned collector oscillator. the mutual inductance of L and L¢ being M . . – + then due to concentration gradient. This field – + causes motion of electrons on p-side of the P N – + junction to n-side and motion of holes on n-side w of junction to p-side. the Electron diffusion Electron drift concentration of electrons is much greater as – + compared to concentration of holes. (a) Two important processes occur during the formation of a p-n junction are (i) diffusion and (ii) drift. Thus a drift current starts. When a p-n junction is formed. Rs Unregulated voltage (VL) IL Load RL Regulated voltage (Vz) Any increase/decrease in the input voltage results in increase/decrease of the voltage drop across Rs without any change in voltage across the zener diode. the holes Hole diffusion Hole diffuse from p side to n side ( p ® n) and Hole drift electrons diffuse from n side to p-side ( n ® p). OR (a) A transistor as an Oscillator: ` The circuit of a tuned collector in CE configuration is shown in figure. while in p n p-type semiconductor. This motion of charge carriers gives rise to diffusion current across the junction. (b) Zener diode is used as voltage regulator. Thus. the zener diode acts as a voltage regulator. the concentration of – + holes is much greater than the concentration Electron of electrons. Due to built in potential barrier an electric field directed from n-region to p-region + + is developed across the junction. (ii) Drift: The drift of charge carriers occurs due to Vo Potential barrier electric field. (i) Diffusion: In n-type semiconductor. The feed back coil L¢ connected to base circuit is mutually coupled with coil L (due to phenomenon of mutual induction). In practice L and L¢ form the primary and secondary coil of the transformer.378 Xam idea Physics—XII 29. Depletion This current is opposite to the direction of layer diffusion current. Fm = q ( v + B ) Total force. 1 f = 2p ( LC) These oscillations induce a small voltage in coil L¢ by mutual induction. When the collector supply voltage is switched on by closing switch S. This may be seen as follows: A phase shift of 180° is created between the voltages of L and L¢ due to transformer action. which is the required condition for a positive feed back. it discharges through coil L . B The equivalent gate for the given circuit is AND gate A B Truth table A 0 0 1 1 ® ® ® ® ® Y = AB B 0 1 0 1 Y 0 0 0 1 30. F = Fe + Fm ® ® ® ® F = q ( E + v ´ B) . thus maintaining oscillations. Fe = qE ® ® ® Magnetic force on particle. A part of this amplifier energy is used to meet losses taking place in oscillatory circuit to maintain oscillations in tank circuit and the balance is radiated out in the form of electromagnetic waves. collector current starts increasing and the capacitor C is charged. Thus the net phase becomes 360° (or zero). its frequency is same as that of resonant LC circuit but its magnitude depends on the number of turns in L¢ and coupling between L and L¢ . Due to positive feed back the energy fed back to the tank circuit is in phase with the generated oscillations.Examination Papers 379 The biasing is provided by emitter resistance R E and potential divider arrangement consisting of resistances R1 and R 2 . The feedback voltage is applied between the base and emitter and appears in the amplified form in the collector circuit. Positive Feed back. (a) Electric force on particle. The feed back applied in tuned collector oscillator circuit is positive. (b) Output y = A + B = A . When the capacitor attains maximum charge. This induced voltage is the feed back voltage. setting up oscillations of natural frequency. A further phase shift of 180° arises between base-emitter and collector circuit due to transistor action in CE configuration. Circuit Operation. The capacitor C1 connected in the base circuit provides low reactance path to the oscillations and the capacitor C E is the emitter by-pass capacitor so that the resistor R E has no effect on the ac operation of the circuit. B = A . ® ® . Let at any instant the normal to the plane of loop make an angle q with the direction of magnetic field B and I be the current in the loop. F3 and F4 respectively. ® ® ® ® N' b sin q ® ® ® ® ® ® Q F1 I q R Axis of loop or normal to loop B F3 P S F4 F1=IlB I (Upward) l oo p q b q ax is N of B I (Downward) F3=IlB According to Fleming’s left hand rule the forces F2 and F4 are equal and opposite but their line of action is same. Therefore each of the forces F2 and F4 acting on these sides has same magnitude F¢ = Blb sin ( 90° . therefore the magnitude of each of forces F1 and F3 is F = IlB sin 90° = IlB. the resultant of F2 and F4 ® ® is zero. Þ Þ Þ F =0 ® ® ® ® ® ® q ( E + v ´ B) = 0 Þ ® ® ® ® E = -( v ´ B ) E =B´ v E = Bv sin q = Bv ( when q = 90° ) E Þ v = (when v.q) = Blb cos q.380 Xam idea Physics—XII If a charge particle enter's perpendicular to both the electric and magnetic fields then it may happen that the electric and magnetic forces cancel each other and so particle will pass undeflected.e. F2 . each side of the loop will experience a force. E and B are mutually perpendicular) B N (b) Torque on a current carrying loop: Consider F2 a rectangular loop PQRS of length l. QR. RS and SP are F1 . Suppose that the forces on sides PQ. breadth b suspended in a uniform magnetic field B . Therefore these forces cancel each other i.q) with the direction of magnetic field. The sides QR and SP make angle ( 90° . The sides PQ and RS of current loop are perpendicular to the magnetic field. We know that a force acts on a current carrying wire placed in a magnetic field. The length of loop = PQ = RS = l and breadth = QR = SP = b. Therefore. ® In such a case. The net force and torque acting on the loop will be determined by the forces acting on all sides of the loop. . the perpendicular distance between F1 and F3 is b sin q.e. t = (Magnitude of one force F ) ´ perpendicular distance = ( BIl ) × ( b sin q) = I (lb) B sin q But lb = area of loop = A (say) \ Torque. The magnetic dipole moment of rectangular current loop = M = NIA ® ® ® \ t = M´B ® ® Direction of torque is perpendicular to direction of area of loop as well as the direction of magnetic field i. It exerts a force on RS. along I A ´ B . therefore the resultant force of F1 and F3 is zero. When the normal to plane of loop makes an angle q with the direction of magnetic field B . but they form a couple called the deflecting couple. t = IAB sin q If the loop contains N-turns.Ig Conversion of Galvanometer into Voltmeter A galvanometer may be converted into voltmeter by connecting high resistance (R) in series with the coil of galvanometer. I g is current in galvanometer for full scale deflection. then t = NI AB sin q ® ® ® ® ® In vector form t = NI A ´ B . If G is resistance of Ammeter galvanometer. then for conversion of galvanometer into ammeter of range I ampere. The force acting on length l of the conductor RS will be . If V volt is the range of voltmeter formed. but their lines of action are different. OR (a) Conversion of Galvanometer into Ammeter A galvanometer may be converted into ammeter by S using very small resistance in parallel with the galvanometer coil. then series resistance V R= -G Ig R G Voltmeter (b) The magnetic field produced by current I1 at any point on conductor RS is m I B1 = 0 2pd This field acts perpendicular to the conductor RS and points into the plane of paper. \ Moment of couple or Torque. the shunt is given by Ig S= G I .Examination Papers ® ® 381 According to Fleming’s left hand rule the forces F1 and F3 acting on sides PQ and RS are equal and opposite. The small resistance connected in G parallel is called a shunt. an equal force is exerted on the wire PQ by the field of conductor RS. 6 e0 5. the flux through each of the six faces of the cube will be same when charge q is placed at its centre. the force between them are attractive. By symmetry. . 6 e0 1 q Thus. 1 q \ fE = . the forces between the two conductors are repulsive. if accelerated with same speed. m I I F12 = 0 1 2 l 2pd P R P R b l F21 I1 Q r a B I2 S I1 Q B I2 a l b F12 S Thus. hence l p < le m Thus. 8. electron has greater de Broglie wavelength. when the currents in the two conductor are in the same direction. In single slit diffraction experiment fringe width is 2 lD b= d . CBSE (All India) Set–II 2. de Broglie wavelength (l) is given as h l= mv Given v p = v e where v p = velocity of proton v e = velocity of electron Since m p > m e From the given relation 1 l µ . When the currents flow in opposite directions.382 Xam idea Physics—XII F21 = I2l B1 sin90° Similarly. electric flux passing through two opposite faces of the cube = 2. Thus electron has greater de Broglie wavelength. therefore. length of wire remains same é æ d öù é æ 2d öù Thus. Intensity of diffraction pattern varies square of slit width.909A = Vrms 220 P= OR Given V = 280 sin 50 pt (a) Here. when the slit gets double. of turns Then. d 2d N N' ( ) . w = 50p 2pn = 50p n = 25Hz V 280 (b) Vrms = 0 = = 197.Examination Papers 383 If d is doubled. 