Page 2 :
PHYSICS, Examination Papers, 2008–2015
Page 3 :
CONTENT, n, , CBSE Examination Paper–2008 (Delhi), , 3, , n, , CBSE Examination Paper–2008 (All India), , 37, , n, , CBSE Examination Paper–2009 (Delhi), , 71, , n, , CBSE Examination Paper–2009 (All India), , 103, , n, , CBSE Examination Paper–2009 (Foreign), , 131, , n, , CBSE Examination Paper–2010 (Delhi), , 161, , n, , CBSE Examination Paper–2010 (All India), , 186, , n, , CBSE Examination Paper–2010 (Foreign), , 213, , n, , CBSE Examination Paper–2011 (Delhi), , 239, , n, , CBSE Examination Paper–2011 (All India), , 273, , n, , CBSE Examination Paper–2011 (Foreign), , 303, , n, , CBSE Examination Paper–2012 (Delhi), , 333, , n, , CBSE Examination Paper–2012 (All India), , 358, , n, , CBSE Examination Paper–2012 (Foreign), , 390, , n, , CBSE Examination Paper–2013 (Delhi), , 419, , n, , CBSE Examination Paper–2013 (All India), , 451, , n, , CBSE Examination Paper–2013 (Foreign), , 480, , n, , CBSE Examination Paper–2014 (Delhi), , 507, , n, , CBSE Examination Paper–2014 (All India), , 535, , n, , CBSE Examination Paper–2014 (Foreign), , 567, , n, , CBSE Examination Paper–2015 (Delhi), , 595, , n, , CBSE Examination Paper–2015 (Ajmer), , 613, , n, , CBSE Examination Paper–2015 (Allahabad), , 633, , n, , CBSE Examination Paper–2015 (Bhubaneshwar), , 654, , n, , CBSE Examination Paper–2015 (Chennai), , 674, , n, , CBSE Examination Paper–2015 (Guwahati), , 695, , n, , CBSE Examination Paper–2015 (Panchkula), , 714, , n, , CBSE Examination Paper–2015 (Patna), , 732, , n, , CBSE Examination Paper–2015 (Foreign), , 750
Page 4 :
CBSE EXAMINATION PAPERS, DELHI–2008, Time allowed : 3 hours, , Maximum marks : 70, , 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, , 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 ?
Page 5 :
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, , 12., , 13., , 14., 15., 16., 17., , (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, Na. Calculate the maximum kinetic energy of, A nucleus 10, 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
Page 6 :
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, , V (Volt), , 20, 30, , 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.
Page 7 :
6, , Xam idea Physics—XII, 3V, , A, , B, 1.5 V, R, , (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., - 0.85 eV, , A, , - 1.5 eV, B, , C, - 3.4 eV, D, - 13.6 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.
Page 8 :
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.
Page 9 :
8, , Xam idea Physics—XII, , 13. Draw the output wave form at X, using the given inputs A, B for the logic circuit shown below., Also identify the gate., A, X, C, , B, , t1, , t2, , t3, , t5, , t6, , t4, A, Input, B, Input, ®, , 15. Derive an expression for the potential energy of an electric dipole of dipole moment p in an, ®, , electric field E., 18. Prove that an ideal inductor does not dissipate power in an a.c. circuit., OR, Derive an expression for the self-inductance of a long air-cored solenoid of length l and number of, turns N., 19. Define conductivity of a conductor. Explain the variation of conductivity with temperature in, (a) good conductors, (b) ionic conductors., 24. 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. Show that the angle of, incidence is equal to the angle of reflection., 25. A coil of number of turns N, area A, is rotated at a constant angular speed w, in a uniform magnetic, field B, and connected to a resistor R. Deduce expressions for :, (i) Maximum emf induced in the coil, (ii) Power dissipation in the coil., , CBSE (Delhi) SET–III, Questions different from Set – I & II, 1., 2., 5., , Name the absorbing material used to control the reaction rate of neutrons in a nuclear reactor., State the reason, why two independent sources of light cannot be considered as coherent sources., An electron and a proton have the same de Broglie wavelength associated with them. How are, their kinetic energy related to each other?, 7. How does the power of a convex lens vary, if the incident red light is replaced by violet light ?, 9. Draw a ray diagram of a compound microscope. Write the expression for its magnifying power., 12. If the output of a 2 input NOR gate is fed as both inputs A and B to another NOR gate, write down, a truth table to find the final output, for all combinations of A, B.
Page 10 :
Examination Papers, , 9, , 15. The oscillating electric field of an electromagnetic wave is given by:, E y = 30 sin [ 2 ´ 1011 t + 300 px ] Vm - 1, (a) Obtain the value of the wavelength of the electromagnetic wave., (b) Write down the expression for the oscillating magnetic field., 16. 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., 20. The energy levels of an element are given below:, - 0.85 eV, , A, , - 1.5 eV, C, - 3.4 eV, B, , D, - 13.6 eV, , Identify, using necessary calculations, the transition, which corresponds to the emission of a, spectral line of wavelength 482 nm:, 24. An inductor of unknown value, a capacitor of 100mF and a resistor of 10 W are connected in series, to a 200 V, 50 Hz a.c. source. It is found that the power factor of the circuit is unity. Calculate the, inductance of the inductor and the current amplitude., 25. Prove that the current density of a metallic conductor is directly proportional to the drift speed of, electrons through the conductor., OR, Define resistivity of a conductor. Plot a graph showing the variation of resistivity with temperature, for a metallic conductor. How does one explain such a behaviour, using the mathematical, expression of the resistivity of a material., 26. A metallic rod of length l is rotated at an angular speed w, normal to a uniform magnetic field B., Derive an expression for the (i) emf induced in the rod (ii) heat dissipation, if the resistance of the, rod is R.
Page 11 :
10, , Xam idea Physics—XII, , Solutions, CBSE (Delhi) SET–I, 1., , Force,, , ®, , ®, , ®, , Fm = q v ´ B, , ®, , ®, , Obviously, the force on charged particle is perpendicular to both velocity v and magnetic field B ?, 2. Wavelength 10 - 2 m belongs to microwaves. It is used in RADAR., 3. 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., 5., , That is kinetic energy of electron and a-particle are in inverse ratio of these masses., The glass lens will disappear in the liquid if the refractive index of liquid is equal to that of glass, i.e., refractive index of liquid = 1× 5., The points A and B are equidistant from the centre of square where charge q = 500 mC is located;, therefore, points A and B are at the same potential i.e., VA = VB ., B, , q, 500 mC, O, , A, , \, 6., , 7., , Work done in moving charge q 0 = 10 mC from A to B is, W = q 0 (VB - VA ) = 0, The basic principle of mechanics is that momentum transfer is maximum when the mass of, colliding particle and target particle are equal. Heavy water has negligible absorption cross-section, for neutrons and its mass is small; so heavy water molecules do not absorb fast neutorns; but, simply slow them., Dl, Fringe width, b =, Þ b µ l for same D and d. When the whole apparatus is immersed in a, d, l l, transparent liquid of refractive index n = 1× 3, the wavelength decreases to l¢ = =, × So, fringe, n 1× 3, 1, times., width decreases to, 1× 3
Page 12 :
Examination Papers, , 11, , Let e be emf and r the internal resistance of each cell. The equation of terminal potential, difference, V = e eff - i r int becomes, …(1), V = 3 e - i r int, where r int is effective (total) internal resistance., From fig., when i = 0, V = 6 × 0 V, \ From (1),, 6, e= =2V, 6 = 3e - 0, Þ, 3, i.e., emf of each cell, e = 2 V, 9. Electric Potential due to an electric dipole at axial point. Consider an electric dipole AB, having, charges - q and + q at points A and B respectively. The separation between the charges is 2l., , 8., , ®, , Electric dipole moment, p = q . 2l, directed from - q to + q., , (r + l), , Consider a point P on the axis of dipole at a distance r from A, 2l, B, mid-point O of dipole., –q, +q, O, The distance of point P from charge + q is BP = r - l, r, The distance of point P from charge - q is AP = r + l, Let V1 and V2 be the potentials at P due to charges + q and - q respectively. Then, q, 1, 1 ( - q), and V2 =, V1 =, 4pe 0 r - l, 4pe 0 r + l, , (r – l), P, , \ Resultant potential at P due to dipole, q, q, 1, 1, V = V1 + V2 =, 4 pe 0 (r - l) 4pe 0 (r + l), é(r + l) - (r - l) ù, é 1, 1, 1 ù, 1, .qê, .q ê, ú, ú=, 2, 2, 4pe 0, ër - l r + l û 4pe 0, úû, êë r - l, q . 2l, 1, =, 4pe 0 r 2 - l 2, =, , As q . 2l = p (dipole moment), p, 1, \, V=, 2, 4pe 0 r - l 2, If point P is far away from the dipole, then r > > l, p, 1, V=, \, 4pe 0 r 2, 10. Magnetic susceptibility: It is defined as the intensity of magnetisation per unit magnetising field,, M, i.e. c m = . It has no unit., H, Iron has positive susceptibility while copper has negative susceptibility., Negative susceptibility of a substance signifies that the substance will be repelled by a strong, magnet or opposite feeble magnetism induced in the substance.
Page 13 :
12, , Xam idea Physics—XII, , 11. (i) Standard equation of magnetic field is, B y = B 0 sin ( wt + kx ) T, Comparing this equation with the given equation, we get, B 0 = 8 ´ 10 - 6 T,, wavelength, l =, , w = 2 ´ 1011 rad s - 1 ,, , k=, , 2p, = 300p, l, , 2p, 1, =, m, 300p 150, , (ii) E 0 = B 0 c = 8 ´ 10 - 6 ´ 3 ´ 108 = 2 × 4 ´ 10 3 Vm - 1 ., ® ® ®, , According to right hand system of E, B, K, the electric field oscillates along negative Z-axis,, so equation is, E Z = - 2 × 4 ´ 10 3 sin ( 2 ´ 1011 t + 300px ) V m - 1, 12. Power dissipated in a.c. circuit, P = Vrms I rms cos f where cos f =, For an ideal capacitor R = 0, \, , R, Z, , R, =0, Z, I rms ´ 0 = 0 (zero)., \ cos f =, , P = Vrms I rms cos f = Vrms, , i.e., power dissipated in an ideal capacitor is zero., OR, Let a circuit contain a resistor of resistance R and an inductor of, inductance L connected in series. The applied voltage is, V = V0 sin wt. Suppose the voltage across resistor is VR and that, across inductor is VL . The voltage VR and current I are in the, same phase, while the voltage VL leads the current by an angle, p, × Thus, VR and VL are mutually perpendicular. The resultant, 2, of VR and VL is the applied voltage i.e.,, , \, , V = ( Ri) 2 + ( X L i) 2, , \, , V, Z = = R 2 + X L2, i, , 23, 10 Ne, , I, , is, , 23, ¾® 11, Na + e - + n, 23, 23, difference, Dm = m N ( 10, Ne) - m N ( 11, Na), , I, VR, , 23, 10 Ne, , Mass, , L, , VL, , Z = R 2 + ( wL) 2, , 13. The equation of b-decay of, , R, , V, , But, VR = Ri, VL = X L i = wLi, where X L = wL is inductive reactance, , Þ, , VL, , V = V0 sin wt, , V = VR2 + VL2, , Impedance,, , VR, , - me
Page 14 :
Examination Papers, , Changing nuclear masses into atomic masses, , {(, , Dm = m, , ) -10m }-{m (, , 23, 10 Ne, , e, , 23, 11 Na, , ) -11m }- m, e, , 13, , e, , 23, 23, Ne) - m ( 11, Na), = m ( 10, , = 22 × 994466 - 22 × 989770, = 0 × 004696 u, \ Maximum K.E., Q = 0 × 004696 u ´ 931× 5 MeV/u, = 4 × 37 MeV, 14., Intrinsic semiconductor, , p-type semiconductor, , (i), , It is a semiconductor in pure form., , (i), , (ii), , Intrinsic charge carriers are electrons and (ii) Majority charge carriers are holes and, holes with equal concentration., minority charge carriers are electrons., , It is a semiconductor doped with p-type, (like Al, In) impurity., , (iii) Current due to charge carriers is feeble (iii) Current due to charge carriers is, (of the order of mA)., significant (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., 15., MI, M2, , F, , Eye, , Advantages: (i) It is free from chromatic aberration., (ii) Its resolving power is greater than refracting telescope due to larger aperture of mirror.
Page 15 :
14, , Xam idea Physics—XII, , 16. Given A = 60° ,, , i=, , 3, 3, A = ´ 60° = 45°, 4, 4, , For minimum deviation i1 = i 2 = i, \ Refractive index of prism, n =, =, , and r1 = r 2 =, , A, = 30°, 2, , sin i, sin r, sin 45° 1 / 2, =, = 2 = 1× 41, sin 30° 1 / 2, , c 3 ´ 108, = 2 × 13 ´ 10 8 ms - 1 ., =, n, 1× 41, 17. The output of NAND gate with inputs A and B is, Y = AB, Speed of light in prism,, , v=, , i.e., output is obtained if either or both inputs are zero., Accordingly the output waveform Y = AB is shown in, Output Y, fig., i.e., output is zero between intervals 0 to t 1 and t 4 to t 5, and in all other intervals it is ‘1’., The output waveform is shown in fig., 18. Given hT = 36 m, hR = 49 m, and R e = 6400 km = 6 × 4 ´ 10 6 m., Maximum LOS distance,, , Output waveform, , t1, , t2, , t3, , t4, , t5, , t6, , d m = 2R e hT + 2R e hR, = 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. Wavefront: A wavefront is a locus of all particles of medium vibrating in the same phase., 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, it is deviated from its path., This phenomenon is called refraction. In transversing from first medium to another medium, the, frequency of wave remains unchanged but its speed and the wavelength both are changed. Let XY, be a surface separating the two media ‘1’ and ‘2’. Let v1 and v 2 be the speeds of waves in these, media., 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. Clearly BB¢ = v1t., 90o, i, i, B', As the wavefront AB advances, it strikes the points between X, Y, r, A, A and B¢ of boundary surface. According to Huygen’s, r, 90o, principle, secondary spherical wavelets originate from these, A', points, which travel with speed v1 in the first medium and, speed v 2 in the second medium.
Page 16 :
Examination Papers, , 15, , First of all secondary wavelet starts from A, which traverses a distance AA¢ ( = v 2 t) in second, medium in time t. In the same time-interval t, the point of wavefront traverses a distance, BB¢ ( = v1t) in first medium and reaches B¢ , from, where the secondary wavelet now starts. Clearly, BB¢ = v1t and AA¢ = v 2 t., 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 incident wavefront AB advances, the secondary wavelets start from points, between A and B¢ , one after the other and will touch A¢ B¢ simultaneously. According to Huygen’s, principle A¢ B¢ is the new position of wavefront AB in the second medium. Hence A¢ B¢ will be the, refracted wavefront., Let the incident wavefront AB and refracted wavefront A¢ B¢ make angles i and r respectively with, refracting surface XY., In right-angled triangle AB¢ B, Ð ABB¢ = 90°, vt, BB¢, ...(1), sin i = sin Ð BAB¢ =, = 1, \, AB¢, AB¢, Similarly in right-angled triangle AA¢ B ¢ , Ð AA¢ B ¢ = 90°, AA¢ v 2 t, ...(2), \, sin r = sin Ð AB¢ A¢ =, =, AB¢ AB¢, Dividing equation (1) by (2), we get, v, sin i, ...(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. This is the second law of refraction,, and is called the Snell’s law., x x x, 20. Consider a metallic rod OA of length l , which is rotating with, x, x, B x, x, x, x, angular velocity w in a uniform magnetic field B , the plane of, x, x x, x, x, x, x, rotation being perpendicular to the magnetic field. A rod may be, xvx, x x x, x, x, x, supposed to be formed of a large number of small elements., x x x xO, A, dx, Consider a small element of length dx at a distance x from centre., x x x x, l, If v is the linear velocity of this element, 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, \, l, , l, , 0, , 0, , e = ò B xw dx = Bw ò x dx, é x2, = Bw ê, êë 2, Current induced in rod I =, , l, , ù, él2, -0, ú = Bw ê, úû 0, êë 2, , e 1 B wl 2, =, ×, R 2 R, , ù 1, ú = B wl, úû 2, , 2
Page 17 :
Xam idea Physics—XII, R, , p, , n, , mA, , Regulated output, , –, +, Semiconductor diode used is Zener diode (but the voltages, quoted in fig. are much more than actual values Vbr is, usually < 6 V)., V, (ii) The circuit diagram for reverse characteristics of Zener diode, –, +, is shown in fig., (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, K, voltage across it remains practically constant for a large, I, IL, change in the current., R, A simple circuit of a voltage regulator using a, IZ, Zener diode is shown in the Fig. The Zener, diode is connected across load such that it is, Vin, VZ, RL V0, reverse biased., The series resistance R absorbs the output, voltage fluctuations so as to maintain constant, voltage across the load., If the input dc voltage increases, the current through R and Zener diode also increases. So, voltage, drop across R increases, without any change in the voltage across zener diode., 22. Given L = 200 mH = 200 ´ 10 - 3 H, C = 500 mF = 500 ´ 10 - 6 F,, , 21. (i), , Unregulated input, , 16, , R =10 W , Vrms =100 V, (i) Angular (resonant) frequency wr at which power factor of the circuit is unity, is given by, 1, 1, 1, wr L =, Þ wr =, =, = 100 rad s - 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, Z = R, V, V 2, \ Current amplitude, I 0 = 0 =, Z, R, 100 2, =, = 10 2A = 14 × 1A, 10, w L 100 ´ 200 ´ 10 - 3, (iii) Q-factor = r =, =2, R, 10, 23. Consider a uniform metallic wire XY of length l and cross-sectional area A. A potential difference, V is applied across the ends X and Y of the wire. This causes an electric field at each point of the, wire of strength, V, ...(1), E= ×, l
Page 18 :
Examination Papers, , 17, , E, , P, Q, X, Y, Due to this electric field, the electrons, gain a drift velocity v d opposite to, Current (I), direction of electric field. If q be the, charge, passing, through, the, vdt, cross-section of wire in t seconds, then I, I, q, ...(2), Current in wire I =, 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, then the total charge flowing, in time t will be equal to the total charge on the electrons present within the cylinder PQ., 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, then the number of free electrons, in the cylinder = n ( A v d t), If charge on each electron is - e ( e = 1× 6 ´ 10 - 19 C), then the total charge flowing through a, , cross-section of the wire, ...(3), q = ( nA v d t) ( - e) = - neAv d t, \ Current flowing in the wire,, q - neAv d t, I= =, t, t, i.e.,, current I = - neAv d, ...(4), This is the relation between electric current and drift velocity. Negative sign shows that the, direction of current is opposite to the drift velocity., Numerically I = neAv d, ...(5), I, Current density, \, J = = nev d, A, Þ, J µ vd ., OR, (i) The circuit arrangement is shown in fig., e,r, e,r, e,r, e,r, c, (ii) Applying Kirchhoff’s second law to the circuit b, abcda, V, - ne - I ( nr ) - IR + e¢ = 0, I, e ¢ - ne, I=, Þ, R + nr, a, d, +, –, R, e ¢ - ne, e, (a) Charging current,, …(a), I=, dc source, R + nr, (b) Potential difference across the combination V is given by, - V - IR + e¢ = 0, Þ, V = e¢ - IR
Page 20 :
Examination Papers, , 19, , Modulation Index: The ratio of amplitude of modulating signal to the amplitude of carrier wave, is called modulation index i.e.,, E, ma = m, Ec, For effective amplitude modulation the modulation index determines the distortions, so its value is, kept £1 for avoiding distortions., 28. (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, ®, , where m 0, , ®, , dl, , ®, , m I dl ´ r, dB = 0, 4p r 3, is permeability of free space. Its value is, m 0 = 4p ´ 10 -7 Wb/A-m., , P, r, , I, , 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 magnetic field dB is perpendicular to the plane containing I dl and r ., Magnetic Field at the centre of circular loop: Consider a circular coil of, radius R carrying current I in anticlockwise direction. Say, O is the centre, of coil, at which magnetic field is to be computed. The coil may be, supposed to be formed of a large number of current elements. Consider a, small current element ‘ab’ of length Dl. 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. Here, q = 90° , because current element at each point of circular path is perpendicular to the radius., Therefore magnetic field produced at O, due to current element ab is, m I Dl, DB = 0, 4p R 2, According to Maxwell’s right hand rule, the direction of magnetic field at O is upward,, perpendicular to the plane of coil. The direction of magnetic field due to all current elements is the, same. Therefore the resultant magnetic field at the centre will be the sum of magnetic fields due to, all current elements. 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)
Page 21 :
20, , Xam idea Physics—XII, , m0 I, × 2pR, 4p R 2, m I, or, B= 0, 2R, If the coil contains N–turns, then S Dl = N. 2pR, m I, m NI, \, B = 0 × 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; but if the current in the coil is clockwise, then the direction of magnetic field, will be perpendicular to the plane of coil downward., (1), B2, (b) Magnetic field due to coil 1 at point O, B, m 0 IR 2, , ®, , B1 =, , ®, , 2 (R 2 + x 2 ) 3/ 2, , along OC1, , R, , B1, , I, C1, , O, , x, , Magnetic field due to coil 2 at point O, m 0 IR 2, , ®, , B2 =, , ®, , ®, , 2 (R 2 + x 2 ) 3/ 2, , x, , along C 2 O, , ®, , Both B 1 and B 2 are mutually perpendicular, so, the net magnetic field at O is, B = B12 + B 22 = 2B1, = 2, , (as B1 = B 2 ), , m 0 IR 2, , C2, (2), , I, , 2 (R 2 + x 2 ) 3/ 2, , As R << x, m 0 2 2 . m 0 I ( pR 2 ), ×, 4p, 2. x 3, x3, m 2 2 m 0 IA, = 0, 4p, x3, , B=, , 2m 0 IR 2, , =, , where A = pR 2 is area of loop., B, tan q = 2, B1, Þ, , tan q =1, p, q=, 4, , Þ, ®, , (Q B 2 = B1 ), , ®, p, with the direction of magnetic field B 1 ., 4, OR, (a) Cyclotron: The cyclotron, devised by Lawrence and Livingston, is a device for accelerating, ions to high speed by the repeated application of accelerating potentials., , \ B is directed at an angle
Page 22 :
Examination Papers, , Principle: The positive ions produced from a source are, accelerated. Due to the presence of perpendicular magnetic, field the ion will move in a circular path. 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., Expression for K.E. attained:, If R be the radius of the path and v max the velocity of the ion, when it leaves the periphery, then, qBR, v max =, m, The kinetic energy of the ion when it leaves, the apparatus is,, S, , 21, , Magnetic Pole, N, S, Dee, , Dee, , Magnetic Pole, S, , Dee-1, , R.F., oscillator, , q2B 2R 2, 1, 2, K. E. = mv max, =, 2, 2m, Dee-2, When charged particle crosses the gap, Beam, between dees it gains KE = q V, In one revolution, it crosses the gap twice,, therefore if it completes n-revolutions before emerging the does, the kinetic energy gained, = 2nqV, q2B 2R 2, Thus, K.E. =, = 2nqV, 2m, Working: The principle of action of the apparatus is shown in fig. The positive ions produced, from a source S at the centre are accelerated by a dee which is at negative potential at that, moment. Due to the presence of perpendicular magnetic field the ion will move in a circular, path inside the dees. 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., The function of electric field is to accelerate the charged particle and so to impart energy to the, charged particle., 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., Expression for Period of Revolution and Frequency:, Suppose the positive ion with charge q moves in a dee with a velocity v, then,, qvB =, , mv 2, r, , or r =, , mv, qB, , ...(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 angular velocity w of the ion is given by,
Page 23 :
22, , Xam idea Physics—XII, , v qB, (from eq. 1), ...(2), =, r m, The time taken by the ion in describing a semi-circle, i.e., in turning through an angle p is,, p pm, ...(3), t= =, w Bq, T, pm, ...(4), =t =, 2, qB, 2p m, ...(5), or, T=, qB, w=, , This is the expression for period of revolution., Clearly the period of revolution is independent of speed of particle., sin i n 2, …(1), 29. (a) Snell’s laws is, =, sin r n1, Critical angle is the angle of incidence in denser medium for which, angle of refraction in rarer medium is 90° i.e.,, i = i c , 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., 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. It is a very long and, thin pipe of quartz ( n = 1× 7) of thickness, Coating n = 1.5, nearly » 10 - 4 m coated all around with, a material of refractive index 1× 5. A, n = 1.7, large number of such fibres held, together form a light pipe and are used, for communication of light signals., When a light ray is incident on one end, at a small angle of incidence, 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. 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., 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.
Page 24 :
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., 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, 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°., According to Malus law, intensity of light transmitted from second polaroid (B, 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, 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,, the intensity decreases and if angle decreases, the intensity increases., (b) Yes, the incident light (of intensity I 0 ) is unpolarised., Reason: If incident light is unpolarised, the intensity of transmitted light through a polaroid is, always I 0 / 2, which is constant., But if incident light is polarised, the intensity variation I = I 0 cos 2 q , necessarily takes place., , 30. (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. To find the, electric field outside the shell, we consider a spherical Gaussian surface of radiusr ( > R),, ®, , concentric with given shell. If E is electric field outside the shell, then by symmetry electric, field strength has same magnitude E 0 on the Gaussian surface and is directed radially, outward. Also the directions of normal at each point is radially outward, so angle between, ®, , ®, , E i and d S is zero at each point. Hence, electric flux through Gaussian surface
Page 25 :
24, , Xam idea Physics—XII, ®, , ®, , = ò E · d S., S, , Q, , = ò E 0 dS cos 0 = E 0 . 4pr, , 2, , EO, , R, r, , dS, , P, , Now, Gaussian surface is outside the given charged shell, so, charge enclosed by Gaussian surface is Q., Hence, 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, electric field outside a charged thin spherical shell is the same as if the whole charge Q, is concentrated at the centre., If s is the surface charge density of the spherical shell, 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. For a positively charged shell, the field lines are, directed radially in outward direction and for negatively charged shell, these are directed in, radially inward direction., , +, , +, , +, , R, O, , +, +, , –, +, , –, , +, , –, , +, , R, O, –, , (a) Positively charged shell, 2, , (b) Given, s =100 mC/m = 100 ´ 10, , –, –, –, –, , (a) Negatively charged shell, -6, , C / m2 ., , Diameter, D = 2R = 2 × 5 m, (i) Charge on sphere, Q = s . 4pR 2 = s . p ( 2R) 2, = (100 ´ 10 - 6 C / m 2 ) ´ 3 ×14 ´ ( 2 × 5 m) 2, = 19 × 625 ´ 10 - 4 C, = 1× 96 ´ 10 - 3 C = 1× 96 mC
Page 26 :
Examination Papers, , 25, , (ii) Electric flux passing through the sphere, 1, 1, f=, ´ (1× 96 ´ 10 - 3 ), ( Q) =, 12, e0, 8 × 86 ´ 10, = 2 × 21 ´ 108 Nm 2 C - 1, OR, (a) Consider an electric dipole placed in a uniform electric, field of strength E in such a way that its dipole moment, ®, , +q, B, , F1 = qE, , p makes an angle q with the direction of E . The, charges of dipole are - q and + q at separation 2l the, dipole moment of electric dipole,, ...(1), p = q.2l, r, ®, Force: The force on charge + q is, F1 = qE, along the, , 2l, , F2 = – qE, , A q, –q, , q, O, , 2l sin q, , ®, , E, , N, , ®, , direction of field E, , ®, , ®, , The force on charge - q is, F2 = qE , opposite to the direction of field E, ®, , ®, , Obviously forces F1 and F2 are equal in magnitude but opposite in direction; hence net force, on electric dipole in uniform electric field is, F = F1 - F2 = qE - qE = 0 (zero), As net force on electric dipole is zero, 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. This couple is called the torque and is denoted by t., \ 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)], ....(2), = pE sin q, Clearly, the magnitude of torque depends on orientation ( q) of the electric dipole relative to, ®, , ®, , electric field. Torque ( t) is a vector quantity whose direction is perpendicular to both p and E ., In vector form, , ®, , ®, , ®, , t =p ´E, , ...(3), , Thus, if an electric dipole is placed in an electric field in oblique orientation, it experiences, no force but experiences a torque. The torque tends to align the dipole moment along the, direction of electric field., Maximum Torque: For maximum torque sin q should be the maximum. As the maximum, value of sin q =1 when q = 90°, \ Maximum Torque, t max = pE
Page 28 :
Examination Papers, , 6., , 27, , 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, the angular separation decreases., , 11., B, , A, , fo, , fe, , Fo, A' Fe', , C1, , Fe, , C2, , B', , in, At, , i, fin, , ty, , Draw backs:, (i) It is not free from chromatic aberration., (ii) The image formed is inverted and fainter., 12. Given focal length f = -10 cm, u = ?, v, magnification, m=- =-2, Þ, v = 2u, u, 1 1 1, From mirror formula = + , we have, f v u, 1, 1 1, - =, +, 10 2u u, 3, 1, =Þ, 2u, 10, 10 ´ 3, Þ, u== - 15 cm, 2, A, 13. Output of first NOR gate, C = AB, B, , C, , Output X = CC, = C = AB, = AB, This is AND operation. Therefore, the output is 1 when both inputs are 1., Accordingly the waveform output is shown in figure., , X
Page 29 :
28, , Xam idea Physics—XII, t1, , t2, , t3, , t4, , t5, , t6, , t1, , t2, , t3, , t4, , t5, , t6, , A, Input, B, , Output, X, , 15. 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. The work done in this process may be supposed to be done in two parts., (i) The work done (W1 ) in bringing the dipole perpendicular to electric field from infinity., (ii) Work done (W2 ) in rotating the dipole such that it finally makes an angle q from the direction of, electric field., (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. The electric, forces an charges +q and - q are qE and qE, along the field direction and opposite to field, direction respectively. As charges + q and - q traverse equal distance under equal and opposite, forces; therefore, net work done in bringing the dipole in the region of electric field, perpendicular to field-direction will be zero, i.e., W1 = 0., +q, , qE, , 2l, qE, , From, infinity, –q, , (ii) Now the dipole is rotated and brought to orientation, making an angle q with the field direction (i.e., q1 = 90°, and q 2 = 0°), therefore, work done, W2 = pE (cos q1 - cos q 2 ), = pE (cos 90° - cos q) = - pE cos q, qE, \ Total work done in bringing the electric dipole from, infinity, i. e., Electric potential energy of electric dipole., U = W1 + W2 = 0 - pE cos q = - pE cos q, In vector form, , ® ®, , U =- p ×E, , +q, , qE, , p, , –q, , …(1)
Page 30 :
Examination Papers, , 18. The power, , 29, , P = Vrms i rms cos f, , R, R, ; For ideal inductor R = 0, \ cos f = = 0, Z, Z, \, P = Vrms i rms cos f = 0 i. e. power dissipated by an ideal inductor in ac circuit is zero., OR, Self Inductance of a long air-cored solenoid:, Consider a long air solenoid having `n' number of turns per unit length. If current in solenoid is I ,, ...(1), then magnetic field within the solenoid, B = m 0 nI, where m 0 = 4p ´ 10 - 7 henry/metre is the permeability of free space., If A is cross-sectional area of solenoid, then effective flux linked with solenoid of length ‘ l’ where, N = nl is the number of turns in length ' l ' of solenoid., \, F = ( nl BA), Substituting the value of B from (1), ...(2), F = nl (m 0 nI ) A = m 0 n 2 AlI, where, , cos f =, , \, , Self-inductance of air solenoid, F, L = = m 0 n 2 Al, I, If N is total number of turns in length l , then, N, n=, l, , ...(3), , A, , 2, , æ N ö, \ Self-inductance L = m 0 ç, ÷ Al, è l ø, m N 2A, ...(4), = 0, l, 19. The reciprocal of resistivity (r) of a material is called its conductivity ( s), i.e.,, 1, s=, r, , l, , S.I. unit of conductivity is mho m -1 (or siemen m -1 )., (i) Conductivity of a metallic conductor s =, , 1 ne 2 t, ., =, r, m, , With rise of temperature, the collision of electrons with fixed lattice, ions/atoms increases so that relaxation time ( t) decreases., Consequently, the conductivity of metals decreases with rise of, temperature. Figure represents the variation of conductivity of, metal with temperature. Initially the variation of conductivity with, temperature is linear and then it is non-linear., (ii) Conductivity of ionic conductor increases with increase of, temperature because with increase of temperature, the ionic bonds, break releasing positive and negative ions which are charge carriers, in ionic conductors., , s, , T
Page 31 :
30, , Xam idea Physics—XII, , 24. Wavefront: A wavefront is a locus of particles of medium all vibrating in the same phase., C, Law of Reflection: Let XY be a reflecting surface at, which a wavefront is being incident obliquely. Let v, A', B, A2, be the speed of the wavefront and at time t = 0, the, D, i A, wavefront touches the surface XY at A. After time t,, 1, r, i, r, i, the point B of wavefront reaches the point B¢ of the X, Y, A, B', surface., According to Huygen’s principle each point of wavefront acts as a source of secondary waves., When the point A of wavefront strikes the reflecting surface, then due to presence of reflecting, surface, it cannot advance further; but the secondary wavelet originating from point A begins to, spread in all directions in the first medium with speed v. As the wavefront AB advances further, its, points A1 , A 2 , A 3 K etc. strike the reflecting surface successively and send spherical secondary, wavelets in the first medium., First of all the secondary wavelet starts from point A and traverses distance AA¢ ( = vt) in first, medium in time t. In the same time t, the point B of wavefront, after travelling a distance BB¢,, reaches point B¢ (of the surface), from where the secondary wavelet now starts. 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¢ . As the incident wavefront AB advances, the secondary wavelets starting from points between, A and B¢, one after the other and will touch A¢ B¢ simultaneously. According to Huygen’s principle, wavefront A¢ B¢ represents the new position of AB, i.e., A¢ B¢ is the reflected wavefront, corresponding to incident wavefront AB., 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.e., both triangles are congruent., \, Ð BAB¢ = Ð AB¢ A¢, i.e., incident wavefront AB and reflected wavefront A¢ B¢ make equal angles with the reflecting, surface XY. As the rays are always normal to the wavefront, therefore the incident and the, reflected rays make equal angles with the normal drawn on the surface XY, i.e.,, angle of incidence i = angle of reflection r, 25., , (i) Suppose initially the plane of coil is perpendicular to the, magnetic field B. When coil rotates with angular speed w,, ®, , then after time t, 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, N-turns, then emf induced in the coil, df, d, e=-N, = - N ( BA cos wt), dt, dt, = + NBA w sin wt, , q = wt, , B, , …(1)
Page 32 :
Examination Papers, , 31, , \ For maximum value of emf e,, sin wt =1, \ Maximum emf induced, e max = NBAw, , e, R, 2, æeö e, …(2), \ Instantaneous power dissipated, 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.e.,, (sin 2 wt) av =, 2, , (ii) If R is resistance of coil, the current induced, I =, , \, , Average power dissipated Pav =, , N 2 B 2 A 2 w2, ×, 2R, , CBSE (Delhi) SET–III, 1., 2., , 5., , Cadmium is the absorbing material for neutrons produced in a nuclear reactor., Coherent sources are defined as the sources in the which initial phase difference remains constant., In the case of two independent sources, 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., h, de Broglie wavelength, l =, 2mE k, Given, \, Þ, i. e.,, , 7., , le = l p, h, 2m e E e, , =, , h, 2m p E p, , Ee m p, =, »1840, E p me, K.E. of electron = 1840 ´ (K.E. of proton), , æ 1, 1, 1, = ( a n g - 1) çç, f, è R1 R 2, and refractive index is maximum for violet light in visible region of spectrum., Power of a lens increases if red light is replaced by violet light because P =, , ö, ÷,, ÷, ø
Page 33 :
32, , Xam idea Physics—XII, , 9., , Eye-piece, uo, , ue, , vo, , D, Objective, , Eye, , B, Fo, , A", A, , A', , Fe', , O, , b, , Fe, , E, , B', , B", , ve, , v0 æ 1, 1, D çç +, u0 è f e ve, v æD Dö, ÷, or, M = - 0 çç +, u 0 è f e v e ÷ø, 12. First gate is NOR gate, its output C = A + B, M =-, , Magnifying power, , ö, ÷, ÷, ø, , Second gate is also NOR gate, its output, Y = C + C = C × C = C = A + B = A + B., This is Boolean expression for OR gate., A, B, , Y, , C, , Its truth table is, A, , B, , Y, , 0, , 0, , 0, , 1, , 0, , 1, , 0, , 1, , 1, , 1, , 1, , 1
Page 34 :
Examination Papers, , 33, , 15. (a) Given equation is, E y = 30 sin ( 2 ´ 1011 t + 300 px ) Vm - 1, Comparing with standard equation, E y = E 0 sin ( wt + kx ) Vm - 1 ,, , we get, , E 0 = 30 Vm - 1 , w = 2 ´ 1011 rad s - 1 , k =, , 2p, = 300 p m - 1, l, , 2p, m, 300p, 1, =, m = 6 × 67 ´ 10 - 3 m, 150, (b) The wave is propagating along X-axis, electric field is oscillating along Y-axis, so according to, \, , l=, , Wavelength,, , ® ® ®, , right hand system of ( E, B, K) the magnetic field must oscillate along Z-axis., E, 30, \, B0 = 0 =, = 10 - 7 T., C 3 ´ 108, \ Equation of oscillating magnetic field is, B Z = B 0 sin ( wt + kx ) T, Þ, Bz = 10 - 7 sin ( 2 ´ 1011 t + 300 px ) T, 16. 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., =, I1, 2R, , hc 6 × 6 ´ 10 - 34 ´ 3 ´ 108, =, J, l, 482 ´ 10 - 9, =, , 6 × 6 ´ 10 - 34 ´ 3 ´ 108, , 482 ´ 10 - 9 ´ 1× 6 ´ 10 - 19, 66 ´ 3000, =, 1027 ´ 16, , eV, , = 2.57 eV, Now,, DE = | - 3 × 4 - ( - 0 × 85) |, = 2 × 55 eV, Hence, transition shown by arrow B corresponds to emission of l = 482 nm., , R, I1, r, C2, , C1
Page 35 :
34, , Xam idea Physics—XII, , X L = XC, , 24. For power factor unity,, Þ, , L=, , 1, 2, , w C, , =, , 1, 2, , ( 2pf ) C, , =, , Þ, , wL =, , 1, wC, , 1, 2 2, , 4p f C, , Given f = 50 Hz, C =100 mF = 100 ´ 10 - 6 F = 10 - 4 F, 1, L=, H = 0 × 10 H, \, 2, 4 ´ (3 ×14) ´ (50) 2 ´ 10 - 4, V, Current amplitude,, I0 = 0, Z, At resonance,, Z =R, V0 200 2, I0 =, =, = 20 2 A, \, R, 10, = 20 ´ 1× 414 A, = 28 × 3 A, 25. 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. A potential difference V is, Current (I), applied across the ends X and Y of, the wire. This causes an electric field, vdt, at each point of the wire of strength, I, I, V, ...(1), E=, V, l, + –, Due to this electric field, the, electrons gain a drift velocity v d opposite to direction of electric field. If q be the charge passing, through the cross-section of wire in t seconds, then, q, ...(2), Current in wire, 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, then the total charge flowing, in time t will be equal to the total charge on the electrons present within the cylinder PQ., 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, then the number of free electrons, in the cylinder = n ( A v d t), If charge on each electron is - e ( e = 1× 6 ´ 10 - 19 C), then the total charge flowing through a, cross-section of the wire, ...(3), q = ( nA v d t) ( - e) = - neAv d t
Page 36 :
Examination Papers, , 35, , \ Current flowing in the wire,, q - neAv d t, I= =, t, t, i.e.,, current I = - neAv d, ...(4), This is the relation between electric current and drift velocity. Negative sign shows that the, direction of current is opposite to the drift velocity., Numerically I = neAv d, ...(5), Current density, \, Þ, We know that,, , I, = nev d, A, J µ vd ., OR, l, R =r, A, J=, , Resistivity r (10–8 Wm), , If l = 1, A = 1 Þ r = R, Thus, resistivity of a material is numerically equal to the resistance of the conductor having unit, length and unit cross-sectional area., The resistivity of a material is found to be dependent on the temperature. Different materials do, not exhibit the same dependance on temperatures. Over a limited range of temperatures, that is not, too large, the resistivity of a metallic conductor is approximately given by,, …(1), r T = r 0 [1 + a (T - T0 )], where r T is the resistivity at a temperature T and r 0 is the same at a reference temperature T0 . a, is called the temperature co-efficient of resistivity., The relation of Eq. (1) implies that a graph of r T plotted against T would be a straight line. At, temperatures much lower than 0°C, the graph, however, deviates considerably from a straight line, (Figure)., 0.4, , I 0.2, , 0, , 50, 100, 150, Temperature T(K), , Resistivity rT of copper as a, function of temperature T, , 26. Consider a metallic rod OA of length l , which is rotating with angular velocity w in a uniform, magnetic field B , the plane of rotation being perpendicular to the magnetic field. A rod may be, supposed to be formed of a large number of small elements. Consider a small element of length dx, at a distance x from centre. If v is the linear velocity of this element, then area swept by the, element per second = v dx
Page 38 :
CBSE EXAMINATION PAPERS, ALL INDIA–2008, Time allowed : 3 hours, , Maximum marks : 70, , 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–, 4pe o, , m 0 = 4p ´ 10 -7 TmA -1, 2, , 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., 2., 3., 4., 5., 6., 7., 8., , Identify the part of the electromagnetic spectrum to which the following wavelengths belong:, (i) 10 - 1 m, (ii) 10 - 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, if that of 27 Al is 3 × 6 fermi?, When current in a coil changes with time, 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. 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, if the distance between the light, source and the cathode of the cell is doubled?, Draw an equipotential surface for a system, consisting of two charges Q, - Q separated by a, distance ‘ r ’ in air.
Page 39 :
Xam idea Physics—XII, , 9., 10., , 11., , 12., , 13., , Define the term : magnetic dipole moment of a current loop. Write the expression for the magnetic, moment when an electron revolves at a speed ‘ v’, around an orbit of radius ‘ r ’ in hydrogen atom., 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, and (ii) width of the source slit is, increased?, A jet plane is travelling west at 450 ms - 1 . If the horizontal component of earth’s magnetic field at, that place is 4 ´ 10 - 4 tesla and the angle of dip is 30°, find the emf induced between the ends of, wings having a span of 30 m., What is meant by the transverse nature of electromagnetic waves ? Draw a diagram showing the, propagation of an electromagnetic wave along the x-direction, indicating clearly the directions of, the oscillating electric and magnetic fields associated with it., 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. What should be the peak voltage, of the modulating signal, in order to have a modulation index of 80%?, ®, , 14. Obtain the expression for the potential energy of an electric dipole of dipole moment p placed in, ®, , an electric field E., 15. The following graph shows the variation of terminal potential difference V, across a combination, of three cells in series to a resistor, versus the current, i:, 6.0, V (volt), , 38, , 3.0, , 1.0, , 2.0, , i (ampere), , (i) Calculate the emf of each cell., (ii) For what current i will the power dissipation of the circuit be maximum ?, 16. State the law of radioactive decay. If N 0 is the number of radioactive nuclei in the sample at some, initial time, t 0 , find out the relation to determine the number N present at a subsequent time. Draw, a plot of N as a function of time., OR, Draw a plot of the binding energy per nucleon as a function of mass number for a large number of, nuclei., Explain the energy release in the process of nuclear fission from the above plot. Write a typical, nuclear reaction in which a large amount of energy is released in the process of nuclear fission., 17. In the figure given below, light rays of blue, green, red wavelengths are incident on an isosceles, right-angled prism. Explain with reason, which ray of light will be transmitted through the face, AC. The refractive index of the prism for red, green, blue light are 1× 39, 1× 424, 1× 476 respectively.
Page 40 :
Examination Papers, , 39, , A, , Red, Green, Blue, 90°, B, , C, , 18. Two wires X, Y have the same resistivity, but their cross-sectional areas are in the ratio 2 : 3 and, lengths in the ratio 1 : 2. They are first connected in series and then in parallel to a d.c. source., Find out the ratio of the drift speeds of the electrons in the two wires for the two cases., 19. 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., 20. Distinguish between isotopes and isobars. Give one example for each of the species. A radioactive, isotope has a half-life of 5 years. How long will it take the activity to reduce to 3 ×125%?, 21. Two signals A, B as given below, are applied as input to (i) AND (ii) NOR and (iii) NAND gates., Draw the output wave-form in each case., V, Input A, t, , 0, , Input B, , t1, , t2, , t3, , t4, , t5, , t6, , t7, , t8, , t1, , t2, , t3, , t4, , t5, , t6, , t7, , t8, , t, , 0, , 22. What does the term ‘LOS communication’ mean ? Name the types of waves that are used for this, communication. Give typical examples, with the help of a suitable figure, of communication, systems that use space wave mode propagation., X, 23. A resistance R = 2 W is connected to one of the gaps in a, R, metre bridge, which uses a wire of length 1 m. An, unknown resistance X > 2 W is connected in the other, G, gap as shown in the figure. The balance point is noticed, J, at ‘ l’ from the positive end of the battery. On, B, A, (100–l) cm, l cm, interchanging R and X, it is found that the balance point, further shifts by 20 cm (away from end A). Neglecting, the end correction, calculate the value of unknown, resistance X used.
Page 41 :
40, , Xam idea Physics—XII, , 24. With a circuit diagram, explain how a zener diode can be used as a voltage regulator., OR, Draw a circuit diagram of a full-wave rectifier. Explain its working principle. Draw the input/output, wave-forms indicating clearly the functions of the two diodes used., 25. Draw a plot showing the variation of power of a lens, with the wavelength of the incident light., A converging lens of refractive index 1× 5 and of focal length 15 cm in air, has the same radii of, curvature for both sides. If it is immersed in a liquid of refractive index 1× 7, find the focal length, of the lens in the liquid., 26. 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, use the expression F = q ( v ´ B) to find the direction of the force F acting, on it., A beam of proton passes undeflected with a horizontal velocity v, through a region of electric and, magnetic fields, mutually perpendicular to each other and perpendicular to the direction of the, beam. If the magnitudes of the electric and magnetic fields are 100 kV/m, 50 mT respectively,, calculate, (i) velocity of the beam v., (ii) force exerted by the beam on a target on the screen, if the proton beam carries a current of 0 × 80 mA., 27. Show that Bohr’s second postulate, ‘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., 28. State the condition for resonance to occur in a series LCR a.c. circuit and derive an expression for, the resonant frequency., Draw a plot showing the variation of the peak current (i m ) with frequency of the a.c. source used., Define the quality factor, Q of the circuit., Calculate the (i) impedance, (ii) wattless current of the given a.c. circuit., C, , R, , 40 V, , 30 V, , 2A, Vrms, , OR, Draw a labelled circuit arrangement showing the windings of primary and secondary coil in a, transformer. Explain the underlying principle and working of a step-up transformer. Write any, two major sources of energy loss in this device., 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. (a) Derive an expression for the energy stored in a parallel plate capacitor C, charged to a potential, difference V., (b) Obtain the equivalent capacitance of the network given below. For a supply of 300 V,, determine the charge and voltage across C 4 .
Page 42 :
Examination Papers, , 41, , 100 pF, C1, 200 pF, C2, , 200 pF, , 200 pF, , C3, , +, , 300 V, , –, , C4, , OR, Explain the principle on which Van de Graaff generator operates. Draw a labelled schematic, sketch and write briefly its working., A Van de Graaff type generator is capable of building up potential difference of 15 ´ 10 6 V. The, dielectric strength of the gas surrounding the electrode is 5 ´ 107 Vm - 1 . What is the minimum, radius of the spherical shell required., 30. Draw a labelled ray diagram of a compound microscope and write an expression for its, magnifying power., The focal length of the objective and eye-lens of a compound microscope are 2 cm, 6 × 25 cm, respectively. The distance between the lenses is 15 cm. (i) How far from the objective lens, will, the object the be kept, so as to obtain the final image at the near point of the eye? (ii) Also, calculate its magnifying power., OR, Draw a labelled ray diagram of an astronomical telescope, in the normal adjustment position and, write the expression for its magnifying power., An astronomical telescope uses an objective lens of focal length 15 m and eye-lens of focal length, 1 cm. What is the angular magnification of the telescope?, If this telescope is used to view moon, 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)., , CBSE (All India) SET–II, Questions different from Set–I, 2., , What is the stopping potential of a photocell, in which electrons with a maximum kinetic energy, of 6 eV are emitted ?, 4. Identify the part of the electromagnetic spectrum to which the following wavelengths belong, (i) 1 mm, (ii) 10 - 11 m., 5. State the reason, why a photodiode is usually operated at a reverse bias., 13. Using Ampere’s circuital law, derive an expression for the magnetic field along the axis of a, toroidal solenoid.
Page 43 :
42, , Xam idea Physics—XII, , 18. Derive an expression for the torque acting on an electric dipole, which is held in a uniform electric, field, when the axis of the dipole makes an angle q with the electric field., 19. Draw a plot showing the variation of power of a lens with the wavelength of the incident light., 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. If it is immersed in a liquid of refractive index 1× 7, calculate the focal, length of the lens in the liquid., 20. 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, use the expression F = q ( v ´ B) to find the direction of the force F acting, on it., A beam of proton passes undeflected with a horizontal velocity v, through a region of electric and, magnetic fields, mutually perpendicular to each other and normal to the direction of the beam. If, the magnitudes of the electric and magnetic fields are 50 kV/m and 50 mT respectively, calculate, (i) velocity v of the beam., (ii) force with which it strikes a target on a screen, if the proton beam current is equal to 0 × 80 mA., X, 26. A resistance R = 5 W is connected to one of the gaps in a, R, metre bridge, which uses a wire of length 1 m. An, unknown resistance X > 5 W is connected in the other, G, gap as shown in the figure. The balance point is noticed, J, at ‘l’ cm from the positive end of the battery. On, B, A, (100–l) cm, l cm, interchanging R and X, it was found that the balance, point further shifts by 20 cm away from end A., Neglecting the end correction, calculate the value of, unknown resistance X used., 27. Draw a labelled circuit diagram of a full-wave rectifier and briefly explain its working principle., OR, Draw a labelled circuit diagram of a transistor amplifier in the common-emitter configuration., Briefly explain, how the input/output signals differ in phase by 180°., , CBSE (All India) SET–III, Questions different from Set–I & Set–II, 2., 4., 5., 6., , A diverging lens of focal length ‘F’ is cut into two identical parts each forming a plano-concave, lens. What is the focal length of each part ?, What is the stopping potential applied to a photocell, if the maximum kinetic energy of electrons, emitted is 5 eV ?, Draw an equipotential surface for a uniform electric field., State the factor, which controls :, (i) wavelength of light, and (ii) intensity of light, emitted by an LED.
Page 44 :
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., 17. In a single slit diffraction experiment, the width of the slit is made double the original width. How, does this affect the size and intensity of the central diffraction band ? Draw a plot of the intensity, distribution., X, 22. A resistance R = 4 W is connected to one of the gaps in a, R, metre bridge, which uses a wire of length 1 m. An, unknown resistance X > 4 W is connected in the other, G, gap as shown in the figure. The balance point is noticed, J, at ‘l’ from the positive end of the battery. On, B, A, (100–l) cm, l cm, interchanging R and X, it is found that the balance point, further shifts by 20 cm (away from end A). Neglecting, the end correction, calculate the value of unknown, resistance X used., 24. 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, use the expression F = q ( v ´ B) to find the direction of the force F acting, on it., A beam of proton passes undeflected with a horizontal velocity v, through a region of electric and, magnetic fields, mutually perpendicular to each other and normal to the direction of the beam. If, the magnitudes of the electric and magnetic fields are 50 kV/m, 100 mT respectively, calculate, (i) velocity v of the beam., (ii) force with which it strikes a target on the screen, if the proton beam current is equal to 0 × 80 mA., 26. Distinguish between paramagnetic and diamagnetic substances. A magnetising field of 1500 A/m, produces a flux of 2 × 4 ´ 10 - 5 weber in a bar of iron of cross-sectional area 0 × 5 cm 2 . Calculate the, permeability and susceptibility of the iron-bar used., , Solutions, CBSE (All India) SET–I, 1., 2., 3., , (i) 10 - 1 m = 10 cm belongs to short radiowaves., (ii) 10 - 12 m = 0 × 01 Å belongs to gamma rays., If the reverse bias across a p-n junction is decreased, the depletion region of p-n junction, decreases., Nuclear radius, R = R 0 A 1/ 3, Þ, R µ A 1/ 3, For Al, A = 27, R Al = 3 × 6 fermi, for Fe A =125, \, Þ, , R Fe æ A Fe, =ç, RAl çè AAl, 5, R Fe = RAl, 3, , 1/ 3, , ö, ÷, ÷, ø, , 1/ 3, , æ 125 ö, =ç, ÷, è 27 ø, , 5, = ´ 3 × 6 fermi = 6 × 0 fermi, 3
Page 45 :
44, , Xam idea Physics—XII, , 4., 5., , The back emf induced in the coil opposes the change in current., 1 1 1, 1 1 1, = - Þ = +, f v u, v f u, Here, u = - f and for a concave lens f = - f, f, 1, 1 1, =- - Þv=\, v, f f, 2, , 6., , That is image will be formed between optical centre and focus of lens; towards the side of the, object., The wavefront is spherical of decreasing radius., , F, , 7., , Stopping potential remains unchanged., Reason: On doubling the distance between the light source and the cathode of the cell, the, intensity of light incident on the photocell becomes one-fourth. As stopping potential does not, depend on intensity, the stopping potential remains unchanged., , 8., +Q, , 9., , –Q, , Magnetic moment of a current loop: The torque on current loop is, t = MB sin q, where q is angle between magnetic moment and magnetic field., t, M=, Þ, B sin q, If B = 1 T, sin q = 1 or q = 90° then M = t., 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., Also,, M = NIA, i.e., magnetic moment of a current loop is the product of number of turns, current flowing in the, loop and area of loop. Its direction is perpendicular to the plane of the loop., Magnetic moment of Revolving Electron,, evr, M=, 2
Page 46 :
Examination Papers, , 45, , b l, =, D d, It is independent of D; therefore, angular separation remains unchanged if screen is moved, lD, away from the slits. But the actual separation between fringes b =, increases, so visibility of, d, fringes increases., (ii) When width of source slit is increased, then the angular fringe width remains unchanged but, s l, fringes becomes less and less sharp; so visibility of fringes decreases. If the condition < is, S d, not satisfied, the interference pattern disappears., 11. The wings of jet plane will cut the vertical component of earth’s magnetic field, so emf is induced, across the wing. The vertical component of earth’s magnetic field., V = H tan q, Given H = 4 × 0 ´ 10 - 4 T, q = 30°, 10. (i) Angular separation b q =, , V = ( 4 × 0 ´ 10 - 4 T) tan 30°, 1, 4, = 4 ´ 10 - 4 ´, =, ´ 10 - 4 T, 3, 3, Induced emf across the wing, e = Vvl, Given v = 450 ms - 1 , l = 30 m, , \, , æ 4, ö, e = çç, ´ 10 - 4 ÷÷ ´ ( 450) ´ 30 V = 3 × 12 V, è 3, ø, 12. Transverse Nature of Electromagnetic Waves:, In an electromagnetic wave, the electric and magnetic field vectors oscillate, perpendicular to the, direction of propagation of wave. This is called transverse nature of electromagnetic wave., \, , ® ®, , ®, , In an electromagnetic wave, the three vectors E, B and K form a right handed system., Accordingly if a wave is propagating along X-axis, the electric field vector oscillates along Y-axis, and magnetic field vector oscillates along Z-axis. Diagram is shown in fig., Y, Envelope of E, E, , B, X, , B, Z, Envelope of B, , 13. High frequency waves require antenna of reasonable length and can travel long distances without, any appreciable power loss; so we need high frequency carrier waves., If E m is the peak value of modulating signal and EC that of carrier wave.
Page 47 :
46, , Xam idea Physics—XII, , ma =, , Modulation index,, , Em, Ec, , Given m a = 80% = 0 × 80, E c = 20 V, \, E m = m a ´ E c = 0 × 80 ´ 20 V = 16 V, 14. 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. The work done in this process may be supposed to be done in two parts., (i) The work done (W1 ) in bringing the dipole perpendicular to electric field from infinity., (ii) Work done (W2 ) in rotating the dipole such that it finally makes an angle q from the direction of, electric field., +q, , qE, , 2l, qE, , From, infinity, –q, , (i) Let us suppose that the electric dipole is brought from, infinity in the region of a uniform electric field such that, ®, , +q, p, , qE, , its dipole moment p always remains perpendicular to, electric field. The electric forces an charges +q and - q, qE, –q, are qE and qE, along the field direction and opposite to, field direction respectively. As charges + q and - q, traverse equal distance under equal and opposite forces;, therefore, net work done in bringing the dipole in the region of electric field perpendicular to, field-direction will be zero, i.e., W1 = 0., (ii) Now the dipole is rotated and brought to orientation making an angle q with the field direction, (i.e., q1 = 90° and q 2 = 0°), therefore, work done, W2 = pE (cos q1 - cos q 2 ), = pE (cos 90° - cos q) = - pE cos q, \ Total work done in bringing the electric dipole from infinity, i. e., Electric potential energy of, electric dipole., U = W1 + W2 = 0 - pE cos q = - pE cos q, In vector form, , ® ®, , U =- p ×E, , 15. (i) Let e be emf and r the internal resistance of each cell., The equation of terminal potential difference, V = e eff - i r int becomes, V = 3 e - i r int, , …(1)
Page 48 :
Examination Papers, , 47, , where r int is effective (total) internal resistance., From fig., when i = 0, V = 6 × 0 V, \ From (1),, 6 = 3e - 0, 6, e= =2V, Þ, 3, i.e., emf of each cell, e = 2 V, (ii) For maximum power dissipation, the effective internal resistance of cells must be equal to, external resistance., From fig., when V = 0, i = 2 × 0 A., \ Equation (1) gives, 0 = 3e - 2 × 0 (r int ), 3e 3 ´ 2, Þ, r int =, =, =3 W, 2×0 2×0, \ For maximum power, external resistance,, R = r int = 3 W, 3´ 2, 3e, Current in circuit,, i=, =, = 1× 0 A, R + r int 3 + 3, Thus, emf of each cell, e = 2 V, and for maximum power dissipation, current in circuit = 1× 0 A, 16. Radioactive decay Law: The rate of decay of radioactive nuclei is directly proportional to the, number of undecayed nuclei at that time., dN, i.e.,, µN, dt, dN, or, = - lN,, dt, where l is the decay constant., Suppose initially the number of atoms in radioactive element is N 0 and N the number of atoms, after time t. According to Rutherford and Soddy law, dN, where l is disintegration constant, = - lN, dt, dN, Þ, = - lt, N, Integrating, ...(1), log e N = - lt + C, where C is a constant of integration., If N 0 is initial number of radioactive nuclei, then at t = 0, N = N 0 ; so, log e N 0 = 0 + C Þ C = log e N 0
Page 49 :
48, , Xam idea Physics—XII, , Substituting this equation in (1), we get, log e N = - lt + log e N 0, Þ, log e N - log e N 0 = - lt, N, Þ, log e, = - lt, N0, N, Þ, = e - lt Þ N = N 0 e - lt, N0, , N0, N, , t, , The graph is shown in figure., OR, The variation of binding energy per nucleon versus mass number is shown in figure., Fe, , 8.6, 8, , Bn, MeV, , 20, , 56, , 180, A, , The binding energy curve indicates that binding energy for nucleon of heavy nuclei is less than, that of middle nuclei. Clearly a heavy nucleus breaks into two lighter nuclei then binding energy per, nucleon will increase and energy will be released in the process. This process is called nuclear fission., 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. 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 - 1 ( 0 × 6775) = 42 × 6°, , As angle of incidence at face AC is 45°,, , C r = sin - 1 ( 0 × 7194) = 46°
Page 50 :
Examination Papers, , 49, , which is smaller than critical angle for red ray but greater than critical angles, for green and blue, rays, therefore, red-way will be transmitted through the face AC., 18. (i) When wires are connected in series :, In series, the current remains the same; so we use the relation i = neAv d ., m, m, Resistivity, r =, n=, Þ, 2, 2, ne t, e tr, \, Þ, , æ m ö, ÷ eAv d, i =ç, ç e 2 tr ÷, è, ø, ie tr, vd =, mA, , For same temperature t is same; v d µ, \, , or, , i=, , m, Av d, e tr, , 1, A, , (v d ) X, A, 3, = Y =, (v d ) Y, AX, 2, , (ii) When wires are connected in parallel : In parallel, the potential difference is the same. In this, case we apply the formula for drift velocity., ær tö, æ e tö V, Þ, vd = ç, vd = ç, ÷E, ÷, è m ø, è mø l, 1, For same temperature t is the same, so v d µ, l, (v d ) X, lY, 2, \, =, =, (v d ) Y lX, 1, 19. Permanent Magnets:The magnets prepared from ferromagnetic materials which retain their, magnetic properties for a long time are called permanent magnets., An efficient way to make a permanent magnet is to place a ferromagnetic rod in a solenoid and pass, a current. The magnetic field of the solenoid magnetises the rod., The materials used for permanent magnet must have the following characteristic properties :, (i) High retentivity so that the magnet may cause strong magnetic field., (ii) High coercivity so that the magnetisation is not wiped out by strong external fields, mechanical, ill-treatment and temperature changes. The loss due to hysteresis is immaterial because the, magnet in this case is never put to cyclic changes., 20., 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., different atomic numbers ( Z ) are called isobars., Examples : 11 H, 12H, 13H, , We know, , R æ 1 ö, =ç, ÷, R0 è 2 ø, , Examples : 13H, 32He, n
Page 52 :
Examination Papers, , 51, , In this communication the space waves (radio or microwaves) travel directly from transmitting, antenna to receiving antenna., dm, dT, , dR, , Earth, , hT, , hR, , Communication System using Space wave mode propagation are (i) LOS communication and, Fig. shows LOS communication system., If transmitting antenna and receiving antenna have heights hT and hR respectively, then Radio, horizon of transmitting antenna,, d T = 2R e hT, where R e is radius of earth and radio horizon of receiving antenna., d R = 2R e hR, \ Maximum line of sight distance, d M = d T + d R, = 2R e hT + 2R e hR, (ii) Television, broadcast, microwave links and satellite communication, The satellite communication is shown in fig. The space wave used is microwave., Communication satellight, , spa, , Transmitting, antenna, , ce w, , ave, , Earth, , Receiving, antenna, , 23. From ‘metre bridge’ formula, R, l, =, X 100 - l, 100 - l, X=, R, Þ, l, (100 - l), …(1), Given R = 2 W \, X=, ´ 2W, l, On interchanging R and X, the balance point is obtained at a distance (l + 20) cm from end A, so, l + 20, l + 20, X, …(2), =, Þ, X=, ´ 2W, R 100 - (l + 20), 80 - l, Equating (1) and (2)
Page 53 :
Xam idea Physics—XII, , (100 - l), l + 20, ´2=, ´2, l, 80 - l, Solving we get l = 40 cm, 100 - l, ´ 2W, l, 100 - 40, X=, ´ 2W, 40, Þ, X =3 W, 24. Zener diode as a Voltage Regulator, The Zener diode makes its use as a voltage regulator due to the following property:, \ Unknown resistance, X =, , When a Zener diode is operated in the breakdown region, 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 Fig. The Zener diode is, connected across load such that it is reverse biased., The series resistance R absorbs the output voltage fluctuations so as to maintain constant voltage, across the load., IL, IZ, Vin, , VZ, , RL, , V0, , Regulated output, , I, , R, Unregulated input, , 52, , If the input dc voltage increases, the current through R and Zener diode also increases. So, voltage, drop across R increases, without any change in the voltage across zener diode., I-V Characteristics, , I (mA), VR, , Zener, voltage, , O, VF, , VZ, , I(mA)
Page 54 :
OR, Full Wave Rectifier : For full wave rectifier we use two junction, diodes. The circuit diagram for full wave rectifier using two junction, diodes is shown in figure., , Input A.C. signal, , Examination Papers, , P1, , I N, , S1, P, B, S, , 53, , 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, then diode I is, forward biased and diode II is reverse biased. Therefore current flows in diode I and 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., Input, wave form, T, , Output wave form, of full wave, rectifier, , 25. Refractive index n = A +, , B, l2, , T, , 2T Time, , 2T Time, , , where l is the wavelength., , æ 1, 1, 1 ö, ÷, = ( n g - 1) çç, ÷, f, è R1 R 2 ø, Clearly, power of a lens µ ( n g - 1). This implies that the power of a lens decreases with increase, æ, ö, 1, of wavelength çç P µ, nearly ÷÷ . The plot is shown in figure., 2, l, è, ø, Given f a =15 cm, n g = 1× 5, n l = 1× 7, Focal length of lens in liquid,, ng -1, 1× 5 - 1, fl =, ´ fa =, ´ 15 cm, ng, 1× 5, 1, P, -1, 1× 7, nl, P=, , Power of a lens, , =, ®, , ®, , 0 × 5 ´ 1× 7, ´ 15 cm = - 63 × 75 cm, 1× 5 - 1× 7, , ®, , F =q v ´ B, , 26., , ®, , ®, , Given v = v $i , B = B $j, \, , ®, , F = q (v i$) ´ ( B $j) = qvB k$, , l
Page 55 :
54, , Xam idea Physics—XII, , 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 =100 kV/m = 100 ´ 10 3 V/m, B = 50 mT = 50 ´ 10 - 3 T, \, , v=, , 100 ´ 10 3, 50 ´ 10, , -3, , = 2 ´ 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, , E, Fe, , v, B, , Fm, , Target, , where n is the number of protons striking the target per second., h, 27. The de Broglie wavelength, …(1), l=, mv, Now for electron in orbit, (for nth orbit), 2pr = n l, Using (1), we get, l, 2pr = n, mv, l, Þ, mvr = n, 2p, This is Bohr’s second postulate. As complete the Broglie wavelength may be in certain fixed, orbits; so non-radiating electron can be only in certain fixed orbits., 28. 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., Phase difference ( f) in series LCR circuit is given by, im, XC - X L, tan f=, lm, R, im, For resonance, f= 0, Þ, XC - X L = 0, 2, or, XC = X L, 1, If wr is resonant frequency, then X C =, f1, fr, f2, wr C, and, \, , X L = wr L, 1, = wr L, wr C, , f, , Þ, , wr =, , 1, LC
Page 56 :
Examination Papers, , Linear resonant frequency, f r =, , 55, , wr, 1, =, 2p 2p LC, , The graph of variation of peak current i m with frequency is shown in fig., Half power frequencies are the frequencies on either side of resonant frequency for which current, reduces to half of its maximum value. In fig. f 1 and f 2 are half power frequencies., Quality Factor ( Q ) : The quality factor is defined as the ratio of resonant frequency to the width, of half power frequencies., fr, wr, w L, i.e.,, Q=, =, = r, R, w2 - w1 f 2 - f 1, (i) Potential difference across capacitance, VC = X C I, V, \ Capacitive reactance, X C = C, I, 40, =, = 20W, 2, V, 30, Resistance,, R= R =, = 15 W, I, 2, , C, , R, , 40 V, , 30 V, 2A, , Vrms, , Z = R 2 + X C2 = (15) 2 + ( 20) 2, , Impedance,, , = 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, I wattless = I sin f = I . 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. (a) and, (b)., Soft iron-core, , Np, , Ns, , Secondary, , (a) Two coils on top of each other, , Primary, , Ns, , Secondary, , Primary, , Np, , (b)Two coils on separate limbs of the core, , Transformer: Transformer is a device by which an alternating voltage may be decreased or, increased. This is based on the principle of mutual-induction.
Page 57 :
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. (i. e., N S > N p )., Working: When alternating current, (A.C. mains), source is connected to the ends of, primary coil, the current changes, Primary, continuously in the primary coil; due to, which the magnetic flux linked with the, secondary coil changes continuously,, 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, NS the number of turns in, secondary coil and f the magnetic flux, (a) Step up, linked with each turn. 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. According to Faraday’s laws the emf induced in the primary, coil, Df, ...(1), ep =-Np, Dt, and emf induced in the secondary coil, Df, ...(2), eS = - NS, Dt, From (1) and (2), eS NS, =, ep Np, , ...(3), , If the resistance of primary coil is negligible, the emf ( e p ) induced in the primary coil, will be, equal to the applied potential difference (V p ) across its ends. Similarly if the secondary circuit is, open, then the potential difference VS across its ends will be equal to the emf ( e S ) induced in it;, therefore, VS e S N S, ...(4), =, =, = r (say), Vp e p N p, where r =, , NS, is called the transformation ratio. If i p and i s are the instantaneous currents in, Np, , primary and secondary coils and there is no loss of energy; 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, , ...(5)
Page 58 :
Examination Papers, , 57, , In step up transformer, N s > N p ® r >1;, VS > V p and i S < i p, , So, , i.e. step up transformer increases the voltage but decreases the current., 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)., (ii) Flux Leakage: Energy is lost due to coupling of primary and secondary coils not being perfect,, i.e., whole of magnetic flux generated in primary coil is not linked with the secondary coil., V, 22, Current is secondary coil, I S = S =, A = 0×1A, Z 220, For an ideal transformer, \, , VS I S = V p I p, V I, Ip = S S, Vp, , Current in primary coil,, , 22 ´ 0 ×1, = 0 × 01 A, 220, A, B, 29. (a) When a capacitor is charged by a battery, work is done, –Q –, + +Q, 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., +, –, VB, Consider a capacitor of capacitance C. Initial charge on VA, +, –, capacitor is zero. Initial potential difference between, +, –, capacitor plates = zero. Let a charge Q be given to it in, +, –, +, –, small steps. When charge is given to capacitor, the potential, difference between its plates increases. Let at any instant, VAB = V, when charge on capacitor be q, 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, which may be obtained by integration. Therefore total work, Q, Q q, W = ò V dq = ò, dq, 0, 0, C, =, , 1 é q2 ù, = ê, ú, C êë 2 úû, , Q, , =, 0, , 1, C, , æ Q2 - 0, ç, ç, 2, è, , ö Q2, ÷=, ÷ 2C, ø, , If V is the final potential difference between capacitor plates, then Q = CV, (CV ) 2 1, 1, \, = CV 2 = QV, W=, 2C, 2, 2
Page 59 :
58, , Xam idea Physics—XII, , This work is stored as electrostatic potential energy of capacitor i.e.,, Electrostatic potential energy, U =, (b), , Q2 1, = CV, 2C 2, , 2, , =, , 1, QV, 2, 100pF, , 100pF, C1, 200pF, , 100pF, , 200pF, , C2, , C3, , +, –, , 200pF, , 200pF, , +, –, , C4, , C4, 200pF, +, –, , 100pF, , +, –, , 200pF, C4, , \, C eq =100 pF, Now, Q = C eq ´ V = 100 ´ 10 - 12 ´ 300 = 3 ´ 10 - 8 coulomb, Q, Potential difference across C 4 =, C4, =, , 3 ´ 10 - 8, 200 ´ 10 - 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., Principle: It is based on the following two electrostatic phenomena:, (i) The charge always resides on the outer surface of a hollow conductor., (ii) The electric discharge in air or a gas takes place readily at the pointed ends of the conductors., Construction. 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. The, lower comb C1 is connected to the positive terminal of a very high voltage source (HTS), ( » 104 volts. ) and the upper comb C 2 is connected to the inner surface of metallic sphere S.
Page 60 :
Examination Papers, , 59, , Working: When comb C1 is given very high, potential, then it produces ions in its vicinity, due to, S, action of sharp points. The positive ions, so, produced, get sprayed on the belt due to the, C2, P2, repulsion between positive ions and comb C1 . These, positive ions are carried upward by the moving belt., The pointed end of C 2 just touches the belt. The, comb C 2 collects positive charge from the belt, which immediately moves to the outer surface of, sphere S. As the belt goes on revolving, it continues, to take (+ ) charge upward, which is collected by, comb C 2 and transferred to outer surface of sphere, S. Thus the outer surface of metallic sphere S gains, A, B, positive charge continuously and its potential rises to, a very high value., When the potential of a metallic sphere gains very, HTS, high value, the dielectric strength of surrounding air, breaks down and its charge begins to leak, to the, C1 P1, 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., Due to a charged shell (radius R), the maximum electric field is at the surface of the shell. If, maximum charge is Q, then, Q, 1, ...(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, , Here V = 15 ´ 10 6 V, E max = 5 ´ 107 Vm - 1, V, \, R min =, E max, =, , 15 ´ 10 6, 5 ´ 107, , = 3 ´ 10 - 1 m = 30 cm., , ...(2)
Page 61 :
60, , Xam idea Physics—XII, , 30., , Eye piece, uo, , ue, , vo, , D, Objective, , Eye, , B, A", A, , Fo, , A', Fe', , O, , E, , Fe, , B', , B", , ve, , Magnifying power of compound microscope is, v æ, Döü, ÷ï, M = - 0 çç1 +, u0 è, be ÷ø ï, ý for final image at distance of distinct vision, L æ, Dö ï, ç1 + ÷, »f 0 çè, f e ÷ø ïþ, v D, L D, for final image at infinity, M =- 0, »u0 f e, f0 fe, Given f 0 = 2 × 0 cm, f e = 6 × 25 cm, L =15 cm, u 0 = ?, (i) When final image is formed at least distance of distinct vision (D = 25 cm) :, For eye lens : Here v e = - 25 cm, 1, 1, 1, \, =, f e ve ue, Þ, , -1 - 4, 1, 1, 1, 1, 1, =, ==, ue ve f e, 25 6 × 25, 25, , or, u e = - 5 cm, As, L = | v 0 | + | u e | Þ | v 0 | = L - | u e | = 15 - 5 = 10 cm, For objective lens:, 1, 1, 1, =, f 0 v0 u0
Page 62 :
Examination Papers, , 61, , 1, 1, 1, 1 1, 2, =, = - =u 0 v 0 f 0 10 2, 5, 5, u 0 = - = - 2 × 5 cm, 2, , Þ, , That is distance of object from objective is 2 × 5 cm., , æ, 10 æ, 25 ö, D ö, ç 1+, ÷ =÷ = - 4 ´ 5 = - 20, ç 1+, 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 - f e = 15 - 6 × 25 = 8 × 75 cm, For objective lens :, 1, 1, 1, =, f 0 v0 u0, M =-, , Magnification,, , v0, u0, , 1, 1, 1, 1, 1 2 - 8 × 75, =, =, - =, u 0 v 0 f 0 8 × 75 2 2 ´ 8 × 75, , Þ, , 2 ´ 8 × 75, 6 × 75, \, u0 = - 2 × 59 cm, |u0 | = 2 × 59 cm, v D, 8 × 75 æ 25 ö, Magnification, M = - 0 ×, =×ç, ÷ = - 13 × 5, 2 × 59 è 6 × 25 ø, u0 f e, OR, f0, Magnifying power m = × It does not change with increase of aperature of objective lens,, fe, because focal length of a lens has no concern with the aperture of lens., u0 = -, , B, , A, , fo, , fe, , Fo, A' Fe', , C1, , B', , in, At, , i, fin, , ty, , Drawbacks:, (i) It is not free from chromatic aberration., (ii) The image formed is inverted and fainter., (a) Given f 0 = 15 m, f e = 1× 0 cm = 1× 0 ´ 10 - 2 m, Angular magnification of telescope,, f, 15, m=- 0 == - 1500, fe, 1× 0 ´ 10 - 2, , C2, , Fe
Page 63 :
62, , Xam idea Physics—XII, , Negative sign shows that the final image is inverted., Let D be diameter of moon, d diameter of image of, moon formed by objective and r the distance of moon, from objective lens, then from Fig., D d, =, r f0, Þ, , Objective lens, , Moon, , D, a, r, , Image of, moon, a, , 6, , 3 × 48 ´ 10, D, d = × f0 =, ´ 15 m, r, 3 × 8 ´ 108, = 0 × 137 m = 13 × 7 cm, , d, , f0, , CBSE (All India) SET–II, 2., , E k = eV0 Þ 6 eV = eV0 Þ V0 = 6V, , The stopping potential V0 = 6 volt (negative)., 4. (i) wavelength 1 mm belongs to the microwaves., (ii) wavelength 10 - 11 m = 0 ×1 Å belongs to gamma rays., 5. 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. So, photodiodes are used to, measure the intensity in reverse bias condition., 13. 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)., r, Let n be the number of turns per unit length of toroid and I the current, flowing through it. The current causes the magnetic field inside the turns, O, of the solenoid. The magnetic lines of force inside the toroid are in the, form of concentric circles. 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., I, I, For points inside the core of toroid, Consider a circle of radius r in the region enclosed by turns of toroid. Now we apply Ampere’s, circuital law to this circular path, i. e.,, ®, , ®, , ®, , ®, , ò B•dl, , ò B•dl, , =m I, = ò Bdl cos 0 = B • 2pr, , 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, , ...(1)
Page 64 :
Examination Papers, , 63, , B 2pr = m 0 ( n 2prI ), Þ, B = m 0 nI, 18. Consider an electric dipole placed in a uniform electric, field of strength E in such a way that its dipole moment, , B, , F1 = qE, , ®, , p makes an angle q with the direction of E . The, charges of dipole are - q and + q at separation 2l the, dipole moment of electric dipole,, ...(1), p = q.2l, ®, , r, Force: The force on charge + q is, F1 = qE, along the, , 2l, , F2 = – qE, , A q, –q, , q, , 2l sin q, , ®, , +q, , O, , E, , N, , ®, , direction of field E, , ®, , ®, , The force on charge - q is, F2 = qE , opposite to the direction of field E, ®, , ®, , Obviously forces F1 and F2 are equal in magnitude but opposite in direction; hence net force on, electric dipole in uniform electric field is, F = F1 - F2 = qE - qE = 0 (zero), As net force on electric dipole is zero, 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. This couple is called the torque and is denoted by t., \ 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)], ....(2), = pE sin q, Clearly, the magnitude of torque depends on orientation ( q) of the electric dipole relative to, ®, , ®, , electric field. Torque ( t) is a vector quantity whose direction is perpendicular to both p and E ., In vector form, , ®, , ®, , ®, , t =p ´E, , ...(3), , Thus, if an electric dipole is placed in an electric field in oblique orientation, it experiences no, force but experiences a torque. The torque tends to align the dipole moment along the direction, of electric field., Maximum Torque: For maximum torque sin q should be the maximum. As the maximum value, of sin q =1 when q = 90°, \ Maximum Torque, t max = pE
Page 65 :
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) çç, ÷, f, R, R, 2 ø, è 1, , 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 çç P µ, nearly ÷÷ ., 2, 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, =, ®, , ®, , P, , l, , 0 × 5 ´ 1× 7, ´ ( -20) cm = + 85 cm, 1× 5 - 1× 7, , (Converging), , ®, , F =q v ´ B, , 20., Given,, , ®, , v = vk$,, , ®, , B = B i$, , ®, , F = q (vk$) ´ ( B i$) = qvBj$, , 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, v=, Þ, 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, , E, Fe, , 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.
Page 66 :
Examination Papers, , 65, , 26. From ‘metre bridge’ formula, R, l, =, X 100 - l, 100 - l, X=, R, Þ, 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., Output, , Input signal, , X=, , Input A.C. signal, , \ Unknown resistance,
Page 67 :
66, , Xam idea Physics—XII, C, , IC, , IB, B, Vi, , E, , Ri, , VEE, –, , Output, voltage, V0, , RL, , IE, +, , +, , –, VCC, , 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, F =Þ, 2 ( n g - 1), For each planoconcave lens, 1, æ 1 1ö, = ( n g - 1) ç - - ÷, F¢, è R ¥ø, R, F¢ = = 2F, Þ, ( n g - 1), , 4., , 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)., , 5., E
Page 68 :
Examination Papers, , 6., , 67, , (i) Wavelength of light emitted depends on the nature of semiconductor., (ii) Intensity of light emitted depends on the forward current., , 11. Gauss Theorem : The net outward electric flux through a closed surface is equal to, , 1, times the, e0, , 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, , òS, , ®, , ®, , ®, , E ·d S =ò, , ®, , E · d S1 + ò, , S1, , S2, , ®, , ®, , E ·d S2 + ò, , òS E dS1 cos 0° + òS, = ò E dS1 + 0 + 0, S, = E ò dS1, , S3, , =, , 1, , 2, , ®, , ®, , E ·d S3, , 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
Page 69 :
68, , Xam idea Physics—XII, , Þ, , E . 2p rl =, , Þ, , E=, , l, , 1, ( ll), e0, , 2pe 0 r, , Thus, the electric field strength due to a line charge is inversely proportional to r., 2l 1, µ × When width of slit ‘a’ is doubled, the, a, a, size of central band becomes half and the intensity is doubled., , 17. The angular size of central diffraction band, 2q =, , Intencity, I0, , –3l/a, , –2l/a, , -l/a, , 0, , l/a, , 2l/a, , 3l/a, , 22. From ‘metre bridge’ formula, 100 - l, R, l, X=, R, =, Þ, l, X 100 - 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=, , ®, , F =q v ´ B, , 24., Given,, , ®, , v = vi$,, , ®, , ®, , B = Bj$, , F = q (vi$) ´ ( Bj$) = qvBk$
Page 70 :
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=, , E, Fe, , v, B, , Fm, , Target, , Dp mv mv mvi mvi, =, =, =, =, Dt Dt q / i, q, ne, , where n is the number of protons striking the target per second., 26. Distinction between diamagnetic, paramagnetic and ferromagnetic substances, Property, , Diamagnetic, , Paramagnetic, , (i) Susceptibility ( c ), , - 1 £ c < 0 (negative and small), , 0 < c < e (positive and small), , (ii) Permeability (m r ), , 0 £ m r < 1 (less than 1), , 1 < m r < 1 + e (slightly greater than, 1), , (iii) Coercivity, , High, , Low, , Gold, , Platinum, , Example, , The magnetic field lines near a diamagnetic substance and a paramagnetic substance are shown, below:, , N, , S, , (i) Field lines near, a Diamagnetic, , N, , S, , (ii) Field lines near, a Paramagnetic
Page 71 :
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
Page 72 :
CBSE EXAMINATION PAPERS, DELHI–2009, Time allowed : 3 hours, , Maximum marks : 70, , 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, , e = 1 × 602 ´ 10 -19 C, 1, = 9 × 109 Nm2C– 2, 4pe o, , h = 6 × 626 ´ 10 -34 Js, m 0 = 4p ´ 10 -7 TmA -1, , 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?, 1, Write the following radiations in ascending order in respect of their frequencies :, 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, , Power (P), , Aperture (A), , L1, , 3D, , 8 cm, , L2, , 6D, , 1 cm, , L3, , 10 D, , 1 cm
Page 73 :
72, 5., 6., , 7., 8., 9., , 10., , 11., 12., 13., , 14., , 15., , Xam idea Physics—XII, , If the angle between the pass axis of polarizer and the analyser is 45°, write the ratio of the, intensities of original light and the transmitted light after passing through the analyser., 1, Photoelectric, The figure shows a plot of three curves a, b, c showing the, current, variation of photocurrent vs. collector plate potential for, I1, I2, three different intensities I 1 , I 2 and I 3 having frequencies, I3, n 1 , n 2 and n 3 respectively incident on a photosenitive, c, b, surface., a, Point out the two curves for which the incident radiations, Collector plate potential, have same frequency but different intensities., 1, What type of wavefront will emerge from a (i) point source, and (ii) distant light source? 1, Two nuclei have mass numbers in the ratio 1 : 2. 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. Predict from the, graph the condition under which ‘V’ becomes equal to ‘E‘., (i) Can two equi-potential surfaces intersect each other? Give reasons., (ii) Two charges - q and + q are located at points A ( 0, 0, - a) and B ( 0, 0, + a) respectively., How much work is done in moving a test charge from point P (7 , 0, 0) to Q ( -3, 0, 0)?, By what percentage will the transmission range of a T.V. 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., 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., ®, , 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., 2, (i) Trace its trajectory., (ii) Does the charge gain kinetic energy as it enters the magnetic, field? Justify your anwer., , B, O, , Z-axis, , Y, , q, X, , The following figure shows the input waveforms ( A , B) and the output wavefrom (Y) of a, gate. Identify the gate, write its truth table and draw its logic symbol., A, , B, , Y, 0, , 1, , 2, , 3, , 4, , 5, , 6, , 7
Page 74 :
Examination Papers, , 16., , State Biot-Savart law., A current I flows in a conductor placed perpendicular to the, plane of the paper. Indicate the direction of the magnetic field, ®, , 17., , 18., , 73, , Z, I, , ®, , dl, due to a small element d l at point P situated at a distance r, O, Y, r, from the element as shown in the figure., P, Why are high frequency carrier waves used for transmission?, X, OR, What is meant by term ‘modulation’? Draw a block diagram of a simple modulator for, obtaining an AM signal., A radioactive nucleus ‘A’ undergoes a series of decays according to the following scheme :, b, , a, , g, , a, , A ¾¾® A1 ¾¾® A2 ¾¾® A 3 ¾¾® A4, , 19., , 20., , 21., , The mass number and atomic number of A are 180 and 72 respectively. What are these, numbers for A4 ?, A thin conducting spherical shell of radius R has charge Q spread uniformly over its surface., Using Gauss’s law, derive an expression for an electric field at a point outside the shell., Draw a graph of electric field E(r) with distance r from the centre of the shell or 0 £ r £ ¥., C1, Three identical capacitors C 1 , C 2 and C 3 of capacitance 6 mF, each are connected to a 12 V battery as shown., +, Find:, 12 V, C3, (i) charge on each capacitor, –, (ii) equivalent capacitance of the network, C2, (iii) energy stored in the network of capacitors, (a) The energy levels of an atom are as shown below. Which of them will result in the, transition of a photon of wavelength 275 nm?, A, , B, , 0 eV, C, , D, , – 2 eV, – 4.5 eV, – 10 eV, , (b) Which transition corresponds to emission of radiation of maximum wavelength?, 22., , 23., , 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, and (ii) less kinetic, enrgy? Justify your answers., In a single slit diffraction experiment, when a tiny circular obstacle is placed in the path of, light from a distant source, a bright spot is seen at the centre of the shadow of the obstacle., Explain why?
Page 75 :
74, , 24., , 25., , 26., , 27., , 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., (a) Define self inductance. Write its S.I. units., (b) Derive an expression for self inductance of a long solenoid of length l, cross-sectional area, A having N number of turns., X, Y, The figure shows experimental set up of a meter, bridge. When the two unknown resistances X and Y, B, are inserted, the null point D is obtained 40 cm from, the end A. When a resistance of 10 W is connected in, G, series with X, the null point shifts by 10 cm. Find the, A, C, D, position of the null point when the 10 W resistance is, instead connected in series with resistance ‘Y’., Determine the values of the resistances X and Y., Derive the expression for force per unit length between two long straight parallel current, carrying conductors. Hence, define one ampere., OR, Explain the principle and working of a cyclotron with the help of a schematic diagram. Write, the expression for cyclotron frequency., a, Three light rays red (R), green (G) and blue ( B) are incident, on a right angled prism ‘abc’ at face ‘ab’. The refractive, indices of the material of the prism for red, green and blue B, wavelengths are 1 × 39, 1 × 44 and 1 × 47 respectively. Out of, the three which colour ray will emerge out of face ‘ac’? G, Justify your answer. Trace the path of these rays after R, 45°, passing through face ‘ab’., b, , c, , 28., , (a) Derive an expression for the average power consumed in a series LCR circuit connected to, a.c. source in which the phase difference between the voltage and the current in the circuit, is f., (b) Define the quality factor in an a.c. circuit. Why should the quality factor have high value, in receiving circuits? Name the factors on which it depends., OR, (a) Derive the relationship between the peak and the rms value of current in an a.c. circuit., (b) Describe briefly, with the help of a labelled diagram, working of a step-up transformer., A step-up transformer converts a low voltage into high voltage. Does it not violate the, principle of conservation of energy? Explain., , 29., , (i) Draw a circuit diagram to study the input and output characteristics of an n-p-n transistor, in its common emitter configuration. Draw the typical input and output characteristics., (ii) Explain, with the help of a circuit diagram, the working of n-p-n transistor as a common, emitter amplifier.
Page 76 :
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., Explain briefly, with the help of a circuit diagram, how a p-n junction diode works as a, half wave rectifier., 30., , 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 . Establish, the relation between the distances of the object, the image and the radius of curvature from, the central point of the spherical surface., Hence, derive the expression of the lens maker’s formula., OR, Draw the labelled ray diagram for the formation of image by a compound microscope., Derive the expression for the total magnification of a compound microscope. Explain why, both the objective and the eye piece of a compound microscope must have short focal lengths., , CBSE (DELHI) SET–II, Questions different from Set–I., 1., , Name the electromagnetic radiation to which waves of wavelength in the range of 10 -2 m, belong. Give one use of this part of EM spectrum., 1, , 2., , What is ground wave propagation?, , 5., , Unpolarized light is incident on a plane surface of refractive index m at angle i. If the reflected, light gets totally polarized, write the relation between the angle i and refractive index m., , 6., , Draw a diagram to show refraction of a plane wavefront incident on a convex lens and hence, draw the refracted wave front., 1, , 8., , The nuclei have mass numbers in the ratio 1 : 3. What is the ratio of their nuclear densities?1, , 11., , The output of a 2-input AND gate is fed to a NOT gate. Give the name of the combination and, its logic symbol. Write down its truth table., 2, , 16., , A radioactive nucleus ‘A’ undergoes a series of decays according to the following scheme: 2, a, , b, , 1, , a, , g, , 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. What are these, numbers for A 4 ?, 19., , The equivalent capacitance of the combination between A and B in the given figure is 4 mF. 3, A, , 20mF, , (i) Calculate capacitance of the capacitor C., , C, , B
Page 77 :
76, , Xam idea Physics—XII, , (ii) Calculate charge on each capacitor if a 12 V battery is connected across terminals A and B., (iii) What will be the potential drop across each capacitor?, 20., , State Gauss’s law in electrostatic. Using this law derive an expression for the electric field due, to a uniformly charged infinite plane sheet., 3, , 22., , An electron and a proton are accelerated through the same potential. Which one and the two, has (i) greater value of de-Broglie wavelength associated with it and (ii) less momentum?, Justify your answer., 3, , CBSE (DELHI) SET–III, Questions different from Set–I and Set–II., 2., , At what angle of incidence should a light beam strike a glass slab of refractive index 3, such, that the reflected and the refracted rays are perpendicular to each other?, 1, , 3., , What is space wave propagation?, , 1, , 5., , Name the part of electromagnetic spectrum which is suitable for :, (i) radar systems used in aircraft navigation, , 1, , (ii) treatment of cancer tumours., 6., , Two nuclei have mass numbers in the ratio 2 : 5. What is the ratio of their nuclear densities? 1, , 7., , Differentiate between a ray and a wavefront., , 12., , (i) Sketch the output wavefrom from an AND gate for the inputs A and B shown in the, figure., 2, A, , 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, name the gate of the, combination so formed., 18., , A radioactive nucleus ‘A’ undergoes a series of decays according to the following scheme : 2, a, , b, , a, , g, , 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. What are these, numbers for A 4 ?, 19., , State Guass’s law in electrostatics. 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 ., 3
Page 78 :
Examination Papers, , 23., , X, , Two parallel plate condition X and Y, have the same area of, plates and same separation between them. X has air between, the plates while Y contains a dielectric medium of Îr = 4., (i) Calculate capacitance of each capacitor if equivalent, capacitance of the combination is 4 mF., (ii) Calculate the potential difference between the plates of X, and Y., (iii) What is the ratio of electrostatic energy stored in X and Y?, , 77, , Y, , 12 V, , 3, , Solutions, CBSE (Delhi) SET–I, 1., 2., 3., , 4., , 5., , 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., Radiowaves, microwaves, UV rays, X-rays., Magnetic field lines can be entirely confined within the core of a toroid because toroid has no, ends. A solenoid is open ended and the field lines inside it which is parallel to the length of, the solenoid, cannot form closed curved inside the solenoid., An astronomical telescope has an eyepiece of shorter aperture and shorter focal length while, an objective of longer aperture and longer focal length., Therefore, we will use L 3 as eyepiece and L 1 as objective., I, Transmitted intensity, I transmitted = 0 cos 2 q, 2, I0/2, , I0, , I0/2 cos2 q, , q, , Polarizer, , Analyzer, , Here q = 45°, \, 6., 7., 8., , I Original, I transmitted, , =, , I0, I transmitted, , =, , 2, 2, , cos 45°, , =, , 4, 1, , Curves a and b have different intensities but same stopping potential, so curves ‘a’ and ‘b’, have same frequency but different intensities., Wavefront from a point source – spherical, Wavefront from a distant light source – plane., Nuclear density is independent of mass number, so ratio 1 : 1.
Page 79 :
78, 9., , 10., , Xam idea Physics—XII, , æ E ö, E, Terminal potential difference, V = IR = ç, ÷R=, r, èR + rø, 1+, R, When R ® 0, V = 0, E, When R = r , V =, 2, When R = ¥ , V = E, The graph is shown in fig., , E, V, E, 2, , O, , r, , R, , (i) No, Reason: At the point of intersection, there will be two different directions of electric field,, which is not possible., (ii) Work done in moving test charge atom P to Q is zero., X, P (7, 0, 0), , A (0, 0, – a), , B (0, 0, a), , –q, , +q, , Z, , Y, Q (–3, 0, 0), , Reason: Test charge is moved along the equatorial line of an electric dipole. As potential at, every point on equatorial line is zero, so work done, W = q0 (VQ - VP ) = q0 ( 0 - 0) = 0., 11., , Transmission range of a TV tower, d = 2hR, If height is increased by 21 %, new height, h ¢ = h +, , h¢, = 1 × 21 = 1 × 1, h, d¢ - d, Dd, % increase in range,, ´ 100% =, ´ 100%, d, d, æ d¢, ö, = ç - 1÷ ´ 100% = (1.1 - 1) ´ 100% = 10%, èd, ø, , If d ¢ is the new range, then, , d¢, =, d, , 21, h = 1 × 21 h, 100
Page 80 :
Examination Papers, , 12., , Consider a metallic conductor XY of length l, and cross-sectional area A. A potential, difference V is applied across the conductor, XY. Due to this potential difference an electric, , 79, , l, E, X, , Y, , ®, , field E is produced in the conductor. The, V, magnitude of electric field strength E = and, l, its direction is from Y to X. This electric field, exerts a force on free electrons; due to which, electrons are accelerated., ®, , ®, , The electric force on electron F = - e E, , vd, , –, , +, V, , (where e = + 1 × 6 ´ 10 -19 coulomb)., , If m is the mass of electron, then its acceleration, ®, , ®, , ®, , F, eE, …(1), =m, m, This acceleration remains constant only for a very short duration, since there are random, forces which deflect the electron in random manner. These deflections may arise due to, (i) ions of metallic crystal vibrate simple harmonically around their mean positions., Different ions vibrate in different directions and may be displaced by different, amounts., (ii) direct collisions of electrons with atoms of metallic crystal lattice., In any way after a short duration t called relaxation time, the motion of electrons, become random. Thus, we can imagine that the electrons are accelerated only for a, short duration. As average velocity of random motion is zero, if we consider the, average motion of an electron, then its initial velocity is zero, so the velocity of electron, a =, , ®, , ®, , ®, , ®, , ®, , after time t (i. e. , drift velocity v d ) is given by the relation v = u + a t (here u = 0,, ®, , ®, , ®, , v = v d , t = t, a, , ®, , = -, , ®, , 13., , ®, , vd =-, , m, , ), , ®, , vd =0-, , Þ, , eE, , eE, m, , et ®, E, m, , t, …(2), , At given temperature, the relaxation time t remains constant, so drift velocity remains, constant., An oscillating electric charge produces oscillating electric field, which produces oscillating, magnetic field; which in turn produces oscillating electric field and so on; thereby producing, an electromagnetic wave propagating in free space.
Page 81 :
80, , Xam idea Physics—XII, Y, Envelope of E, E, , B, X, , B, Z, Envelope of B, , 14., , Z, , (i) The trajectory is shown in fig., (ii) No, ®, , X', , B, , ®, , ®, , Reason: Magnetic force F m = q v ´ B, , ®, , ®, , r, , Force ( F m ) is perpendicular to velocity v , so work done by the, , 15., , O, , magnetic force on charge is zero; so charge does not gain, kinetic energy on entering the magnetic field., , v, , The logic gate is NAND gate., , X, , Y, , q, , A, Y, B, , Truth Table, Inputs, , 16., , Output, , A, , B, , Y, , 0, , 0, , 1, , 0, , 1, , 1, , 1, , 0, , 1, , 1, , 1, , 0, , 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 =, , ®, , ®, , m 0 I dl ´ r, , dl, , 4p, , I, , r3, where m 0 is permeability of free space. Its value is, m 0 = 4p ´ 10 -7 Wb/A-m., , q, , P, r
Page 82 :
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 magnetic field dB is perpendicular to the plane, , Z, , containing I dl and r ., , I, , ®, , ®, , B, , The direction of current element is along Z-direction and that, ®, , dl, , of r along Y-direction, so magnetic field, ® m ( I dl k$) ´ (r j$) m Idl, B= 0, = 0, ( -i$) ., 3, 4p r 2, 4p, r, 17., , X', , Y, , X, , That is magnetic field is directed along negative X-direction., Use of high frequency carrier wave in transmission of signals:, (i) High frequencey carrier wave reduces the size of antenna as h =, , l, l, or ×, 2, 4, , (ii) High frequency carrier wave radiates more power in space as P µ n 2 ., (iii) High frequency carrier wave avoids mixing up of message signals., OR, Meaning of Modulation: The original low frequency message/information signal cannot be, transmitted over long distances. Therefore, at the transmitter end, information contained in, the low frequency message signal, is susperimposed on a high frequency carrier signal by a, process known as modulation., m(t), Am sin wmt, (modilating, signal), , Square, law, device, , +, , Band pass, filter centered, at wc, , AM wave, , Bx(t), + C x2(t), , c(t), = Ac sin wct, (Carrier), , Block diagram of simple modulator, , 18., , 19., , b - 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, mass number of A4 is 172 and atomic number is 69., 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. To find, the electric field outside the shell, we consider a spherical Gaussian surface of radius r ( > R),, ®, , concentric with given shell. If E is electric field outside the shell, then by symmetry electric, field strength has same magnitude E0 on the Gaussian surface and is directed radially
Page 83 :
82, , Xam idea Physics—XII, , outward. Also the directions of normal at each point is radially outward,, ®, , ®, , so angle between E i and d S is zero at each point. Hence,, electric flux through Gaussian surface = ò, = ò E0 dS cos 0 = E0 . 4pr, , ®, S, , ®, , Q, , E·d S, , EO, , R, r, , 2, , dS, , P, , Now, Gaussian surface is outside the given charged shell,, so charge enclosed by Gaussian surface is Q., Hence, by Gauss's theorem, ®, ®, 1, òS E 0 · d E = e ´ charged enclosed, 0, E0 4pr 2 =, , Þ, , 1, 1 Q, ´ Q Þ E0 =, e0, 4pe 0 r 2, , Thus, electric field outside a charged thin spherical, shell is the same as if the whole charge Q is, concentrated at the centre., If s is the surface charge density of the spherical, shell, then, , E, , R, , r, , f = 4 pR 2 s C, \, , 20., , E0 =, , (i), , 1 4 pR 2 s R 2 s, =, 4pe 0, r2, e 0r 2, , The graph is shown in fig., Capacitors C 1 and C 2 are in series across a 12 V supply while there exists p.d. 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 , 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., , (a), , The energy (E) of a photon of wavelength ( l) is given by
Page 84 :
Examination Papers, , E=, , 83, , -34, ´ 3 ´ 10 8, hc 6 × 626 ´ 10, =, J, l, 275 ´ 10 -9, , =, , 6 × 626 ´ 10 -34 ´ 3 ´ 10 8, 275 ´ 10 -9 ´ 1 × 6 ´ 10 -19, , eV = 4 × 5 eV, , From fig. this transition corresponds to B since for transition B., , 22., , E = 0 - ( - 4 × 5 eV) = 4 × 5 eV, hc 1, (b) Energy of Photon Emitted, E =, µ, l, l, For minimum wavelength of emission, the energy is minimum., This transition A corresponds to emission of radiation of maximum wavelength., h, (i ) de-Broglie wavelength, l =, 2mqV, lp =, \, , h, ,, 2mp qp V, lp, la, , =, , 2ma qaV, , ma qa, ., mp qp, , As ma = 4mp and qa = 2qp, lp, = 4´2 = 8, \, la, , \, , lp > la, , i.e., Proton has greater de-Broglie wavelengths., Kinetic energy, K = qV, Kp q p, e 1, =, =, =, Ka q a 2 e 2, , (ii), , 23., , h, , la =, , Kp < Ka, i.e., proton has less kinetic energy., The waves diffracted at the edge of circular obstacle produce constructive interference at the, centre of the shadow; producing a bright spot., Difference between Interference and Diffraction, Interference, , Diffraction, , 1., , All bright fringes are of equal intensity., , The central maximum has maximum intensity, and the intensity of secondary maxima goes on, decreasing with increase of order., , 2., , Fringe width of all fringes are equal., , The width of central maximum is maximum, and goes on decreasing with increase of order of, secondary maxima.
Page 85 :
84, 24., , Xam idea Physics—XII, , (a), , (b), , The self inductance is defined on the magnetic flux linked with the coil when unit, current flows through it., Or, The self inductance is defined as the emf induced in the coil, when the rate of change of, current in the coil is 1 ampere/second., The SI unit of self-inductance is henry (H)., Self Inductance of a long air-cored solenoid:, Consider a long air solenoid having `n' number of turns per unit length. If current in, solenoid is I , then magnetic field within the solenoid, B = m 0 nI, ...(1), where m 0 = 4p ´ 10 - 7 henry/metre is the permeability of free space., If A is cross-sectional area of solenoid, then effective flux linked with solenoid of, length `l '; F = NBA where N = nl is the number of turns in length ' l ' of solenoid., \, F = (nl BA), Substituting the value of B from (1), ...(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 , then, N, n=, l, N ö2, \ Self-inductance L = m 0 æç, ÷ Al, è l ø, =, , 25., , A, , m0 N 2A, , l, , l, Let r = resistance per cm length of bridge wire, X, 40r, X 2, =, Þ, =, Y (100 - 40) r, Y 3, , …(1), , When a resistance of 10 W is connected in series with X, the null point is obtained at, ( 40 + 10) = 50 cm., X + 10 50r, X + 10, \, =, Þ, =1, Y, 50r, Y, Þ, 3, Y= X, 2, , From (1),, \, , From (2),, , \, , Þ, , Y = X + 10, 3, X = X + 10 Þ X = 20 W, 2, , Y = 20 + 10 = 30 W, , When a resistance 10 W is inserted in series with Y, let the balancing length be l2 ., l2, X, \, =, Y + 10 (100 - l2 ), , …(2)
Page 86 :
Examination Papers, , Þ, , l2, 20, =, 30 + 10 (100 - l2 ), , Þ, , l2, 100 - l2, , =, , 85, , l, 2, , This gives null point length; l2 = 33 × 3 cm., 26., , Force per unit length between two long straight parallel conductors:, 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. It has been, DL, observed experimentally that when, a, B, DF, DF, the currents in the wire are in the, a B, DL, same direction, they experience an, b, I2, I1, attractive force (fig. a) and when they I1, carry currents in opposite directions,, they experience a repulsive force (fig. 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. (fig.)., Consider a current–element ‘ab’ of length DL of wire RS. The magnetic field produced by, current-carrying conductor PQ at the location of other wire RS, m I, ...(1), B1 = 0 1, 2p r, According to Maxwell’s right hand rule or right hand palm rule no. 1, the direction of B1 will, be perpendicular to the plane of paper and directed downward. Due to this magnetic field,, each element of other wire experiences a force. The direction of current element is, perpendicular to the magnetic field; 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, , ...(2), , According to Fleming’s left hand rule, the direction of magnetic force will be towards PQ i.e., the force will be attractive., On the other hand if the currents I 1 and I 2 in wires are in opposite directions, the force will, be repulsive. The magnitude of force in each case remains the same.
Page 87 :
86, , Xam idea Physics—XII, , Definition of Ampere: In S.I. system of fundamental unit of current ‘ampere' has been, defined assuming the force between the two current carrying wires as standard., 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 , r = 1 m, then, m, f = 0 = 2 ´ 10 - 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., OR, The cyclotron, devised by Lawrence and Livingston, is a, device for accelerating ions to high speed by the repeated, application of accelerating potentials., Principle: The positive ions produced from a source, are accelerated. Due to the presence of perpendicular, magnetic field the ion will move in a circular path. 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., Expression for K.E. attained:, If R be the radius of the path and v max the, velocity of the ion when it leaves the, periphery, then, qBR, v max =, m, Beam, The kinetic energy of the ion when it leaves, the apparatus is,, K. E. =, , Magnetic Pole, N, S, Dee, , Dee, , Magnetic Pole, S, , Dee-1, , R.F., oscillator, , S, , Dee-2, , q2 B2 R 2, 1, 2, mv max =, 2, 2m, , When charged particle crosses the gap between dees it gains KE = q V, In one revolution, it crosses the gap twice, therefore if it completes n-revolutions before, emerging the does, the kinetic energy gained, = 2nqV, q2 B2 R 2, , = 2nqV, 2m, Working: The principle of action of the apparatus is shown in fig. The positive ions produced, from a source S at the centre are accelerated by a dee which is at negative potential at that, moment. Due to the presence of perpendicular magnetic field the ion will move in a circular, , Thus, , K.E. =
Page 88 :
Examination Papers, , 87, , path inside the dees. 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., The function of electric field is to accelerate the charged particle and so to impart energy to, the charged particle., 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., Expression for Period of Revolution and Frequency:, Suppose the positive ion with charge q moves in a dee with a velocity v, then,, qvB =, , mv 2, r, , or r =, , mv, qB, , ...(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 angular velocity w of the ion is given by,, v qB, (from equation 1), ...(2), w= =, r m, The time taken by the ion in describing a semi-circle, i.e., in turning through an angle p is,, p pm, ...(3), t= =, w Bq, Thus the time is independent of the speed of the ion i.e., although the speed of the ion goes on, increasing with increase in the radius (from eq. 1) when it moves from one dee to the other,, yet it takes the same time in each dee., m, From eq. (3) it is clear that for a particular ion,, being known, B can be calculated for, q, 27., , producing resonance with the high frequency alternating potential., a, Angle of incidence at face ac for all three, colours,, 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 . This condition is satisfied for red colour, (m = 1 × 39). So only red ray will be transmitted,, Blue and Green rays will be totally reflected., , B, G, R, R, , 45°, b, , c, B, , G
Page 89 :
88, 28., , Xam idea Physics—XII, , (a) In series LCR circuit., Voltage, V = V0 sin wt, Current in circuit, I = I 0 sin ( wt + f), Instantaneous Power, 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 - cos ( 2wt + f)], 2, 2, Average value of cos ( 2wt + f) over a complete cycle is zero i.e., cos ( 2wt + f) = 0., \ 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., w LI, 1, L, L, Q= r, =, ., = wr, RI, R, LC R, Þ, Also,, , 1, R, , L, C, wr, Q=, w2 - w1, Q=, , where w2 - w1 = band width of resonant curve. Smaller is the band width, larger is the, quality factor and selectivity (or sharpness of resonance) of the circuit., That is why in receiving circuits, quality factor must be very high. The quality factor, depends on the values of resistance, inductance and capacitance of the circuit., OR, (a) Relationship between Peak and RMS Value of Current:, Current I = I 0 sin wt, I 2 = I 02 sin 2 wt, Mean square current over full cycle, ( I 2 ) mean = I 02 (sin 2 wt) mean, Mean value of sin 2 wt over full cycle is, T, , i.e.,, , 2, , (sin wt) mean, , ò, = 0, , 1, ., 2, , sin 2 wt dt, T, , ò0, , dt, , =, , 1, 2
Page 90 :
Examination Papers, , \, , ( I 2 ) mean = I 02 ., , 89, , 1, 2, , \ Root mean square current, I rms = ( I 2 ) mean =, or, , I0, 2, , I rms = 0 × 707 I 0, , This is required relation., (A.C. mains), , (b) Transformer: Transformer is a device by, which an alternating voltage may be, decreased or increased. This is based on, the principle of mutual induction., , Primary, , Primary, , 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. (i. e. , N S > N p )., , laminated, iron core, , Core, Secondary, , Working: When alternating current, Step up, source is connected to the ends of, Secondary, primary coil, the current changes, Transformer, continuously in the 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., Let N p be the number of turns in primary coil, NS the number of turns in secondary coil, and f the magnetic flux linked with each turn. 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., According to Faraday’s laws the emf induced in the primary coil, Df, ...(1), e p = - Np, Dt, and emf induced in the secondary coil, Df, e S = - NS, Dt, , ...(2), , From (1) and (2), eS, ep, , =, , NS, Np, , ...(3), , If the resistance of primary coil is negligible, the emf ( e p ) induced in the primary coil, will, be equal to the applied potential difference (Vp ) across its ends. Similarly if the secondary, circuit is open, then the potential difference VS across its ends will be equal to the emf ( e S ), induced in it; therefore
Page 91 :
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, Power in primary = Power in secondary, , For about 100% efficiency,, , Vp ip = VS iS, iS Vp N p 1, =, =, =, ip VS N S r, , \, , ...(5), , In step up transformer, Ns > N p ® r > 1 ;, VS > Vp and iS < ip, , So, , i.e. step up transformer increases the voltage., , mA, , – IB, , +, , –, , mA, VBB, , +, , +, , VBE, –, , Rh1, , –, , +, , IC, C, , +, , +, , B, , VCE, E, IE, , – VCC, , –, , =, 10, , VCE = 5V, , CE, , (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, , Common Emitter Characteristics:, , IB, , 29., , 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., (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., , V, , VBE
Page 92 :
Examination Papers, , 91, , (ii) The circuit of common emitter amplifier using n-p-n transistor is shown below:, IC, C1, , RB, , C, , C2, RL, , IB, B, , Vi, , +, –, , +, , E, VBB, , VCC, , IE, , –, , V0, , Output, waveform, , 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), , Forward, bias, , Vr, VZ, , VF, Reverse, bias, I (mA), , 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.
Page 93 :
92, , Xam idea Physics—XII, , Half Wave Rectifier: The circuit diagram for junction diode as half wave rectifier is shown in, Fig., , p1, , s1, RL, , p2, , s2, B, (a), , 30., , +, , n, , Input, , Output voltage, VDC, , Input A.C. signal, , A, p, , 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)., 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., N, , S, , n1, Rarer medium ‘1’, , A, r, , O, , C, , I, , i, , r, , n2, , h, R, , M, , B, , P, , v, , Denser, medium `2', , u, S', , Let angle of incidence of ray OA be i and angle of refraction be r i.e., Ð OAC = i and Ð NAB = r, Let, 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, , ...(1), ...(2)
Page 94 :
Examination Papers, , sin i, , From Snell’s law, , sin r, , =, , n2, n1, , 93, , ...(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, hö, æ h, æ h, n1 ç, ÷ = n2 ç, ÷, è PC PO ø, è PC PI ø, n1, n, n, n, or, - 1 = 2 - 2, PC PO PC 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, , -, , n2, , R, v, n2 - n1, R, , 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
Page 96 :
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, , ue, , vo, , D, Objective, , Eye, , B, A", A, , Fo, Fe', , O, , A', , B', , B", , ve, Compound microscope, , E, , Fe
Page 97 :
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, ...(1), Magnifying power, M= ., a, , B, , E, , A, D, , 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, 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 ), b = tan b =, , and from figure, , Hence magnifying power, M =, , b A ¢ B¢ /( - ue ) A ¢ B¢ D, =, ×, =, AB / ( - D), AB ue, a, , By sign conventions, magnification of objective lens, v0, A ¢ B¢, =, AB, ( - u0 ), \, , M=-, , Using lens formula, , u0, , ×, , D, ue, , Magnifying power, M=-, , ... (2), , 1 1 1, = - for eye-lens, (i.e. using f = f e , v = - v e , u = - ue ),, f v u, , 1, 1, 1, =, f e - v e ( - ue ), , we get, , or, , v0, , M=-, , or, , 1, 1, 1, =, +, ue f e v e, , æ1, 1ö, Dç + ÷, u0 è f e v e ø, v0, , 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.
Page 98 :
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, , 16., , Output, , A, , B, , Y, , 0, , 0, , 1, , 1, , 0, , 1, , 0, , 1, , 1, , 1, , 1, , 0, , b-, , 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
Page 99 :
98, , Xam idea Physics—XII, , or, Þ, 16C = 80, C = 5 mF, (ii) Charge on each capacitor, 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, V2 = =, = 9×6 V, C, 5 mF, 20., , Gauss Theorem: The net outward electric flux through a closed surface is equal to, , 1, times, e0, , the net charge enclosed within the surface i.e.,, ® ®, 1, òS E · dS = e 0 Sq, Let electric charge be uniformly distributed over the surface of a thin, non-conducting, infinite sheet. Let the surface charge density (i.e., charge per unit surface area) be s. We have, to calculate the electric field strength at any point distance r from the sheet of charge., q=90o, dS3, o, S1 E 90 S3, , S2, , dS1, E, q=0o, , dS2, E, q=0o, B, , A, Sheet, , To calculate the electric field strength near the 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). 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., Total electric flux, ®, , òSE, or, , ®, , òSE, , ®, , •dS =, ®, , •dS, , ®, , òSE1, , =ò, , S1, , ®, , • d S1 +, , ®, , òSE2, , E dS1 cos 0° +, , ®, , • d S2 +, , òS2, , ®, , òSE3, , ®, , • d S3, , 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 ., As s is charge per unit area of sheet and a is the intersecting area, the charge enclosed by, Gaussian surface = s a
Page 100 :
Examination Papers, , 99, , According to Gauss’s theorem,, 1, Total electric flux =, ´ (total charge enclosed by the surface), e0, i.e.,, \, , 22., , 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. If the surface charge, density s is negative the electric field is directed towards the surface charge., h, de-Broglie wavelength,, l=, 2mqV, For electron q = e , m = me ,, , le =, , h, 2me . eV, , For proton, q = e , m = mp ,, , lp =, , h, 2mp eV, , \, , mp, le, =, lp, me, (i) As me > mp , l e > l p i.e., electron has greater de-Broglie wavelength., Momentum, p = 2meV µ m, (ii) As me < mp ; p e < p p so electron has less momentum., , CBSE (Delhi) SET–III, 2., , The reflected and refracted rays are mutually perpendicular at polarising angle; so from, Brewster’s law, ip = tan -1 (n) = tan -1 ( 3 ) = 60° ., , 3., , 5., 6., 7., , Space Wave Propagation: It is the straight line propagation of electromagnetic wave from, transmitting antenna to receiving antenna both installed on the ground., Alternatively, space wave propagation is the line of sight (LOS) communication., (i) Microwave, (ii) g-rays., Nuclear density is independent of mass number so ratio is 1 : 1., A wavefront is a surface of constant phase. A ray is a perpendicular line drawn at any point, on wavefront and represents the direction of propagation of the wave.
Page 101 :
100, , Xam idea Physics—XII, , 12., , (i) Output waveform of AND gate is shown in fig. (output is 1 when both inputs are 1)., , 0, , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , (ii) NAND gate., 18., , b - 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., 19., , Gauss Theorem: The net outward electric flux through a closed surface is equal to, , 1, times, e0, , the net charge enclosed within the surface i.e.,, ® ®, 1, òS E · dS = e Sq, 0, Electric field due to infinitely long, thin and uniformly charged straight wire: Consider an, infinitely long line charge having linear charge density l coulomb metre - 1 (linear charge, density means charge per unit length). To find the electric field strength at a distance r, we, consider a 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 S2 and (iii) Flat surface S 3 ., E, , dS1, E, E, , E, , E, , S1, , 90o, , 90o, , r, , dS2, , dS3, , + + + + + + + + + + + + + + + + + + +, S2, , S3, , By symmetry the electric field has the same magnitude E at each point of curved surface S1, and is directed radially outward., ®, , We consider small elements of surfaces S1 , S2 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).
Page 104 :
CBSE EXAMINATION PAPERS, ALL INDIA–2009, Time allowed : 3 hours, , Maximum marks : 70, , 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, , e = 1 × 602 ´ 10 -19 C, 1, = 9 × 109 Nm2C– 2, 4pe o, , h = 6 × 626 ´ 10 -34 Js, m 0 = 4p ´ 10 -7 TmA -1, , 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., 2., 3., 4., 5., 6., 7., 8., , 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. 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 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. What is the focal length of the, combination?, 1, The stopping potential in an experiment on photoelectric effect is 1 × 5 V. What is the, maximum kinetic energy of the photoelectrons emitted?, 1, Two nuclei have mass numbers iin the ratio 1 : 8. What is the ratio of their nuclear radii? 1, Give the logic symbol of NOR gate.
Page 105 :
104, , Xam idea Physics—XII, , 9., , Draw 3 equipotential surfaces corresponding to a field that uniformly increases in, magnitude but remains constant along Z-direction. How are these surfaces different from, that of a constant electric field along Z-direction?, 2, Define electric flux. Write its S.I. units., A charge q is enclosed by a spherical surface of radius R. If the radius if reduced to half, how, would the electric flux through the surface change, 2, Define refractive index of a transparent medium., A ray of light passes through a triangular prism. Plot a graph showing the variation of the, angle of deviation with the angle of incidence., 2, Calculate the current drawn from the battery in the given network., 2, , 10., , 11., , 12., , R5 = 2 W, , R1 = 1 W, , R2 = 5 W R3 = 4 W, , R4 = 2 W, + –, 4V, , 13., , 14., , 15., , 16., , 17., , 18., , 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. Why?, (b) The small ozone layer on top of the stratosphere is crucial for human survival. Why?, Define current sensitivity and voltage sensitivity of a galvanometer., Increasing the current sensitivity may not necessarily increase the voltage sensitivity of a, galvanometer. Justify., 2, Define the term ‘linearly polarised light.’, When does the intensity of transmitted light become maximum, 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. It is then cut into, two equal parts. These parts are then connected in parallel across a 3 × 0 volt battery. Find the, current drawn from the battery., 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., (b) Plot a graph showing the variation of potential energy of a pair of nucleons as a function, of their separtion., 2, Write the function of (i) Transducer and (ii) Repeater in the context of communication, system., OR, Write two factors justifying the need of modulation for transmission of a signal., 2
Page 106 :
Examination Papers, , 19., , 20., , 105, , A positive point charge ( +q) is kept in the vicinity of an uncharged conducting plate. Sketch, electric field lines originating from the point on to the surface of the plate., Derive the expression for the electric field at the surface of a charged conductor., 3, OR, A parallel plate capacitor is charged by a battery. After some time the battery is disconnected, and a dielectric slab of dielectric constant K is inserted between the plates. How would (i) the, capacitance, (ii) the electric field between the plates and (iii) the energy stored in the, capacitor, be affected? Justify your answer., 2, X, (i) State the principle of working of a meter bridge., R, (ii) In a meter bridge balance point is found at a distance l1, with resistance R and S as shown in the figure., S, When an unknown resistance X is connected in parallel, G, l1, with the resistance S, the balance point shifts to a distance, B, 3 A, l . Find the expression for X in terms of l1 , l2 and S., 2, , +, , –, , 21., , 4V, (i) State Faraday’s law of electromagnetic induction., (ii) A jet plane is travelling towards west at a speed of 1800 km/h. What is the voltage, difference developed between the ends of the wing having a span of 25 m, if the Earth’s, magnetic field at the location has a magnitude of 5 ´ 10 -4 T and the dip angle of 30°?, , 22., , In Young’s double slit experiment, 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. Another source of monochromatic light produces the interference, pattern in which the two consecutive bright fringes are separated by 7 × 2 mm. Find the, wavelength of light from the second source., 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. 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., 3, Distinguish between sky wave and space wave propagation. Give a brief description with the, help of suitable diagrams indicating how these waves are propagated., With the help of a suitable diagram, explain the formation of depletion region in a p-n, junction. How does its width change when the junction is (i) forward biased, and (ii) reverse, biased?, 3, Give a circuit diagram of a common emitter amplifier using an n-p-n transistor. Draw the, input and output waveforms of the signal. Write the expression for its voltage gain., 3, Draw a plot showing the variation of binding energy per nucleon versus the mass number A., Explain with the help of this plot the release of energy in the processes of nuclear fission and, fusion., 3, Draw a schematic sketch of a cyclotron. Explain briefly how it works and how its is used to, accelerate the charged particles., (i) Show that time period of ions in a cyclotron is independent of both the speed and radius, of circular path., , 23., 24., 25., , 26., 27., , 28.
Page 107 :
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., Deduce the expression for the force per unit length between them., Depict the pattern of magnetic field lines around them., (b) A rectangular current carrying loop EFGH is kept in, a uniform magnetic field as shown in the fig., 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, (b) zero?, 5, H, , 29., , (a) What are eddy currents? Write their two applications., (b) Figure shows a rectangular conducting loop PQRS in, which arm RS of length ' l' is movable. The loop is kept in a, uniform magnetic field ‘B’ directed downward, perpendicular to the plane of the loop. The arm RS is, moved with a uniform speed ‘v’., , P, , l, , G, , R, , v, , Deduce an expression for :, Q, S, (i) the emf induced across the arm ‘RS’,, (ii) the external force required to move the arm, and, (iii) the power dissipated as heat., 5, OR, (a) State Lenz’s law. Give one example to illustrate this law. “The Lenz’s law is a consequence, of the principle of conservation of energy.” Justify this statement., (b) Deduce an expression for the mutual inductance of two long co-axial solenoids but, having different radii and different number of turns., 5, 30. (a) (i) Draw a labelled ray diagram to show the formation of image in an astronomical, telescope for a distant object., (ii) Write three distinct advantages of a reflecting type telescope over a refracting type, telescope., (b) A convex lens of focal length 10 cm is placed coaxially 5 cm away form a concave lens of, focal length 10 cm. If an object is placed 30 cm in front of the convex lens, find the position, of the final image formed by the combined system., 5, OR, (a) With the help of a suitable ray diagram, derive the mirror formula for a concave mirror., (b) The near point of a hypermetropic person is 50 cm from the eye. 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
Page 108 :
Examination Papers, , 107, , CBSE (All India) SET–II, Questions different from Set–I., 1., 5., 6., 7., 8., 9., 21., , 22., , 23., , 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. What is the focal length of the, combination?, 1, Give the logic symbol of NAND gate., 1, Two nuclei have mass numbers in the ratio 8 : 125. What is the ratio of their nuclear radii? 1, The maximum kinetic energy of a photoelectron is 3 eV. What is its stopping potential?, 1, (i) State the principle on which the working of an optical fiber is based., (ii) What are the necessary conditions for this phenomenon to occur?, 2, (i) State the law that gives the polarity of the induced emf., (ii) A 15 × 0 mF capacitor is connected to 220 V, 50 Hz source. Find the capacitive reactance and, the rms current., (a) In a single slit diffraction experiment, a slit of which ‘d’ is illuminated by red light of, wavelength 650 nm. For what value of ‘d’ will:, (i) the first minimum fall at an angle of diffraction of 30°, 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, 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 - s respectively., 3, OR, (a) A charge +Q is placed on a large spherical conducting shell of radius R. Another small, conducting sphere of radius r carrying charge ‘q’ is introdcued inside the large shell and is, placed at its centre. Find the potential difference between two points, one lying on the, sphere and the other on the shell., (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., 3, , 25., , (i) With the help of circuit diagrams distinguish between forward biasing and reverse, biasing of a p-n junction diode., (ii) Draw V-I characteristics of a p-n junction diode in (a) forward bias, (b) reverse bias., 3, , CBSE (All India) Set–III, Questions different from Set–I and Set–II., 1., , Define the term ‘potential energy’ of charge ‘q’ at a distance ‘r’ in an external electric field. 1
Page 109 :
108, , Xam idea Physics—XII, , 4., , The stopping potential in an experiment on photoelectric effect is 2 V. 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. What is the focal length of the, combination?, 1, Give the logic symbol of AND gate., 1, Two nuclei have mass numbers in the ratio 27 : 125. 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., 2, A wire of 20 W resistance is gradually stretched to double its original length. It is then cut into, two equal parts. These parts are then connected in parallel across a 4 × 0 volt battery. Find the, current drawn from the battery., 2, In Young’s double slit experiment, 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. Another source of monochromatic light produces the interference, pattern in which the two consecutive bright fringes are separated by 8 mm. Find the, wavelength of light from the second source., 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. Draw its I-V characteristics., 3, Define the activity of a radionuclide. Write its S.I. units. Give a plot of the activity of a, radioactive species versus time., How long will a radioactive isotope, whose half life is T years, take for its activity to reduce to, 1/8th of its initial value?, 3, , 5., 7., 8., 11., 16., , 22., , 25., 27., , Solutions, CBSE (All India) SET–I, 1., 2., , Zero,, X-Rays, Frequency range : 3 ´ 10 16 Hz – 3 ´ 10 19 Hz., , 3., 4., , ®, , ®, , Magnetic field is parallel or antiparallel to velocity of electron i.e., angle between v and B is, 0° or 180°., Angular separation between fringes, l, bq =, d, where l = wavelength, d = separation between coherent sources, b q is independent of, distance between the slits and screen; so angular separation (b q ) will remain unchanged.
Page 110 :
Examination Papers, , 5., , Net power of lens combination P = + 6 D - 2 D = + 4 D, 1 1, Focal length, F = = m = 25 cm, \, P 4, , 6., , Kmax = eVs = e (1 × 5 V) = 1 × 5 eV, , = 1 × 5 ´ 1 × 6 ´ 10 -19 J = 2 × 4 ´ 10 -, , 7., , J, , Nuclear radius, R = R 0 A 1/ 3, æA ö, =ç 1÷, R 2 è A2 ø, R1, , \, 8., , 19, , 109, , 1/ 3, , 1 1/ 3 1, = æç ö÷, =, è 8ø, 2, , Symbol of NOR gate., A, Y, B, , 9., , ®, , For increasing electric field, , For constant electric field E, d1, , d2, , V, , d1, , d2, 2V, , E, V, , 10., , 2V, d 1 = d2, , 3V, E, , 3V, d1 > d2, , Difference: For constant electric field, the equipotential surfaces are equidistant for same, potential difference between these surfaces; while for increasing electric field, the separation, between these surfaces decreases, in the direction of increasing field, for the same potential, difference between them., Electric Flux: The total number of electric lines of force diverging normally from a surface is, called the electric flux through that surface., S.I. unit of electric flux is volt metre., ®, , Electric flux through surface element D S, ®, , = EDS cos q, where E is electric field strength., Electric flux through entire closed surface is, ®, , ®, , ®, , is Df = E. D S, , DS cos q, E, , q, DS, , ®, , f = ò E .dS, S, , As electric flux through the surface =, , 11., , 1, ´q, e0, , On decreasing the radius of the spherical surface to half there will be no effect on the electric, flux., 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.e., n = ., v
Page 111 :
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.e.,, sin i, n=, sin r, , Angle of deviation,d, , Y, , d, dm, , i1, , 12., , i, i2, Angle of incidence,i, , x, , The equivalent circuit is shown in fig., The five resistors form a balanced Wheatstone’s bridge. Since, R1 R4, =, R5 R 3, So, R 2 is ineffective., The effective resistance of R 1 and R5 in series,, 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 =, , R1, 1W, A, , R5, 2W, 5W, 2W, , B, , R2, 4W, , R4, , R3, , +, , –, 4V, , E, 4, = = 2A, R AB 2, , 13. (a) The visible radiations and radiowaves can penetrate the earth's atmosphere but X-rays are, absorbed by the atmosphere., (b) The ozone layer absorbs ultraviolet and other low wavelength radiations which are, harmful to living cells of human bodies and plants; hence ozone layer is crucial for human, survival., 14., , Current sensitivity : It is defined as the deflection of coil per unit current flowing in it., q NAB, Current Sensitivity,, Sq = =, I, C
Page 112 :
Examination Papers, , 15., , 111, , Voltage sensitivity : It is defined on the deflection of coil per unit potential difference across, its ends., q NAB, Voltage Sensitivty,, SV = =, V, GC, where G is resistance of galvanometer., Justification: When number of turns N is doubled, then the current sensitivity ( µ N) is doubled;, but at the same time, the resistance of galvanometer coil (G) will also be doubled, so voltage, N, sensitivity æç SV µ ö÷ will remain unchanged; hence inreasing current sensitivity does not, è, Gø, necessarily increase the voltage sensitivity., 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., 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., This is maximum when sin 2q = 1 or q = 45° ., 16., , When length of a given wire is made n-times by strecting it, its resistance becomes n 2 times, i.e., R ¢ = n 2 R = ( 2) 2 ´ 15 = 60 W, Resistance of each half part =, , 60, = 30 W, 2, , When both parts are connected in parallel, final resistance =, , 30, = 15 W, 2, , Current drawn from battery,, V, I=, R, 3×0, =, = 0×2A, 15, 17., , (a) The mass of a nucleons in ground state is always less than the total mass of its, constituents neutrons and protons; because this mass difference appears in the form of, binding energy to hold the nucleons inside the nucleus.
Page 113 :
112, , Xam idea Physics—XII, , (b) Part AB represents repulsive force and part BCD represents attractive force., , A, +100, , MeV, , 0, , Repulsive, B, D, Attractive, , –100, , C 1, , 2, , 3, , 4, , r (fm), , 18., , 19., , (i) Transducer: A device which convert one form of energy into the other., (ii) Repeater: A repeater picks up the signal from the transmitter, amplifies and retransmits it, to the receiver sometimes with a change in a carrier frequency., OR, Need of Modulation: Modulation is needed for (i) Practicable size of antenna. (ii) More, effective power radiation by an antenna., The electric field lines are shown in fig., Electric field on the Surface of a Charged Conductor:, Let electric charge be uniformly distributed over the surface of, +q, a thin, non-conducting infinite sheet. Let the surface charge, density (i.e., charge per unit surface area) be s. We have to, calculate the electric field strength at any point distance r from, –, –, –, –, –, the sheet of charge., To calculate the electric field, strength near the sheet, 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). 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., o
Page 114 :
Examination Papers, , 113, , Total electric flux, ®, , òSE, or, , ®, , òSE, , ®, , •dS, , ®, , •dS, , =, =ò, , ®, , òSE1, S1, , ®, , • d S1 +, , ®, , òSE2, , E dS1 cos 0° +, , ®, , •d S2 +, , òS2, , ®, , òSE3, , ®, , •dS 3, , 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 ., \, As s is charge per unit area of sheet and a is the intersecting area, the charge enclosed by, Gaussian surface = s a, According to Gauss’s theorem,, 1, Total electric flux =, ´ (total charge enclosed by the surface), e0, i.e.,, \, , 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. If the surface charge, density s is negative the electric field is directed towards the surface charge., OR, K e0 A, (i) The capacitance of capacitor increases to K times (since C =, µ K), d, 1, (ii) The potential difference between the plates becomes times., K, Q, Reason: V = ; Q same, C increases to K times, C, V, \, V¢ =, K, V, 1, As E = and V is decreased; therefore, electric field decreases to times., d, K, (iii) Energy stored by the capacitor, U =, , Q2, ., 2C, , As Q = constant, C is increased, and so energy stored by capacitor decreases to, , 1, times., K, , 20. (i) Metre Bridge: Meter bridge is based on the principle of Wheatstone’s bridge., The resistance of wire is divided into two resistances P and Q. R is known resistance and, S is unknown resistance., lr, P R, R, At balance, = Þ, =, Q S, (100 - l) r S
Page 115 :
114, , Xam idea Physics—XII, , Þ, , æ 100 - l ö, unknown resistance, S = ç, ÷R, è l ø, , (ii) In first case, l1, R, =, S 100 - l1, , …(1), , When X and S are in parallel, let resistance, XS, S¢ =, X+S, In second case, l2, R, =, æ XS ö 100 - l2, ç, ÷, è X + Sø, , Resistance box, , ... (2), , Dividing (2) by (1), we get, X + S l1 æ 100 - l1 ö, =, ç, ÷, X, l2 è 100 - l2 ø, S, X=, Þ, l2 æ 100 - l1 ö, ç, ÷ -1, l1 è 100 - l2 ø, 21., , Resistance wire, (S), , (R), D, G, A, , (P), , B, , (Rh), + –, Cell, , (Q), , C, , (100–l) cm, , l cm, , K, , Rheostat, , (i) Faraday’s Laws of Electromagnetic Induction:, (a) Whenever there is a change in magnetic flux linked with of a coil, an emf is induced in, the coil. The induced emf is proportional to the rate of change of magnetic flux linked, with the coil., Df, i.e., e µ, Dt, (b) emf induced in the coil opposes the change in flux, i.e.,, Df, Df, e µÞe =-k, Dt, Dt, where k is a constant of proportionality., Df, In S.I. system f is in weber, t in second, e in volt, then k = 1, so e = Dt, Df, If the coil contains N-turns, then e = - N, Dt, (ii) The wing of horizontal travelling plane will cut the vertical component of earth’s, magnetic field, so emf is induced across the wing. The vertical component of earth's field, is given by, V = Be sin q;, 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, l = 25 m, q = 30°,
Page 116 :
Examination Papers, , v = 1800 km/h = 1800 ´, \, , 5, m/s = 500 m/s, 18, , e = (5 × 0 ´ 10 -, , 4, , ´ sin 30° ) ´ 500 ´ 25, , = (5 × 0 ´ 10 -, , 4, , ´ 0 × 5) ´ 500 ´ 25 = 3 × 1 V, , b1 =, , 22., , b2 =, \, , b2, , Þ, , l2 =, , b1, , =, , 115, , l 1D, , …(1), , d, l 2D, , …(2), , d, l2, l1, b2, b1, , l1, , Given b 1 = 8 × 1 mm, b 2 = 7 × 2 mm, 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, we obtain an interference, pattern consisting of a central white fringe having on both sides symmetrically a few, coloured fringes and then uniform illumination., 23., , The Schematic arrangement of Geiger-Marsdon Experiment (also known as Rutherford, Scattering Experiment) is shown in fig., ZnS screen, Lead bricks, , Source of, a-particles, , Gold foil, Beam of, a-particles, f, , Most of, a-particles, , (~ 10– 4 m thick), , Detector, , Observations: (i) Only a small fraction of number of a-particles rebound back. This, shows that the number of a-particles undergoing head on collision is very small. The, conclusion is that the entire positive charge of atom is concentrated in a small volume, called the nucleus.
Page 117 :
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. 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., 24., , Distinction between sky wave and Space wave Propagation:, , Ionospheric, layers, , Earth, , Sky wave propagation, , Sky wave propagation is achieved by ionospheric reflection of radiowaves, while space wave, propagation is direct, line of sight propagation from the transmitted to the receiver., Sky Wave Propagation: In sky wave propagation, the radiowaves transmitted from antenna, get reflected from the ionosphre and thereby reach the receiving antenna., Space Wave Propagation: In space wave propagation the radiowaves transmitted from, antenna reach the receiving antenna through a line of sight (straight) propagation. The range, of such a tranmission is limited by the curvature of the earth., dM, dT, , hT, , Earth, , Space wave propagation, , hR
Page 118 :
Examination Papers, , 25., , VB, Formatium of Depletion Layer: 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, 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. This sets up a potential, difference and hence an internal electric field Ei, across the junctions. The field Ei is directed from n-region to p-region., This field stops the further diffusion of charge carriers. Thus the layers, ( » 10 - 4 cm to 10 - 6 cm) on either side of the junction becomes free from, , mobile charge carriers and hence is called the depletion layer. The symbol, of p-n junction diode is shown in Fig., 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 reducing the potential, barrier, and hence the width of depletion layer, R, p, decreases., , 117, , n, , p, , Ei, , n, , E, , 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. This results in an, increase in barrier voltage and hence the width of, depletion layer increases., , Ei, , R, , p, , E, , n, , –, , 26., , +, K, (c) Reverse current, , Common emitter amplifier using n-p-n transistor, The circuit of common emitter amplifier using n-p-n transistor is shown below:, IC, C1, , RB, , C, IB, , C2, RL, , B, Vi, , +, –, , E, VBB, , IE, , +, VCC, –, , V0, , Output, waveform
Page 119 :
118, , Xam idea Physics—XII, , The phase difference between input signal and output voltage is 180°., The input and output waveforms are shown in, Input waveform, figure., R, Voltage gain Av = b L ×, Ri, , The variation of binding energy per nucleon versus mass number is shown in figure., 9.0, , Binding Energy per Nucleon (in MeV), , 27., , Output waveform, , O16, 8.0C12, F18, He4 N14, 7.0, , Fe56, U238, , 6.0, Li7, , 5.0, 4.0, 3.0, 2.0, , H2, , 1.0, 0.0, , 0, , 20, , 40, , 60, , 80 100 120 140 160 180 200 220 240, Mass Number, , Inferences from graph, 1. The nuclei having mass number below 20 and above 180 have relatively small binding, energy and hence they are unstable., 2. The nuclei having mass number 56 and about 56 have maximum binding energy – 5·8, MeV and so they are most stable., 3. Some nuclei have peaks, e.g., 2 He 4 , 6 C12 , 8 O16 ; this indicates that these nuclei are, relatively more stable than their neighbours., Explanation: When a heavy nucleus (A ³ 235 say) breaks into two lighter nuclei (nuclear, fission), the binding energy per nucleon increases i.e, nucleons get more tightly bound. This, implies that energy would be released in nuclear fission., When two very light nuclei ( A £ 10) join to form a heavy nucleus, the binding is energy per, nucleon of fused heavier nucleus more than the binding energy per nucleon of lighter nuclei,, so again energy would be released in nuclear fusion.
Page 120 :
Examination Papers, , 28., , Working: The principle of action of the apparatus is shown, in fig. The positive ions produced from a source S at the, centre are accelerated by a dee which is at negative, potential at that moment. Due to the presence of, perpendicular magnetic field the ion will move in a circular, path inside the dees. 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, Beam, bombarded., , 119, , Magnetic Pole, N, S, Dee, , Dee, , Magnetic Pole, S, , Dee-1, , R.F., oscillator, , S, , Dee-2, , The function of electric field is to accelerate the charged particle and so to impart energy to, the charged particle., 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., (i) Expression for Period of Revolution, Suppose the positive ion with charge q moves in a dee with a velocity v, then,, qvB =, , mv 2, r, , or r =, , mv, qB, , ...(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 angular velocity w of the ion is given by,, v qB, (from equation 1), ...(2), w= =, r m, The time taken by the ion in describing a semi-circle, i.e., in turning through an angle p is,, p pm, ...(3), t= =, w Bq, Thus the time is independent of the speed of the ion., (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. This is called resonance condition., Now for the cyclotron to work, the applied alternating potential should also have the, same semi-periodic time (T / 2) as that taken by the ion to cross either dee, i.e.,, T, pm, ...(4), =t =, 2, qB
Page 121 :
120, , Xam idea Physics—XII, , or, , T=, , 2p m, , ...(5), , qB, , This is the expression for period of revolution., Obviously, period of revolution is independent of speed of charged particle and radius of, circular path., 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. It has been, observed experimentally that when the currents in the wire are in the same direction, they, experience an attractive force (fig. a) and when they carry currents in opposite directions,, they experience a repulsive force (fig. b)., Let the conductors PQ and RS carry currents I 1 and I 2 in same direction and placed at, separation r. (fig.)., Consider a current–element ‘ab’ of length DL of wire RS. The magnetic field produced by, current-carrying conductor PQ at the location of other wire RS, m I, ...(1), B1 = 0 1, 2p r, According to Maxwell’s right hand rule or right hand palm rule, no. 1, the direction of B1 will be perpendicular to the plane of paper, and directed downward. Due to this magnetic field, each element, of other wire experiences a force. The direction of current element, is perpendicular to the magnetic field; 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, , a, , I1, , B, I2, , Q, , S, r, , ...(2), , According to Fleming’s left hand rule, the direction of magnetic force will be towards PQ, i.e. the force will be attractive., (i) The magnetic field lies due to two current carrying, parallel wires are shown in figure., 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 . As wire of loop carrying, r, opposite current is nearer, so the net froce acting on, the loop is repulsive., , B, I1, I2
Page 122 :
Examination Papers, , 121, , (b) (i) Direction of magnetic moment m of the current loop is perpendicular to the plane of, paper and directed downward., ®, , ®, , (ii) Torque acting on the current loop is (a) maximum when m is perpendicular to B., ®, , ®, , (b) Minimum when M is parallel to B., 29., , (a) Eddy currents: When a metallic plate is placed in a time varying magnetic field, the, magnetic flux linked with the plate changes, the induced currents are set up in the plate;, these currents are called eddy currents. These currents are sometimes so strong, that the, metallic plate becomes red hot., Application of Eddy Currents:, 1. Induction Furnace: In induction furnance, the metal to be heated is placed in a rapidly, varying magnetic field produced by high frequency alternating current. Strong eddy, currents are set up in the metal produce so much heat that the metal melts. This, process is used in extracting a metal from its ore. The arrangement of heating the metal, by means of strong induced currents is called the induction furnace., 2. Induction Motor : The eddy currents may be used to rotate the rotor. Its principle is :, When a metallic cylinder (or rotor) is placed in a rotating magnetic field, eddy currents, are produced in it. According to Lenz’s law, these currents tend to reduce to relative, motion between the cylinder and the field. The cylinder, therefore, begins to rotate in, the direction of the field. This is the principle of induction motion., 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, S, P, (ii) Force on arm RS = BIl, e, B2 l 2v, æ Blv ö, l=, = B æç ö÷ . l = B ç, ÷, èR¢ø, è R¢ ø, R¢, where R ¢ = resistance., (iii) Power dissipated on heat, P = Fv =, , B2 l 2v 2, B2 l 2v, 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.”, Example: When the north pole of a coil is brought near a closed coil,, the direction of current induced in the coil is such as to oppose the, S, N, N, approach of north pole. For this the nearer face of coil behaves as, north pole. This necessitates an anticlockwise current in the coil,, (a), when seen from the magnet side [fig. (a)]
Page 123 :
122, , Xam idea Physics—XII, , Similarly when north pole of the magnet is moved away from the, coil, the direction of current in the coil will be such as to attract the, S, S, N, magnet. For this the nearer face of coil behaves as south pole. This, necessitates a clockwise current in the coil, when seen from the, (b), magnet side (fig. b)., Conservation of Energy in Lenz’s Law: Thus, in each case whenever there is a relative, motion between a coil and the magnet, a force begins to act which opposes the relative, motion. Therefore to maintain the relative motion, a mechanical work must be done. This, work appears in the form of electric energy of coil. Thus Lenz’s law is based on conservation, of energy., (b) Suppose two co-axial solenoids S1 and S2 of, S1, radii r1 and r2 , number of turns N 1 and N 2, each of length ‘l’, Suppose I 1 is the current in outer solenoid;, magnetic field at the axis, B1 = m 0 n1 I 1 where, n1 = number of turns/meter of outer, S2, solenoid., Magnetic flux linked with inner solenoid, ( S 2 )., f 2 = ( N 2 ) B1 A1 = N 2 (m 0 n1 I 1 ) pr12, = m 0 n1 N 2 pr12 I 1, \ Mutual inductance of two solenoid system, f, M = 2 = m 0 n1 N 2 pr12, I1, But, 30., , (a) (i), , n1 =, , N1, l, , \ M=, , B, , A, , m 0 N 1 N 2 pr12, l, fo, , fe, , Fo, A' Fe', , C1, , C2, , Fe, , B', , 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
Page 124 :
123, , Examination Papers, , (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, , For concave lens u2 = 15 - 5 = + 10 cm, f 2 = - 10 cm, 1, 1, 1, 1, 1, 1, 1, 1, \, =, Þ, =, +, =+, f 2 v 2 u2, v 2 f 2 u2, 10 10, , v 1 = 15 cm, , Þ, , That is final image is formed at infinity., OR, (a) Mirror Formula: M 1 M 2 is a concave mirror having, pole P, focus F and centre of curvature C., A, An object AB is placed in front of mirror with point B, O, on the principal axis. The image formed by mirror is, A ¢ B¢ . The perpendicular dropped from point of, B, incidence D on principal axis is DN., In DABC and DA ¢ B¢ C, (each equal, ÐABC = ÐA ¢ B¢ C, to 90°), (opposite, ÐACB = ÐA ¢ CB¢, angles), Both triangles are similar., 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), we get, BC, FN, =, B¢ C B¢ F, , v2 = µ, , M1, D, , C, , B' F i, r, I, , N, , A', , M2, u, , …(1), , …(2), , …(3), , P
Page 125 :
124, , Xam idea Physics—XII, , If aperture of mirror is very small, the point N will be very near to P, so FN = FP, PB - PC, BC, FP, FP, or, …(4), \, =, =, B¢ C B¢ F, PC - PB¢ PB¢ - PF, By sign convention, Distance of object from mirror PB = - u, Distance of image from mirror PB¢ = - v, Focal length of mirror PF = - f, Radius of curvature of mirror PC = - R = - 2 f, Substituting these values in (4), we get, - u - (- 2 f ), -f, =, Þ, - 2 f - ( - v) - v - ( - f ), Þ, , 2f, , 2, , - vf = - uf + uv + 2 f, , 2, , - 2 fv, , - u + 2f, - 2f + v, or, , =, , -f, -v+ f, , f v + uf = uv, , Dividing both sides by uvf , 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., That is, u = - 25 cm, v = - 50 cm, 1, 1, 1, \, =+, f, 50 25, Þ, , f = 50 cm = 0 × 50 m, 1, 1, Power, P = =, D= 2D, f 0 × 50, , CBSE (All India) SET–II, 1., 5., , At every point on equatorial axis, the potential is zero, so work done W = q DV = 0 (zero)., Net power, P = P1 + P2 = + 4 D - 2 D = + 2 D, 1 1, \, F = = = 0 × 50 m = 50 cm, P 2, , 6., , Logic symbol of NAND gate.
Page 126 :
Examination Papers, , 125, , A, Y, , B, , 7., , Nuclear radius R = R 0 A 1/ 3, æA ö, =ç 1÷, R 2 è A2 ø, R1, , \, 8., , (Ek ) max = eVs, Stopping potential, Vs =, , 1/ 3, , (Ek ) max, e, , 8 ö, = æç, ÷, è 125 ø, , =, , 3 eV, e, , 1/ 3, , =, , 2, 5, , =3V, , (i) The working of optical fiber is based on total internal reflection., Statement: When a light ray goes from denser to rarer medium at an angle greater than, critical angle, the ray is totally reflected in first (denser) medium., This phenomenon is called 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. e. , i > C)., 21. (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., 1, (ii) Capacitive reactance, XC =, 2pfC, , 9., , =, , 1, 2 ´ 3 × 14 ´ 50 ´ 15 ´ 10 -6, , RMS Current, I =, =, 22., , W = 212 × 1 W, , E, XC, 220, = 1 × 03 A, 212 × 1, , (a) (i) For nth minima, d sin q = nl, Given l = 650 nm = 650 ´ 10 -9 m, n = 1, 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, d sin q = ( 2n + 1), n = 1, q = 30°, l = 650 ´ 10 -9 m, , l, 2
Page 127 :
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, the wavelets from lesser and lesser part of slit produce, constructive interference., 23., , Let electric charge be uniformly distributed over the surface of a thin, non-conducting infinite, sheet. Let the surface charge density (i.e., charge per unit surface area) be s. We have to, calculate the electric field strength at any point distance r from the sheet of charge., To calculate the electric field, strength near the sheet, 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). 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., Total electric flux, o, , ®, , ®, , ò E •dS, S, , ®, , òSE, , or, , ®, , •dS, , =, , ®, , ®, , ò E • d S1 +, S1, , =ò, , S1, , ®, , ®, , ò E •d S2 +, S2, , E dS1 cos 0° +, , òS2, , ®, , ®, , ò E •dS 3, S3, , 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, the charge enclosed by, Gaussian surface = s a, +, –, According to Gauss’s theorem,, +, –, 1, Total electric flux =, ´ (total charge enclosed by the surface), +, –, e0, 1, (sa), e0, , i.e.,, , 2Ea =, , \, , s, E=, ×, 2e 0, ®, , E1 =, , ®, , s $ ®, s $, i, E2 =, i, 2e 0, 2e 0, , ®, , ®, , E = E1 + E2 =, , s $, i, e0, , +, +, +, E1, +, +, +, +, +s, , E2, , –, –, –, –, –, –, –, –s
Page 128 :
Examination Papers, , 127, , OR, (a) For external points, the charge may be supposed to be concentrated at the centre, so, 1 æQ + qö, V ( R) =, ç, ÷, 4pe 0 è R ø, +Q, V(r), , For internal points, the potential is same as on the surface, 1 æq Qö, \, V (r) =, ç + ÷, 4pe 0 è r R ø, \ Potential difference, V (r) - V ( R) =, =, , 25., , R, , r, +q, , O, , V(R), , 1 é q Q Q + qù, + 4pe 0 êë r R, R úû, 1, 1 1, qé - ù, 4pe 0 êë r R úû, , (b) When both shells are connected by a conducting wire, whole charge of inner shell will, flow to outer shell., Device Working on this principle is Van de Graaff Generator., (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, 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. The, external field E is much stronger than internal field Ei ., Ei, , Ei, , n, , p, , R, , p, +, , (a) No current, , E, , p, , n, , n, +, , –, , –, , K, , (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, so that the external field is established to, help the internal field Ei as shown in Fig. 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, therefore the current flow stops., Ei, , E, p, , R, , p, –, , n, , n, +, , K, (c) Reverse current, , –, +, Reverse biasing
Page 129 :
128, , Xam idea Physics—XII, , (ii) V-I Characteristics of (a) forward bias and (b) reverse bias:, I (mA), , (a) Forward, bias, , Avalanche, breakdown, , F, O, , (–) V, , R, , (b) Reverse, bias, , (+), , V, , I (µA), , CBSE (All India) SET–III, 1., , 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)., , 4., , Maximum kinetic energy, (Ek ) max = eVs = e ( 2 V) = 2 eV = 2 ´ 1 × 6 ´ 10 -19 J = 3 × 2 ´ 10 -, , 5., , Net power, P = P1 + P2 = + 5 - 2 × 5 = + 2 × 5 D, 1, 1, m = 0 × 4 m = 40 cm, \: Focal length, F = =, P 2 ×5, , 7., , Symbol of AND gate., , 19, , J, , A, Y, , B, , 8., 11., , æA ö, =ç 1÷, R 2 è A2 ø, R1, , 1/ 3, , 27 ö 1/ 3 3, = æç, =, ÷, è 125 ø, 5, , (i) Relation between refractive index (n) and critical angle (C) is, 1, n=, sin C, (ii) Yes, critical angle depends on wavelength or colour of light; it increases with increase of, wavelength being maximum for red and minimum for violet., , 16., , Resistance of stretched wire R ¢ = n 2 R = ( 2) 2 ´ 20 = 80 W, When wire is cut into two equal parts, resistance of each part = 40 W, 40 ´ 40, When there parts are connected in parallel, net resistance Rnet =, = 20 W, 40 + 40, V, 4×0, Current, I =, =, A = 0×2A, Rnet, 20
Page 130 :
Examination Papers, , b=, \, Þ, , Dl, d, , b1, b2, , =, , l2 =, =, , l2, b2, b1, , l1, , 8 mm, 10 mm, , ´ 600 nm = 480 nm, , On replacing monochromatic light by white light; the interference pattern will contain, central white fringe surrounded on either side by few coloured fringes., Zener diode as a Voltage Regulator, I, IL, R, The Zener diode makes its use as a voltage, regulator due to the following property :, IZ, When a Zener diode is operated in the, breakdown region, the voltage across it, Vin, VZ, RL V0, 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. The, Zener diode is connected across load such that it is reverse biased., The series resistance R absorbs the output voltage fluctuations so as to maintain constant, voltage across the load., If the input dc voltage increases, the current through R and Zener diode also increases. So,, voltage drop across R increases, without any change in the voltage across zener diode., I-V Characteristics, Unregulated input, , 25., , l1, , I (mA), VR, , Zener, voltage, , O, VF, , VZ, , I(mA), , Regulated output, , 22., , 129
Page 132 :
CBSE EXAMINATION PAPERS, FOREIGN–2009, Time allowed: 3 hours, , Maximum marks: 70, , 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, , e = 1 × 602 ´ 10 -19 C, 1, = 9 × 109 Nm2C– 2, 4pe o, , h = 6 × 626 ´ 10 -34 Js, m 0 = 4p ´ 10 -7 TmA -1, , 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., 2., , 3., 4., 5., 6., , 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. Which of these, quantities is constant along the conductor:, Current, current density, drift speed, electric field?, 1, Name the electromagnetic radiations which are produced when high energy electrons are, bombarded on a metal target., 1, Draw the wavefront coming out of a convex lens when a point source of light is placed at its, focus., 1, Unpolarised light of intensity I is passed through a polaroid. What is the intensity of the light, transmitted by the polaroid?, 1, Why are coherent soruces required to create interference of light?, 1
Page 133 :
132, , Xam idea Physics—XII, , 7., , In the Rutherford scattering experiment the distance of closest approach for an a-particle is, d 0 . If a-particle is replaced by a proton, 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., 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, B and C lie in a uniform electric field (E) of 5 ´ 10 3 NC -1 as shown in the, , 8., 9., , 10., , figure. Find the potential difference between A and C., A, , 2, , B, 5 cm, , 3 cm, E, C, , 11., 12., , 13., , 14., , OR, The sum of two point charges is 7 m C. They repel each other with a force of 1 N when kept 30, cm apart in free space. Calculate the value of each charge., 2, Name the electromagnetic radiations having the wavelength range from 1 mm to 700 nm., Give its two important applications., 2, A wire of length L is bent round in the form of a coil having N turns of same radius. If a, steady current I flows through it in a clockwise direction, find the magnitude and direction of, the magnetic field produced at its centre., 2, Derive an expression for the de-Broglie wavelength associated with an electron accelerated, through a potential V. Draw a schematic diagram of a localised-wave describing the wave, nature of the moving electron., 2, Figure shows variation of stopping potential (V0 ) with the frequency ( n) for two, photosensitive materials M1 and M2 ., V0, , M1, , M2, , 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., 2, 15., , The energy of the electron in the ground state of hydrogen atom is - 13 × 6 eV ., (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
Page 134 :
Examination Papers, , 16., , Draw the logic symbol of the gate whose truth table is given below:, Input, , 17., 18., 19., , 20., , Output, , A, , B, , Y, , 0, , 0, , 1, , 0, , 1, , 0, , 1, , 0, , 0, , 1, , 1, , 0, , If this logic gate is connected to NOT gate, what will be output when (i) A = 0, B = 0 and, (ii) A = 1, 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?, 2, (i) How are eddy currents reduced in a metallic core?, (ii) Give two uses of eddy currents., 2, Define the term ‘electric dipole moment.’ Is it scalar or vector?, 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. Use Kirchhoff’s rules to show that no current flows in the given, circuit., e1 = 2 V, , e2 = 2 V, , r1, , r2, 2V, , 21. (a) State the Principle of working of a potentiometer., (b) Figure shows the circuit diagram of a potentiometer, for determining the emf ‘e’ of a cell of negligible, internal resistance., , 22., , 133, , A, , R1, , J, B, , (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, 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. Indicate the direction of the torque, acting on the loop.