11. Hence. VC – VA = 4E (ii) The direction of electric field is always towards the decreasing potential. Vrms = 220 V. V0 = 280.99V 2 2 V 197. it makes the intensity four times. dl = -ò E dl cos180° = E ò dl = E ´ 4 = 4E Hence. So. electric potential at C would be more VC > VA . (i) P = 200 W. we may directly move from A to C VC . N ´ ê2pç ÷ú = N ¢ ê2pç ÷ú 2 ë è øû ë è 2 øû N Þ N¢ = 2 1 2 Now. work done is independent of path followed . if accelerated with same speed. Thus size of central maxima is reduced to half. (i) As. the width of central maxima is halved. magnetic moment (m) = NIA where N = No.VA = Potential difference between A and C. m A = NIA A = NI ( pr A ) = NIpd 2 4 NI 1 2 Similarly m B = N ¢I AB = pr B = ( NIpd 2 ) 2 2 Þ R= 220 ´ 220 = 242W 200 16. We know. F = 50 Hz We know V 2 rms V 2 rms or R = R P (ii) P = Irms Vrms P 200 Irms = = 0.95 A I rms = rms = 40 W 2 17. C® ® A C A C A = -ò E .99 V = 4. the image formed by lens L2 is at infinity. So. The equivalent diagram of given electric circuit is 6V 0. 23.2 A B C 5W 15W 2V 30W 10W D 10W 0. It means that the object for L2 lies at its focus.2) = 6 – 2 Þ R = 5W When R = unknown value of resistance VAB = 5 × 0. f 1 = 15 cm v1 = 60 cm So. lens formula 1 1 1 .2 + R(0.= v1 u1 f 1 1 1 1 = + v1 15 -20 \ Þ mB 2 = mA 1 u1 = – 20 cm. distance between L1 and L2 is d1 = v1 + u 2 = 60 + 15 = 75 cm As the image formed by lens L 2 lies at infinity then. Ek = 5MeV ( 2e) ( Ze) 1 Ek = 4pe 0 r0 . Given Z = 75. u2 = 15 cm Hence. the distance between L 2 and L 3 does not matter.2 = 1volt 27. Given f 1 = f 2 = f 3 = 15cm We have.384 Xam idea Physics—XII 1 mB 2 2 = = Þ mA 1 1 4 19. the distance between L 2 and L 3 can have any value.2 A R A Effective resistance between A and B 1 1 1 1 = + + Þ R ¢ = 5W R ¢ 10 30 15 Applying Kirchhoff's Law 5 × 0. Hence.E of a-particle is converted into its potential energy.2) + 10(0. Let r0 be the distance of closest approach where the K. 51 = -13. We know æ 1 1 1 ö ÷ = Rç ç n2 n2 ÷ l A ø è B Here n A = 4. The fringe width is lD d If D (distance between slits and screen) is doubled.E is doubled.0. the distance of closest approach is halved if K. The frequency of electromagnetic waves does not change while travelling through a medium.7 ´ 10 -14 m As r0 µ Hence. – 1.6 ´ 10 -10 8 ´ 10 6 = 2.6 eV When electron undergoes transition from E A = . R = 1. 3. from equation (i) 13. then fringe width will be doubled.097 × 107 ç ç 2 . The total flux passing through the cube is q f= e0 b= 4. OR -13. . n B = 3.6 – 0.. E1 = -13.6 ´ 10 -19 ) 2 5 ´ 10 ´ 1.2 l 4 è3 l = 1875 nm ö ÷ ÷ ø m -1 CBSE (All India) Set–III 1.(i) En = eV n2 For n = 1.6 ´ 10 1 Ek 6 -19 = 18 ´ 75 ´ 1.6 nB 2 Þ nB = 3 Hence.Examination Papers 385 or r0 = r0 = 1 ( 2e)( Ze) 4pe 0 E k 9 ´ 10 9 ´ 75 ´ 2 ´ (1. electron transits from n = 4 to n = 3 which corresponds to Paschen series of hydrogen atom.85 = Þ nA = 4 nA 2 Similarly..85 eV Then. .0974 ´ 107 Then Þ æ 1 1 1 = 1.6 As we know. Base Current and Collector Current: Under forward bias of emitter-base junction. giving rise to a feeble base current ( I B ). +1V C B Action of n-p-n transistor 0V An NPN transistor is equivalent to two P-N junction diodes placed back to back with their very thin P-regions connected together. In the process an equal number of electrons leave the negative terminal of battery VCC and enter the positive terminal of battery VEE . This causes a current in collector circuit. Consequently the internal potential barrier at emitter-base junction neutralises while the width of depletion layer at collector-base junction increases. are attracted towards the positive terminal of collector battery VCC . the electrons in emitter and holes in base are compelled to move towards the junction. The hole released in the base region compensates the loss of hole neutralised by electrons. The two batteries VEE and VCC represent emitter supply and collector supply respectively. due to reverse biasing of collector-base junction. Depletion layer n E IE e– – IE + e– – + VEE O – + VCC B IB IB e– – + IC p n C IC The electrons crossing the base and entering the collector. The electron released is attracted by positive terminal of emitter battery VEE . thus the depletion layer of emitter-base junction is eliminated. The magnetic field at the centre due a circular coil of N turns and radius r carrying current I is m NI B= 0 2r The magnetic moment of the coil is m = NIA = NI ´ pr 2 10. The emitter-base junction is forward biased and the base-collector junction is reverse biased.386 Xam idea Physics—XII 9. The emitter is reverse biased and the collector is forward biased. In . The circuit diagram –2V E for the operation of NPN transistor is shown in fig. As soon as a hole (in P-region) combines with an electron. a covalent bond of crystal atom of base region breaks releasing an electron-hole pair. As the base region is very thin. The given transistor is p-n-p transistor. called the collector current. most electrons (about 98%) starting from emitter region cross the base region and reach the collector while only a few of them (about 2%) combine with an equal number of holes of base-region and get neutralised. Its direction in external circuit is from emitter to base. 12.. Emitter Current: When electrons enter the emitter battery VEE from the base causing base current or electrons enter the collector battery VCC from the collector causing collector current.68 A 220 Þ I rms = P Vrms OR We know. The process continues. the emitter current I E is the sum of base current I B and the collector current I C . This current is called the leakage current. (ii) leakage current ( I leakage ) due to minority charge carriers. i.98A = 25 2 [where R = 25W] V = V0 sin wt V = 70 sin 100 pt w = 100 p [Q w = 2pn] (a) V0 = 70 volt. Relation between Emitter. Vrms = 220 V. Base and Collector Currents: Applying Kirchhoff’s I law at terminal O . e. This is the fundamental relation between currents in the bipolar transistor circuit. I c = I nc + I leakage . causing the emitter current. collector current is formed of two components: (i) Current ( I nc ) due to flow of electrons (majority charge carriers) moving from emitter to collector. . f = 60 Hz We have 2 Vrms Þ R 220 ´ 220 R= = 322.Examination Papers 387 addition to this the collector current is also due to flow of minority charge carriers under reverse bias of base-collector junction. Thus. an equal number of electrons enter from emitter battery VEE to emitter.6 W 150 P= R= 2 Vrms P (ii) P = I rms Vrms I rms = 150 = 0. we get I E = I B + IC That is. Given 2pn = 100p n = 50 Hz V 70 (b) Vrms = 0 = 2 2 I rms = Vrms 70 / 2 = R 25 70 = 1. (i) P = 150 W. 22.e. The image formed by lens L.. Hence. dl = -ò E dl cos180° = E ò dl = E ´ 4 = 4E So. its object should be at focus. the electric potential will be more at C. Thus lens L2 should produce image at infinity. It is necessary to compensate for attenuation of signal in communication system. amplifies and retransmits it to receiver sometimes with a change in carrier frequency. Then. Given f 1 = f 2 = f 3 = 10 cm. (i) From the given diagram. picks up the signal from the transmitter. the distance between L 2 and L3 can have any value. i. The equivalent diagram of the given electrical circuit is as follows. 10V 0. (i) Transducer: Any device which converts one form of energy into another. focus of lens L2. Hence. u1 = – 15 cm 1 1 1 By lens formula . potential difference between A and C C® ® A C A C A VC – VA = -ò E . 26.= v1 u1 f 1 1 1 1 3 -2 1 = = Þ = + v1 10 -15 30 30 v1 = . (ii) Repeater: It is a combination of a receiver and a transmitter.e. is at 15 cm on the right side of lens L1which lies at 10 cm left of lens L2. Hence.388 Xam idea Physics—XII 16.. (i.VA = 4E (ii) The direction of electric field is in the decreasing potential.2 A A B 5W 10W 5V 30W 10W C 15W R 0. Hence. VC > VA 20. Hence. the distance between lens L2 and L3 does not matter. distance between L1 and L2 = 30 cm + 10 cm = 40 cm As the image formed by Lens L2 lies at infinity.2 A D . VC . (iii) Amplification: It is the process of increasing the amplitude of a signal using an electronic circuit. amplifier).30 cm Since. A repeater. for L3. the object must be at infinity. So for L2. final image is formed by lens L3 at focus. Examination Papers 389 The effective resistance between A and D 1 1 1 1 3 +1 + 2 = = + + ¢ R 10 30 15 30 Þ R¢ = 5 W Applying Kirchhoff Law. to find unknown R.2 = 1 volt .2 = + 10 – 5 R 1 + + 3 = +5 Þ R = 5W 5 Hence VAD = 5 ´ 0. 5 × 0.2 + R × 0.2 + 15 × 0. K2 . What are the values of (i) horizontal component of Earth’s magnetic field and (ii) angle of dip at this place? 3. × × × × × 5. orients itself vertically at a certain place on the Earth. Show on a graph the variation of the de Broglie wavelength (l) associated with an electron. 8. which (i) maintain the Earth’s warmth and (ii) are used in aircraft navigation. with the square root of accelerating potential (V). while the key K2 remain open and the key K1 closed? S Justify. How does focal length of a lens change when red light incident × P Q on it is replaced by violet light? Give reason for your answer. how will the position of the null point be affected it J A B (i) ‘X’ increases the value of resistance R in the set-up by keeping the key K1 closed and the e key K2 open? G (ii) ‘Y’ decreases the value of resistance S in the + – set-up. Is it a scalar or a vector? 9. Write the relationship between the size of a nucleus and its mass × S R number (A). 10. × × × × × 6. Predict the direction of the induced current in the loop. Two students ‘X’ and ‘Y’ perform an experiment on E K1 + – R potentiometer separately using the circuit given: Keeping other parameters unchanged. Maximum marks: 70 CBSE (Foreign) Set–I 1. Name the electromagnetic waves. The closed loop (PQRS) of wire is moved into a uniform magnetic field at right angles to the plane of the paper as shown in the figure. free to rotate in a vertical plane. If the length of the conductor is tripled by gradually stretching it. 4. how will (i) drift speed of electrons and (ii) resistance of the conductor be affected? Justify your answer.CBSE EXAMINATION PAPERS FOREIGN–2012 Time allowed: 3 hours General Instructions: As given in CBSE Examination Papers Delhi–2011. Why is the potential inside a hollow spherical charged conductor constant and has the same value as on its surface? 2. × × × × × × 7. Define dipole moment of an electric dipole. A conductor of length ‘l’ is connected to a dc source of potential ‘V’. A magnetic needle. keeping ‘V’ constant. Trace the path of a ray of light passing through a glass prism (ABC) as shown in the figure. Transmitting antenna m(t) Message signal X Y 16. Write two characteristic features to distinguish between n-type and p-type semiconductors. It experiences a torque of 0. It is subjected to a uniform ® magnetic field B directed perpendicular to its velocity. State Gauss’s law in electrostatic. A cube with each side ‘a’ is kept in an electric field given by ® $. In a given sample. Show that it describes a circular path. A short bar magnet of magnetic moment 0. C= 2 mF and R = 10 W. (as is shown in the figure) where C is a positive dimensional constant. The half lives of A and B are respectively 100 years and 50 years. find out of the value of the angle of emergence from the prism. OR How does a light emitting diode (LED) work? Give two advantages of LED’s over the conventional incandescent lamps. two radioisotopes. 13. Calculate the quality factor of a series LCR circuit with L = 2-0 H. Explain briefly how electromagnetic waves are produced by an oscillating charge.Examination Papers ® 391 11.9 J/T is place with its axis at 30° to a uniform magnetic field. Write the expression for its radius. A and B. 12. A 60° B C 17. How is the frequency of the em waves produced related to that of the oscillating charge? 14. If the refractive index of glass is 3. 18. Find the time after which the amounts of A and B become equal. (i) Calculate the magnitude of the magnetic field.063 J. A particle of charge ‘q’ and mass ‘m’ is moving with velocity V . Mention the significance of quality factor in LCR circuit. are initially present in the ratio of 1:4. Identify the boxes ‘X’ and ‘Y’ and write the their functions. Find out E =C´i . (ii) In which orientation will the bar magnet be in stable equilibrium in the magnetic field? 19. Figure shows a block diagram of a transmitter. 15. Draw a labelled diagram of a moving coil galvanometer and explain its working. and (ii) the net charge inside the cube A capacitor of 200pF is charged by a 300V battery. (ii) the equivalent resistance of the combination. e1. Write two characteristic features observed in photoelectric effect which support the photon picture of electromagnetic radiation. For an amplitude modulated wave. 23. e 2 and internal resistance r1 and r2 respectively are connected in parallel as shown in the figure. Determine the modulation index m. show that the total energy (E) of the electron in the stationary states can be expressed as the sum of kinetic energy (K) and potential energy (U). 22.m.r2 (iii) the potential difference between the points A and B. Give its physical significance. the maximum amplitude is found to be 10V while the minimum amplitude is 2V. Using Bohr’s postulates for hydrogen atom. of the combination. . (i) the electric flux through the cube.r1 Two cells of emfs e1 . 26. The battery is then disconnected and the charged capacitor is connected to another uncharged capacitor of 100 pF. Hence derive Snell’s law of refraction.Use Huygens’ geometrical construction to show the propagation of a plane wavefront from a rarer medium (i) to a denser medium (ii) undergoing refraction. I I Deduce the expressions for A B (i) the equivalent e.392 Xam idea Physics—XII y a a x z 20. 25. where K = –2U. Define a wavelength.f. 24. Hence deduce the expression for the total energy in the nth energy level of hydrogen atom. Deduce the relation = + for two thin lenses kept in f f1 f 2 contact coaxially. Calculate the difference between the final energy stored in the combined system and the initial energy stored in the single capacitor. What is the function of radial magnetic field inside the coil? 1 1 1 Define power of a lens. State clearly how this graph can be used to determine (i) Planck’s constant and (ii) work function of the material. Define modulation index. Draw a graph between the frequency of incident radiation (v) and the maximum kinetic energy of the electrons emitted from the surface of a photosensitive material. 21. Write its units. 27. and e2. how would this affect the size and intensity of the central diffraction band? Justify. (b) Describe briefly any two energy losses. (b) A conducting rod held horizontally along East-West direction is dropped from rest from a certain height near the Earth’s surface. Show for which regions in the plot. giving the reasons for their occurrence in actual transformers. Explain clearly why the secondary maxima go on becoming weaker with increasing n. Show that the intensity at a point where the path difference is l 3 is I 0 / 4. (b) Draw a plot of the transfer characteristic (V0 versus Vi) for a base-biased transistor in CE configuration. . (a) What is the effect on the interference fringes in a Young’s double slit experiment when (i) the separation between the two slits in decreased? (ii) the width of the source slit is increased? (iii) the monochromatic source is replaced by a source of white light? Justify your answer in each case. (a) Draw the circuit for studying the input and output characteristics of an n-p-n transistor in CE configuration. Describe briefly with the help of a circuit diagram how a zener diode works to obtain a constant dc voltage from the unregulated dc output of a rectifier. with the help of a labelled diagram. OR (a) State the principle of a step-up transformer. Explain. why does the refracted light have the same frequency as that of the incident light? 28. (ii) reflecting by a concave mirror. Show. how from the output characteristics. OR (a) Obtain the conditions for the bright and dark fringes in diffraction pattern due to a single narrow slit illuminated by a monochromatic source. Why should there be an induced emf across the ends of the rod? Draw a plot showing the instantaneous variation of emf as a function of time from the instant it begins to fall. (b) The intensity at the central maxima in Young’s double slit experimental set-up is I0. 29. 