Page 135 :
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’. Indicate the direction of the magnetic dipole moment.3, 23. Depict the field-line pattern due to a current carrying solenoid of finite length., (i) In what way do these lines differ from those due to an electric dipole?, (ii) Why can’t two magnetic field lines intersect each other?, 3, 24. State the conditions under which total internal reflection occurs., One face of a prism with a refracting angle of 30° is coated with silver. A ray incident on, another face at an angle of 45° is refracted and reflected from the silver coated face and, retraces its path. Find the refractive index of the material of the prism., 3, 25. (a) Why do we not encounter diffraction effects of light in everyday observations?, (b) In the observed diffraction pattern due to a single slit, how will the width of central, maximum be affected if, , 26., , 27., 28., , 29., , (i) the width of the slit is doubled;, (ii) the wavelength of the light used is increased?, Justify your answer in each case., 3, (a) What is meant by half life of a radioactive element?, (b) The half life of a radioactive substance is 30 s. Calculate, (i) the decay constant, and, (ii) time taken for the sample to decay by 3/4th of the initial value., 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., 3, State the working principle of an AC generator with the help of a labelled diagram., Derive an expression for the instantaneous value of the emf induced in coil., 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., Deduce the expressions for the secondary voltage and secondary current in terms of the, number of turns of primary and secondary windings., How is the power transmission and distribution over long distances done with the use of, transformers?, 5, (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., (b) Define power of a lens and give its S.I. units., If a convex lens of focal length 50 cm is placed in contact coaxially with a concave lens of, focal length 20 cm, what is the power of the combination?, 5, OR, Draw a labelled ray diagram to show the image formation by an astronomical telescope., Derive the expression for its magnifying power in normal adjustment. Write two basic, features which can distinguish between a telescope and a compound microscope., 5
Page 136 :
Examination Papers, , 135, , 30. (a) Explain the formation of ‘depletion layer' and ‘barrier potential’ in a p-n junction., (b) With the help of a labelled circuit diagram explain the use of a p-n junction diode as a full, wave rectifier. Draw the input and output waveforms., 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. Describe briefly its working., Explain how a transistor in active state exhibits a low resistance at its emitter base junction, and high resistance at its base collector junction., Draw a circuit diagram and explain the operation of a transistor as a switch., 5, , CBSE (Foreign) SET–II, Questions different from Set–I., 3., 7., 11., 12., , Define self-inductance. Give its S.I. units., 1, Sketch the shape of wavefront emerging from a point source of light and also make the rays. 1, A logic gate is obtained by applying output of OR gate to a NOT gate. Name the gate so, formed. Write the symbol and truth table of this gate., 2, Y, A point charge is moving with a constant velocity, B, perpendicular to a uniform magnetic field as shown in the, figure. 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. (a) Draw a graph showing variation of photo-electric current ( I ) with anode potential (V ) for, different intensities of incident radiation. Name the characteristic of the incident radiation, that is kept constant in this experiment., (b) If the potential difference used to accelerate electrons is doubled, by what factor does the, de-Broglie wavelength associated with the electrons change?, 2, 14., , Three points A, B and C lie in a uniform electric field (E) of, 5 ´ 10 3 NC -1 as shown in the figure. Find the potential, difference between A and C., , 2, , A, , B, 10 cm, , 6 cm, , E, , C, , OR, The sum of two point charges is 9 µC. They repel each other is force of 2 N when kept 30 cm, apart in free space. Calculate the value of each charge., 2
Page 137 :
136, , Xam idea Physics—XII, , 23., , 26., , (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., 3, (a) Define the terms (i) ‘amplitude modulation’ and (ii) ‘modulation index’., , 27., , (b) If a low frequency signal in the audio frequency range is to be transmitted over long, distances, explain briefly the need of translating this signal to high frequencies before, transmission., 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., , 3, , CBSE (Foreign) SET–III, Questions different from Set–I and Set–II., 1., 4., 6., 8., 9., , A metal plate is introdcued between the plates of a charged parallel plate capacitor. What is, its effect on the capacitance of the capacitor?, 1, Define mutual inductance. Give its S.I. units., 1, Name the electromagnetic radiation which can be produced by klystron or a magnetron, valve., 1, Define a wavefront., 1, P1 P2, Figure shows two large metal plates P1 and P2, tightly held against each other, and placed between two equal and unlike point charges perpendicular to the, line joining them., (i) What will happen to the plates when they are released?, Q, Q, +, –, (ii) Draw the pattern of the electric field lines for the system., 2, , 10., , 13., , A 800 pF capacitor is charged by a 100 V battery. After some time the battery is disconnected., The capacitor is then connected to another 800 pF capacitor. What is the electrostatic energy, stored?, 2, OR, The sum of two point charges is 7 m C. They repel each other with a force of 1 N when kept, 30 cm apart in free space. Calculate the value of each charge., 2, -3, Name the elecromagnetic radiations having the wavelength range from 1 nm to 10 nm., Give its two important applications., 2
Page 138 :
Examination Papers, , 14., 16., , 137, , A logic gate is obtained by applying output of AND gate to a NOT gate. Name the gate so, formed. Write the symbol and truth table of this gate., 2, The graph below shows variation of photo-electric current with collector plate potential for, different frequencies of incident radiations., Photo-electric, current, , n1, , n2, , n3, Collector plate potential, , 24., , (i) Which physical parameter is kept constant for the three curves?, (ii) Which frequency ( n 1 , 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. Identify whether specimens X and Y are diamagnetic, paramagnetic or, ferromagnetic., , Y, X, , (iii) How is the magnetic permeability of specimen X different from that of specimen Y?, 26., , (a) What is meant by half life of a radioactive element?, (b) The half life of a radioactive substance is 30 s. Calculate:, (i) the decay constant, and, (ii) time taken for the sample to decay by 3/4th of the initial value., , 3, , 3
Page 139 :
138, , Xam idea Physics—XII, , Solutions, CBSE (Foreign) SET–I, 1., , 2., 3., 4., , Surface of a conductor is an equipotential surface and field lines are always directed from, higher to lower potential, so field lines in the vicinity of a conductor must be normal to the, surface of conductor., Current is constant along a conductor., X-rays., Wavefront is plane as shown in fig., , F, , Plane wavefront, , 5., 6., 7., , I, ×, 2, Coherent sources are required for sustained interference. If sources are incoherent, the, intensity at a point will go on changing with time., 1 ( Ze) ( 2e), (for a-particle, q = 2e), Ek =, 4pe 0, d0, Intensity of light transmitted through the polaroid =, , Ek¢ =, Ek¢, Ek, 8., , 9., , 10., , =, , 1 ( Ze) ( e), (for proton, 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., Mathematically, the induced emf is given by, Df, e=Dt, (i) No effect on capacitance if foil is electrically neutral., (ii) If foil is connected to upper plate with a conducting wire, the effective separation between, plates becomes half, so capacitance is doubled., The line joining B to C is perpendicular to electric field, so potential of B = potential of C i.e.,, VB = VC ., Distance AB = 4 cm
Page 141 :
140, , Xam idea Physics—XII, , If V volt is accelerating potential of electron, then Kinetic energy,, EK = eV, \ Equation (ii) gives, h, l=, 2meV, , ...(iii), , This is the required expression for de Broglie wavelength associated with electron, accelerated to potential of V volt., The diagram of wave packet describing the motion of a moving electron is shown., , 14., , 15., , æ hö, (i) The slope of stopping potential (V0 ) versus frequency ( n) is equal to ç ÷ which is universal, èeø, constant, so slope is same for both lines., (ii) K.E. = hn - hn 0, As threshold frequency n 0 is lesser for M 1 , so K.E. will be greater for M 1 for same frequency n., (i) Negative sign shows that electron in ground state is bound in H-atom due to attractive, force between electron and nucleus., (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 = - 3 × 4 eV, 4, Energy required to take electron from ground state to first excited state, DE = E2 - E1, = - 13 × 6 eV - ( - 3 × 4 eV), = 10 × 2 eV, , 16., , The given truth table is of NOR gate. The logic symbol is shown in fig., A, Y, B, , When it is connected to a NOT gate, the gate becomes OR gate., (i) A = 0, B = 0 gives output 0., (ii) A = 1, B = 1 gives output 1.
Page 142 :
Examination Papers, , 141, , The combination is shown in fig., A, , A+B, , A+B, , B, , 17. (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., (b) TV signals have high frequency range 100 to 200 MHz. Ionospheric layers do not reflect, back such high frequency signals. Hence, sky waves cannot be used for transmission of TV, signals., 18., 19., , (i) A metallic core cuts the path of eddy currents, this reducing the strength of eddy currents., (ii) Eddy currents are used in (a) induction furnace (b) induction motor., 2l, The electric dipole moment is a vector quantity whose, +q, magnitude is equal to the product of charge on one dipole and –q, p, distance between them. Its direction is from - q to + q., ®, , ®, , p = q2 l, , i.e.,, , Electric dipole moment is a vector quantity., Derivation of electric field at a point of equatorial plane:, E1, Consider a point P on broad side on the position of, dipole formed of charges + q and - q at separation 2l., E = E1cosq + E2cosq, From figure,, AP = BP = r 2 + l, \, , ®, , E1 =, ®, E2, , =, , q, 1, 2, 4p e 0 r + l, q, 1, 2, 4p e 0 r + l, , E, , 2, , 2, , \, , E = 2E1 cos q = 2 ´, =, , P, , E2sinq, , E2, , r, 2, , OB, l, =, PB, r2 + l, , cos q =, , q, , along B to P, along P to A, A, , \ 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, , E1sinq, , 2, , =, , l, 2, , (r + l 2 ) 1/ 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, , q, O, , –q, l, , +q, l, , 2l, , B
Page 143 :
142, , Xam idea Physics—XII, , \, , E=, , p, 1, 2, 4p e 0 (r + l 2 ) 3 / 2, , ...(iii), , If dipole is infinitesimal and point P is far away, we have l < < r , so l, , 2, , may be neglected as, , 2, , 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, , i.e. electric field strength due to a short dipole at broadside on position, E=, , 20., , ®, p, 1, parallel to BA, 4p e 0 r 3, , ...(iv), , Its direction is parallel to the axis of dipole from positive to negative charge., Kirchhoff’s Rules:, (i) First law (or junction law): The algebraic sum of currents meeting at any junction is, zero,, i.e.,, SI = 0, This law is based on conservation of 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, i.e.,, SV = 0, This law is based on conservation of energy., I 2V, Numerical: Applying Kirchhoff’s second law SV = 0 to given, r1, d, a, closed circuit along the path abcda., + 2 - Ir2 - Ir1 + 2 = 0, I (r1 + r2 ) = 4, 4, I=, r1 + r2, , Þ, , b, , c, , r2, , 2V, , This current is non-zero, so given question is wrong., However, if terminals of one of the batteries are interchanged,, then current is zero., , I, , 2V, , a, , r1, , d, , - 2 - Ir2 - Ir1 + 2 = 0, Þ, , I=0, b, 2V, , r2, , c
Page 144 :
Examination Papers, , 143, , 21. (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., B1, +, , K, –, , Rh, , P2, , +, A, , P1, J, , –, B, G, , e, , + –, C, R, , HR, , (), K1, , Suppose A and r are resepectively the area of cross-section and specific resistance of the, material of the wire. Let V be the potential difference across the portion of the wire of, length l whose resistance is R. If I is the current flowing through the wire, then from Ohm’s, law;, , As, , V=IR, rl, R=, A, , \, , V=Ir, , or, , V µI, , l, A, , æ where K = Ir ö, ç, ÷, è, Aø, (if I and A are constant), , i.e., potential difference across any portion of potentiometer wire is directly proportional to, length of the wire of that portion., V, Here,, = K = is called potential gradient,, l, i.e., the fall of potential per unit length of wire., (b), , (i) The purpose of high resistance R 2 is to reduce the current through the, galvanometer. When jockey is far from balance point, this saves the galvanometer and, the cell (of emf e) from being damaged., (ii) When resistance R 1 is decreased, the potential gradient of potentiometer wire, increases, so balance point ( J ) shifts to longer length of wire., (iii) (1) The balance point is not obtained because maximum emf across potentiometer, wire is 2 V., (2) When key ( K) is closed, the terminal potential difference of cell is zero; so, balance point cannot be between Aand B. (Since V = kl Þ l = 0 for V = 0)
Page 145 :
N, , ®, , F1, I, F3, , ®, , ®, , P, S, F4, , p, , F1=IlB, I, (Upward), , ®, , 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, ®, , q, , N, b, , q, , B, , ®, , forces F2 and F4 acting on these sides has same, , I (Downward), N', , magnitude, F ¢ = Blb sin ( 90° - q) = Blb cos q., According to Fleming’s left hand rule the forces, ®, , B, R, , 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, ®, , q, , Q, , 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, , ®, , Axis of loop, or normal to loop, , F2, , loo, , Torque on a current carrying loop: Consider a, rectangular loop PQRS of length l, breadth b, , of, , 22., , is, , Xam idea Physics—XII, , ax, , 144, , b sin q, F3=IlB, , ®, , 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., \ Moment of couple or Torque,, 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,, \, t = IAB sin q, If the loop contains N-turns, then t = NI AB sin q
Page 146 :
Examination Papers, ®, , ®, , 145, , ®, , In vector form t = NI A ´ B., Direction of torque is perpendicular to direction of area of loop as well as the direction of, ®, , ®, , magnetic field i.e., along I A ´ B., 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., 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, ...(i), I=, =, \, 2 pr / v 2 pr, Area of current loop (electron orbit), A = pr 2, Magnetic moment due to orbital motion,, ev, evr, M l = IA =, ( pr 2 ) =, 2 pr, 2, , ...(ii), , This equation gives the magnetic dipole moment of a revolving electron. The direction of, magnetic moment is along the axis., Relation between magnetic moment and angular momentum, Orbital angular momentum of electron, L = me v r, , ...(iii), , where me is mass of electron,, Dividing (ii) by (iii), we get, M l ev r / 2, e, =, =, L, me v r, 2me, Magnetic moment, M l =, , e, L, 2m e, , …(iv), , This is expression of magnetic moment of revolving electron in terms of angular momentum, of electron., In vector form, ®, , Ml = -, , e ®, L, 2me, , ...(v)
Page 147 :
146, , Xam idea Physics—XII, , 23., , Axis, , Axis, I, Field fines of a current carrying solenoid, , 24., , Field fines of an electric dipole, , (i) Difference: Field lines of a solenoid form continuous current loops, 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., (ii) Two magnetic field lines cannot intersect because at the point of intersection, these will, be two directions of magnetic field which is impossible., (i) The conditions for total internal reflection are, (a) The ray must travel from a denser into a rarer medium., (b) The angle of incidence i > critical angle C., (ii) Numerical: Given A = 30°, When ray is incident normally on the other face, it retraces its path, so, r2 = 0., As r1 + r2 = A Þ r1 = A - r2 = 30° - 0 = 30°, i1 = 45°, Refractive index, n =, Þ, , 25., , 26., , n=, , sin i1, sin r1, , =, , 30°, 45°, , r1, , sin 45°, sin 30°, , 1/ 2, = 2 = 1 × 414, 1/2, , (a) We do not encounter diffraction effects of light in everyday observations. To observe, diffraction, 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, 1, (b) (i) If the width of slit is doubled, the (angular) width of central fringe æç µ ö÷ is halved., è aø, (ii) When wavelength of light used is increased (b q µ l), the width of central fringes increases., (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., (b) Give T = 30 s, 0 × 6931 0 × 6931 -1, (i) Decay constant, l =, =, s = 0 × 0231s -1, T, 30
Page 148 :
Examination Papers, , (ii), , N æ 1 ön, =ç ÷, N0 è 2 ø, , 1-, , 3 æ 1 ön, 1 2, 1 n, = ç ÷ or æç ö÷ = æç ö÷, è 2ø, è 2ø, 4 è 2ø, , t, = 2 or t = 2T = 2 ´ 30 = 60 s, T, Detection: Detection is the process of recovering the modulating signal from the modulated, carrier wave., Explanation of Detection with the help of a block diagram:, This gives, , 27., , Þ, , 147, , n=, , AM wave, , Envelope, Detector, , Rectifier, , (a), , (b), , time, , m(t), Out put, , (c), , time, , time, Output, , AM input wave, , 28., , Rectified wave, , The modulated carrier wave contains frequencies wc ± wm . The detection means to obtain, message signal m(t) of frequency wm . The method is shown in the form of a block diagram., The modulated signal is passed through a rectifier. It produces rectified wave [fig. (b)];' the, envelope of which is the message signal., The rectified wave is passed through an envelope detector, whose output is the required, message signal m(t)., AC generator: A dynamo or generator is a device which converts mechanical energy into, electrical energy. It is based on the principle of electromagnetic induction., Construction: It consists of the four main parts:, (i) Field Magnet: It produces the magnetic field. In the case of a low power dynamo, the, magnetic field is generated by a permanent magnet, while in the case of large power, dynamo, the magnetic field is produced by an electromagnet., (ii) Armature: It consists of a large number of turns of insulated wire in the soft iron drum or, ring. It can revolve round an axle between the two poles of the field magnet. 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., (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., (iv) Brushes: These are two flexible metal plates or carbon rods ( B1 and B2 ) which are fixed, and constantly touch the revolving rings. The output current in external load R L is taken, through these brushes.
Page 149 :
148, , Xam idea Physics—XII, w, Working: When the armature coil is rotated in the, Armature coil, strong magnetic field, the magnetic flux linked, b, with the coil changes and the current is induced in, the coil, its direction being given by Fleming’s right, c, hand rule. Considering the armature to be in, vertical position and as it rotates in anticlockwise, N B, S, Field magnet, direction, the wire ab moves upward and cd, a, downward, so that the direction of induced current, is shown in fig. In the external circuit, the current, d, B1, flows along B1 R L B2 . The direction of current, R1, remains unchanged during the first half turn of, armature. During the second half revolution, the, Slip rings, Brushes Load, RL, wire ab moves downward and cd upward, 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., If N is the number of turns in coil, f the frequency of rotation, A area of coil and B the, magnetic induction, then induced emf, df d, e== {NBA (cos 2p f t )} = 2p NBA f sin 2p f t, dt dt, Obviously, the emf produced is alternating and hence the current is also alternating., Current produced by an ac generator cannot be measured by moving coil ammeter; because, the average value of ac over full cycle is zero., The source of energy generation is the mechanical energy of rotation of armature coil., When plane of armature coil is parallel to magnetic field, then sin wt = 1, so emf is maximum,, the maximum value is e 0 = NBAw., OR, Transformer: Transformer is a device by which an alternating voltage may be decreased or, increased. This is based on the principle of mutual-induction., (A.C. mains), Construction: It consists of laminated core of, soft iron, on which two coils of insulated, Primary, copper wire are separately wound. These coils, are kept insulated from each other and from, the iron-core, but are coupled through mutual, Primary, induction. The number of turns in these coils, are different. Out of these coils one coil is, Core, laminated, called primary coil and other is called the, iron core, Secondary, secondary coil. The terminals of primary coils, are connected to AC mains and the terminals, of the secondary coil are connected to external, Step up, circuit in which alternating current of desired, Secondary, voltage is required., Step up Transformer: It transforms the, Transformer, alternating low voltage to alternating high
Page 150 :
Examination Papers, , 149, , voltage and in this the number of turns in secondary coil is more than that in primary coil., (i. e. , N S > N p )., Working: When alternating current source is connected to the ends of primary coil, the, current changes continuously in the 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., Let N p be the number of turns in primary coil, NS the number of turns in secondary coil and f, the magnetic flux linked with each turn. 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. According to, Faraday’s laws the emf induced in the primary coil, Df, ...(i), e p = - Np, Dt, and emf induced in the secondary coil, Df, e S = - NS, Dt, , ...(ii), , From (1) and (ii), eS, ep, , =, , NS, , ...(iii), , Np, , If the resistance of primary coil is negligible, the emf ( e p ) induced in the primary coil, will be, equal to the applied potential difference (Vp ) across its ends. Similarly if the secondary circuit, is open, then the potential difference VS across its ends will be equal to the emf ( e S ) induced, in it; therefore, VS e S N S, ...(iv), =, =, = r (say), Vp e p N p, where r =, , NS, Np, , 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, , In step up transformer,, , ...(v), , Ns > N p ® r > 1 ;, , VS > Vp and iS < ip, i.e., step up transformer increases the voltage., Power Transmission Over Long Distances, The power (electrical energy) is transmitted to long distances by the use of transformers. The, voltage output of the generator is stepped up by means of step up transformer, this steps, down the current, so power loss I 2 R is significantly reduced. At the receiving station the, voltage is stepped down to 220 V for domestic supply., So
Page 151 :
150, , Xam idea Physics—XII, , L, 29. (a) Lens Maker’s Formula: Suppose, L is a thin lens. 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 . The optical, P1 C P2, I, I', O, v, centre of lens is C and X ¢ X is, u, v', principal axis. The radii of, curvature of the surfaces of the, lens are R 1 and R 2 and their, 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 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, ...(i), =, v¢, u, R1, , 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, ...(ii), = 1, v (v¢ - t), R2, For a thin lens t is negligible as compared to v ¢ , therefore from (ii), n1 n2, n - n1, ...(iii), =- 2, v (v¢ ), R2, Adding equations (i) and (iii), we get, n1, v, , -, , æ 1, 1 ö, = (n2 - n1 ) ç, ÷, u, è R1 R2 ø, , n1, , or, , öæ 1, 1 1 æ n2, 1 ö, - =ç, - 1÷ ç, ÷, v u è n1, ø è R1 R2 ø, , i.e., , æ 1, 1 1, 1 ö, - = ( 1 n2 - 1) ç, ÷, v u, è R1 R2 ø, , where, , 1 n2, , =, , n2, n1, , ...(iv), , 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
Page 152 :
Examination Papers, , 151, , Therefore from equation (iv), æ 1, 1 1, 1 ö, - = ( 1 n2 - 1) ç, ÷, f ¥, è R1 R2 ø, æ 1, 1, 1 ö, = ( 1 n2 - 1) ç, ÷, f, è R1 R2 ø, , i.e., , ...(v), , 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 (v) may be written as, æ 1, 1, 1 ö, = (n - 1) ç, ÷, f, è R1 R2 ø, , ...(vi), , (b) Power of a Lens: The power of a lens is its ability to deviate the rays towards its principal, axis. It is defined as the reciprocal of focal length in metres., 1, 100, Power of a lens, P =, diopters =, diopters, f ( in metres), f ( in cm), The SI unit for power of a lens is dioptre (D)., Numerical:, P1 =, , Power of convex lens,, , Power of concave lens, P2 =, \, , 1, 1, = 2×0 D, D=, 0 × 50, F1, , 1, 1, D=, = -5×0 D, F2, - 0 × 20, , Power of combination of lenses in contact, P = P1 + P2 = 2 × 0 D - 5 × 0 D = - 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, by the object directly, when the final image and the object both, lie at infinite distance from the eye., , F0, a, , b
Page 153 :
152, , Xam idea Physics—XII, , b, a, , Magnifying power, m =, , (1), , As angles a and b are small, therefore, a » tan a and b » tan b., tan b, From equation (1),, m=, tan a, In DA ¢B ¢ C 2 ,, , tan b =, , A ¢B ¢, C2 B¢, , In DA ¢B ¢ C 1 ,, , tan a =, , A ¢B ¢, C 1B¢, , Put in equation (2),, , m=, , or, , m=, , (2), , A ¢B ¢ C 1 B ¢ C 1 B ¢, ´, =, C 2 B ¢ A ¢B ¢ C 2 B ¢, f0, - fe, , where C 1 B ¢ = f 0 = focal length of objective lens,, C 2 B ¢ = - f e = focal length of eye lens., Negative sign of m indicates that final image is inverted., The diameter of objective is kept large to increase (i) intensity of image, (ii) resolving, power of telescope., , Distinction, Telescope, , 30., , Compound Microscopes, , 1., , Objective lens is of large focal length and Both objective and eye lenses are of small focal, eye lens is of small focal length., lengths but focal length of eye lens is larger, than that of objective lens., , 2., , Objective is of very large aperature., , Objective is of small aperture., , (a) Formation of Depletion Layer and Potential Barrier, VB, 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, Ei, is an excess of positively charged ions in n-region, p, Depletion, and an excess of negatively charged ions in, layer, p-region. This sets up a potential difference called, potential barrier and hence an internal electric field Ei across the, junctions. The field Ei is directed from n-region to p-region. This field, stops the further diffusion of charge carriers. Thus the layers, ( » 10 - 4 cm to 10 - 6 cm) on either side of the junction becomes free, , n, , p, , n
Page 154 :
Examination Papers, , 153, , from mobile charge carriers and hence is called the depletion layer. The symbol of p-n, junction diode is shown in Figure., P1, Input A.C. signal, , (b) Full Wave Rectifier: For full wave rectifier we, use two junction diodes. The circuit diagram, for full wave rectifier using two junction diodes, is shown in figure., , S1, , I n, p, , B, , RL, , 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, then diode I is, S2, P2, p II n, forward biased and diode II is reverse biased., Therefore current flows in diode, I and not in diode II. 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. In next half, cycle, the terminal S1 is negative, relative to S and S2 is positive, relative to S. Then diode I is Output wave form, reverse biased and diode II is, of full wave, 2T Time, rectifier, T, forward, biased., Therefore, current flows in diode II and, there is no current in diode I. The direction of current i2 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, E, , Input, , n-p-n, , C, , B, , R1, , R2, Output, , – +, VEE, , –, , +, VCC, , n-p-n Comman base configuration, , Base Current and Collector Current: Under forward bias of emitter-base junction, the, electrons in emitter and holes in base are compelled to move towards the junction, thus the, depletion layer of emitter-base junction is eliminated. As the base region is very thin, 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. 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. The
Page 155 :
154, , Xam idea Physics—XII, , electron, released, is, Depletion, attracted, by, positive, layer, terminal of emitter battery, n, p, n, VEE , giving rise to a feeble, base current ( I B ). Its, E, C, direction in external circuit, IE, IC, is from emitter to base. The, e–, hole released in the base, region compensates the loss, B, +, –, of hole neutralised by I, IB, IC, e–, E, electrons., –, +, IB, The electrons crossing the, e–, base and entering the, collector, due to reverse, –, –, +, +, O, VEE, VCC, biasing of collector-base, junction, are attracted towards the positive terminal of collector battery VCC . 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, called the collector, current. In addition to this the collector current is also due to flow of minority charge carriers, under reverse bias of base-collector junction. This current is called the leakage current., Thus, collector current is formed of two components:, (i) Current ( I nc ) due to flow of electrons (majority charge carriers) moving from emitter to, collector., (ii) leakage current ( I leakage ) due to minority charge carriers, i. e. , I c = I nc + I leakage ., 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, an equal number of electrons enter from emitter battery VEE to emitter, causing the, emitter current. The process continues., Relation between Emitter, Base and Collector Currents:, Applying Kirchhoff’s I law at terminal O , we get, I E = I B + IC, That is, the emitter current I E is the sum of base current I B and the collector current I C . This is the, fundamental relation between currents in the bipolar transistor circuit., Transistor as a Switch, A switch is a device which can turn ON and OFF current is an electrical circuit., A transistor can be used to turn current ON or OFF rapidly in electrical circuits., Operation: The circuit diagram of n-p-n transistor in CE configuration working as a switch is, shown in fig. VBB and VCC are two dc supplies which bias base-emitter and emitter collecter, junctions respectively., Let VBB be the input supply voltage. This is also input dc voltage (VC ). The dc output voltage, is taken across collector-emitter terminals, R L is the load resistance in output circuit.
Page 156 :
Examination Papers, , IC, , C, RB, , IB, , B, , VCE, +, , E, Vi, –, , VBB, , 1, , VO, , RL, , VBE, , +, , 155, , –, , 2, VCC, , Applying Kirchhoff’s second law to input and output meshes (1) and (2), we get, VBB = I B R B + VBE, and, , …(i), , VCC = I C R L + VCE, , …(ii), , We have VBB = Vi and VCE = V0 , so above equations take the form, Vi = VBE + I B R B, , …(iii), , and, …(iv)Let us see the change in, V0 ( = VCE ) = VCC - I C R L, V0 due to a change in Vi . In case of Si transistor; the barrier voltage across base-emitter, junction is 0 × 6 V. Therefore, when Vi is less than 0 × 6 V, there is no collector current ( I C = 0),, so transistor will be in cut off state. Hence, from (iv) with I C = 0; V0 = VCC ., When Vi becomes greater than 0 × 6 V, I C begins to flow and increase with increase of Vi ., Thus, from (iv), V0 decreases upto Vi = 1 V; the increase in I C is linear and so decrease in, output voltage V0 is linear., Beyond Vi = 1 V, the change in collector current and, hence in output voltage V0 is non-linear and the, transistor goes into saturation. With further increase, in Vi , the output voltage further decrease towards zero, (though it never becomes zero)., If we plot V0 versus Vi , we get the graph as shown in, fig. [This characteristics curve is also called transfer, characteristic curve of base biased transistor.], , Cut off, , Active, region region, , V0, Saturation, region, , AV, , The curve shows that there are non-linear regions., (i) between cut off state and active state and (ii), Vi, 0.6 V, 1.0 V, between active state and saturation state; thus, showing that the transitions (i) from cut off to active state and from active to saturation state, are not sharply defined.
Page 157 :
156, , Xam idea Physics—XII, , Now we are in the position to explain the action of transistor as a switch. When transistor is, non-conducting ( I C = 0), it is said to be ‘switched off’ but when it is conducting (I C is not, zero); it is said to be ‘switched ON’., As long as input voltage Vi is low and unable to overcome the barrier voltage of the emitter, base junction, V0 is high (I C = 0 and V0 = VCC ), 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 - I C R L ) is, low, very near to zero, so the transistor is ‘switched ON’. Thus we can say low input switches, the transistor is OFF state and high input switches it ON., The switching circuits are designed in such a way that the transistor does not remain in active, state., , CBSE (Foreign) SET–II, 3., 7., , When the current in a coil is changed, a back emf is induced in the same coil. This, phenomenon is called self-induction., The rays coming out of the convex lens, when point source is at focus, are parallel, so, wavefront is plane (figure)., , Parallel, rays, , (b) Plane wavefront, , 11., , Name of gate formed is NOR gate symbol., Truth Table, Inputs, A, Y, B, , 12., , For particle to move undeviated,, ®, , ®, , ®, , qE +q v´ B =0, Þ, , ®, , ®, , ®, , E=- v ´ B, , Output, , A, , B, , Y, , 0, , 0, , 1, , 1, , 0, , 0, , 0, , 1, , 0, , 1, , 1, , 0
Page 158 :
Examination Papers, ®, , Given v = - v i$ ,, , ®, , B = - B k$, , ®, , E = - ( -vi$) ´ ( - Bk$) = - vB ( - j$) = vBj$, , \, 13., , i.e., magnitude of electric field is vB and its direction is along positive Y-axis., (a) The frequency of incident radiation was kept constant., I, (b) de-Broglie wavelength,, h, 1, l=, µ, 2mqV, V, If potential difference V is doubled, the de-Broglie, 1, times., wavelength is decreased to, 2, , 14., , 157, , I3, I2, , n0, , q1 + q2 = 9 ´ 10 -6 C, , After OR, , I3 > I2 > I1, , I1, , …(1), , 1 q1 q2, =2, 4pe 0 r 2, Þ, , q1 q2 =, , 2 ´ ( 0 × 30) 2, 9 ´ 10, , 9, , = 20 ´ 10 -12, , ( q1 - q2 ) = ( q1 + q2 ) 2 - 4q1 q2 = 81 ´ 10 -12 - 80 ´ 10 -12 = 1 ´ 10 -12, Solving (1) and (2), 16., , …(2), , q1 = 5 mC, q2 = 4 mC, , Microwaves:, Uses: 1. In Radar for aircraft navigation., 2. In microwave ovens to heat the food., , 23., , (i) A diamagnetic substance is attracted towards a region of weaker magnetic field., (ii) (a) Permanent magnets are made of steel which is characterised by high retentivity and, high coercivity., (b) Electromagnets are made of soft iron which is characterised by high retentivity and low, coercivity., , 26., , (a) (i) Amplitude Modulation, In amplitude modulation, the amplitude of modulated (carrier) wave varies in, accordance with amplitude of information (signal) wave. When amplitude of, information increases, the amplitude of modulated wave increases and vice versa. In, this case the amplitude of modulated wave is not constant., (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.e., amplitude modulation index,, E, ma = m, Ec
Page 159 :
158, , Xam idea Physics—XII, , For modulated wave,, E, - Emin, ma = max, Emax + Emin, (b) The modulation is needed due to, (i) Transmission of audiofrequency electrical signals need long impracticable antenna., (ii) The power radiated at audio frequency is quite small, hence transmission is quite, lossy., (iii) The various information signals transmitted at low frequency get mixed and hence can, not be distinguished., 27., , (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., (b), , (i) Given T = 20 s, Decay constant, l =, (ii) Fraction decayed =, , 0 . 6931, T, , =, , 0 . 6931, 20, , s -1 = 0 × 0346 s -1 = 0.0346 s–1, , 7, 8, , Fraction remained, , N, 7 1, =1- =, 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, n = 3,, , Þ, Þ, , æ1 ö, ç ÷, è 2ø, , 3, , 1 n, = æç ö÷, è 2ø, , t = 3T = 3 ´ 20 s = 60 s, , CBSE (Foreign) SET–III, 1., , By introducing the metal plate between the plates of charged capacitor, the capacitance of, capacitor increases., Reason: It t is thickness of metal plate, then, e0 A, C=, 1, d - t æç1 - ö÷, è, Kø, e A, For metal plate K = ¥, C = 0, ( d - t), Obviously, effective separation between plates is decreased from d to ( d - t).
Page 160 :
Examination Papers, , 4., 6., 8.i, , 9., , 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., Electromagnetic radiation produced by a Klystron or a Magnetron valve is microwave., Wavefront: The locus of particles of a medium vibrating in the same phase is called a, wavefront. From a point source, the wavefront is spherical; while for a line source the, wavefront is cylindrical. The distant wavefront is plane., P1 P2, (i) Charges induced on outer surfaces of P1 and P2 are -Q and +Q, –, +, respectively., When plates are released, they will tend to move away from one, another; plate P1 moving towards +Q and P2 towards -Q due to, attraction., , P1, , +Q, , P2, , –, , +, , –, , +, , –, , +, , –, , +, , –, , +, , Common potential, V =, =, , –Q, , C 1V1 + 0, C1 + C2, ( 800 pF) ´ 100, 800 + 800, , V = 50 V, , Net capacitance, 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., , +Q, , –, –, –, –, –, –, –, –, , +, +, +, +, +, +, +, +, , –Q +Q, , (ii) The field pattern is shown in fig., , 10., , 159, , X-rays, Uses: (i) To study crystal structure., (ii) to detect fracture in bones, stone in gall bladder and kidney etc., , –Q
Page 161 :
160, , Xam idea Physics—XII, , 14., , NAND gate., Truth Table, A, Y, , B, , 16., 24., , A, , B, , Y, , 0, , 0, , 1, , 1, , 0, , 1, , 0, , 1, , 1, , 1, , 1, , 0, , Intensity of incident radiations was kept constant. Frequency n 1 is highest., (i) Angle of dip decreases from 90° to 0° as one goes from magnetic pole to magnetic equator, of earth., , (ii) X is diamagnetic and Y is ferromagnetic., 0 × 6931 -1, 26. (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
Page 162 :
CBSE EXAMINATION PAPERS, DELHI–2010, Time allowed: 3 hours, , Maximum marks: 70, , 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, , CBSE (Delhi) SET–I, 1. In which orientation, a dipole placed in a uniform electric fields is in (i) stable, (ii) unstable, equilibrium?, 2. Which part of electromagnetic spectrum has largest penetrating power?, 3. A plot of magnetic flux ( f) versus current ( I ) is shown in the figure for, A, two inductors A and B. Which of the two has larger value of self, inductance?, f, B, , 4. Figure shows three point charges, +2q, - q and +3q. Two charges, +2q and -q are enclosed within a surface ‘S’. What is the electric, flux due to this configuration through the surface ‘S’?, , I, +2q, +3q, –q, , S
Page 163 :
162, , Xam idea Physics—XII, , 5. A glass lens of refractive index 1× 45 disappears when immersed in a liquid. What is the value of, refractive index of the liquid?, 6. What is the ratio of radii of the orbits corresponding to first excited state and ground state in a, hydrogen atom?, 7. A wire of resistance 8 R is bent in the form of a circle. What is the, A, B, effective resistance between the ends of a diameter AB ?, 8. State the conditions for the phenomenon of total internal reflection to, occur., 9. Explain the function of a repeater in a communication system., 10., (i) Write two characteristics of a material used for making permanent magnets., (ii) Why is core of an electromagnet made of ferromagnetic materials?, OR, Draw magnetic field lines when a (i) diamagnetic, (ii) paramagnetic substance is placed in an, external magnetic field. Which magnetic property distinguishes this behaviour of the field lines, due to the two substances?, 11. Draw the circuit diagram of an illuminated photodiode in reverse bias. How is photodiode used to, measure light intensity?, Lamp, C, 12. An electric lamp having coil of negligible inductance connected in series, with a capacitor and an AC source is glowing with certain brightness., How does the brightness of the lamp change on reducing the (i), capacitance, and (ii) the frequency? Justify your answer., Source, 13. 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., 14. The radii of curvature of the faces of a double convex lens are 10 cm and 15 cm. If focal length of, the lens is 12 cm, find the refractive index of the material of the lens., 15. An electron is accelerated through a potential difference of 100 volts. What is the de-Broglie, wavelength associated with it? To which part of the electromagnetic spectrum does this value of, wavelength correspond?, 16. A heavy nucleus X of mass number 240 and binding energy per nucleon 7.6 MeV is split into two, fragments Y and Z of mass numbers 110 and 130. The binding energy of nucleons in Y and Z is, 8.5 MeV per nucleon. Calculate the energy Q released per fission in MeV., 17. (a) The bluish colour predominates in clear sky., (b) Violet colour is seen at the bottom of the spectrum when white light is dispersed by a prism., State reason to explain these observations.
Page 164 :
Examination Papers, , 163, , 18. 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 ). On what, factors does the (i) slope and (ii) intercept of the lines depend?, 19. A parallel plate capacitor is charged by a battery. After sometime the battery is disconnected and a, dielectric slab with its thickness equal to the plate separation is inserted between the plates. How, will (i) the capacitance of the capacitor, (ii) potential difference between the plates and (iii) the, energy stored in the capacitor be affected?, Justify your answer in each case., 20. Write the principle of working of a potentiometer. Describe briefly, with the help of a circuit, diagram, how a potentiometer is used to determine the internal resistance of a given cell., ®, , I1, 21. 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., 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., I, l, Give reasons to explain that the loop will experience a net force but no, torque. Write the expression for this force acting on the loop., 22. (a) Depict the equipotential surfaces for a system of two identical positive point charges placed a, distance ‘d’ apart., (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 ., 23. 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. Write the expression for Brewster, angle in terms of the refractive index of denser medium., 24., (i) Define ‘activity’ of a radioactive material and write its S.I. unit., (ii) Plot a graph showing variation of activity of a given radioactive sample with time., (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, what are their, corresponding values for D ?, 25. A long straight wire of a circular cross-section of radius ‘a’ carries a steady current ‘I’. The, current is uniformly distributed across the cross-section. 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., OR, State the underlying principle of working of a moving coil galvanometer. Write two reasons why a, galvanometer can not be used as such to measure current in a given circuit. Name any two factors, on which the current sensitivity of a galvanometer depends., 26. What is space wave propagation? Give two examples of communication system which use space, wave mode., A TV tower is 80 tall. Calculate the maximum distance upto which the signal transmitted from the, tower can be received.
Page 165 :
164, , Xam idea Physics—XII, , 27. In a meter bridge, the null point is found at a distance of 40 cm from A. If a resistance of 12 W is, connected in parallel with S, the null point occurs at 50 × 0 cm from A. Determine the values of, R and S., 12W, , S, , R, G, A, , B, , 28. Describe briefly, with the help of a labelled diagram, the basic elements of an AC generator. State, its underlying principle. Show diagrammatically how an alternating emf is generated by a loop of, wire rotating in a magnetic field. Write the expression for the instantaneous value of the emf, induced in the rotating loop., OR, A series LCR circuit is connected to an ac source having voltage v = v m sin wt. Derive the, expression for the instantaneous current J and its phase relationship to the applied voltage., Obtain the condition for resonance to occur. Define ‘power factor’. State the conditions under, which it is (i) maximum and (ii) minimum., 29. State Huygen’s principle. Show, with the help of a suitable diagram, how this principle is used to, obtain the diffraction pattern by a single slit., Draw a plot of intensity distribution and explain clearly why the secondary maxima become, weaker with increasing order ( n) of the secondary maxima., OR, Draw a ray diagram to show the working of a compound microscope. Deduce an expression for, the total magnification when the final image is formed at the near point., In a compound microscope, an object is placed at a distance of 1× 5 cm from the objective of focal, length 1× 25 cm. If the eye piece has a focal length of 5 cm and the final image is formed at the, near point, estimate the magnifying power of the microscope., 30., (a) Explain the formation of depletion layer and potential barrier in a p-n junction., (b) In the figure given below, the input waveform is converted into the output wave from a, device ‘X’. Name the device and draw its circuit diagram., Device ‘X’, Input, , Output, , (c) Identify the logic gate represented by the circuit as shown and write its truth table., A, Y, B
Page 166 :
Examination Papers, , 165, , OR, (a) With the help of the circuit diagram explain the working principle of a transistor amplifier as, an oscillator., (b) Distinguish between a conductor, a semiconductor and an insulator on the basis of energy, band diagrams., , CBSE (Delhi) SET–II, Questions uncommon to Set–I, 3. The radius of innermost electron orbit of a hydrogen atom is 5 × 3 ´ 10 -11 m. What is the radius of, orbit in the second excited state?, 6. Which part of electromagnetic spectrum is absorbed from sunlight by ozone layer?, 9., (i) When primary coil P is moved towards secondary coil S (as shown in the figure below) the, galvanometer shows momentary deflection. What is can be done to have larger deflection in, the galvanometer with the same battery?, (ii) State the related law., P, , S, , –, G, , +, V, , 10. What is the range of frequencies used for T.V. transmission? What is common between these, waves and light waves?, 2, 11. A biconvex lens has a focal length times the radius of curvature of either surface. Calculate the, 3, refractive index of lens material., 14., (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. An electron is accelerated through a potential difference of 144 volts. What is the de-Broglie, wavelength associated with it? To which part of the electromagnetic spectrum does this, wavelength correspond?, 19. A parallel plate capacitor, each with plate area A and separation d, is charged to a potential, difference V. The battery used to charge it remains connected. A dielectric slab of thickness d and, dielectric constant k is now placed between the plates. What change, if any, 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., 20., (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. What is the maximum distance between them for satisfactory, communication in line of sight mode?
Page 167 :
166, , Xam idea Physics—XII, , 22. In a meter bridge, the null point is found at a distance of 60.0 cm from A. If now a resistance of, 5 W is connected in series with S, the null point occurs at 50 cm. Determine the values of R and S., S, , R, G, A, , B, , CBSE (Delhi) SET–III, Questions uncommon to Set–I, 4., 5., 7., 11., , Which part of electromagnetic spectrum is used in radar systems?, Calculate the speed of light in a medium whose critical angle is 30°., Write the expression for Bohr’s radius in hydrogen atom., What is the range of frequencies used in satellite communication? What is common between these, waves and light waves?, 12. A coil Q is connected to low voltage bulb B and placed near another coil P is shown in the figure., Give reason to explain the following observations:, B, , Q, , 13., 14., , 15., , 20., , A.C Source, , P, , (a) The bulb ‘B’ lights., (b) Bulb gets dimmer if the coil Q is moved towards left., Find the radius of curvature of the convex surface of a plano-convex lens, whose focal length is, 0 × 3 m and the refractive index of the material of the lens is 1× 5., An electron is accelerated through a potential difference of 64 volts. 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., (ii) Give the formula that can be used to determine refractive index of material of a prism in, minimum deviation condition., In a metre bridge, the null point is found at a distance of l1 cm from A. If now a resistance of X is, connected in parallel with S, the null point occurs at l2 cm. Obtain a formula for X in terms of, l1 , l2 and S.
Page 168 :
Examination Papers, , 167, , X, , S, , R, G, A, , B, , 27. A parallel plate capacitor is charged to a potential difference V by a d.c. source. The capacitor is, then disconnected from the source. If the distance between the plates is doubled, state with reason, how the following will change:, (i) electric field between the plates., (ii) capacitance, and, (iii) energy stored in the capacitor., , Solutions, CBSE (Delhi) Set-I, (i) In stable equilibrium the dipole moment is parallel to the direction of electric field (i.e., q = 0)., , 1., , (ii) In unstable equilibrium P.E. is maximum, so q = p so dipole moment is antiparallel to electric field., 2. g-rays have largest penetrating power., 3. f = L I, For same current fA > fB , so L A > L B, i.e., Inductor A has larger value of self-inductance., 1, 4. Electric flux, f =, ´ (net charge enclosed by surface S), e0, q, 1, =, ´ ( 2q - q) =, e0, e0, 5. For disappearance of glass lens in liquid, refractive index of liquid, = refractive index of lens = 1× 45, 6. r n =, , e0 h2 n2, 2, , µ n2, , pme, For I excited state, n = 2, For ground state, n =1, , \, , r2 4, =, r1 1
Page 169 :
168, , Xam idea Physics—XII, , 7. Two parts each of resistance 4R are connected in parallel; so effective resistance across ends of, diameter AB is, 4R ´ 4R, =, = 2R, 4R + 4R, 8. Conditions for total internal reflection are:, (i) Light must travel from denser to rarer medium., (ii) Angle of incidence must be greater than critical angle (C)., 9. A repeater is a combination of a receiver and a transmitter. Repeaters are used to increase the, range of communication of signals. A repeater picks up the signal from the transmitter, amplifiers, and retransmits it to the receiver, sometimes with a change in carrier frequency. A typical example, of repeater station is a communication satellite., 10., (i) For permanent magnet the material must have high retentivity and high coercivity (e.g.,, steel)., (ii) Ferromagnetic material has high retentivity, so when current is passed in ferromagnetic, material it gains sufficient magnesium immediately on passing a current through it., 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., 11. It is a reversed biased p-n junction, illuminated by radiation. When p-n junction is reversed biased, with no current, a very small reverse saturated current flows across the junction called the dark, current. When the junction is illuminated with light, electron-hole pairs are created at the junction,, due to which additional current begins to flow across the junction; the current is solely due to, minority charge carriers., Light, , p, , n, mA, p, , n, Symbol, , –, , +, (a), , (b), , Photodiode
Page 170 :
Examination Papers, , 12., , (i) When capacitance is reduced, capacitive reactance X C =, , 1, wC, , 169, Lamp, , C, , increases, hence impedance of circuit, Z = R 2 + X C2, increases and so current I =, , V, decreases. As a result the brightness, Z, , of the bulb is reduced., , Source, , 1, increases and hence, 2pnC, impedance of circuit increases, so current decreases. As a result brightness of bulb is reduced., 13. In ascending power of frequencies: radiowaves, microwaves, ultraviolet rays, X-rays and gamma, rays., Uses of Electromagnetic Spectrum, (i) g-rays are highly penetrating, they can penetrate thick iron blocks. Due to high energy, they, are used to produce nuclear reactions. g-rays are produced in nuclear reactions. In medicine,, they are used to destroy cancer cells., (ii) X-rays are used in medical diagnostics to detect fractures in bones, tuberculosis of lungs,, presence of stone in gallbladder and kidney. They are used in engineering to check flaws in, bridges. In physics X-rays are used to study crystal structure., (iii) Radiowaves are used for broadcasting programmes to distant places. According to frequency, range, 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, namely; klystrons, magnetrons and gunn, diodes. Their frequency range is 3 GHz to 300 Ghz. They are used in radar systems used in air, craft navigation and microwave users in houses., 14. Given R1 = 10 cm, R 2 = - 15 cm, f =12 cm, (ii) When frequency in decreases; capacitive reactance X C =, , Refractive index n = ?, Lens-maker’s formula is, æ 1, 1, 1 ö, ÷, = ( n - 1) çç, ÷, f, R, R, 2 ø, è 1, 1, 1ö, æ1, = ( n - 1) ç + ÷, Þ, 12, è 10 15 ø, 5, = ( n - 1) ´, 30, 30 1, 30, or n = 1 +, Þ, n -1 = ´, Þ n = 1 + 0 × 5 = 1× 5, 5 12, 60, 15. l =, , h, 2meV, , =, , 6 × 63 ´ 10 -34, 2 ´ 9 ×1 ´ 10 -31 ´ 1× 6 ´ 10 -19 ´ 100, , m
Page 171 :
170, , Xam idea Physics—XII, , = 1× 227 ´ 10 -10 m = 1× 227 Å, This wavelength corresponds to X-ray region of em spectrum., 16. Q = ( M y + M z ) c 2 - M ´ c 2, = 8 × 5 ´ 240 MeV – 7 × 6 ´ 240 MeV, = (8 × 5 - 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 çç i. e., I µ ÷÷ ., l4 ø, è, In visible light blue colour has minimum wavelength, so it is scattered most, that is why, bluish colour predominates in a clear sky., (b) While light consists of infinite wavelengths starting from 400 nm (violet) to 750 nm (red)., The refractive index of proton is maximum for violet and minimum for red; so prism, separates constituent colours of white light and causes maximum deviation for violet colour., That is why violet colour is seen at the bottom of spectrum when white light is dispersed, through a prism., 2, 1, 18. The graph of stopping potential Vs and frequency (n) for two, metals 1 and 2 is shown in fig., Vs, h, (i) Slope of graph tan q = and depends on h and e., q, q, e, n, (ii) Intersect of lines depend on the work function., 2 (w2/e), 19., (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., Q, V, Reason: V = ; Q same, C increases to K times; V ¢ =, C, K, V, 1, (iii) As E = and V is decreased; therefore, electric field decreases to times. Energy stored by, d, K, Q2, . As Q = constant, C is increased, and so energy stored by capacitor, the capacitor, U =, 2C, 1, decreases to times., K, 20., Principle: 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, i.e.,, V µl, Method: (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, 17.
Page 172 :
Examination Papers, , 171, , potentiometer wire, the deflection in, B1, K, the galvanometer is on both sides., Suppose that in this position the, +, –, Rh, potential gradient on the wire is k., (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. Let this, e, + –, HR, position be P1 and AP1 = l1 ., C, In this situation the cell is in, R (), open circuit, therefore the, K1, 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 l, ö, æ e, ö, r =ç, - 1 ÷ R = çç 1 - 1 ÷÷ R, è V, ø, è l2, ø, From this formula r may be calculated., 21. Magnetic moment due to a planar square loop of side l carrying current I is, \ Internal resistance of cell,, , ®, , ®, , m =I A, , For square loop A = l, , ®, , l, , m = I l 2 n$, , \, , m, , l, , 2, , where n$ is unit vector normal to loop., Magnetic field due to current carrying wire at the location of loop is directed downward, perpendicular to plane of loop., I1, y, , l, P, , Q, I, , x, , l, z, S, , R
Page 173 :
172, , Xam idea Physics—XII, , Force on QR and SP are equal and opposite, so net force on these sides is zero., ®, , ®, , From on side RS, FRS, ®, , ®, , ®, , ®, , FPQ = I l ´ B1 li$, , Force no side PQ,, , u I, = Ili$ ´ 0 1 ( -k$), 2pl, m 0 II 1 $, =, j;, 2p, m I, = m 0 l ( -i$) + 0 1 ( -k$), 2p ( 2l), ®, , m 0 II 1 $, j;, 4p, ® ® ®, m II, Torque t = r ´ F = - l $j ´ 0 1 j$ ;, 2p, m, II, æ, 0, 1 $ö, + ( -21$j) ´ ç, j ÷ = zero, è 4p, ø, , Net force F = FPQ ´ FRS =, , That is loop experiences a repulsive force but no torque., 22., (a) Equipotential surfaces due to two identical charges is shown in fig., , (b) Potential energy of a system of two charges in an external electric field., ®, , ®, , Suppose q1 and q 2 are two charges brought from infinity at locations r1 and r 2 respectively 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., ®, , ®, , 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 . dr =, r, dr, ., (, ), ¥, 4pe 0 ò¥, r2, r, 1, 1, =q1 q 2 ò 12, dr, ¥, 4pe 0, r2
Page 174 :
Examination Papers, r /2, , ér -1 ù, 1, q1 q 2 ê, =ú, 4pe 0, êë -1 ûú ¥, ®, , =, , 173, , 1 q1 q 2, 4pe 0 r12, , ®, , where r12 = | r 2 - 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 - r1 |, , 23. 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., Unpolarised, light, , ip, , n, , r', , Polarised light, Partially polarised light, , r, , If unpolarised light falls on a transparent surface of refractive index n at a certain angle i p , called, polarising angle, 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., Brewster’s law: n = tan i p ., 24., , (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 - 2D ,, , b-, , ¾®, , (i) Given A - 4 = 176 Þ Mass number of D, A =180, (ii) Z - 1 = 71 Þ Atomic number of D, Z = 72., , A-4, Z - 1 D2
Page 175 :
174, , Xam idea Physics—XII, , 25. Magnetic Field due to a straight thick wire of, uniform cross-section: Consider an infinitely long, cylinderical wire of radius a, carrying current I., 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., 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., 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. 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, , B 2pr = m 0 ´ I, m I, Þ, B= 0, 2pr, This expression is same as the magnetic field due to a long current carrying straight wire., This shows that for external points the current flowing in wire may be, supposed to be concerned at the axis of cylinder., Magnetic Field at Internal Points: Consider a circular path of radius, r ( < a), passing through internal point Q, concentric with circular, cross-section of the wire. In this case the assumed circular path encloses, only a path of current carrying circular cross-section of the wire., , \ Current enclosed by path, = current per unit cross-section ´ cross section of assumed circular path, æ I, = i ´ pr 2 = çç, è pa 2, , ö, Ir 2, ÷÷ ´ pr 2 =, a2, ø, , \ By Ampere’s circuital law, ® ®, , ò B × dl = m 0 ´ current closed by path
Page 176 :
Examination Papers, , Þ, , B × 2pr = m 0 ´, B=, , 175, , Ir 2, a2, , m 0 Ir, , 2pa 2, Clearly, magnetic field strength inside the current carrying wire is directly proportional to distance, of the point from the axis of wire., At surface of cylinder r = a, 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., 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, , Maximum value of magnetic field is at the surface given by B outside =, , m 0I, ×, 2pa, , OR, Principle: When current ( I ) is passed in the coil, 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, N is, the number of turns in a coil., When the magnetic field is radial, as in the case of cylindrical pole pieces and soft iron core, then, in every position of coil the plane of the coil, is parallel to the magnetic field lines, so that q = 90°, and sin 90° = 1, Deflecting torque,, t = NIAB, A galvanometer cannot be used as such to measure current due to following two reasons., (i) A galvanometer has a finite large resistance and is connected in series in the circuit, so it will, increase the resistance of circuit and hence change the value of current in the circuit., (ii) A galvanometer is a very sensitive device, it gives a full scale deflection for the current of the, order of microampere, hence if connected as such it will not measure current of the order of, ampere., Current sensitivity of galvanometer depends on, (i) Number of turns N: It increases with increase of number of turns., (ii) Area of coil A: It increases with increase of area of coil., (iii) Strength of magnetic poles (B): It increases with increase of strength of poles.
Page 177 :
176, , Xam idea Physics—XII, , (iv) Torsional rigidity of suspension: It increases with decrease of torsional rigidity of, suspension., 26. Space wave propagation is a straight line propagation of electromagnetic wave from transmitting, antenna to recieving antenna both installed in the ground., Maximum coverage distance, d = 2R e h, = 2 ´ 6400 ´ 10 3 ´ 80 = 32 ´ 10 3 m = 32 km, 27. In first case l1 = 40 cm, l1, R, =, S 100 - 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, so, 12S, …(ii), S¢ =, 12 + S, R, 50, …(iii), \, =, = 1 Þ S¢ = R, S ¢ 100 - 50, 3, From (i), S= R, 2, Substituting this value in (ii), 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), 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 , S =6W, 28. AC generator consists of the four main parts:, (i) Field Magnet: It produces the magnetic field. In the case of a low power dynamo, the, magnetic field is generated by a permanent magnet, while in the case of large power dynamo,, the magnetic field is produced by an electromagnet., (ii) Armature: It consists of a large number of turns of insulated wire in the soft iron drum or, ring. It can revolve round an axle between the two poles of the field magnet. 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.
Page 178 :
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. These rings are fixed to the shaft which rotates the armature coil so that, the rings also rotate along with the armature., (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 output current in external load R L is taken through, these brushes., Principle: When the armature coil is rotated in the strong magnetic field, the magnetic flux, linked with the coil changes and the current is induced in the coil, its direction being given by, Fleming’s right hand rule. The direction of current remains unchanged during the first half turn of, armature. During the second half revolution,, w, the direction of current is reversed. Thus, the, Armature coil, direction of induced emf and current changes, b, in the external circuit after each half, c, revolution., If N is the number of turns in coil, f the, N B, S, Field magnet, frequency of rotation, A area of coil and B the, a, magnetic induction, 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, inductance L and capacitance C are connected in series and an alternating, source of voltage V = V0 sin wt is applied across it. (fig. a) On account of being in series, the, current (i ) flowing through all of them is the same., C, , R, , L, , VL, , VR, , VC, , V=V0 sin wt, (a), , Suppose the voltage across resistance R is VR , voltage across inductance L is VL and voltage, across capacitance C is VC . The voltage VR and current i are in the same phase, the voltage VL will, lead the current by angle 90° while the voltage VC will lag behind the current by angle 90° (fig., b). Clearly VC and VL are in opposite directions, therefore their resultant potential difference, = VC - VL (if VC > VC )., Thus VR and (VC - VL ) are mutually perpendicular and the phase difference between them is 90°., As applied voltage across the circuit is V, the resultant of VR and (VC - VL ) will also be V. From, fig., V, , 2, , = VR 2 + (VC - VL ) 2 Þ V = VR 2 + (VC - VL ) 2, , ...(i)
Page 179 :
178, , Xam idea Physics—XII, , But, where X C, \, \, , VR = R i , VC = X C i and VL = X L i, 1, =, = capacitance reactance and X L = wL = inductive reactance, wC, , ...(ii), , V = ( R i) 2 + ( X C i - X L i) 2, V, Impedance of circuit, Z = = R 2 + ( X C - X L ) 2, i, , i.e., Instantaneous current I =, , æ 1, ö, Z = R 2 + (X C - X L ) 2 = R 2 + ç, - wL ÷, w, C, è, ø, V0 sin ( wt + f), æ 1, ö, R2 + ç, - wL ÷, è wC, ø, , 2, , 2, , 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., Phase difference ( f) in series LCR circuit is given by, X - XL, tan f= C, R, For resonance, f= 0, Þ, XC - X L = 0, or, XC = X L, 1, If wr is resonant frequency, 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, i.e., cos f = ., Z, For maximum power, cos f =1 or, Z =R, i.e., circuit is purely resistive., For minimum power, cos f = 0 or R = 0, i.e., circuit should be free from ohmic resistance., 29. Principle:, (i) Every point on a given wavefront may be regarded as a source of new disturbance., (ii) The new disturbances from each point spread out in all directions with the velocity of light, and are called the secondary wavelets., (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., Let us illustrate this principle by the following example:
Page 180 :
Examination Papers, , 179, , Let AB shown in the fig. be the section of a wavefront in a homogeneous isotropic medium at, t = 0. We have to find the position of the wavefront at time t using Huygens’ principle. Let v, be the velocity of light in the given medium., (a) Take the number of points 1, 2, 3, … on the wavefront AB. These points are the sources, of secondary wavelets., (b) At time t the radius of these secondary wavelets is vt. Taking each point as centre, draw, circles of radius vt., (c) Draw a tangent A1 B1 common to all these circles in the forward direction., 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, we get diffraction, pattern on a screen placed behind the slit., 2, 2, The diffraction pattern contains bright and, dark bands, the intensity of central band is, 3, 3, maximum and goes on decreasing on both, S, 4, sides., 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. According to Fresnel the diffraction, B2, B, (a), (b), pattern is the result of superposition of a, 1, large number of waves, starting from, different points of illuminated slit., 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 path difference between rays diffracted at points A and B,, D = BP - AP = BN, In D ANB , ÐANB = 90°, \ and ÐBAN = q, BN, or BN = AB sin q, sin q =, \, AB, As AB = width of slit = a, \ Path difference,, ....(i), D = a sin q, To find the effect of all coherent waves at P, we have to sum up their contribution, each with a, different phase. This was done by Fresnel by rigorous calculations, but the main features may, be explained by simple arguments given below:, At the central point C of the screen, the angle q is zero. Hence the waves starting from all, points of slit arrive in the same phase. This gives maximum intensity at the central point C., If point P on screen is such that the path difference between rays starting from edges A and B, is l , then path difference, l, a sin q = l Þ sin q =, a
Page 181 :
180, , Xam idea Physics—XII, , sin q = q =, , If angle q is small,, , l, a, , ...(ii), , I0, , –4l, a, , –3l, a, , –2l, a, , –l, a, , l, a, , 0, , 2l, a, , 3l, a, , 4l, a, , The intensity of secondary maxima decreases with increase of order n because with, increasing n, the contribution of slit decreases., For n = 2, it is one-fifth, for n = 3, it is one-seventh and so on., OR, Eyepiece, uo, , ue, , vo, , D, Objective, , L1, , B, A", A, , Eye, , Fo, Fe', , O, , A', , B', , B", , ve, , Magnifying power of microscope,, v æ, Dö, M = - 0 çç1 + ÷÷, u0 è, fe ø, , E, , Fe
Page 182 :
Examination Papers, , 181, , Given u 0 = - 1× 5 cm, f 0 = 125 cm, f e = 5 cm, 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, , I n, p, B, , S, , RL A, Output, , P2 S2 p II n, , Output, , P1 S1, , Input signal, , (a) Formation of depletion layer and potential barrier:, VB, 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, +, +, +, Ei, charges to neutralise each other. 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. This sets up a potential difference called, potential barrier and hence an internal electric field Ei across the, p, n, junctions. The field Ei is directed from n-region to p-region. This field, stops the further diffusion of charge carriers. Thus the layers, ( » 10 - 4 cm to 10 - 6 cm) on either side of the junction becomes free, from mobile charge carriers and hence is called the depletion layer. The symbol of p-n, junction diode is shown in Fig., (b) The box contains the circuit of full wave rectifier., , Input A.C. signal, , 30., , v 0 = 7 × 5 cm, 7 ×5 æ, 25 ö, M =ç1 +, ÷ = - 5 ´ 6 = - 30, 1× 5 è, 5 ø, , (c) Logic gate is AND gate. Its truth table is, A, , B, , Y, , 0, , 0, , 0, , 1, , 0, , 0, , 0, , 1, , 0, , 1, , 1, , 1
Page 183 :
Xam idea Physics—XII, , OR, (a) Principle: An oscillator converts dc into ac. 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., (b) If the valence and conduction bands overlap, the, substance is referred as a conductor., If the valence and conduction bands have a forbidden, gap more than 3 eV, the substance is an insulator., If the valence and condition bands have a small, forbidden gap (=1 eV), the substance is a, semiconductor., , 1, , L, , +, –, , npn, , L, , C, +, –, , 182, , CBSE (Delhi) SET–II, 3. Radius of nth orbit of hydrogen atom, rn =, , e0 h2 n2, , 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. Ozone layers absorbs ultraviolet rays., 9., (i) For larger deflection to coil P should be moved at a faster rate., (ii) Faraday law: The induced emf is directly proportional to rate of change of magnetic flux, linked with the circuit., 10. 76–890 MHz., Speed of waves is same for TV waves and light waves., æ 1, 1, 1 ö, ÷, 11., = ( n - 1) çç, ÷, f, è R1 R 2 ø, For biconvex lens R1 = + R, R 2 = - R, 2, Given f = R, 3, 3, 3, 3 7, æ2ö, = ( n - 1) ç ÷ Þ ( n - 1) = Þ n = 1 + =, \, 2R, R, 4, 4 4, è ø
Page 184 :
Examination Papers, , 14., , 183, , (i) The light is scattered by air molecules. According to Lord Rayleigh the intensity of scattered, light, 1, 1, I µ, Þ Iµ, 4, (wavelength), l4, As l blue < l red , accordingly blue colour is scattered the most and red the least, so sky, appears blue., At the time of sunrise and sunset, blue colour is scattered the most and red colour enters our, eyes, so sunrise and sunset appear red., (ii) Refractive index of prism material is maximum for violet and minimum for red colour., , 17. l =, , h, 2meV, , =, , 6 × 63 ´ 10 -34, 2 ´ 9 ×1 ´ 10, , -31, , ´ 1× 6 ´ 10, , -19, , = 1 ´ 10 -10 m = 1 Å, ´ 144, , This corresponds to X-rays., 19., (i) The charge Q = CV , V = same, C = increases; there, charge on plates increases., V, (ii) A electric field E = , and V = constant, d = constant; therefore, electric field strength, d, remains the same., (iii) The capacitance of capacitor increases as K >1., 20., (i) The line of sight (LOS) mode is limited to frequencies above 40 MHz, 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., For receiving signals beyond horizon, the receiving antenna must be very high to intercept, the LOS wave., (ii) Maximum LOS distance, d m = 2R e hT + 2R e hR, Given hT = 32 m, hR = 50 m ; 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, , 6 × 4 + 10 ´ 10 3, , 3, , 6×4, , 3, , 6 × 4 = 18 ´ 10 ´ 2 × 53 m, 3, , = 45 × 5 ´ 10 m = 45 × 5 km, 22., , R, l, =, S (100 - l), Given balancing length l = 60 × 0 cm., R 60, R 3, =, Þ, =, S 40, S 2, When a resistance of 5 W is connected in series with S, l¢ = 50 cm, R, 50, R, =, Þ, =1, \, S + 5 100 - 50, S +5, , …(i)
Page 185 :
184, , Xam idea Physics—XII, , or, R =S + 5, Solving (i) and (ii), R =15 W, S =10 W, , …(ii), , CBSE (Delhi) SET-III, 4. Microwaves are used in operating a RADAR., 1, 1, 5. n =, =, =2, sin C sin 30°, Speed of light in medium v =, , c 3 ´ 108, = 1× 5 ´ 10 8 m / s, =, n, 2, , 1× 5 ´ 108 m/s, e0 h2, , = 0 × 529 ´ 10 -10 m, pme 2, 11. Range of frequencies used for satellite communication, 5 × 925 - 6 × 425 GHz (Uplink), 3 × 7 - 4 × 2 GHz (Downlink), Speed of wave is same for these waves and light waves., 12., (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., (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., 13. For a plano-convex lens R1 = ¥, R 2 = - R, 7. Bohr’s radius, r1 =, , æ 1, 1, 1 ö, ÷ gives, = ( n - 1) çç, ÷, f, è R1 R 2 ø, 1, æ1 1ö, = ( n - 1) ç + ÷, f, èµ Rø, 1 n -1, =, Þ R = ( n - 1) f, f, R, , \, , or, , Given f = 0 × 3 m, n = 1× 5, \, R = (1× 5 - 1) ´ 0 × 3 m = 0 ×15 m = 15 cm, 14. de-Broglie wavelength, l=, , h, 2meV, , 6 × 28 ´ 10 -34, , =, =, , 2 ´ 9 ×1 ´ 10 -31 ´ 1× 6 ´ 10 -19 ´ V, 12 × 27, , Here V = 64 V, , V, , ´ 10 -10 m =, , 12 × 27, V, , Å
Page 187 :
CBSE EXAMINATION PAPERS, ALL INDIA–2010, Time allowed: 3 hours, , Maximum marks: 70, , 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–, 4pe o, , m 0 = 4p ´ 10 -7 TmA -1, 2, , 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. Name the physical quantity whose S.I. unit is JC -1 . Is it a scalar or a vector quantity?, 2. A beam of a particles projected along +x-axis, experiences a force due to a magnetic field along, the +y-axis. What is the direction of the magnetic field?, x, , e particle, z, y, , 3. Define self-inductance of a coil. Write its SI units., 4. A converging lens is kept co-axially in contact with a diverging lens – both the lenses being of, equal focal lengths. What is the focal length of the combination?, 5. Define ionisation energy. What is its value for a hydrogen atom?
Page 188 :
Examination Papers, , 187, , 6. Two conducting wires X and Y of same diameter but different materials are joined in series across, a battery. If the number density of electrons in X is twice that in Y, find the ratio of drift velocity, of electrons in the two wires., 7. Name the part of electromagnetic spectrum whose wavelength lies in the range of 10 -10 m. Give, its one use., 8. When light travels from a rarer to a denser medium, the speed decreases. Does this decrease in, speed imply a decrease in the energy carried by the light wave? Justify your answer., 9. Deduce the expression for the magnetic dipole moment of an electron orbiting around the central, nucleus., 10. A spherical conducting shell of inner radius r1 and outer radius r 2 has a charge ‘Q’. A charge ‘q’, is placed at the centre of the shell., (a) What is the surface charge density on the (i) inner surface, (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., 11. Draw a sketch of a plane electromagnetic wave propagating along the z-direction. Depict clearly, the directions of electric and magnetic fields varying sinusoidally with z., ®, s, 12. Show that the electric field at the surface of a charged conductor is given by E =, n$, where s is, e0, the surface charge density and n$ is a unit vector normal to the surface in the outward direction., 13. Two identical loops, one of copper and the other of aluminium, are rotated with the same angular, speed in the same magnetic field. Compare (i) the induced emf and (ii) the current produced in the, two coils. Justify your answer., 14. An a-particle and a proton are accelerated from rest by the same potential. Find the ratio of their, de-Broglie wavelengths., 15. Write two factors justifying the need of modulating a signal., A carrier wave of peak voltage 12 V is used to transmit a message signal. What should be the peak, voltage of the modulating signal in order to have a modulation index of 75%?, 16. Write Einstein’s photoelectric equation. State clearly the three salient features observed in, photoelectric effect, which can be explained on the basis of the above equation., 17. Draw a plot of potential energy of a pair of nucleons as a function of their separation. Write two, important conclusions which you can draw regarding the nature of nuclear forces., OR, Draw a plot of the binding energy per nucleon as a function of mass number for a large number of, nuclei, 2 £ A £ 240. 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; so binding, energy per nucleon remains constant.], 18., (i) Identify the logic gates marked P and Q in the given logic circuit., A, P, , C, , B, , Q, , X, , B, , (ii) Write down the output at X for the inputs A = 0, B = 0 and A = 1, B = 1.
Page 189 :
188, , Xam idea Physics—XII, , 19. 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. Why is there an upper limit to frequency of waves used in this mode?, +, –, 20. Write any two factors on which internal resistance of a cell depends., V, The reading on a high resistance voltmeter, when a cell is connected, across it, is 2 × 2 V. When the terminals of the cell are also connected to a, +, –, resistance of 5 W as shown in the circuit, the voltmeter reading drops to, 1× 8 V. Find the internal resistance of the cell., R = 5W, C2, , 21. A network of four capacitors each of 12 mF capacitance is, connected to a 500 V supply as shown in the figure. Determine (a), equivalent capacitance of the network and (b) charge on each, capacitor., , K, , C1, , C3, , C4, A, , 22., , 500 V, , B, , (i) Draw a neat labelled ray diagram of an astronomical telescope in normal adjustment. Explain, briefly its working., (ii) An astronomical telescope uses two lenses of powers 10 D and 1 D. What is its magnifying, power in normal adjustment?, , OR, (i) Draw a neat labelled ray diagram of a compound microscope. Explain briefly its working., (ii) Why must both the objective and the eye-piece of a compound microscope have short focal, lengths?, 23. In Young’s double slit experiment, the two slits 0 ×15 mm apart are illuminated by monochromatic, light of wavelength 450 nm. The screen is 1.0 m away from the slits., (a) Find the distance of the second (i) bright fringe, (ii) dark fringe from the central maximum., (b) How will the fringe pattern change if the screen is moved away from the slits?, 24. State Kirchhoff’s rules. Use these rules to write the expressions for the currents I 1 , I 2 and I 3 in, the circuit diagram shown., A, , C, , E, , I1, , E1 = 2V, , r1 = 4W, B, , I2, , E2 = 1V, , I3, , E3 = 4V, , r2 = 3W, D, r3 = 2W, F
Page 190 :
Examination Papers, , 25., , (a) Write symbolically the b - decay process of, , 189, , 32, P., 15, , (b) Derive an expression for the average life of a radionuclide. Give its relationship with the, half-life., 26. How does an unpolarised light get polarised when passed through polaroid?, Two polaroids are set in crossed positions. A third polaroid is placed between the two making an, angle q with the pass axis of the first polaroid. Write the expression of the intensity of light, transmitted from the second polaroid. In what orientations will the transmitted intensity be, (i) minimum and (ii) maximum?, 27. An illuminated object and a screen are placed 90 cm apart. Determine the focal length and nature, of the lens required to produce a clear image on the screen, twice the size of the object., 28., (a) With the help of a diagram, explain the principle and working of a moving coil, galvanometer., (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., (b) Use this expression to define S. I. unit of current., B, (c) A long straight wire AB carries a current I. A proton P travels, with a speed v, parallel to the wire, at a distance d from it in a, I, direction opposite to the current as shown in the figure. What is, P, Proton, d, the force experienced by the proton and what is its direction?, A, , 29. State Faraday’s law of electromagnetic induction., 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. The field, extends from x = 0 to x = b and is zero for x > b. Assume, that only the arm PQ possesses resistance r. 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,, obtain the expressions for the flux and the induced emf., Sketch the variations of these quantities with distance, 0 £ x £ 2b., , S, P, , Q, R, x=0, , x=b, , x = 2b, , OR, Draw a schematic diagram of a step-up transformer. Explain its working principle. Deduce the, expression for the secondary to primary voltage in terms of the number of turns in the two coils., In an ideal transformer, 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?
Page 191 :
190, , Xam idea Physics—XII, , 30., , (a) Draw the circuit diagrams of a p-n junction diode in (i) forward bias, (ii) reverse bias. 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., OR, (a) Draw the circuit arrangement for studying the input and output characteristics of an n-p-n, transistor in CE configuration. With the help of these characteristics define (i) input, resistance, (ii) current amplification factor., (b) Describe briefly with the help of a circuit diagram how an n-p-n transistor is used to produce, self-sustained oscillations., , CBSE (All India) SET–II, Questions uncommon to Set–I, 1. 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, and, (ii) the highest permitted energy level to the first permitted level., x, , 4. A beam of electrons projected along +X axis, experiences a force due, to a magnetic field along the +y-axis. What is the direction of the, magnetic field?, , e, z, y, , 5. Which of the following has the shortest wavelength:, Microwaves, Ultraviolet rays, X-rays, 12. 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. Explain in which loop do you expect the, induced emf to be constant during the passage out of the field region. The magnetic field is normal, to the loops., , v, , v, , B
Page 192 :
Examination Papers, , 19. A network of four capacitors each of 15 mF capacitance is connected, to a 500 V supply as shown in the figure. Determine (a) equivalent, capacitance of the network and (b) charge on each capacitor., , 191, , C2, , C1, , C3, , C4, 500 V, , 20. Write any two factors on which internal resistance of a cell depends., The reading on a high resistance voltmeter, when a cell is connected, across it, is 2 × 0 V. When the terminals of the cell are also connected to a, resistance of 3 W as shown in the circuit, the voltmeter reading drops to, 1× 5 V. Find the internal resistance of the cell., , +, , +, , R = 3W, , V, , –, , –, , K, , 22. In Young’s double slit experiment, the two slits 0 ×12 mm apart are illuminated by monochromatic, light of wavelength 420 nm. The screen is 1× 0 m away from this slits., (a) Find the distance of the second (i) bright fringe, (ii) dark fringe from the central maximum., (b) How will the fringe pattern change if the screen is moved away from ths slits?, 23. State Kirchhoff’s rules. Apply Kirchhoff’s rules to the loops ACBPA and ACBQA to write the, expressions for the currents I 1 , I 2 and I 3 in the network., E1 = 6V, P, I1, 0.5W, A, , B, 1W, I2, , I3, , Q, E2 = 10V, C, R = 12W, , 27. The image obtained with a convex lens is erect and its length is four times the length of the object., If the focal length of the lens is 20 cm, calculate the object and image distances., , CBSE (All India) SET–III, Questions uncommon to Set–I and Set–II, 5. Arrange the following in descending order of wavelength:, X-rays, Radio waves, Blue light, Infrared light, 8. The ground state energy of hydrogen atom is – 13 × 6 eV. What are the kinetic and potential, energies of electron in this state?
Page 193 :
192, , Xam idea Physics—XII, , 21. A convex lens is used to obtain a magnified image of an object on a screen 10 m from the lens. If, the magnification is 19, find the focal length of the lens., +, –, 24. Write any two factors on which internal resistance of a cell depends. The, V, reading on a high resistance voltmeter, when a cell is connected across it,, is 2 × 5 V. When the terminals of the cell are also connected to a resistance, +, –, of 5 W are shown in the circuit, the voltmeter reading drops to 2 × 0 V., Find the internal resistance of the cell., R = 5W, , K, , 25. In Young’s double slit experiment, the two slits 0 × 20 mm apart are illuminated by monochromatic, light of wavelength 600 nm. The screen is 1× 0 m away from the slits., (a) Find the distance of the second (i) bright fringe, (ii) dark fringe from the central maximum., (b) How will the fringe pattern change if the screen is moved away from the slits?, 26. State Kirchhoff’s rules. Apply these rules to the loops PRSP and PRQP to write the expressions, for the currents I 1 , I 2 and I 3 in the given circuit., 200 W, , S, , R, , 5V, 60 W, I2, , P, , 20W, (milliammeter), , I3, I1, , 4V, , Q, , Solutions, CBSE (AI) Set–I, 1. Electric potential. It is a scalar quantity., 2. By Fleming’s left hand rule magnetic field must be along negative Z-axis, 3. The self inductance is defined on the magnetic flux linked with the coil when unit current flows, through it., Or, The self inductance is defined as the emf induced in the coil, when the rate of change of current in, the coil is 1 ampere/second., The unit of self-inductance is henry (H)., 4. Let focal length of converging and diverging lenses be + f and - f respectively., 1, 1, Power of converging lens P1 = , Power of diverging lens P2 = f, f, 1 1, Power of combination P = P1 + P2 = - = 0, \, f f
Page 194 :
Examination Papers, , \, , Focal length of combination F =, , 193, , 1 1, = = ¥ (infinite), P 0, , F = ¥., 5. The minimum energy required to remove an electron from atom to infinitely for away is called the, ionisation energy. The ionisation energy for hydrogen atom is 13.6 eV., 6. In series current is same i X = iY ., For same diameter, cross-sectional area is same., \, n x eAv x = n y eAv y, ny, vx n y, 1, \, Þ, =, =, =, v y n x 2n y 2, , v x : v y =1 : 2, , 7. X-ray; used to study crystal structure, 8. No; when light travels from a rarer to denser medium, its frequency remains unchanged., According to quantities theory, the energy of a light beam depends on frequency and not on speed, 9. Consider an electron revolving around a nucleus ( N) in circular path of, L, radius r with speed v. 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), A = pr 2, Magnetic moment due to orbital motion,, ev, evr, M l = IA =, ( pr 2 ) =, 2pr, 2, 10., (a) Charge Q resides on outer surface of spherical conducting shell., Due to charge q placed at centre, charge induced on inner surface, is –q and on outer surface it is +q. So, total charge on inner, surface -q and on outer surface it is Q + q., q, Surface charge density on inner surface = 4pr12, Q+ q, Surface charge density on outer surface =, 4pr 22, , r2, , Q +q, , q -q, O r1, , (b) For external points, whole charge acts at centre, so electric field at distance x > r 2 ,, 1 Q+q, E( x ) =, ., 4pe 0 x 2, , P
Page 195 :
194, , Xam idea Physics—XII, , 11. Electric field is along x-axis and magnetic field is along y-axis, x, , B, , E, , z, B, y, , 12. Let a charge Q be given to a conductor, this charge under electrostatic equilibrium will redistribute, and the electric field inside the conductor is zero (i. e., Ein = 0)., Let us consider a point P at which electric field, strength is to be calculated, just outside the surface of, A, dS3, the conductor. 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 . Now consider a small cylindrical, C dS1, E, S2, box CD having one base C passing through P ; the, dS S1, other base D lying inside the conductor and the curved, surface being perpendicular to the surface of the, B, conductor., Let the area of each flat base be a. As the surface of the conductor is equipotential surface, the, electric field strength E at P, just outside the surface of the conductor is perpendicular to the, surface of the conductor in the neighbourhood of P., The flux of electric field through the curved surface of the box is zero, since there is no, component of electric field E normal to curved surface. Also the flux of electric field through the, base D is zero, as electric field strength inside the conductor is zero. Therefore the resultant flux of, electric field through the entire surface of the box is same as the flux through the face C. This may, be analytically seen as:, If S1 and S 2 are flat surfaces at C and D and S 3 is curved surface, then, Total electric flux, , òS, , ®, , ®, , E× d S = ò, , S1, , =ò, , S1, , ®, , ®, , E× d S 1 + ò, , S2, , E dS1 cos 0 +, , ®, , ®, , E× d S 2 + ò, , S3, , òS, , ®, 2, , ®, , ®, , ®, , E× d S 3, , 0× d S 2 +, , òS, , 3, , E dS 3 cos 90°, , = ò E dS1 = Ea, S, , As the charge enclosed by the cylinder is ( sa) coulomb, we have, using Gauss’s theorem,, 1, Total electric flux =, ´ charge enclosed, e0, 1, s, ...(i), or, Ea =, E=, Þ, ( sa), 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. This is known as Coulomb’s law. The, electric field strength is directed radially away from the conductor if s is positive and towards the, conductor if s is negative.
Page 196 :
Examination Papers, , 195, , If n$ is unit vector normal to surface in outward direction, then, ®, 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., df, d, 13., (i) Induced emf e = = - ( BA cos wt), dt, dt, = BAw sin wt, As B, A, w are same for both loops so induced emf is same for both loops., e, (ii) Current induced I =, R, As resistance R is less for copper loop, so current induced is larger in copper loop., h, h, 14. de Broglie wave length l =, =, 2mE, 2mqV, h, , For a-particle,, , la =, , For proton,, , lp =, , \, , mpqp, la, =, lp, ma qa, , But, \, , 2m a q a V, h, 2m p q p V, , ma, q, = 4, a = 2, mp, qp, la, 1 1, 1, =, × =, lp, 4 2 2 2, , 15. The modulation is needed due to, (i) Transmission of audio frequency electrical signals need long impracticable antenna., (ii) The power radiated at audio frequency is quite small, hence transmission is quite lossy., (iii) The various information signals transmitted at low frequency get mixed and hence can not be, distinguished., Modulation index,, E, ma = m, Ec, \, , Peak voltage of modulating signal,, 75, em = m a ´ E c =, ´ 12 = 9 V, 100, 16. Einstein's photoelectric equation is E k = hn - W for a single photon ejecting a single electron., (i) Explanation of frequency law: When frequency of incident photon (n), increases, the kinetic, energy of emitted electron increases. Intensity has no effect on kinetic energy of photoelectrons.
Page 197 :
196, , Xam idea Physics—XII, , (ii) Explanation of intensity law: When intensity of incident light increases, the number of, incident photons increases, as one photon ejects one electron; the increase in intensity will, increase the number of ejected electrons. In other words, photocurrent will increase with, increase of intensity. Frequency has no effect on photocurrent., (iii) Explanation of no time lag law: When the energy of incident photon is greater than work, function, the photoelectron is immediately ejected. Thus there is no time lag between, incidence of light and emission of photoelectrons., 17. Part AB represents repulsive force and Part BCD represents attractive force., , A, +100, , MeV, , Repulsive, B, , 0, Attractive, , –100, , C, , 2, , 1, , 3, , 4, , r (fm), , Conclusions:, (i) Nuclear forces are attractive and stronger, then electrostatic force., (ii) Nuclear forces are charge-independent., OR, The variation of binding energy per nucleon versus mass number is shown in figure., , Binding Energy per Nucleon (in MeV), , 9.0, O16, 8.0C12, F18, He4 N14, 7.0, , Fe56, U238, , 6.0, Li7, , 5.0, 4.0, 3.0, 2.0, , H2, , 1.0, 0.0, , 0, , 20, , 40, , 60, , 80 100 120 140 160 180 200 220 240, Mass Number
Page 198 :
Examination Papers, , 197, , Since nuclear forces are short-ranged, every nucleon interacts with their neighbours only., Therefore, binding energy per nucleon remains constant., 18., (i) P is ‘NAND’ gate and Q is ‘OR’ gate., A, C, P, (ii) C = A × B \ X = C + B = A × B + B, When A = 0, B = 0, C = A × B = 0 × 0 = 1, , B, , Q, , B, , \, X =1 + 0 =1, When A = 1, B = 1, C = A × B =1 ×1 = 1 = 0, \, X = 0 + 1 =1, 19. The mode of propagation used by short wave, broadcast services having frequency range from a few, MHz upto 30 MHz is sky wave propagation., The diagram is shown in fig., There is an upper limit of frequency because for, frequency higher than 30 MHz the radiowaves, penetrate through the ionosphere and escape., , 20. The internal resistance of a cell depends on, (i) distance (l) between electrodes., (ii) area ( A) of immersed part of electrode, and, (iii) nature and concentration of electrolyte., Given, E = 2 × 3 V, V = 1× 8 V, R = 5 W, æE ö, r = ç - 1÷ R, èV, ø, 10, æ 2×2 ö, \, r =ç, - 1÷ ´ 5 W = W = 1× 1 W, 9, è 1× 8, ø, 21., , Ionospheric, layers, , Sky wave communication, , (a) C1 , C 2 and C 3 are in series, 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 mF, C 4 is in parallel with C ¢, 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, , X
Page 199 :
198, , Xam idea Physics—XII, , 22., , (i) Working: Suppose AB is the point object whose end A is on the axis of telescope. The, objective lies ( L1 ) forms image A ¢B ¢ of object AB at the second principal focus F0 . The image, A ¢B ¢ is real, inverted and diminished. For normal adjustment position this image also lies at, first focus ( Fe ) of eye lens ( L 2 ). This image acts as an object for eye lens and final image is, formed at infinity. The final image A ¢¢B ¢¢ (say), is magnified and inverted., B, , fo, , A, , fe, , Fo, A' Fe', , C1, , C2, , Fe, , B', , in, At, , (ii) M = -, , i, fin, , ty, , f0, P, 10 D, = - 10, =- e =fe, P0, 1D, OR, , (i), Eyepiece, uo, , ue, , vo, , D, Objective, , Eye, , B, A", A, , Fo, Fe', , O, , A', , E, , Fe, , B', , B", , ve, , Working: Suppose a small object AB is placed slightly away from the first focus F0 ¢ of the, objective lens. The objective lens forms the real, inverted and magnified image A¢ B¢ , which, acts as an object for eyepiece. The eyepiece is so adjusted that the image A¢ B¢ lies between, the first focus Fe ¢ and the eyepiece E. The eyepiece forms its image A¢ ¢ B¢ ¢ which is virtual,, erect and magnified. 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., Magnifying power of compound microscope is
Page 202 :
Examination Papers, P1, I0, , I1 =, I0/2, , P3, q, , 201, , P2, I2, I3, , Let intensity of incident unpolarised light on first polaroid be I 0 ., I, Intensity of light transmitted through 1st polaroid P1 is I 1 = 0 ., 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 - 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, 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 |, , Screen, , F, , I = 20, v, , or, (numerically), …(ii), | v| =2|u|, From (i) and (ii), | u | = 30 cm, | v | = 60 cm, By sign convention u = - 30 cm, v = 60 cm, 1 1 1, gives, = f v u, 1+ 2, 1, 1, =, +, =, 60 30, 60, (convex lens), Þ, f = 20 cm, 28. (a) Priciple and working: When current ( I ) is passed in the coil, 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,, N is the number of turns in a coil.
Page 203 :
202, , Xam idea Physics—XII, H, , Suspension wire, M, , T1, T2, , Coil, , NIBl, b, , N, , S, , N, , S, , N, , S, , NIBl, Coiled strip, , (a), , (b), Magnetic lines of, force of radial magnetic field, , (c), , When the magnetic field is radial, as in the case of cylindrical pole pieces and soft iron core,, then in every position of coil the plane of the coil, is parallel to the magnetic field lines, so, that q = 90° and sin 90° = 1, Deflecting torque,, t = NIAB, If C is the torsional rigidity of the wire and q is the twist of suspension strip, then restoring, torque = C q, For equilibrium, deflecting torque = restoring torque, i.e., NIAB = C q, NAB, ...(i), q=, I, \, C, i.e., qµI, deflection of coil is directly proportional to current flowing in the coil and hence we can, construct a linear scale., (b) Importance (or function) of uniform radial magnetic field: In radial magnetic field, sin q =1, so torque is t = NIAB. This makes the deflection ( q) proportional to current. In other, words, the radial magnetic field makes the scale linear., 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., (c) A voltmeter is used to measure p.d. across a resistance in an electrical circuit. It is connected, in parallel across the resistance. If a voltmeter has very high resistance, it will not affect the, resistance of circuit, hence reading will be true. That is why while using a moving coil, galvanometer on a voltmeter, a high resistance in series is required., An ammeter is used to measure current in circuit, hence it is connected in series with the, circuit. If an ammeter has very low resistance it will not affect the circuit - resistance and so, reading will be true. That is why while using a moving coil galvanometer as an ammeter, a, shunt (small resistance in parallel) is used.
Page 204 :
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. It has been observed, experimentally that when the currents in the wire are in the same direction, they experience, an attractive force (fig. a) and when they carry currents in opposite directions, they experience a, repulsive force (fig. b)., Let the conductors PQ and RS carry currents I 1 and I 2 in same direction and placed at, separation r. (fig.)., Consider a current–element ‘ab’ of length DL of wire RS. The magnetic field produced by, current-carrying conductor PQ at the location of other wire RS, m I, ...(i), B1 = 0 1, 2p r, According to Maxwell’s right, hand rule or right hand palm rule, no. 1, the direction of B1 will be, perpendicular to the plane of, paper and directed downward., Due to this magnetic field, each, element, of, other, wire, experiences a force. The, direction of current element is, perpendicular to the magnetic, field; therefore the magnetic, force on element ab of length DL, DF = B1 I 2 DL sin 90° =, , P, , R, , P, , R, , b, , I2, DL, , DF, I1, Q, , a, , B, , B, I2, , I1, , S, , Q, , a, DL, b, , DF, , S, , r, , 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, , ...(ii), , (b) Definition S.I. unit of Current (Ampere): In S.I. system of fundamental unit of current, ‘ampere' has been defined assuming the force between the two current carrying wires as, standard., 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, r =1 m, then, m, f = 0 = 2 ´ 10 - 7 N/m, 2p
Page 205 :
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., (c) Magnetic field due to current carrying wire is perpendicular to plane of paper – downward., ®, m I, i.e.,, B = - 0 k$, 2pd, ®, , ®, , ®, , Force F = q v ´ B, , æ m I ö m evI $, = e ( -v$j) ´ ç - 0 k$ ÷ = 0, i, 2pd, è 2pd ø, m evI, That is the magnetic force has magnitude 0, and is directed along positive x-axis ie., in, 2pd, ®, , the plane of paper perpendicular to direction of v and to the right., 29. Faraday’s law of electromagnetic induction states that whenever there is a change in magnetic flux, linked with of a coil, an emf is induced in the coil. The induced emf is proportional to the rate of, change of magnetic flux linked with the coil., Df, i.e., e µ, Dt, Df, If the coil contains N-turns, then e = - N, Dt, Let length of conductor PQ = l, As x = 0, magnetic flux f = 0., When PQ moves a small distance from x to x + dx , then magnetic flux linked = BdA = Bldx, The magnetic field is from x = 0 to x = b, so final magnetic flux = SBldx = Bl S dx, (increasing), = Blb, 1, Mean magnetic flux from x = 0 to x = b is Blb., 2, The magnetic flux from x = b to x = 2b is zero., , S, P, , Q, R, x=0, , Induced emf, e = -, , df, d, = - ( Bldx ), dt, dt, , dx, x=b, , x = 2b
Page 206 :
Examination Papers, , = - Bl, , 205, , dx, = - Blv, dt, , dx, = velocity of arm PQ from x = 0 to x = b., dt, During return from x = 2b to x = b, the induced emf is zero; but now area is decreasing so, magnetic flux is decreasing, and induced emf will be in opposite direction., e = Blv, Graph is shown in figure., , where v =, , From x = b to x = 0, f, x=0, , e, x=b, , x = 2b, , 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, the, current changes continuously in the 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., Let N p be the number of turns in primary coil,, NS the number of turns in secondary coil and f, the magnetic flux linked with each turn. 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. 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.C. mains), Primary, , Primary, laminated, iron core, , Core, Secondary, , Step up, Secondary, , Transformer, , ...(i), , ...(ii), , ...(iii), , If the resistance of primary coil is negligible, the emf ( e p ) induced in the primary coil, will be, equal to the applied potential difference (V p ) across its ends. Similarly if the secondary circuit is, open, then the potential difference VS across its ends will be equal to the emf ( e S ) induced in it;, therefore
Page 207 :
206, , Xam idea Physics—XII, , VS e S N S, =, =, = r (say), Vp e p N p, where r =, , ...(iv), , NS, is called the transformation ratio. If i p and i s are the instantaneous currents in, Np, , primary and secondary coils and there is no loss of energy; then, For about 100% efficiency, 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, , \, , ...(v), , In step up transformer, N s > N p ® r >1;, VS > V p and i S < i p, , 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., Step up transformer is used at power house to transmit power at high voltage 11000 V or 33000 V. The, current in wires at this voltage is quite small, so power loss I 2 R is negligible. At town, the step down, transformer is used to supply power at 220 V. This saves enormous electrical energy., 30., (a) (i) Forward Bias:, Ei, , E, p, , R, , p, +, , n, , n, –, , +, , K, , –, , Forward current, , (ii) Reverse Bias:, Ei, , E, p, , R, , p, –, , n, , n, +, , Reverse current, , K, , –, +, Reverse biasing
Page 208 :
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, , 60 40, , 20, O, , Vbr, , 0.2 0.4 0.6 0.8, 10, , V (volt), , 20, 30, , 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.
Page 209 :
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, , – IB, , +, , –, , mA, VBB, , C, , +, , +, Rh1, , –, , +, , +, , B, , VBE, –, , +, , IC, VCE, E, IE, , – VCC, , –, , =, 10, , V, , 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 = çç, ÷, è DI B øVCE = constant, CE, , IB, , 208, , 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 = çç, ÷, è DI E øVC = constant, 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 = çç, ÷, è DI B øV = constant, C, , 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)
Page 210 :
Examination Papers, , (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), , 209, , 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, ÷÷ = Rhc, 1. E I = Rhc çç, 2, 22 ø 4, è1, æ 1, 1 ö, ÷÷ = Rhc, E II = Rhc çç, 2, ¥2 ø, è1, E, 3, Ratio I =, E II 4, ®, , ®, , ®, , ®, , 4. Fm = q v ´ B Þ Fm $j = - evi$ ´ B, , For validity of this equation, the direction of magnetic field must be along z-axis, since, (i$ ´ k$) = - $j, 5. X-rays has shortest wavelength.
Page 211 :
210, , Xam idea Physics—XII, x, , x, , x, , x, , x, , x, , x, , 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, , x, x, x, x, x, , x, x, x, x, x, (b), , 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 = 1W ú, ø, è 1× 5, ø, ë èV, û, , C1, , C3, , C4, 500 V, , 22. Given d = 0 ×12 mm = 0 ×12 ´ 10 -3 m, l = 420 nm = 420 ´ 10 -9 m, D = 1× 0 m, æ nDl ö 2 ´ 1× 0 ´ 420 ´ 10, y2 = ç, ÷=, è d ø, 0 ×12 ´ 10 -3, , -9, , = 7 ´ 10 -3 m = 7 mm, , hö l æ, 1 ö 1× 0 ´ 420 ´ 10 -9, æ, y¢2 = ç n - ÷ = 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, , 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, , v, , C2, , Þ, , (a) (i), , x, x, x, x, x, , P, I1, 0.5W, A, , B, 1W, I2, , I3, , Q, E2 = 10V, C, R = 12W
Page 212 :
Examination Papers, , 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., , 27. Given, , I, O, , F, , F', u, v, , 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, , Potential energy, U = Total energy, , 1 e2, 4pe 0 r, , E =K +U =-, , …(i), …(ii), , 1 e2, 4pe 0 2r, , …(iii), , 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, , 211
Page 214 :
CBSE EXAMINATION PAPERS, FOREIGN–2010, Time allowed: 3 hours, , Maximum marks: 70, , 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, , 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?
Page 215 :
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.
Page 216 :
Examination Papers, , 19., , 20., , 21., , 22., , 23., , 215, , (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., (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, in the two cases,, becomes the same., 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. It is observed that the first, minimum is at a distance of 2.5 mm from the centre of the screen. Find, (a) the width of the slit., (b) the distance of the second maximum from the centre of the screen., (c) the width of the central maximum., OR, A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference, fringes in a Young’s double slit experiment. 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., Draw a schematic arrangement of the Geiger – Marsden experiment for studying a-particle, scattering by a thin foil of gold. Dsecribe briefly, by drawing trajectories of the scattered, a-particles, how this study can be used to estimate the size of the nucleus., (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. What is its focal length in water? (Refractive, index of air-water = 1× 33, refractive index of air-glass = 1× 5)., A network of resistors is connected to a 16 V battery of internal resistance of 1 W as shown in the, figure., 4W, A, , 12 W, B, , 1W, , C, , 4W, , 16 V, , D, , 6W, , 1W, , (a) Compute the equivalent resistance of the network., (b) Obtain the voltage drops VAB and VCD ., 24., (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, where d is the separation between the plates. Find the, 2, expression for the capacitance when the slab is inserted between the plates., 25. Draw a schematic diagram of a reflecting telescope (Cassegrain). Write two important advantages, that the reflecting telescope has over a refracting type.
Page 217 :
216, , Xam idea Physics—XII, , 26. Define the terms ‘threshold frequency’ and ‘stopping potential’ in the study of photoelectric, emission., Explain briefly the reasons why wave theory of light is not able to explain the observed features in, photoelectric effect., 27., (a) State briefly any two reasons explaining the need for modulating a signal., (b) Draw a labelled block diagram of a simple modulator for obtaining an AM signal., 28., (a) State Ampere’s circuital law., (b) Use it to derive an expression for magnetic field insdie, along the axis of an air cored, solenoid., (c) Sketch the magnetic field lines for a finite solenoid. How are these field lines different from, the electric field lines from an electric dipole?, OR, (a) Using Biot-Savart Law, deduce an expression for the magnetic field on the axis of a circular, current loop., (b) Draw the magnetic field lines due to a current carrying loop., (c) A straight wire carrying a current of 12 A is bent into a, semi-circular arc of radius 2 × 0 cm as shown. What is the, ®, , 29., , 30., , O, , magnetic field B at O due to (i) straight segments (ii) the, semi-circular arc?, 5, 50, (a) A resistor of 400 W, 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., Find the voltage (rms) across the resistor, the inductor and the capacitor. Is the algebraic sum, of these voltages more than the source voltage? If yes, resolve the paradox. (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’. On closing the switch, show that the sum of the energies in the capacitor, and inductor is constant in time in the free oscillations of the LC circuits., 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. It supplies a load with 20 A. The ratio of, the primary winding to the secondary is 10: 1. If the transformer is 90% efficient, calculate:, (i) the power output,, (ii) the voltage, and, (iii) the current in the secondary., (a) Draw I-V characteristics of a Zener diode., (b) Explain with the help of a circuit diagram, the use of a Zener diode as a voltage-regulator., (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. Explain giving reason., OR, (a) Draw the circuit diagram of a base-biased n-p-n transistor in C-E configuration. Explain how, this circuit is used to obtain the transfer characteristic (Vo - Vi characteristics). How do we, explain the working of a transistor as a switch using the characteristic?
Page 218 :
Examination Papers, , 217, , Collector current (IC) in mA, , (b) The typical output characteristics ( I C - VCE ) of an n-p-n transistor in C-E configuration is, shown in the figure. Calculate (i) the output resistance r 0 and (ii) the current amplification, factor b ac ., Base current (IB), , 10, 8.5, , 60 mA, , 8, , 50 mA, 6, , 40 mA, , 4, , 30 mA, 20 mA, , 2, , 10 mA, 0, , 2, , 4, 6, 8, 10, 12, 14 16, Collector to emitter voltage (VCE) in volts, , CBSE (Foreign) SET–II, Questions uncommon to Set–I, 4. A circular loop is moved through the region of uniform magnetic, field. Find the direction of induced current (clockwise or, anticlockwise) when the loop moves:, (i) into the field, and, (ii) out of the field., B, 8. Name the electromagnetic radiation used to destroy cancer cells and, write its frequency range., 10. How are infra-red rays produced? Write their two important uses., 11. State the principle on which the working of a meter bridge is based. Under what condition is the, error in determining the unknown resistance minimized?, 12. 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. Show p, the trajectory followed by the two particles in the magnetic field. Find, the ratio of the radii of the circular paths which the particles may e, describe., B, 13. Wrtie the truth table for the following circuit. Name the gate that this, circuit represents., A, Y, B
Page 219 :
218, , Xam idea Physics—XII, , 18. 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., E1, E2, 21. Two cells E1 and E 2 of EMF’s 5 V and 9 V and internal resistances, 0.3 W, 1.2 W, of 0.3 W and 1.2 W respectively are connected to a network of, 5V 9V, resistances as shown in the figure. Calculate the value of current, flowing through the 3 W resistance., 6W, 4.5 W, 3W, , 22., , 23., , (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., (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, in the two cases, becomes the same., (a) How is the focal length of a spherical mirror affected when it is immersed in water., (b) A convex lens has 10 cm focal length in air. What is its focal length in water? (Refractive, index of air-water = 1× 33, refractive index of air-glass = 1× 5)., , CBSE (Foreign) SET–III, Questions uncommon to Set–I and Set–II, 5. To which part of the electromagnetic spectrum do the waves emitted by radioactive nuclei belong?, What is its frequency range?, 6. A rectangular loop of wire is pulled to the right, away from the long straight, wire through which a steady current I flows upwards. What is the direction, of induced current in the loop?, I, , 9. Draw the output wavefrom for the following gate. Also, name the gate., 1, A 0, , t1 t2, , t3, , t4, , t5, A, , 1, B 0, , Y, , B, t1, , t2, , t3, , t4, , t5, , 13. How are microwaves produced? Write their two important uses., 15. Draw a ray diagram to show the formation of the image in a far-sighted (hypermetropic) eye., Show with the help of a ray diagarm how this defect is corrected., 19., (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.
Page 220 :
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, in the two cases, becomes the same., 20. Calculate the steady current through the 2 W resistor in the circuit shown below., 2W, A, , B, 3W, 2 mF, , 6V, , 4W, , 2.8 W, , Solutions, CBSE (Foreign) SET–I, 1., , Q, mV - m, 6e 0, , 2. Diamagnetic substances: These are the substances in which feeble magnetism is produced in a, direction opposite to the applied magnetic field. These substances are repelled by a strong magnet., ®, , These substances have small negative values of magnetism M and susceptibility c and positive, low value of relative permeability m r , i.e.,, 1 £ c £ 0, 0 < m r < 1, The examples of diamagnetic substances are bismuth, antimony, copper, lead, water, nitrogen (at, STP) and sodium chloride., 3. Short radiowaves l <10 m or v > 30 MHz are used in satellite communication., 4. Path (d) is followed by electric field line., Reason: There are no electric field lines within a metallic sphere and field lines are normal at each, point of the surface., 5. The light is scattered by air molecules. According to Lord Rayleigh the intensity of scattered light, 1, 1, I µ, Þ Iµ, 4, (wavelength), l4, As l blue < l red , accordingly blue colour is scattered the most and red the least, so sky appears, blue., 6. Davission-Germer experiment shows wave nature of electrons., The phenomenon of diffraction of electron beam was observed in this experiment to produce, magnetic field upward., 7. The induced current always opposes the change in magnetic flux. Loop abc is entering the, magnetic field; so magnetic flux linked with loop tends to increase, so current induced in loop abc, is anticlockwise to produce magnetic field upward to oppose the increase in flux. Loop defg is
Page 221 :
220, , Xam idea Physics—XII, , leaving the magnetic field; so flux linked with it tends to decrease, the induced current will be, clockwise to produce magnetic field downward to oppose the decrease in magnetic flux., 8. Quantum Condition: The stationary orbits are those in which angular momentum of electron is an, h, integral multiple of, i.e.,, 2p, h, …(ii), mvr = n, ,, n =1, 2, 3, K, 2p, Integer n is called the principal quantum number. This equation is called Bohr’s quantum, condition., 9. Sky wave propagation., Above a frequency of few MHz, the electromagnetic wave penetrate and escape., 10. Y = A + B = A + B, The equivalent gate is OR gate., The Truth Table., Input, , Output, , A, , B, , Y, , 0, , 0, , 0, , 1, , 0, , 1, , 0, , 1, , 1, , 1, , 1, , 1, , 11. 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. It is, given by, ®, vd, , ®, , eE, =t, m, where m = mass of electron, e = charge of electron, E = electric field applied, Becomes 4 times., OR, Mobility of an ion is defined as the drift velocity per unit electric field i.e.,, v, et, m= d =, E, m, 2, Its unit is m /Vs., When temperature increases, covalent bonds of neighbouring atoms break and charge carrier, become free to cause conductive, so resistivity of semi-conductor decreases with rise of, temperature., 12. In fact the number of protons and number of neutrons are same before and after a nuclear reaction,, but the binding energies of nuclei present before and after a nuclear reaction are different. This, difference is called the mass defect. This mass defect appears as energy of reaction. In this sense a, nuclear reaction is an example of mass-energy interconversion.
Page 222 :
Examination Papers, , 221, , 13. Myopia or shortsightedness: Myopia is the defect of eye in which a person can see only nearby, objects, but fails to see the far away objects distinctly. This defect is due to, Corrective, lens, , Eye lens, Parallel rays, from object, , F, , Retina, I, , I, Eye, lens, (a) Image formation, by myopic eye, , (b) Corrected myopia, , (a) decrease in focal length of the eye lens., (b) Spreading of the eye-sphere., Due to these reasons the image is formed in front of the retina., 14. Radius of charged particle in magnetic field, mv, r=, qB, m, µ for same v and B., 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. It is a device to measure the potential, K, –, difference across a circuit element, +, Rh, Driver battery, accurately. 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. For the working of potentiometer, Secondary, E, circuit, emf of battery B1 > emf of battery B 2 ., G, When a steady current is passed through a, + –, E, potentiometer wire AB, there is a fall of, potential along the wire from A to B. The fall of potential per unit length along potentiometer wire is, called the potential gradient. If L is length of wire AB and V is the potential difference across it then, V, Potential gradient k =, L, The S.I. unit of potential gradient is volt/metre., 16. Unit of resolution of microscope, 1× 22l, dq =, 2n sin q, 1, Resolving power µ, unit of resolution, n sin q, µ, l
Page 223 :
222, , Xam idea Physics—XII, , (i) When wavelength l decreases, resolving power increases., (ii) When diameter of obejctive lens decreases, q decreases; so resolving power decreases., r1, 17. 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. 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, the, magnetic field produced within this solenoid., m N I, ...(ii), B1 = 0 1 1, l, The flux linked with each turn of inner solenoid coil is f2 = B1 A 2 , where A 2 is the cross-sectional, area of inner solenoid. The total flux linkage with inner coil of N 2 -turns., 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, 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, then, M = m 0 n1 N 2 pr 22 ., 18. Production of X-rays: When high energetic electrons strike a metallic target of high atomic, weight and high melting point, X-rays are produced. In production of X-rays mechanical energy of, electrons is converted with electromagneitic energy of X-rays., Uses: X-rays are used in medical diagnostics to detect fractures in bones, tuberculosis of lungs,, presence of stone in gallbladder and kidney. They are used in engineering to check flaws in, bridges. In physics X-rays are used to study crystal structure., 19., (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., E, (b) Let C be capacitance of each capacitor., E V, or, C, In series arrangement net capacitance C s = ., V, V, 2, E, R, In parallel arrangement, net capacitance, C p = 2C, 1, Energy stored U = CV 2, 2, If VS and VP are potential differences applied across series and parallel arrangements, then, given, Us =U p, 1, 1, Þ, C S VS2 = C p V p2, 2, 2, Vp, Cs, C/2 1, =, =, =, Þ, Vs, Cp, 2C, 2
Page 224 :
Examination Papers, , 20., , 223, , (a) Given l = 500 nm = 5 ´ 10 -7 m, D =1 m, If a is width of slit, then for first minimum, l, sin q1 =, a, y, For small q1 , 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, y n = ç n + ÷, 2ø a, è, For second maximum, n = 2, 1 ö 1 ´ 5 ´ 10 -7, æ, \, = 5 ´ 10 -4 m = 0 × 5 mm, ( y 2 ) max = ç 2 + ÷, 2 ø 2 × 5 ´ 10 -3, è, æ 2Dl ö, (c) Width of central maximuim, 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, there must be a difference of 1 in order of l1 and l 2 ., As, l1 > l 2 ,, n1 < n 2, If, n1 = n, n 2 = n + 1, nDl1 ( n + 1) Dl 2, =, Þ, \, ( y n ) l1 = ( y n + 1 ) l2, d, d, l2, 520 nm, 520, or n =, =, =4, Þ, nl1 = ( n + 1) l 2, Þ n=, l1 - l 2 ( 650 - 520) nm, 130, \ Least distance,, , y min =, , nDl1 4 ´ 1 ´ 650 ´ 10 - 9, =, d, 1 ´ 10 - 3, , = 2 × 6 ´ 10 - 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
Page 225 :
224, , Xam idea Physics—XII, , 21. At the suggestion of Rutherford, in 1911, H. Geiger, and E. Marsden performed an important, experiment called Geiger-Marsden experiment (or Rutherford’s scattering experiment). It consists, of, 1. Source of a-particles : The radioactive source polonium emits high energetic alpha ( a –), particles. Therefore, polonium is used as a source of a-particles. This source is placed in an, enclosure containing a hole and a few slits A1 , A 2 , K etc. are placed in front of the hole. This, arrangement provides a fine beam of a-particles., 2. Thin gold foil : It is a gold foil* of thickness nearly 10 - 6 m, a-particles are scattered by this, foil. The foil taken is thin to avoid multiple scattering of a-particles, i. e., to ensure that, a-particle be deflected by a single collision with a gold atom., 3. Scintillation counter : By this the number of a-particles scattered in a given direction may, be counted. The entire apparatus is placed in a vacuum chamber to prevent any energy loss of, a-particles due to their collisions with air molecules., Method: When a-particle beam falls on gold foil, the a-particles are scattered due to collision, with gold atoms. This scattering takes place in all possible directions. The number of a-particles, scattered in any direction is counted by scintillation counter., Observations and Conclusions, (i) Most of a-particles pass through the gold foil undeflected. This implies that “most part of the, atom is hollow.”, (ii) a-particles are scattered through all angles. Some a-particles (nearly 1 in 2000), suffer, scattering through angles more than 90°, while a still smaller number (nearly 1 in 8000), retrace their path. This implies that when fast moving positively charged a-particles come, near gold-atom, then a few of them experience such a strong repulsive force that they turn, back. On this basis Rutherford concluded that whole of positive charge of atom is, concentrated in a small central core, called the nucleus., ZnS, Screen, , Incident, beam of, a-particles, Nucleus, Detector, , The distance of closest approach of a-particle gives the estimate of nuclear size. 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
Page 226 :
Examination Papers, , Ek =, r0 =, , 225, , 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 - 14 m, while size of atom is of, 10 - 14, 1, the order of 10 - 10 m; therefore the size of nucleus is about, times the size of, =, - 10 10, 000, 10, atom., (iii) The negative charges (electrons) do not influence the, scattering process. This implies that nearly whole mass, f, of atom is concentrated in nucleus., , 22., , (a) Wavelength of light has no effect on focal length of a spherical mirror., 4, (b) Given, fa = 20 cm, ng = 1.5, n l =, 3, ng -1, 1× 5 - 1, ´ 20 cm » 80 cm, fl =, ´ fa =, ng, ö, æ 1× 5, - 1÷, ç, -1, è 1× 33 ø, nl, , 23., , (a) R AB =, R BC, , 4´ 4, = 2 W,, 4+4, 12 ´ 6, =1W, RCD =, =4W, 12 + 6, , Equivalent resistance of network, R AD = R AB + R BC + RCD, = 2 + 1 + 4 = 7 W,, E, 16, (b) Current in circuit I =, =, =2A, R + r 7 +1, VAB = R AB I = 2 ´ 2 = 4 V, VCD = RCD . I = 4 ´ 2 = 8 V, 24., , (a) Initial electric field between the plates, E 0 =, , q, s q/ A, =, =, e0, e0, A e0, , After introduction of dielectric; the permittivity of medium becomes Ke 0 ;, E, q, so final electric field between the plates, E =, = 0, AK e 0, K, 1, i.e., electric field reduces to times., K
Page 227 :
226, , Xam idea Physics—XII, , (b) Consider a parallel plate capacitor, area of each plate being A,, the separation between the plates being d. Let a dielectric slab of, dielectric constant K and thickness t < d be placed between the, plates. The thickness of air between the plates is ( d - t ). If, charges on plates are + Q and - Q, 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, 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)., Q, Q, (d - t ) +, t, = E1 ( d - t ) + E 2 t =, e0 A, K e0 A, \ VAB =, , Q é, t ù, d -t + ú, e 0 A êë, K û, , \ Capacitance of capacitor, C =, , or,, , C=, , Here, t =, \, , e0 A, t, d -t +, K, , =, , Q, Q, =, Q, t ö, VAB, æ, ç d -t +, ÷, e0 A è, K ø, , e0 A, 1 ö, æ, d -t ç 1 - ÷, K ø, è, , d, 2, , C=, , e0 A, e0 A, =, dæ, 1ö dæ, 1ö, d - ç1 - ÷, ç1 + ÷, Kø, 2è Kø 2è, , 25., , Objective, mirror, , Secondary, mirror, Eyepiece, , Advantages : (i) It is free from chromatic aberration., (ii) Its resolving power is greater than refracting telescope due to larger aperture of mirror.
Page 228 :
Examination Papers, , 227, , 26. Threshold Frequency : The minimum frequency of incident light which is just capable of, ejecting electrons from a metal is called the threshold frequency. It is denoted by v 0 ., 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 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., (i) According to wave theory, the light propagates in the form of wavefronts and the energy is, distributed uniformly over the wavefronts. With increase of intensity of light, the amplitude, of waves and the energy stored by waves will increase. These waves will then, provide more, energy to electrons of metal; consequently the energy of electrons will increase., Thus, according to wave theory, the kinetic energy of photoelectrons must depend on the, intensity of incident light; but according to experimental observations, the kinetic energy, of photoelectrons does not depend on the intensity of incident light., (ii) According to wave theory, 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, but according to experimental observations, the light of frequency less than, threshold frequency can not emit electrons; whatever the intensity of incident light may be., (iii) According to wave theory, the energy transferred by light waves will not go to a particular, electron, but it will be distributed uniformly to all electrons present in the illuminated surface., Therefore, electrons will take some time to collect the necessary energy for their emission., The time for emission will be more for light of less intensity and vice versa. But experimental, observations show that the emission of electrons take place instantaneously after the light is, incident on the metal; whatever the intensity of light may be., 27., (a) The modulation is needed due to, (i) Transmission of audiofrequency electrical signals need long impracticable antenna., (ii) The power radiated at audio frequency is quite small, hence transmission is quite lossy., (iii) The various information signals transmitted at low frequency get mixed and hence can, not be distinguished., (b) Amplitude Modulation:, The block diagram is shown in fig., x(t), , m(t), +, , Am sin wmt, c(t), (Modulating signal ), AC sin wct, (carrier wave), , 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, 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 , 2wm , 2wc and allows wave of frequency wc , wc - wm and wc + wm . The output of, Bandpass filter is AM wave.
Page 229 :
228, , Xam idea Physics—XII, , 28., , (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.e.,, ® ®, , ò B . 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, carrying current. 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. When an, d c s, r, electric current is passed through the solenoid,, a magnetic field is produced within the, solenoid. If the solenoid is long and the, B, p, q, successive insulated copper turns have no, l, gaps, then the magnetic field within the, B, solenoid is uniform; with practically no, magnetic field outside it. The reason is that the, solenoid may be supposed to be formed of a, large number of circular current elements. 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., 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, 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., Let I be the current flowing in the solenoid, then by right hand rule, the magnetic field is, parallel to the axis of the solenoid., Field outside the solenoid: Consider a closed path abcd. 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., ®, , Field Inside the solenoid: Consider a closed path pqrs. The line integral of magnetic field B, along path pqrs is
Page 230 :
Examination Papers, , òpqrs, , ®, , ®, , B · dl = ò, , ®, , ®, , ®, , B · dl + ò, , pq, , qr, , ®, , B · dl + ò, , rs, , ®, , ®, , B • dl + ò, , sp, , ®, , ®, , B · dl, , 229, , ...(i), , ®, , ®, , For path pq, B and dl are along the same direction,, ®, , òpq, , \, , ®, , B · dl = ò B dl = Bl, ®, , ( pq = l say), , ®, , For paths qr and sp, B and d l are mutually perpendicular., ®, , òqr, , \, , ®, , ®, , B · dl = ò, , ®, , B · d l = ò B dl cos 90° = 0, , sp, , For path rs, B = 0 (since field is zero outside a solenoid), , òrs, , \, , ®, , ®, , B · dl = 0, , In view of these, equation (i) gives, , òpqrs, , ®, , ®, , B · dl = ò, ®, , ®, pq, , ®, , B · dl = Bl, , ...(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., (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., OR, (a) Magnetic field at the axis of a circular loop:, Consider a circular loop of radius R carrying current I,, 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. Consider a small, element of length dl of the coil at point A. The, ®, , magnitude of the magnetic induction dB at point P due, to this element is given by
Page 231 :
230, , Xam idea Physics—XII, , dB =, , m 0 I dl sin a, 4p, r2, , ...(i), , ®, , ®, , ®, , The direction of dB is perpendicular to the plane containing dl and r and is given by right, ®, , ®, , hand screw rule. As the angle between I dl and r is 90°, the magnitude of the magnetic, ®, , induction dB is given by,, m I dl sin 90° m 0 I dl, dB = 0, =, ×, 4p, r2, 4p r 2, ®, , ...(ii), ®, , ®, , ®, , In figure dB has been represented by P Q. The vector dB or ( P Q) can be resolved into two, components, namely PM and PN along and perpendicular to the axis respectively. 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; PM and PN ¢ along the axis and perpendicular to the axis, respectively. Thus if we consider the magnetic induction produced by the whole of the, circular coil, then by symmetry the components of magnetic induction perpendicular to the, axis will be cancelled out, while those parallel to the axis will be added up. 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,, 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, , ...(iii), , ...(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, , Therefore, B =, =, , m 0 IR, 4p ( R 2 + x 2 ) 3 / 2, m 0 IR 2, 2 (R 2 + x 2 ) 3/ 2, , If the coil contains N turns, then, , ( 2 p R), , tesla., , ...(v)
Page 232 :
Examination Papers, , B=, , m 0 NIR 2, 2 (R 2 + x 2 ) 3/ 2, , ...(vi), , tesla., , (b) Magnetic field lines due to a circular current loop:, , (c) Magnetic field due to a current carrying element., ®, , ®, , ®, , m I dl ´ r, dB= 0, 4p r 3, (i) For straight segments q = 0 or p, ®, , Þ, , ®, , dl ´ r = dl r sin 0 n$ = 0, , \, , B1 = 0, , (ii) For semicircular arc S dl = pr , q =, , p, 2, , p, ® ®, I S dl sin, m 0 SI d l ´ r m 0, 2 n$ = m 0 Ipr n$, =, =, \, 3, 2, 4p, 4p, 4p r 2, r, r, m I, = 0 ,, 4r, directed perpendicular to plane of paper downward., 5, 50, 50, 29. R = 400 W, L = H, C = mF = ´ 10 -6 F, p, p, p, E =140 sin 140pt, Comparing with V = V0 sin wt, 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, ®, B2, , 231
Page 233 :
232, , Xam idea Physics—XII, , Capacitive reactance, X C =, , 1, =, wC, , 1, = 200 W, æ 50 ö, -6, 100p ´ ç ÷ ´ 10, è pø, , Impedance Z = R 2 + ( X L - X C ) 2 = ( 400) 2 + (500 - 200) 2 = 500 W, Vrms 100, =, = 0×2 A, Z, 500, Voltage across resistor, VR = RI = 400 ´ 0 × 2 = 80 V, Voltage across inductor, VL = X L I = 500 ´ 0 × 2 = 100 V, Voltage across capacitor, VC = X C I = 200 ´ 0 × 2 = 40 V, Algebraic sum of voltages = 80 + 100 - 40 = 160 V, Yes algebraic sum of voltages (160 V) is more than the source voltage (100 V)., , rms current in circuit, I =, , This is because VR and (VL - VC ) are not in phase but (VL - VC ) leads VR by an angle, , p, ×, 2, , V = VR2 + (VL - VC ) 2 = (80) 2 + ( 60) 2 = 100 V, This resolves the paradox., 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 Varying, x x, magnetic, changes, so induced currents are induced in the body of, x x field, x x, conductor, which causes heating of conductor. The x x, x x, currents induced in the conductor are called the eddy x x x x x x x, currents. In varying magnetic field, the free electrons of, conductor experience Lorentz force and traverse closed paths; which are equivalent to small, current loops. These currents are the eddy currents; they cause heating effect and sometimes, the conductor becomes red-hot., Eddy current losses may be reduced by using laminated soft iron cores in galvanometers,, transformers, etc. and making holes in conductor., (b) Given V p = 2 × 5 kV = 2 × 5 ´ 10 -3 V, I p = 20 A, Input power, Pin = V p I p = 2 × 5 ´ 10 3 ´ 20 = 50 ´ 10 3 W = 50 kW, Turn ratio, r =, (i) h =, , Vs, 1, = , Efficiency h = 90% = 0 × 9, V p 10, , Pout, Pin, , Power output, 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, 250, Vs
Page 234 :
Examination Papers, , (a) I-V Characteristics of Zener diode:, I (mA), 100, 80, 60, 40, 20, 100 80, Vbr, , 60 40, , 20, O, , 0.2 0.4 0.6 0.8, 10, , V (volt), , 20, 30, , I (mA), , Regulated output, , (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, 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 Fig. The Zener, diode is connected across load such that it is reverse biased., The series resistance R absorbs the output voltage fluctuations so as to maintain constant, voltage across the load., 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. 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., Let I be the current drawn from supply, I Z the current through Zener diode and I L the current, through load. Then obviously, I = I Z + I L or I Z = I - I L, If R Z is Zener diode resistance, then, V0 = VZ = I Z × R Z = I L R L, Applying Kirchhoff’s law to the mesh containing resistance R, Zener diode and supply, voltage Vin , we have, Unregulated input, , 30., , 233
Page 235 :
Xam idea Physics—XII, , RI + VZ = Vin, i.e.,, …(i), VZ = Vin - RI ., When the input voltage Vin is lower than the Zener voltage VZ of diode, there is no current, conduction, i.e.,, I Z = 0., This implies, V0 = Vin ., As input voltage Vin is increased so that it becomes equal to VZ , the breakdown point is, reached and the voltage across the diode VZ = (Vin - RI ) becomes constant., 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., Thus in breakdown region, we have, …(ii), V0 = VZ = Vin - RI, Fig. represents the plot of output voltage V0 versus, Regulated output voltage, input voltage Vin . It is clear from the graph that the, VZ, output voltage remains constant when the diode is in, Zener region., It may be pointed out that for maintaining constant, regulated output, 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., (c) A photodiode is used in reverse bias, although in forward bias current in more, then current in, reverse bias because in reverse bias it is easier to observe change in current with change in, light intensity., OR, æ DVCE ö, ÷, (b) (i) Output resistance r 0 = çç, ÷, è DI C ø I = constant, Output voltage, V0, , 234, , B, , Taking I B = 60 mA = constant, For DVCE = (16 - 4) V= 12 V, Corresponding DI C = (8 × 5 - 8) mA = 0 × 5 mA, 12, r0 =, W = 24 ´ 10 3 W = 24 kW, \, -3, 0 × 5 ´ 10, (ii) Current amplification factor, æ DI ö, b ac = çç c ÷÷, è DI B øVCE = cosntant, Taking VCE = 8 V, DI C = (5 × 8 - 3 × 8) mA = 2 mA, DI B = ( 40 - 30) mA = 10 mA, 2 mA 2 ´ 10 -3, \, b ac =, =, = 200, 10 mA 10 ´ 10 -6
Page 236 :
Examination Papers, , 235, , CBSE (Foreign) SET–II, 4. (i) anticlockwise (ii) clockwise, 8. g-rays., Frequency range 1019 Hz—10 22 Hz., 10. Infrared rays produced by hot bodies and molecules. These waves are used for long distance, photography and for therapeutic purposes., 11. Working of meter bridge is based on Wheatstone bridge., The error may be minimised by taking balancing length near the middle of the bridge., ×, ×, ×, ×, ×, 12. Trajectories are shown in fig., ×, ×, ×, ×, ×, mv, As r =, ®r µ m, ×, ×, ×, ×, ×, p, qB, ×, ×, ×, ×, ×, e, , Ratio of radii of electron path and proton path., re m e, =, rp mp, , B, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , As mass of proton m p » 1840 ´ mass of electron ( m e ), me, 1, », \, m p 1840, re, 1, =, r p 1840, , \, , 13. The output Y = A + B = A × B = AB, That is equivalent gate is ‘AND’ gate., The symbol and truth table are shown in fig., Truth Table, A, , Y, , B, , A, , B, , Y, , 0, , 0, , 0, , 1, , 0, , 0, , 0, , 1, , 0, , 1, , 1, , 1, , 18. Myopia or shortsightedness: Myopia is the defect of eye in which a person can see only nearby, objects, but fails to see the far away objects distinctly. This defect is due to, Corrective, lens, , Eye lens, Parallel rays, from object, , F, , Retina, I, , I, Eye, lens, (a) Image formation, by myopic eye, , (b) Corrected myopia
Page 237 :
236, , Xam idea Physics—XII, , (a) decrease in focal length of the eye lens., (b) spreading of the eye-sphere., Due to these reasons the image is formed in front of the retina., Remedy: To eliminate this defect a concave lens of suitable focal length is used. 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. If a myopic eye has a far point at F, then the parallel rays from, infinity will be incident on concave lens and form its vertical image at F. This image will act as an, object for eye lens and the final image ( I ) will be formed at the retina [Fig (b)]. Clearly, 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., 21. Net emf of circuit E = E 2 - E1 = 9 - 5 = 4V, Net resistance of circuit,, R eq = 4 × 5 +, , 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, ÷÷ = 1 V, V = IR = 0 × 5 ´ çç, è6 +3ø, Curernt in circuit, I =, , \ Current in R1 = 3 W resistance, V 1, I1 =, = A, R1 3, 22., , (b) C1 = C, C 2 = 2C, In series combination of C1 and C 2 , equivalent capacitance, CC, C ´ 2C 2, CS = 1 2 =, = C, C1 + C 2 C + 2C 3, In parallel combination of C1 and C 2 , equivalent capacitance,, C p = C1 + C 2, = C + 2C = 3C, 1, Energy stored,, U series = C sVs2, 2, 1, U parallel = C p V p2, 2, Given, U series = U parallel, 1, 1, \, C sVs2 = C p V p2, 2, 2, Vp, C s æ 2C / 3 ö 2, Þ, =, =ç, ÷=, Vs, C p è 3C ø 9, , 23., , (a) No change. The focal length of a concave mirror does not depend on the nature of the, medium.
Page 238 :
Examination Papers, , 237, , (b) Given, fa = 10 cm, ng = 1.5, nl = 1.33, n g -1, 1.5 -1, fe =, ´ fa =, ´ 10 cm, ng, 1.5, 1, -1, 1.33, nl, 0.5 ´ 1.33 ´ 10, cm = 39.12 cm, =, 0.17, , CBSE (Foreign) SET–III, 5. g-rays, Frequency range 1019 Hz—10 22 Hz., 6. Direction of induced current in loop is clockwise., Reason: Induced current opposes the motion of loop away from wire; as similar currents attracts,, so in nearer side of loop the current will be upward i.e., in loop current in clockwise., 9. Output waveform is shown in fig., Y = AB, The gate is NAND gate., 14. The paths of proton ( p) and deuteron ( d) are shown in figure., mv, Radius, r =, qB, m pv, rp =, eB, md v, rd =, eB, rp mp 1, =, =, rd m d 2, , t1 t2, , p, d, , t3, , t4, , t5, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , 13. Microwaves are produced by special vacuum tubes, namely; klystrons, magnetrons and gunn, diodes. Their frequency range is 3 GHz to 300 Ghz., They are used in radar systems used in air craft navigation and microwave users in houses., 15. 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. This defect is due to, (a) Increase in focal length of eye lens., (b) Contraction of eye-sphere., Due to these reasons the image of a nearby object is formed behind the retina., Remedy: The near point of hypermetropic eye is displaced from D = 25 cm to some distant point., To eliminate this defect a convex lens of suitable focal length is used. 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., Suppose the near point of a normal eye is at N and that of a hypermetropic eye is at O. The, corrective convex lens forms the image of near point ( N) at point O, then this image will act as the
Page 239 :
238, , Xam idea Physics—XII, , object for eye lens and the final image ( I ) will be formed at the retina. Therefore the corrective, lens enables to form the distinct image of near point ( N) at retina., Retina, Normal near point, , Normal, near point, I, , N, , O, , Near point of, defective eye, , D = 25 cm, (a) Image formation, by hypermetropic eye, , 19., , I, , N, , Corrective, convex lens, (b) Corrected hypermetropia, , (b) Let C1 = C, C 2 = 3C, In series combination C s =, , C1C 2, 3 ´ 3C 3, =, = C, C1 + C 2 C + 3C 4, , In parallel combination, C p = C1 + C 2 = C + 3C = 4C, Given, Us =U p, 1, 1, Þ, C sVs2 = C p V p2, 2, 2, Vp, C s 3C / 4 3, Þ, =, =, =, Vs, Cp, 4C, 16, 20. In steady state there is no current in capacitor branch, so, equivalent circuit is shown in fig., Net resistance of circuit,, 2´3, R eq =, + 2 × 8 = 1× 2 + 2 × 8 = 4 W, 2+3, Net emf, E = 6 V, Current in circuit, I =, , E, 6, = = 1× 5 A, R eq 4, , 2W, A, , B, , I1, I, 3W, I, 6V, , Potential difference across parallel combination of 2 W and 3 W resistances., V ¢ = IR ¢ = 1× 5 ´ 1× 2 = 1× 8 V, Current in R1 = 2 W resistance, V ¢ 1× 8, I1 =, =, = 0×9 A, R1, 2, , 2.8 W
Page 240 :
CBSE EXAMINATION PAPERS, DELHI–2011, Time allowed: 3 hours, , Maximum marks: 70, , 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–, 4pe o, , m 0 = 4p ´ 10 -7 TmA -1, 2, , 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. A point charge Q is placed at point O as shown in the figure. Is the potential difference VA – VB, positive, negative, or zero, if Q is (i) positive (ii) negative?, Q, O, , A, , B, , 2. A plane electromagnetic wave travels in vacuum along z-direction. What can you say about the, direction of electric and magnetic field vectors ?, 3. A resistance R is connected across a cell of emf e and internal resistance r. A potentiometer now, measures the potential difference between the terminals of the cell as V. Write the expression for, ‘r’ in terms of e, V and R., 4. The permeability of a magnetic material is 0.9983. Name the type of magnetic materials it, represents., 5. Show graphically, the variation of the de-Broglie wavelength (l) with the potential (V) through, which an electron is accelerated from rest.
Page 241 :
240, , Xam idea Physics—XII, , 6. In a transistor, doping level in base is increased slightly. How will it affect (i) collector current and, (ii) base current?, 7. Define the term ‘wattless current’., 8. When monochromatic light travels from one medium to another, its wavelength changes but, frequency remains the same. Explain., 9. Two uniformly large parallel thin plates having charge densities +s and –s are kept in the X-Z plane, at a distance ‘d’ apart. Sketch an equipotential surface due to electric field between the plates. If a, particle of mass m and charge ‘–q’ remains stationary between the plates, what is the magnitude, and direction of this field?, OR, Two small identical electrical dipoles AB and CD, each of dipole moment ‘p’ are kept at an angle, of 120° as shown in the figure. Whatr is the resultant dipole moment of this combination? If this, system is subjected to electric field ( E) directed along + X direction, 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 magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip, down at 60° with the horizontal. The horizontal component of the earth’s magnetic field at the place, is known at to be 0.4 G. Determine the magnitude of the earth’s magnetic field at the place., 11. Figure shows two identical capacitors, C1 and C2, each of 1 mF capacitance connected to a battery, of 6 V. Initially switch ‘S’ is closed. 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. 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. Two convex lenses of same focal length but of aperture A1 and A2 (A2 < A1), are used as the, objective lenses in two astronomical telescopes having identical eyepieces. What is the ratio of, their resolving power? Which telescope will you prefer and why? Give reason.
Page 242 :
Examination Papers, , 241, , 13. Draw the output waveform at X, using the given inputs A and B for the logic circuit shown below., Also, identify the logic operation performed by this circuit., , A, , A, , Y, , X, , B, , B, , t1, , t2, , t3, , t4, , t5, , t6, , t7, , 14. Name the semiconductor device that can be used to regulate an unregulated dc power supply., With the help of I-V characteristics of this device, explain its working principle., 15. How are infrared waves produced? Why are these referred to as ‘heat waves’? Write their one, important use., 16. Draw the transfer characteristic curve of a base biased transistor in CE configuration. Explain clearly, how the active region of the Vo versus Vi curve in a transistor is used as an amplifier., 17., (i) Define modulation index., (ii) Why is the amplitude of modulating signal kept less than the amplitude of carrier wave?, 18. A current is induced in coil C1 due to the motion of current carrying coil C2. (a) Write any two, ways by which a large deflection can be obtained in the galvanometer G. (b) Suggest an, alternative device to demonstrate the induced current in place of a galvanometer., C2, , C1, , G, , K, , 19. Define the terms (i) drift velocity, (ii) relaxation time., A conductor of length L is connected to a dc source of emf e. If this conductor is replaced by, another conductor of same material and same area of cross-section but of length 3L, how will the, drift velocity change?, 20. 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. Draw a graph showing the variation of, electric field with r, for r > R and r < R., 21. An electron and a photon each have a wavelength 1.00 nm. Find, (i) their momenta,
Page 243 :
242, , Xam idea Physics—XII, , 22., , 23., , 24., , 25., , (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., 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. Deduce the expression for the fringe width., OR, Use Huygen’s principle to verify the laws of refraction., (a) Describe briefly, with the help of suitable diagram, how the transverse nature of light can be, demonstrated by the phenomenon of polarization., (b) When unpolarized light passes from air to a transparent medium, under what condition does the, reflected light get polarized?, The energy levels of a hypothetical atom are shown below. 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, , – 2 eV, B, , C, , – 4.5 eV, , D, , – 10 eV, , 26. State the law of radioactive decay., 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 . Depict in the plot the number of undecayed nuclei at (i), t = 3T1/ 2 and (ii) t = 5T1/ 2 ., 27. In the circuit shown, R1 = 4 W, R2 = R3 = 15 W, R4 = 30 W and E = 10 V. Calculate the equivalent, resistance of the circuit and the current in each resistor., R1, I1, , A, I2, R2, , E, , I1, , I4, R4, , B, , I3, R3
Page 244 :
Examination Papers, , 243, , 28. State Biot-Savart law, giving the mathematical expression for it., Use this law to derive the expression for the magnetic field due to a circular coil carrying current, at a point along its axis., How does a circular loop carrying current behave as a magnet?, OR, With the help of a labelled diagram, state the underlying principle of a cyclotron. Explain clearly, how it works to accelerate the charged particles., Show that cyclotron frequency is independent of energy of the particle. Is there an upper limit on, the energy acquired by the particle? Give reason., 29., (a) Draw a ray diagram to show refraction of a ray of monochromatic light passing through a glass, prism., Deduce the expression for the refractive index of glass in terms of angle of prism and angle of, minimum deviation., (b) Explain briefly how the phenomenon of total internal reflection is used in fibre optics., OR, (a) Obtain lens makers formula using the expression, n 2 n1 ( n 2 - 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), is incident on the convex side of spherical refracting surface, of radius of curvature R., (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. Using this diagram, derive the magnification formula for the, image formed., 30., (i) With the help of a labelled diagram, describe briefly the underlying principle and working of, a step up transformer., (ii) Write any two sources of energy loss in a transformer., (iii) A step up transformer converts a low input voltage into a high output voltage. Does it violate, law of conservation of energy? Explain., OR, Derive an expression for the impedance of a series LCR circuit connected to an AC supply of, variable frequency., Plot a graph showing variation of current with the frequency of the applied voltage., Explain briefly how the phenomenon of resonance in the circuit can be used in the tuning, mechanism of a radio or a TV set., , CBSE (Delhi) SET–II, Questions uncommon to Set–I, 2. The susceptibility of a magnetic material is 1.9 × 10–5. Name the type of magnetic materials it, represents.
Page 245 :
244, , Xam idea Physics—XII, , 4. A plane electromagnetic wave travels in vacuum along x-direction. What can you say about the, direction of electric and magnetic field vectors?, 10. A magnet is quickly moved in the direction indicated by an arrow between two coils C1 and C2 as, shown in the figure. What will be the direction of induced current in each coil as seen from the, magnet? Justify your answer., C1, , C2, N, , S, , A, A, , 11. Figure shows two identical capacitors C1 and C2 each of 1 mF capacitance, connected to a battery, of 5 V. Initially switch ‘S’ is closed. 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. How will the (i) charge and (ii) potential difference between the plates of the, capacitors be affected after the slabs are inserted?, S, , 5V, , 2mF, , C1, , 2m F, , C2, , 13. Draw the output wave form at X, using the given inputs A and B for the logic circuit shown below., Also, identify the logic operation performed by this circuit., , A, B, , Y, , X, , 16. How is forward biasing different from reverse biasing in a p-n junction diode?
Page 246 :
Examination Papers, , 245, , 20. In the circuit shown, R1 = 4 W, R2 = R3 = 5 W, R4 = 10 W and E = 6 V. Work out the equivalent, resistance of the circuit and the current in each resistor., R1, I1, , A, I4, , I2, R2, , E, , I1, , R4, , I3, R3, , B, , 23. An electron and a photon each have a wave length of 1.50 nm. Find (i) their momenta, (ii) the, energy of the photon and (iii) kinetic energy of the electron., 24. State the law of radioactive decay., 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 ., 2, Depict in the plot the number of undecayed nuclei at (i) t = 2 T1 and (ii) t = 4T1 ., 2, , 2, , CBSE (Delhi) SET–III, Questions uncommon to Set–I & II, 2. A plane electromagnetic wave travels in vacuum along y-direction. What can you say about the, direction of electric and magnetic field vectors?, 7. The susceptibility of a magnetic materials is –4.2 × 10 -6 . Name the type of magnetic materials, it represents., 9. Explain how a depletion region is formed in a junction diode., 14. Figure shows two identical capacitors C1 and C2 each of 1.5 mF capacitance, connected to a, battery of 2 V. Initially switch ‘S’ is closed. 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. How will the (i) charge and (ii) potential difference between the plates of the, capacitors be affected after the slabs are inserted?, S, , 2V, , 1.5 mF, , C1 1.5 mF, , C
Page 247 :
246, , Xam idea Physics—XII, , 16. Write the truth table for the logic circuit shown below and identify the logic operation performed, by this circuit., A, , C, Y, , B, , D, , 17. Predict the polarity of the capacitor when the two magnets are quickly moved in the directions, market by arrows., A, B, S, , N, , S, , N, , 23. In the circuit shown, R1 = 2 W, R 2 = R 3 = 10 W, R 4 = 20 W and E = 6 V. Work out the equivalent, resistance of the circuit and the current in each resistor., R1, I1, , A, I2, R2, , E, , I1, , I4, R4, , I3, R3, , B, , 25. An electron and a photon each have a wavelength of 2 nm. Find, (i) their momenta, (ii) the energy of the photon, (iii) the kinetic energy of the electron.
Page 248 :
Examination Papers, , 247, , Solutions, CBSE (Delhi) SET–I, 1. The potential due to a point charge decreases with increase of distance. So, in case (i) VA - VB is, positive., For case (ii) VA - VB is negative., 2. Electric field vector along X-axis, Magnetic field vector along Y-axis., æe ö, 3. r = ç -1÷ R, èV ø, 4. mr < 1, so magnetic material is diamagnetic., 1, 5. l µ , so graph is shown in figure below., V, , l, , V, , 6. When doping level in base is increased slightly, (i) collector current decreases slightly and, (ii) base current increases slightly., 7. Current flowing in an ac circuit without any net dissipation of power is called wattless current., 8. 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. The equipotential surface is at a distance d/2, +s, from either plate in X-Z plane. For a particle, Equipotential, of charge (–q) at rest between the plates, then, surface, (i) weight mg acts, vertically downward, (ii) electric force qE acts vertically upward., d/2, –s, so mg = qE, mg, E=, , vertically downward, i.e., along (–) Y-axis., q, , + + + + + + +, , V=0, , + + + + + + +, , y, x, , d, z
Page 249 :
248, , Xam idea Physics—XII, , OR, Resultant dipole moment, ®, pr, , D, –q, , = p 2 + p 2 + 2 pp cos120 o, = 2 p + 2 p 2 cos120 o, , Y, A +q, , 2, , 120°, X’, , X, , æ 1ö, = 2 p + (2 p ) ´ ç - ÷, è 2ø, 2, , 2, , +q, B –q C, , = 2 p 2 - p 2 = p,, , Y’, , making an angle 60° with Y-axis or 30° with X-axis., ®, , ®, , ®, , ®, , ®, , ®, , Torque, t = P ´ E ( t is perpendicular to both P and E ), 1, = p E sin 30° = pE., 2, Direction of torque is along negative Z-direction., 10. Angle of dip, q = 60°, H = 0.4 G = 0.4 × 10–4 T, If Be is earth’s magnetic field, then, H = Be cosq., H, 0.4 ´ 10 -4 T 0.4 ´ 10 -4 T, Þ Be =, =, =, cos q, 0 ×5, cos 60 o, , ®, , p, , 60° p, 120°, , = 0.8 × 10–4 T = 0.8 G, 11. When switch S is closed, p.d. across each capacitor is 6V, V1 = V2 = 6 V, C1 = C 2 = 1 m C, S, , \ Charge on each capacitor, q1 = q 2 ( = CV ) = (1 mF) ´ (6 V) = 6 mC, When switch S is opened, the p.d. across C1 remains 6 V, 6 V, while the charge on capacitor C2 remains 6 mC. After, insertion of dielectric between the plates of each capacitor,, the new capacitance of each capacitor becomes, C1¢ = C ¢2 = 3 ´ 1 mF = 3 mF, (i) Charge on capacitor C1, q1¢ = C1¢ V1 = (3 mF) ´ 6 V = 18 mC, Charge on capacitor C2 remains 6mC, (ii) Potential difference across C1 remains 6 V., Potential difference across C2 becomes, q, 6 mC, =2V, V2¢ = 2 =, C ¢2 6 mF, , 1 mF, , C1, , 1 mF, , C2
Page 250 :
Examination Papers, , 12. Resolving power R =, \, , R1 A1, =, R2 A2, , 249, , A, , where A is aperture, 122l, , fo, = same for both. We prefer telescope of higher resolving, fe, power to view the fine details of the object, i.e., telescope having convex lens of aperture A1., A ×B, = A ×B = A + B = A + B, 13. Y = A + B =, (A × B ) + A × B, Magnification of telescope, m =, , That is logic operation is OR., , A, , B, , Output Waveform, 14. Zener diode can be used to regulate an unregulated dc power supply., Forward biased, , Vz, , I, (mA), VF (volt), , VR (volt), Reverse, biased, , I (mA), , V-I characteristics of zener diode, , Principle: In reverse breakdown (zener) region, 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., 15. Infrared waves are produced by hot bodies and molecules., They are referred as heat waves because they are readily absorbed by water molecules in most, materials, which increases their thermal motion, so they heat up the material., Use: For therapeutic purpose and long distance photography., 16. In the active region, a small increase of Vi results in a large (almost linear) increase in I C . This, results in an increase in the voltage drop across RC .
Page 251 :
250, , Xam idea Physics—XII, , Active, region, , V0, Cut off, region, , Saturation region, q, AV, , O, , 17., , Vi, , (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.e.,, amplitude modulation index,, E, ma = m, Ec, For modulated wave,, E, - E min, m a = max, E max + E min, , E m < E c , to avoid distortion, (ii) The amplitude of modulating signal is kept less than the amplitude of carrier wave to avoid, distortion., 18., (a) The deflection in galvanometer may be made large by, (i) moving coil C2 towards C1 with high speed., (ii) by placing a soft iron laminated core at the centre of coil C1., (b) The induced current can be demonstrated by connecting a torch bulb (in place of, galvanometer) in Coil C1. Due to induced current the bulb begins to glow., 19., (i) Drift Velocity: When a potential difference is applied across a conductor, the free electrons, drift towards the direction of positive potential. The small average velocity of free electrons, along the direction of positive potential is called the drift velocity., (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., et e 1, Drift velocity, Vd =, a, m L L, When length L is made 3L, drift velocity becomes one-third., 20. 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. To find the electric field, outside the shell, we consider a spherical Gaussian surface of radius r ( > R), concentric with, ®, , given shell. If E is electric field outside the shell, then by symmetry electric field strength has
Page 252 :
Examination Papers, , 251, , same magnitude E 0 on the Gaussian surface and is directed radially outward. Also the directions, of normal at each point is radially outward, so angle between, ®, , ®, , E i and d S is zero at each point. Hence, electric flux, , through Gaussian surface = ò, , ®, E0, S, , Q, , ®, , EO, , R, , · d S., , r, , dS, , P, , = ò E 0 dS cos 0 = E 0 . 4pr 2, Now, Gaussian surface is outside the given charged shell, so, charge enclosed by Gaussian surface is Q., Hence, by Gauss's theorem, ®, ®, 1, ò S E 0 · d E = e 0 ´ charged enclosed, 1, 1 Q, ´ Q Þ E0 =, Þ, E 0 4pr 2 =, e0, 4pe 0 r 2, Thus, electric field outside a charged thin spherical shell is the same as if the whole charge Q is, concentrated at the centre., If s is the surface charge density of the spherical shell, 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. Thus a hollow charged conductor is equivalent to a charged spherical, shell. To find the electric field inside the shell, we consider a spherical Gaussian surface of radius, ®, , r ( < R), concentric with the given shell. If E is the electric field inside the shell, then by symmetry, electric field strength has the same magnitude Ei on the Gaussian surface and is directed radially, ®, , outward. Also the directions of normal at each point is radially outward, so angle between E i and, ®, , d S is zero at each point. Hence, electric flux through Gaussian surface, ®, , ®, , = ò Ei . d S, S, , = ò Ei dS cos 0 = Ei . 4pr 2, Now, Gaussian surface is inside the given charged shell,, so charge enclosed by Gaussian surface is zero., Hence, by Gauss’s theorem, ®, , ®, , òS Ei . dS, Þ, , Ei 4pr 2 =, , 1, ´0, e0, , =, , 1, ´ charge enclosed, e0, Þ Ei = 0, , E, , R, , r
Page 253 :
252, , Xam idea Physics—XII, , 21. Given l = 1× 00 nm = 1× 00 ´ 10 - 9 m, (i) Momenta of electron and photon are equal; given by, p=, , h 6 × 63 ´ 10 - 34, =, = 6 × 63 ´ 10 - 25 kg ms - 1, l 1× 00 ´ 10 - 9, , (ii) Energy of photon, E = hn = h ×, , c h, = c, l l, , = pc = 6 × 63 ´ 10 - 25 ´ 3 ´ 108 J = 19 × 89 ´ 10 - 17 J, =, , 19 × 89 ´ 10 - 17, 1× 6 ´ 10, , - 19, , eV = 1× 24 ´ 10 3 eV = 1 × 24 keV, p2, 1, me v 2 =, 2, 2m e, , (iii) Kinetic energy of electron E k =, =, , ( 6 × 63 ´ 10 - 25 ) 2, 2 ´ 9 ×1 ´ 10 - 31, , = 2 × 42 ´ 10 - 19 J =, , J, 2 × 42 ´ 10 - 19, 1× 6 ´ 10 - 19, , eV = 1 × 51 eV, , 22., , LOS, , Three propagation modes of em waves, , Frequency Range of Different Services, (i) Standard AM broadcast 540–1600 kHz
Page 254 :
Examination Papers, , 253, , (ii) Television 54–890 MHz, (iii) Satellite Communication 5.925–6.425 GHz uplink, 3.7–4.2 GHz downlink., 23. 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., Conditions: (i) The sources should be monochromatic and originating from common single source., (ii) The amplitudes of the waves should be equal., Expression for Fringe Width: Let S1 and S 2 be two coherent sources separated by a distance d., Let the distance of the screen from the coherent sources be D. Let M be the foot of the, perpendicular drawn from O, the midpoint of S1 and S 2 on the screen. Obviously point M is, equidistant from S1 and S 2 . Therefore the path difference between the two waves at point M is, zero. Thus the point M has the maximum intensity. Consider a point P on the screen at a distance y, from M. Draw S1 N perpendicular from S1 on S 2 P., P, S1, S, , d, , O, S2, , q, , y, q, M, , N, , D, , The path difference between two waves reaching at P from S1 and S 2 is D = S 2 P - S1 P » S 2 N, As D > > 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 ,, sin q = 2, S1S 2, MP, In DMOP,, 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., OM, D, yd, ...(i), Path difference D = S 2 P - S1 P = S 2 N =, \, 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.e. D = nl, yd, \, = nl , n = 0, 1, 2, 3, K, D
Page 255 :
254, , Xam idea Physics—XII, , nDl, ., d, This equation gives the distance of nth bright fringe from the point M. Therefore writing, y n for y, we get, nDl, ...(ii), yn =, ., d, (ii) Positions of dark fringes (or minima): For dark fringe or minimum intensity at P, the path, l, difference must be an odd number multiple of half wavelength. i.e. D = ( 2n - 1), 2, y.d, l, where n =1, 2, 3, ....., = ( 2n - 1), \, D, 2, ( 2n - 1) lD, 1 Dl, or, y=, = (n - ), ., 2d, 2 d, This equation gives the distance of nth dark fringe from point M. Therefore writing y n for y,, we get, 1 Dl, ...(iii), yn = ( n - ), 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. It is denoted by w., For Bright Fringes: If y n + 1 and y n denote the distances of two consecutive bright fringes, \, , y=, , from M, then we have, Dl, nDl, and y n =, d, d, Dl nDl Dl, ...(iv), \ Fringe width, b = y n + 1 - y n = ( n + 1), =, ., 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, then we have, 1 Dl, 1 Dl, yn + 1 = ( n + ), , yn = ( n - ), 2 d, 2 d, \ Fringe width, b = y n + 1 - y n, 1 Dl, 1 Dl, = (n + ), - (n - ), 2 d, 2 d, Dl, 1, 1 Dl, =, (n + - n + ) =, d, 2, 2, d, Thus, 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, , ...(v)
Page 256 :
Examination Papers, , 255, , OR, The assumptions of Huygen’s theory are: (i) A source sends waves in all possible directions. The, locus of particles of a medium vibrating in the same phase is called a wavefront. From a point, source, the wavefront is spherical; while for a line source the wavefront is cylindrical. The distant, wavefront is plane., (ii) Each point of a wavefront acts as a source of secondary wavelets. The envelope of all wavelets, at a given instant gives the position of a new wavefront., Rectilinear Propagation of Light: According to Newton’s corpuscular theory, the path of light is, a straight line, but according to wave theory the rectilinear propagation of light is only, approximate., 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, it is deviated from its path., This phenomenon is called refraction. In transversing from first medium to another medium, the, frequency of wave remains unchanged but its speed and the wavelength both are changed. Let XY, be a surface separating the two media ‘1’ and ‘2’. Let v1 and v 2 be the speeds of waves in these, media., 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. Clearly BB¢ = v1t. As the wavefront AB advances, it strikes the points, between A and B¢ of boundary surface. According to Huygen’s principle, secondary spherical wavelets, originate from these points, which travel with speed v1 in the first medium and speed v 2 in the, second medium., First of all secondary wavelet starts from A, which traverses a distance AA¢ ( = v 2 t) in second, medium in time t. In the same time-interval t, the point of wavefront traverses a distance, BB¢ ( = v1t) in first medium and reaches B¢ , from, where the secondary wavelet now starts. Clearly, BB¢ = v1t and AA¢ = v 2 t., 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 incident wavefront AB advances, the secondary wavelets start from points, between A and B¢ , one after the other and will touch A¢ B¢ simultaneously. According to Huygen’s, principle A¢ B¢ is the new position of wavefront AB in the second medium. Hence A¢ B¢ will be the, refracted wavefront., First law: As AB, A¢ B ¢ and surface XY are in the plane of paper, therefore the perpendicular, drawn on them will be in the same plane. As the lines drawn normal to wavefront denote the rays,, therefore we may say that the incident ray, refracted ray and the normal at the point of incidence, all lie in the same plane., This is the first law of refraction., Second law: Let the incident wavefront AB and refracted wavefront A¢ B¢ make angles i and r, respectively with refracting surface XY., In right-angled triangle AB¢ B, Ð ABB¢ = 90°, vt, BB¢, ...(i), sin i = sin Ð BAB¢ =, = 1, \, AB¢, AB¢, Similarly in right-angled triangle AA¢ B ¢ , Ð AA¢ B ¢ = 90°
Page 257 :
256, , Xam idea Physics—XII, , \, , sin r = sin Ð AB¢ A¢ =, , AA¢ v 2 t, =, AB¢ AB¢, , Dividing equation (i) by (ii), we get, v, sin i, = 1 = constant, sin r, v2, , ...(ii), , ..(iii), , As the rays are always normal to the wavefront, therefore, B, the incident and refracted rays make angles i and r with the, o, 90, i, normal drawn on the surface XY i.e. i and r are the angle of, i, B', Y, incidence and angle of refraction respectively. 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. This is the second, law of refraction, and is called the Snell’s law., 24., (a) Light from a source S is allowed to fall normally on the flat surface of a thin plate of a, tourmaline crystal, cut parallel to its axis. Only a part of this light is transmitted through A., If now the plate A is rotated, the character of transmitted light remains unchanged. Now, another similar plate B is placed at some distance from A such that the axis of B is parallel to, that of A. If the light transmitted through A is passed through B, the light is almost completely, transmitted through B and no change is observed in the light coming out of B., Polarised, light, , Unpolarised, light, , Polarised, light, , S, A, Unpolarised, light, , (a), , B, , Polarised, light, No light, , S, (b), , If now the crystal A is kept fixed and B is gradually rotated in its own plane, the intensity of, light emerging out of B decreases and becomes zero when the axis of B is perpendicular to, that of A. If B is further rotated, the intensity begins to increase and becomes maximum when, the axes of A and B are again parallel., Thus, 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., From this experiment, it is obvious that light waves are transverse and not longitudinal; because,, if they were longitudinal, the rotation of crystal B would not produce any change in the intensity, of light., (b) The reflected ray is totally plane polarised, when reflected and refracted rays are, perpendicular to each other.
Page 258 :
Examination Papers, , 257, , 25. Energy of photon wavelength 275 nm, E=, , 6 × 63 ´ 10 -34 ´ 3 ´ 108, hc, eV = 4.5 eV., =, l 275 ´ 10 -9 ´ 1× 6 ´ 10 -19, , This corresponds to transition ‘B’., hc, hc, (i) DE = Þ l =, l, DE, For maximum wavelength DE should be minimum. This corresponds to transition A., (ii) For minimum wavelength DE should be maximum. This corresponds to transition D., 26. The number of nuclei undergoing the decay per unit time, at any instant, is proportional to the, total number of nuclei present in the sample at that instant., , No, , N = No e–lt, N, No, 2, , No, No 8, 32, , T1/2, , 2T1/2, , 3T1/2, , 4T1/2, , 5T1/2, , t, , No, N, and at t = 5T1/2, it is o ., 8, 32, 27. Given R1 = 4W, R2 = R3 = 15W, R4 = 30W, E = 10 V., Equivalent Resistance:, R2, R3 and R4 are in parallel, 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, , A, , F, I1, , R1, , C, I4, , I2, , I3, , E, R2, , G, , I1, , B, , R4, , R3, , D
Page 259 :
258, , Xam idea Physics—XII, , R1 is in series with R, so equivalent resistance, Req = R + R1 = 6 + 4 = 10 W., Currents:, E, 10, I1 =, = =1 A, R eq 10, , ...(i), , This current is divided at A into three parts I2, I3 and I4., \, I2 + I3 + I4 = 1 A, ....(ii), Also, I2 R2 = I3R3 = I4R4, I2 × 15 = I3 × 15 = I4 ×30, Þ, I2 = I3 = 2I4, ...(iii), Þ, Substituting values of I2, I3 in (ii), we get, 2I4 + 2I4 + I4 = 1 A, Þ I4 = 0.2 A, \, I2 = I3 = 2 × 0.2 = 0.4 A, Thus, I1 = 1A, I2 = I3 = 0.4A and I4 = 0.2A, 28. 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, ®, , where m 0, , ®, , ®, , m I dl ´ r, dB = 0, 4p r 3, is permeability of free space. Its value is m 0 = 4p ´ 10 -7 Wb/A-m., , 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 magnetic field dB is perpendicular to the plane containing I dl and r ., Magnetic field at the axis of a circular loop: Consider a circular loop of radius R carrying, current I, 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. Consider a small element of length dl of, ®, , the coil at point A. The magnitude of the magnetic induction dB at point P due to this element is, given by, m I dl sin a, ...(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. As the angle between I dl and r is 90°, the magnitude of the magnetic induction dB is, given by,
Page 260 :
Examination Papers, , dB =, , m 0 I dl sin 90° m 0 I dl, =, ×, 4p, r2, 4p r 2, , ®, , ®, , 259, , ...(ii), ®, , ®, , In figure dB has been represented by P Q. The vector dB or ( P Q), can be resolved into two components, namely PM and PN along, and perpendicular to the axis respectively. Now consider another, small current element `of length dI ¢ at A ¢ . The magnetic induction, ®, , ®, , dl, , q, , P, r, , dB due to this element has been represented by P Q¢ whose I, m I dl ¢, and which can also be resolved into two, magnitude is 0, 4p r 2, components; PM and PN ¢ along the axis and perpendicular to the axis respectively. Thus if we, consider the magnetic induction produced by the whole of the circular coil, then by symmetry the, components of magnetic induction perpendicular to the axis will be cancelled out, while those, ®, , parallel to the axis will be added up. 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,, 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, , ...(iii), , dl, , ...(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, , Therefore,, , B=, =, , ...(v), , m 0 IR, 4p ( R 2 + x 2 ) 3 / 2, m 0 IR 2, 2 (R 2 + x 2 ) 3/ 2, , ( 2 p R), , tesla., , If the coil contains N turns, then, B=, , m 0 NIR 2, 2 (R 2 + x 2 ) 3/ 2, , tesla., , ...(vi)
Page 261 :
260, , Xam idea Physics—XII, , A magnetic needle placed at the center and axis of a circular coil shows deflection. This implies, that a circular coil behaves as a magnet., Infact every current carrying coil is equivalent to a magnetic dipole (or magnet) of magnetic, moment m = IA = I × pr 2 ., 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., 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. Due to the presence of perpendicular magnetic field, the ion will move in a circular path inside the dees. 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, 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.F., oscillator, bombarded., The function of electric field is to accelerate, Dee-2, the charged particle and so to impart energy to, the charged particle., 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., Expression for Period of Revolution and Frequency:, Suppose the positive ion with charge q moves in a dee with a velocity v, then,, qvB =, , mv 2, r, , or r =, , mv, qB, , ...(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., The angular velocity w of the ion is given by,, v qB, (from eq. i), ...(ii), w= =, r m, The time taken by the ion in describing a semi-circle, i.e., in turning through an angle p is,, p pm, ...(iii), t= =, w Bq, Thus the time is independent of the speed of the ion i.e., although the speed of the ion goes on, increasing with increase in the radius (from eq. i) when it moves from one dee to the other, yet it, takes the same time in each dee.