30. its working. (a) State the principle on which AC generator works. (i) passing through a biconvex lens. the information about the current amplification factor ( bac) can be obtained. OR Why is a zener diode considered as a special purpose semiconductor diode? Draw the I–V characteristic of a zener diode and explain briefly how reverse current suddenly increases at the breakdown voltage. the transistor can operate as a switch. (b) When the width of the slit is made double. (b) When monochromatic light is incident on a surface separating two media.Examination Papers 393 OR (a) Use Huygens’ geometrical construction to show the behaviour of a plane wavefront. Draw a labelled diagram and explain its working. 394 Xam idea Physics—XII CBSE (Foreign) Set–II Questions uncommon to Set–I 1. A and B. Calculate the torque acting on the needle. Identify the boxes ‘X’ and ‘Y’ and write their functions. A capacitor of 150 pF is charged by a 220 V battery. 12. . In a given sample. where V is the accelerating potential for the electron. two radioactive nuclei. Figure shows a block diagram of a detector for amplitude modulated signal. Mention the significance of quality factor in LCR circuit. Calculate the difference between the final energy stored in the combined system and the initial energy stored in the single capacitor. The battery is then disconnected and the charged capacitor is connected to another uncharged capacitor of 50 pF. The prism is made of glass with critical angle ic= 40°.8 × 10–2 J T–1 is placed at 30° with the direction of uniform magnetic field of magnitude 3 × 10–2 T. Trace the path of ray (P) of light passing through the glass prism as shown in the figure. 14. 7. 26. A magnetised needle of magnetic moment 4. What is the angle of dip at a place where the horizontal and vertical components of the Earth’s magnetic field are equal? 2. × × × × × × × × × P × × S × × × × × × × × × × × × × R × × × × × × × × Q × × × × × × × × 8. The half lives of A and B are respectively 25 years. Predict the direction of the induced current in the loop. Why is there no work done in moving a charge from one point to another on an equipotential surface? 10. The closed loop (PQRS) of wire is moved out of a uniform magnetic field at right angles to the plane of the paper as shown in the figure. Find the time after which the amounts of A and B become equal. Show on a graph variation of the de-Broglie wavelength (l) associated with the electron versus 1 / V . are initially present in the ratio of 4:1. A P 45° B C 13. Calculate the quality factor of a series LCR circuit with L = 4.0 H. C = 1 mF and R = 20 W. 15. 21. If the length of the conductor is doubled by gradually stretching it. A conductor of length ‘l’ is connected to a dc source of potential ‘V’. 2. Calculate the difference between the final energy stored in the combined system and the initial energy stored in the single capacitor. The battery is then disconnected and the charged capacitor is connected to another uncharged capacitor of 100 pF. how will (i) drift speed of electrons and (ii) resistance of the conductor be affected? Justify your answer. 15. What are isotopes? Give one example. Given that the prism is made of glass with critical angle 40°. Why do the equipotential due to a uniform electric field not intersect each other? 9.0 H. 5. A P 45° B C 11. Define modulation index. trace the path of the ray P incident normal to the face AC.5 J/T is placed with its axis is 30° to a uniform magnetic of 0. A right angle prism is placed as shown in the figure. A small magnet is pivoted to move freely in the magnetic meridian. At what place on the surface of the Earth will the magnet be vertical? 7. 14. Determine the modulation index m. Write the relation between de-Broglie wavelength (l) associated with the electron and its kinetic energy E. Mention the significance of quality factor in LCR circuit. Give its physical significance.1 T.Examination Papers 395 CBSE (Foreign) Set–III Questions uncommon to Set – I & II 1. For an amplitude modulated wave. the maximum amplitude is 4V. keeping ‘V’ constant. C = 4 mF and R = 20W. A short bar magnet of magnetic moment 0. . Calculate (i) the magnitude of the torque experienced and (ii) the direction on which it acts. A capacitor of 400pF is charged by a 100V battery. Calculate the quality factor of a series LCR circuit with L = 4. 27. l (ii) R =r . 2.396 Xam idea Physics—XII Solutions CBSE (Foreign) Set–I 1. the direction of induced current is such that it opposes its own cause of production. æ 1 1 1 ö 5. P = q (2 a ) eVt 1 (i) We know that v d = µ ml l When length is tripled. we can say by replacing red light with violet light. (a) 0° (b) 90° 3. We know l = Å V \ l V = constant The nature of the graph between l and V is hyperbola. The relationship is 1 3 R = R0 A where R = Radius of nucleus A = Mass number 1. So. By Lenz’s law. there is no potential difference between any two points inside or on the surface of the conductor. V 8.22 7. 6. l¢ = 3 l A . It is a vector quantity. We know = (m -1)ç ÷ f è R1 R 2 ø fµ 1 and m v > m R (m -1) The increase in refractive index would result in decrease of focal length of lens. The induced current opposes the increase in magnetic flux. the drift velocity becomes one-third. Hence the direction of induced current is PSRQP (anticlockwise). ® ® 9. Hence. Electric field intensity is zero inside the hollow spherical charged conductor. no work is done in moving a test charge inside the conductor and on its surface. Dipole moment of an electric dipole is the product of either of charge and the length of dipole. (i) Infrared rays (ii) Microwaves. 4. decreases the focal length of the lens used. Therefore. When a particle of charge ‘q’ of mass ‘m’ is directed to move perpendicular to the uniform magnetic field ‘B’ with velocity ‘ v ’ The force on the charge ® ® ® × × × × × × × × × × q F × × r F × × × × v × v × × × × × ® F = q ( v ´ B) This magnetic force acts always perpendicular to the velocity of charged particle. We have N = N 0 e . Q= R 1 = 10 P path followed by the charge. we can write N A = N 0 e . qv B = Þ r= r qB Here r = radius of the circular 1 12. These oscillating electric and magnetic field. so potential gradient decreases.lA t A N B = 4N 0 e . mv mv 2 Then. 13. Hence magnitude of velocity remains constant but direction changes continuously. Consequently the path of the charged particle in a perpendicular magnetic field becomes circular.lt For radio isotopes A and B.. An oscillating change produces an oscillating electric field in space which further produces an oscillating magnetic field which in turn is a source of electric field.. Hence a greater length of wire would be needed for balancing the same potential difference. keep on regenerating each other and an electromagnetic wave is produced The frequency of em wave = Frequency of oscillating charge 14.(ii) Let t be the time after which N A = N B t A = t B = t( say) . 11.. An oscillating or accelerated is supposed to be source of an electromagnetic wave.Examination Papers 397 New resistance l¢ 3l = r´ = 9R R¢ = r ¢ A A3 R' = 9R Hence. We have.lB t B . So the null point would shift towards B.. (ii) By decreasing resistance S. (i) By increasing resistance R the current through AB decreases.(i) . potential gradient does not change. the new resistance will be 9 times the original. As K2 is open so there is no effect of S on null point. hence. the current through AB remains the same. L C 2 = 100 2 ´ 10 -6 It signifies the sharpness of resonance. 10. The magnetic force (q v B) provides the necessary centripetal force to move along a circular path. Given n g = 3 i=0 At the interface AC.lA t = 4 N 0 e . n–type Semiconductor (i) It is formed by doping pentavalent impurities. hence r = 0 At the interface AB. OR A light emitting diode is simply a forward biased p-n junction which emits spontaneous light radiation. X ® Amplitude Modulator Y ® Power Amplifier Function of X: The original message signal has very small energy and dies out very soon if transmitted directly as such. (ii) Fast action and no warm up time required. Function of Y: The signal cannot be transmitted as such because they get weaken after travelling long distance. use of power amplifier provides them necessary power before feeding the signal to the transmitting antenna. A 16.l A t) log e e é log 2 log e 2 ù log e 2 Þ 2 log e 2 = ê e Ql = út TA1 / 2 ú T ê ë TB1 / 2 û 1 1 ö 2 -1 ö Þ 2 =æ Þ 2 =æ ç ÷t ç ÷t è 50 100 ø è 100 ø Þ t = 200 years 15. At the junction. By Snell’s Law ng sin i = sin r na But sin i = sin 0° = 0. i = 30° Applying Snell’s Law sin 30° n a 1 = = Þ sin e = 3 sin 30° sin e ng 3 17. .398 Xam idea Physics—XII \ Þ N 0 e . energy is released in the form of photons due to the recombination of the excess minority charge carrier with the majority charge carrier. these signals are modulated by mixing with very high frequency waves (carrier wave) by modulator power. Advantages (i) Low operational voltage and less powder.lB t Þ 4 = e lB t . 30° e 60° B C Þ e = 60° (iii) The electrons are majority carriers and holes are (ii) The holes are majority carriers and electrons are minority carriers ( ne >> nh ). Hence. Hence. p–type Semiconductor (i) It is doped with trivalent impurities. minority carriers (nh >> ne ).lA t log e 4 = (l B t . The charge would be shared between them.063 = 0. dS = 2aC dS cos 0° = 2 aC ´ a 2 = 2a 3 C f2 = aC ´ a 2 cos 180° = -a 3 C f = 2a 3 C + ( -a 3 C) = a 3 C Nm 2 C -1 (iii) Net charge (q) = e 0 ´ f = a 3 C e 0 coulomb q = a 3 C e 0 coulomb. dS = s q e0 Ù n f2 a a f1 Ù n (ii) Net flux f = f1 + f2 ® ® where f1 = E . (i) We know t = M ´ B t = M B sin q 0.9 × B × sin 30° or B = 0. The energy (U) = –MB cos q When q = 0° Þ U = –MB = Minimum energy Hence. Q = q + q¢ .14 T (ii) The position of minimum energy corresponds to position of stable equilibrium. (i) Gauss’s Law in electrostatics states that the total electric flux through a closed surface 1 enclosing a charge is equal to times the magnitude of that charge. = C C¢ q ® charge on capacitor (first) q' ® charge on capacitor (second) C = 200 pF. it is the state of stable equilibrium. when the bar magnet is placed parallel to the magnetic field.Examination Papers ® ® ® 399 18. Initial energy of capacitor (Ui ) = CV 2 2 1 Ui = ´ 200 ´ 10 -12 ´ (300) 2 = 9 × 10–6J 2 Charge on capacitor ‘Q’ = CV = 200 × 10–12 × 300 = 6 × 10–8 C When both capacitors are connected then let V be common potential difference across the two capacitors. f = ò E . e0 ® ® or 19. C ¢ = 100 pF q q¢ = Þ q = 2q¢ -12 200 ´ 10 100 ´ 10 -12 Q 60 nC Then = 20 nC Q = 2q¢ + q¢ = 3q¢ Þ q' = = 3 3 . q q¢ Hence. 1 20. given by t = NIAB sin q where q is the angle between the normal to plane of coil and the magnetic field of strength B. torque t acts on the coil. Principle and working: When current ( I ) is passed in the coil. H Suspension wire M Coil NIBl b T1 T2 N S N S N S NIBl Coiled strip (b) Magnetic lines of force of radial magnetic field (a) (c) The pole pieces of the permanent magnet are cylindrical so that the magnetic field is radial at any position of the coil. The levelling screws are also provided at the base of the instrument. A mirror ( M ) is fixed on the phosphor bronze strip by means of which the deflection of the coil is measured by the lamp and scale arrangement.400 Xam idea Physics—XII and q = 2q' = 40 nC q 2 q¢ 2 + 2C 2C ¢ Hence. A soft iron core in cylindrical form is placed between the coil. . N is the number of turns in a coil. The other end of coil is connected to a loosely coiled strip. which serves as the other terminal (T2 ).9 ´ 10 -6 = -3 ´ 10 -6 J Þ DU = 3 ´ 10 -6 J (in magnitude) 21.Ui = 6 ´ 10 -6 . total final energy U f = 1 ( 40 ´ 10 -9 ) 2 1 ( 20 ´ 10 -9 ) 2 Uf = ´ + ´ 2 200 ´ 10 -12 2 100 ´ 10 -12 U f = 6 ´ 10 -6 J Energy difference ( DU ) = U f . One end of coil is attached to suspension wire which also serves as one terminal (T1 ) of galvanometer. The other end of the suspension is attached to a torsion head which can be rotated to set the coil in zero position. Moving coil galvanometer: A galvanometer is used to detect current in a circuit. Construction: It consists of a rectangular coil wound on a non-conducting metallic frame and is suspended by phosphor bronze strip between the pole-pieces ( N and S ) of a strong permanent magnet. . 1 1 1 1 . then in every position of coil the plane of the coil. so that q = 90° and sin 90° = 1 Deflecting torque.= . then restoring torque =C q For equilibrium.(iv) . NIAB = C q NAB .= + v u f1 f 2 .e. t = NIAB If C is the torsional rigidity of the wire and q is the twist of suspension strip. the lenses are thin. The lens A produces an image at I1 which serves as a virtual object for lens B which produces final image at I.Examination Papers 401 When the magnetic field is radial.. we have ... 22.(i) \ q= I C i. The optical centres (P) of the lenses A and B is co-incident.. A B O P v I I1 u v1 An object is placed at point O. we have 1 1 1 . then 1 1 1 .= v1 u f 1 1 1 1 For lens B.= v u f . For lens A..(iii) If two lenses are considered as equivalent to a single lens of focal length f.(ii) v v1 f 2 Adding equations (i) and (ii)...(i) . Power of lens: It is the reciprocal of focal length of a lens. 1 P = (f is in metre) f Unit of power of lens: Diopter.. Given. deflecting torque = restoring torque i.e.. qµI deflection of coil is directly proportional to current flowing in the coil and hence we can construct a linear scale. as in the case of cylindrical pole pieces and soft iron core. is parallel to the magnetic field lines. A min 10 . Maximum amplitude. For cell e1 e -V Then. A max + A min 10 + 2 12 3 25. I = I1 + I 2 . V = e1 . the Planck constant can be calculated by the slope of the current q X D( KE) n0 h= n(Hz) Dv (ii) Work function is the minimum energy required to eject the photo-electron from the metal surface.. where v0 = Threshold frequency 24. of electron (eV) . A m= m Ac 23. (i) From this graph.2 8 2 Modulation index = max = = = . (b) Photons are electrically neutral and are not affected by presence of electric and magnetic fields. for cell e 2 I2 = 2 r2 Putting these values in equation (i) e -V e 2 -V I= 1 + r1 r2 or e ö æe æ1 1 ö I = ç 1 + 2 ÷ -V ç + ÷ è r1 r 2 ø è r1 r 2 ø K. Physical significance: It signifies the level of distortion or noise. Here. Amax = Ac + Am = 10 V Minimum aptitude.E. Amin = Ac – Am = 2 V A . r1 Let V = Potential difference between A and B. we can write 1 1 1 = + f f1 f 2 (a) All photons of light of a particular frequency ‘n’ have same energy and momentum whatever the intensity of radiation may be. f = hv 0 .402 Xam idea Physics—XII From equation (iii) and equation (iv). Modulation index: It is the ratio of peak value of modulating signal to the peak value of carrier wave. r2 I B ..I 1 r1 Þ I 1 = 1 r1 e -V Similarly. A lower value of modulation index indicates a lower distortion in the transmitted signal.(i) I1 I A I2 . Iç ÷ è r1 + r 2 ø è r1 + r 2 ø Comparing the above equation with the equivalent circuit of emf ‘eeq’ and internal resistance ‘req’ then. .. For electron revolving in an orbit of radius r hydrogen (z = 1) atom with speed v.(iii) Total energy = KE + PE æ 1 e2 1 e2 ö ÷ En = +ç÷ 2 4e 0 r ç 4 pe r 0 è ø En = 1 e2 2 4 pe 0 r . we have En = me 4 8n 2 e 0 2 h 2 r = n2 h2 e0 2 . ....(ii) The potential energy is due to the presence of charge (+e) on the nucleus and is given by PE = Potential × charge = 1 e 1 e2 .Examination Papers 403 æ e r + e 2 r1 ö æ r1r 2 ö .(iii) V = e eq . mv 2 1 e2 = r 4pe 0 r 2 Þ KE = 1 1 e2 mv 2 = 2 2 4pe 0 r (from first postulate of Bohr’s atom model) ..(ii) V =ç 1 2 ÷ .Ir eq Then e r + e 2 r1 rr (i) e eq = 1 2 (ii) r eq = 1 2 r1 + r 2 r1 + r 2 or (iii) The potential difference between A and B V = e eq ....I r eq 26.. Energy of an electron in the stationary orbits of the hydrogen atom can be obtained by adding its kinetic and potential energies.(iv) According to Bohr’s second postulate nh nh mvr = Þ v= 2p 2pmr From equation (i) m n2h2 1 e2 = r 4p 2 m 2 r 2 4pe 0 r 2 Þ mpe 2 Substituting the value of r in equation (iv).. (–e) = 4pe 0 r 4pe 0 r .(v) ... from. B i X A A' 90o i r 90 o B' r Y Suppose a plane wavefront AB in first medium is incident obliquely on the boundary surface XY and its end A touches the surface at A at time t = 0 while the other end B reaches the surface at point B¢ after time-interval t. The envelope of all wavelets at a given instant gives the position of a new wavefront. Proof of Snell’s law of Refraction using Huygen’s wave theory : When a wave starting from one homogeneous medium enters the another homogeneous medium. which traverses a distance AA¢ ( = v 2 t) in second medium in time t. Huygen’s Principle : There are some phenomena like interference. A¢ B ¢ and surface XY are in the plane of paper. Let v1 and v 2 be the speeds of waves in these media. the secondary wavelets start from points between A and B¢ . one after the other and will touch A¢ B¢ simultaneously. The locus of particles of a medium vibrating in the same phase is called a wavefront. the frequency of wave remains unchanged but its speed and the wavelength both are changed. For a point source. Clearly BB¢ = v1t. As the wavefront AB advances. Each point of a wavefront acts as a source of secondary wavelets. As the incident wavefront AB advances. This is the first law of refraction. In transversing from first medium to another medium. The distant wavefront is plane. . diffraction and polarisation which could not be explained by Newton’s corpuscular theory. it is deviated from its