Page 262 :
Examination Papers, , 261, , m, being known, B can be calculated for producing, q, resonance with the high frequency alternating potential., 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 called resonance condition., Now for the cyclotron to work, the applied alternating potential should also have the same, semi-periodic time (T / 2) as that taken by the ion to cross either dee, i.e.,, T, pm, ...(iv), =t =, 2, qB, 2p m, ...(v), or, T=, qB, From eq. (iii) it is clear that for a particular ion,, , This is the expression for period of revolution., Obviously, period of revolution is independent of speed of charged particle and radius of circular, path., \ Frequency of revolution of particles, qB, 1, f = =, T 2pm, This frequency is called the cyclotron frequency. Clearly the cyclotron frequency is independent, of speed of particle., P, 29., (a) Let PQR be the principal section of the prism. The, refracting angle of the prism is A., 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 . The refractive, index of material of prism for this ray is n. 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. The angle of, Q, R, refraction for this face is r1 . The refracted ray, FG becomes incident on face PR and is refracted, away from the normal GN 2 and emerges in the, direction GH. The angle of incidence on this, face is r 2 (into prism) and angle of refraction, d, (into air) is i 2 . The incident ray EF and emergent, ray GH when produced meet at O. The angle, dm, between these two rays is called angle of d, deviation ‘d‘., Ð OFG = i1 - r1 and Ð OGF = i 2 - r 2, In D FOG, d is exterior angle, x, i1, i, i2, \ d = Ð OFG + Ð OGF = (i1 - r1 ) + (i 2 - r 2 ), i, ...(i), = (i1 + i 2 ) - (r1 + r 2 ), The normals FN1 and GN 2 on faces PQ and PR respectively, when produced meet at N. Let, Ð FNG = q
Page 263 :
262, , Xam idea Physics—XII, , In D FGN,, r1 + r 2 + q = 180°, In quadrilateral PFNG, Ð PFN = 90° , Ð PGN = 90°, \, A + 90° + q + 90° = 360° or A + q = 180°, Comparing (ii) and (iii),, r1 + r 2 = A, Substituting this value in (i), we get, d = i1 + i 2 - A, or, i1 + i 2 = A + d, sin i1 sin i 2, From Snell’s law n =, =, sin r1 sin r 2, , ...(ii), ...(iii), ...(iv), ...(v), ...(vi), ...(vii), , Minimum Deviation: From equation (v), it is clear that the angle of deviation depends upon, the angle of incidence i1 . As the path of light is reversible, therefore if angle of incidence be, i 2 , then angle of emergence will be i1 . Thus for two angles of incidence i1 and i 2 , there will be, one angle of deviation., If we determine experimentally, the angles of deviation corresponding to different angles of, incidence and then plot i (on X-axis) and d (on Y-axis), we get a curve as shown in figure., Clearly if angle of incidence is gradually increased, from a small value, the angle of deviation, first decreases, becomes minimum for a particular angle of incidence and then begins to, increase. Obviously for one angle of deviation ( d), there are two angles of incidences i1 and i 2 ,, but for one and only one particular value of angle of incidence (i), the angle of deviation is, the minimum. This minimum angle of deviation is represented by d m . For minimum, deviation i1 and i 2 become coincident, i.e.,i1 = i 2 = i (say), So from (vii), r1 = r 2 = r (say), Hence from (iv) and (vi), 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,, 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. When a light ray is, incident on one end at a small angle of incidence, 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. 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.
Page 264 :
Examination Papers, , 263, , Coating n = 1.5, , n = 1.7, , A, , B, , OR, L, (a) Lens Maker’s Formula:, Suppose L is a thin lens. 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 . The, u, v', optical centre of lens is C and, X ¢ X is principal axis. The, radii of curvature of the, surfaces of the lens are R1 and R 2 and their 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 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, n 2 n1 n 2 - n1, ...(i), =, v¢, u, R1, 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 n 2 ), to first medium (refractive index n1 ), therefore from refraction formula at spherical surface, n1, n2, n - n2, ...(ii), = 1, v (v¢ - t), R2, For a thin lens t is negligible as compared to v¢ , therefore from (ii), n1 n 2, n - n1, ...(iii), =- 2, v (v¢ ), R2, Adding equations (i) and (iii), we get, æ 1, n1 n1, 1, = ( n 2 - n1 ) çç, u, v, è R1 R 2, or, , öæ 1, 1 1 æ n2, 1, - = çç, - 1÷÷ çç, v u è n1, ø è R1 R 2, , ö, ÷, ÷, ø, , ö, ÷, ÷, ø
Page 265 :
264, , Xam idea Physics—XII, , æ 1, 1 1, 1 ö, ÷, ...(iv), - = ( 1 n 2 - 1) çç, ÷, v u, è R1 R 2 ø, n, where 1 n 2 = 2 is refractive index of second medium (i.e. medium of lens) with respect to, n1, 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 (iv), æ 1, 1 1, 1 ö, ÷, - = ( 1 n 2 - 1) çç, ÷, f ¥, è R1 R 2 ø, i.e., , æ 1, 1, 1 ö, ÷, ...(v), = ( 1 n 2 - 1) çç, ÷, f, è R1 R 2 ø, 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 n 2 = n, therefore, equation (v) may be written as, æ 1, 1, 1 ö, ÷, ...(vi), = ( n - 1) çç, ÷, 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., i.e., , A', M, h', , A, h, C, , F, Q, , Magnification: m =, , i, i, , B, u, , i, P, , B', v, , Size of image (A ¢B ¢), Size of object (AB), , From fig. ÐAPB = ÐBPQ = i, Also, ÐBPQ = ÐA ¢PB ¢ = i, AB, In D APB, tan i =, BP, A ¢B ¢, In D A ¢PB ¢, tan i =, B ¢P, , ...(1), ...(2)
Page 266 :
Examination Papers, , 265, , From (1) and (2), , 30., , AB A ¢B ¢, =, BP B ¢P, A ¢B ¢ B ¢P, Þ Magnification, 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. (i. e., N S > N p )., Working: When alternating current source is connected to the ends of primary coil, the, current changes continuously in the 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., Let N p be the number of turns in primary coil, NS the number of turns in secondary coil and f, the magnetic flux linked with each turn. 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. According to, Faraday’s laws the emf induced in the primary coil, Df, ...(i), ep =-Np, Dt, and emf induced in the secondary coil, Df, ...(ii), eS = - NS, Dt, From (i) and (ii), eS NS, ...(iii), =, ep Np, If the resistance of primary coil is negligible, the emf ( e p ) induced in the primary coil, will be, equal to the applied potential difference (V p ) across its ends. Similarly if the secondary circuit, is open, then the potential difference VS across its ends will be equal to the emf ( e S ) induced, in it; therefore, VS e S N S, ...(iv), =, =, = r (say), Vp e p N p, where r =, , NS, is called the transformation ratio. If i p and i s are the instantaneous currents in, Np, , primary and secondary coils and there is no loss of energy; then, For about 100% efficiency,, Power in primary = Power in secondary, V p i p = VS i S, iS V p N p 1, ...(v), \, =, =, =, i p VS N S r
Page 267 :
266, , Xam idea Physics—XII, , In step up transformer,, , N s > N p ® r >1;, , So, VS > V p and i S < i p, i.e., step up transformer increases the voltage., Soft iron-core, , Ns, , Secondary, , Primary, , Np, , 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)., (b) Flux Leakage: Energy is lost due to coupling of primary and secondary coils not being, perfect, i.e., whole of magnetic flux generated in primary coil is not linked with the, secondary coil., (iii) 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., OR, (a) Expression for Impedance in LCR series circuit: Suppose resistance R, inductance L and, capacitance C are connected in series and an alternating source of voltage V = V0 sin wt is, applied across it. (fig. a) On account of being in series, the current (i ) flowing through all of, them is the same., R, , L, , VL, , VR, , C, , VL, 90o, , V=V0 sin wt, (a), , VC, , (VC–VL), , 90, , o, , VR, f, , i, , VC, (b), , Suppose the voltage across resistance R is VR , voltage across inductance L is VL and voltage, across capacitance C is VC . The voltage VR and current i are in the same phase, the voltage VL, will lead the current by angle 90° while the voltage VC will lag behind the current by angle, 90° (fig. b). Clearly VC and VL are in opposite directions, therefore their resultant potential, difference = VC - VL (if VC > VC )., Thus VR and (VC - VL ) are mutually perpendicular and the phase difference between them is, 90°. As applied voltage across the circuit is V, the resultant of VR and (VC - VL ) will also be, V. From fig.
Page 268 :
Examination Papers, , V, But, where X C, \, , 2, , = VR 2 + (VC - VL ) 2 Þ V = VR 2 + (VC - VL ) 2, , 267, ...(i), , VR = R i , VC = X C i and VL = X L i, 1, =, = capacitance reactance and X L = wL = inductive reactance, wC, , ...(ii), , V = ( R i) 2 + ( X C i - X L i) 2, , \ Impedance of circuit, Z =, i.e., Instantaneous current I =, , V, = R 2 + (X C - X L ) 2, i, , æ 1, ö, Z = R 2 + (X C - X L ) 2 = R 2 + ç, - wL ÷, è wC, ø, V0 sin ( wt + f), æ 1, ö, R2 + ç, - wL ÷, è wC, ø, , 2, , 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, so X C - 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.e., X L = X C . Alternatively when the frequency of applied a.c. source becomes, 1, ., fr, 2p LC, Graph is shown in fig., im = V/R, i, , fr, , f
Page 269 :
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. The circuit, remains connected with an areal coil through the phenomenon of mutual inductance. Suppose a, radio or TV station is transmitted a programme at frequency f, then waves produce alternating, voltage of frequency f, in areal, due to which an emf of same frequency is induced in LC circuit,, 1, When capacitor C is in circuit is varied then for a particular value of capacitance C, f =, ,, 2p LC, the resonance occurs and maximum current flows in the circuit; so the radio or TV gets tunned., , CBSE (Delhi) SET–II, 2. Susceptibility is small positive, so material is paramagnetic., 4. Electric field along Y-direction. Magnetic field along Z-direction., r r r, (so that K, E, B form a right handed system), 10. According to Lenz’s law, the direction of induced current is such that it opposes the relative, motion between coil and magnet., The near face of coil C1 will become S-pole, so the direction of current in coil C1 will be, clockwise., The near face of coil C2 will also become S-pole to oppose the approach of magnet, so the current, in coil C2 will also be clockwise., 11. 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.d. across C1 remains 5V and charge on C2 remains 10 mC., New capacitance of C1 and C2 after filling dielectric, C1¢ = C 2¢ = 5 ´ 2 = 10 mF., Charge on C1 , q1¢ = C1¢ V1 = (10 mF) ´ 5 V = 50 mC, Charge on C2, q¢2 = q 2 = 10 mC, (ii) Potential difference across, C1 = V´1 = V1 = 5V, Potential difference across C2,, q¢ 10 mC, V2¢ = 2 =, =1 V, C ¢2 10 mF, 13. Y = AB, X = AB × AB = AB = AB, That is the logic operation is AND operation. Therefore output is 1 only, when both inputs are 1., The output waveform is shown in figure.
Page 270 :
Examination Papers, , 269, , A, B, , t1, , t2, , t3, , t4, , t5, , t6, , t7, , Output X, , 16. 1. Forward Bias:, (i) Within the junction diode the direction of applied voltage is opposite to that of built-in, potential., (ii) The current is due to diffusion of majority charge carriers through the junction and is of, the order of milliamperes., (iii) The diode offers very small resistance in the forward bias., 2. Reverse Bias:, (i) The direction of applied voltage and barrier potential is same., (ii) The current is due to leakage of minority charge carriers through the junction and is very, small of the order of mA., (iii) The diode offers very large resistance in reverse bias., 1, 1, 1, 1 1 1 1 2 + 2 +1, 20. Effective resistance of R2, R3 and R4 in parallel, is =, =, +, +, = + +, 10, R R 2 R 3 R 4 5 5 10, R=2W, Þ, R1 is in series with R, so equivalent resistance., Req = R1 + R = 4 + 2 = 6 W, E, 6, Current in R1 , I 1 =, = = 1A, R eq 6, Also, 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), equation (iii) gives, Þ I4 =, , 2I 4 + 2I 4 + I 4 = 1, , ...(ii), ...(iii), , ...(iv), 1, A = 0.2 A, 5, , I 2 = I 3 = 2 ´ 0.2 = 0.4 A, \ I 1 = 1A, I 2 = I 3 = 0.4A, I 4 = 0.2 A, –9, , ...(i), , 23. Given l = 1.50 nm = 1.50 × 10 m., (i) As momentum of electron = momentum of photon, h 6.63 ´ 10 -34, = =, = 4.42 ´ 10 -25 kg m s -1, -9, l 150, . ´ 10, , ...(v)
Page 271 :
270, , Xam idea Physics—XII, , hc, = pc, l, = 4.42 × 10–25 × 3 × 108 J, = 13.26 × 10–17 J, , (ii) Energy of photon, E =, , =, , 13.26 ´ 10 -17, , 1.6 ´ 10 -19, (iii) Kinetic energy of electron, , eV = 828.7 eV, , P 2 ( 4.42 ´ 10 -25 ) 2, J, =, 2m e, 2 ´ 9 ×1´ 10 -31, = 1.07 × 10–19 J, , EK =, , =, , 1.0.7 ´ 10 -19, , eV = 0.67 eV, 1.6 ´ 10 -19, 24. The number of nuclei undergoing the decay per unit time, at any instant, is proportional to the, total number of nuclei present in the sample at that instant. This is the law of radioactive decay., , No, , N = No e–lt, N, No, 2, No, 4, , No, 16, T1/2, , 2T1/2, , 3T1/2, , 4T1/2, , t, , Number of undecayed nuclei at t = 2T1/2 is, , No, N, and at t = 4T1/2, it is o ., 4, 16, , CBSE (Delhi) SET–III, 2. Electric field along X-direction., Magnetic field along Z-direction., 7. Susceptibility of material is negative, so given material is diamagnetic., 9. Formation of depletion layer:, 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 positively charged ions in n-region and an excess of negatively
Page 272 :
Examination Papers, , 271, , charged ions in p-region. This sets up a potential difference called potential barrier and hence an, internal electric field Ei across the junctions. The field Ei is directed from n-region to p-region., This field stops the further diffusion of charge carriers. Thus the layers ( » 10 - 4 cm to 10 - 6 cm), on either side of the junction becomes free from mobile charge carriers and hence is called the, depletion layer., 14. (i) Charge on C1 , q1¢ = 6 mC, Charge on C 2 , q¢2 = 3 mC, q ¢ 3 mC, (ii) P.D. across C1 , V1¢ = 2 V, P.D. across C 2 , V2¢ = 2 =, =1 V, C ¢2 3 mF, 16. C = A , D = B, Y = CD = A B = A + B = A + B, The logic circuit performs OR-operation., Truth table., A, , B, , Y, , 0, , 0, , 0, , 1, , 0, , 1, , 0, , 1, , 1, , 1, , 1, , 1, , 17. Current induced in coil will oppose the approach of magnet; therefore, left face of coil will act as, N-pole and right face as S-pole. For this the current in coil will be anticlockwise as seen from left,, therefore, the plate A of capacitor will be positive and plate B will be negative., A, S, , N, , B, , S, , N, , 23. R1 = 2 W, R 2 = R 3 = 10 W, R 4 = 20 W, E = 6 V, Equivalent Resistance: R 2 = R 3 and R 4 are in parallel, 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, I1 =, = =1A, R eq 6, Also,, Þ, Þ, and, , I2R2 = I3R3 = I4R4, I2.10 = I3.10 = I4.20, I2 = I3 = 2I4, I2 + I3 + I4 = 1 A, , L, C, , ...(i), ...(ii)
Page 273 :
272, , Xam idea Physics—XII, , Solving (i) and (ii), I2 = I3 = 0.4A, I4 = 0.2 A, 25. l = 2 nm = 2 × 10–9 m, (i) Momentum of electron = Momentum of photon = p, =, (ii) Energy of photon, E =, , h 6.63 ´ 10 -34, = 3.315 × 10–25 kg m s–1, =, l, 2 ´ 10 -9, , hc, = pc, l, , = 3.315 ´ 10 -25 ´ 3 ´ 108 J, = 9.945 ´ 10 -17 J =, , 9.945 ´ 10 -17, , 1.6 ´ 10 -19, = 6.216 × 102 eV = 0.62 keV, (iii) Kinetic energy of electrons, EK =, , p2, (3.315 ´ 10 -25 ) 2, 1, m eV 2 =, =, J, 2, 2m e, 2 ´ 9.1´ 10 -31, = 0.604 × 10–19 J = 0.38 eV, , eV
Page 274 :
CBSE EXAMINATION PAPERS, ALL INDIA–2011, Time allowed: 3 hours, , Maximum marks: 70, , 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–, 4pe o, , m 0 = 4p ´ 10 -7 TmA -1, 2, , 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. Define electric dipole moment. Write its S.I. unit., 2. 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., What is the potential at the centre of the sphere?, 4. How are radio waves produced?, 5. Write any two characteristic properties of nuclear force., 6. Two bar magnets are quickly moved towards a metallic loop connected across a capacitor ‘C’, as shown in the figure. Predict the polarity of the capacitor., C, S, , N, , S, , N
Page 275 :
274, , Xam idea Physics—XII, , 7. What happens to the width of depletion layer of a p-n junction when it is (i) forward biased,, (ii) reverse biased?, 8. Define the term ‘stopping potential’ in relation to photoelectric effect., 9. A thin straight infinitely long conducting wire having charge density l is enclosed by a cylindrical, surface of radius r and length l, its axis coinciding with the length of the wire. Find the expression, for the electric flux through the surface of the cylinder., æ 1 ö, ÷÷, where r is the distance, 10. Plot a graph showing the variation of coulomb force (F) versus çç, èr 2 ø, between the two charges of each pair of charges: (1 mC, 2 mC) and (2 mC – 3 mC). Interpret the, graphs obtained., 11. Write the expression for Lorentz magnetic force on a particle of charge ‘q’ moving with velocity, ®, , ®, , v in a magnetic field B . Show that no work is done by this force on the charged particle., OR, A steady current (I1) flows through a long straight wire. Another wire carrying steady current (I2), in the same direction is kept close and parallel to the first wire. Show with the help of a diagram, how the magnetic field due to the current I1 exerts a magnetic force on the second wire. Write the, expression for this force., 12. What are eddy currents? Write any two applications of eddy currents., 13. What is sky wave communication? Why is this mode of propagation restricted to the frequencies, only up to few MHz?, 14. In the given circuit, assuming point A to be at zero potential, use Kirchhoff’s rules to determine, the potential at point B., 1A, , R, , D, , 2V, , 2A, , R, R11, , 2W, , A, , 1V, , C, , 2A, , 15. A parallel plate capacitor is being charged by a time varying current. Explain briefly how, Ampere’s circuital law is generalized to incorporate the effect due to the displacement current., 16. Net capacitance of three identical capacitors in series is 1 mF. 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., 17. Using the curve for the binding energy per nucleon as a function of mass number A, state clearly, how the release in energy in the processes of nuclear fission and nuclear fusion can be explained., 18. In the meter bridge experiment, balance point was observed at J with AJ = l.
Page 276 :
Examination Papers, , 275, , (i) The values of R and X were doubled and then interchanged. What would be the new position, of balance point?, (ii) If the galvanometer and battery are interchanged at the balance position, 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 1.5 is dipped, in turn, in (i) a medium of, refractive index 1.65, (ii) a medium of refractive index 1.33., (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. Draw a plot showing the variation of photoelectric current with collector plate potential for two, different frequencies, v1 > v 2 , of incident radiation having the same intensity. In which case will, the stopping potential be higher? Justify your answer., 21. Write briefly any two factors which demonstrate the need for modulating a signal., Draw a suitable diagram to show amplitude modulation using a sinusoidal signal as the modulating, signal., 22. 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., (b) a convex mirror always produces a virtual image independent of the location of the object., (c) an object placed between the pole and focus of a concave mirror produces a virtual and, enlarged image., 23. Draw a labelled diagram of a full wave rectifier circuit. State its working principle. Show the, input-output waveforms., 24., (a) Using de Broglie’s hypothesis, explain with the help of a suitable diagram, Bohr’s second, postulate of quantization of energy levels in a hydrogen atom., (b) The ground state energy of hydrogen atom is –13.6 eV. What are the kinetic and potential, energies of the electron in this state?, 25. You are given a circuit below. Write its truth table. Hence, identify the logic operation carried out, by this circuit. Draw the logic symbol of the gate it corresponds to., A, , X, Z, , B, , Y
Page 277 :
276, , Xam idea Physics—XII, , 26. A compound microscope uses an objective lens of focal length 4 cm and eyepiece lens of focal, length 10 cm. An object is placed at 6 cm from the objective lens. Calculate the magnifying power, of the compound microscope. Also calculate the length of the microscope., OR, A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an, eyepiece lens of focal length 1.0 cm is used, find the angular magnification of the telescope., If this telescope is used to view the moon, what is the diameter of the image of the moon formed, by the objective lens? The diameter of the moon is 3.42 × 106 m and the radius of the lunar orbit is, 3.8 × 108 m., 27. Two heating elements of resistance R1 and R2 when operated at a constant supply of voltage, V,, consume powers P1 and P2 respectively. Deduce the expressions for the power of their, combination when they are, in turn, connected in (i) series and (ii) parallel across the same voltage, supply., 28., (a) State the principle of the working of a moving coil galvanometer, giving its labelled diagram., (b) “Increasing the current sensitivity of a galvanometer may not necessarily increase its voltage, sensitivity.” Justify this statement., (c) Outline the necessary steps to convert a galvanometer of resistance RG into an ammeter of a, given range., OR, (a) Using Ampere’s circuital law, obtain the expression for the magnetic field due to a long, solenoid at a point inside the solenoid on its axis., (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., (c) How is the magnetic field inside a given solenoid made strong?, 29. State the working of a.c. generator with the help of a labelled diagram., The coil of an a.c. generator having N turns, each of area A, is rotated with a constant angular, velocity w . Deduce the expression for the alternating e.m.f. generated in the coil., What is the source of energy generation in this device?, OR, (a) Show that in an a.c. circuit containing a pure inductor, the voltage is ahead of current by p/2, in phase., (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., (i) Write the expression for the instantaneous value of the e.m.f. induced in the wire., (ii) What is the direction of the e.m.f.?, (iii) Which end of the wire is at the higher potential?, 30. State the importance of coherent sources in the phenomenon of interference., In Young’s double slit experiment to produce interference pattern, obtain the conditions for, constructive and destructive interference. Hence, deduce the expression for the fringe width., How does the fringe width get affected, if the entire experimental apparatus of Young is immersed, in water?
Page 278 :
Examination Papers, , 277, , OR, (a) State Huygen’s principle. 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., (b) Show that the angular width of the first diffraction fringe is half of that of the central fringe., (c) If a monochromatic source of light is replaced by white light, what change would you, observe in the diffraction pattern?, , CBSE (All India) SET–II, Questions uncommon to Set–I, 1. A hollow metal sphere of radius 10 cm is charged such that the potential on its surface is 5V., What is the potential at the centre of the sphere?, 2. How are X-rays produced?, 4. Where on the surface of Earth is the angle of dip zero?, 9. Net capacitance of three identical capacitors in series is 2 mF. 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., 12. State the principle of working of a transformer. Can a transformer be used to step up or step down, a d.c. voltage? Justify your answer., 14. In the given circuit, assuming point A to be at zero potential, use Kirchhoff’s rules to determine, the potential at point B., 1A, , R, , D, , 3A, , 4V, B, , R, R11, , 2W, , A, , 2V, , C, , 3A, , 17. What is ground wave communication? On what factors does the maximum range of propagation, in this mode depend?, 21. You are given a circuit below. Write its truth table. Hence, identify the logic operation carried out, by this circuit. Draw the logic symbol of the gate it corresponds to., A, , X, Z, , B, , Y
Page 279 :
278, , Xam idea Physics—XII, , CBSE (All India) SET–III, Questions uncommon to Set – I & II, 1. A hollow metal sphere of radius 6 cm is charged such that the potential on its surface is 12 V., What is the potential at the centre of the sphere?, 3. How are microwaves produced?, 7. Where on the surface of Earth is the vertical component of Earth’s magnetic field zero?, 12. Mention various energy losses in a transformer., 14. In the given circuit, assuming point A to be at zero potential, use Kirchhoff’s rules to determine, the potential at point B., 1A, , R, , 6V, , 4A, , D, , R1, , 2W, , A, , 3V, , 4A, , C, , 16. What is space wave communication? Write the range of frequencies suitable for space wave, communication., 20. A converging lens has a focal length of 20 cm in air. It is made of a material of refractive index, 1,6. It is immersed in a liquid of refractive index 1.3. Calculate its new focal length., 22. You are given a circuit below. Write its truth table. Hence, identify the logic operation carried out, by this circuit. Draw the logic symbol of the gate it corresponds to., A, , X, Z, , B, , Y
Page 280 :
Examination Papers, , 279, , Solutions, CBSE (All India) SET-I, 1. 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., The SI unit of electric dipole moment is coulomb metre (Cm)., 2. Angle of dip is 90° at the poles., 3. Potential at centre of sphere = 10 V., 4. Radio waves are produced by the accelerated motion of charges in conducting wires., 5., (i) Nuclear forces are short range attractive forces., (ii) Nuclear forces are charge - independent., 6. Current induced in coil will oppose the approach of magnet; therefore, left face of coil will act as, N-pole and right face as S-pole. For this the current in coil will be anticlockwise as seen from left,, therefore, the plate A of capacitor will be positive and plate B will be negative. The upper plate is, positive and lower plate is negative., A, S, , 7., , N, , B, , S, , N, , (i) When forward biased, the width of depletion layer decreases., (ii) When reverse biased, the width of depletion layer increases., 8. The minimum retarding (negative) potential of anode of a photoelectric tube for which, photoelectric current stops or becomes zero is called the stopping potential., 9. Charge enclosed by the cylindrical surface = ll, 1, × charge enclosed., By Gauss Theorem, electric flux =, eo, 1, =, ( l l), eo, q1 q 2, 1, 10. F =, ., 4pe o, r2, 1, 1, is a straight line of slope, The graph between F and, q1 q 2 passing through origin., 2, 4, pe, r, o
Page 281 :
280, , Xam idea Physics—XII, 1/r2, F, , Repulsive, , Attractive, , F, 1/r2, , Since, magnitude of the slope is more for attraction, therefore, attractive force is greater than, repulsive force., ® ®, , 11. Lorentz magnetic force, Fm = q v ´ B, r ®, ® ®, Work done, W = F m × S = ò Fm . v dt =, , ® ®, , ®, , ò q ( v ´ B ) . v dt, , ® ® ®, , As, , ( v ´ B) . v = 0, , \, , Work, W = 0, , 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. 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, they experience an attractive, b, I2, I1, force and when they carry currents in I1, opposite directions, they experience a, repulsive force., 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. Consider a current–element ‘ab’ of length DL of wire RS. 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
Page 282 :
Examination Papers, , 281, , 12. Eddy currents: When a metallic plate is placed in a time varying magnetic field, the magnetic, flux linked with the plate changes, the induced currents are set up in the plate; these currents are, called eddy currents. These currents are sometimes so strong, that the metallic plate becomes red, hot. In transformer frames there is a huge loss of energy due to production of eddy currents, so, these currents are undesirable in transformer., Application of Eddy Currents:, 1. Induction Furnace: In induction furnance, the metal to be heated is placed in a rapidly, varying magnetic field produced by high frequency alternating current. Strong eddy currents, are set up in the metal produce so much heat that the metal melts. This process is used in, extracting a metal from its ore. The arrangement of heating the metal by means of strong, induced currents is called the induction furnace., 2. Induction Motor: The eddy currents may be used to rotate the rotor. Its principle is: When a, metallic cylinder (or rotor) is placed in a rotating magnetic field, eddy currents are produced, in it. According to Lenz’s law, these currents tend to reduce to relative motion between the, cylinder and the field. The cylinder, therefore, begins to rotate in the direction of the field., This is the principle of induction motion., 13. 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., For frequencies higher than few MHz, the sky waves penetrate the ionosphere and are not, reflected back., 14. Current in 2 W resistor is 1 A. Applying Kirchhoff’s II law along the path ACDB,, VA + 1 + 2× 1 – 2 = VB., As VA = 0, \ VB = 1 + 2 – 2 = 1 V., IA, R, R, , B, , 2W, , 1V, C, , A, , 15. Displacement current and generalised Ampere’s, Circuital Law: Consider a parallel plate capacitor,, being charged by a battery. A time varying current is, flowing through the capacitor. If we consider only the, conduction current I, then we apply Ampere’s, Circuital Law to two closed loops C1 and C2, then we, get, , òC, and, , òC, , Ø Ø, , 1, , B × dl = m 0 I, Ø Ø, , B × dl = 0, 2, , ...(i), ...(ii), , C1, I, , C2, I
Page 283 :
282, , Xam idea Physics—XII, , Since there cannot be any conduction current in region between the capacitor plates. As C1 and C2, are very close, we must expect, , ò, , Ø Ø, , C1, , Ø Ø, , B × dl = ò C B ×dl, , ...(iii), , 2, , But this condition is violated by equations (i) and (ii). Hence Ampere’s Circuital Law seems to be, inconsistent in this case. Therefore, Maxwell postulated the existence of displacement current, which is produced by time varying electric field. If s (t) is the surface charge density on capacitor, s (t) q (t), plates and q(t) is the charge, then time varying electric field E (t) =, , where A is area of, =, e0, Ae 0, each plate., dE, 1 dg(t), =, dt Ae 0 dt, dg(t), dE, or, = e0 A, dt, dt, This is expression for displacement current (ld)., Applying Kirchhoff’s first law at power P, we get I = Id, Hence, equation (i) and (ii) take the forms, , ò, , Ø Ø, , C1, , B × dl = m 0 I and, , ò, , Ø Ø, , C2, , B × dl = m 0 I d = m 0 I, , The total current is the sum of the conduction current and displacement, current. Thus, modified form of Ampere’s circuital law is, Ø Ø, dE ö, æ, ò B. dl = m 0 ( I + I d ) = m 0 çè I + e 0 A dt ÷ø, But, EA = Electric flux fE, Ø Ø, df E, \, B, ò . dl = m 0 ( I + m 0 e 0 dt ), 16. Let C be the capacitance of each capacitor, then in series, 1, 1 1 1 3, = + + =, Cs C C C C, , Id, , I, P, , ( ), , or, C = 3Cs = 3 × 1 mF = 3 mF, When these capacitors are connected in parallel, net capacitance, Cp = 3 C = 3 ´ 3 = 9 mF, When these two combinations are connected to same source the potential difference across each, combination is same., Ratio of energy stored,, 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
Page 284 :
Examination Papers, , 283, , 17. The variation of binding energy per nucleon versus mass number is shown in figure., , Binding Energy per Nucleon (in MeV), , 9.0, O16, 8.0C12, F18, He4 N14, 7.0, , Fe56, U238, , 6.0, Li7, , 5.0, 4.0, 3.0, 2.0, , H2, , 1.0, 0.0, , 0, , 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),, the binding energy per nucleon increases i.e, nucleons get more tightly bound. This implies that, energy would be released in nuclear fission., When two very light nuclei ( A £10) join to form a heavy nucleus, the binding is energy per, nucleon of fused heavier nucleus more than the binding energy per nucleon of lighter nuclei, so, again energy would be released in nuclear fusion., R, rl, 18., (i), =, X r (100 - l), R, l, ...(i), =, X 100 - l, When both R and X are doubled and then interchanged, the new balance length becomes l¢, given by, 2X, l¢, =, 2R (100 - l¢), X, l¢, ...(ii), =, Þ, R 100 - l¢, Þ, , From (i) and (ii),, l¢, 100 - l, =, l, 100 - l¢, Þ, l¢ = (100 - l), (ii) If galvanometer and battery are interchanged, there is no effect on the balance point.
Page 285 :
284, , Xam idea Physics—XII, , 19. 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 = 1.5, n l = 1.65, ng, 15, ., =, < 1, so f l and f a are of opposite sign, so convex lens in liquid nl = 1.65 behaves as, n l 1.65, a diverging lens, (ii) ng = 1.5, nl = 1.33, ng, 15, ., \, =, >1, n l 133, ., so fl and fa are of same sign, so convex lens in liquid (nl = 1.33) behaves as a convergent lens, 1.5 -1, (b) (i) Focal length, f 1 =, f a = -5.5f a (Focal length becomes negative and its magnitude, 1.5, -1, 1.65, increases), 1.5 -1, (ii) Focal length, f 2 =, f a = 4f a (Focal length increases), 1.5, -1, 1.33, 20. The plots are shown in fig. The stopping potential (Vs) is higher for radiations of frequency n1 ., Stopping potential is directly proportional to the frequency of incident radiation., I, , n1 = n2, L = Constant, , Saturation, photocurrent, , n1, n2, (Vs)1, , (Vs)2, , Collector potential, , 21. The modulation is needed due to, (i) Transmission of audio frequency electrical signals need long impracticable antenna., (ii) The power radiated at audio frequency is quite small, hence transmission is quite lossy., (iii) The various information signals transmitted at low frequency get mixed and hence can not be, distinguished.
Page 286 :
Examination Papers, , 285, , EC, , (a) Carrier wave ec = Ec sin wct, Em, , (b) Modulating wave (em), , Envelope, , Em, Ec, , (c) Modulated wave e(t), , 22. Mirror equation is, , 1 1 1, = +, f v u, , or, , 1 1 1, = v f u, , (a) For a concave mirror, f is negative, i.e., f < 0., For a real object (on the left of mirror)., 1, 1, 1, 1 1 1, For u between f and 2f implies lies between and, i.e.,, > > (as u, f are negative), u, f, 2f, 2f u f, 1, 1, 1, or, <- <2f, u, f, 1 1 1 1, or, < - <0, f 2f f u, 1 1, or, < <0, 2f v, 1, i.e.,, is negative., v, This implies that v is negative and greater than 2f . This means that the image lies beyond 2f, and it is real., (b) For a convex mirror, f is positive i.e., f > 0., For a real object on the left u is negative., 1 1 1, 1 1 1, = + implies = f v u, v f u, 1, As u is negative and f is positive; must be positive, so v must be positive i. e., image lies, v, behind the mirror. Hence, image is virtual whatever the value of u may be.
Page 287 :
286, , Xam idea Physics—XII, , (c) For a mirror,, 1 1 1, = v f u, , ...(i), , For a concave mirror, f is negative f < 0, As u is also negative, so f < u < 0, 1 1, This implies, - > 0, f u, 1, Then from (1) > 0 or v is positive., v, i.e., image is on the right and hence virtual., f, v, Magnification, m = - = u, u-f, As u is negative and f is positive, magnification m =, , |f |, >1, | f | -| u |, , i.e., image is enlarged., 23. Full Wave Rectifier, D1 diode, , Input, RL, Tapped, secondary, , Output, , D2 diode, , 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., Input, wave form, T, , Output wave form, of full wave, rectifier, , T, , 2T Time, , 2T Time, , 24. (a) According to de Broglie’s hypothesis,, h, ...(i), l=, mv, According to de Broglie’s condition of stationary orbits, the stationary orbits are those which, contain complete de-Broglie wavelength., ...(ii), 2pr = nl
Page 288 :
Examination Papers, , Substituting value of l from (ii) in (i), we get, h, 2pr = n, mv, h, ...(iii), Þ, mvr = n, 2p, This is Bohr’s postulate of quantisation of energy levels., , n=3, , (b) Kinetic energy,, , K=, , 1, 1, e2, ×, mv 2 =, 2, 4pe 0 2r, , …(i), , U =-, , 1 e2, 4pe 0 r, , …(ii), , Total energy E = K + U = -, , 1 e2, 4pe 0 2r, , …(iii), , Potential energy,, , Comparing equations (i), (ii), (iii), we have, K = - E and U = 2E, Given, , E = - 13 × 6 eV, , \ Kinetic energy,, Potential energy, , (in ground state), , K = 13 × 6 eV, U = 2 ´ ( - 13 × 6 eV) = - 27 × 2 eV, , 25. The output Y = A + B = A × B = AB, That is equivalent gate is ‘AND’ gate., The logic symbol and truth table are shown:, Truth Table, , A, B, , Y, , A, , B, , Y, , 0, , 0, , 0, , 1, , 0, , 0, , 0, , 1, , 0, , 1, , 1, , 1, , 287
Page 289 :
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 = - o çç1 + ÷÷, 1u o 1 è f e ø, From lens formula, , Þ, \, , 1, 1, 1, =, f o vo uo, 1, 1, 1 1 1 3 -2, =, =, +, = - =, v o f o u o 4 6 12, , vo = 12 cm, 12 æ 25 ö, m = – ç1 + ÷ = -2 ´ 3.5 = -7, 6 è 10 ø, , Negative sign shows that the image is inverted., Length of microscope L = v o + u e, 1, 1 1, For eye lens, =, f e ve ue, 1, 1 1, 1 1, =, =, - =- ue ve f e, 25 10, 50, u e = - ( cm) = -7 ×14, Þ, 7, \, L = v o + u e = 12 + 7 ×14 = 19 ×14 cm, , ( v e = D = - 25 cm, u e = ?), , OR, Given f 0 = 15 m, f e = 1× 0 cm = 1× 0 ´ 10, , -2, , m, , Angular magnification of telescope,, f, 15, m= 0 =, = 1500, fe, 1× 0 ´ 10 - 2, Let D be diameter of moon, d diameter of image of moon formed by objective and r the distance, of moon from objective lens, then from Fig., D d, =, r f0, Þ, 27., , d=, , 3 × 48 ´ 10 6, D, × f0 =, ´ 15 m = 0 × 137 m = 13 × 7 cm, r, 3 × 8 ´ 108, , (i) In series combinations, Net resistance, R = R1 + R2, As heating elements are operated at same voltage V, we have, V, V2, V2, and R 2 =, R = 2 , R1 =, P, P1, P2, , ...(i)
Page 290 :
Examination Papers, , 289, , \ From equation (i), 1 1, 1, V2, V2 V2, =, +, 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. (a) Principle: It works on the principle that a current carrying coil when kept inside a uniform, magnetic field, can experience a torque., When current ( I ) is passed in the coil, 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,, N is the number of turns in a coil., When the magnetic field is radial, as in the case of cylindrical pole pieces and soft iron core,, then in every position of coil the plane of the coil, is parallel to the magnetic field lines, so that, q = 90° and sin 90° = 1, Deflecting torque,, t = NIAB, If C is the torsional rigidity of the wire and q is the twist of suspension strip, then restoring, torque = C q, For equilibrium, deflecting torque = restoring torque, i.e., NIAB = C q, NAB, ...(i), q=, I, \, C, i.e., qµI, deflection of coil is directly proportional to current flowing in the coil and hence we can, construct a linear scale., H, , Suspension wire, M, , T1, T2, , Coil, , NIBl, b, , N, , S, , N, , S, , N, , S, , NIBl, Coiled strip, , (a), , (b), Magnetic lines of, force of radial magnetic field, , (c)
Page 291 :
290, , Xam idea Physics—XII, , æ q ö NAB, (b) Current sensitivity, S I = ç ÷ =, C, èIø, æ q ö NAB, Voltage sensitivity, SV = ç ÷ =, è V ø GC, Dividing (ii) by (i), SV 1, =, SI G, , Þ, , SV =, , ... (i), ... (ii), , 1, SI, G, , Clearly the voltage sensitivity depends on current sensitivity and the resistance of, galvanometer. If we increase current sensitivity and resistance G is larger, then it is not, certain that voltage sensitivity will be increased. Thus, the increase of current sensitivity does, not imply the increase of voltage sensitivity., (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’., The resistance of an ammeter is to be made as low, as possible so that it may read current without any, appreciable error. Therefore to convert a, (i – ig), S, galvanometer into ammeter a shunt resistance., (i.e. small resistance in parallel) is connected across the coil of galvanometer., Let G be the resistance of galvanometer and I g the current required for full scale deflection., Suppose this galvanometer is to converted into ammeter of range I ampere and the value of, shunt required is S. If I S is current in shunt, then from fig., ...(i), I = I g + I S Þ I S = (I - I g ), Also potential difference across A and B, (VAB ) = I S . S = I g . G, Substituting value of I S from (i), we get, or, (I - I g ) S = I g G, or, IS - I g S = I g G, or, IS = I g (S + G ), S, ...(ii), or, Ig =, I, S +G, GI g, i.e. shunt required, S =, ...(iii), I - Ig, This is the working equation of conversion of galvanometer into ammeter., 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, carrying current. 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. When an electric current is passed, through the solenoid, a magnetic field is produced within the solenoid. If the solenoid is long
Page 292 :
Examination Papers, , 291, , and the successive insulated copper turns have, d c s, r, no gaps, then the magnetic field within the, solenoid is uniform; with practically no, magnetic field outside it. The reason is that the, B, p, q, solenoid may be supposed to be formed of a, l, large number of circular current elements. 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. 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., 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, 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., Let I be the current flowing in the solenoid, then by right hand rule, the magnetic field is, parallel to the axis of the solenoid., Field outside the solenoid: Consider a closed path abcd. 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., ®, , Field Inside the solenoid: Consider a closed path pqrs. The line integral of magnetic field B, along path pqrs is, , òpqrs, , ®, , ®, , B · dl = ò, , ®, , ®, , B · dl + ò, , pq, , qr, , ®, , ®, , B · dl + ò, , rs, , ®, , ®, , B • dl + ò, , sp, , ®, , ®, , For path pq, B and dl are along the same direction,, \, , òpq, , ®, , ®, , B · dl = ò B dl = Bl, ®, , ( pq = l say), , ®, , For paths qr and sp, B and d l are mutually perpendicular., \, , òqr, , ®, , ®, , B · dl = ò, , sp, , ®, , ®, , B · d l = ò B dl cos 90° = 0, , ®, , ®, , B · dl, , ...(i)
Page 293 :
292, , Xam idea Physics—XII, , For path rs, B = 0 (since field is zero outside a solenoid), \, , òrs, , ®, , ®, , B · dl = 0, , In view of these, equation (i) gives, , òpqrs, By Ampere’s law, , ®, , ®, , B · dl = ò, ®, , ®, pq, , ®, , B · dl = Bl, , ...(ii), , ®, , ò B · dl = m 0 ´ net current enclosed by path, , \, Bl = m 0 ( nl I ), \ B = m 0 nI, (b) In a toroid, magnetic lines do not exist outside the body./Toroid is closed whereas solenoid is, open on both sides./Magnetic field is uniform inside a toroid whereas for a solenoid, it is, different at the two ends and cenre., (Any one), Q, S, , Ä, , Ä, , Ä, , Ä, , Ä, , Ä, , Ä, , P, , Ä, , Ä, , O, , Ä, , Ä, , Ä, , (c) The magnetic field lines of toroid are circular having common centre. Inside a given solenoid,, the magnetic field may be made strong by, (i) passing large current and, (ii) using laminated coil of soft iron., w, 29. Working: When the armature coil is rotated in the, Armature coil, strong magnetic field, the magnetic flux linked with, b, the coil changes and the current is induced in the, c, coil, its direction being given by Fleming’s right, hand rule. Considering the armature to be in vertical, position and as it rotates in anticlockwise direction,, N B, S, Field magnet, the wire ab moves upward and cd downward, so that, a, the direction of induced current is shown in fig. In, d, the external circuit, the current flows along B1 R L B 2 ., B1, The direction of current remains unchanged during, R1, the first half turn of armature. During the second half, Brushes Load, Slip rings, revolution, the wire ab moves downward and cd, RL, upward, 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.
Page 294 :
Examination Papers, , 293, , If N is the number of turns in coil, f the frequency of rotation, A area of coil and B the magnetic, induction, 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., Current produced by an ac generator cannot be measured by moving coil ammeter; because the, average value of ac over full cycle is zero., The source of energy generation is the mechanical energy of rotation of armature coil., OR, (a) AC circuit containing pure inductance: Consider a coil of self-inductance L and negligible, ohmic resistance. An alternating potential difference is applied across its ends. The magnitude, and direction of AC changes periodically, due to which there is a continual change in, magnetic flux linked with the coil. Therefore according to Faraday’s law, an induced emf is, produced in the coil, which opposes the applied voltage. As a result the current in the circuit is, reduced. That is inductance acts like a resistance in ac circuit. The instantaneous value of, alternating voltage applied, ...(i), V = V0 sin wt, di, If i is the instantaneous current in the circuit and , the rate of change of current in the circuit, dt, at that instant, then instantaneous induced emf, di, e=-L, dt, According to Kirchhoff’s second law in closed circuit., 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), L, dt, L, Integrating with respect to time ‘t’,, 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, ö, æ p, = - 0 sin ç - wt ÷, wL, 2, ø, è, or, , i=, , V0, p ö, æ, sin ç wt - ÷, wL, 2 ø, è, , This is required expression for, p, æ, or, i = i 0 sin ç wt 2, è, , current, ö, ÷, ø, , …(ii), , ...(iii)
Page 295 :
294, , Xam idea Physics—XII, , V0, wL, is the peak value of alternating current, where, , i0 =, , ...(iv), , p, Also comparing (i) and (iii), we note that current lags behind the applied voltage by an angle ., 2, (b) (i) Induced emf e = B H vL, Where BH is horizontal component of earth’s magnetic field directed from S to N., (ii) West to east., (iii) East-send will be at higher potential., 30. If coherent sources are not taken, the phase difference between two interfering waves, will change, continuously and a sustained interference pattern will not be obtained. Thus, coherent sources, provide sustained interference pattern., Suppose two coherent waves travel in the same direction along a straight line, the frequency of, w, and amplitudes of electric field are a1 and a 2 respectively. If at any time t, the, each wave is, 2p, electric fields of waves at a point are y1 and y 2 respectively and phase difference is f, then, equation of waves may be expressed as, ...(i), y1 = a1 sin wt, ...(ii), y 2 = a 2 sin ( wt + f), According to Young’s principle of superposition, the resultant displacement at that point will be, ...(iii), y = y1 + y 2, Substituting values of y1 and y 2 from (i) and (ii) in (iii), 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), ...(iv), = ( a1 + a 2 cos f) sin wt + ( a 2 sin f) cos wt, Let, ...(v), a1 + a 2 cos f = A cos q, And, ...(vi), a 2 sin f = A sin q, where A and q are new constants., Then equation (iv) gives, ...(vii), y = A cos q sin wt + A sin q cos wt = A sin ( wt + q), This is the equation of the resultant disturbance. Clearly the amplitude of resultant disturbance is A, and phase difference from first wave is q. The values of A and q are determined by (v) and (vi)., Squaring (v) and (vi) and then adding, 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), , As cos 2 q + sin 2 q = 1, we get, 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
Page 296 :
Examination Papers, , \, , Amplitude, A = a12 + a 22 + 2a1 a 2 cos f, , 295, , ...(viii), , As the intensity of a wave is proportional to its amplitude i.e. 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. In interference, the frequencies of two waves are same and medium is same, therefore for convenience, we may, take, K =1, then the units of intensity I will not be watt / m 2 but arbitrary., \ Intensity of resultant wave, I = A 2 = a12 + a 22 + 2a1 a 2 cos f, , ...(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., Constructive Interference: For maximum intensity at any point cos f = + 1, phase difference f = 0, 2p , 4p , 6p ........., = 2np ( n = 0, 1, 2, ....), The maximum intensity,, , or, , I max = a12 + a 22 + 2a1 a 2 = ( a1 + a 2 ) 2, , ...(x), ...(xi), , l, l, ...(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 ., Destructive Interference: For minimum intensity at any point cos f = -1, Path difference, , D=, , phase difference, f = p , 3p , 5p , 7p ...., = ( 2n - 1) p , n = 1, 2, 3 ...., In this case the minimum intensity,, , or, , I min = a12 + a 22 - 2a1 a 2 = ( a1 - a 2 ) 2, , ...(xiii), ...(xiv), , l, ´ Phase difference, 2p, l, l, ...(xv), =, ´ ( 2n - 1) p = ( 2n - 1), 2p, 2, Clearly, 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 ., , Path difference,, , D=
Page 297 :
296, , Xam idea Physics—XII, I, , Imax, , Imin, –2l, , –l, , l, , O, x, , 2l, , From equations (xii) and (xvi) it is clear that the intensity 2a1 a 2 is transferred from positions of, minima to maxima. This implies that the interference is based on conservation of energy., Variation of Intensity of light with position x is shown in fig., Expression for Fringe Width: Let S1 and S 2 be two coherent sources separated by a distance d., Let the distance of the screen from the coherent sources be D. Let M be the foot of the, perpendicular drawn from O, the midpoint of S1 and S 2 on the screen. Obviously point M is, equidistant from S1 and S 2 . Therefore the path difference between the two waves at point M is, zero. Thus the point M has the maximum intensity. Consider a point P on the screen at a distance y, from M. Draw S1 N perpendicular from S1 on S 2 P., P, S1, S, , d, , O, S2, , q, , y, q, M, , N, , D, , The path difference between two waves reaching at P from S1 and S 2 is D = S 2 P - S1 P » S 2 N, As D > > 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 ,, sin q = 2, S1S 2, MP, In DMOP,, 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., \, OM, D, yd, ...(i), Path difference D = S 2 P - S1 P = S 2 N =, \, 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.e. D = nl
Page 298 :
Examination Papers, , 297, , yd, = nl , n = 0, 1, 2, 3, K, D, nDl, \, y=, ., d, This equation gives the distance of nth bright fringe from the point M. Therefore writing, y n for y, we get, nDl, ...(ii), yn =, ., d, (ii) Positions of dark fringes (or minima): For dark fringe or minimum intensity at P, the path, l, difference must be an odd number multiple of half wavelength. i.e. D = ( 2n - 1), 2, y.d, l, where n =1, 2, 3, ....., \, = ( 2n - 1), D, 2, ( 2n - 1) lD, 1 Dl, or, y=, = (n - ), ., 2d, 2 d, This equation gives the distance of nth dark fringe from point M. Therefore writing y n for y,, we get, 1 Dl, ...(iii), yn = ( n - ), 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. It is denoted by w., For Bright Fringes: If y n + 1 and y n denote the distances of two consecutive bright fringes, \, , from M, then we have, Dl, nDl, y n + 1 = ( n + 1), and y n =, d, d, \ Fringe width,, , Dl nDl Dl, ...(iv), =, ., d, d, d, and y n are the distances of two consecutive dark fringes from, , b = y n + 1 - y n = ( n + 1), , For Dark Fringes: If y n + 1, M, then we have, , 1 Dl, 1 Dl, yn + 1 = ( n + ), , yn = ( n - ), 2 d, 2 d, \ Fringe width, b = y n + 1 - y n, 1 Dl, 1 Dl, = (n + ), - (n - ), 2 d, 2 d, Dl, 1, 1 Dl, =, (n + - n + ) =, d, 2, 2, d, Thus, fringe width is the same for bright and dark fringes equal to, Dl, b=, d, When entire apparatus is immersed in water, the fringe-width decreases ( l w < l a )., , ...(v)
Page 299 :
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. 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., (ii) The new disturbances from each point spread out in all directions with the velocity of, light and are called the secondary wavelets., (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., Diffraction of light at a single slit: When monochromatic light is made incident on a single, slit, we get diffraction pattern on a screen placed behind the slit. The diffraction pattern, contains bright and dark bands, the intensity of central band is maximum and goes on, decreasing on both sides., Explanation: Let AB be a slit of width ‘a’ and a parallel beam of monochromatic light is, incident on it. According to Fresnel the diffraction pattern is the result of superposition of a, large number of waves, starting from different points of illuminated slit., , P, q, , A, Light from, source, , M1, , q, , O, M2 90o N, , C, q, , B, , 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 path difference between rays diffracted at points A and B,, D = BP - AP = BN, In D ANB , ÐANB = 90° and ÐBAN = q, BN, or BN = AB sin q, sin q =, \, AB, As AB = width of slit = a, \ Path difference,, ...(i), D = a sin q, To find the effect of all coherent waves at P, we have to sum up their contribution, each with a, different phase. This was done by Fresnel by rigorous calculations, but the main features may, be explained by simple arguments given below:
Page 300 :
Examination Papers, , 299, , At the central point C of the screen, the angle q is zero. Hence the waves starting from all, points of slit arrive in the same phase. This gives maximum intensity at the central point C., If point P on screen is such that the path difference between rays starting from edges A and B, is l , then path difference, l, a sin q = l Þ sin q =, a, l, ...(ii), If angle q is small,, sin q = q =, a, a, Minima: Now we divide the slit into two equal halves AO and OB, each of width . Now for, 2, a, every point, M 1 in AO, there is a corresponding point M 2 in OB, 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 = . 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. Thus equation (2) gives the angle of diffraction at, which intensity falls to zero. Similarly it may be shown that the intensity is zero for, nl, sin q =, , with n as integer. Thus the general condition of minima is, a, ...(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
Page 301 :
300, , Xam idea Physics—XII, , 2l, a, 2l l l, Angular width of first maximum, (b I ) =, - =, a a a, bI, 1, =, (b q ) C 2, , (b) Angular width of central maximum (b q ) C =, , Hence, the fringe width of the first diffraction fringe is half that of the central fringe., (c) If monochromatic light is replaced by white light, each diffraction band gets splited into, the number of coloured bands, the angular width of violet is least and that of red is, maximum., , CBSE (All India) SET–II, 1. 5 V., 2. X-rays are produced by sudden deceleration or acceleration of electrons./In an X-ray tube., 4. Angle of dip is zero at equator of earth’s surface., C, 9. In series, C S =, 3, In parallel, CP = 3C, CP, =9, CS, Þ, , 12., , 14., , Þ, 17., , C P = 9 ´ 2 mF = 18 mF, 1, C sV 2, Es, C, 2 1, 2, Required ratio =, =, = s = = ., Ep 1, C p 18 9, C pV 2, 2, Working of a transformer is based on the principle of mutual induction. Transformer cannot step, up or step down a dc voltage., Reason: No change in magnetic flux., Explanation: When dc voltage source is applied across a primary coil of a transformer, the, current in primary coil remains same, so there is no change in magnetic flux and hence no voltage, is induced across the secondary coil., Current in 2W resistor, I = 2 A, Applying Kirchhoff’s law along the path ACDB, VA + 2 + 2 ´ 2 - 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., The maximum range of propagation in this mode depends on, (i) transmitted power and, (ii) frequency (less than a few MHz)
Page 302 :
Examination Papers, , 301, , 21. The circuit represents OR gate., Truth table, , A, Y, B, , A, , B, , X=A, , Y=B, , Z, , 0, , 0, , 1, , 1, , 0, , 0, , 1, , 1, , 0, , 1, , 1, , 0, , 0, , 1, , 1, , 1, , 1, , 0, , 0, , 1, , CBSE (All India) SET–III, 1. 12 V, 3. Microwaves are produced by special vacuum tubes, namely, klystrons, Gunn diodes, magnetrons,, etc., 7. At magnetic equator., 12. Energy losses in a transformer are due to, (i) Flux leakage, (ii) Joule heating in resistance of winding, (iii) Eddy currents, (iv) Hysteresis, 14. Current in 2 W resistor is 3 A, VA + 3 + 3 × 2 – 6 = VB, Þ, VB = VA + 3 = 0 + 3 = 3 V, 16. When a wave propagates in a straight line, from the transmitting antenna to the receiving antenna,, its mode of propagation is called space wave communication., Frequency range: Above 40 MHz., a n g -1, 20. 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
Page 303 :
302, , Xam idea Physics—XII, , 22. y = A + B = A . B = AB, The circuit represents AND gate., Logic symbol:, , A, y, B, , Truth table:, A, 0, , B, 0, , Y, 0, , 1, , 0, , 0, , 0, , 1, , 0, , 1, , 1, , 1
Page 304 :
CBSE EXAMINATION PAPERS, FOREIGN–2011, Time allowed: 3 hours, , Maximum marks: 70, , 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–, 4pe o, , m 0 = 4p ´ 10 -7 TmA -1, 2, , 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. Define the term ‘threshold frequency’ in relations to photoelectric effects., 2. The peak value of e.m.f. in a.c. is E 0 . Write its (i) rms (ii) average value over a complete cycle., 3. Two insulated charged copper spheres A and B if identical size have charges q A and q B, respectively. 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. What are the new charges on, A and B?, 4. A narrow beam of protons and deuterons, each having the same momentum, enters a region of, uniform magnetic field directed perpendicular to their direction of momentum. What would be the, ratio of the circular paths described by them?, 5. What is the function of 'Repeater' in communication system?, 6. Draw the logic circuit of NAND gate and write its truth table., 7. How is the mean life of a radioactive sample related to its half life?, 8. Write two uses of microwaves.
Page 305 :
304, , Xam idea Physics—XII, , 9. Calculate the amount of work done in rotating a dipole, of dipole moment 3 ´ 10 -8 cm, from its, position of stable equilibrium to the position of unstable equilibrium, in a uniform electric field of, intensity 10 4 N/C., 10. Plot a graph showing temperature dependence of resistivity for a typical semiconductor. How is, this behaviour explained?, 11. When four hydrogen nuclei combine to form a helium nucleus, estimate the amount of energy in, MeV released in this process of fusion (Neglect the masses of electrons and neutrinos) Given:, (i) mass of 11 H = 1.007825 u, (ii) mass of helium nucleus = 4.002603 u, 1u = 931 MeV / c 2, 12. For an amplitude modulated wave, the maximum amplitude is found to be 10 V while the, minimum amplitude is 2.V. Calculate the modulation index. Why is modulation index generally, kept less than one?, 13. Draw a block diagram showing the important components in a communication system. What is the, function of a transducer?, 14. 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., (ii) Where on the surface of the Earth is the vertical component of the Earth's magnetic field, zero?, 15. Show how the equation for Ampere's circuital law, viz., , ò, , ® ®, , B . dl = m I, , 16. Current in a circuit falls steadily from 5.0 A to 0.0 A in 100 ms. If an average e.m.f. of 200 V is, induced, calculate the self-inductance of the circuit., 17., (a) You are required to select a carbon resistor of resistance 47 kW ± 10% from a large collection., 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., 18. Two identical parallel plate (air) capacitors C1 and C 2 have capacitances C each. The between, their plates is now filled with dielectrics as shown. If the two capacitors still have equal, capacitance, obtain the relation between dielectric constants K, K1 and K 2 ., C1, l/2, K, , K1, , l, K2, , d, , d, , C2
Page 306 :
Examination Papers, , 305, , 19. State the principal of the device that can build up high voltages of the order of a few million volts., Draw its labelled diagram. 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 why?, 20. Light of wavelength 2000 Å falls on a metal surface of work functions 4.2 eV. 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 6.5 eV, what will be the energy of, emitted electrons?, 21. Name the important process that occur during the formation of a p-n junction. Explain briefly,, with the help of a suitable diagram, how a p-n junction is formed. Define the term 'barrier, potential'., 22. The intensity at the central maxima (O) in a Young's double slit experiment is I 0. If the distance OP, I, equals one-third of the fringe width of the pattern, show that the intensity at point P would be 0 ., 4, P, S1, , y, , d, , O, S2, D, , OR, In the experiment on diffraction due to a single slit, show that, (i) the intensity of diffraction fringes decreases as the order (n) increases., (ii) angular width of the central maximum is twice that of the first order secondary maximum., 23. Find the position of the image formed of the object 'O’ by the lens combination given in the, figure., f = +10, , –10, , + 30 cm, , O, 30 cm, 5 cm, , 10 cm, , 24. Draw transfer characteristics of common emitter n-p-n transistor. Point out the region in which the, transistor operates as an amplifier. Define the following terms used in transistor amplifiers:, (i) Input resistance, (ii) Output resistance, (iii) Current amplification factor, 25., (i) Light passes through two polaroids P1 and P2 with axis of P2 making an angle q with the pass, axis of P1 . For what value of q is the intensity of emergent light zero?
Page 307 :
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 . Find a value of b for which the intensity of light emerging from P2 is 0 ,, 8, where I 0 is the intensity of light on the polaroid P1 ., 26. Using the postulates of Bohr's model of hydrogen atom, 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 )., 27. State the underlying principal of potentiometer., Describe briefly, giving the necessary circuit diagram, how a potentiometer is used to measure the, internal resistance of a given cell., 28., (a) Show that a planar loop carrying a current I, having N closely wound turns and area of, ®, , ®, , cross-section A, possesses a magnetic moment m = N I A ., ®, , 29., , 30., , (b) When this loop is placed in a magnetic field B, find out the expression for the torque acting, on it., (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, explain its working principal and deduce the, expression for the kinetic energy of the ions accelerated., (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 the nature and magnitude of the magnetic, force between. 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 )., (b) Show that mutual inductance of solenoid 1 due to solenoid 2, M 12 , is the same as that of 2 due, to 1 i.e., M 21 ., (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. Calculate the number of turns and, the current in the secondary in order to get output power at 220 V., 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). Deduce an expression for the average power dissipated over a cycle., (b) For circuits used for transporting electric power, a low power factor implies large power loss, in transmission. Explain., (c) Determine the current quality factor at resonance for a series LCR circuit with L = 1.00 mH,, 1.00 nF and R =100 W connected to an a.c. source having peak voltage of 100 V., (i) A plane wavefront approaches a plane surface separating two media. If medium 'one’ is, optically denser and medium 'two’ is optically rarer, using Huygens’ principle, explain and, show how a refracted wavefront is constructed., (ii) Hence verify Snell's law., (iii) When a light wave travels from rarer to denser medium, the speed decreases. Does it imply, reduction its energy? Explain.
Page 308 :
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. Trace the path of ray passing through the prism. Hence, derive an, expression for the refractive index of the material of the prism in terms of the angle minimum, deviation and its refracting angle., a, (ii) Three light rays red (R), green (G) and blue (B) are, incident on the right angled prism abc at face ab. The, B, refractive indices of the material of the prism for red, G, green and blue wavelengths are respectively 1.39, 1.44 R, and 1.47. Trace the paths of these rays reasoning out, 45°, the difference in their behaviour., b, c, , CBSE (Foreign) SET–II, Questions uncommon to Set-I., 1. The current flowing through a pure inductor of inductance 4 mH is i =12 cos 300 t ampere. What is, (i) rms and (ii) average value of the current for a complete cycle?, 2. Show the variation of photocurrent with collector plate potential for different frequencies but same, intensity of incident radiation., 3. Two insulated charged copper spheres A and B of identical size have charges q A and q B, respectively. When they are brought in contact with each other and finally separated, what are the, new charges on them?, 4. What is the function of a transmitter in a communication system?, 5. Write two uses of infrared rays., 7. Draw the logic circuit of AND gate and write its truth table., 10. Calculate the amount of work done in rotating a dipole, of dipole moment 5 ´ 10 -8 cm, from its, position of stable equilibrium to the position of unstable equilibrium, in electric field of intensity, 10 4 N/C., 12. For an amplitude wave, the maximum amplitude is found to be 12 V while the minimum the, amplitude is 2V. Calculate the modulation index. Why is modulation index generally kept low?, 13. You are given an air filled parallel plate capacitor C1 . The space between its plates is now filled, with slabs of dielectric constants K1 and K 2 as shown in C 2 . Find the capacitances of the, capacitor C 2 if area of the plates is A distance between the plates is d., d/2, , C2, , K1, , K2, d, , d
Page 309 :
308, , Xam idea Physics—XII, , 15. Current in a circuit steadily from 2.0 A to 0.0 A in 10 ms. If an average e.m.f. of 200V is induced,, calculate the self-inductance of the circuit., 27. Light of wavelength 2500 Å falls on a metal surface of work function 3.5 V. 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 5.5 eV, what will be the energy of, emitted electrons?, , CBSE (Foreign) SET–III, Questions uncommon to Set-I & II., 2. Show the variation of photocurrent with collector plate potential for different intensity but same, frequency of incident radiation., 3. Two insulated charged copper spheres A and B of identical size have charges q A and -3q A, respectively. When they are brought in contact with each other and then separated, what are the, new charges on them?, 4. The current flowing through a pure inductor of inductance 2 mH is i =15 cos 300 t ampere. What is, the (i) rms and (ii) average value of current for a complete cycle?, 6. Draw the logic circuit of NOT gate and write its truth table., 7. What is the function of a Receiver in a communication system?, 8. Write two uses of X-rays., 9. Calculate the amount of work done in rotating a dipole, of dipole moment 2 ´ 10 -8 cm, from its, position of stable equilibrium to the position of unstable equilibrium, in uniform electric field of, intensity 5 ´ 10 4 N/C., 12. You are given an air filled parallel plate capacitor C1 . The, space between its plates is now filled with slabs of, dielectric constants K1 and K 2 as shown in C 2 . Find the, capacitances of the capacitor C 2 if area of the plates is A, and distance between the plates is d., , C1, , C2, K1, , K2, , d, , d, , 14. Current in a circuit falls steadily from 3.0 to 0.0 A in 300 ms. If an average e.m.f. of 200 V is, induced, the self-inductance of the circuit., 15. For an amplitude modulated wave the maximum amplitude is found to be 15 V While the, minimum amplitude is 3V. Calculate the modulation index. Why is modulation index generally, kept less than one?, 26. Light of wavelength 2400 Å falls on a metal surface of work function 3.6 eV. 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 5.5 eV, what will be the energy of, emitted electrons?
Page 310 :
Examination Papers, , 309, , Solutions, CBSE (Foreign) SET–I, 1. Threshold frequency is defined as the minimum frequency of incident radiation which can cause, photoelectric emission. It is different for different metal ., 2. 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. New charge on A is A and New charge on B is A, 2, 4, 4. Charge on deutron ( q d ) = charge on proton ( q p ), qd = q p, Radius of circular path (r ) =, rµ, so,, , rp, rd, , 1, q, =, , 2, æ, çQ qvB = mv, ç, r, è, , p, Bq, , ö, ÷, ÷, ø, , [for constant momentum (P)], qd q p, =, =1, qp qp, , Hence,, r p : r d =1 :1, 5. A repeater which is a combination of a transmitter, on amplifier and a receiver which picks up, signal from the transmitter, amplifies and retransmits it to the receiver., 6. Logic circuit of NAND gate:, A, Y, B, , Truth table, Input, , Output, , A, , B, , Y, , 0, , 0, , 1, , 0, , 1, , 1, , 1, , 0, , 1, , 1, , 1, , 0, , 7. Mean life ( t) and half life (T1/ 2 ) are related as:, T, t = 1/ 2, 0.6931
Page 311 :
310, , Xam idea Physics—XII, , 8. Uses of microwaves:, (i) In long distance communications., (ii) In radar, 9. P = 3 ´ 10 -8 cm ; 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 - cos q 2 ), = (3 ´ 10 -8 ) (10 4 ) [cos 0° - cos180° ], = 3 ´ 10 -4 [1 - ( -1)], W = 6 ´ 10 -8 J, 10. 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. But the effect of increase in n, has higher impact than decrease of t. So, resistivity decreases with increase in temperature., 11. Energy released = Dm ´ 931 MeV, Dm = 4m ( 11 H) - m ( 42 He), Energy released ( Q) = [ 4. m ( 11 H) - m ( 42 He)] ´ 931 MeV, = [ 4 ´ 1.007825 - 4.002603] ´ 931 MeV = 26.72 MeV., 12. A max = 10V, A min = 2 V, Modulation index =, , A max - A min 10 - 2 8, =, = = 0 × 67, A max + A min 10 + 2 12, , Generally, the modulation index is kept less then one to avoid distortion., 13. Block diagram of communication system:, Communication System, Information Message, source, signal, , Transmitter, , Transmitted, Received, User of, Channel, Receiver, signal, signal, Message information, signal, , Noise, , Function of a transducer is to convert one form of energy into another form.
Page 312 :
Examination Papers, , 14., , 311, , (i) Magnetic lines of force form continuous closed loops because a magnet is always a dipole, and as a result, the net magnetic flux of a magnet is always zero., (ii) When a diamagnetic substance is placed in an external magnetic field, a feeble magnetism is, induced in opposite direction. So, 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., 15. Displacement current and generalised Ampere’s, I, Circuital Law: Consider a parallel plate capacitor,, being charged by a battery. A time varying current is, flowing through the capacitor. If we consider only the, conduction current I, then we apply Ampere’s, Circuital Law to two closed loops C1 and C2, then we, get, Ø Ø, , òCB × dl = m, 1, , 0, , C2, I, , ...(i), , I, , Ø Ø, , òCB × dl = 0, , and, , C1, , ...(ii), , 2, , Since there cannot be any conduction current in region between the capacitor plates. As C1 and C2, are very close, we must expect, , ò, , Ø Ø, , C1, , Ø Ø, , B × dl = ò C B ×dl, , ...(iii), , 2, , But this condition is violated by equations (i) and (ii). Hence Ampere’s Circuital Law seems to be, inconsistent in this case. Therefore, Maxwell postulated the existence of displacement current, which is produced by time varying electric field. If s (t) is the surface charge density on capacitor, s (t) q (t), , where A is area of, plates and q(t) is the charge, then time varying electric field E (t) =, =, e0, Ae 0, each plate., dg(t), dE, 1 dg(t), dE, or, =, = e0 A, dt Ae 0 dt, dt, dt, This is expression for displacement current (ld)., Applying Kirchhoff’s first law at power P, we get I = Id, Hence, equation (i) and (ii) take the forms, , ò, , Ø Ø, , C1, , B × dl = m 0 I and, , ò, , Ø Ø, , C2, , B × dl = m 0 I d = m 0 I
Page 313 :
312, , Xam idea Physics—XII, , The total current is the sum of the conduction current and displacement current. Thus, modified, form of Ampere’s circuital law is, Ø Ø, dE ö, æ, ò B. dl = m 0 ( I + I d ) = m 0 çè I + e 0 A dt ÷ø, But, EA = Electric flux fE, Ø Ø, df E, \, ò B. dl = m 0 ( I + m 0 e 0 dt ), 16. Change in current ( DI ) = ( 0 × 0 - 5 × 0) A = - 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., , Id, , I, P, , ( ), , (f = LI), , ( 200) . (100 ´ 10 -3 ), Dt, =( - 5 × 0), DI, , L = 4×0 H, 17. (a) Resistance = 47 kW ± 10% = 47 ´ 10 3 W ± 10%, Sequence of colour should be:, Yellow, Violet, Orange and Silver, (b) (i) Very low temperature coefficient of resistance., (ii) High resistivity, 18. Let A ® area of each plate., Î A, Let initially C1 = C = 0 = C 2, d, After inserting respective dielectric slabs:, C1¢ = KC, Î ( A / 2) K 2 Î0 ( A / 2), and, C ¢2 = K1 0, +, d, d, Î0 A, =, ( K1 + K 2 ), 2d, C, C ¢2 = ( K1 + K 2 ), 2, From (i) and (ii), C1¢ = C ¢2, C, KC = ( K1 + K 2 ), 2, , …(i), , …(ii)
Page 314 :
Examination Papers, , 1, ( K1 + K 2 ), 2, 19. This device is Van de Graaff generator., Principle: Suppose we have a large spherical conducting shell of, radius R, carrying charge Q. The charge spreads uniformly over, whole surface of the shell. Now suppose a small conducting, sphere of radius ‘ r ’ is introduced inside the spherical shell and, placed at its centre, so that both the sphere and shell have same, centre O. The electric field in the region inside the small sphere, and large shell is due to charge +q only, so electric field strength, at a distance x from the centre O is, q, 1, E=, , directed radially outward, 4pe 0 x 2, , 313, , K=, , Q, R, O, , r, +q, , P, x, , E, , The potential difference between the sphere and the shell, V (r ) - U ( R) = - ò, =-ò, , r, R, r, R, , ®, , ®, , E .d x, , r, , q, q, q é x -1 ù, 1, dx = ú =, ê, 2, 4pe 0, 4pe 0 x, 4pe 0 êë -1 ûú, R, , æ1 1 ö, ç - ÷, èr R ø, , This is independent of charge Q on the large, spherical shell. As r < R ; V (r ) - U ( R) is positive. As, charge flows from higher to lower potentials, S, therefore, if we connect the small sphere and large, C2, shell by a conducting wire, the charge flows from, P2, sphere to outer shell whatsoever the charge on outer, shell may be. This forms the principle of Van de, Graaff generator. The maximum charge that may be, given to outer shell which may cause discharge in, air., Working: When comb C1 is given very high, potential, then it produces ions in its vicinity, due to, action of sharp points. The positive ions, so produced,, A, B, get sprayed on the belt due to the repulsion between, positive ions and comb C1 . These positive ions are, carried upward by the moving belt. The pointed end, of C 2 just touches the belt. The comb C 2 collects, HTS, positive charge from the belt which immediately, moves to the outer surface of sphere S. As the belt, C1 P1, goes on revolving, it continues to take (+ ) charge, upward, which is collected by comb C 2 and, transferred to outer surface of sphere S. Thus the, outer surface of metallic sphere S gains positive charge continuously and its potential rises to a very, high value.
Page 315 :
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
Page 316 :
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, y, tan q = 1, D, y1, Þ, q=, D, l, Þ, D = y1 = y 2, a, lD, ., Hence, whole width on secondary maxima on one side is, d
Page 317 :
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, , Central max, D, y2, , 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 ¥, I1, , O, , 30 cm, , 5 cm, , 10 cm, , I3
Page 318 :
Examination Papers, , 24., , 317, , Active, region, V0 Cut off, 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 = çç, ÷, è DI E øVCB = constant, 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.,, r 0 = çç, ÷, è DI C ø I E = constant, 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 = çç, ÷, è DI E øVC = constant, 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 = çç, ÷, è DI B øVC = constant, 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)
Page 319 :
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 êë, 2, úû, 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=, , =-, , 1 Ze 2, 4p e 0 r, , Total energy, E = KE + PE, =, , 1 Z e 2 æç, 1 Z e2, + ç 4p e 0 r, 4p e 0 2r, è, , E =-, , 1 Z e2, 4p e 0 2r, , ö, ÷, ÷, ø
Page 321 :
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 æç - Rc ö÷, n=, ni2 çè n f2 ÷ø, é1, 1 ù, ú, n = Rc ê, ê n f2 ni2 ú, û, ë, 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, , 28., , æ l, ö, ö, æ e, \ Internal resistance of cell, r = ç, - 1 ÷ R = çç 1 - 1 ÷÷ R, V, l, ø, è, è 2, ø, From this formula r may be calculated., (a) Torque ( t) on the loop is given by:, ®, , ®, , ®, , t = NI A ´ B, , which can be written as,
Page 322 :
Examination Papers, ®, , ®, , 321, , ®, , t =M ´ B, , ®, , where, M is the magnetic dipole moment given by, ®, , ®, , 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, , N, , 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, , F1, I, F3, , ®, , P, S, F4, , q, , ®, , and SP are F1 , F2 , F3 and F4 respectively., , N, b, , q, , The sides QR and SP make angle ( 90° - q), with the direction of magnetic field. Therefore, ®, , B, , I (Downward), , ®, , N', , each of the forces F2 and F4 acting on these, , b sin q, , sides, has, same, magnitude, F¢ = Blb sin ( 90° - q) = Blb cos q. According, ®, , of, , I, (Upward), , lo o, , p, , F1=IlB, , ax, is, , ®, , B, R, , ®, , ®, , q, , 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, , Axis of loop, or normal to loop, , F2, , F3=IlB, , ®, , 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.
Page 323 :
322, , Xam idea Physics—XII, , \ Moment of couple or Torque,, 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, t = IAB sin q, If the loop contains N-turns, then t = NI AB sin q, ®, , ®, , ®, , In vector form t = NI A ´ B., Direction of torque is perpendicular to direction of area of loop as well as the direction of, ®, , ®, , magnetic field i.e., along I A ´ B ., (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., Value of R is given by:, 15, V, R=, -G =, - 50, Ig, 5 ´ 10 -3, = 3000 - 50, = 29 50 W = 2.95 kW., (a), , OR, Principle: The positive ions produced, from a source are accelerated. Due to, Dee-1, the presence of perpendicular magnetic, field the ion will move in a circular, path. The phenomenon is continued till, S, R.F., 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., Expression for K.E. attained:, If R be the radius of the path and v max the velocity of the ion when it leaves the periphery,, then, qBR, v max =, m, The kinetic energy of the ion when it leaves the apparatus is,, 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, it crosses the gap twice, therefore if it completes n-revolutions before, emerging the does, the kinetic energy gained, K. E. =
Page 324 :
Examination Papers, , Thus, , 323, , = 2nqV, q2B 2R 2, K.E. =, = 2nqV, 2m, , (b) I 1 = 2 A , I 2 = 5 A , 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., , (a) Suppose there are two coils C1 and C 2 . The current I 1 is flowing in primary coil C1 ; due to, which an effective magnetic flux F2 is linked with secondary coil C 2 . By experiments, ...(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. From (i), F, M= 2, I1, If I 1 = 1 ampere, M = F2, i.e., the mutual inductance between two coils is numerically equal to the effective flux, linkage with secondary coil, when current flowing in primary coil is 1 ampere., 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., Number of turns per unit length in order, N, l, (primary) solenoid, n = 1 × If I 1 is the, l, current flowing in primary solenoid, the magnetic field produced within this solenoid., m N I, ...(ii), B1 = 0 1 1, l, The flux linked with each turn of inner solenoid coil is f2 = B1 A 2 , where A 2 is the, cross-sectional area of inner solenoid. The total flux linkage with inner coil of N 2 -turns., 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, 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,, then M = m 0 n1 N 2 pr 22 ., (b) Due to current I 1 through solenoid of radius r1 , flux linked with second solenoid, N 2 f2 = M 21 I 1, , …(i)
Page 325 :
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, , Hence, , m 0 N1 I 1, ., l, , as l >> r 2 and r 2 >> r1, , Using (1), m 0 N1 N 2, pr12, l, which is same as expression of mutual inductance derived in part (a) above., \, M 21 = M 12, (c) V p = 2200V , I 0 = 5A, N p = 4000, Vs = 220 V , N s = ? I s = ?, Vs I p N s, =, =, Vp I s N p, M 21 =, , 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,, P = Vi, = Vm sin wt . i 0 sin ( wt + f) = Vm i m sin wt sin ( wt + f), 1, = Vmi m 2 sin wt . sin ( wt + f), 2, From trigonometric formula, 2 sin A sin B = cos ( A - B) - cos ( A + B), 1, P = Vmi m [cos ( wt + f - wt) - cos ( wt + f + wt)], \ Instantaneous power,, 2, 1, … (i), = Vm i m [cos f - cos ( 2wt + f)], 2, Average power for complete cycle
Page 326 :
Examination Papers, , 325, , ®, , 1, Vm i m [cos f - cos ( 2wt + f) ], 2, where cos ( 2wt + f) is the mean value of cos ( 2wt + f) over complete cycle. But for a, complete cycle, cos ( 2wt + f) = 0, P=, , \ Average power,, ® 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 . If cos f is small, then current considerably increases when, voltage is constant. Power loss, we know is I 2 R. Hence, power loss increases., (c) I v = ? , Q = ?, L = 1 × 00 mH = 1 ´ 10 -3 H, C = 1.00 nF = 1 ´ 10 -9 F, R =100 W , E 0 = 100 V, I0 =, , \, , E0, 2, , =, , E0, Z, , 1 ö, æ, R 2 + ç wL ÷, wC ø, è, V 100, I= =, R 100, I 0 =1 A, I, 1, 2 1.44, Iv = 0 =, =, =, = 0 × 707 A, 2, 2, 2, 2, , 1 ü, ì, ï at resonance wL =, ï, í, wC ý, ïî Hence Z = R, ïþ, , I v = 0 × 707 A, Q=, , 30., , 1, R, , L, 1 1 × 0 ´ 10 -3, 1, =, =, ´ 10 3 = 10, 9, C 100 1 × 0 ´ 10, 100, , Q =10, (i) When a wave starting from one homogeneous, medium enters the another homogeneous medium, it, is deviated from its path. This phenomenon is called, refraction. In transversing from first medium to, another medium, the frequency of wave remains, unchanged but its speed and the wavelength both are, changed. Let XY be a surface separating the two, media ‘1’ and ‘2’. Let v1 and v 2 be the speeds of, waves in these media., , B, i, X, , 90o, i, , B', r, , A, , o, , 90, A', , r, , Y
Page 327 :
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. Clearly BB¢ = v1t. As the wavefront AB advances, it, strikes the points between A and B¢ of boundary surface. According to Huygen’s principle,, secondary spherical wavelets originate from these points, which travel with speed v1 in the first, medium and speed v 2 in the second medium., First of all secondary wavelet starts from A, which traverses a distance AA¢ ( = v 2 t) in second, medium in time t. In the same time-interval t, the point of wavefront traverses a distance, BB¢ ( = v1t) in first medium and reaches B¢ , from, where the secondary wavelet now starts., Clearly BB¢ = v1t and AA¢ = v 2 t., 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 incident wavefront AB advances, the secondary wavelets start from, points between A and B¢ , one after the other and will touch A¢ B¢ simultaneously. According, to Huygen’s principle A¢ B¢ is the new position of wavefront AB in the second medium. Hence, A¢ B¢ will be the refracted wavefront., (ii) Proof of Snell’s law of Refraction using Huygen’s wave theory: When a wave starting, from one homogeneous, First law: As AB, A¢ B ¢ and surface XY are in the plane of paper, therefore the perpendicular, drawn on them will be in the same plane. As the lines drawn normal to wavefront denote the, rays, therefore we may say that the incident ray, refracted ray and the normal at the point of, incidence all lie in the same plane., This is the first law of refraction., Second law: Let the incident wavefront AB and refracted wavefront A¢ B¢ make angles i and r, respectively with refracting surface XY., In right-angled triangle AB¢ B, Ð ABB¢ = 90°, vt, BB¢, ...(i), sin i = sin Ð BAB¢ =, = 1, \, AB¢, AB¢, Similarly in right-angled triangle AA¢ B ¢ , Ð AA¢ B ¢ = 90°, AA¢ v 2 t, ...(ii), sin r = sin Ð AB¢ A¢ =, \, =, AB¢ AB¢, Dividing equation (i) by (ii), we get, v, sin i, ...(iii), = 1 = constant, sin r, v2, As the rays are always normal to the wavefront, therefore the incident and refracted rays, make angles i and r with the normal drawn on the surface XY i.e. i and r are the angle of, incidence and angle of refraction respectively. 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. This is the second law of, refraction, and is called the Snell’s law., (iii) No. Because energy of wave depends on its frequency and not on its speed.