path. The assumptions of Huygen’s theory are : (i) A source sends waves in all possible directions. Wavefront : A wavefront is a locus of all particles of medium vibrating in the same phase. the wavefront is spherical. the point of wavefront traverses a distance BB¢ ( = v1t) in first medium and reaches B¢ . therefore the perpendicular drawn on them will be in the same plane. secondary spherical wavelets originate from these points. therefore we may say that the incident ray. where the secondary wavelet now starts. Hence A¢ B¢ will be the refracted wavefront. They were explained by wave theory first proposed by Huygen. Clearly BB¢ = v1t and AA¢ = v 2 t. which travel with speed v1 in the first medium and speed v 2 in the second medium. refracted ray and the normal at the point of incidence all lie in the same plane. Second law : Let the incident wavefront AB and refracted wavefront A¢ B¢ make angles i and r respectively with refracting surface XY. This phenomenon is called refraction. Let XY be a surface separating the two media ‘1’ and ‘2’. while for a line source the wavefront is cylindrical. In the same time-interval t. First of all secondary wavelet starts from A. According to Huygen’s principle. According to Huygen’s principle A¢ B¢ is the new position of wavefront AB in the second medium. it strikes the points between A and B¢ of boundary surface.404 Xam idea Physics—XII 27. Assuming A as centre. we draw a spherical arc of radius AA¢ ( = v 2 t) and draw tangent B¢ A¢ on this arc from B¢ . As the lines drawn normal to wavefront denote the rays. First law : As AB. therefore the incident and refracted rays make angles i and r with the normal drawn on the surface XY i. When a light wave travels from rarer to denser medium. For a point source.(iii) As the rays are always normal to the wavefront. Ð AA¢ B ¢ = 90° AA¢ v 2 t \ sin r = sin Ð AB¢ A¢ = = AB¢ AB¢ Dividing equation (i) by (ii).. This is the second law of refraction.. i and r are the angle of incidence and angle of refraction respectively. This is because energy of wave depends on its frequency and not on its speed.. while for a line source the wavefront is cylindrical. It does not imply reduction its energy. Rectilinear Propagation of Light : According to Newton’s corpuscular theory.. we get v sin i = 1 = constant sin r v2 .(ii) .. and is called the Snell’s law. (i) The action of a convex lens : A plane wavefront becomes spherical convergent wavefront after refraction.(i) .Examination Papers 405 In right-angled triangle AB¢ B. The envelope of all wavelets at a given instant gives the position of a new wavefront. but according to wave theory the rectilinear propagation of light is only approximate. Plane wavefront Spherical wavefront Lens . The distant wavefront is plane. According to equation (3) : The ratio of sine of angle of incidence and the sine of angle of refraction is a constant and is equal to the ratio of velocities of waves in the two media. the wavefront is spherical. diffraction and polarisation which could not be explained by Newton’s corpuscular theory. the path of light is a straight line. The locus of particles of a medium vibrating in the same phase is called a wavefront. OR (a) Wave Nature of Light : Huygen’s Theory There are some phenomena like interference. (ii) Each point of a wavefront acts as a source of secondary wavelets.e. the speed decreases. Ð ABB¢ = 90° vt BB¢ \ sin i = sin Ð BAB¢ = = 1 AB¢ AB¢ Similarly in right-angled triangle AA¢ B ¢ . They were explained by wave theory first proposed by Huygen. The assumptions of Huygen’s theory are : (i) A source sends waves in all possible directions.. Fig. starting from different points of illuminated slit. When the source slit is made wide which does not fullfil the above condition and interference pattern not visible. the phase difference. lD (a) (i) Fringe width (b) = d If d decreases. light reflects and refracts due to the interaction of incident light with the atoms of the medium. Explanation : Let AB be a slit of width ‘a’ and a parallel beam of monochromatic light is incident on it. (b) Intensity at a point is f I = I 0 cos 2 2 l 2p l 2p At path difference . b increases (ii) For interference fringe. (b) As frequency of light is the characteristic of its source. Let q be the angle of diffraction for waves reaching at point P of screen and AN the perpendicular dropped from A on wave diffracted from B. the intensity of central band is maximum and goes on decreasing on both sides. we get diffraction pattern on a screen placed behind the slit. fringe pattern gets less sharp or faint. frequency remains same. On the either side of the central white fringe the coloured bands (few coloured maxima and minima) will appear. The diffraction pattern contains bright and dark bands. f = . If the source slit width increases. The path difference between rays diffracted at points A and B. D = distance of source from slits. According to Fresnel the diffraction pattern is the result of superposition of a large number of waves. These atoms always take up the frequency of the incident light which forces them to vibrate and emit light of same frequency. (iii) The central fringes are white.AP = BN . = 3 l 3 3 I 1 2 p p æ ö æ ö \ I = I 0 cos 2 ç ÷ = I 0 cos 2 ç ÷ = 0 2è 3 ø è3ø 4 OR (a) Diffraction of light at a single slit : When monochromatic light is made incident on a single slit. the condition is s l < D d where s = size of source. Hence. Fig. This is because fringes of different colours overlap.406 Xam idea Physics—XII (ii) Action of concave mirror : A plane wavefront becomes spherical convergent after reflection. Plane wavefront Spherical wavefront mirror 28. D = BP . but the main features may be explained by simple arguments given below : P A Light from source M1 q C q O M2 90o N B q At the central point C of the screen. . Similarly it may be shown that the intensity is zero for nl sin q = .. Thus the general condition of minima is a .(ii) a a Minima : Now we divide the slit into two equal halves AO and OB. each with a different phase. we have to sum up their contribution..Examination Papers 407 In D ANB . This was done by Fresnel by rigorous calculations.. Hence the waves starting from all points of slit arrive in the same phase.(i) D = a sin q To find the effect of all coherent waves at P.. Now for 2 a every point... Thus equation (2) gives the angle of diffraction at which intensity falls to zero. M 1 in AO. with n as integer. If point P on screen is such that the path difference between rays starting from edges A and B is l . This gives maximum intensity at the central point C. there is a corresponding point M 2 in OB. sin q = q = . then path difference l a sin q = l Þ sin q = a l If angle q is small. This means that the contributions from the two halves of slit AO and OB are 2 2 opposite in phase and so cancel each other.. the angle q is zero. Then 2 path difference between waves arriving at P and starting from M 1 and M 2 will be a l sin q = .(iii) a sin q = nl Secondary Maxima : Let us now consider angle q such that . ÐANB = 90° \ and ÐBAN = q BN or BN = AB sin q \ sin q = AB As AB = width of slit = a \ Path difference. such that M 1 M 2 = . each of width . while in the case of large power dynamo. (ii) Armature: It consists of a large number of turns of insulated wire in the soft iron drum or ring. sin q = q = . width of central maxima becomes half. intensity gets four times. the magnetic field is produced by an electromagnet... If we take the first two of parts of slit. For n = 2. Clearly there will be a maxima between first two minima.. (iv) Brushes: These are two flexible metal plates or carbon rods ( B1 and B 2 ) which are fixed and constantly touch the revolving rings. the magnetic field is generated by a permanent magnet.408 Xam idea Physics—XII 29. the contribution of slit decreases. but this maxima will be of much weaker intensity than central maximum. It width of slit is doubled. The drum or ring serves the two purposes: (i) It serves as a support to coils and (ii) It increases the magnetic field due to air core being replaced by an iron core. In general the position of nth maxima will be given by 1 ö . It can revolve round an axle between the two poles of the field magnet. In a similar manner we can show that there are secondary maxima between any two consecutive minima. (a) AC generator: A dynamo or generator is a device which converts mechanical energy into electrical energy. 2. The output current in external load R L is taken through these brushes. It is based on the principle of electromagnetic induction. 