Page 328 :
Examination Papers, , 327, , OR, (i) Let PQR be the principal section of the prism. The refracting angle of the prism is A., P, A ray of monochromatic light EF is incident on, face PQ at angle of incidence i1 . The refractive, A, index of material of prism for this ray is n. 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. The angle of, q 2 G, refraction for this face is r1 . 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 incidence on this face is r 2 (into prism) and angle of refraction, (into air) is i 2 . The incident ray EF and emergent ray GH when produced meet at O. The angle, between these two rays is called angle of deviation ‘d‘., Ð OFG = i1 - r1, and Ð OGF = i 2 - r 2, In D FOG, d is exterior angle, \, d = Ð OFG + Ð OGF = (i1 - r1 ) + (i 2 - r 2 ), ...(i), = (i1 + i 2 ) - (r1 + r 2 ), The normals FN1 and GN 2 on faces PQ and PR respectively, when produced meet at N. Let, Ð FNG = q, In D FGN,, ...(ii), r1 + r 2 + q = 180°, In quadrilateral PFNG, Ð PFN = 90° , Ð PGN = 90°, ...(iii), \, A + 90° + q + 90° = 360° or A + q = 180°, Comparing (ii) and (iii),, ...(iv), r1 + r 2 = A, Substituting this value in (i), we get, ...(v), d = i1 + i 2 - A, or, ...(vi), i1 + i 2 = A + d, sin i1 sin i 2, ...(vii), From Snell’s law n =, =, sin r1 sin r 2, For minimum deviation i1 and i 2 become coincident, i.e.,i1 = i 2 = i (say), So from (vii), r1 = r 2 = r (say), Hence from (iv) and (vi), 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,, n=, =, sin r, æ A ö, sin ç, ÷, è 2 ø
Page 329 :
328, , Xam idea Physics—XII, a, , (ii), , a, B, Blue, , G, R, , Green, R, , Red, 90°, , 45°, , 45°, , b, , b, , c, , Angle of incidence at face ac for all three colours,, i = 45°, Refractive index corresponding to critical angle 45° is, 1, m=, = 2 = 1× 414, sin 45°, , c, B, , G, , The ray will be transmitted through face ‘ ac’ if i < i c . This condition is satisfied for red colour, (m = 1× 39). So only red ray will be transmitted, Blue and Green rays will be totally reflected., , CBSE (Foreign) SET–II, 1. Peak value (i 0 ) = 12 A, (i) i rms =, (ii) i av, , i0, , 2, =0A, , =, , 12, 2, , =, , 12, 2, , ´, , 2, 2, , =6 2A, , n03, , Photo current, , 2., , n 02, n01, , V03, , V02, , V01, , n03 >n02 >n01, Collector potential, , qA + qB, ., 2, 4. 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., , 3., , The charge on both spheres would be =
Page 330 :
Examination Papers, , 329, , 5., , (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., 7. Logic circuit of AND gate:, Truth Table, Inputs, A, Y, B, , Output, , A, , B, , Y, , 0, , 0, , 0, , 0, , 1, , 0, , 1, , 0, , 0, , 1, , 1, , 1, , 10. Dipole moment P = 5 ´ 10 -4 Cm., Electric field ( E) =10 4 N/C, At stable equilibrium q1 = 0°, At unstable equilibrium q 2 = 180°, Work done = - PE (cos q 2 - cos q1 ), = - (5 ´ 10 -8 ) (10 4 ) (cos 180 - cos 0), = 10 ´ 10 -4 J, 12. A max = 12 V, A min = 2 V, , A max - A min, A max + A min, 12 - 2 10, =, = = 0 × 714, 12 + 2 4, , Modulation index =, , To minimize distortion of signal due to noise signal from atmosphere and electrical disturbances, modulation index is kept less than one., e A, 13. 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
Page 332 :
Examination Papers, , 331, , 4. 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. Logic circuit of NOT gate, A, Input, , Output, , Truth Table, Input, A, , Output, Y, , 0, , 1, , 1, , 0, , 7. The function of receiver is to receive the modulated wave and it demodulates, amplifies and, reproduces the original message signal from it., 8. (i) In the detection of fracture, deformity of the bones/skeletal system., (ii) In study of crystal structure., 9. Dipole moment ( P) = 2 ´ 10 -8 Cm, Electronic field ( E) = 5 ´ 10 4 N/C, q1 = 0° , q 2 = 180°, Work done = PE (cos q1 - cos q 2 ), , = ( 2 ´ 10 -8 ) (5 ´ 10 4 ) [cos 0 - cos 180], = (10 ´ 10 -4 ) [1 - ( -1)), = 20 ´ 10 -4 J., , 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. Charge in current ( D I ) = ( 0 × 0 - 3 × 0) A = - 3 × 0 A, , 12. C1 =, , Time taken ( D t) = 300 ms = 300 ´ 10 -3 s, Df, Induced emf ( e) = Dt, - D ( LI ), =, Dt
Page 333 :
332, , Xam idea Physics—XII, , DI, DI, Dt, L = -e., DI, e=-L, , =, , - [ 200] [300 ´ 10 -3 ], [ -3 × 0 ], , L = 20. 0 H, 15. A max = 15 V, A min = 3 V, - A min, A, Modulation index ( m) = max, A max + A min, 15 - 3 12, =, = = 0 × 67, 15 + 3 18, The modulation index is kept low to reduce distortion., 26. l = 2400 Å = 2400 ´ 10 -10 m, work function (W0 ) = 3 × 6 eV, hc, K.E. =, - W0, l, ù, é( 6 × 63 ´ 10 -34 ) (3 ´ 108 ), 1, =ê, ´, - 3 × 6ú eV, -10, -19, 2400 ´ 10, 1 × 6 ´ 10, úû, êë, = (5 × 18 - 3 × 6) eV, = 1 × 58 eV, (i) K.E. of fastest electron = 1 × 58 eV, (ii) K.E. of slowest electron = 0 eV, If the same light (having energy 5. 18 eV) falls on the surface of work function 5. 5 eV, then, no photoelectron will emit.
Page 334 :
CBSE EXAMINATION PAPERS, DELHI–2012, Time allowed: 3 hours, , Maximum marks: 70, , General Instructions: As given in CBSE Examination Papers Delhi–2011., , CBSE (Delhi) Set–I, 1. When electrons drift in a metal from lower to higher potential, does it mean that all the free, electrons of the metal are moving in the same direction?, 2. The horizontal component of the earth’s magnetic field at a place is B and angle of dip is 60°., What is the value of vertical component of earth’s magnetic field at equator?, 3. Show on a graph, the variation of resistivity with temperature for a typical semiconductor., 4. Why should electrostatic field be zero inside a conductor?, 5. Name the physical quantity which remains same for microwaves of wavelength 1 mm and UV, radiations of 1600 Å in vacuum., 6. 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. 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., 1, I, 2, , 8. State de-Broglie hypothesis., 9. A ray of light, incident on an equilateral glass prism (m g = 3 ) moves parallel to the base line of, the prism inside it. Find the angle of incidence for this ray., 10. Distinguish between ‘Analog and Digital signals’., OR, Mention the function of any two of the following used in communication system:, (i) Transducer, (ii) Repeater, (iii) Transmitter, (iv) Bandpass Filter, 11. A cell of emf E and internal resistance r is connected to two external resistances R1 and R2 and a, perfect ammeter. The current in the circuit is measured in four different situations:, (i) without any external resistance in the circuit., (ii) with resistance R1 only, (iii) with R1 and R2 in series combination, (iv) with R1 and R2 in parallel combination., The currents measured in the four cases are 0.42 A, 1.05 A, 1.4 A and 4.2 A, but not necessarily in, that order. Identify the currents corresponding to the four cases mentioned above.
Page 335 :
334, , Xam idea Physics—XII, , 12. The susceptibility of a magnetic material is –2.6 × 10–5. Identify the type of magnetic material and, state its two properties., 13. 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., Q, I, P, I, , 14. When an ideal capacitor is charged by a dc battery, no current flows. However, when an ac source, is used, the current flows continuously. How does one explain this, based on the concept of, displacement current?, 15. 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., 16. Define self-inductance of a coil. Show that magnetic energy required to build up the current I in a coil of, 1, self inductance L is given by LI2., 2, 17. The current in the forward bias is known to be more (~mA) than the current in the reverse bias, (~mA). What is the reason, then, to operate the photodiode in reverse bias?, 18. 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, about an axis, passing through the centre and perpendicular to the plane of the ring. A constant and uniform, magnetic field B parallel to the axis is present everywhere. Deduce the expression for the emf, between the centre and the metallic ring., 19. The figure shows a series LCR circuit with L = 5.0 H, C = 80 mF, R = 40 W connected to a variable, frequency 240V source. Calculate., R, , C, L, , (i) The angular frequency of the source which drives the circuit at resonance., (ii) The current at the resonating frequency., (iii) The rms potential drop across the capacitor at resonance.
Page 336 :
Examination Papers, , 20. A rectangular loop of wire of size 4 cm × 10 cm carries a steady current of, 2A. A straight long wire carrying 5A current is kept near the loop as, shown. If the loop and the wire are coplanar, 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., , I = 5A, , 2A, , 335, , 4 cm, , 10 cm, , 2A, , 21., , (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., (b) The electron in hydrogen atom is initially in the third excited state., What is the maximum number of spectral lines which can be emitted when it finally moves to, the ground state?, 22. In the figure a long uniform potentiometer wire AB is having a constant potential gradient along its, length. 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. Find (i) e1 / e 2 and (ii) position of, null point for the cell e1 ., How is the sensitivity of a potentiometer increased?, , 300 cm, 120 cm, A, , e1, , B, , e2, e1, , e2, , OR, Using Kirchoff’s rules determine the value of unknown resistance R in the circuit so that no, current flows through 4 W resistance. Also find the potential difference between A and D., F, , 1W, , E, , D, 4W, R, , 1W, 6V, I, A, , 9V, , B, , 3V, , C
Page 337 :
336, , Xam idea Physics—XII, , 23., , 24., , 25., , 26., , 27., , 28., , 29., , 30., , (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., Write any two factor which justify the need for modulating a signal., Draw a diagram showing an amplitude modulated wave by superposing a modulating signal over, a sinusoidal carrier wave., Write Einstein’s photoelectric equation. State clearly how this equation is obtained using the, photon picture of electromagnetic radiation., Write the three salient features observed in photoelectric effect which can be explained using this, equation., (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, the intensity, of light at a point on the screen where path difference is l, is K units. Find out the intensity of, light at a point where path difference is l / 3., Use Huygen’s principle to explain the formation of diffraction pattern due to a single slit, illuminated by a monochromatic source of light., When the width of slit is made double the original width, 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., Draw a schematic diagram and explain the working of this device., Is there any restriction on the upper limit of the high voltages set up in this machine? Explain., OR, (a) Define electric flux. Write its S.I. units., (b) Using Gauss’s law, prove that the electric field at a point due to a uniformly charged infinite, plane sheet is independent of the distance from it., (c) How is the field directed if (i) the sheet is positively charged, (ii) negatively charged?, Define magnifying power of a telescope. Write its expression., A small telescope has an objective lens of focal length 150 cm and an eye piece of focal length, 5 cm. If this telescope is used to view a 100 m high tower 3 km away, find the height of the, final image when it is formed 25 cm away from the eye piece., 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 1.25 cm and 5 cm, respectively. Find the position of the object relative to the objective in order to obtain an angular, magnification of 30 in normal adjustment., Draw a simple circuit of a CE transistor amplifier. Explain its working. Show that the voltage gain, b R, AV, of the amplifier is given by A V = - ac L ,where b ac is the current gain, RL is the load, ri, resistance and ri is the input resistance of the transistor. What is the significance of the negative, sign in the expression for the voltage gain?
Page 338 :
Examination Papers, , 337, , OR, (a) Draw the circuit diagram of a full wave rectifier using p-n junction diode. Explain its working, and show the output, input waveforms., (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. Why must electrostatic field be normal to the surface at every point of a charged conductor?, 6. Predict the direction of induced current in a metal ring when the ring is moved towards a straight, conductor with constant speed v. The conductor is carrying current I in the direction shown in the, figure., I, , v, , 10. Derive the expression for the self inductance of a long solenoid of cross sectional area A and, length l, having n turns per unit length., 14. The susceptibility of a magnetic material is 2.6 × 10–5. Identify the type of magnetic material and, state its two properties., 16. Two identical circular loops, P and Q, each of radius r and carrying currents I and 2I respectively, are lying in parallel planes such that they have a common axis. 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., 2r, , P, , O, , I, , Q, , 2I
Page 339 :
338, , Xam idea Physics—XII, , 20. A series LCR circuit with L = 4.0 H, 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;, (ii) the current at the resonating frequency;, (iii) the rms potential drop across the inductor at resonance., 21., (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, the intensity of light at a point on the screen where, path difference is l, is K units. Find out the intensity of light at a, 2l, point where path difference is ., 3, 22. A rectangular loop of wire of size 2 cm × 5 cm carries a steady current of, 1 A. A straight long wire carrying 4 A current is kept near the loop as, shown in the figure. If the loop and the wire are coplanar, 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., , 2 cm, , 5 cm, , 1A, , 1cm, , 27. Name the three different modes of propagation of electromagnetic waves. Explain, using a proper, diagram the mode of propagation used in the frequency range above 40 MHz., , CBSE (Delhi) Set–III, Questions uncommon to Set–I & II, 6. Why is electrostatic potential constant throughout the volume of the conductor and has the same, value (as inside) on its surface?, 8. Predict the direction of induced current in metal rings 1 and 2 when current I in the wire is steadily, decreasing?, 1, I, 2
Page 340 :
Examination Papers, , 339, , 9. The relative magnetic permeability of a magnetic material is 800. Identify the nature of magnetic, material and state its two properties., 12. Define mutual inductance between two long coaxial solenoids. 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, P and Q, each of radius r and carrying equal currents are kept in the, parallel planes having a common axis passing through O. 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., Find the magnitude of the net magnetic field at O., 2r, , P, , O, , Q, , I, , I, , 21. Name the three different modes of propagation of electromagnetic waves. Explain, using a proper, diagram the mode of propagation used in the frequency range from a few MHz to 40 MHz., 23. A rectangular loop of wire of size 2.5 cm × 4 cm carries a steady current of 1 A. A straight wire, carrying 2 A current is kept near the loop as shown. If the loop and the wire are coplanar, 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., 2A, , 2.5 cm, , 1A, , 4 cm, , 2cm, , 25. The figure shows a series LCR circuit with L = 10.0 H, C = 40 mF, R = 60 W connected to a, variable frequency 240 V source, calculate, (i) the angular frequency of the source which drives the circuit at resonance,, (ii) the current at the resonating frequency,, (iii) the rms potential drop across the inductor at resonance., R, , C, L
Page 341 :
340, , Xam idea Physics—XII, , Solutions, CBSE (Delhi) Set–I, 1. No., 2. Zero, 3., r, , t, , 4. The charge inside the conductor is zero, so electric field is zero., 5. Velocity (c = 3×108 m/s), This is because both are electromagnetic waves., 6. When m L = m g, where m L = Refractive index of liquid and mg = Refractive index of glass, 7., 1, , Clockwise, , 2, , Anti clockwise, , 8. According to hypothesis of de Broglie "The atomic particles of matter moving with a given, velocity, can display the wave like properties.", h, (mathematically), i.e.,, l=, mv, 9., , 60°, i, , r 30°, 60°, , From the figure, we see, r = 30°
Page 342 :
Examination Papers, , 341, , We know, sin i, sin r, , Þ, , n 21 =, , Þ, , sin i = 3 sin 30° = 3 ´, , Þ, , 3=, , sin i, sin 30°, , 1, 2, , Þ, i = 60°, 10. Analog signals: They are the continuous variations of voltage or current., 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., (ii) Repeater: It is a combination of a receiver and a transmitter., (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., (iv) Bandpass filter: A bandpass filter blocks lower and higher frequencies and allows only a, band of frequencies to pass through., e, 11., (i) i =, r, where e = emf, r = Internal resistance, In this situation, effective resistance of circuit is minimum so current is maximum., So, i = 4.2 A, e, (ii) i =, R1 + r, Here, effective resistance is more than (i) and (iv) but less than (iii), So, i = 1.05 A, e, (iii) i =, r + R1 + R 2, In this situation effective resistance is maximum so current is minimum., So, i = 0.42 A, e, (iv) i =, R R, r+ 1 2, R1 + R 2, In this situation, the effective resistance is more than (i) but less than (ii) and (iii)., Hence, i = 1.4 A, 12. The magnetic material having negative susceptibility is diamagnetic in nature., Two properties:, (i) This material expels the magnetic field lines., (ii) They have the tendency to move from stronger to weaker part of the external magnetic, field.
Page 343 :
342, , Xam idea Physics—XII, , 13., Q, I, , BP, B, BQ, q = 45°, , P, I, , Bp ® directed vertically upward, BQ ® horizontally directed, \, , B = B 2p + BQ2, , We have, , B P = BQ =, , m 0I, Þ, 2R, , B = 2 Bp =, , 2, , m 0I, 2R, , Þ, , B=, , m 0I, , 2R, The net magnetic field is directed at angle of 45° with either of the fields., 14. When an ideal capacitor is charged by dc battery, charge flows (momentarily) till the capacitor, gets fully charged., dq, keep on flowing in the, When an ac source is connected then conduction current i c =, dt, connecting wire. Due to changing current, charge deposited on the plates of the capacitor changes, with time. 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., As we know, displacement current, df, and, id = ic at all instants., id = e 0 E, dt, 1, 1, and V µ, 15. Here e µ, 2, r, r, Y, 3.0, 2.5, , E, V, , e or V, , 2.0, 1.5, 1.0, , V, , 0.5, , E, 0.5, , 1.0, , 1.5, r, , 2.0, , 2.5, , 3.0, , X
Page 344 :
Examination Papers, , 343, , 16. Self inductance of a coil is numerically equal to the magnetic flux linked with the coil when the, current through coil is 1A., Energy stored in an inductor:, Consider a source of emf connected to an inductor L. As the current starts growing, 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, then the amount of, work done by the source, is given by, di, dw = e i dt = L i dt = Li di, dt, 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, I, , I, , I, , éi 2 ù, W = ò Li di = L ò i di = L ê ú, êë 2 úû 0, 0, 0, 1, W = LI2, 2, This work done gets stored in the inductor in the form of energy., 1, \, U = LI2, 2, 17. Consider the case of n-type semiconductor. The majority carrier (electron) density is larger than, the minority hole density, i.e., n >> p., On illumination, the no. of both types of carriers would equally increase in number as, n' = n + Dn, p' = p + Dp, But Dn = Dp and n >> p, Dp, Dn, <<, (fractional change in minority, Hence, the fractional change in majority carrier, i.e.,, 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., Hence photodiodes are used in reverse bias condition for measuring light intensity., df, 18. 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. Let q be the angle between the rod and the radius, of the circle at P at time t., ×, ×, ×, ×, ×
Page 346 :
Examination Papers, , ni = 2 to, nf = 1, Total no. of transitions = 6, n=4, n=3, n=2, , n=1, , 22., , 23., , (i) Let k = potential gradient in V/cm, e1 + e 2 = 300k, e1 – e 2 = 120k, e, 7, We can solve, 1 =, e2 3, (ii) From equation (i), e1 + e 2 = 300k, 3, e1 + e1 = 300k, Þ, e1 = 210k, \, 7, Therefore, balancing length for cell e1 is 210 cm., (iii) By decreasing potential gradient, OR, Applying Kirchhoff's loop rule for loop ABEFA,, –9 + 6 + 4 × 0 + 2I = 0, I = 1.5 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., (ii) We have, R = R0 A, , 1, , 3, , Let m = mass of nucleon, , ...(i), ...(ii), , ...(i), , 345
Page 347 :
346, , Xam idea Physics—XII, , mA, mA, m, =, =, 3, 4, 4, 1, 3, 4 æ, p R0 A, p R 03, pç R 0 A 3 ö÷, 3, 3, ø, 3 è, Hence r is independent of A., 24. Two factors justifying the need for modulation:, (i) Practical size of antenna, (ii) To avoid mixing up of signals from different transmitters., \ Density (r) =, , 0.5, , 1, , Envelope, 1.5, , 0.5, , 1, , 1.5, , Em, Ec, , (amplitude modulated wave), 25. Einstein photo electric equation is hn = f0 + K max ,, 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 ., (ii) impart maximum K.E. to the emitted electron (Kmax.), \, hn = f0 + K max, Three salient features are:, (i) Cut-off potential of the emitted electrons is proportional to n., f, (ii) Photo electric emission of electrons is possible only when v > v0 = 0, h, (iii) Max. K.E. is independent of intensity of incident radiations., 26., (a) This is because coherent sources are needed to ensure that the positions maxima and minima, do not change with time., (b) We know, I = 4 I 0 cos 2, , f, , 2, , for path difference l, phase difference f = 2p, Intensity of light = K, Hence, K = 4I 0 cos 2 p = 4I 0, l, 2p, For path difference , 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
Page 348 :
Examination Papers, , 347, , 27. 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., S1, The point O is maxima because contribution from, each half of the slit S1S2 is in phase, i.e., the path a S, O, difference is zero., S2, At point P, (i) If S2P – S1P = nl Þ the point P would be, D, minima., Slit, Screen, l, (ii) If S 2 P – S1 P = ( 2n + 1) Þ the point would be, 2, maxima but with decreasing intensity., 2lD, The width of central maxima =, a, When the width of the slit is made double the original width, then the size of central maxima, will be reduced to half and intensity will be four times., 28. Van de Graaff, in 1931 designed an electrostatic generator which is capable of producing very, high potential of the order of 10 MV., Principle: Suppose we have a large spherical conducting shell of radius R, carrying charge Q. The, charge spreads uniformly over whole surface of the shell. Now suppose a small conducting sphere, of radius ‘ r ’ is introduced inside the spherical shell and placed at its centre, so that both the sphere, and shell have same centre O. The electric field in the region inside the small sphere and large shell, is due to charge +q only, so electric field strength at a distance x from the centre O is, q, 1, E=, , directed radially outward, 4pe 0 x 2, The potential difference between the sphere and the shell, V (r ) - U ( R) = - ò, =-ò, , r, R, r, R, , ®, , ®, , E .d x, , q, q, 1, dx = 4pe 0 x 2, 4pe 0, , r, , é x -1 ù, q, ú =, ê, 4, pe, 1, úû R, êë, 0, , æ1 1 ö, ç - ÷, èr R ø, , This is independent of charge Q on the large spherical shell. As r < R ; V (r ) - U ( R) is positive. As, charge flows from higher to lower potentials therefore, if we connect the small sphere and large, shell by a conducting wire, the charge flows from sphere to outer shell whatsoever the charge on, outer shell may be. This forms the principle of Van de Graaff generator. The maximum charge that, may be given to outer shell which may cause discharge in air., Construction. 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. The lower comb C1, is connected to the positive terminal of a very high voltage source (HTS) ( » 104 volts. ) and the, upper comb C 2 is connected to the inner surface of metallic sphere S.
Page 349 :
348, , Xam idea Physics—XII, , Working: When comb C1 is given very high, potential, then it produces ions in its vicinity, due to, S, action of sharp points. The positive ions, so produced,, C2, get sprayed on the belt due to the repulsion between, P2, positive ions and comb C1 . These positive ions are, carried upward by the moving belt. The pointed end, of C 2 just touches the belt. The comb C 2 collects, positive charge from the belt which immediately, moves to the outer surface of sphere S. As the belt, goes on revolving, it continues to take (+ ) charge, upward, which is collected by comb C 2 and, transferred to outer surface of sphere S. Thus the, outer surface of metallic sphere S gains positive, A, B, charge continuously and its potential rises to a very, high value., When the potential of a metallic sphere gains very, HTS, high value, the dielectric strength of surrounding air, breaks down and its charge begins to leak, to the, C1 P1, 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., Yes, the high voltages can be built up to the breakdown field of the surrounding medium., OR, (a) Electric flux: It is defined as the total number of electric field lines passing through an area, normal to them:, ®, , ®, , Also, f= ò E . d s, , The SI unit is Nm2/C or volt-metre., (b) Gauss’s Theorem, It states that the total electric flux through a closed surface is equal to, , 1, times the net charge, e0, , enclosed by the surface, i.e.,, , ®, , òE, s, , ®, , .dS =, , 1, Sq, e0, , Let electric charge be uniformly distributed over the surface of a thin, non-conducting infinite, sheet. Let the surface charge density (i.e., charge per unit surface area) be s. We have to, calculate the electric field strength at any point distance r from the sheet of charge.
Page 350 :
Examination Papers, , 349, , q=90o, dS3, o, S1 E 90 S3, , S2, , dS1, E, q=0o, , dS2, E, o, q=0, B, , A, Sheet, , To calculate the electric field strength near the 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). 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., Total electric flux, ®, , ò SE, or, , ®, , ò SE, , ®, , •dS, , =, , ®, , ò SE, 1, , ®, , •dS, , =ò, , S1, , ®, , • d S1 +, , ®, , ò SE, 2, , E dS1 cos 0° +, , ®, , • d S2 +, , òS, , 2, , ®, , ò SE, 3, , ®, , • d S3, , 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 ., As s is charge per unit area of sheet and a is the intersecting area, the charge enclosed by, Gaussian surface = s a, According to Gauss’s theorem,, 1, Total electric flux =, ´ (total charge enclosed by the surface), e0, 1, i.e.,, 2Ea =, ( sa), e0, s, \, E=, ×, 2e 0, , 29., , 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. If the surface charge, density s is negative the electric field is directed towards the surface charge., (c) (i) Away from the charged sheet., (ii) Towards the plane sheet., (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., f æ f ö, b f, or, m= = 0, m = 0 ç1 + e ÷, a fe, fe è, Dø
Page 351 :
350, , Xam idea Physics—XII, , (b) Using, the lens equation for objective lens., 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, , Hence, magnification due to the objective lens, m0 =, , v 0 150 ´ 10 -2 m 10 -2, ~, =, –, 3000 m, 20, u0, , m 0 = 0.05 ´ 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 =, , \ Magnification due to eyepiece m e =, , -25, cm, 6, , -25, =6, 25, 6, , Hence, total magnification m = m e ´ m 0, m = 6 × 5 × 10–4 = 30 × 10–4, Hence, 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. The microscope does the same for a small object kept just beyond the focus, of its objective., (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., (iii) The objective of a telescope has large focal length and large aperture while the, corresponding for a microscope have very small values., (b) Given fo = 1.25 cm, fe = 5 cm, Angular magnification m = 30, Now, m = me × mo
Page 352 :
Examination Papers, , 351, , In normal adjustment, angular magnification of eyepiece, d, 25, =5, me = = +, fe, 5, Hence m 0 = 6, v, But m 0 = 0, u0, v, -6 = 0, Þ, u0, , Þ, , v 0 = - 6u 0, , Applying lens equation to the objective lens, 1, 1, 1, 1, 1, 1, =, Þ, =, 1.25 - 6u 0 u 0, f 0 v0 u0, Þ, u 0 = -1. 46 cm = – 1.5 cm, 30. Circuit diagram of CE transistor Amplifier., IC, C1, , RB, , C, , IB, , C2, RL, , B, Vi, , +, –, , E, , VBB, , IE, , +, VCC, –, , V0, Output, waveform, , When an ac input signal Vi (to be amplified) is superimposed on the bias VBB, the output, which is, measured between collector and ground, increases., We first assume that Vi = 0. Then applying Kirchhoff's law to the output loop., VCC = VCE + I C R L, Similarly the input loop gives, VBB = VBE + I B R B, When Vi is not zero, 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, 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
Page 353 :
Xam idea Physics—XII, , R, V0 DVCE, -b ac DI B R L, =, = -b ac L, =, rDI B, r, Vi, rDI B, Negative sign in the expression shows that output voltage and input voltage have phase difference, of p., AV =, , OR, (a), P1 N1, , p1, Input AC signal, to be rectified, , 352, , i1, , s1, D1, Centre tap, , +, A, , s2, P2 N2, , i2, , RL, , Output, , p2, B, , D2, –, , Working: The AC input voltage across secondary s1 and s 2 changes polarity after each half, cycle. Suppose during the first half cycle of input AC signal, the terminal s1 is positive, relative to centre tap O and s 2 is negative relative to O. Then diode D1 is forward biased and, diode D2 is reverse biased. Therefore, diode D1 conducts while diode D2 does not. The, direction of current (i1 ) due to diode D1 in load resistance R L is directed from A to B. In next, half cycle, the terminal s1 is negative and s 2 is positive relative to centre tap O. The diode D1, is reverse biased and diode D2 is forward biased. Therefore, diode D2 conducts while D1, does not. The direction of current (i 2 ) due to diode D2 in load resistance R L is still from A to, B. Thus the current in load resistance R L is in the same direction for both half cycles of input, AC voltage. Thus for input AC signal the output current is a continuous series of, unidirectional pulses., , Waveform, at P1, , Waveform, at P2, , 2T, O, , T, , T, 2, , 3, T, 2, , t, , O, T, 2, Due to, D1, , Output, waveform O, (across RL), , Due to, D2, T, 2, , 3, T, 2, , T, Due to, D1, T, , 3, T, 2, , 2T, , t, , Due to, D2, 2T, , t, , In a full wave rectifier, if input frequency is f hertz, then output frequency will be 2f hertz, because for each cycle of input, two positive half cycles of output are obtained.
Page 354 :
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. In the static situation, E has to ensure that the free charges on the surface do not experience any, force., 6. Clockwise, 10. Magnetic field inside the solenoid, B = m 0 nI, n ® number of turns per unit length, I ® current flowing,, 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., 14. The material having positive susceptibility is paramagnetic material., Properties, (i) They have tendency to move from a region of weak magnetic field to strong magnetic field,, i.e., they get weakly attracted to a magnet., (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., 16., 2r, , Clockwise, , Clockwise, , BP, , P, I Current, , O, , BQ, , Q, , X-axis, , Q, 2I Current
Page 355 :
354, , Xam idea Physics—XII, , m 0r 2 I, , ®, , | BP| =, ®, , | BQ| =, ®, , 2, , 3, , 2, , 2(r + r ), m 0 ( 2I )r 2, 2(r 2 + r 2 ), , ®, , 3, , ®, , =, 2, , =, 2, , | B| = | B Q | – | B p | =, , u0 I, , Pointing towards P, , 4 2r, m 0I, , Pointing towards Q, , 4 2r, m 0I, , 4 2r, , So, magnetic field at point O has a magnitude, 20., , 1, , (i) w =, , LC, , =, , 1, 4 ´ 100 ´ 10 -6, , m 0I, 4 2r, , = 50 rad/s, , 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., (b) We know that, I = I 1 + I 2 + 2 I 1 I 2 cos f, (ii) I =, , 21., , I = 4I 0 cos 2, , Þ, , f, , (I1 = I2 = I0), , 2, , For path difference l, phase difference f = 2p, Thus,, K = 4I 0 cos 2 p = 4I 0, 2l, When path difference =, 3, 2p, Phase difference, f =, 3, Hence, intensity, I ' = 4I 0 cos 2, 22., , 2, , 4p, K, æ1ö, = 4I 0 ç ÷ = I 0 =, 3, 4, è2ø, , (i) Torque = MB sin q, ®, , ®, , As M and B are parallel. So t = 0., (ii) Force acting on the loop, m ii æ1 1 ö, F = 0 1 2 l çç - ÷÷, 2p è r1 r 2 ø, æ 1, ö, 1, ÷, = 2 × 10–7 × 4 × 1 × 5 × 10–2çç, -2, -2 ÷, 3 ´ 10 ø, è 10, 2, 80, = 40 ´ 10 -7 ´ N = ´ 10 -7 N = 26.66 × 10–7N = 2.67 mN, 3, 3, Direction: Towards the conductor or attractive.
Page 356 :
Examination Papers, , 355, , 27. Three mode of propagation of electromagnetic waves are: (i) Ground waves, (ii) Sky waves,, (iii) Space waves, (a) LOS (Line of Sight) Communication, dM, dT, , dR, , Earth, , hT, , hR, , (b) Satellite communication, Communication satellite, , ave, ce w, a, p, S, Transmitting, antenna, , Earth, , Receiving, antenna, , Above 40 MHz, 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., Space waves are used for the line of sight (LOS) communication as well as satellite, communication., , CBSE (Delhi) Set–III, 6. Electric field intensity is zero throughout the volume and the potential just inside, has to be equal, to potential on the surface., Magnetic field along X-direction., 8. In coil 1, clockwise, In coil 2, anticlockwise, 9. The magnetic material is paramagnetic in nature., For properties, The material having positive susceptibility is paramagnetic material., Properties, (i) They have tendency to move from a region of weak magnetic field to strong magnetic field,, i.e., they get weakly attracted to a magnet., (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., 12. Mutual Inductance: It is defined as the magnetic flux linked to the second coil when unit current is, flowing in the first coil.
Page 358 :
Examination Papers, , 357, , The Ionospheric layers act as a reflector for a certain range of frequencies (from few MHz to, 40MHz). Hence these waves reach from the transmitting antenna to the receiving antenna via, their reflection from the ionosphere., ®, , ®, , 23. (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 çç - ÷÷, 2p è r1 r 2 ø, æ, ö, 1, 1, ÷, = 2 × 10–7 × 2 × 1 × 4 × 10–2 çç, -2, -2 ÷, 4.5 ´ 10 ø, è 2 ´ 10, é 4.5 - 2 ù 8 ´ 5 ´ 10 -7, = 4.44 × 10–7N, = 16 × 10–7 × ê, ú=, 9, 2, ´, 4, ., 5, ë, û, Direction of force is towards conductor (attractive)., 1, 1, = 50 rad/s, 25., (i) w =, =, 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
Page 359 :
CBSE EXAMINATION PAPERS, ALL INDIA–2012, Time allowed: 3 hours, General Instructions: As given in CBSE Examination Papers Delhi–2011., , Maximum marks: 70, , CBSE (AIl India) Set–I, 1. Two wires of equal length, one of copper and the other of manganin have the same resistance., Which wire is thicker?, 2. 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. 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. A bar magnet is moved in the direction indicated by the arrow between two coils PQ and CD., Predict the directions of induced current in each coil, P, , C, , Q, , N, , D, , S, , A, , A, , 5. For the same value of angle incidence, the angles of refraction in three media A, B and C are 15°,, 25° and 35° respectively. In which medium would the velocity of light be minimum?, 6. A proton and an electron have same kinetic energy. Which one has greater de-Broglie wavelength, and why?, 7. Mention the two characteristic properties of the material suitable for making core of a transformer., 8. A charge ‘q’ is placed at the centre of a cube of side l. What is the electric flux passing through, each face of the cube?, 9. 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. (i) Calculate the potential difference between, A and C. (ii) At which point (of the two) is the electric potential more and why?, (2, 3), B, , (2, 0), C, , E, , (6, 0), A
Page 360 :
Examination Papers, , 359, , 10. An electric dipole is held in a uniform electric field., (i) Show that the net force acting on it is zero., (ii) The dipole is aligned parallel to the field. Find the work done in rotating it through the angle, of 180°, 11. State the underlying principle of a transformer. How is the large scale transmission of electric, energy over long distances done with the use of transformers?, 12. A capacitor of capacitance of ‘C’ is being charged by connecting it across a dc source along with, an ammeter. Will the ammeter show a momentary deflection during the process of charging? If so,, 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., 13. An object AB is kept in front of a concave mirror as shown in the figure., A, , B, , C, , F, , (i) Complete the ray diagram showing the image formation of the object., (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. Draw a labelled ray diagram of a reflecting telescope. Mention its two advantages over the, refracting telescope., 15. Describe briefly with the help of a circuit diagram, 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., 16. In the given block diagram of a receiver, identify the boxes labelled as X and Y and write their, functions., Receiving, antenna, , Received, signal, , Amplifier, , X, , Detector, , Y, , 17. A light bulb is rated 100 W for 220 V ac supply of 50 Hz. Calculate, (i) the resistance of the bulb;, (ii) the rms current through the bulb, , Output
Page 361 :
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. Find, (i) the frequency of the source., (ii) the rms current through the resistor., 18. A circular coil of N turns and radius R carries a current I. It is unwound and rewound to make, another coil of radius R/2, current I remaining the same. Calculate the ratio of the magnetic, moments of the new coil and the original coil., 19. Deduce the expression for the electrostatic energy stored in a capacitor of capacitance ‘C’ and, having charge ‘Q’., 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. 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. What would be the potential difference between points B and E?, 8V, , 0.2 A, , B, , 15W, , 5W, C, , 10W, , 30W, , 10W, , 3V, R, , 0.2 A E, , D, , 21. You are given three lenses L1, L2 and L3 each of focal length 20 cm. An object is kept at 40 cm in, front of L1, as shown. The final real image is formed at the focus ‘I’ of L3. Find the separations, between L1, L2 and L3., L1, , L2, , L3, , I, 40cm, , 20cm, , 22. Define the terms (i) ‘cut-off voltage’ and (ii) ‘threshold frequency’ in relation to the phenomenon, of photoelectric effect., 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., 23. A series LCR circuit is connected to an ac source. Using the phasor diagram, derive the expression, for the impedance of the circuit. Plot a graph to show the variation of current with frequency of, the source, explaining the nature of its variation., 24. Mention three ‘different modes of propagation used in communication system. Explain with the, help of a diagram how long distance communication can be achieved by ionospheric reflection of, radio waves.
Page 362 :
Examination Papers, , 361, , 25. Draw a plot of potential energy of a pair of nucleons as a function of their separations. Mark the, regions where the nuclear force is (i) attractive and (ii) repulsive. Write any two characteristic, features of nuclear forces., 26. In a Geiger–Marsden experiment, calculate the distance of closest approach to the nucleus of, Z = 80, when an a-particle of 8 MeV energy impinges on it before it comes momentarily to rest, and reverses its direction., 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 – 13.6 eV. If an electron makes a transition from an, energy level – 0.85 eV to –3.4 eV, calculate the wavelength of the spectral line emitted. To which, series of hydrogen spectrum does this wavelength belong?, 27. Define relaxation time of the free electrons drifting in a conductor. 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., 28. (a) In Young’s double slit experiment, derive the condition for (i) constructive interference and, (ii) destructive interference at a point on the screen., (b) A beam of light consisting of two wavelengths, 800 nm and 600 nm is used to obtain the, interference fringes in a Young’s double slit experiment on a screen placed 1.4 m away. If the, two slits are separated by 0.28 mm, calculate the least distance from the central bright, maximum where the bright fringes of the two wavelengths coincide., OR, (a) How does an unpolarised light incident on a polaroid get polarised?, Describe briefly, with the help of a necessary diagram, the polarisation of light by reflection, from a transparent medium., (b) Two polaroids ‘A’ and ‘B’ are kept in crossed position. 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., (a) Describe briefly, with the help of a diagram, the role of the two important processes involved, in the formation of a p-n junction., (b) Name the device which is used as a voltage-regulator. Draw the necessary circuit diagram, and explain its working., OR, (a) Explain briefly the principle on which a transistor-amplifier works as an oscillator. Draw the, necessary circuit diagram and explain its working., (b) Identify the equivalent gate for the following circuit and write its truth table., A, A', Y, B', B
Page 363 :
362, , Xam idea Physics—XII, , 30., , ®, , (a) Write the expression for the force, F , acting on a charged particle of charge ‘q’, moving with, ®, , ®, , ®, , a velocity v in the presence of both electric field E and magnetic field B . Obtain the, condition under which the particle moves undeflected through the fields., (b) A rectangular loop of size l × b carrying a steady current I is placed in a uniform magnetic, ®, , ®, , ®, , ®, , ®, , ®, , field B. Prove that the torque t acting on the loop is given by t = m × B, where m is the, magnetic moment of the loop., OR, (a) Explain, giving reasons, the basic difference in converting a galvanometer into (i) a voltmeter, and (ii) an ammeter., (b) Two long straight parallel conductors carrying steady current I1 and I2 are separated by a, distance ‘d’. Explain briefly, with the help of a suitable diagram, how the magnetic field due, to one conductor acts on the other. Hence deduce the expression for the force acting between, the conductors. Mention the nature of this force., , CBSE (All India) Set–II, Questions uncommon to Set–I, 2. A charge q is placed at the centre of a cube of side l. What is the electric flux passing through two, opposite faces of the cube?, 5. A proton and an electron have same velocity. Which one has greater de-Broglie wavelength and, why?, 8. In a single-slit diffraction experiment, the width of the slit is made double the original width. How, does this affect the size and intensity of the central diffraction band?, 11. A test charge ‘q’ is moved without acceleration, (1, 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. (i) Calculate the potential difference, between A and C. (ii) At which point (of the two), is the electric potential more and why?, (1, 0), (5, 0), 16. A light bulb is rated 200 W for 220 V ac supply of, C, A, 50 Hz. Calculate, (i) the resistance of the bulb;, (ii) the rms current through the bulb., OR, An alternating voltage given by V = 280 sin 50 pt is connected across a pure resistor of 40 W. Find, (i) the frequency of the source., (ii) the rms current through the resistor., 17. A circular coil of ‘N’ turns and diameter ‘d’ carries a current ‘I’. It is unwound and rewound to, make another coil of diameter ‘2d’, current ‘I’ remaining the same. Calculate the ratio of the, magnetic moments of the new coil and the original coil.
Page 364 :
Examination Papers, , 363, , 19. You are given three lenses L1, L2 and L3 each of focal length 15 cm. An object is kept at 20 cm in, front of L1, as shown. The final real image is formed at the focus ‘I’ of L3. Find the separations, between L1, L2 and L3., L1, , L2, , L3, , I, 20cm, , 15cm, , 23. 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. What would be the potential difference between points A and B?, 6V, , 0.2 A, , B, , 10W, , 5W, C, , 30W, , 15W, , 5W, , 2V, 0.2 A, , R, , A, , D, , 27. In a Geiger– Marsden experiment, calculate the distance of closest approach to the nucleus of, Z = 75, when an a-particle of 5 MeV energy impinges on it before it comes momentarily to rest, and reverses its direction., 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 –13.6 eV. If an electron makes a transition from an, energy level – 0.85 eV to – 1.51 eV, calculate the wavelength of the spectral line emitted. To, which series of hydrogen spectrum does this wavelength belong?, , CBSE (All India) Set–III, Questions uncommon to Set – I & II, 1. How does the fringe width, in Young’s double-slit experiment, change when the distance of, separation between the slits and screen is doubled?, 3. A charge ‘q’ is placed at the centre of a cube. What is the electric flux passing through the cube?, 1, 4. The speed of an electromagnetic wave in a material medium is given by v =, , m being the, me, permeability of the medium and e its permittivity. How does its frequency change?, 9. A circular coil of closely wound N turns and radius r carries a current I. Write the expressions for, the following:, (i) the magnetic field at its centre, (ii) the magnetic moment of this coil
Page 365 :
364, , Xam idea Physics—XII, , 10. Describe briefly with the help of a circuit diagram, the paths of current carriers in an n-p-n, transistor with emitter-base junction forward biased and base-collector junction reverse biased., 12. A light bulb is rated 150 W for 220 V ac supply of 60 Hz. Calculate, (i) the resistance of the bulb;, (ii) the rms current through the bulb., (–2, 2), OR, B, An alternating voltage given by V = 70 sin 100 p t is, E, connected across a pure resistor of 25 W. Find, (i) the frequency of the source., (ii) the rms current through the resistor., (–2, –1), (2, –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. (i) Calculate, the potential difference between A and C. (ii) At which point (of the two) is the electric potential, more and why?, 20. Explain briefly the following terms used in communication system:, (i) Transducer, (ii) Repeater, (iii) Amplification, 22. You are given three lenses L1, L2 and L3 each of focal length 10 cm. An object is kept at 15 cm in, front of L1, as shown. The final real image is formed at the focus ‘I’ of L3. Find the separations, between L1, L2 and L3., L1, , L2, , L3, , I, 15cm, , 10cm, , 26. 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. What would be the potential difference between points A and D?, 10V, , 0.2 A, , A, , 15W, 10W, , B, 30W, , 5W, , 5V, R, , 0.2 A D, , C, 10W
Page 366 :
Examination Papers, , 365, , Solutions, CBSE (All India) SET–I, 1. We know, R =r, , l, ;, A, , For copper,, R Cu = r Cu, , lCu, A Cu, , ...(i), , For manganin,, Rm = r m, , lm, Am, , ...(ii), , But r m > r Cu ,, and lm = lCu ,, 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., Þ, 2. Both electric field and magnetic fields are electromagnetic waves. There waves are perpendicular, to each other and perpendicular to the direction of propagation., b Dl / d l, 3. Angular separation is q = =, =, D, D, d, Since q is independent of D, Angular separation would remain same., 4., Q, , P, 1, A, , D, , C, , N, , S, , 2, A, , In the figure, N pole is receding away coil (1), so in coil (1), 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), so current in coils will flow clockwise as seen from the side of magnet. The direction of, current will be from P to Q in coil (1) and from C to D in coil (2)., sin i c, 5. From Snell's law n =, =, sin r v, For given i, v µ sin r ; r is minimum in medium A, so velocity of light is minimum in medium A.
Page 367 :
366, , Xam idea Physics—XII, , 6. We know, de Broglie wavelength, h, l=, 2m E k, For same K.E. of electron and proton., As, m p > me, Þ, le > l p, Hence wavelength of electron is greater than that of proton., 7. Two characteristic properties:, (i) Low hysteresis loss, (ii) Low coercivity, 8. By Gauss's Theorem in electrostatics, ® ®, q, f = ò E . ds =, e0, Total flux through all six faces would be f' = 6f, where f = Flux through one face., q, q, Hence,, 6f =, Þ, f=, e0, 6e 0, 9., , (i) From the given diagram,, potential difference between A and C, C® ®, , VC – VA = -ò E . dl cos180° = E × 4 = 4E, A, , 10., , Hence, VC – VA = 4E, (ii) VC > VA, Because direction of electric field is in decreasing potential., (i) The dipole moment of dipole is, , B, , ®, , +q, , | P | = q ´ ( 2a), , ®, +qE, , ®, , Force on –q at A = –qE, , 2a, , ®, , Force on +q at B = + q E, , ®, , ®, , ®, –qE, –q, , N, Net force on the dipole = qE – qE = 0, A, (ii) Work done on dipole, W = DU = PE(cos q1 - cos q 2 ), = PE(cos 0°- cos180° ), W = 2 PE, 11. 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. At the power generating station, the step up transformers step up the output voltage which, reduces the current through the cables and hence reduce resistive power loss. Then, at the, consumer end, a step down transformer step down the voltage.
Page 368 :
Examination Papers, , 367, , Hence, the large scale transmission of electric energy over long distances done with the use of, transformers is taken place., 12. Yes, because of the production of displacement current between the plate of capacitor on account, of changing electric field., Current inside the capacitor, d fE, I D = e0, dt, 13., (i) Image formed will be inverted diminished between C and F., A, , D, B' F, , B, , C, , P, , A', , (ii) No change in position of image and its intensity will get reduced., 14., , Objective, mirror, , Secondary, mirror, Eyepiece, , Advantages:, (i) It is free from chromatic aberration., (ii) Its resolving power is greater than refracting telescope due to larger aperture of mirror., 15. The emitter-base junction is given a small forward bias, while base collector junction is given a, large reverse bias., Under the forward biasing of emitter-base region, the positive, p, n, p, holes of P - region move towards the base. 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. As soon as a hole combines with the electron, 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 . The holes, entering the collector move under the aiding reverse bias, towards terminal C. As a hole reaches terminal C, an electron, +, –, +, –, VEE, VCC, leaves the negative terminal of battery VCC and neutralises the, hole. This causes the collector current I C . Both these currents, I B and I C combine to form the emitter current I e i. e.,, I E = I B + IC
Page 369 :
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., 16., Receiving, antenna, , Received, signal, , Amplifier, , X, , Detector, , Y, , Output, , Here, X ® IF stage (Intermediate Frequency stage), Its function is to change the carrier frequency to lower frequency., Y ® Amplifier, Its function is to amplify the signal because the detected signal may not be strong enough to use, by the user., 17., , V 2 rms 220 ´ 220, = 48 W, =, P, 100, P 100, (b) I rms =, = 0.45 A, Vrms 220, (a) R =, , OR, (i) V =140 sin 314t, Q V = V0 sin wt, Hence w = 314, (ii) Vrms =, , V0, , =, , Þ, Þ, , 2pn = 314, n = 50 Hz, , 140, , = 98.9 V, 2, V, 98 . 9, = 1.97 A, I = rms =, R, 50, 18. As length of wire remains the same, R, N1 ´ 2pR = N 2 ´ 2p, 2, N2 = 2N, \, Magnetic moment of a coil, m = NAI, For the coil of radius R, magnetic moment, m1 = N1 IA1 = N1 I pR 2, 2, , For the coil of radius, , R, , 2,, , magnetic moment, , m 2 = N 2 IA 2 =, , 2N1 I pR 2, N IpR 2, = 1, 4, 2
Page 370 :
Examination Papers, , Now,, , m2, = 1:2, m1, , 19. When a capacitor is charged by a battery, 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., Consider a capacitor of capacitance C. Initial charge on, capacitor is zero. Initial potential difference between, capacitor plates = zero. Let a charge Q be given to it in small, steps. When charge is given to capacitor, the potential, difference between its plates increases. Let at any instant, , A, , VA, , 369, , B, , + +Q, +, +, +, +, +, +, +, , –Q –, –, –, –, –, –, –, –, , VB, , VAB = V, , q, when charge on capacitor be q, the potential difference 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, which may be obtained by integration. Therefore total work, Q, Q q, W = ò V dq = ò, dq, 0, 0, C, 1 é q2 ù, = ê, ú, C êë 2 ûú, , Q, , =, 0, , 1, C, , æ Q2 0 ö Q2, ç, - ÷=, ç 2 2 ÷ 2C, è, ø, , If V is the final potential difference between capacitor plates, then Q = CV, (CV ) 2 1, 1, = CV 2 = QV, 2C, 2, 2, This work is stored as electrostatic potential energy of capacitor i.e.,, Q2 1, 1, Electrostatic potential energy,, = CV 2 = QV, U=, 2C 2, 2, When battery is disconnected, Q2, Q02, U, (i) Energy stored will be decreased. The energy becomes, U = 0 =, = 0, 2C 2KC 0, K, 1, Thus, energy is reduced to times the initial energy., K, E, (ii) In the presence of dielectric the electric field is reduced to E = 0 ., K, 20. Here, 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 × 0.2 + R × 0.2 + 15 × 0.2 = 8 – 3, \, , W=
Page 371 :
370, , Xam idea Physics—XII, , Þ, Hence, , R = 5W, , VBE = IR BE, = 0.2 × 5 = 1 volt, 21. Given f 1 = f 2 = f 3 = 20 cm, For lens L1, u1 = – 40 cm, 1 1, 1, By lens formula, - =, v1 u1 f 1, 1, 1, 1, Þ v1 = 40 cm, =, Þ, v1 20 -40, For lens L3, f3 = 20 cm, v3 = 20 cm, u3 = ?, 1, 1, 1, By lens formula,, =, v3 u 3 f 3, 1, 1, 1, =, Þ, 20 u 3 20, 1, = 0 Þ u3 = ¥, Þ, u3, , Intercept = – f/e, , Thus lens L2 should produce image at infinity., Hence, for L2, its objective should be at focus. 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.e., focus of lens L2., Hence, the distance between L1 and L2 = 40 + 20 = 60 cm., As the image formed by lens L2 lies at infinity, then, the distance between lens L2 and L3, does not matter., Hence, the distance between L2 and L3 can have any value., 22., (i) Cut off or stopping potential is that minimum value, Y, of negative potential at anode which just stops the, photo electric current., (ii) For a given material, there is a minimum frequency, h, Slope = e, of light below which no photo electric emission, will take place, this frequency is called as threshold, frequency., V0, X, By Einstein's photo electric equation, n0, n, hC, - f = hn - hn 0, KE max =, l, eV0 = hn - hn 0, h, h, V0 = n - n 0, e, e, Clearly, V0 – n graph is a straight line.
Page 372 :
Examination Papers, , 371, , 23. Expression for Impedance in LCR series circuit: Suppose resistance R, inductance L and, capacitance C are connected in series and an alternating source of voltage V = V0 sin wt is applied, across it. (fig. a) On account of being in series, the current (i ) flowing through all of them is the, same., R, , L, , C, , VL, 90o, , VL, , VR, , VC, , 90, , (VC–VL), , V=V0 sin wt, (a), , o, , VR, , i, , f, , VC, (b), , Suppose the voltage across resistance R is VR , voltage across inductance L is VL and voltage across, capacitance C is VC . The voltage VR and current i are in the same phase, the voltage VL will lead the, current by angle 90° while the voltage VC will lag behind the current by angle 90° (fig. b). Clearly, VC and VL are in opposite directions, therefore their resultant potential difference, = VC - VL (if VC > VC )., Thus VR and (VC - VL ) are mutually perpendicular and the phase difference between them is 90°., As applied voltage across the circuit is V, the resultant of VR and (VC - VL ) will also be V. From fig., V, , 2, , = VR 2 + (VC - VL ) 2 Þ V = VR 2 + (VC - VL ) 2, , ...(i), , ...(ii), VR = R i , VC = X C i and VL = X L i, 1, where X C =, = capacitance reactance and X L = wL = inductive reactance, wC, , But, , \, , V = ( R i) 2 + ( X C i - X L i) 2, , \ Impedance of circuit,, , Z=, , i.e., Instantaneous current I =, , V, = R 2 + (X C - X L ) 2, i, , æ 1, ö, Z = R 2 + (X C - X L ) 2 = R 2 + ç, - wL ÷, è wC, ø, V0 sin ( wt + f), æ 1, ö, R2 + ç, - wL ÷, w, C, è, ø, , 2, , 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., With increase in frequency, current first increases and then, decreases. At resonant frequency, the current amplitude is, maximum., , im, lm, im, 2, , f1, , fr, f, , f2
Page 373 :
372, , Xam idea Physics—XII, , 24. 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), emitted from the transmitting, antenna, reach the receiving antenna after reflection from the ionosphere which acts as a, reflector for radio waves., 25. Part AB represents repulsive force and Part BCD represents attractive force., , A, +100, , MeV, , Repulsive, B, , 0, D, Attractive, , –100, , C, , 1, , 2, r (fm), , 3, , 4
Page 374 :
Examination Papers, , 373, , Conclusions:, (1) Nuclear forces are attractive and stronger, then electrostatic force., (2) Nuclear forces are charge-independent., 26. Let r 0 be the distance of closest approach where the K.E. of a-particle is converted into its, potential energy., Given, Z = 80, Ek = 8MeV, 1 ( Ze) ( 2e), K=, 4pe 0, r0, Þ, , r0 =, r0 =, =, , 1 ( Ze)( 2e), 2Ze 2, =, 4pe 0, K, 4pe 0 K, 9 ´ 10 9 ´ 2 ´ 80 ´ (1.6 ´ 10 -19 ) 2, 8 ´ 10 6 ´ (1.6 ´ 10 -19 ), 18 ´ 80 ´ 1.6 ´ 10 -10, 8 ´ 10 6, , m = 2.88 × 10–14 m, , 1, K, If kinetic energy (K) of a-particle is doubled, the distance of closest approach will become half., OR, As we know:, -13.6, ...(i), En =, eV, n2, For n =1, E1 = -13.6 eV, When electron undergoes transitions from E A = -0.85eV to E B = -3.4eV, Then, from equation (i), -13.6, – 0.85 =, Þ, nA = 4, n A2, -13.6, Similarly,, – 3.4 =, Þ, nB = 2, n B2, As, , r0 µ, , Hence electron transits from n = 4 to n = 2. It corresponds to Balmer series., æ 1, 1, 1 ö÷, = Rç, We know,, 2, 2 ÷, ç, l, è nB nA ø, Here,, , n A = 4, n B = 2, R = 1.097 ´ 107 m–1, , æ 1, 1, 1 ö, ÷÷ Þ, l = 4862Å, = 1.097 ´ 107 çç, 2, l, 42 ø, è2, 27. Relaxation time of free electrons drifting in a conductor is the average time elapsed between two, successive collisions., The relation between t andv d is, Then,
Page 375 :
374, , Xam idea Physics—XII, ®, , ®, vd, , eE, =t, m, Let a conductor of length 'l' and area of cross-section ‘A’ with electron density n., 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, then electric field ( E) =, Then current becomes, n A e 2 Vt, I=, ml, V, ml, Then,, =, I ne 2 tA, By Ohm's Law R =, , E, , V, l, , A, , l, , V, I, , æ m öl, ÷ =r l, R =ç, ç ne 2 t ÷ A, A, è, ø, m, r=, \, ne 2 t, 28., (a) Conditions of Constructive and Destructive Interference:, , V, , Suppose two coherent waves travel in the same direction along a straight line, the frequency, w, of each wave is, and amplitudes of electric field are a1 and a 2 respectively. If at any time, 2p, t, the electric fields of waves at a point are y1 and y 2 respectively and phase difference is f,, then equation of waves may be expressed as, y1 = a1 sin wt, , ...(i), , y 2 = a 2 sin ( wt + f), , ...(ii), , According to Young’s principle of superposition, the resultant displacement at that point, will be, y = y1 + y 2, , ...(iii), , Substituting values of y1 and y 2 from (i) and (ii) in (iii), 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), = ( a1 + a 2 cos f) sin wt + ( a 2 sin f) cos wt, , Let, , a1 + a 2 cos f = A cos q, , ...(iv), ...(v)
Page 376 :
Examination Papers, , and, a 2 sin f = A sin q, where A and q are new constants., Then equation (iv) gives, y = A cos q sin wt + A sin q cos wt = A sin ( wt + q), , 375, ...(vi), ...(vii), , This is the equation of the resultant disturbance. Clearly the amplitude of resultant, disturbance is A and phase difference from first wave is q. The values of A and q are, determined by (v) and (vi). Squaring (v) and (vi) and then adding, 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), , As cos 2 q + sin 2 q = 1, we get, 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, A = a12 + a 22 + 2a1 a 2 cos f, , ...(viii), , 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, , ...(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., Constructive Interference : For maximum intensity at any point cos f = + 1, or, , phase difference, f = 0, 2p , 4p , 6p ........., = 2np ( n = 0, 1, 2, ....), , ...(x), , The maximum intensity,, I max = a12 + a 22 + 2a1 a 2 = ( a1 + a 2 ) 2, Path difference, , D=, , l, l, ´ Phase difference =, ´ 2 np = nl, 2p, 2p, , ...(xi), ...(xii), , 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 ., Destructive Interference : For minimum intensity at any point cos f = -1, or, , phase difference, f = p , 3p , 5p , 7p ...., , = ( 2n - 1) p , n = 1, 2, 3 ...., In this case the minimum intensity,, I min = a12 + a 22 - 2a1 a 2 = ( a1 - a 2 ) 2, , ...(xiii), ...(xiv)
Page 377 :
376, , Xam idea Physics—XII, , l, ´ Phase difference, 2p, l, l, =, ´ ( 2n - 1) p = ( 2n - 1), 2p, 2, , D=, , Path difference,, , ...(xv), , Clearly, 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 - a 2 ) which is less than the sum of intensities of the individual waves by an, amount 2a1 a 2 ., 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. This implies that the interference is based on conservation of energy., Variation of Intensity of light with position x is shown in fig., (b) Given l1 = 800 nm = 800 ´ 10 -9 m, l 2 = 600nm = 600 ´ 10 -9 m, D = 1.4 m, d = 0.28 mm = 0.28 × 10–3m, For least distance of coincidence of fringes, there must be a difference of 1 in order of l1 and l 2 ., As, l1 > l 2 ,, n1 < n 2, If n1 = n,, 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 - l 2 800 - 600, nDl1, y min =, d, =, , 3 ´ 1.4 ´ 800 ´ 10 -9, 0.28 ´ 10 -3, , = 12000 × 10–6, = 12 × 10–3m
Page 378 :
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. So, when an unpolarised, light falls on a polaroid, it lets only those of its electric vectors that are oscillating along a, direction perpendicular to its aligned molecules to pass through it. The incident light thus gets, linearly polarised., D, B, , A, Incident, ray, , ip r', , Reflected ray, 90°, , O, r, , E, , Refracted ray, , C, , Whenever unpolarised light is incident on the boundary between two transparent media, the, reflected light gets partially or completely polarised. When reflected light is perpendicular to, the refracted light, the reflected light is a completely polarised light., (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 . cos 2 ( 90 - q), 2, I0, I, =, cos 2 q . sin 2 q = 0 (cos q . sin q) 2, 2, 2, According to question, I0, I, (cos q sin q ) 2 = 0, 2, 8, I0, 2, , 2, , I, æ 2 sin q cos q ö, ç, ÷ = 0, 2, 8, è, ø, sin 2q = 1, , or,, , 2q = 90°, , Þ, , q = 45°, , The third polaroid is placed at q = 45°.
Page 379 :
378, , Xam idea Physics—XII, , 29., , (a) Two important processes occur during the formation of a p-n junction are (i) diffusion and, (ii) drift., (i) Diffusion: In n-type semiconductor, the, Electron diffusion, Electron drift, concentration of electrons is much greater as, –, +, compared to concentration of holes; while in, p, n, p-type semiconductor, the concentration of, – +, holes is much greater than the concentration, Electron, of electrons. When a p-n junction is formed,, – +, then due to concentration gradient, 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)., This motion of charge carriers gives rise to, diffusion current across the junction., (ii) Drift: The drift of charge carriers occurs due to, Vo Potential barrier, electric field. Due to built in potential barrier an, electric field directed from n-region to p-region, + +, is developed across the junction. 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. Thus a drift current starts., Depletion, This current is opposite to the direction of, layer, diffusion current., (b) Zener diode is used as voltage regulator., , 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. Thus, the zener diode acts as, a voltage regulator., OR, (a) A transistor as an Oscillator:, The circuit of a tuned collector in CE configuration is shown in figure., 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 feed back coil L¢ connected to base circuit is mutually coupled with coil L (due to, phenomenon of mutual induction); the mutual inductance of L and L¢ being M. In practice, L and L¢ form the primary and secondary coil of the transformer.
Page 380 :
Examination Papers, , 379, , The biasing is provided by emitter resistance R E and potential divider arrangement, consisting of resistances R1 and R 2 . 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., 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), 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., (b) Output y = A + B = A . B = A . B, The equivalent gate for the given circuit is AND gate, A, , Y = AB, , B, Truth table, , 30., , A, , B, , Y, , 0, , 0, , 0, , 0, , 1, , 0, , 1, , 0, , 0, , 1, , 1, , 1, , ®, , ®, , (a) Electric force on particle, Fe = qE, ®, , ® ®, , Magnetic force on particle, Fm = q ( v + B ), Total force,, , ®, , ®, , ®, , F = Fe + Fm, , ®, , ® ® ®, , F = q ( E + v ´ B)
Page 381 :
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., ®, , F =0, , In such a case,, , ® ® ®, , Þ, , q ( E + v ´ B) = 0 Þ, ®, , Þ, , ®, , ®, , ®, , ®, , E = -( v ´ B ), , ®, , E =B´ v, , ®, , E = Bv sin q = Bv ( when q = 90° ), E, v = (when v, E and B are mutually perpendicular), Þ, B, N, (b) Torque on a current carrying loop: Consider, F2, a rectangular loop PQRS of length l, breadth b, ®, , ®, , ®, , ®, , F1, I, F3, P, S, F4, , q, , ®, , N, b, , q, , and SP make angle ( 90° - q) with the direction, of magnetic field. Therefore each of the forces, , B, , I (Downward), , ®, , N', , F2 and F4 acting on these sides has same, magnitude, , of, , I, (Upward), , loo, , p, , F1=IlB, , F1 , F2 , F3 and F4 respectively. The sides QR, , ®, , B, R, , ®, , 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, , q, , Q, , 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, , Axis of loop, or normal to loop, , is, , Þ, , ax, , 380, , b sin q, F3=IlB, , F¢ = Blb sin ( 90° - q) = Blb cos q., ®, , ®, , According 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.
Page 382 :
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, 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., \ Moment of couple or Torque,, 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, t = IAB sin q, If the loop contains N-turns, then t = NI AB sin q, ®, , ®, , ®, , In vector form t = NI A ´ B., 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.e., along I A ´ B ., 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. The small resistance connected in, G, parallel is called a shunt. If G is resistance of, Ammeter, galvanometer, I g is current in galvanometer for full, scale deflection, then for conversion of galvanometer into ammeter of range I ampere, the, shunt is given by, Ig, S=, G, I - 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. If V volt is the range of voltmeter formed,, 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. It exerts, a force on RS. The force acting on length l of the conductor RS will be
Page 383 :
382, , Xam idea Physics—XII, , F21 = I2l B1 sin90°, Similarly, an equal force is exerted on the wire PQ by the field of conductor RS., m I I, F12 = 0 1 2 l, 2pd, P, , R, , P, , R, , b, , I2, l, , F21, I1, Q, , a, , B, , B, I2, , I1, , S, , Q, , a, l, b, , F12, , S, , r, , Thus, when the currents in the two conductor are in the same direction, the force between them are, attractive. When the currents flow in opposite directions, the forces between the two conductors, are repulsive., , CBSE (All India) Set–II, 2. By symmetry, the flux through each of the six faces of the cube will be same when charge q is, placed at its centre., 1 q, \, fE = ., 6 e0, 1 q, Thus, electric flux passing through two opposite faces of the cube = 2. ., 6 e0, 5. 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 , hence, l p < le, m, Thus, electron has greater de Broglie wavelength, if accelerated with same speed., 8. In single slit diffraction experiment fringe width is, 2 lD, b=, d
Page 385 :
384, , Xam idea Physics—XII, , 1, mB 2 2, = =, Þ, mA 1 1, 4, 19. Given f 1 = f 2 = f 3 = 15cm, We have, lens formula, 1 1, 1, \, - =, v1 u1 f 1, 1, 1, 1, = +, Þ, v1 15 -20, , mB 2, =, mA 1, , u1 = – 20 cm, f 1 = 15 cm, v1 = 60 cm, , So, the image formed by lens L2 is at infinity., It means that the object for L2 lies at its focus., So,, u2 = 15 cm, Hence, 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, the distance between L 2 and L 3 does, not matter. Hence, the distance between L 2 and L 3 can have any value., 23. The equivalent diagram of given electric circuit is, 6V, , 0.2 A, , B, , C, , 10W, , 5W, 30W, , 15W, 2V, , 10W, , 0.2 A, , R, , A, , D, , Effective resistance between A and B, 1, 1 1 1, = + +, Þ, R ¢ = 5W, R ¢ 10 30 15, Applying Kirchhoff's Law, 5 × 0.2 + R(0.2) + 10(0.2) = 6 – 2, Þ, R = 5W, When R = unknown value of resistance, VAB = 5 × 0.2 = 1volt, 27. Let r0 be the distance of closest approach where the K.E of a-particle is converted into its potential, energy., Given Z = 75, Ek = 5MeV, ( 2e) ( Ze), 1, Ek =, 4pe 0, r0
Page 386 :
Examination Papers, , or, , r0 =, r0 =, , As, , r0 µ, , 385, , 1 ( 2e)( Ze), 4pe 0 E k, 9 ´ 10 9 ´ 75 ´ 2 ´ (1.6 ´ 10 -19 ) 2, 6, , 5 ´ 10 ´ 1.6 ´ 10, , -19, , =, , 18 ´ 75 ´ 1.6 ´ 10 -10, 8 ´ 10, , 6, , = 2.7 ´ 10 -14 m, , 1, Ek, , Hence, the distance of closest approach is halved if K.E is doubled., OR, -13.6, ...(i), As we know,, En =, eV, n2, For n =1, E1 = -13.6 eV, When electron undergoes transition from E A = - 0.85 eV, Then, from equation (i), 13.6, – 0.85 = Þ, nA = 4, nA 2, Similarly, – 1.51 =, , -13.6, nB 2, , Þ, , nB = 3, , Hence, electron transits from n = 4 to n = 3 which corresponds to Paschen series of hydrogen atom., We know, æ 1, 1, 1 ö÷, = Rç, ç n2 n2 ÷, l, A ø, è B, Here n A = 4, n B = 3, R = 1.0974 ´ 107, Then, Þ, , æ 1, 1, 1, = 1.097 × 107 çç, 2, l, 42, è3, l = 1875 nm, , ö, ÷÷, ø, , m -1, , CBSE (All India) Set–III, 1. The fringe width is, , lD, d, If D (distance between slits and screen) is doubled, then fringe width will be doubled., 3. The total flux passing through the cube is, q, f=, e0, b=, , 4. The frequency of electromagnetic waves does not change while travelling through a medium.
Page 387 :
386, , Xam idea Physics—XII, , 9. 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 given transistor is p-n-p transistor. The emitter is reverse biased and the, collector is forward biased., , +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. The circuit diagram, –2V E, for the operation of NPN transistor is shown in fig. The two batteries, VEE and VCC represent emitter supply and collector supply respectively. The emitter-base, junction is forward biased and the base-collector junction is reverse biased. Consequently the, internal potential barrier at emitter-base junction neutralises while the width of depletion layer at, collector-base junction increases., , Base Current and Collector Current: Under forward bias of emitter-base junction, the electrons, in emitter and holes in base are compelled to move towards the junction, thus the depletion layer, of emitter-base junction is eliminated. As the base region is very thin, 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. 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. The electron released is attracted by positive terminal of, emitter battery VEE , giving rise to a feeble base current ( I B ). Its direction in external circuit is, from emitter to base. The hole released in the base region compensates the loss of hole neutralised, by electrons., Depletion, layer, n, , p, , n, , E, , C, , IE, , IC, e–, B, , –, , +, IB, , IE, +, , IC, , e–, –, , IB, e–, –, , +, VEE, , O, , –, , +, VCC, , The electrons crossing the base and entering the collector, due to reverse biasing of collector-base, junction, are attracted towards the positive terminal of collector battery VCC . 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, called the collector current. In
Page 388 :
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. This current is called the leakage current., Thus, collector current is formed of two components:, (i) Current ( I nc ) due to flow of electrons (majority charge carriers) moving from emitter to, collector., (ii) leakage current ( I leakage ) due to minority charge carriers, i. e., I c = I nc + I leakage ., 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, an, equal number of electrons enter from emitter battery VEE to emitter, causing the emitter current., The process continues., Relation between Emitter, Base and Collector Currents:, Applying Kirchhoff’s I law at terminal O , we get, I E = I B + IC, That is, the emitter current I E is the sum of base current I B and the collector current I C . This is, the fundamental relation between currents in the bipolar transistor circuit., 12., (i) P = 150 W, Vrms = 220 V, f = 60 Hz, We have, , 2, Vrms, Þ, R, 220 ´ 220, R=, = 322.6 W, 150, , P=, , (ii) P = I rms Vrms, I rms =, , Þ, , I rms =, , R=, , 2, Vrms, P, , P, Vrms, , 150, = 0.68 A, 220, OR, , We know,, , V = V0 sin wt, , Given, , V = 70 sin 100 pt, , (a) V0 = 70 volt,, 2pn = 100p, , w =100 p, [Q w = 2pn], , n = 50 Hz, V, 70, (b) Vrms = 0 =, 2, 2, Vrms 70 / 2, =, R, 25, 70, = 1.98A, =, 25 2, , I rms =, , [where R = 25W]
Page 389 :
388, , Xam idea Physics—XII, , 16., , (i) From the given diagram,, potential difference between A and C, C® ®, , C, , C, , A, , A, , A, , VC – VA = -ò E . dl = -ò E dl cos180° = E ò dl = E ´ 4 = 4E, So, VC - VA = 4E, (ii) The direction of electric field is in the decreasing potential. Hence, the electric potential will, be more at C., Hence,, VC > VA, 20., (i) Transducer: Any device which converts one form of energy into another., (ii) Repeater: It is a combination of a receiver and a transmitter. A repeater, picks up the signal, from the transmitter, amplifies and retransmits it to receiver sometimes with a change in, carrier frequency., (iii) Amplification: It is the process of increasing the amplitude of a signal using an electronic, circuit. (i.e., amplifier). It is necessary to compensate for attenuation of signal in, communication system., 22. Given f 1 = f 2 = f 3 = 10 cm, u1 = – 15 cm, 1 1, 1, By lens formula, - =, v1 u1 f 1, 1, 1, 1, 3 -2, 1, =, =, Þ, = +, v1 10 -15, 30, 30, v1 = - 30 cm, Since, final image is formed by lens L3 at focus. Hence, for L3, the object must be at infinity. Thus, lens L2 should produce image at infinity. So for L2, its object should be at focus., The image formed by lens L, is at 15 cm on the right side of lens L1which lies at 10 cm left of lens, L2, i.e., focus of lens L2., Hence, distance between L1 and L2 = 30 cm + 10 cm = 40 cm, As the image formed by Lens L2 lies at infinity. Then, the distance between lens L2 and L3 does, not matter., Hence, the distance between L 2 and L3 can have any value., 26. The equivalent diagram of the given electrical circuit is as follows., 10V, , 0.2 A, , A, , B, , 15W, , 5W, 10W, , 5V, , 30W, 10W, , R, , 0.2 A D, , C
Page 390 :
Examination Papers, , The effective resistance between A and D, 1 1 1 1, 3 +1 + 2, =, = + +, ¢, R 10 30 15, 30, Þ, R ¢ = 5W, Applying Kirchhoff Law, to find unknown R., 5 × 0.2 + R × 0.2 + 15 × 0.2 = + 10 – 5, R, 1 + + 3 = +5, Þ, R = 5W, 5, Hence VAD = 5 ´ 0.2 = 1 volt, , 389
Page 391 :
CBSE EXAMINATION PAPERS, FOREIGN–2012, Time allowed: 3 hours, , Maximum marks: 70, , General Instructions: As given in CBSE Examination Papers Delhi–2011., , CBSE (Foreign) Set–I, 1. Why is the potential inside a hollow spherical charged conductor constant and has the same value, as on its surface?, 2. A magnetic needle, free to rotate in a vertical plane, orients itself vertically at a certain place on, the Earth. What are the values of (i) horizontal component of Earth’s magnetic field and (ii) angle, of dip at this place?, 3. 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. Predict the direction of the induced current in the loop., 4. Name the electromagnetic waves, which (i) maintain the Earth’s warmth and (ii) are used in, aircraft navigation., ×, ×, ×, ×, ×, 5. 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., ×, ×, ×, ×, ×, 6. Write the relationship between the size of a nucleus and its mass ×, S, R, number (A)., ×, ×, ×, ×, ×, ×, 7. Show on a graph the variation of the de Broglie wavelength (l), associated with an electron, with the square root of accelerating potential (V)., 8. Define dipole moment of an electric dipole. Is it a scalar or a vector?, 9. A conductor of length ‘l’ is connected to a dc source of potential ‘V’. If the length of the conductor, is tripled by gradually stretching it, keeping ‘V’ constant, how will (i) drift speed of electrons and (ii), resistance of the conductor be affected? Justify your answer., 10. Two students ‘X’ and ‘Y’ perform an experiment on, E, K1, + –, R, potentiometer separately using the circuit given:, Keeping other parameters unchanged, 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, while the key K2 remain open and the, key K1 closed?, S, Justify., K2
Page 392 :
Examination Papers, , 391, , ®, , 11. A particle of charge ‘q’ and mass ‘m’ is moving with velocity V . It is subjected to a uniform, ®, , magnetic field B directed perpendicular to its velocity. Show that it describes a circular path., Write the expression for its radius., 12. Calculate the quality factor of a series LCR circuit with L = 2-0 H, C= 2 mF and R = 10 W. Mention, the significance of quality factor in LCR circuit., 13. Explain briefly how electromagnetic waves are produced by an oscillating charge. How is the, frequency of the em waves produced related to that of the oscillating charge?, 14. In a given sample, two radioisotopes, A and B, are initially present in the ratio of 1:4. The half, lives of A and B are respectively 100 years and 50 years. Find the time after which the amounts of, A and B become equal., 15. Figure shows a block diagram of a transmitter. Identify the boxes ‘X’ and ‘Y’ and write the their, functions., Transmitting, antenna, , m(t), Message signal, , X, , Y, , 16. Trace the path of a ray of light passing through a glass prism (ABC) as shown in the figure. If the, refractive index of glass is 3, find out of the value of the angle of emergence from the prism., A, , 60°, B, , C, , 17. Write two characteristic features to distinguish between n-type and p-type semiconductors., OR, How does a light emitting diode (LED) work? Give two advantages of LED’s over the, conventional incandescent lamps., 18. A short bar magnet of magnetic moment 0.9 J/T is place with its axis at 30° to a uniform magnetic, field. It experiences a torque of 0.063 J., (i) Calculate the magnitude of the magnetic field., (ii) In which orientation will the bar magnet be in stable equilibrium in the magnetic field?, 19. State Gauss’s law in electrostatic. A cube with each side ‘a’ is kept in an electric field given by, ®, , E = C ´ i$, (as is shown in the figure) where C is a positive dimensional constant. Find out
Page 393 :
392, , Xam idea Physics—XII, y, , a, , a, , x, , z, , 20., , 21., 22., , 23., , 24., , 25., , 26., , 27., , (i) the electric flux through the cube, and, (ii) the net charge inside the cube, A capacitor of 200pF is charged by a 300V battery. The battery is then disconnected and the, charged capacitor is connected to another uncharged capacitor of 100 pF. Calculate the difference, between the final energy stored in the combined system and the initial energy stored in the single, capacitor., Draw a labelled diagram of a moving coil galvanometer and explain its working. What is the, function of radial magnetic field inside the coil?, 1 1 1, Define power of a lens. Write its units. Deduce the relation = +, for two thin lenses kept in, f f1 f 2, contact coaxially., Write two characteristic features observed in photoelectric effect which support the photon picture, of electromagnetic radiation., 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. State clearly how this graph, can be used to determine (i) Planck’s constant and (ii) work function of the material., Define modulation index. Give its physical significance. For an amplitude modulated wave, the, maximum amplitude is found to be 10V while the minimum amplitude is 2V. Determine the, modulation index m., e1,r1, Two cells of emfs e1 , e 2 and internal resistance r1, and r2 respectively are connected in parallel as, shown in the figure., I, I, Deduce the expressions for, A, B, (i) the equivalent e.m.f. of the combination,, (ii) the equivalent resistance of the combination, and, e2,r2, (iii) the potential difference between the points A, and B., Using Bohr’s postulates for hydrogen atom, 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), where, K = –2U. Hence deduce the expression for the total energy in the nth energy level of hydrogen, atom., Define a wavelength.Use Huygens’ geometrical construction to show the propagation of a plane, wavefront from a rarer medium (i) to a denser medium (ii) undergoing refraction., Hence derive Snell’s law of refraction.
Page 394 :
Examination Papers, , 393, , OR, (a) Use Huygens’ geometrical construction to show the behaviour of a plane wavefront., (i) passing through a biconvex lens;, (ii) reflecting by a concave mirror., (b) When monochromatic light is incident on a surface separating two media, why does the, refracted light have the same frequency as that of the incident light?, 28., (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., (b) The intensity at the central maxima in Young’s double slit experimental set-up is I0. Show, that the intensity at a point where the path difference is l 3 is I 0 / 4., 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., Explain clearly why the secondary maxima go on becoming weaker with increasing n., (b) When the width of the slit is made double, how would this affect the size and intensity of the, central diffraction band? Justify., 29., (a) State the principle on which AC generator works. Draw a labelled diagram and explain its, working., (b) A conducting rod held horizontally along East-West direction is dropped from rest from a, certain height near the Earth’s surface. 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., OR, (a) State the principle of a step-up transformer. Explain, with the help of a labelled diagram, its, working., (b) Describe briefly any two energy losses, giving the reasons for their occurrence in actual, transformers., 30., (a) Draw the circuit for studying the input and output characteristics of an n-p-n transistor in CE, configuration. Show, how from the output characteristics, the information about the current, amplification factor ( bac) can be obtained., (b) Draw a plot of the transfer characteristic (V0 versus Vi) for a base-biased transistor in CE, configuration. Show for which regions in the plot, the transistor can operate as a switch., 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., 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.
Page 395 :
394, , Xam idea Physics—XII, , CBSE (Foreign) Set–II, Questions uncommon to Set–I, 1. What is the angle of dip at a place where the horizontal and vertical components of the Earth’s, magnetic field are equal?, 2. Show on a graph variation of the de-Broglie wavelength (l) associated with the electron versus, 1/ V , where V is the accelerating potential for the electron., 7. 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. Predict the direction of the induced current in the loop., ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, Q, , ×, , ×, , P, ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , S, , R, , 8. Why is there no work done in moving a charge from one point to another on an equipotential, surface?, 10. A magnetised needle of magnetic moment 4.8 × 10–2 J T–1 is placed at 30° with the direction of, uniform magnetic field of magnitude 3 × 10–2 T. Calculate the torque acting on the needle., 12. Trace the path of ray (P) of light passing through the glass prism as shown in the figure. The, prism is made of glass with critical angle ic= 40°., A, P, , 45°, B, , C, , 13. In a given sample, two radioactive nuclei, A and B, are initially present in the ratio of 4:1. The half, lives of A and B are respectively 25 years. Find the time after which the amounts of A and B, become equal., 14. Calculate the quality factor of a series LCR circuit with L = 4.0 H, C = 1 mF and R = 20 W., Mention the significance of quality factor in LCR circuit., 15. Figure shows a block diagram of a detector for amplitude modulated signal. Identify the boxes ‘X’, and ‘Y’ and write their functions., 26. A capacitor of 150 pF is charged by a 220 V battery. The battery is then disconnected and the, charged capacitor is connected to another uncharged capacitor of 50 pF. Calculate the difference, between the final energy stored in the combined system and the initial energy stored in the single, capacitor.
Page 396 :
Examination Papers, , 395, , CBSE (Foreign) Set–III, Questions uncommon to Set – I & II, 1. Write the relation between de-Broglie wavelength (l) associated with the electron and its kinetic, energy E., 2. What are isotopes? Give one example., 5. 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. Why do the equipotential due to a uniform electric field not intersect each other?, 9. A right angle prism is placed as shown in the figure. Given that the prism is made of glass with, critical angle 40°, trace the path of the ray P incident normal to the face AC., A, , P, , 45°, B, , C, , 11. A short bar magnet of magnetic moment 0.5 J/T is placed with its axis is 30° to a uniform, magnetic of 0.1 T. Calculate (i) the magnitude of the torque experienced and (ii) the direction on, which it acts., 14. A conductor of length ‘l’ is connected to a dc source of potential ‘V’. If the length of the, conductor is doubled by gradually stretching it, keeping ‘V’ constant, how will (i) drift speed of, electrons and (ii) resistance of the conductor be affected? Justify your answer., 15. Calculate the quality factor of a series LCR circuit with L = 4.0 H, C = 4 mF and R = 20W. Mention, the significance of quality factor in LCR circuit., 21. Define modulation index. Give its physical significance., For an amplitude modulated wave, the maximum amplitude is 4V. Determine the modulation, index m., 27. A capacitor of 400pF is charged by a 100V battery. The battery is then disconnected and the, charged capacitor is connected to another uncharged capacitor of 100 pF. Calculate the difference, between the final energy stored in the combined system and the initial energy stored in the single, capacitor.
Page 397 :
396, , Xam idea Physics—XII, , Solutions, CBSE (Foreign) Set–I, 1. Electric field intensity is zero inside the hollow spherical charged conductor. So, no work is done, in moving a test charge inside the conductor and on its surface. Therefore, there is no potential, difference between any two points inside or on the surface of the conductor., 2. (a) 0°, , (b) 90°, , 3. By Lenz’s law, the direction of induced current is such that it opposes its own cause of, production. The induced current opposes the increase in magnetic flux. Hence the direction of, induced current is PSRQP (anticlockwise)., 4., (i) Infrared rays, (ii) Microwaves., æ 1, 1, 1 ö, 5. 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. Hence, we can, say by replacing red light with violet light, decreases the focal length of the lens used., 6. The relationship is, R = R0 A, , 1, , 3, , where, , R = Radius of nucleus, A = Mass number, 1.22, Å, 7. We know l =, V, l V = constant, \, The nature of the graph between l and V is hyperbola., , V, , 8. Dipole moment of an electric dipole is the product of either of charge and the length of dipole. It, is a vector quantity., ®, , 9., , ®, , P = q (2 a ), eVt 1, (i) We know that v d = µ, ml l, When length is tripled, the drift velocity becomes one-third., l, (ii) R =r , l¢ = 3 l, A
Page 398 :
Examination Papers, , 397, , New resistance, 3l, l¢, = 9R, = r´, R¢ = r, ¢, A3, A, R' = 9R, Hence, the new resistance will be 9 times the original., 10., (i) By increasing resistance R the current through AB decreases, so potential gradient decreases., Hence a greater length of wire would be needed for balancing the same potential difference., So the null point would shift towards B., (ii) By decreasing resistance S, the current through AB remains the same, potential gradient does, not change. As K2 is open so there is no effect of S on null point., 11. 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, ®, , ®, , ®, , F = q ( v ´ B), , This magnetic force acts always perpendicular to the velocity, of charged particle. Hence magnitude of velocity remains, constant but direction changes continuously. Consequently, the path of the charged particle in a perpendicular magnetic, field becomes circular. The magnetic force (q v B) provides, the necessary centripetal force to move along a circular path., mv 2, mv, Then,, qv B =, r=, Þ, r, qB, , ×, , ×, , ×, , ×, , ×, , ×, , ×, r, , ×, , v, , ×, , ×, , ×, , ×, , P, , ×, , ×, , ×, , ×, , ×, , Here r = radius of the circular, 1, 12. We have,, Q=, R, 1, =, 10, , F, F, q, , ×, , ×, , ×, , v, ×, , ×, , path followed by the charge., L, C, 2, = 100, 2 ´ 10 -6, It signifies the sharpness of resonance., 13. An oscillating or accelerated is supposed to be source of an electromagnetic wave. 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. These oscillating electric and magnetic, field, hence, keep on regenerating each other and an electromagnetic wave is produced, The frequency of em wave = Frequency of oscillating charge, 14. We have, N = N 0 e - lt, For radio isotopes A and B, we can write, N A = N 0 e - lA t A, N B = 4N 0 e, , - lB t B, , Let t be the time after which N A = N B, t A = t B = t( say), , ...(i), ...(ii)
Page 399 :
398, , Xam idea Physics—XII, , \, , N 0 e - lA t = 4 N 0 e - lB t, , Þ, , 4 = e lB t - lA t, , log e 4 = (l B t - 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, T, TA1 / 2 úû, êë TB1 / 2, 1, 1 ö, 2 -1 ö, 2 = æç 2 = æç, Þ, Þ, ÷t, ÷t, è 50 100 ø, è 100 ø, Þ, t = 200 years, 15. 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. Hence, these signals are modulated by mixing with very high, frequency waves (carrier wave) by modulator power., Function of Y: The signal cannot be transmitted as such because they get weaken after travelling, long distance. Hence, use of power amplifier provides them necessary power before feeding the, signal to the transmitting antenna., A, 16. Given n g = 3, Þ, , i=0, At the interface AC,, By Snell’s Law, ng, sin i, =, sin r, na, But sin i = sin 0° = 0, hence r = 0, At the interface AB, i = 30°, Applying Snell’s Law, sin 30° n a, 1, =, =, Þ sin e = 3 sin 30°, sin e, ng, 3, , 30°, e, 60°, B, , Þ, , C, , e = 60°, , 17., n–type Semiconductor, (i) It is formed by doping pentavalent impurities., , p–type Semiconductor, (i) It is doped with trivalent impurities., , (iii) The electrons are majority carriers and holes are (ii) The holes are majority carriers and electrons are, minority carriers (nh >> ne )., minority carriers ( ne >> nh )., , OR, A light emitting diode is simply a forward biased p-n junction which emits spontaneous light, radiation. At the junction, energy is released in the form of photons due to the recombination of, the excess minority charge carrier with the majority charge carrier., Advantages, (i) Low operational voltage and less powder., (ii) Fast action and no warm up time required.
Page 400 :
Examination Papers, ®, , 18., , ®, , 399, , ®, , (i) We know t = M ´ B, , t = M B sin q, 0.063 = 0.9 × B × sin 30°, or, B = 0.14 T, (ii) The position of minimum energy corresponds to position of stable equilibrium., The energy (U) = –MB cos q, When q = 0° Þ U = –MB = Minimum energy, Hence, when the bar magnet is placed parallel to the magnetic field, it is the state of stable, equilibrium., (i) Gauss’s Law in electrostatics states that the total electric flux through a closed surface, 1, times the magnitude of that charge., enclosing a charge is equal to, e0, or, , 19., , ® ®, , f = ò E . dS =, s, , q, e0, , (ii) Net flux f = f1 + f2, ® ®, , Ù, n, , where f1 = E . dS, = 2aC dS cos 0° = 2 aC ´ a 2 = 2a 3 C, f2 = aC ´ a 2 cos 180° = -a 3 C, , Ù, n, f2, , a, , f1, a, , 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., 1, 20. 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., The charge would be shared between them., q q¢, Hence, Q = q + q¢ ,, =, C C¢, q ® charge on capacitor (first), q' ® charge on capacitor (second), C = 200 pF,, C ¢ = 100 pF, q, q¢, =, Þ q = 2q¢, -12, 200 ´ 10, 100 ´ 10 -12, Q 60 n C, = 20 nC, Then, q' = =, Q = 2q¢ + q¢ = 3q¢ Þ, 3, 3
Page 401 :
400, , Xam idea Physics—XII, , and, , q = 2q' = 40 nC, , Hence, total final energy U f =, , q 2 q¢ 2, +, 2C 2C ¢, , 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 - Ui = 6 ´ 10 -6 - 9 ´ 10 -6 = -3 ´ 10 -6 J, , Þ, DU = 3 ´ 10 -6 J (in magnitude), 21. 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., A soft iron core in cylindrical form is placed between the coil., One end of coil is attached to suspension wire which also serves as one terminal (T1 ) of, galvanometer. The other end of coil is connected to a loosely coiled strip, which serves as the, other terminal (T2 ). The other end of the suspension is attached to a torsion head which can be, rotated to set the coil in zero position. 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. The, levelling screws are also provided at the base of the instrument., H, , Suspension wire, M, , T1, T2, , Coil, , NIBl, b, , N, , S, , N, , S, , N, , S, , NIBl, Coiled strip, , (a), , (b), Magnetic lines of, force of radial magnetic field, , (c), , The pole pieces of the permanent magnet are cylindrical so that the magnetic field is radial at any, position of the coil., Principle and working: When current ( I ) is passed in the coil, 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, N is, the number of turns in a coil.
Page 402 :
Examination Papers, , 401, , When the magnetic field is radial, as in the case of cylindrical pole pieces and soft iron core, then, in every position of coil the plane of the coil, is parallel to the magnetic field lines, so that q = 90°, and sin 90° = 1, Deflecting torque,, t = NIAB, If C is the torsional rigidity of the wire and q is the twist of suspension strip, then restoring torque, =C q, For equilibrium, deflecting torque = restoring torque, i.e., NIAB = C q, NAB, ...(i), q=, I, \, C, i.e., qµI, deflection of coil is directly proportional to current flowing in the coil and hence we can, construct a linear scale., 22. 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., A, , B, , P, , O, , I, , I1, , v, , u, , v1, , An object is placed at point O. The lens A produces an image at I1 which serves as a virtual object, for lens B which produces final image at I., Given, the lenses are thin. The optical centres (P) of the lenses A and B is co-incident., For lens A, we have, 1 1 1, ...(i), - =, v1 u f 1, 1 1, 1, ...(ii), For lens B, we have - =, v v1 f 2, Adding equations (i) and (ii),, 1 1 1 1, - = +, v u f1 f 2, , ...(iii), , If two lenses are considered as equivalent to a single lens of focal length f, then, 1 1 1, ...(iv), - =, v u f
Page 403 :
402, , Xam idea Physics—XII, , From equation (iii) and equation (iv), 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., (b) Photons are electrically neutral and are not, affected by presence of electric and magnetic, fields,, (i) From this graph, 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., f = hv 0 , where v0 = Threshold frequency, 24. Modulation index: It is the ratio of peak value of modulating signal to the peak value of carrier, wave., A, m= m, Ac, K.E. of electron (eV), , 23., , Physical significance: It signifies the level of distortion or noise. A lower value of modulation, index indicates a lower distortion in the transmitted signal., Maximum amplitude, Amax = Ac + Am, = 10 V, Minimum aptitude, Amin = Ac – Am = 2 V, A, - A min, 10 - 2, 8, 2, =, =, = ., Modulation index = max, A max + A min, 10 + 2 12 3, 25. Here,, , I = I1 + I 2, , Let V = Potential difference between A and B., For cell e1, e -V, Then, V = e1 - I 1 r1 Þ I 1 = 1, r1, e -V, Similarly, for cell e 2, I2 = 2, r2, Putting these values in equation (i), e -V e 2 -V, I= 1, +, r1, r2, or, , , r1, , ...(i), , e ö, æe, æ1 1 ö, I = ç 1 + 2 ÷ -V ç + ÷, è r1 r 2 ø, è r1 r 2 ø, , I1, I, , I, , A, , B, I2, , r2
Page 405 :
404, , Xam idea Physics—XII, , 27. Wavefront : A wavefront is a locus of all particles of medium vibrating in the same phase., Huygen’s Principle : There are some phenomena like interference, diffraction and polarisation, which could not be explained by Newton’s corpuscular theory. 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. The, locus of particles of a medium vibrating in the same phase is called a wavefront. For a point, source, the wavefront is spherical; while for a line source the wavefront is cylindrical. The distant, wavefront is plane., Each point of a wavefront acts as a source of secondary wavelets. 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, it is, B, deviated from its path. This phenomenon is called, o, 90, i, i, refraction. In transversing from first medium to another X, B', Y, r, A, medium, the frequency of wave remains unchanged but its, o, r, 90, speed and the wavelength both are changed. Let XY be a, surface separating the two media ‘1’ and ‘2’. Let v1 and v 2, A', be the speeds of waves in these media., 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. Clearly BB¢ = v1t. As the wavefront AB advances, it strikes the, points between A and B¢ of boundary surface. According to Huygen’s principle, secondary spherical, wavelets originate from these points, which travel with speed v1 in the first medium and speed v 2, in the second medium., First of all secondary wavelet starts from A, which traverses a distance AA¢ ( = v 2 t) in second medium, in time t. In the same time-interval t, the point of wavefront traverses a distance BB¢ ( = v1t) in first, medium and reaches B¢ , from, where the secondary wavelet now starts. Clearly BB¢ = v1t and, AA¢ = v 2 t., 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 incident wavefront AB advances, the secondary wavelets start from points, between A and B¢ , one after the other and will touch A¢ B¢ simultaneously. According to Huygen’s, principle A¢ B¢ is the new position of wavefront AB in the second medium. Hence A¢ B¢ will be the, refracted wavefront., First law : As AB, A¢ B ¢ and surface XY are in the plane of paper, therefore the perpendicular, drawn on them will be in the same plane. As the lines drawn normal to wavefront denote the rays,, therefore we may say that the incident ray, refracted ray and the normal at the point of incidence, all lie in the same plane., This is the first law of refraction., Second law : Let the incident wavefront AB and refracted wavefront A¢ B¢ make angles i and r, respectively with refracting surface XY.
Page 406 :
Examination Papers, , In right-angled triangle AB¢ B, Ð ABB¢ = 90°, vt, BB¢, \, sin i = sin Ð BAB¢ =, = 1, AB¢, AB¢, Similarly in right-angled triangle AA¢ B ¢ , Ð AA¢ B ¢ = 90°, AA¢ v 2 t, sin r = sin Ð AB¢ A¢ =, =, \, AB¢ AB¢, Dividing equation (i) by (ii), we get, v, sin i, = 1 = constant, sin r, v2, , 405, , ...(i), , ...(ii), , ...(iii), , As the rays are always normal to the wavefront, therefore the incident and refracted rays make, angles i and r with the normal drawn on the surface XY i.e. i and r are the angle of incidence and, angle of refraction respectively. 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. This is the second law of refraction,, and is called the Snell’s law., When a light wave travels from rarer to denser medium, the speed decreases. It does not imply, reduction its energy. This is because energy of wave depends on its frequency and not on its, speed., OR, (a) Wave Nature of Light : Huygen’s Theory, There are some phenomena like interference, diffraction and polarisation which could not be, explained by Newton’s corpuscular theory. 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. The locus of particles of a medium, vibrating in the same phase is called a wavefront. For a point source, the wavefront is, spherical; while for a line source the wavefront is cylindrical. The distant wavefront is, plane., (ii) Each point of a wavefront acts as a source of secondary wavelets. The envelope of all, wavelets at a given instant gives the position of a new wavefront., Rectilinear Propagation of Light : According to Newton’s corpuscular theory, the path of, light is a straight line, but according to wave theory the rectilinear propagation of light is only, approximate., (i) The action of a convex lens : A plane wavefront becomes spherical convergent, wavefront after refraction. Fig., Spherical wavefront, , Plane wavefront, , Lens
Page 407 :
406, , Xam idea Physics—XII, , (ii) Action of concave mirror : A plane wavefront becomes spherical convergent after, reflection. Fig., Plane wavefront, , Spherical wavefront, , mirror, , 28., , (b) As frequency of light is the characteristic of its source, light reflects and refracts due to the, interaction of incident light with the atoms of the medium. These atoms always take up the, frequency of the incident light which forces them to vibrate and emit light of same frequency., Hence, frequency remains same., lD, (a) (i) Fringe width (b) =, d, If d decreases, b increases, (ii) For interference fringe, the condition is, s l, <, D d, where s = size of source, D = distance of source from slits., If the source slit width increases, fringe pattern gets less sharp or faint., When the source slit is made wide which does not fullfil the above condition and, interference pattern not visible., (iii) The central fringes are white. On the either side of the central white fringe the coloured, bands (few coloured maxima and minima) will appear. This is because fringes of, different colours overlap., (b) Intensity at a point is, f, I = I 0 cos 2, 2, l, 2p l 2p, At path difference , the phase difference, f = . =, 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, we get diffraction pattern on a screen placed behind the slit. The diffraction pattern, contains bright and dark bands, the intensity of central band is maximum and goes on, decreasing on both sides., Explanation : Let AB be a slit of width ‘a’ and a parallel beam of monochromatic light is, incident on it. According to Fresnel the diffraction pattern is the result of superposition of a, large number of waves, starting from different points of illuminated slit., 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 path difference between rays diffracted at points A and B,, D = BP - AP = BN
Page 408 :
Examination Papers, , 407, , In D ANB , ÐANB = 90°, \ and ÐBAN = q, BN, or BN = AB sin q, \, sin q =, AB, As AB = width of slit = a, \ Path difference,, ....(i), D = a sin q, To find the effect of all coherent waves at P, we have to sum up their contribution, each with a, different phase. This was done by Fresnel by rigorous calculations, but the main features may, be explained by simple arguments given below :, , P, q, , A, Light from, source, , M1, , q, , O, M2 90o N, , C, q, , B, , At the central point C of the screen, the angle q is zero. Hence the waves starting from all, points of slit arrive in the same phase. This gives maximum intensity at the central point C., If point P on screen is such that the path difference between rays starting from edges A and B, is l , then path difference, l, a sin q = l Þ sin q =, a, l, If angle q is small, sin q = q =, ...(ii), a, a, Minima : Now we divide the slit into two equal halves AO and OB, each of width . Now for, 2, a, every point, M 1 in AO, there is a corresponding point M 2 in OB, 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 = . 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. Thus equation (2) gives the angle of diffraction at, which intensity falls to zero. Similarly it may be shown that the intensity is zero for, nl, sin q =, , with n as integer. Thus the general condition of minima is, a, ...(iii), a sin q = nl, Secondary Maxima : Let us now consider angle q such that
Page 409 :
408, , Xam idea Physics—XII, , 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. If we take the first two of parts of slit, 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., 2, The remaining third part will contribute to the intensity at a point between two minima., Clearly there will be a maxima between first two minima, but this maxima will be of much, weaker intensity than central maximum. This is called first secondary maxima. In a similar, manner we can show that there are secondary maxima between any two consecutive minima;, and the intensity of maxima will go on decreasing with increase of order of maxima. In, general the position of nth maxima will be given by, 1, ...(iv), a sin q = æç n + ö÷ l ,, [ n =1, 2, 3, 4, .... ], 2 ø, è, The intensity of secondary maxima decrease with increase of order n because with increasing, n, the contribution of slit decreases., For n = 2, it is one-fifth, for n = 3, it is one-seventh and so on., 2Dl, (b) Width of central Maxima ' b' =, a, a ® size of slit, If size of slit is doubled, width of central maxima becomes half. Intensity varies as square of, slit width. It width of slit is doubled, intensity gets four times., (a) AC generator: A dynamo or generator is a device which converts mechanical energy into, electrical energy. It is based on the principle of electromagnetic induction., Construction: It consists of the four main parts:, (i) Field Magnet: It produces the magnetic field. In the case of a low power dynamo, the, magnetic field is generated by a permanent magnet, while in the case of large power, dynamo, the magnetic field is produced by an electromagnet., (ii) Armature: It consists of a large number of turns of insulated wire in the soft iron drum, or ring. It can revolve round an axle between the two poles of the field magnet. 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., (iii) Slip Rings: The slip rings R1 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., (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 output current in external load R L is taken, through these brushes., sin q = q =, , 29.
Page 410 :
Examination Papers, , 409, , Working: When the armature coil is rotated in the strong magnetic field, the magnetic flux, linked with the coil changes and the current is induced in the coil, its direction being given by, Fleming’s right hand rule. Considering the, w, armature to be in vertical position and as it, Armature coil, rotates in anticlockwise direction, the wire ab, b, moves upward and cd downward, so that the, c, direction of induced current is shown in fig. In, the external circuit, the current flows along, N B, S, B1 R L B 2 . The direction of current remains, Field magnet, unchanged during the first half turn of, a, armature. During the second half revolution,, d, the wire ab moves downward and cd upward,, B1, so the direction of current is reversed and in, R1, external circuit it flows along B 2 R L B1 . Thus, Brushes Load, Slip rings, the direction of induced emf and current, RL, changes in the external circuit after each half, R2, B2, revolution., Expression for Induced emf: If N is the number of turns in coil, f the frequency of rotation,, A area of coil and B the magnetic induction, 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., Current produced by an ac generator cannot be measured by moving coil ammeter; because, the average value of ac over full cycle is zero., The source of energy generation is the mechanical energy of rotation of armature coil., (b) As the earth’s magnetic field lines are cut by the falling, rod, the change in magnetic flux takes place. This change, in flux induces an emf across the ends of the rod., Since the rod is falling under gravity,, e, ( Q u = 0), v = gt, Induced emf, e = Blv, e = Blgt, \, e µt, t, OR, (a) Transformer: Transformer is a device by which an alternating voltage may be decreased or, increased. This is based on the principle of mutual-induction., Construction: It consists of laminated core of soft iron, on which two coils of insulated, copper wire are separately wound. These coils are kept insulated from each other and from, the iron-core, but are coupled through mutual induction. 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. 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. Transformers are of two types:
Page 411 :
410, , Xam idea Physics—XII, , 1. 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., (i. e., N S > N p )., 2. 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. e., N S < N p ), (A.C. mains), Primary, , laminated, iron core, , Primary, , Primary, , Secondary, , Secondary, , (a) Step up, , (b) Step down, , Core, , Secondary, Transformer, , Working: When alternating current source is connected to the ends of primary coil, the, current changes continuously in the 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., Let N p be the number of turns in primary coil, NS the number of turns in secondary coil and f, the magnetic flux linked with each turn. 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. According to, Faraday’s laws the emf induced in the primary coil, Df, ...(i), ep =-Np, Dt, and emf induced in the secondary coil, Df, ...(ii), eS = - NS, Dt, From (i) and (ii), eS NS, ...(iii), =, ep Np, If the resistance of primary coil is negligible, the emf ( e p ) induced in the primary coil, will be, equal to the applied potential difference (V p ) across its ends. Similarly if the secondary circuit, is open, then the potential difference VS across its ends will be equal to the emf ( e S ) induced, in it; therefore, VS e S N S, ...(iv), =, =, = r (say), Vp e p N p
Page 412 :
Examination Papers, , where r =, , 411, , NS, is called the transformation ratio. If i p and i s are the instantaneous currents in, Np, , primary and secondary coils and there is no loss of energy; then, For about 100% efficiency, 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, , \, , ...(v), , In step up transformer,, , N s > N p ® r >1;, , So, , VS > V p and i S < i p, , i.e. step up transformer increases the voltage., (b), , (i) Flux leakage: There is always some flux leakage, that is, 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., (ii) Eddy currents: The alternating magnetic flux induces eddy currents in the iron core and, causes heating., , (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, , – IB, , +, , –, , µA, VBB, , +, , +, , VBE, –, , Rh1, , –, , +, , IC, C, +, , B, , +, , VCE, E, IE, , –, , –, , VCC, , CE, , =, 10, , V, , 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 base current is changed to some other, fixed value and the observations of I C versus VCE are, VCE = 5V, repeated. Fig. represents the output characteristics of a, V, common-emitter circuit., The characteristic curves show:, (i) When collector-emitter voltage VCE is, VBE, increased from zero, 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 value of VCE upto which, collector current IC changes is called the knee voltage Vknee ., IB, , 30.
Page 413 :
Xam idea Physics—XII, , Collector current in (IC) mA, , 412, , 12.5, Base current (IB), , 10, 7.5, , 60 µA, 50 µA, 40 µA, , 5, , 30 µA, , 2.5, , 20 µA, 10 µA, , 0.5, , 1, , 1.5, , 2, , 2.5, , 3, , 3.5, , 4, , 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, the region where collector current, ( I C ) is almost independent of VCE ., Now we choose any two characteristic curves for given values of I B and find the two, corresponding values of I C ., æ DI ö ( I ) - ( I C ) 1, Then, b =ç C ÷ = C 2, è DI B ø ( I B ) 2 - ( I B ) 1, From graph ( I C ) 1 = 5 × 2 mA, ( I C ) 2 = 7 × 3 mA, ( I B ) 1 = 30 mA,, ( I B ) 2 = 40 mA, (7 × 3 - 5 × 2) mA 2 ×1 ´ 10 - 3, b=, =, = 210, ( 40 - 30) mA, 10 ´ 10 - 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., A transistor can be used to turn current ON or OFF rapidly in electrical circuits., IC, , C, RB, , IB, , B, , VCE, +, , E, Vi, –, , VBB, , 1, , VO, , RL, , VBE, , +, , –, , 2, VCC, , DI C, ., DI B
Page 414 :
Examination Papers, , 413, , Operation: The circuit diagram of n-p-n transistor in CE configuration working as a switch is, shown in fig. VBB and VCC are two dc supplies which bias base-emitter and emitter collecter, junctions respectively., Let VBB be the input supply voltage. This is also input dc voltage (VC ). The dc output voltage, is taken across collector-emitter terminals, R L is the load resistance in output circuit., Applying Kirchhoff’s second law to input and output meshes (1) and (2), we get, …(i), VBB = I B R B + VBE, and, …(ii), VCC = I C R L + VCE, We have VBB = Vi and VCE = V0 , so above equations take the form, …(iii), Vi = VBE + I B R B, and, …(iv), V0 ( = VCE ) = VCC - I C R L, Let us see the change in V0 due to a change in Vi . In case of Si transistor; the barrier voltage, across base-emitter junction is 0 × 6 V. Therefore, when Vi is less than 0 × 6 V, there is no, collector current ( I C = 0), so transistor will be in cut off state. Hence, from (iv) with, I C = 0; V0 = VCC ., When Vi becomes greater than 0 × 6 V, I C begins to, Cut off, Active, flow and increase with increase of Vi . Thus, from, region region, (iv), V0 decreases upto Vi =1 V; the increase in I C is, linear and so decrease in output voltage V0 is linear., Beyond Vi =1 V, the change in collector current and, V0, hence in output voltage V0 is non-linear and the, Saturation, region, transistor goes into saturation. With further increase, in Vi , the output voltage further decrease towards, zero (though it never becomes zero)., AV, If we plot V0 versus Vi , we get the graph as shown in, fig. [This characteristics curve is also called transfer, characteristic curve of base biased transistor.], Vi, 1V, The curve shows that there are non-linear regions., 0.6 V, (i) between cut off state and active state and (ii) between active state and saturation state; thus, showing that the transitions (i) from cut off to active state and from active to saturation state, are not sharply defined., Now we are in the position to explain the action of transistor as a switch. When transistor is, non-conducting ( I C = 0), it is said to be ‘switched off’ but when it is conducting (I C is not, zero); it is said to be ‘switched ON’., As long as input voltage Vi is low and unable to overcome the barrier voltage of the emitter, base junction, V0 is high (I C = 0 and V0 = VCC ), 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 - I C R L ) is, low, very near to zero, so the transistor is ‘switched ON’. Thus we can say low input switches, the transistor is OFF state and high input switches it ON., The switching circuits are designed in such a way that the transistor does not remain in active, state., The transistor can operate as a switch in cut off region and saturation region.
Page 415 :
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., We know that reverse current is due to the flow of electrons, (minority carriers) from p ® n and holes from n ® p. As, the reverse bias voltage is increased, the electric field at the, junction becomes significant. When the reverse bias voltage, V = Vz, 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. These electrons causes high, current at breakdown., 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. Now, 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. Thus, the zener diode acts as a, voltage regulator., , BV, = tan d, BH, , Given, BV = B H then tan d =1, Angle of dip, d = 45°, h, 2. l =, 2 meV, , Reverse bias, Vz, , Forward bias, V(V), , I (µA), , Rs, Unregulated, voltage, (VL), , CBSE (Foreign) Set–II, 1. We know, , I (mA), , IL, Load, RL, , Regulated, voltage, (Vz)
Page 416 :
Examination Papers, ×, , ×, , 415, ×, , 7. From the given figure, it is clear that the magnetic flux decreases. By, P, Q, Lenz’s law, the induced current opposes the change in flux by producing ×, ×, ×, the magnetic field in the same direction as the external magnetic field., R, By right hand thumb rule, the induced current will flow in a clockwise × S, ×, ×, direction. i.e., PQRSP., 8. The potential difference between any two points of equipotential surface is zero. We have, W, (Q dv = 0), = dv, q, Thus W = 0, the work done in moving a charge on an equipotential is zero., 10. We have,, t = M B sin q, t ® torque acting on magnetic needle, M ® Magnetic moment, B ® Magnetic field strength, Then, t = 4.8 × 10–2 × 3 × 10–2 sin 30°, 1, = 4.8 × 10–2 × 3 × 10–2 ×, 2, –4, t = 7.2 × 10 Nm, 12. From the figure, the incident ray is normal to the surface of, prism. So, incident angle = 0°, Then, angle of refraction will be zero. It means that the ray of, light will pass through the prism undeviated, reaches the other, end of prism., The second angle of incident = 45° (greater than critical angle, of prism). The ray of light undergoes the phenomenon of total, internal reflection and continues in the same manner., 13. We have N = N 0 e - lt, where, , For radio isotopes A and B, we can write, N A = 4 N 0 e - lA t A, N B = N 0 e - lB t B, Let t be the time after which N A = N B, t A = t B = t( say), \, , 4 N 0 e - lA t = N 0 e - lA t, , Þ, Þ, , 4 = 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 ÷ø, è 1/2, 2 -1 ö, 2 = æç, ÷t, è 50 ø, , Þ, Þ, , Þ, , 1 1, 2 = æç - ö÷ t, 25, 50 ø, è, , Þ, , t =100 years., , 45°, 45°, 45°, 45° 45°, 45°
Page 417 :
416, , Xam idea Physics—XII, , 14. We have, the quality factor ‘Q’ of a series LCR, 1 L, Here,, R = 20, L = 4H, C = 10–6F, Q=, C, R, 1, 4, = 100, Q=, 20 10 -6, The quality factor measures the sharpness of resonance of an LCR circuit., 15. Here, X ® Rectifier, Y ® Envelope detector, Rectifier: It allows only the positive half of the AM input wave to go onwards., It separates the message signal from the carrier wave., 26. Given, 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, common potential (V) = 1, =, C1 + C 2 C1 + C 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 = 2.25 × 10–6 J, 2, Change in energy, U 2 - U1 = 2.25 ´ 10 -6 – 3 × 10–6 = – 0.75 × 10–6 J, Final energy,, , CBSE (Foreign) Set–III, 1. 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. Isotopes are the two or more nuclides having the same atomic number (Z) but different mass, number (A)., One example of isotope is 11 H , 12 H , 13 H. (i.e., isotopes of hydrogen), 5. Poles of the earth., 7. This is because, at the point of intersection there will be two values of electric potential which is, not possible.
Page 418 :
Examination Papers, C, , 417, , 9. From the given figure, incident angle = 0, Then angle of refraction will be zero. Thus the light ray, will pass through the prism undeviated, and reaches to the, other end of prism., P, 45°, The second angle of incident = 45° which is greater than, the critical angle of prism., 45°, Hence the light ray undergoes the phenomenon of total, internal reflection making angle 90° with the incident ray., A, B, 11., (i) We know t = MB sin q, where t = Torque acting on a short bar magnet, P', M = Magnetic moment, B = Magnetic field, t = 0.5 × 0.1 × sin 30°, 1, = 0.5 × 0.1 × = 2.5 × 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 ÷., è, ø, 14., (i) We know that drift speed, eVt 1, vd = µ, ml l, When length of the conductor is doubled, drift velocity gets halved., rl, (ii) As, R=, A, A, Now, l¢ = 2l and A ¢ =, 2, r( 2l), rl, \ New resistance, R ¢ =, =4, A, A, 2, Thus, resistance becomes four times., 15. We have, the quality factor of a series LCR., 1 L, where R = 20 W, Q=, 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.
Page 419 :
418, , Xam idea Physics—XII, , 21. Modulation index: It is the ratio of peak value of modulating signal to the peak value of carrier, wave., A, m= m, Ac, It signifies the level of distortion., Maximum amplitude, Amax = Ac + Am = 8V, Minimum amplitude, Amin = Ac – Am = 4V, A, - A min, m = max, A max + A min, 8 -4, 4 1, =, =, =, 8 + 4 12 3, 27. Given, C1 = 400, 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, common potential (V) = 1, =, C1 + C 2 C1 + C 2, Initial energy, U1 =, , =, , 400 ´ 10 -12 ´ 100, ( 400 + 100) ´ 10 -12, , = 80 V, , 1, (C1 + C 2 )V 2, 2, 1, = ( 400 + 100) ´ 10 -12 ´ (80) 2 = 1.6 × 10–6 J, 2, Difference in energy, , Fixed energy, U2 =, , U 2 - U1 = 1. 6 ´ 10 -6 J - 2 ´ 10 -6 J, = – 0.4 × 10–6 J
Page 420 :
CBSE Examination Papers, Delhi–2013, SET–I, Time allowed : 3 hours, , Maximum marks: 70, , General Instructions:, (i) All questions are compulsory., (ii) There are 29 questions in total. Questions 1 to 8 are very short answer questions and carry one mark each., (iii) Questions 9 to 16 carry two marks each, questions 17 to 25 carry three marks each, question 26 is, value-based question carry four marks and questions 27 to 29 carry five marks each., (iv) 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., (v) Use of calculators is not permitted. However, you may use log tables if necessary., (vi) You may use the following values of physical constants wherever necessary., c = 3 ´ 10 8 m/s, h = 6.63 ´ 10 -34 Js, e = 1.6 ´ 10 -19 C, m 0 = 4p ´ 10 -7 TmA -1, 1, = 9 ´ 10 9 Nm 2 / C 2, 4pe, me = 9.1 ´ 10 -31 kg, 1. What are permanent magnets? Give one example., 2. What is the geometrical shape of equipotential surfaces due to a single isolated charge?, 3. Which of the following waves can be polarized (i) Heat waves (ii) Sound waves? Give reason to, support your answer., 4. A capacitor has been charged by a dc source. What are the magnitudes of conduction and, displacement currents, when it is fully charged?, 5. Write the relationship between angle of incidence ‘i’, angle of prism ‘A’ and angle of minimum, deviation for a triangular prism.
Page 421 :
420, , Xam idea Physics—XII, , 6. The given graph shows the variation of photo-electric current (I) versus applied voltage (V) for two, different photosensitive materials and for two different intensities of the incident radiation. Identify, the pairs of curves that correspond to different materials but same intensity radiation., , I, , 3, 1, , 2, 4, V, , 7. A 10 V battery of negligible internal resistance is connected across a 200 V battery and a resistance of, 38 W as shown in the figure. Find the value of the current in circuit., 10 V, , 38 W, , 200 V, , 8. The emf of a cell always greater than its terminal voltage. Why? Give reason., 9., , (a), , Write the necessary conditions for the phenomenon of total internal reflection to occur., , (b) Write the relation between the refractive index and critical angle for a given pair of optical, media., 10. State Lenz’s Law., A metallic rod held horizontally along east-west direction, is allowed to fall under gravity. Will there, be an emf induced at its ends? Justify your answer., 11. A convex lens of focal length 25 cm is placed coaxially in contact with a concave lens of focal length, 20 cm. Determine the power of the combination. Will the system be converging or diverging in, nature?, 12. An ammeter of resistance 0.80 W can measure current upto 1.0 A., (i) What must be the value of shunt resistane to enable the ammeter to measure current upto 5.0 A?, (ii) What is the combined resistance of the ammeter and the shunt?
Page 422 :
Examination Papers, , 421, , 13. In the given circuit diagram, a voltmeter ‘V’ is connected aross a lamp ’L’. How would (i) the, brightness of the lamp and (ii) voltmeter reading ‘V’ be affected, if the value of resistance ‘R’ is, decreased? Justify your answer., , V, , L, +, , 9V, , –, , R, , 14., , (a), , ®, , An em wave is travelling in a medium with a velocity v = v i$. Draw a sketch showing the, propagation of the em wave, indicating the direction of the oscillating electric and magnetic, fields., , (b) How are the magnitudes of the electric and magnetic fields related to the velocity of the em, wave?, 15. Block diagram of a receiver is shown in the figure:, Receiving, antenna, , Received, signal, , (a), , Amplifier, , Detector, , X, , Y, , Output, , Identity ‘X’ and ‘Y’., , (b) Write their functions., 16. Explain, with the help of a circuit diagram, the working of a photo-diode. Write briefly how it is used, to detect the optical signals., OR, Mention the important considerations required while fabricating a p-n junction diode to be used as a, Light Emitting Diode (LED). What should be the order of band gap of an LED if it is required to emit, light in the visible range?, 17. Write three important factors which justify the need of modulating a message signal. Show, diagrammatically how an amplitude modulated wave is obtained when a modulating signal is, superimposed on a carrier wave.
Page 423 :
422, , Xam idea Physics—XII, , 18. A capacitor of unknown capacitance is connected across a battery of V volts. The charge stored in it is, 360 mC. When potential across the capacitor is reduced by 120 V, the charge stored in it becomes, 120 mC., Calculate:, (i) The potential V and the unknown capacitance C., (ii) What will be the charge stored in the capacitor, if the voltage applied had increased by 120 V?, OR, A hollow cylindrical box of length 1 m and area of cross-section 25 cm2 is placed in a three, dimensional coordinate system as shown in the figure. The electric field in the region is given by, ®, , E =50 xi$, where E is in NC–1 and x is in metres. Find, (i) Net flux through the cylinder., (ii) Charge enclosed by the cylinder., Y, , O, , 1m, , X, , Z, , 19., , (a), , In a typical nuclear reaction, e.g., 2, 1H, , + 21 H ¾, ¾® 32 He + n + 3.27MeV,, , although number of nucleons is conserved, yet energy is released. How? Explain., (b) Show that nuclear density in a given nucleus is independent of mass number A., 20., , (a), , Why photoelectric effect can not be explained on the basis of wave nature of light? Give reasons., , (b) Write the basic features of photon picture of electromagnetic radiation on which Einstein’s, photoelectric equation is based., 21. A metallic rod of length ‘l’ is rotated with a frequency v with one end hinged at the centre and the, other end at the circumference of a circular metallic ring of radius r, about an axis passing through the, centre and perpendicular to the plane of the rinig. A constant uniform magnetic field B parallel to the, axis is present every where. Using Lorentz force, explain how emf is induced between the centre and, the metallic ring and hence obtain the expression for it., 22. Output characteristics of an n-p-n transistor in CE configuration is shown in the figure. Determine:, (i) dynamic output resistance, (ii) dc current gain and, (iii) ac current gain at an operating point VCE =10 V, when I B =30mA.
Page 424 :
Examination Papers, , 423, , Y, 6, , 50 mA, , 5, , 40 mA, , 4, , 30 mA, , IC(mA), , 3.5, 3, , 20 mA, , 2, , 10 mA, , 1, , IB = 0, , 0, , 2, , 4, , 6, , 8, , 10, , 12, , 14, , 16, , 18, , X, , VCE(V), , 23. Using Bohr’s postulates, obtain the expression for the total energy of the electron in the stationary, states of the hydrogen atom. Hence draw the energy level diagram showing how the line spectra, corresponding to Balmer series occur due to transition between energy levels., 24., , (a), , In what way is diffraction from each slit related to the interference pattern in a double slit, experiment?, , (b) Two wavelengths of sodium light 590 nm and 596 nm are used, in turn, to study the diffraction, taking place at a single slit of aperture 2 × 10–4 m. The distance between the slit and the screen is, 1.5 m. Calculate the separation between the positions of the first maxima of the diffraction, pattern obtained in the two cases., 25. In a series LCR circuit connected to an ac source of variable frequency and voltage v = v m sin wt,, draw a plot showing the variation of current (I) with angular frequency (w) for two different values of, resistance R1 and R 2( R1 > R 2 ). Write the condition under which the phenomenon of resonance, occurs. For which value of the resistance out of the two curves, a sharper resonance is produced?, Define Q-factor of the circuit and give its significance., 26. While travelling back to his residence in the car, Dr. Pathak was caught up in a thunderstorm., It became very dark. He stopped driving the car and waited for thunderstorm to stop., Suddenly he noticed a child walking alone on the road. He asked the boy to come inside the car, till the thunderstorm stopped. Dr. Pathak dropped the boy at his residence. The boy insisted, that Dr. Pathak should meet his parents. The parents expressed their gratitude to Dr. Pathak, for his concern for safety of the child., Answer the following question based on the above information:, (a) Why is it safer to sit inside a car during a thunderstorm?, (b) Which two values are displayed by Dr. Pathak in his actions?, (c), , Which values are reflected in parents’ response to Dr. Pathak?, , (d) Give an example of a similar action on your part in the past from everyday life.
Page 425 :
424, 27., , Xam idea Physics—XII, , (a), , Draw a ray diagram showing the image formation by a compound microscope. Hence obtain, expression for total magnification when the image is formed at infinity., , (b) Distinguish between myopia and hypermetropia. Show diagrammatically how these defects can, be corrected., OR, (a), , State Huygen’s principle. Using this principle draw a diagram to show how a plane wave front, incident at the interface of the two media gets refracted when it propagates from a rarer to a, denser medium. Hence verifiy Snell’s law of refraction., , (b) When monochromatic light travels from a rarer to a denser medium, explain the following,, giving reasons:, (i) Is the frequency of reflected and refracted light same as the frequency of incident light?, (ii) Does the decrease in speed imply a reduction in the energy carried by light wave?, 28., , (a), , State the working principle of a potentiometer. With the help of the circuit diagram, explain how, a potentiometer is used to compare the emf’s of two primary cells. Obtain the required, expression used for comparing the emfs., , (b) Write two possible causes for one sided deflection in a potentiometer experiment., OR, (a), , State Kirchhoff’s rules for an electric network. Using Kirchhoff’s rules, obtain the balance, condition in terms of the resistances of four arms of Wheatsone bridge., , (b) In the meterbridge experimental set up, shown in the figure, the null point ‘D’ is obtained at a, distance of 40 cm from end A of the meterbridge wire. If a resistance of 10 W is connected in, series with R1, null point is obtained at AD = 60 cm. Calculate the value of R1 and R2., R1, , R2, , B, , G, A, , D, , C, K, , 29., , (a), , Derive the expression for the torque on a rectangular current carrying loop suspended in a, uniform magnetic field., , (b) A proton and a deuteron having equal momenta enter in a region of uniform magnetic field at, right angle to the direction of the field. Depict their trajectories in the field., OR, (a) A small compass needle of magnetic moment ‘m’ is free to turn about an axis perpendicular to, the direction of uniform magnetic field ‘B’. The moment of inertia of the needle about the axis is, ‘I’. The needle is slightly disturbed from its stable position and then released. Prove that it, executes simple harmonic motion. Hence deduce the expression for its time period.
Page 426 :
Examination Papers, , 425, , (b) A compass needle, free to turn in a vertical plane orients itself with its axis vertical at a certain, place on the earth. Find out the values of (i) horizontal component of earth’s magnetic field and, (ii) angle of dip at the place., , SET–II, Questions Uncommon to Set–I, 1. A cell of emf ‘E’ and internal resistance ‘r’ draws a current ‘I’. Write the relation between terminal, voltage ‘V’ in terms of E, I and r., 2. Which of the following substances are diamagnetic?, Bi, Al, Na, Cu, Ca and Ni, 3. A heating element is marked 210 V, 630 W. What is the value of the current drawn by the element, when connected to a 210 V dc source?, 10. An ammeter of resistance 1 W can measure current upto 1.0 A (i) What must be the value of the shunt, resistance to enable the ammeter to measure upto 5.0 A? (ii) What is the combined resistance of the, ammeter and the shunt?, 14. A convex lens of focal length 20 cm is placed coaxially in contact with a concave lens of focal length, 25 cm. Determine the power of the combination. Will the system be converging or diverging in, nature?, 19. Using Bohr’s postulates, obtain the expression for (i) kinetic energy and (ii) potential energy of the, electron in stationary state of hydrogen atom., Draw the energy level diagram showing how the transitions between energy levels result in the, appearance of Lymann Series., 22. Figure shows a rectangular loop conducting PQRS in which the arm PQ is free to move. A uniform, magnetic field acts in the direction perpendicular to the plane of the loop. Arm PQ is moved with a, velocity v towards the arm RS. Assuming that the arms QR, RS and SP have negligible resistances and, the moving arm PQ has the resistance r, obtain the expression for (i) the current in the loop (ii) the, force and (iii) the power required to move the arm PQ., S, P, V, Q, R, , 23. Distinguish between ‘sky waves’ and ‘space waves’ modes of propagation in communication system., (a) Why is sky wave mode propagation restricted to frequencies upto 40 MHz?, (b) Give two examples where space wave mode of propagation is used.
Page 427 :
426, , Xam idea Physics—XII, , SET–III, Questions Uncommon to Set-I and II, 6. A 5 V battery of negligible internal resistance is connected across a 200 V battery and a resistance of, 39 W as shown in the figure. Find the value of the current., 5V, , 39 W, , 200 V, , 9. An ammeter of resistance 0.6 W can measure current upto 1.0 A. Calculate (i) The shunt resistance, required to enable the ammeter to measure current upto 5.0 A (ii) The combined resistance of the, ammeter and the shunt., 15. A convex lens of focal length 30 cm is placed coaxially in contact with a concave lens of focal length, 40 cm. Determine the power of the combination. Will the system be converging or diverging in, nature?, 19. (a) Write two characteristic features distinguishing the diffraction pattern from the interference, fringes obtained in Young’s double slit experiment., (b) Two wavelengths of sodium light 590 nm and 596 nm are used, in turn, to study the diffraction, taking place due to a single slit of aperture 1´ 10 – 4 m. The distance between the slit and the, , 21., , screen is 1.8 m. Calculate the separation between the positions of the first maxima of the, diffraction pattern obtained in the two cases., (a) In a nuclear reaction, 3, 2 He, , + 32 He ¾, ¾® 42 He + 11 H + 11 H + 12.86MeV, though the number of nucleons is conserved, on both sides of the reaction, yet the energy is released. How? Explain., (b) Draw a plot of potential energy between a pair of nucleons as a function of their separation. Mark, the regions where potential energy is (i) positive and (ii) negative., 25., , (a) Using Bohr’s postulates, obtain the expression for total energy of the electron in the nth orbit of, hydrogen atom., (b) What is the significance of negative sign in the expression for the energy?, (c), , Draw the energy level diagram showing how the line spectra corresponding to Paschen series, occur due to transition between energy levels., zzz
Page 428 :
Examination Papers, , 427, , Solutions, CBSE (Delhi) Set–I, 1. Substances that retain their ferromagnetic property for a long period of time at room temperature, are called permanent magnets., Examples : Steel, alnico, cobalt and ticonal., 2. Spherical shape with point charges q > 0 or q < 0 at the centre., 3. (i) Heat waves: As heat waves are transverse or electromagnetic in nature., 4. During charging of a capacitor by a d.c. source,, Ic =Id =, , e 0 dj E, dt, , When capacitor is fully charged than (j E ) = Max, So, , Ic =Id = 0, , 5. Angle of incidence i =, , A + dm, 2, , Concept : i + e = A + d, But at minimum angle of deviation, i = e, 6. Pairs of curves for different materials and same intensity radiations are graphs, (1, 3) and ( 2, 4)., Concept : Stopping potential depends on (i) nature of material (ii) frequency of radiation., 7. If cells are in oppositions, Enet, , = E1 –E2, = (200 – 10) V = 190 V, , Current I =, 8., , E net 190, = 5A, =, R eq, 38, , (i) In an open circuit, the emf of a cell and terminal voltage are same., (ii) In closed circuit, a current is drawn from the source, so, V = E –Ir, it is true/valid, because, each cell has some finite resistance., , 9. (a) (i) Light ray must travel from denser medium, into rarer medium., (ii) The angle of incidence in denser medium, must be greater than the critical angle., (b) Relation for pair of optical media with, refractive indices n1 and n2 (n2 > n1)., , Rarer, medium, Denser, medium, , n1, n2, ic
Page 429 :
428, , Xam idea Physics—XII, , From snell’s law =, , sin i c, sin 90, , \ sin ic =, , o, , = n21 = 2n1 =, , n1, n2, , n1, n2, , 10. The polarity of the induced emf at the open ends of a closed loop is such that it tends to produce a, current which opposes the change in magnetic flux that produced it., , N, –ve, +ve, , Yes, an emf will be induced at its ends., Justification: As the metallic rod falls down, the magnetic flux due to vertical component of, Earth's magnetic field keeps on changing., 11. Convex lens and concave lens are in contact as shown in fig., Power of convex lens P1, , =, , 1, 100, =, + f 1 (in m), f 1 (in cm), , L1, , L2, , 100, + 25, = 4D, , =, , Power of convex lens P2, , =, , 1, 100, =, - 20, + f2, f1= 25cm, , = – 5D, Power of combination P, , = P1 + P2, = 4D + (–5D), = – 1D, , System of lenses is diverging in nature., 12. (i) Fig. shows an ammeter of resistance 0.80 W that measure current of 1.0 A., A, I1 = 1 A, RA = 0.80W, , If a shunt ‘S’ is connected in parallel, a current (I – I1) flows through ‘S’., S, , I=5A, , I1, , A, , I – I1, , f2= –20cm
Page 430 :
Examination Papers, , 429, , For parallel combination of resistors, I1. RA = (I – I1) S, , (½), , 1 × 0.80 = (5 – 1) S, \S=, , 0.8, = 0.2 W, 4, , (ii) Combined resistance of the ammeter and the shunt, 1, 1, 1, +, =, R RA S, Þ, , R=, , R A ´ S 0.8 ´ 0.2, =, R A + S 0.8 + 0.2, , = 0.16 W, 13., , (i) If the value of the resistance R is reduced, the current in the forward biased input circuit, increases. The emitter current IE and the collector current IC (= IE – IB) both increase. Hence,, the brightness of the lamp increases., (ii) Due to increase in IC, the potential drop across lamp L increases and hence the voltmeter, reading V increases., ®, , ®, , 14. The direction of propagation of electromagnetic wave is given by E × B, Y, Envelope of E, E, , B, X, , B, Z, Envelope of B, Ù, , Ù, , Ù, , (a) i = j × k., (b) The speed of electromagnetic wave c =, 15., , (1), E0, B0, , (1), , (a) Box ‘x’ represents IF stage (intermediate frequency stage), Box ‘y’ represents Amplifier., (b) Function of IF stage - It changes the electromagnetic wave of high frequency to a lower, frequency for further detection in detector., Function of amplifier - The detected signal may not be strong enough to be made use of, then, amplifier is required. Its function is to enhance the power of the signals upto a required level.
Page 431 :
430, , Xam idea Physics—XII, , 16., hn, , ®, E, , mA, p-side, , n-side, , R, Circuit diagram of photodiode, , Working: In diode (any type of diode), an electric field ‘E’ exists across the junction from n-side, to p-side, when light with energy hn greater than energy gap Eg. illuminates the junction. Then, electron- hole pairs are generated due to absorption of photons, in or near the depletion region of, the diode. Due to existing electric field, electrons and holes get separated. The free electrons are, collected on n-side and holes are collected on p-side, giving rise to an emf., Due to this generated emf, an electric current of mA order flows through the external resistance., Detection of Optical Signals:, It is easier to observe the change in the current with change in the light intensity if a reverse bias is, applied. Thus, photodiode can be used as a photodetector to detect optical signals., OR, Important consideration in the fabrication of LED:, (a) (i) light emitting diode is a heavily doped p-n junction., (ii) The reverse breakdown voltages of LEDs are very low, typically around 5V., (b) The order of band gap of an LED to emit light in the visible range is about 3 eV to 1.8 eV., 17. Three factors for the need of modulating a message signal:, l, ) (i) Size of the antenna or aerial: about, 4, ælö, (ii) Effective power radiated by an antenna: Proportional to ç ÷, è lø, (iii) Mixing up of signals from different transmitters., , 2
Page 432 :
Examination Papers, , Diagrammatic representation:, +1, , C(t) O, , t, , –1, , +1, , Cm(t) O, , t, , –1, , +1, , t, , Cm(t) O, , –1, , 18., –, , +, C, , (i) If unknown capacitor of capacitance ‘C’ is connected to a battery of ‘V’ volts,, Q = CV, CV = 360 mC, (1), Þ, On reducing the potential/voltage by 120V, So, Q¢ = C (V -120), C (V – 120) = 120 mC, (2), Þ, On solving equation (1) and (2), 360mC 120 mC, =, V, V -120, Þ, V = 180V, Unknown capacitance from equation (1), Q = CV, 360 mC = C × 180V, , 431
Page 433 :
432, , Xam idea Physics—XII, , 360 mC, =2, 180V, C = 2 mF, (ii) Charge on the capacitor, if voltage is increased by 120V, Q = C (V + 120), = 2 (180 +120), Q = 600mC, OR, Þ, , C=, , Y, L, , R, , O, X, , 1m, 1m, , Z, ® ®, , (i) Electric flux through a surface j = E. S, , Flux through the left surface j L = – |E| |S|, = – 50x. |S|, Since x = 1m,, –4, j L = –50×1×25×10, = –1250×10–4, = –0.125Nm2 C–1, Flux through the right surface, j R = |E| |S|, Since x = 2m,, j R = 50x |S|, = 50 × 2 × 25 × 10–4, = 2500 × 10–4, = 0.250Nm2 C–1, Net flux through the cylinder, j net = j R + j L, = 0.250 – 0.125, = 0.125Nm2 C–1, (ii) Charge inside the cylinder, by Gauss’s Theorem, q, j net =, e0, = q = e 0 j Net, = 8.854 × 10, , –12, , × 0.125, , = 8.854 × 10, , –12, , ×, , = 1.107 ×10, , –12, , C, , 1, 8
Page 434 :
Examination Papers, , 433, , 19. (a) In nuclear reaction, 2, 1H, , +, , 2, 1H, , ¾, ¾®, , 3, 2 He, , + n + 3.27 MeV, , Cause of the energy released:, (i) Binding energy per nucleon of, , 3, 2 He, , becomes more than the (BE/A) of, , 2, 1 H., , (ii) Mass defect between the reactant and product nuclei, DE = DmC2, , [ ( H) - m( He) + m (n)]C, , = 2m, , 2, 1, , 3, 2, , 2, , (b) The radius of nucleus of mass number A is given by R = R0A1/ 3, 4, 4, Volume of the nucleus V = pR 3 = pR 30 A, 3, 3, Density of the matter in the nucleus, A(u), Mass, =, r=, 4, Volume, pR 30 A, 3, 1, 3, =, r=, 3, 4, 3, pR 0 4pR 0, 3, 20., , The expression of the density is independent of mass number A., (a) The observed characteristics of photoelectric effect could not be explained on the basis of, wave theory of light due to the following reasons., (i) According to wave theory, the light propagates in the form of wavefronts and the energy, is distributed uniformly over the wavefronts. With increase of intensity of light, the, amplitude of waves and the energy stored by waves will increase. These waves will then,, provide more energy to electrons of metal; consequently, the energy of electrons will, increase., Thus, according to wave theory, the kinetic energy of photoelectrons must depend on, the intensity of incident light; but according to experimental observations, the kinetic, energy of photoelectrons does not depend on the intensity of incident light., (ii) According to wave theory, 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, but according to experimental observations, the light of frequency, less than threshold frequency cannot emit electrons; whatever the intensity of incident, light may be., (iii) According to wave theory, the energy transferred by light waves will not go to a, particular electron, but it will be distributed uniformly to all electrons present in the, illuminated surface. Therefore, electrons will take some time to collect the necessary, energy for their emission. The time for emission will be more for light of less intensity, and vice versa. But experimental observations show that the emission of electrons take, place instantaneously after the light is incident on the metal; whatever the intensity of, light may be.
Page 435 :
434, , Xam idea Physics—XII, , (b) Features of the photons:, h, (i) Photons are particles of light having energy E = hn and momentum p = , where h is, l, planck constant., (ii) Photons travel with the speed of light in vacuum, independent of the frame of reference., (iii) Intensity of light depends on the number of photons crossing unit area in a unit time., 21. As the rod is rotated, free electrons in the rod move towards the outer end due to Lorentz force, and get distributed over the ring. Thus, the resulting separation of charges produces an emf across, the ends of the rod. At a certain value of emf, there is no more flow of electrons and a steady state, is reached., Expression for Induced emf in a Rotating Rod, Consider a metallic rod OA of length l , which is rotating with, angular velocity w in a uniform magnetic field B , the plane of, rotation being perpendicular to the magnetic field. A rod may be, supposed to be formed of a large number of small elements., Consider a small element of length dx at a distance x from centre., If v is the linear velocity of this element, then area swept by the, element per second = v dx, 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, l, , l, , 0, , 0, , x, x, , x, x, , x, , x, x, x x x, x x x, x x, x x, x, , e = ò B xw dx = Bw ò x dx, é x2, = Bw ê, êë 2, , l, , él2, ù, -0, ú = Bw ê, êë 2, úû 0, , ù 1, ú= Bwl, úû 2, , 2, , e 1 B wl 2, =, ×, R 2 R, It circuit is closed, power dissipated,, , Current induced in rod I =, , e 2 B 2 w2 l 2, =, R, 4R, (i) Dynamic output resistance is given by, æ DV ö, ro = çç CE ÷÷, è DIC ø I B, =, , 22., , For IB = 30 mA, DV CE = (12 - 8) = 4V and DI C = (3.6 – 3.4) = 0.2mA, 4V, 4, \ ro =, =, = 2 × 104 ohm, 0.2 m A 0.2 ´ 10 -3, (ii) dc current gain, I, bdc = c, Ib, At VCE = 10 V and Ib = 30 m A, the value of Ic = 3.5m A, , x x x, x, x B x, x, x x, x x x, v, x x, x, x, x, xO, dx, x, l, x x x x x, x x x x, x, x x x, , w, , A
Page 437 :
436, , Xam idea Physics—XII, , or,, , r=, , or,, , rn =, , e0 h2 n2, p m Ze 2, e0 h2 n2, , …(iii), , p m Ze 2, , Substituting value of r n in equation (ii), we get, En = -, , or,, , 1, 4p e 0, , Ze 2, æ e h2n2, 2ç 0, ç p mZ e 2, è, , En = -, , Z 2 Rhc, n2, , ö, ÷, ÷, ø, , =-, , m Z 2 e4, 8 e0 h2 n2, , , where R =, , me 4, 8 e 20 ch 3, , R is called Rydberg constant., For hydrogen atom Z =1,, -Rch, En =, n2, For Balmer series nf = 2, while ni = 3, 4, 5, ...................¥, n=¥, , –0.85 eV n = 4, –1.5 eV n = 3, , –3.4 eV n = 2, –13.6 eV n = 1, , Balmer series, , Ground state, Energy level diagram of Balmer series, , 24., , (a) When a plane wavefront of monochromatic, light illuminates, the slit LN, each point in the, slit LN becomes the source of secondary, wavelets., The secondary wavelets originating from, different points superpose on each, while, travelling towards the point C and point P; at, angle q. However the superposition of the, secondary wavelets produces a diffraction, pattern of varying intensity, as shown in fig., , To P, L, , From S, , M, , N, , q, , To C
Page 438 :
437, , Examination Papers, Intensity, Io, , O, Angle q, , (b) For maxima other than central maxima, 1ö, æ, a.q, = çn+ ÷l, 2ø, è, y, and q, =, D, y, 1ö, æ, = çn+ ÷l, \ a., D, 2ø, è, a = 2×10–4m, For light of wavelength l1 = 590 nm, y, æ 1ö, 2 ´ 10 -14 ´ 1 = ç1 + ÷ × 590, 1.5 è 2 ø, , P, y, q, , C, , D = 1.5m, , -9, , 3 590 ´ 10 ´ 1.5, ×, 2, 2 ´ 10 -4, = 6.64 mm, For light of wavelength l 2 =596 mm, y, æ 1ö, 2 ´ 10 -4 ´ 2 = ç1 + ÷ × 596 nm, 1.5, è 2ø, y1 =, , 3 596 ´ 10 -9 ´ 1.5, ×, 2, 2 ´ 10 -4, = 6.705 mm, Separation between two positions of first maxima, Dy = y 2 – y1, , Þ, , y2 =, , = 6.705 – 6.64, = 0.065 mm, 25., , At certain frequency w, the flow of current through the series combination, vm, Im =, 2, R + (X L - X C ) 2, 1, Where X L = wL, and X C =, wC, , Screen
Page 439 :
438, , Xam idea Physics—XII, , Condition of resonance – If system (LCR) of, natural frequency w0 is driven by an energy, source at a frequency w, the amplitude of the, current flow increases, however the amplitude, Im, of the current rises to its maximum value, if, frequency of the energy source becomes, exactly equal to the natural frequency., R2 < R1, For resistance R 2 < R1 , series LCR shows a, sharp resonance., R1, Q-factor – The ratio of reactance (either, wo, inductive or capacitive) at natural frequency to, 0.5, 1.0, 1.5, the resistance of the current is called Q - factor., w (rad/s), X, w L, Q= L = 0, R, R, Significance: (i) If resistance R is low or inductance L is large then Q – factor is large and the, circuit is more selective., (ii) If resonance is less sharp, tunning of the circuit will not be good., 26., (a) On the basis of electrostatic screening, no electric field exists inside the charged conducting, body. During lightening a shower of the charged particles falls on the earth. So it would be, safer to sit inside the car., (b) Dr. Pathak knows the result of lightening during thunderstorm; so he displayed two actions;, (i) Shows love, kindness and sympathy to the child., (ii) Keeping in view the safety of the child, he allow the boy to sit in the car till the, thunderstorm stopped., (c) Parent meets Dr. Pathak; and express their gratitude and heart felt thank for providing the, safety to the child from lightening and thunderstorm., (d) Many of us have read in the newspaper that the person either working in the field or in open, space have lost their life during thunderstorm, so the persons belonging to villages must be, given advices that they should remain inside the houses (or closed caves etc.) during, thunderstorm., 27., (a) Difference between myopia and hypermetropia, Myopia, , Hypermetropia, , 1., , The eye ball is elongated., , 1., , The eye ball is shortened., , 2., , Person cannot see distant objects clearly., , 2., , Person cannot see near objects clearly., , • Myopic eye is corrected by interposing a concave lens between eye and object., • Hypermetropia is corrected by interposing a convex lens between eye and object., Corrective, lens, , Normal near point, , O, , I, , N, , Near point of, defective eye, , Corrective, convex lens, Corrected hypermetropia, , F, , Retina, I, Eye, lens, Corrected myopia
Page 440 :
Examination Papers, , 439, , (b) If image A ¢ B¢ is exactly at the focus of the eyepiece, then image A ¢¢ B¢¢ is formed at infinity., If the object AB is very close to the focus of the objective lens of focal length fo, then, magnification Mo by the objective lens, L, Me =, fo, where L is tube length (or distance between lenses Lo and Le), Magnification Me by the eyepiece, D, Me =, fe, where D = Least distance of distinct vision, Total magnification, m = MoMe, æ L öæ D ö, = çç ÷÷ çç ÷÷, è f o øè f e ø, OR, (a) Law of Reflection : Let XY be a reflecting surface at which a wavefront is being incident, obliquely. Let v be the speed of the wavefront and at time t = 0, the wavefront touches the, surface XY at A. After time t, the point B of wavefront reaches the point B¢ of the surface., According to Huygen’s principle each point of wavefront acts as a source of secondary, waves. When the point A of wavefront strikes the reflecting surface, then due to presence of, reflecting surface, it cannot advance further; but the secondary wavelet originating from point, A begins to spread in all directions in the first medium with speed v. As the wavefront AB, advances further, its points A1 , A 2 , A 3 K etc. strike the reflecting surface successively and, send spherical secondary wavelets in the first medium., C, , A2, i A, , i, , i, , X, A, , B, , A', D, , 1, , r, , r, Y, B', , First of all the secondary wavelet starts from point A and traverses distance AA¢ ( = vt) in first, medium in time t. In the same time t, the point B of wavefront, after travelling a distance BB¢,, reaches point B¢ (of the surface), from where the secondary wavelet now starts. 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¢ . As the incident wavefront AB advances, the secondary wavelets starting from points, between A and B¢, one after the other and will touch A¢ B¢ simultaneously. According to, Huygen’s principle wavefront A¢ B¢ represents the new position of AB, i.e., A¢ B¢ is the, reflected wavefront corresponding to incident wavefront AB., Now in right-angled triangles ABB¢ and AA¢ B¢, Ð ABB¢ = Ð AA¢ B¢ (both are equal to 90°), side BB¢ = side AA¢, (both are equal to vt), and side AB¢ is common, i.e., both triangles are congruent., \, Ð BAB¢ = Ð AB¢ A¢
Page 441 :
440, , Xam idea Physics—XII, , i.e., incident wavefront AB and reflected wavefront A¢ B¢ make equal angles with the reflecting, surface XY. As the rays are always normal to the wavefront, therefore the incident and the, reflected rays make equal angles with the normal drawn on the surface XY, i.e.,, angle of incidence i = angle of reflection r, This is the second law of reflection., Since AB, A¢ B¢ and XY are all in the plane of paper, therefore the perpendiculars dropped on, them will also be in the same plane. Therefore we conclude that the incident ray, reflected, ray and the normal at the point of incidence, all lie in the same plane. This is the first law of, reflection. Thus Huygen’s principle explains both the laws of reflection., (b) (i) If the radiation of certain frequency interact with the atoms/molecules of the matter, they, start to vibrate with the same frequency under forced oscillations., Thus, the frequency of the scattered light (Under reflection and refraction) equals to the, frequency of incident radiation., (ii) No, energy carried by the wave depends on the amplitude of the wave, but not on the, speed of the wave., 28., (a) Working Principle of Potentiometer, Principle. Consider a long resistance wire AB of uniform cross-section. Its one end A is, connected to the positive terminal of battery B1 whose negative terminal is connected to the, other end B of the wire through key K and a rheostat ( Rh). The battery B1 connected in circuit, is called the driver battery and this circuit is called the primary circuit. By the help of this, circuit a definite potential difference is applied across the wire AB; the potential falls, continuously along the wire from A to B. The fall of potential per unit length of wire is, called the potential gradient. It is denoted by ‘k’. A cell e is connected such that its positive, terminal is connected to end A and the negative terminal to a jockey J through the, galvanometer G. This circuit is called the secondary circuit., In primary circuit the rheostat ( Rh) is so adjusted that the deflection in galvanometer is on one, side when jockey is touched on wire at point A and on the other side when jockey is touched, on wire at point B., The jockey is moved and touched to the potentiometer wire and the position is found where, galvanometer gives no deflection. Such a point P is called null deflection point., VAB is the potential difference between points A and B and L metre be the length of wire, then, the potential gradient, V, k = AB, L, If the length of wire AP in the null deflection position be l, then the potential difference, between points A and P,, VAP = kl, \ The emf of cell,, e = VAP = kl, In this way the emf of a cell may be determined by a potentiometer., Comparison of emf’s of two cells: First of all the ends of potentiometer are connected to a, battery B1 , key K and rheostat Rh such that the positive terminal of battery B1 is connected to, end A of the wire. This completes the primary circuit., Now the positive terminals of the cells C1 and C 2 whose emfs are to be compared are, connected to A and the negative terminals to the jockey J through a two-way key and a, galvanometer (fig). This is the secondary circuit.
Page 442 :
Examination Papers, B, , 441, , K, , 1, Method: (i) By closing key K, a, +, –, potential difference is established, Rh, and rheostat is so adjusted that, when jockey J is made to touch at, P2, P1, –, ends A and B of wire, the deflection +, B, J, A, in galvanometer is on both sides., G, I1, Suppose in this position the, + –, potential gradient is k., e1 C1, 1, (ii) Now plug is inserted between, 3, HR, 2, the terminals 1 and 3 so that, e2 –, cell C1 is included in the, + C2, I2, secondary circuit and jockey J, is slided on the wire at P1 (say) to obtain the null point. The distance of P1 from A is, measured. Suppose this length is l1 i. e. AP1 = l1, The emf of cell C1 ,, ...(i), \, e1 = kl1, (iii) Now plug is taken off between the terminals 1 and 3 and inserted in between the, terminals 2 and 3 to bring cell C 2 in the circuit. Jockey is slided on wire and null, deflection position P2 is noted. Suppose distance of P2 from A is l2 i. e. AP2 = l2, The emf of cell C 2 ,, ...(ii), \, e 2 = kl2, e1 l1, ...(iii), Dividing (i) by (ii), we get, =, e 2 l2, , (b), (i), (ii), , (a), , Thus emf’s of cells may be compared. Out of these cells if one is standard cell, then the, emf of other cell may be calculated., Possible causes for one side deflection:, If emf e1 (or e 2 ) is more than the emf driver cell (auxiliary battery), then we have one sided, deflection., when the positive end of the potentiometer wire is connected to negative terminal of the cell, whose emf is to be determined., OR, Kirchhoff’s Rule, (i) At any junction, the sum of the currents entering the junction is equal to the sum of the, currents leaving the junction., (ii) The algebraic sum of the charges in potential around any closed loop involving resistors, and cells in the loop is zero., B, Condition of balance of a Wheatstone bridge:, I1, The circuit diagram of Wheatstone bridge is shown in, P, Q, K2, fig., I1, P, Q, R and S are four resistance forming a closed, G, C, bridge, called Wheatstone bridge. A battery is A, I2, connected across A and C, while a galvanometer is, connected between B and D. At balance, there is no, R, S, current in galvanometer., I2, I, Derivation of Formula: Let the current given by, D, battery in the balanced position be I. This current on, + –, reaching point A is divided into two parts I 1 and I 2 . As, E, , K1
Page 443 :
442, , Xam idea Physics—XII, , there is no current in galvanometer in balanced state, current in resistances P and Q is I 1, and in resistances R and S it is I 2 ., Applying Kirchhoff’s I law at point A, ...(i), I - I 1 - I 2 = 0 or I = I 1 + I 2, Applying Kirchhoff’s II law to closed mesh ABDA, - I 1 P + I 2 R = 0 or I 1 P = I 2 R ...(ii), Applying Kirchhoff’s II law to mesh BCDB, - I 1 Q + I 2 S = 0 or I 1 Q = I 2 S ...(iii), Dividing equation (ii) by (iii), we get, I1P I 2 R, P R, or, ...(iv), =, =, I1Q I 2S, Q S, This is the condition of balance of Wheatstone bridge., (b), R1, , R2, , B, G, l, , A, , D, , y, , D¢, , C, , For null point at D, balance length l 1 = 40 cm, R1 AD, 40, 2, =, =, =, So,, R 2 DC (100 - 40) 3, , ...(1), , If resistance 10 W is connected in series of R1 , then balance length AD¢ > AD i.e. balance, point shifts by length ‘y’ towards C i.e., AD = 60 cm., R1 +10 AD¢, 60, 3, =, =, =, R2, D¢ C 100 - 60 2, R1 10, 3, +, =, R2 R2, 2, From equations (1) and (2), we have, 2 10 3, +, =, 3 R2 2, 10 3 2 9 - 4 5, = – =, =, R2 2 3, 6, 6, Þ, , R2 =, , 10 ´ 6, = 12 ohm, 5, , From equation (1), we have, R1 2, 12 ´ 2, =, = 8 ohm, Þ, R1 =, 12 3, 3, , ...(2)
Page 444 :
Examination Papers, , 29 (a), , 443, , F2, B, C, B, B, , B, , m, , n, , D, A, , F1, , A rectangular loop ABCD of dimensions l and b, carrying a steady current is placed in uniform, magnetic field as shown in fig; such that normal of the plane is at angle q with the magnetic field, lines., The force FBC and FAD on arms BC and AD are equal, opposite and along the axis of the coil, so, they cancel each other., The forces FAB and FCD are also equal and opposite, but are not collinear, so they constitute a, couple, and the magnitude of the torque can be given as, b, b, t = FAB. sin q + FCD. sin q, 2, 2, Since, \, , |FAB| = |FCD| = BIl, t = BIl × b sin q, = BI (lb) sin q, = BI A sin q, , [A = l b = area of the rectangle], , Since magnetic moment m = I |A|, t = mB sin q, ® ® ®, In vector from t = m ´ B, (b) If a charge particle enters right angle to the direction of magnetic field, it follows a circular, trajectory, and radius can be given as, m v2, r, mv, P, r=, =, qB qB, q vB=, , Þ, , Since momentum are equal, and they have equal charges., So,, , rp : rd = 1:1, , OR, r, (a) If magnetic compass of dipole moment m is placed at angle q in uniform magnetic field, and, released it experiences a restoring torque.
Page 446 :
Examination Papers, , 445, , CBSE (Delhi) Set–II, Questions uncommon to Set–I, 1. The terminal voltage V < E, so V = E – Ir, V, I, , (e, r), , 2. Diamagnetic substances are (i) Bi (ii) Cu., 3. If marked voltage is equal to the voltage of the source, then on element., P = VI, 630 = 210 × I, 630, = 3A, \, I=, 210, 10., S, , I2, I, , A, , I1, , R1 = 1.0 A, , (i) From Ohm’s Law, , I 1 R1 = (I 2 - I 1 ) ´ S, , 1 × 1 = ( 5 – 1) × S, 1, S = W = 0.25W, Þ, 4, S ´ R1, (ii) Combined resistance, Req =, S + R1, 0.25 ´ 1 1, = = 0.2 W, 0.25 + 1 5, 1, 100, 14. Power of convex lens, P1 =, =, f (in m) f (in cm), Req =, , 100, = 5D, 20, 100, 100, Power of concave lens, P2 =, =, = – 4D, f (in m) -25, , L1, , L2, , P=, , Power of the combination, P, , f1 = 20cm, , f2 = 25cm, , = P1 + P2, = 5D + ( – 4D), = 1D, , The value of power of combination is positive, so the system acts as converging lens in nature.