3l 2a which is midway between two dark bands given by l 2l sin q = q = and sin q = q = a a Let us now divide the slit into three parts. This is called first secondary maxima. 4. 3. for n = 3. (iii) Slip Rings: The slip rings R1 and R 2 are the two metal rings to which the ends of armature coil are connected.. it is one-seventh and so on. Intensity varies as square of slit width. it is one-fifth. .. 2Dl (b) Width of central Maxima ' b' = a a ® size of slit If size of slit is doubled. Construction: It consists of the four main parts: (i) Field Magnet: It produces the magnetic field. the path difference between rays diffracted from the extreme ends of the first two parts 2 2 3l a sin q = a ´ =l 3 3 2a l Then the first two parts will have a path difference of and cancel the effect of each other. ] ç n + ÷ l. 2 ø è The intensity of secondary maxima decrease with increase of order n because with increasing n. and the intensity of maxima will go on decreasing with increase of order of maxima.(iv) a sin q = æ [ n = 1. These rings are fixed to the shaft which rotates the armature coil so that the rings also rotate along with the armature. 2 The remaining third part will contribute to the intensity at a point between two minima. In the case of a low power dynamo. A area of coil and B the magnetic induction. the current flows along N B S B1 R L B 2 . In the external circuit. the change in magnetic flux takes place. These coils are kept insulated from each other and from the iron-core. its direction being given by Fleming’s right hand rule. on which two coils of insulated copper wire are separately wound. The source of energy generation is the mechanical energy of rotation of armature coil. This is based on the principle of mutual-induction. because the average value of ac over full cycle is zero. The number of turns in these coils are different. Out of these coils one coil is called primary coil and other is called the secondary coil. e ( Q u = 0) v = gt Induced emf. Considering the w armature to be in vertical position and as it Armature coil rotates in anticlockwise direction. (b) As the earth’s magnetic field lines are cut by the falling rod. e = Blv e = Blgt \ e µt t OR (a) Transformer: Transformer is a device by which an alternating voltage may be decreased or increased. but are coupled through mutual induction. f the frequency of rotation. Construction: It consists of laminated core of soft iron. Current produced by an ac generator cannot be measured by moving coil ammeter. During the second half revolution. The direction of current remains Field magnet unchanged during the first half turn of a armature. the magnetic flux linked with the coil changes and the current is induced in the coil. the emf produced is alternating and hence the current is also alternating. The terminals of primary coils are connected to AC mains and the terminals of the secondary coil are connected to external circuit in which alternating current of desired voltage is required. d the wire ab moves downward and cd upward.Examination Papers 409 Working: When the armature coil is rotated in the strong magnetic field. Thus Slip rings Brushes Load the direction of induced emf and current RL changes in the external circuit after each half R2 B2 revolution. Transformers are of two types: . Expression for Induced emf: If N is the number of turns in coil. This change in flux induces an emf across the ends of the rod. so that the c direction of induced current is shown in fig. then induced emf df d e== {NBA (cos 2p f t )} dt dt = 2p NBA f sin 2p f t Obviously. Since the rod is falling under gravity. the wire ab b moves upward and cd downward. B1 so the direction of current is reversed and in R1 external circuit it flows along B 2 R L B1 . According to Faraday’s laws the emf induced in the primary coil Df .. e.C.(i) ep =-Np Dt and emf induced in the secondary coil Df . Step up Transformer: It transforms the alternating low voltage to alternating high voltage and in this the number of turns in secondary coil is more than that in primary coil..(iii) = ep Np If the resistance of primary coil is negligible. Similarly if the secondary circuit is open. Let N p be the number of turns in primary coil... due to which the magnetic flux linked with the secondary coil changes continuously. therefore the alternating emf of same frequency is developed across the secondary. e. mains) Primary Primary laminated iron core Core Secondary Primary Secondary (a) Step up Secondary Transformer (b) Step down Working: When alternating current source is connected to the ends of primary coil.(ii) eS = . will be equal to the applied potential difference (V p ) across its ends... therefore VS e S N S . (i.NS Dt From (i) and (ii) eS NS . N S > N p ). then the potential difference VS across its ends will be equal to the emf ( e S ) induced in it...410 Xam idea Physics—XII 1. 2. the current changes continuously in the primary coil. NS the number of turns in secondary coil and f the magnetic flux linked with each turn. N S < N p ) (A.. the emf ( e p ) induced in the primary coil.. We assume that there is no leakage of flux so that the flux linked with each turn of primary coil and secondary coil is the same.(iv) = = = r (say) Vp e p N p . Step down Transformer: It transforms the alternating high voltage to alternating low voltage and in this the number of turns in secondary coil is less than that in primary coil (i. Fig. So (b) i. Power in primary = Power in secondary V p i p = VS i S iS V p N p 1 = = = i p VS N S r N s > N p ® r > 1. CE = 10 IB V . If i p and i s are the instantaneous currents in Np primary and secondary coils and there is no loss of energy.. VS > V p and i S < i p \ . the collector current IC increases as VCE increases from 0 to 1 V only and then the collector current becomes almost constant and independent of VCE. The characteristic curves show: (i) When collector-emitter voltage VCE is VBE increased from zero. 30. not all of the flux due to primary passes through the secondary due to poor design of the core or the air gaps in the core.Examination Papers 411 where r = NS is called the transformation ratio. + µA VBB + – Rh1 + VBE – B E IE – IB C + VCE – – mA – IC + + VCC Output characteristics: These characteristics are obtained by plotting collector current IC versus collector-emitter voltage VCE at a fixed value of base current I B .. The value of VCE upto which collector current IC changes is called the knee voltage Vknee . (ii) Eddy currents: The alternating magnetic flux induces eddy currents in the iron core and causes heating.(v) In step up transformer. (a) Characteristic Curves: The circuit diagram for determining the static characteristic curves of an n-p-n transistor in common-emitter configuration is shown in figure. that is. The base current is changed to some other fixed value and the observations of I C versus VCE are VCE = 5V repeated. (i) Flux leakage: There is always some flux leakage.e. represents the output characteristics of a V common-emitter circuit. then For about 100% efficiency. step up transformer increases the voltage. æ DI ö ( I ) .3 b= = = 210 ( 40 .5 0.5 3 3.( I C ) 1 Then b =ç C ÷ = C 2 è DI B ø ( I B ) 2 .30) mA 10 ´ 10 . Now we choose any two characteristic curves for given values of I B and find the two corresponding values of I C . C RB IB B E + Vi – VBB 1 VBE 2 VCC RL VCE + VO – IC DI C .5 4 Base current (IB) 60 µA 50 µA 40 µA 30 µA 20 µA 10 µA Collector to emitter voltage (FCE) in volts Determination of Current Gain æ DI ö Current gain b = ç C ÷ è DI B øV CE We take the active region of output characteristics i.5 × 2) mA 2 ×1 ´ 10 .5 10 7. the region where collector current ( I C ) is almost independent of VCE .5 5 2.5 1 1. ( I B ) 2 = 40 mA (7 × 3 . DI B . A transistor can be used to turn current ON or OFF rapidly in electrical circuits.( I B ) 1 From graph ( I C ) 1 = 5 × 2 mA.6 Using any two curves from output characteristics current amplification factor b ac = (b) A switch is a device which can turn ON and OFF current is an electrical circuit.412 Xam idea Physics—XII Collector current in (IC) mA 12.5 2 2. ( I C ) 2 = 7 × 3 mA ( I B ) 1 = 30 mA. the barrier voltage across base-emitter junction is 0 × 6 V. . Thus. very near to zero. The switching circuits are designed in such a way that the transistor does not remain in active state. there is no collector current ( I C = 0). the change in collector current and V0 hence in output voltage V0 is non-linear and the Saturation region transistor goes into saturation. when Vi is less than 0 × 6 V. thus showing that the transitions (i) from cut off to active state and from active to saturation state are not sharply defined. so the transistor is ‘switched OFF’ and if it is high enough to derive the transistor into saturation ( I C is high and so V0 ( = VCC . the output voltage further decrease towards zero (though it never becomes zero). we get the graph as shown in fig. it is said to be ‘switched off’ but when it is conducting ( I C is not zero). AV If we plot V0 versus Vi . The transistor can operate as a switch in cut off region and saturation region.Examination Papers 413 Operation: The circuit diagram of n-p-n transistor in CE configuration working as a switch is shown in fig.] Vi 1V The curve shows that there are non-linear regions. Beyond Vi = 1 V. we get …(i) VBB = I B R B + VBE and …(ii) VCC = I C R L + VCE We have VBB = Vi and VCE = V0 . When Vi becomes greater than 0 × 6 V. so the transistor is ‘switched ON’. Thus we can say low input switches the transistor is OFF state and high input switches it ON.I C R L ) is low. Now we are in the position to explain the action of transistor as a switch. the increase in I C is linear and so decrease in output voltage V0 is linear. Therefore. so above equations take the form …(iii) Vi = VBE + I B R B and …(iv) V0 ( = VCE ) = VCC . As long as input voltage Vi is low and unable to overcome the barrier voltage of the emitter base junction. The dc output voltage is taken across collector-emitter terminals. V0 decreases upto Vi = 1 V. R L is the load resistance in output circuit. so transistor will be in cut off state. VBB and VCC are two dc supplies which bias base-emitter and emitter collecter junctions respectively. With further increase in Vi . This is also input dc voltage (VC ).6 V (i) between cut off state and active state and (ii) between active state and saturation state. In case of Si transistor. Hence. I C begins to Cut off Active flow and increase with increase of Vi . Let VBB be the input supply voltage. V0 is high ( I C = 0 and V0 = VCC ). from (iv) with I C = 0. When transistor is non-conducting ( I C = 0). it is said to be ‘switched ON’. [This characteristics curve is also called transfer characteristic curve of base biased transistor. 0. V0 = VCC . Applying Kirchhoff’s second law to input and output meshes (1) and (2).I C R L Let us see the change in V0 due to a change in Vi . from region region (iv). We know that reverse current is due to the flow of electrons (minority carriers) from p ® n and holes from n ® p.414 Xam idea Physics—XII OR A zener diode is considered as a special purpose semiconductor diode because it is designed to operate under reverse bias in the breakdown region. I (mA) Reverse bias Vz Forward bias V(V) I (µA) Rs Unregulated voltage (VL) IL Load RL Regulated voltage (Vz) CBSE (Foreign) Set–II 1. Thus. the zener diode acts as a voltage regulator. then the electric field strength is high enough to pull valence electrons from the host atoms on the p-side which are accelerated to n-side. l = 2 meV . We know BV = tan d BH Given BV = B H then tan d = 1 Angle of dip. Now. d = 45° h 2. When the reverse bias voltage V = Vz. any increase/decrease in the input voltage results in increase/decrease of the voltage drop across R s without any change in voltage across the zener diode. As the reverse bias voltage is increased. Working: The unregulated dc voltage output of a rectifier is connected to the zener diode through a series resistance R s such that the zener diode is reverse biased. the electric field at the junction becomes significant. These electrons causes high current at breakdown. R By right hand thumb rule.÷t 25 50 ø è t = 100 years. the induced current will flow in a clockwise × S × × direction. angle of refraction will be zero. By P Q Lenz’s law.lA t A N B = N 0 e . . i. We have N = N 0 e .. It means that the ray of light will pass through the prism undeviated. the work done in moving a charge on an equipotential is zero. 8. the induced current opposes the change in flux by producing × × × the magnetic field in the same direction as the external magnetic field. the incident ray is normal to the surface of prism.e.lB t + lA t log e 4 = ( -l B t + l A t) log e e æ log e 2 log e 2 ö ÷t 2 log e 2 = ç ç TA TB1 / 2 ÷ è 1/2 ø 2 -1 ö 2 =æ ç ÷t è 50 ø where 45° 45° 45° 45° 45° 45° Þ Þ 1 1 ö 2 =æ ç . So. From the figure.lA t = N 0 e . it is clear that the magnetic flux decreases. we can write N A = 4 N 0 e .lt For radio isotopes A and B. From the given figure.8 × 10–2 × 3 × 10–2 × 2 –4 t = 7.lA t 4 = e . 13. We have W (Q dv = 0) = dv q Thus W = 0.8 × 10–2 × 3 × 10–2 sin 30° 1 = 4. 10.lB t B Let t be the time after which N A = N B t A = t B = t( say) \ Þ Þ Þ Þ 4 N 0 e . PQRSP. t = M B sin q t ® torque acting on magnetic needle M ® Magnetic moment B ® Magnetic field strength Then t = 4. reaches the other end of prism.2 × 10 Nm 12. We have.Examination Papers × × 415 × 7. The potential difference between any two points of equipotential surface is zero. The ray of light undergoes the phenomenon of total internal reflection and continues in the same manner. The second angle of incident = 45° (greater than critical angle of prism). incident angle = 0° Then. U1 = 2. at the point of intersection there will be two values of electric potential which is not possible. Here X ® Rectifier Y ® Envelope detector Rectifier: It allows only the positive half of the AM input wave to go onwards. C1 = 150 pF = 150 × 10–12F V1 = 200 V Initial energy of the first capacitor 1 1 U1 = C1V12 = ´ 150 ´ 10 -12 ´ ( 200) 2 2 2 –6 = 3 × 10 J When another uncharged capacitor of capacitance C2 = 50 pF = 50 × 10–12F is connected across q + q 2 C1V1 + 0 the first capacitor. 15. This is because.75 × 10–6 J CBSE (Foreign) Set–III 1. common potential (V) = 1 = C1 + C 2 C1 + C 2 = Final energy. C = 10–6F Q= C R 1 4 Q= = 100 20 10 -6 The quality factor measures the sharpness of resonance of an LCR circuit. 1 H .25 × 10–6 J 2 Change in energy. We have. R = 20. L = 4H. isotopes of hydrogen) 5. It separates the message signal from the carrier wave. 1 H . 26.e.25 ´ 10 -6 – 3 × 10–6 = – 0. . Isotopes are the two or more nuclides having the same atomic number (Z) but different mass number (A). Poles of the earth. 7. U 2 . 150 ´ 10 -12 ´ 200 (150 + 50) ´ 10 -12 = 150 V 1 U 2 = (C1 + C 2 )V 2 2 1 = (150 + 50) ´ 10 -12 ´ (150) 2 = 2. the quality factor ‘Q’ of a series LCR 1 L Here.416 Xam idea Physics—XII 14. (i. Relation between the de Broglie wavelength (l) associated with the electron and its kinetic energy h where m = Mass of electron l= 2mE E = KE of electron 2. Given. 2 3 One example of isotope is 1 1 H .. (i) We know t = MB sin q where t = Torque acting on a short bar magnet P' M = Magnetic moment B = Magnetic field t = 0. incident angle = 0 Then angle of refraction will be zero. 45° Hence the light ray undergoes the phenomenon of total internal reflection making angle 90° with the incident ray. From the given figure. è ø 14. A B 11. P 45° The second angle of incident = 45° which is greater than the critical angle of prism. (i) We know that drift speed eVt 1 vd = µ ml l When length of the conductor is doubled. Thus the light ray will pass through the prism undeviated.5 × 10–2 Nm 2 (ii) The direction of torque is always perpendicular to the plane containing magnetic moment and æ ® ®ö magnetic field ç M ´ B ÷.Examination Papers C 417 9. R ¢ = =4 A A 2 Thus. 1 L Q= where R = 20 W R C C=4m F L=4H 1 4 1 Q= = ´ 10 3 20 4 ´ 10 -6 20 Q = 50 The quality factor measures the sharpness of resonance of an LCR circuit.1 × sin 30° 1 = 0.5 × 0. rl (ii) As R= A A Now.5 × 0. drift velocity gets halved. the quality factor of a series LCR. and reaches to the other end of prism. l¢ = 2l and A ¢ = 2 r( 2l) rl \ New resistance. We have. 15. . resistance becomes four times.1 × = 2. Amax = Ac + Am = 8V Minimum amplitude.4 × 10–6 J . Modulation index: It is the ratio of peak value of modulating signal to the peak value of carrier wave. C1 = 400. U1 = = Fixed energy. 6 ´ 10 -6 J .U1 = 1. pF = 400 × 10–12 F V1 =100 V 1 C1 V1 2 2 1 = ´ 400 ´ 10 -12 ´ (100) 2 2 = 2 × 10–6 J When another uncharged capacitor of capacitance C2 = 100 pF is connected across the first q + q 2 C1V1 + 0 capacitor.6 × 10–6 J 2 Difference in energy U 2 . A m= m Ac It signifies the level of distortion.A min m = max A max + A min 8 -4 4 1 = = = 8 + 4 12 3 27. Maximum amplitude. common potential (V) = 1 = C1 + C 2 C1 + C 2 Initial energy. Given.418 Xam idea Physics—XII 21.2 ´ 10 -6 J = – 0. Amin = Ac – Am = 4V A . U2 = 400 ´ 10 -12 ´ 100 ( 400 + 100) ´ 10 -12 = 80 V 1 (C1 + C 2 )V 2 2 1 = ( 400 + 100) ´ 10 -12 ´ (80) 2 = 1.
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