Page 448 :
Examination Papers, , 22., , (i) Current in the loop PQRS,, e, I=, r, df, Blv, Since e = = Blv So, I =, dt, r, (ii) The force required to keep the arm PQ in, constant motion, , 447, , P, S, , R, Q, , F = BIl, æ Blv ö, = Bç, ÷l, è r ø, F =, , B 2 l 2v, r, , (iii) Power required to move the arm PQ, P = | F |v|, æ B 2 l 2v ö, ÷v, =ç, ç r ÷, ø, è, æ B 2 l 2v 2, =ç, ç, r, è, , ö, ÷, ÷, ø, , 23. Sky wave — Sky wave propagation can be achieved by ionospheric reflection of radio waves, back towards the earth, in the frequency range from few MHz up to 30 to 40 MHz., Space wave — In this mode of propagation, radio wave travels in a straight line from transmitting, antenna to the receiving antenna., (a) Sky wave propagation is restricted to frequency upto 40 MHz, because the radio waves of, frequencies more than 40 MHz penetrates into the ionosphere., (b) Space wave propagation (LOS) is used in, (i) Television broadcast, (ii) Microwave link, (iii) Satellite communication, , CBSE (Delhi) Set–III, Questions uncommon to Set–I and II, E -E, 6. Current, I = 2 1, R, 200 - 5, =, 39, 195, =, = 5A, 39, , [E2 > E1], , E1 = 5V, , I, , 39 W, , 200 V = E2
Page 449 :
448, , Xam idea Physics—XII, S, , 9., , I = 5.0 A, , I1= 1.0 A, , A, RA = 0.6 W, , (i) From Ohm’s law, I1 RA= (I - I1)S, 1.0 × 0.6 = (5 – 1) × S, 0.6, = 0.15 W, S =, Þ, 4, RA S, RA + S, 0.6 ´ 015, ., 0.09, =, =, 0.6 + 015, ., 0.75, 3, =, 25, = 0.12 W, 1, 100, =, 15. Power of convex lens, P1 =, f 1 ( in m) f 1 ( in cm), 100 10, P1 =, =, D, 30, 3, 100, 5, 100, =, Power of concave lens, P2 =, =- D, f 2 ( in cm) -40, 2, (ii) Combined resistance Req, , =, , f1 = 30cm, , f2 = 40cm, , Power of the combination, P = P1 + P2, 10 æ 5 ö, + ç- ÷, =, 3 è 2ø, 20 -15, 5, =, D, =, 6, 6, 5, P =, D, 6, Sign of the power of the combination is positive, so the system behave as a convex converging, lens., 19. (a) Distinguish between diffraction and interference, S., No., , Interference, , Diffraction, , (i) Intensity of all bright fringes is same., , Intensity of bright fringes decreases on either, side of central maxima., , (ii) Size of all fringes is same., , Size of the central maxima is nearly twice the, size of other maxima.
Page 452 :
CBSE Examination Papers, All India–2013, SET–I, Time allowed : 3 hours, , Maximum marks: 70, , General Instructions: As given in CBSE Examination Paper Delhi–2013., 1. Two charges of magnitudes – 2Q and + Q are located at points (a, 0) and (4a, 0) respectively. What is, the electric flux due to these charges through a sphere of radius ‘3a’ with its centre at the origin?, 2. How does the mutual inductance of a pair of coils change when, (i) distance between the coils is increased and, (ii) number of turns in the coils is increased?, 3. The graph shown in the figure represents a plot of current versus voltage for a given semiconductor., Identify the region, if any, over which the semiconductor has a negative resistance., , Current (mA), , B, , C, A, , 0, Voltage (V), , 4. Two identical cells, each of emf E, having negligible internal resistance, are connected in parallel with, each other across an external resistance R. What is the current through this resistance?, 5. The motion of copper plate is damped when it is allowed to oscillate between the two poles of a, magnet. What is the cause of this damping?, 6. Define the activity of a given radioactive substance. Write its S.I. unit., 7. Welders wear special goggles or face masks with glass windows to protect their eyes from, electromagnetic radiations. Name the radiations and write the range of their frequency., 8. Write the expression for the de Broglie wavelength associated with a charged particle having charge, ‘q’ and mass ‘m’, when it is accelerated by a potential V., 9. Draw typical output characteristics of an n-p-n transistor in CE configuration. Show how these, characteristics can be used to determine output resistance.
Page 453 :
452, , Xam idea Physics—XII, , 10. A parallel beam of light of 500 nm falls on a narrow slit and the resulting diffraction pattern is, observed on a screen 1 m away. It is observed that the first minimum is at a distance of 2.5 mm from, the centre of the screen. Calculate the width of the slit., 11. A slab of material of dielectric constant K has the same area as that of the plates of a parallel plate, capacitor but has the thickness d/2, where d is the separation between the plates. Find out the, expression for its capacitance when the slab is inserted between the plates of the capacitor., 12. A capacitor, made of two parallel plates each of plate area A and separation d, is being charged by an, external ac source. Show that the displacement current inside the capacitor is the same as the current, charging the capacitor., 13. Explain the term ‘drift velocity’ of electrons in a conductor. Hence obtain the expression for the, current through a conductor in terms of ‘drift velocity’., OR, Describe briefly, with the help of a circuit diagram, how a potentiometer is used to determine the, internal resistance of a cell., 14. A convex lens of focal length f 1 is kept in contact with a concave lens of focal length f 2 . Find the focal, length of the combination., 15. In the block diagram of a simple modulator for obtaining an AM signal, shown in the figure, identify, the boxes A and B. Write their functions., +, , Modulating, signal, , x(t), , y(t), , A, , AM, , B, , Wave, , Carrier, wave, , 16. In the circuit shown in the figure, identify the equivalent gate of the circuit and make its truth table., A', , A, , Y, B, , 17., , (a), , B', , For a given a.c., i = i m sin wt, show that the average power dissipated in a resistor R over a, 1, complete cycle is i m2 R., 2, , (b) A light bulb is rated at 100 W for a 220 V a.c. supply. Calculate the resistance of the bulb., 18. A rectangular conductor LMNO is placed in a uniform magnetic field of 0.5 T. The field is directed, perpendicular to the plane of the conductor. When the arm MN of length of 20 cm is moved towards, left with a velocity of 10 ms–1, calculate the emf induced in the arm. Given the resistance of the arm to, be 5 W (assuming that other arms are of negligible resistance) find the value of the current in the arm., B, , M, , L, V, , O, , l, , N
Page 454 :
Examination Papers, , 453, , OR, A wheel with 8 metallic spokes each 50 cm long is rotated with a speed of 120 rev/min in a plane, normal to the horizontal component of the Earth’s magnetic field. The Earth’s magnetic field at the, plane is 0.4 G and the angle of dip is 60°. Calculate the emf induced between the axle and the rim of, the wheel. How will the value of emf be affected if the number of spokes were increased?, 19. Define the current sensitivity of a galvanometer. Write its S.I. unit., Figure shows two circuits each having a galvanometer and a battery of 3 V., When the galvanometers in each arrangement do not show any deflection, obtain the ratio R1 / R 2 ., , 4W, , R1, , 6W, , 12W, , G 60W, , 6W, , 3.0 V, , 9W, , R2, , 8W, , G, 3.0 V, , 1.20W, , 20. A wire AB is carrying a steady current of 12 A and is lying on the table. Another wire CD carrying 5A, is held directly above AB at a height of 1 mm. Find the mass per unit length of the wire CD so that it, remains suspended at its position when left free. Give the direction of the current flowing in CD with, respect to that in AB. [Take the value of g = 10 ms–2], 21. Draw V – I characteristics of a p–n junction diode. Answer the following questions, giving reasons:, (i), , Why is the current under reverse bias almost independent of the applied potential upto a critical, voltage?, , (ii) Why does the reverse current show a sudden increase at the critical voltage?, Name any semiconductor device which operates under the reverse bias in the breakdown region., 22. Draw a labelled ray diagram of a refracting telescope. Define its magnifying power and write the, expression for it., Write two important limitations of a refracting telescope over a reflecting type telescope., 23. Write Einstein’s photoelectric equation and point out any two characteristic properties of photons on, which this equation is based., Briefly explain the three observed features which can be explained by this equation., 24. Name the type of waves which are used for line of sight (LOS) communication. What is the range of, their frequencies?, A transmitting antenna at the top of a tower has a height of 20 m and the height of the receiving, antenna is 45 m. Calculate the maximum distance between them for satisfactory communication in, LOS mode. (Radius of the Earth = 6.4 × 106 m)
Page 455 :
454, 25., , Xam idea Physics—XII, , (a), , What is linearly polarized light? Describe briefly using a diagram how sunlight is polarised., , (b) Unpolarised light is incident on a polaroid. How would the intensity of transmitted light change, when the polaroid is rotated?, 26. One day Chetan’s mother developed a severe stomach ache all of a sudden. She was rushed to the, doctor who suggested for an immediate endoscopy test and gave an estimate of expenditure for the, same. Chetan immediately contacted his class teacher and shared the information with her. The class, teacher arranged for the money and rushed to the hospital. On realising that Chetan belonged to a, below average income group family, even the doctor offered concession for the test fee. The test was, conducted successfully., Answer the following questions based on the above information:, (a) Which principle in optics is made use of in endoscopy?, (b) Briefly explain the values reflected in the action taken by the teacher., 27., , (c), , In what way do you appreciate the response of the doctor on the given situation?, , (a), , Using Biot-Savart’s law, derive the expression for the magnetic field in the vector form at a point, on the axis of a circular current loop., , (b) What does a toroid consist of ? Find out the expression for the magnetic field inside a toroid for N, turns of the coil having the average radius r and carrying a current I. Show that the magnetic field, in the open space inside and exterior to the toroid is zero., OR, (a), , Draw a schematic sketch of a cyclotron. Explain clearly the role of crossed electric and magnetic, field in accelerating the charge. Hence derive the expression for the kinetic energy acquired by, the particles., , (b) An a–particle and a proton are released from the centre of the cyclotron and made to accelerate., (i), (ii), 28., , (a), , Can both be accelerated at the same cyclotron frequency? Give reason to justify your, answer., When they are accelerated in turn, which of the two will have higher velocity at the, exit slit of the dees?, , Define electric dipole moment. Is it a scalar or a vector? Derive the expression for the electric, field of a dipole at a point on the equatorial plane of the dipole., , (b) Draw the equipotential surfaces due to an electric dipole. Locate the points where the potential, due to the dipole is zero., OR, Using Gauss’ law deduce the expression for the electric field due to a uniformly charged spherical, conducting shell of radius R at a point (i) outside and (ii) inside the shell., Plot a graph showing variation of electric field as a function of r > R and r < R. (r being the distance, from the centre of the shell), 29. Using Bohr’s postulates, derive the expression for the frequency of radiation emitted when electron in, hydrogen atom undergoes transition from higher energy state (quantum number n i ) to the lower state,, (n f )., When electron in hydrogen atom jumps from energy state n i = 4 to n f =3, 2, 1, identify the spectral, series to which the emission lines belong., OR
Page 456 :
Examination Papers, , (a), , 455, , Draw the plot of binding energy per nucleon (BE/A) as a function of mass number A. Write two, important conclusions that can be drawn regarding the nature of nuclear force., , (b) Use this graph to explain the release of energy in both the processes of nuclear fusion and fission., (c), , Write the basic nuclear process of neutron undergoing b–decay. Why is the detection of, neutrinos found very difficult?, , SET–II, Questions Uncommon to Set–I, 4. Two charges of magnitudes – 3Q and + 2Q are located at points (a, 0) and (4a, 0) respectively. What is, the electric flux due to these charges through a sphere of radius ‘5a’ with its centre at the origin?, 7. A light metal disc on the top of an electromagnet is thrown up as the current is switched on. Why?, Give reason., 9. In the circuit shown in the figure, identify the equivalent gate of the circuit and make its truth table., A, , Y1, , Y, , B, Y2, , 13. A parallel beam of light of 600 nm falls on a narrow slit and the resulting diffraction pattern is, observed on a screen 1.2 m away. It is observed that the first minimum is at a distance of 3 mm from, the centre of the screen. Calculate the width of the slit., 19. A wire AB is carrying a steady current of 10 A and is lying on the table. Another wire CD carrying 6A, is held directly above AB at a height of 2 mm. Find the mass per unit length of the wire CD so that it, remains suspended at its position when left free. Give the direction of the current flowing in CD with, respect to that in AB. [Take the value of g = 10 ms–2], 23. Name the type of waves which are used for line of sight (LOS) communication. What is the range of, their frequencies?, A transmitting antenna at the top of a tower has a height of 45 m and the height of the receiving, antenna is 80 m. Calculate the maximum distance between them for satisfactory communication in, LOS mode. (Radius of the Earth = 6.4 ´ 10 6 m)., , SET–III, Questions Uncommon to Set-I and II, 3. Two charges of magnitudes + 4Q and – Q are located at points (a, 0) and (– 3a, 0) respectively. What, is the electric flux due to these charges through a sphere of radius ‘2a’ with its centre at the origin?, 4. The motion of copper plate is damped when it is allowed to oscillate between the two poles of a, magnet. If slots are cut in the plate, how will the damping be affected?
Page 457 :
456, , Xam idea Physics—XII, , 6. How does the mutual inductance of a pair of coils change when, (i) distance between the coils is decreased and, (ii) number of turns in the coils is decreased?, 9. In the circuit shown in the figure, identify the equivalent gate of the circuit and make its truth table., A, , A¢, , Y¢, , B, , Y, , B¢, , 15. A slab of material of dielectric constant K has the same area as that of the plates of a parallel plate, capacitor but has the thickness 2d/3, where d is the separation between the plates. Find out the, expression for its capacitance when the slab is inserted between the plates of the capacitor., 25. Name the type of waves which are used for line of sight (LOS) communication. What is the range of, their frequencies?, A transmitting antenna at the top of a tower has a height of 45 m and the receiving antenna is on the, ground. Calculate the maximum distance between them for satisfactory communication in LOS, mode. (Radius of the Earth = 6.4 ´ 10 6 m)., zzz
Page 458 :
Examination Papers, , 457, , Solutions, SET–I, 1. Electric flux, j =, , – 2Q, e0, , Concept: (i) Mark the position of the charges on number line., (ii) Draw a sphere of radius 3a about the origin and observe that which charge is inside the, sphere, and then use Gauss theorem., –2Q, (0, 0), , 2., , +Q, , (a, 0), , (4a, 0), , (i) Mutual inductance decreases., (ii) Mutual inductance increases., Concept: (i) If distance between two coils is increased as shown in figure,, d, , C1, , C2, , It causes decrease in magnetic flux linked with the coil C2. Hence induced emf in coil C2, – dj 2, . Hence mutual inductance decreases., decreases by relation e 2 =, dt, (ii) From relation M 21 =m 0 n1 n 2 Al, if number of turns in one of the coils or both increases,, means mutual inductance will increase., 3. In region BC i.e., the region showing negative slope., Concept: In figure draw two horizontal lines, as marked by doted lines and use the formula:, æ + DV ö, ÷÷ in the region, B to C., R = çç, è + DI ø, E, R, Concept: (i) emf of combination of two (or more cells) remain same., , 4. Current, I =, , e, @, eeq, e, , = e
Page 459 :
458, , Xam idea Physics—XII, , (ii) Internal resistance is negligible i.e., zero., e eq, e, So, I =, =, R + r eq R, , (req = 0), , 5. As the plate oscillate, the changing magnetic flux through the plate produces a strong eddy current, in the direction, which opposes the cause., Also, copper being diamagnetic substance, it gets magnetised in the opposite direction, so the, plate motion gets damped., 6. The rate of decay of a radioactive substance is called activity of that substance., It is negative of the rate of decay of the radioactive substance., N¢ – N, Activity, R =, t¢ – t, DN, =–, Dt, æ DN ö, dN, ÷÷ = –, = lim çç –, Dt ® 0 è D t ø, dt, S.I. unit of activity (i) becquerel (Bq)., (ii) decay per second., 7. Ultraviolet radiations, OR, Radiations above violet (in VIBGYOR), Frequency range 1015 – 1017 Hz., , N, , N¢, , O, , t, , t¢, , t, , Hint: Frequecy of visible light is of the order of 1014 Hz., h, 8. de Broglie wavelength l = h =, p, 2mqV, Hint:, , W = K = qV =, , p2, or p = 2mqV, 2m, , 9., Y, , Ib, IC(in mA), , Ib, , 4, , 3, , Ib, , 2, , Ib, , O, VCE (in volts), , 1, , X, , The reciprocal of the slope of the linear part of the output characteristics represents the output, resistance.
Page 460 :
Examination Papers, , IC, , O, VCE, , æ DV, r 0 = çç CE, è DI C, , ö, ÷÷ (as shown in fig.), øIB, 4, , 10., , y1, , q, , a, , O, Central, maxima, , D, , Screen, , Hint: From condition of diffraction,, sin q = n l (for minima), 1ö, æ, for maxima, =ç n + ÷ l, 2ø, è, Provided n =1, 2, 3..., and n = 0 for central maxima, From condition of minima,, a sin q = l, ( n =1), Since the value of l is of nm, so, a. q = l, y, a. =l, D, lD, a=, \, y, a=, , 500 ´ 10 –9 ´ 1, 2.5 ´ 10 –3, , = 2 ´ 10 – 4 m, , m, , arc ù, é, êëangle = radius úû, , 459
Page 461 :
460, , Xam idea Physics—XII, , 11., Area = A, Separation 'd', , Vacuum, , Dielectric, , d, 2, , Capacitance with dielectric of thickness ‘t’, e0 A, C=, t, d –t +, K, d, Put, t=, 2, e0 A, C=, d d, d– +, 2 2K, e0 A, =, d d, +, 2 2K, e0 A, =, d æ 1ö, ç1 + ÷, 2 è Kø, =, , 2e 0 AK, d ( K + 1), , 12., +q, , –q, , E, Ic, , In Fig. conduction current is flowing in the wires, causes charge on the plates, So, , Ic =, , dq, dt, , (1), , According to Maxwell, displacement current between plates, I d = e0, , dj E, , where fE = Electric flux, dt, , Using Gauss’s Theorem, if one of the plate is inside the tiffin type Gaussian surface, fE =, , q, e0, , (2)
Page 462 :
Examination Papers, , So, , I d = e0, Id =, , dæ q, ç, dt çè e 0, , 461, , ö, ÷÷, ø, , dq, dt, , (3), , From equation (2) and (3),, Both conduction current and displacement currents are equal., 13. The modified velocity gained by the accelerating electrons in uniform electric field inside the, conductor is called drift velocity., , Vd, , The average velocity, acquired by free electrons along the length of a metallic conductor, due to, existing electric field is called drift velocity., Let n be the number density of free electrons in a conductor of length l and area of cross-section, ‘A’., Total charge in the conductor, Q = Ne, =( n Al) e, Time taken at average velocity v d is t =, So, by definition, I =, , l, vd, , Q ( n Al) e, =, t æ l ö, ÷÷, çç, è vd ø, , I = neA v d, OR, B, +, , K, –, , Rh, , P2, , +, A, , P1, J, , –, B, G, , e, + –, C, , HR, , RB ( ), K1, , If key k1 is closed, (while key k 2 is open), galvanometer shows null deflection at balancing, length l1 .
Page 463 :
462, , Xam idea Physics—XII, , So,, (1), E = k l1, If both keys k1 and k 2 are closed and R is the resistance of resistance box, galvanometer now, shows null deflection at balancing length l2 (l2 < l1 )., So,, (2), V = k l2, æE ö, From relation, r = R ç –1÷, èV ø, ö, æl, r = R çç 1 – 1÷÷, ø, è l2, , We have,, 14., , +f 1, , P, , O, , I1, , O, , I1, , –u, , I, , v', v, , For convex lens of focal length (+ f1), 1 1 1, + = –, f 1 v¢ u, , (1), , For concave lens of focal length (– f 2 ), 1 1 1, –, = –, f 2 v v¢, , (2), , Adding equation (1) and (2), 1 1 1 1, –, = –, f1 f 2 v u, , (3), , For an equivalent lens (using lens formula), 1 1 1, = – , where f is the focal length of combination., f v u, From equation (3) and (4),, 1 1 1, = –, f f1 f 2, , (4)
Page 464 :
Examination Papers, , 463, , 15. Boxes ‘A’ and ‘B’ represents, (i) square law device, (ii) band pass filter, (i) Square law device is a non linear device and produces the output, y (t) = Bx (t) + Cx (t) 2, where B and C are constants., (ii) It rejects dc and sinusoids of frequencies wm , 2wm and 2wc and finally produces amplitude, modulated wave by retaining frequncies wc , wc – wm and wc + wm ., 16., , A, , P, `A, y, , R, `B, , Q, , B, , Gates, P, Q and R act as NOT, NOT and NAND gates respectively., y= A.B =A + B, , So, , The combination acts as OR Gate., Truth Table of the combination, A, , B, , A, , B, , A. B, , A. B, , Y, , 0, , 0, , 1, , 1, , 1, , 0, , 0, , 1, , 0, , 0, , 1, , 0, , 1, , 1, , 0, , 1, , 1, , 0, , 0, , 1, , 1, , 1, , 1, , 0, , 0, , 0, , 1, , 1, , Truth Table of OR Gate, , 17. (a), , A, , B, , Y, , 0, , 0, , 0, , 1, , 0, , 1, , 0, , 1, , 1, , 1, , 1, , 1, , From graph of i 2 – t, Average power consumed in resistor R, T, 1, . ò i 2 R dt, Pav =, T, 0, ò dt, =, =, , 0, 2, im R, , T, , 2, im, R, 2T, , T, , ò0, T, , sin 2 wt dt, , ò0 (1 – cos 2 wt) dt, , lm, , 2, , i2, , ...(1), O, , T, 2, , T, t
Page 465 :
464, , Xam idea Physics—XII, , =, , 2, T, im, R é T, dt – ò cos 2 wt dt ù, ò, ê, úû, 0, 2T ë 0, , =, , 2, im, R, [T – 0], 2T, , =, , 2, im, R, 2, , (2), , (b) In case of ac, Pav, , 2, 2, Veff, Vrms, =, =, R, R, , 2, Vrms, Pav, 220 ´ 220, =, 100, = 484W, 18. Induced emf in a moving rod in a magnetic field is given by, e = – Blv, Since the rod is moving to the left so, e = + Blv, = 0.5 ´ 0.2 ´ 10, =1n, e 1, Current in the rod I = = = 0.2A, R 5, OR, If a rod of length ‘l’ rotates with angular speed w in uniform magnetic field ‘B’, 1, e = Bl 2 w, 2, In case of earth’s magnetic field B H =| B e | cos d, and, BV =| B e | sin d, 1, \, e = | B e | cos d. l 2 w, 2, 1, = ´ 0.4 ´ 10 – 4 cos 60°´ ( 0.5) 2 ´ 2pn, 2, æ 120 rev ö, 1, 1, ÷÷, = ´ 0.4 ´ 10 – 4 ´ ´ ( 0.5) 2 ´ 2p ´ çç, 2, 2, è 60 s ø, , Þ R=, , =10 – 5 ´ 0.25 ´ 2 ´ 314, . ´2, = 314, . ´ 10 – 5 volt, Induced emf is independent of the number of spokes i.e., it remain same., 19. Ratio of deflection produced in the galvanometer to the current flowing through it., q, Current sensitivity Si =, I, S.I. unit of current sensitivity Si is division/ampere or radian/ampere.
Page 466 :
Examination Papers, , 465, , For balanced Wheatone bridge, if no current flows through the galvanometer, 4 6, 4´ 9, = Þ R1 =, =6W, R1 9, 6, For another current, 6 R2, 6´8, =, Þ R2 =, =4W, 12 8, 12, R1 6 3, \, = =, R2 4 2, 20. Concepts: (i) Current carrying conductors repel each other, if current flows in the opposite, direction., Repulsion, C, , D, , Weight, , F, , C, , D, , Weight, , Attraction, , A, , B, , A, , B, , (ii) Attract each other if current flows in the same direction., If wire CD remain suspended above AB then, Frepulsion = Weight, m 0 I1 I 2 l, = mg, 2 pr, where r = Separation between the wires, m m 0 I1 I 2, =, l, 2 prg, =, , 2 ´ 10 – 7 ´ 12 ´ 5, 1´ 10 – 3 ´ 10, , =1.2 ´ 10 – 3 kg / m, Current in CD should be in opposite direction to that in AB., 21., , Forward, bias, , l (mA), Avalanche, breakdown, (–), , F, , (+), , V, V, 0.65 V for Si, Reverse 0.2 V for Ge, bias, R, l (mA), O, , V-I Characteristics, , (i) In the reverse biasing, the current of order of m A is due to movement/drifting of minority, charge carriers from one region to another through the junction., A small applied voltage is sufficient to sweep the minority charge carriers through the, junction. So reverse current is almost independent of critical voltage.
Page 467 :
466, , Xam idea Physics—XII, , (ii) At critical voltage (or breakdown voltage), a large number of covalent bonds break, resulting, in the increase of large number of charge carriers. Hence current increases at critical voltage., Semiconductor device that is used in reverse biasing is zener diode., B, , Objective, , a, A, , ue, , f0, , Distant Object, , 22., , Eye piece, , a A", , F'e, , F0, A', b, , C1, , C2, , F'e, , B', L2, , B", , ve, , It is defined as the ratio of the angle (b) subtended by the final image on the eye to the angle( a), subtended by the object on eye., tan b æ b ö, M=, =ç ÷, tan a è a ø, Magnifying power M =, , – f0, fe, , (for comfortable view), , –f0 æ fe ö, (for strained eye), ç1 + ÷, fe è, Dø, Limitations: (i) Image is not free from chromatic aberration and spherical aberration., (ii) Aperture of the objective lens should be large for high resolving power., 23. If radiation of frequency (n) greater than threshold frequency (n 0 ) irradiate the metal surface,, electrons emitted out from the metal. So Einstein’s photoelectric equation can be given as, 1, 2, K max = m v max, = hn – h n 0, 2, Characteristic properties of photons:, =, , (i) Energy of photon is directly proportional to the frequency (or inversely proportional to the, wavelength)., (ii) In photon-electron collision, total energy and momentum of the system of two constituents, remains constant., (iii) In the interaction of photons with the free electrons, the entire energy of photon is absorbed., Features of photoelectric effects, (i) Explanation of frequency law: When frequency of incident photon (n), increases, the, kinetic energy of emitted electron increases. Intensity has no effect on kinetic energy of, photoelectrons.
Page 468 :
Examination Papers, , 467, , Photo electric current, , (ii) Explanation of intensity law: When intensity of incident light increases, the number of, incident photons increases, as one photon ejects one electron; the increase in intensity will, increase the number of ejected electrons. In other words, photocurrent will increase with, increase of intensity. Frequency has no effect on photocurrent., (iii) Explanation of no time lag law: When the energy of incident photon is greater than work, function, the photoelectron is immediately ejected. Thus, there is no time lag between, incidence of light and emission of photoelectrons., , Intensity, , 24. Name of the wave may be (i) space wave (ii) radiowave (iii) microwave., Frequency for LOS communication must be more than 40 MHz., Maximum distance between transmitting antenna and receiving antenna is, d max = 2RhT – 2RhR, = 2 ´ 6.4 ´ 10 6 ´ 20 + 2 ´ 6.4 ´ 10 6 ´ 45, = (16 + 24) ´ 10 3 m, 25., , = 40 km, (a) Molecules in air behave like a dipole radiator. When the sunlight falls on a molecule, dipole, molecule does not scatter energy along the dipole axis, however the electric field vector of, light wave vibrates just in one direction perpendicular to the direction of the propagation. The, light wave having direction of electric field vector in a plane is said to be linearly polarised., In figure, a dipole molecule is lying along x-axis. Molecules behave like dipole radiators and, scatter no energy along the dipole axis., Linearly, polarised, , Unpolarised, 90°, X, 90°, , Linearly, polarised, , The unpolarised light travelling along x-axis strikes on the dipole molecule get scattered, along y and z directions. Light traversing along y and z directions is plane polarised light.
Page 469 :
468, , Xam idea Physics—XII, , (b) In figure unpolarised light falls on the polaroid, and transmitted light has electric vibrations in, the plane consisting of polaroid axis and direction of wave propagation as shown in Fig., , Intensity (Io), , 26., , 27., , Intensity, , (Io), 2, , If polaroid is rotated the plane of polarisation will change, however the intensity of, transmitted light remain uncharged., Coating n = 1.5, (a) Total internal reflection: If a light ray, enters at one end of an optic fibre, coated with a material of low, n = 1.7, refractive index, it refracted and, strikes the walls at angle greater than, critical angle., Thus light rays shows multiple, reflections, without being absorbed at, A, B, the side walls., (b) The teacher knows that Chetan, belongs to a below average income, group family, so he/she immediatelly arranged the money required to be paid as test fee., His/her caring and helping attitude towards the others resulted in timely help to Chetan’s, mother., Such helping attitude on the part of the person living in the society make it a better society to, live in., (c) Seeing the situation of Chetan’s family and helping attitude of class teacher, doctor took the, sympathetic view of the situation, and give the reduction in fee, which is highly appreciable., Such professional ethics of doctor in the society would be an immense help to the person’s, belonging to below average income groups., (a) Magnetic field at the axis of a circular loop: Consider, a circular loop of radius R carrying current I, with its, a, plane perpendicular to the plane of paper. Let P be a, a, point of observation on the axis of this circular loop at, a distance x from its centre O. Consider a small, element of length dl of the coil at point A. The, ®, , magnitude of the magnetic induction dB at point P due, to this element is given by, ® m 0 I dl sin a, ...(i), =, dB 4p, r2, ®, , ®, , ®, , The direction of dB is perpendicular to the plane containing dl and r and is given by right, ®, , ®, , hand screw rule. As the angle between I dl and r is 90°, the magnitude of the magnetic, ®, , induction dB is given by,
Page 470 :
Examination Papers, , ® m 0 I dl sin 90° m 0 I dl, =, =, ×, dB, 4p, r2, 4p r 2, , 469, , ...(ii), , If we consider the magnetic induction produced by the whole of the circular coil, then by, symmetry the components of magnetic induction perpendicular to the axis will be cancelled, , ®, , out, while those parallel to the axis will be added up. 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,, m I dI, ®, sin a, = 0, dB x, 4p r 2, But sin a =, , ...(iii), , R, and r = ( R 2 + x 2 ) 1/ 2, r, , ®, m I dl R m 0 IR, m 0 IR, dB x = 0, × =, dl =, dl, 2, 3, r 4p r, 4p r, 4p ( R 2 + x 2 ) 3 / 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, , ò dl = length of the loop = 2pR, , But, , B=, , Therefore,, , m 0 IR, 4p ( R 2 + x 2 ) 3 / 2, , ®, B = B x i$ =, , ...(v), ( 2 p R), , m 0 IR 2, 2, , 2 3/ 2, , 2 (R + x ), , i$., , (b) A long solenoid on bending in the form of closed ring is called a toroidal solenoid., Figure shows a toroidal solenoid of average radius ‘r’ and of N turns., (i) For points inside the core of toroid, Current I, flowing through it, set up a magnetic field within the, core., According to Ampere’s circuital law, O, , P, r, , ® ®, , ò B . dl = m 0 I, where ‘I’ is the current in the toroid., Net current = NI, ® ®, , \, , ò B . dl = m 0 NI, , Þ, , | B| 2 p r = m 0 NI, m NI, | B| = 0, 2 pr, , Þ, , =m 0 n I, , N ù, é, êëQ n = 2pr úû, , I, , I
Page 471 :
470, , Xam idea Physics—XII, , (ii) For points in the open space inside the toroid: No current flows through the, Amperian loop, so I = 0, ® ®, , ò B . dl = m 0 I = 0, Þ, | B| inside = 0, (iii) For points in the open space exterior to the toroid : The net current entering the plane of, the toroid is exactly cancelled by the net current leaving the plane of the toroid., ® ®, , ò B . dl = 0, Þ, , | B| exterior = 0, , OR, Role of electric field (i) Electric field accelerates the charge particle passing through the gap with, the help of electric oscillator., (ii) Electric oscillator imparts the energy to charged particle till it comes out from the exit slit., Role of magnetic field, As the accelerated charge particle enters normally to the uniform magnetic field, it exerts a, magnetic force in the form of centripetal force and charge particle moves on a semicircular path of, increasing radii in each dee ( D1 or D2 )., Magnetic field, out of paper, , Deflecting plate, , Exit port, Charged, particle, , D1, D2, , Electric Oscillator, , mv 2, r, q Br, v=, Þ, m, 1, Kinetic energy K = mv 2, 2, So, , qv B =, , =, , q 2 B 2r 2, 2m, , (1), , (2)
Page 473 :
472, , Xam idea Physics—XII, ®, , ®, , To find the resultant of E1 and E 2 , we resolve them along and perpendicular to AB., ®, , ®, , Component of E1 along AB = E1 cos q , parallel to BA, ®, , Component of E1 perpendicular to AB = E1 sin q along O to P, ®, , ®, , Component of E 2 along AB = E 2 cos q , parallel to BA, ®, , Component of E 2 perpendicular to AB = E 2 sin q along P to O, ®, , ®, , Clearly components of E1 and E 2 perpendicular to AB : E1 sin q and E 2 sin q being equal, ®, , ®, , and opposite cancel each other, while the components of E1 and E 2 along AB : E1 cos q, and E 2 cos q, being in the same direction add up and give the resultant electric field whose, ®, , direction is parallel to BA ., \ Resultant electric field at P is E = E1 cos q + E 2 cos q, q, 1, But, E1 = E 2 =, 2, 4p e 0 (r + l 2 ), and, \, , cos q =, , OB, l, =, 2, PB, r +l, , E = 2E1 cos q = 2 ´, =, , 2, , l, , =, (r, , 2, , + l 2 ) 1/ 2, , q, 1, l, ×, 2, 2, 2, 4p e 0 (r + l ) (r + l 2 ) 1/ 2, , 2ql, 1, 2, 4p e 0 (r + l 2 ) 3 / 2, , But q.2l = p = electric dipole moment, p, 1, \, E=, 2, 4p e 0 (r + l 2 ) 3 / 2, , ...(iii), , If dipole is infinitesimal and point P is far away, we have l < < r , so l, , 2, , may be neglected as, , 2, , compared to r and so equation (3) gives, p, p, 1, 1, E=, =, 2, 3, /, 2, 4p e 0 (r ), 4p e 0 r 3, i.e. electric field strength due to a short dipole at broadside on position, E=, , ®, p, 1, , parallel to BA, 4p e 0 r 3, , ...(iv), , (b), , +Q, , –Q, , Electric potential is zero at all points in the plane passing through the dipole equator.
Page 474 :
473, , Examination Papers, , OR, (i) 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. To find the electric field outside the shell, we, consider a spherical Gaussian surface of radius r ( > R),, , Q, , EO, , R, , dS, , P, , r, , ®, , concentric with given shell. If E is electric field outside, the shell, then by symmetry electric field strength has, same magnitude E 0 on the Gaussian surface and is, directed radially outward. Also the directions of normal at each point is radially outward, so, ®, , ®, , angle between E i and d S is zero at each point. Hence, electric flux through Gaussian, surface ò, , ®, , S, , ®, , = E 0 · d S., , ò = E 0 dS cos 0 = E 0 . 4pr, , 2, , Now, Gaussian surface is outside the given charged shell, so charge enclosed by Gaussian, surface is Q., Hence, by Gauss's theorem, ®, ®, 1, ò S = E 0 · d S = e 0 ´ charged enclosed, 1, 1 Q, ´ Q Þ E0 =, E 0 4pr 2 =, Þ, e0, 4pe 0 r 2, Thus, electric field outside a charged thin spherical shell is the same as if the whole charge, Q is concentrated at the centre., If s is the surface charge density of the spherical shell, then, Q = 4pR 2 s C, \, , E0 =, , 1 4pR 2 s R 2 s, =, 4pe 0 r 2, e 0r 2, , (ii) Electric field inside the shell (hollow charged conducting, sphere): The charge resides on the surface of a conductor. Thus, a hollow charged conductor is equivalent to a charged spherical, shell. To find the electric field inside the shell, we consider a, spherical Gaussian surface of radius r ( < R), concentric with the, , Ei, r, , dS, R, , ®, , given shell. If E is the electric field inside the shell, then by, symmetry electric field strength has the same magnitude Ei on, the Gaussian surface and is directed radially outward. Also the, , ®, , ®, , directions of normal at each point is radially outward, so angle between E i and d S is zero at, each point. Hence, electric flux through Gaussian surface, ®, , ®, , = ò Ei . d S, S, , = ò Ei dS cos 0 = Ei . 4pr 2, Now, Gaussian surface is inside the given charged shell, so charge enclosed by Gaussian surface, is zero.
Page 477 :
476, , Xam idea Physics—XII, , Conclusions:, (i) Nuclear forces non-central and short ranged force., (ii) Nuclear forces between proton-neutron and neutron-neutron are strong and attractive in, nature., (b) Explanation of Nuclear Fission: When a heavy nucleus (A ³ 235 say) breaks into two, lighter nuclei (nuclear fission), the binding energy per nucleon increases i.e, nucleons get, more tightly bound. This implies that energy would be released in nuclear fission., Explanation of Nuclear Fusion: When two very light nuclei ( A £10) join to form a heavy, nucleus, the binding is energy per nucleon of fused heavier nucleus more than the binding, energy per nucleon of lighter nuclei, so again energy would be released in nuclear fusion., (c) During decay of a neutron, we have, 1, 0n, , ® 11p +, , 0, –1 b + n, , The detection of neutrinos is very difficult because it shows weak interactions with other, particles., , SET–II, 4. j T =, , –Q, e0, , (0, 0), , +2Q, , –3Q, , (a, 0), , (4a, 0), , Draw a sphere of radius 5a about origin ‘O’ and on applying Gauss’s Theorem., SQi, j (T) =, e0, 7. A metal disc is placed on the top of a magnet, as the electric current flows through the coil, an, induced current in the form of eddies flows through the metal plate, the lower face attains the, same polarity, and hence the metal disc is thrown up., 9., , (i) AND Gate, (ii) Truth Table, A, , C, , Y, B, , D, (b)
Page 478 :
Examination Papers, A, , B, , y=A + B, , 0, , 0, , 0, , 1, , 0, , 0, , 0, , 1, , 0, , 1, , 1, , 1, , 477, , 13. d . q = n l, æyö, dç ÷= nl, èDø, For first minima n =1, lD, d=, y, =, , To P, , q, , d, , To C, , 1.2 ´ 600 ´ 10 – 9, 3 ´ 10 – 3, , = 2.4 ´ 10 – 4 m, 19. In stable equilibrium, Frepulsion = mg, m 0 I1 I 2 l, = mg, 2 pr, \, =, , æ m ö m 0 I1 I 2, ç ÷=, 2 p rg, è l ø, , C, , D, , A, , B, , 2 ´ 10 – 7 ´ 10 ´ 6, 2 ´ 10 – 3 ´ 10, , = 6 ´ 10 – 4 kg / m, Current in both wires should be opposite, so both conductors repel each other., 23. Maximum distance between transmitting antenna and receiving antenna, d max = 2RhT + 2R hR, = 2 ´ 6.4 ´ 10 6 ´ 80 + 2 ´ 6.4 ´ 10 6 ´ 45, = ( 1024 + 576 ) ´ 10 3, = (32 + 24) ´ 10 3, = 56 ´ 10 3 m
Page 479 :
478, , Xam idea Physics—XII, , Set–III, 3. fT =, , 4Q, e0, , –Q, , For sphere of radius ‘2a’ charge –Q is outside the, æ 4Q ö, ÷÷, Gaussian surface. So fT = çç, è e0 ø, 4. Decreases, 6. (i) Mutual inductance increases., (ii) Mutal inductance decreases., 9. (i) AND Gate, (ii), Truth Table, , 4Q, O, , (–3a, 0), , A, , B, , y=A + B, , 0, , 0, , 0, , 1, , 0, , 0, , 0, , 1, , 0, , 1, , 1, , 1, , (a, 0), , A¢, , A, , y¢, B, , y, , B¢, , Extended Truth Table, A, , B, , A, , B, , y ¢= A + B, , y=A + B, , 0, , 0, , 1, , 1, , 1, , 0, , 1, , 0, , 0, , 1, , 1, , 0, , 0, , 1, , 1, , 0, , 1, , 0, , 1, , 1, , 0, , 0, , 0, , 1, , 15. Capacitance of parallel plate capacitor filled with dielectric:, e0 A, C=, t, ( d – t) +, k, 2d, Since thickness t =, 3, , + + + + + +, d, , t, – – – – – –
Page 481 :
CBSE Examination Papers, Foreign–2013, SET–I, Time allowed : 3 hours, , Maximum marks: 70, , General Instructions: As given in CBSE Examination Paper Delhi–2013., 1. Define electric dipole moment. Is it scalar or vector?, 2. On what factors does the magnitude of the emf induced in the circuit due to magnetic flux depend?, 3. Write the function of a transmitter in a communication system., 4. A ray of monochromatic light passes from medium (1) to medium (2). If the angle of incidence in, medium (1) is q and the corresponding angle of refraction in medium (2) is q / 2, which of the two, media is optically denser? Give reason., 5. Why are broadcast frequencies (carrier waves) sufficiently spaced in amplitude modulated wave?, 6. Plot a graph showing the variation of resistance of a conducting wire as a function of its radius,, keeping the length of the wire and its temperature as constant., 7. Two materials Si and Cu, are cooled from 300 K to 60 K. What will be the effect on their resistivity?, 8. A long straight wire carries a steady current I along the positive y-axis in a coordinate system. A, ®, , particle of charge + Q is moving with a velocity v along the x-axis. In which direction will the, particle experience a force?, 9. Calculate the value of the current drawn from a 5 V battery in the circuit as shown., 10 W, , A, , 5W, , B, , 10 W, , C, , 20 W, , D, , 10 W, + –, 5V, , 10. Two concentric metallic spherical shells of radii R and 2R are given charges Q1 and Q2 respectively., The surface charge densities on the outer surfaces of the shells are equal. Determine the ratio Q1 : Q2., 11. Three rays of light, red (R), green (G) and blue (B), are incident on the face AB of a right angled, prism, as shown in the figure. The refractive indices of the material of the prism for red, green and, blue are 1.39, 1.44 and 1.47 respectively. Which one of the three rays will emerge out of the prism?, Give reason to support your answer.
Page 482 :
Examination Papers, , 481, , A, B, G, R, 45°, B, , C, , 12. A resistor ‘R’ and an element ‘X’ are connected in series to an ac source of voltage. The voltage is, found to lead the current in phase by p/4. If ‘X’ is replaced by another element ‘Y’, the voltage lags, behind the current by p/4., (i), Identify elements ‘X’ and ‘Y’., (ii) When both ‘X’ and ‘Y’ are connected in series with ‘R’ to the same source, will the power, dissipated in the circuit be maximum or minimum? Justify your answer., OR, A series LCR circuit is connected to an ac source (200 V, 50 Hz). The voltages across the resistor,, capacitor and inductor are respectively 200 V, 250 V and 250 V., (i), The algebraic sum of the voltages across the three elements is greater than the voltage of the, source. How is this paradox resolved?, (ii), , Given the value of the resistance of R is 40W, calculate the current in the circuit., o, , 13. Ultraviolet light of wavelength 2271 A from 100 W mercury source irradiates a photocell made of, molybdenum metal. If the stopping potential is –1.3 V, estimate the work function of the metal. How, would the photocell respond when the source is replaced by another source of high intensity, o, , (~ 105 Wm -2 ) red light of wavelength 6328 A Justify your answer., 14. The circuit shown in the figure has two oppositely connected ideal diodes connected in parallel. Find, the current flowing through each diode in the circuit., 2W, , D1, , D2, , 3W, , 4W, , – +, 12V, , 15. An electron and a proton, each have de Broglie wavelength of 1.00 nm., (a) Find the ratio of their momenta., (b) Compare the kinetic energy of the proton with that of the electron., 16. A parallel plate capacitor, each of plate area A and separation ‘d’ between the two plates, is charged, with charges + Q and – Q on the two plates. Deduce the expression for the energy stored in the, capacitor., 17. In an experiment on a-particle scattering by a thin foil of gold, draw a plot showing the number of, particles scattered versus the scattering angle q., Why is it that a very small fraction of the particles are scattered at q > 90º?, Write two important conclusions that can be drawn regarding the structure of the atom from the study, of this experiment.
Page 483 :
482, , Xam idea Physics—XII, , OR, Derive the expression for the law of radiactive decay of a given sample having initially N0 nuclei, decaying to the number N present at any subsequent time t., Plot a graph showing the variation of the number of nuclei versus the time t lapsed., Mark a point on the plot in terms of T1/ 2 value when the number present N = N0 /16., 18. Give reasons for the following:, Calculate:, (i) For ground wave transmission, size of antenna should be comparable to the wavelength of the, signal, e.g. ~l/4., (ii) Audio signals converted into electromagnetic waves are not transmitted as such directly., (iii) The amplitude of modulating signal is kept less than that of the carrier wave., 19., , (a), , An infinitely long positively charged straight wire has a linear charge density l cm–1. An, electron is revolving around the wire as its centre with a constant velocity in a circular plane, perpendicular to the wire. Deduce the expression for its kinetic energy., , (b) Plot a graph of the kinetic energy as a function of charge density l., 20. (a), , (b), 21. (i), (ii), (iii), 22., , ®, , ®, , (i) A circular loop of area A, carrying a current I is placed in a uniform magnetic field B. Write, ®, , the expression for the torque t acting on it in a vector form., (ii) If the loop is free to turn, what would be its orientation of stable equilibrium? Show that in, this orientation, the flux of net field (external field + the field produced by the loop) is, maximum., Find out the expression for the magnetic field due to a long solenoid carrying a current I and, having n number of turns per unit length., Draw a schematic labelled ray diagram of a reflecting type telescope., Write two important advantages justifying why reflecting type telescopes are preferred over, refracting telescopes., The objective of a telescope is of larger focal length and of larger aperture (compared to the, eyepiece). Why? Give reasons., , (a) State, with the help of a suitable diagram, the principle on which the working of a meter bridge, is based., (b) Answer the following:, (i), Why are the connections between resistors in a meter bridge made of thick copper strips?, (ii) Why is it generally preferred to obtain the balance point near the middle of the bridge, wire in meter bridge experiments?, , 23. (a) How are electromagnetic waves produced by oscillating charges?, (b) State clearly how a microwave oven works to heat up a food item containing water molecules., (c) Why are microwaves found useful for the radar systems in aircraft navigation?, 24. (a) The energy levels of a hypothetical hydrogen-like atom are shown in the figure. Find out the, transition, from the ones shown in the figure, which will result in the emission of a photon of, wavelength 275 nm.
Page 484 :
Examination Papers, , A, , B, , 483, , 0 eV, , C, , D, , –2 eV, , –4.5 eV, , –10 eV, , (b) Which of these transitions corresponds to the emission of radiation of (i) maximum and (ii), minimum wavelength?, 25. When unpolarised light is incident on the boundary separating the two transparent media, explain,, with the help of a suitable diagram, the conditions under which the reflected light gets polarised., Hence define Brewster’s angle and write its relationship in terms of the relative refractive index of the, two media., 26. Kamal’s uncle was advised by his doctor to undergo an MRI scan test of his chest and gave him an, estimate of the cost. Not knowing much about the significance of this test and finding it to be too, expensive he first hesitated. When Kamal learnt about this, he decided to take help of his family,, friends and neighbours and arranged for the cost. He convinced his uncle to undergo this test so as to, enable the doctor to diagnose the disease. he got the test done and the resulting information greatly, helped the doctor to give him proper treatment., (a) What, according to you, are the values displayed by Kamal, his family, friends and neighbours?, (b) Assuming that the MRI scan test involved a magnetic field of 0.1 T, find the maximum and, minimum values of the force that this field could exert on a proton moving with a speed of, 104 ms–1. State the condition under which the force can be minimum., 27. Using phasor diagram for a series LCR circuit connected to an ac source of voltage v = v0 sin wt,, derive the relation for the current flowing in the circuit and the phase angle between the voltage across, the resistor and the net voltage in the circuit., Draw a plot showing the variation of the current I as a function of angular frequency ‘w’ of the applied, ac source for the two cases of a series combination of (i) inductance L1, capacitance C1 and resistance, R1 and (ii) inductance L2, capacitance C2 and resistance R2 where R2 > R1., Write the relation between L1, C1 and L2,C2 at resonance. Which one, of the two, would be better, suited for fine tuning in a receiver set? Give reason., OR, (a) Define the term ‘mutual inductance’., Deduce the expression for the mutual inductance of two long coaxial solenoids having different, radii and different number of turns., , 28., , (b) A coil is mechanically rotated with constant angular speed w in a uniform magnetic field which, is perpendicular to the axis of rotation of the coil. The plane of the coil is initially held, perpendicular to the field. Plot a graph showing variation of (i) magnetic flux f and (ii) the, induced emf in the coil as a function of wi., (a) A monochromatic source of light of wavelength l illuminates a narrow slit of width d to produce, a diffraction pattern on the screen. Obtain the conditions when secondary wavelets originating, from the slit interfere to produce maxima and minima on the screen.
Page 485 :
484, , Xam idea Physics—XII, , (b) How would the diffraction pattern be affected when, (i) the width of the slit is decreased?, (ii) the monochromatic source of light is replaced by white light?, OR, , 29., , (a), , A thin convex lens having two surfaces of radii of curvature R1 and R2 is made of a material of, refractive index m2. It is kept in a medium of refractive index m1. Derive, with the help of a ray, diagram, the lens maker formula when a point object placed on the principal axis in front of the, radius of curvature R1 produces an image I on the other side of the lens., Distinguish between an intrinsic semiconductor and a p-type semiconductor. Give reason why a, p-type semiconductor is electrically neutral, although nh >> ne., , (b) Explain, how the heavy doping of both p-and n-sides of a p-n junction diode results in the, electric field of the junction being extremely high even with a reverse bias voltage of a few volts., Explain, with the help of a circuit diagram, how this property is used in voltage regulator., OR, Draw the circuit arrangement for studying the input and output characteristics of n-p-n, transistor in CE configuration., Draw the typical nature of these input and output characteristics. Explain how these are, obtained., Define the terms (i) input resistance and (ii) current amplification factor., , SET–II, Questions Uncommon to Set–I, 1., , ®, , ®, , When a charged particle moving with velocity v is subjected to magnetic field B, the force acting on it, is non-zero. Would the particle gain any energy?, , 5. Two materials, Ge and Al, are cooled from 300K to 60 K. What will be the effect on their resistivity?, 11. Two concentric metallic spherical shells of radii R and 3R are given charges Q1 and Q2 respectively., The surface charge densities on the outer surfaces of the shells are equal. Determine the ratio Q1 : Q2., 12. In the circuit shown in the figure, the galvanometer ‘G’ gives zero deflection. If the batteries A and B, have negligible internal resistance, find the value of the resistor R., 500W, G, , 12V, , B, , R, , A, , 2V, , 15. The circuit shown in the figure contains two diodes each with a forward resistance of 50 W and, infinite backward resistance. Calculate the current in the 100 W resistance?
Page 486 :
Examination Papers, , D1, , 150W, , D2, , 50W, , –, , 100W, , +, , 485, , 6V, , 21. Write the generalised expression for the Ampere’s circuital law in terms of the conduction current and, the displacement current. Mention the situation when there is:, (i) only conduction current and no displacement current., (ii) displacement current and no conduction current., 22., , (a), , Draw a labelled ray diagram of a compound microscope., , (b) Derive an expression for its magnifying power., (c), , Why is objective of a microscope of short aperture and short focal length? Give reason., , SET–III, Questions Uncommon to Set-I and Set-II, 4. Two materials, Ag and GaAs, are cooled from 300 K to 60 K. What will be the effect on the, resistivity?, ®, , ®, , 5. In a certain region of space, electric field E and magnetic field B are perpendicular to each other. An, ®, , ®, , electron enters in the region perpendicular to the directions of both B and E and moves undeflected., Find the velocity of the electron., 10. What will be the value of current through the 2W resistance for the circuit shown in the figure? Give, reason to support your answer., , 10 V, , 10 W, , 5W, , 20 V, , 2W, , 13. The circuit shown in the figure contains two diodes each with a forward resistance of 50 W and, infinite backward resistance. Calculate the current in the 100 W resistance.
Page 487 :
486, , Xam idea Physics—XII, , +, , D1, , 200W, , D2, , 50W, , –, , 100W, , 6V, , 18., , 19., , (a), , Why are infra-red radiations referred to as heat waves? Name the radiations which are next to, these radiations in the electromagnetic spectrum having (i) shorter wavelength (ii) longer, wavelength., (b) State the conditions under which a microwave oven heats up a food item containing water, molecules., (a) Draw a labelled ray diagram of a refraction type telescope in normal adjustment., (b) Give its two shortcomings over reflection type telescope., (c) Why is eyepiece of a telescope of short focal length, while objective is of large focal length?, Explain., zzz
Page 488 :
Examination Papers, , 487, , Solutions, CBSE (Foreign) SET–I, 1. The product of the magnitude of either charge and shortest distance between the opposite charges., It is a vector quantity and its direction is from negative charge towards positive charge., –q, , ®, P, , +q, , 2. Depends on the time rate of change in magnetic flux (or simply change in magnetic flux), Df, | e| =, Dt, 3. Processes the incoming message signal on suitable carrier waves. So as to make it suitable for, transmission through a channel., 4. Medium (2) is optically denser., Reason - angle of refraction is less than the angle of incidence., OR, Angle of refraction decreases, as the refractive index of the medium decreases., q, m 1 sin q = m 2 sin, (m 2 > m 1 ), 2, 5. To avoid mixing up of signals from different transmitters. This can be done by modulating the signals, on high frequency carrier waves, e.g. frequency band for satellite, communication is 5.925– 6.425 GHz., 6. Resistance of a conductor of length l, and radius r is given by, l, R =r 2, pr, , R, , r, , 7. In silicon, the resistivity increases, In copper, the resistivity decreases., , r, , r, , T (K), , T (K)
Page 489 :
488, , Xam idea Physics—XII, , r, $ = +qvB(j), $, 8. From relation F =qvB[ $i ´ (–k)], , ®, × B, ×, , ®, v, , ×, , r, Magnetic force F along + y axis., , ×, ×, , OR, From Fleming’s left hand rule, thumb points along + y direction, so, the direction of magnetic force will be along + y axis (or in the, direction of flow of current)., , 9. In case of balanced Wheatstone bridge, no current flows through the resistor 10W between points B, and C., 10W, The resistance of arm ACD, R S1 = 10 + 20 = 30W, The resistance of arm ABD, R S2 = 5 + 10 = 15W, R S ´ R S2, 5W, 10W, 20W, B, D, Equivalent resistance Req = 1, R S1 + R S2, A, C, =, , 30 ´ 15, 30 + 15, , =, , 30 ´ 15, 45, , 10W, , =10W, Current drawn from the source,, V, 5 1, I=, = = A = 0 × 5A, R eq 10 2, , + –, 5V, , 10. Surface charge density s = constant, Charge Q1 = 4pR 2 s, Charge Q2 = 4p ( 2R 2 ) s, \, , R, , 2, , Q1, 4pR s, 1, =, =, Q2 4p(2R) 2 s 4, , 2R, , 11. If angle of incidence ‘i’ is less than the critical angle of glass-air interface, AC then it will emerge out., 1, Critical angle sin iC =, m, 1, 1, B, = 2 = 1.414, \m =, =, sin iC sin 45 o, Since m R = 1.39, mG = 1.44 and mB = 1.47, so from, equation (1) angle of incidence for red colour iC > 45º, while angle of incidence for blue and green colours iC < 45º,, hence blue and green colour rays will emerge out., , 12., , A, , G, 45°, R, B, , C, , p, (i) In R– L series combination, voltage leads the current by phase f = . It means element x is an, 4, p, inductor. In R.C. series combination, voltage lags behind the current by phase f = . So element, 4, y is a capacitor.
Page 491 :
490, , Xam idea Physics—XII, , Since frequency of red light is less than threshold frequency so photocell will not respond to red light,, howsoever high (105 W/m2) be the intensity of light., 14. (i) Diode D1 is reverse biased, so it offers an infinite resistance. So, no current flows in the branch of diode D1., , 2W, , D2, , (ii) Diode D2 is forward biased, and offers no resistance in the, circuit. So current in the branch, , D1, , 3W, , I=, , 15., , (i), , le =, So,, Þ, , 12V, V, =2A, =, R eq 2W + 4W, , le p p 1, =, =, l p pe 1, pp, , 1, = = 1:1, pe 1, , Ke =, , Ke, , 12V, , h, h, and l p = , l e = l p = 1.00 nm, Pe, Pp, , (ii) From relation K =, , Kp, , 4W, , p2, 1, mv 2 =, 2, 2m, , p 2p, pe2, and K p =, 2m e, 2m p, , =, , p 2p, 2m p, , ´, , 2m e, pe, , 2, , =, , me, mp, , Since me <<< mp. So Kp <<< Ke., Kp, . ´ 10 -31, 91, = 5.4 × 10–4, =, K e 1.67 ´ 10 -27, 16. Q and – q are charges on the plates and produces a uniform, s, electric field E =, between the plates and a potential, e0, q, difference V =, . . . (i), C, , +Q, , –Q, , dq, , If a charge dq is transported in steps from negative charged plate to +ve charged plate, till charges, rises to +Q and –Q, then, Workdone dW = dq. V, , . . . (ii), , From equations (1) and (2), q, dW = dq ( ), C, Total electrostatic potential energy stored can be given as
Page 492 :
Examination Papers, , 491, , Q, , Q2, q, . dq =, C, 2C, 0, , U =W = ò, U=, , Q2, 2C, , where l is disintegration constant., The rate of change in N in time Dt® 0, can be, dN, expressed as, = -lN, N, On integrating both sides, N, , ò, , N0, , t, , dN, = -ò ldt, N, 0, , where N0 is initial undecayed nuclei., N, (ln N), = -lt, N0, ln, , N, = -lt, N0, , N = N 0 e - lt, , No. of undecayed nuclei, , No. of scattered, a particles, , 17. A small fraction of the alpha particles, 108, scattered at angle q > 90º is due to the, following reasons., 106, (i) If impact parameter ‘b’ reduces to zero,, 104, coulomb force increases, and hence alpha, particles are scattered at angle q > 90º, and, 102, only one alpha particle is scattered at angle, 180º., 10, Conclusions:, 0, (i) Entire positive charge and most of the, 45°, 90°, 135°, 180°, Scattering angle q, mass of the atom is concentrated in the, nucleus with the electrons some distance, away., (ii) Size of the nucleus is about 10–15 m to 10–14 m, while size of the atom 10–10 m, so the electrons, are at distance 104 m to 105m from the nucleus, and being large empty space in the atom, most a, particles go through the empty space., OR, Let N be the number of undecayed nuclei in the, N0, sample at time t and DN nuclei undergo decay in, time Dt. Then,, -DN, µN, Dt, N, -DN, = lN, N–DN, Dt, O, , t, , t+ Dt, , Time t
Page 493 :
492, , Xam idea Physics—XII, , N0, in terms of T1/2 is shown in fig., 16, , No. of undecayed nuclei, , Mark of N =, , No, , No, 2, No, 4, No, 8, No, 16, , P, T1, , 2T1, , 3T1, , 4T1, , 2, , 2, , 2, , 2, , Time t, , 18. Ground wave propagation is possible for radio waves of frequency band 540kHz – 1600 kHz (or max., 2 MHz), If a base band signal of frequency 20 kHz, wavelength of the wave must be l =15 km, obviously, antenna of size l = 3.75 km is not possible to construct and operate., If base band signals are translated into high frequency radio wave of frequency n >1MHz, then, antenna of few metre can be constructed and can be used in sending the information along the ground., (i), , Low size of the antenna is required for waves of short wavelength and hence comparable., , (ii), , Audio signals of large wavelength cannot be send directly, because large size antenna is, required., V, Modulation index m = m , and its value should be less than 1, other distortion will produce in, VC, , (iii), , the wave., 19., , (a) Infinitely long charged wire produces a radical electric, field., l, ... (1), E=, 2pe0r, The revolving electron experience an electrostatic force, and provides necessarily centripetal force., eE =, , mv 2, r, , ... (2), , e. l, mv 2, =, 2pe 0r, r, Þ mv 2 =, , el, 2pe 0, , Kinetic energy of the electron, 1, el, K = mv 2 =, 2, 4pe 0, , e
Page 494 :
Examination Papers, , 493, , (b), , KE, , l, , ® ® ®, ® ®, (i) Torque acting on the current loop t = m ´ B = I (A ´ B), , 20. (a), , ® ®, (ii) If magnetic moment m = I A is in the direction of, external magnetic field i.e., q = 0 0 ., , ®, A, , q, , ®, B, , I, , ®, ® ®, Magnetic flux fB( B ext + BC ) × A, é ®, m Iù, fmax = ê| B ext |+ 0 ú| A| cos 0 o, 2r úû, êë, I, , where r is radius of the loop., ®®, On applying Ampere’s circuital law f B × dl= m 0 [Total current], , (b), , Þ, , ®®, f, ò B × dl +, PQ, , ®®, ò B × dl +, QR, , ®®, ò B × dl +, , ®®, ò B × dl = m 0 [ nlI ], , RS, , SP, , S, , l, , R, f, , P, , Q, , As no magnetic field exists in direction QR, RS and SP, so, l, , ò | B| dl + 0 + 0 + 0 = m 0 nlI, 0, , Þ | B| l = m 0 nlI ‘, Þ, B = mr 0 I, , ®, B
Page 495 :
494, , Xam idea Physics—XII, , 21. (i), , Secondary, convex mirror, , Eyepiece, Objective, concave mirror, , (ii), , (iii), , Advantages : (a) Parabolic mirror is used to remove the spherical aberration., (b) No chromatic aberration in mirror., (c) Light mechanical support is required, because mirror weighs much less than a lens of, equivalent optical quality., F, In normal adjustment, magnifying power of the telescope M = 0 ., Fe, (a) If focal length of the objecctive lens is large in comparison to the eyepiece, magnifying, power increases., a, ., (b) Resolving power of the telescope RP =, 1.22l, , To increase the resolving power of the telescope, large aperture of the objective lens is required., RB, , S, , 22., , (a) If ratio of arms resistors in wheatstone, bridge is constant, then no current flows, through, the, galvanometer, (or, B, bridgewire)., G, (b) (i) The resistivity of copper is several, A, C, times less than the resistivity of the, D, experimental alloy wire. As such, area of thick copper strips is more,, so copper strips almost offer zero, + –, K, resistance in the circuit., (ii) If any one resistance in wheatstone bridge is either very small (or very large) in respect of, other, then balance point might be very close to terminal A or terminal B. So generally, balance point is taken in the middle of the bridge wire., , 23., , (a) If a charge particle oscillate with some frequency, produces an oscillating electric field in, space, which produces an oscillating magnetic field, which inturn, is a source of electric field,, and so on. Thus oscillating electric fields and magnetic fields regenerate each other, and an, electromagnetic wave propagates in the space., (b) In microwave oven, the frequency of the microwaves is selected to match the resonant, frequency of water molecules so that energy from the waves get transferred efficiently to the, kinetic energy of the molecules. This kinetic energy raises the temperature of any food, containing water.
Page 496 :
Examination Papers, , 495, , (c) Microwaves are short wavelength radio waves, with frequency of order of GHz. Due to short, wavelength, they have high penetrating power with respect to atmsophere and less diffraction in the, atmospheric layers. So these waves are suitable for the radar systems used in aircraft navigation., 24., , (a) The energy of photon of wavelength (275nm) in terms of eV can be given as, æ lc ö, E = ç ÷ eV, è el ø, =, , 6.6 ´ 10 -34 ´ 3 ´ 108, , 1.6 ´ 10 -19 ´ 275 ´ 10 -9, 19.8, =, ´ 10 2 eV = 4.5 eV, 1.6 ´ 275, The energy of photon in transition B DE = [0eV- ( -4.5 eV]= 4.5 eV., Hence transition ‘B’ is possible., lc, DE, (i) Maximum wavelength of photon is possible for transition having minimum DE, so, transition ‘A’ is possible with DE= 2eV., (ii) Minimum wavelength of the photon is possible for transition having maximum energy, difference. So transition D is possible with DE= 10 eV., , (b) The wavelength of the photon in a transition is given by l =, , ef, le, , ct, ed, , Normal, Incident, , R, , 25. When unpolarised light incident on the water, molecules, the oscillating electrons in the water, produces the reflected wave., The double arrows are parallel to the direction of, the reflected wave, so they do not contribute to the, reflected wave; while dots are perpendicular to the, plane of incidence and they get reflected and, becomes linearly polarised light. This polarisation, can be checked through an analyser., The angle of incidence of unpolarised light at which, (i) refracted wave and reflected wave becomes, perpendicular to each other or (ii) reflected wave, becomes totally polarised wave is called Brewster’s, angle., p, Let ip be the angle of polarisation and i p + r =, 2, , i, , i, , r, , Refracted, , Then from Snell’s law, m=, , sin i p, sin r, , =, , sin i p, sin( 90º - i p ), , = tan i p, , 26. (a) Values displayed by Kamal:, (i) Being educated person knows about MRI (magnetic resonance imaging), (ii), , Took prompt decisions to take the help of his family, friends and neighbours and arranged the, cost of MRI., (iii) He shows his empathy, helping attitude and caring nature for his uncle.
Page 499 :
498, , Xam idea Physics—XII, , (b) The plane of the coil is in yz plane and perpendicular to the x-axis i.e., direction of magnetic, field., y, , z, , x, , Ù, n, , B, z, , A maximum magnetic flux F max = B| A|., As the coil rotates with angular speed w , magnetic flux at any instant t, (or at angle w t), F =| B||A| cosw t, 28. (a) When plane wavefront coming from distant source, illuminate the slit of size (=d), each point within the, slit becomes the source of secondary wavelets, and, these wavelets superpose on each other to generate, the maxima and minima on the screen; path, difference between the rays, directing to the point P, on the screen can be given as, In D ABT, BT D, sin q =, =, AB d, path difference D = d sin q, Condition of Minima:, , To P, q, , A, , q, , From m, source, T, , To C, , q, , B, , d, If set AB is divided into the equal halves (or in even parts) each of size , for every point in part, 2, AM, there is a point in part MB that contribute the secondary wavelets out of phase (i.e., 180º)., So net contribution from two halves becomes zero and hence intensity falls to zero for path, difference D = nl., \ d sin q = nl, d. q = nl where n is integer except n = 0., Condition of maxima:, If slit AB is divided into three equal parts (or in odd parts). First two thirds of the slit having a, l, path difference, between them cancel each other, and only the remaining one third of the set, 2, 1, contributes to the intensity at the point between two minima, so for path difference D = ( n + ) l., 2, 1, We have d sin q = ( n + ) l., 2, where n is integer except n = 0.
Page 500 :
Examination Papers, , 499, , (b) (i) Effect of the width of the slit —, For given monochromatic waves, if slit width is decreased, the fringe pattern becomes broader., y, y, 1, d. n = ( n + ) l, [q = n ], D, 2, D, Þ d. y n = constant, 1, Þ yn µ, d, (ii) If monochromatic source of light is replaced by white light, instead of white fringes we have, few coloured fringes on either side of central white fringe, and then uniform illumination on the, screen., y n a l (l for VIBGYOR), OR, A, N1, , O, , C2, , N2, , D, , B, , I, , C1, , I1, , Ray of light ON1 strikes the convex lens ABCD of radii R1 and R2. First refraction occurs at face, ABC and forms the image I1 of the object O. The image I1 acts as a virtual object for surface ADC that, forms the image I., On applying the condition of refractions on surfaces ABC and ADC., , and, , m 1 m 2 m 2 -m 1, +, =, OB BI 1, BC1, , . . . (1), , -m 2 m 1 m 2 - m 1, +, =, DI 1 DI, DC 2, , . . . (2), , For thin lens BI1 = DI1, and on adding equation (1) and (2)., m1 m1, 1, 1, +, = (m 2 - m 1 ) (, +, ), OB DI, BC1 DC 2, Suppose object is at infinity and image at focus then, OB = ¥ and DI = f, æ 1, m1, 1, = (m 2 - m 1 ) çç, +, f, è BC1 DC 2, , ö, ÷÷, ø, , 1 æm2 ö æ 1, 1, +, = çç, -1÷÷ çç, f è m 1 ø è BC1 DC 2, , ö, ÷÷, ø, , \
Page 501 :
500, , Xam idea Physics—XII, , On applying sign convention for convex lens BC1 = + R1 and DC2 = – R2., \, 29., , æ, m, ççQ m 21 = 2, m1, è, , 1, 1, 1, = (m 21 -1) ( - ), f, R1 R 2, , ö, ÷÷, ø, , (a), S.No., , p-type semiconductor, , Intrinsic semiconductor, , (i), , It is a semiconductor in pure form., , It is a semiconductor doped with p-type (like Al,, In) impurity., , (ii), , Intrinsic charge carriers are electrons and holes Majority charge carriers are holes and minority, with equal concentration., charge carriers are electrons., , (iii), , Current due to charge carriers is feeble (of the Current due to charge carriers is significant (of, order of mA)., 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., (b) If p-type and n-type semiconductor are heavily doped. Then due to diffusion of electrons from, n-region to p-region, and of holes from p-region to n-region, a depletion region formed of size, of order less than 1 mm. The electric field directing from n-region to p-region produces a, reverse bias voltage of about 5V and electric field becomes very large., E=, , DV 5 V, =, » 5 ´ 10 6 V/m, Dx 1mm, , Explanation of voltage regulator., A, If reverse bias voltage V reaches the breakdown, voltage VZ of zener diode, there is a large change, R, in the current. After that (just above VZ there is a, large change in the current by almost insignificant, change in revese bias voltage. This means diode, voltage remains constant., RL, For example: If unregulated voltage is supplied at, terminals A and B, and input voltage increases, the B, current through resistor RZ and diode also, increases. This current increases the voltage across RZ without any change in the voltage across, diode. Thus, we have a regulated voltage across load reistor RL., OR, n-p-n, , IB, –, , +, , IC, –, , C, , +, , mA, , mA, RB, , VBB, , +, , +, – V, BE, , IE, , RC, , VCE –, VCC
Page 502 :
Examination Papers, , 501, , VCE=10.0 V, , Input characteristics: The variation of the base, current Ib with the base-emitter voltage Vbe at, constant Vce., , 100, 80, IB(mA), , Input characteristics can be drawn, when the, 60, transistor is in active state, so Vce is kept enough, 40, large. So base collector junction remain in reverse, biased mode i.e., Vce >> 0.7 V., 20, Output characteristics:, O, Output characteristic is obtained by measuring the, VBE (V), 0.2 0.4 0.6 0.8 1.0, collector current IC with the variation in Vce keeping, base current Ib constant., Input resistance: It is defined as the ratio of change in base - emitter voltage Vbe to the resulting, change in base current Ib at constant Vce., æ DV ö, r ac = çç be ÷÷, è DI b ø Vce, Y, , Collector current (IC) in mA, , 10, , Base current (IB), 60 mA, , 8, , 50 mA, 6, , 40 mA, 30 mA, , 4, , 20 mA, 2, , 10 mA, , 0, , 2, , 4, , 6, , 8, , 10, , 12, , 14, , 16, , X, , Collector to emitter voltage (VCE) in volts, , Current Amplification: It is defined as the ratio of change in collector current IC to the change in, base current Ib at a constant Vce, when the transistor is in active state., æ DI ö, b ac = çç C ÷÷, è DI b øVce, , SET–II, Questions Uncommon to Set–I, 1., , No (i) This is because the charge particle moves on a circular path., ®, r ®, (ii) F = q (v ´ B)
Page 504 :
Examination Papers, D1, , R1=150W, , D2, , R2=50W, , 15. Resistance of diode D1 in forward biasing RZ = 50 W., Resistance of diode D2 in reverse biasing = ¥., As no current flows through the diode D2, so current from the, source flows through diode D1., 6.0, e, I=, =, R net 50 + 150 + 100, , I1, , 503, , I2, I, R3=100W, , 6.0, =, = 0.02 A, 300, , 6V, , I = 0.02 A flows through 100 W resistor., 21. Generalised Ampere’s circuital Law —, ® ®, df, f B× dl = m 0 I C + m 0 e 0 E, dt, Line integral of magnetic field over closed loop is equal to m0 times sum of conduction current and, displacement current., (i) In case of steady electric field in a conducting wire, electric field does not change with time,, ® ®, , conduction current exists in the wire but displacement current may be zero. So, f B× dl = m 0 I C, , (ii) In large region of space, where there is no conduction current, but there is only a displacement, ® ®, , current due to time varing electric field (or flux). So, f B× dl = m 0 e 0, 22., , (a), , dfE, dt, , Labelled diagram of compound microscope., The objective lens form image A ¢ B¢ near the first focal point of eyepiece., , u, , A, B², , fo, , fe, , h, , B¢, B, , b Fe, , Fo, , E, , h¢, A¢, , Objective lens, , Eyepiece, , D, , A², , (b) Angular magnification of objective lens m0 = linear magnification, Since tan b =, , h¢, h, , h¢ h, =, L f0, m0 =, , h¢ L, =, h f0, , . . . (1)
Page 505 :
504, , Xam idea Physics—XII, , where L is the distance between second focal point of the objective and first focal point of, eyepiece., If the final image A ¢¢ B¢¢ is formed at the near point., æ Dö, Angular magnification me = çç1+ ÷÷, è fe ø, D, If the final image A ¢¢ B¢¢ is formed at infinity, then angular magnification me =, . . . (2), fe, Thus, total magnification of the compound microscope, M = m 0 × me, L D, = ´, f0 fe, (c) Aperture and focal length increase or decrease the resolving power of the compound microscope., Resolving power of microscope is given by, 2n sin q, R.P. =, 1.22 l, (i) On decreasing the aperture (diameter) of the objective lens, value of sin q decreases, and, hence resolving power decreases., (ii) On decreasing the focal length of the objective lens, value of sin q increases and hence, resolving power increases., , SET–III, Questions Uncommon to Set-I and Set-II, 4., , (i) In case of conductor Ag, the resistivity decreases with the fall of temperature from 300 K to, 60 K., (ii) In case of semiconductor (or their alloys GaAs) the resistivity increases with the fall of, temperature from 300 K to 60 K., ®, , ®, , 5. Net force on electron moving in the combined electric field E and magnetic field B is, ®, , ®, , ® ®, , F = -e[ E+ v ´ B], ®, , Since electron moves undeflected then F = 0., ®, , ® ®, , E + v ´ B=0, , Þ | E| = | v | ´ | B|, | E|, Þ |v| =, | B|
Page 506 :
Examination Papers, , 10., , B, , F, , C, , 505, , G, , I1, I2, 10 W, , 5W, , 10 V, , I1, A, , I1, , 20 V, , 2W, E, , D, , I2, , H, , No current will flow through 2W resistor, because in a closed loop, total p.d. must be zero. So, 10 V – 5I1 = 0, . . . (1), 20 V – 10I2 = 0, . . . (2), and resistor 2W is not part of any loop ABCD and EFGH., 13. Diode D1 offers a resistance of 50 W and D2 an infinite resistance. It means diode D1 conducts and, diode D2 does not conduct. So for conducting path through diodes,, D1, emf, 200W, I=, Net resistance, 6, =, D2, 50 + 200 + 100, 50W, 6, =, 350, = 0.0171 A, 100W, , A current of 0.171 A will flow through 100 W resistor., 18., , (a), , 19., , (a), , 6V, , Infrared waves are produced by hot bodies and molecules,, so are referred to as heat waves., (i) Em wave having short wavelength than infrared waves are visible, UV, X-rays and g-rays., (ii) Em wave having longer wavelength than infrared waves are microwaves, short radio waves,, television and FM radio., (b) Refer to Ans. of 23(b) of Set-I., Refraction type telescope in normal adjustment., fo, , fe, , B¢, , O, a, , h¢, , E, , A¢, , Objective lens, , Eyepiece, , b
Page 507 :
506, , Xam idea Physics—XII, , (b) Shortcomings —, (i) Very difficult and expensive to make large sized lenses, which can form images that are, free from any kind of chromatic aberration and distortions., (ii) By lenses of diameter 1.02 m and above tend to be very heavy and therefore, difficult to, make and support by their edges., , (c) If eyepiece of short focal length and objective lens of large focal length are used in constructing, the telescope, magnifying power increases, as per relation M =, , f0, ., fe, , This telescope of high magnifying power can see path of stars of actual separation of one minute, arc and even less values of arc., zzz
Page 508 :
CBSE, Examination, Paper Delhi-2014, Time allowed : 3 hours, , Maximum marks : 70, , General Instructions:, (i) All questions are compulsory., (ii) Question numbers 1 to 8 are very short answer questions and carry 1 mark each., (iii) Question numbers 9 to 18 are short answer questions and carry 2 marks each., (iv) Question numbers 19 to 27 are also short answer questions and carry 3 marks each., (v) Question numbers 28 to 30 are long answer questions and carry 5 marks each., (vi) Use log tables, if necessary. Use of calculators is not allowed., , SET–I, 1. Define the term ‘Mobility’ of charge carries in a conductor. Write its SI unit., 2. The carrier wave is given by, C(t) = 2 sin (8p t) volt., The modulating signal is a square wave as shown. Find modulation index., 1, m(t) in volt, , 1, , 2, , t in second, , 3. “For any charge configuration, equipotential surface through a point is normal to the electric field.”, Justify., 4. Two spherical bobs, one metallic and the other of glass, of the same size are allowed to fall freely, from the same height above the ground. Which of the two would reach earlier and why?, 5. Show variation of resistivity of copper as a function of temperature in a graph., 6. A convex lens is placed in contact with a plane mirror. A point object at a distance of 20 cm on the, axis of this combination has its image coinciding with itself. What is the focal length of the lens?, ®, , 7. Write the expression, in a vector form, for the Lorentz magnetic force F due to a charge moving with, ®, , ®, , velocity V in a magnetic field B. What is the direction of the magnetic force?
Page 509 :
508, , Xam idea Physics—XII, , 8. The figure given below shows the block diagram of a generalised communication system. Identify the, element labelled ‘X’ and write its function., , User, , X, Information, Source, , Transmitter, , Receiver, , Message, Signal, , Message, Signal, , 9. Out of the two magnetic materials, ‘A’ has relative permeability slightly greater than unity while ‘B’, has less than unity. Identify the nature of the materials ‘A’ and ‘B’. Will their susceptibilities be, positive or negative?, ®, , 10. Given a uniform electric field E = 5 × 103 i$ N/C, find the flux of this field through a square of 10 cm, on a side whose plane is parallel to the y-z plane. What would be the flux through the same square if, the plane makes a 30° angle with the x-axis?, 11. For a single slit of width “a”, the first minimum of the interference pattern of a monochromatic light, l, l, of wavelength l occurs at an angle of . At the same angle of , we get a maximum for two narrow, a, a, slits separated by a distance “a”. Explain., 12. Write the truth table for the combination of the gates shown. Name the gates used., A, , A, R, B, , S, Y', , Y, , OR, Identify the logic gates marked ‘P’ and ‘Q’ in the given circuit. Write the truth table for the combination., A, B, , P, Q, , X, , 13. State Kirchhoff’s rules. Explain briefly how these rules are justified., 14. A capacitor ‘C’, a variable resistor ‘R’ and a bulb ‘B’ are connected in series to the ac mains in circuits, as shown. The bulb glows with some brightness. How will the glow of the bulb change if (i) a, dielectric slab is introduced between the plates of the capacitor, keeping resistance R to be the same;, (ii) the resistor R is increased keeping the same capacitance?, B, R, , C, , Mains, , 15. State the underlying principle of a cyclotron. Write briefly how this machine is used to accelerate, charged particles to high energies.
Page 510 :
Examination Papers, , 509, , 16. An electric dipole of length 4 cm, when placed with its axis making an angle of 60° with a uniform, electric field, experiences a torque of 4 3 Nm. Calculate the potential energy of the dipole, if it has, charge ±8 nC., 17. A proton and a deuteron are accelerated through the same accelerating potential. Which one of the two has, (i) greater value of de-Broglie wavelength associated with it, and, (ii) less momentum?, Give reasons to justify your answer., 18., , (i) Monochromatic light of frequency 6.0 × 1014 Hz is produced by a laser. The power emitted is, 2.0 × 10–3 W. Estimate the number of photons emitted per second on an average by the source., (ii) Draw a plot showing the variation of photoelectric current versus the intensity of incident, radiation on a photosensitive surface., , 19. A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. Upto which, energy level the hydrogen atoms would be excited?, Calculate the wavelengths of the first member of Lyman and first member of Balmer series., 20. When Sunita, a class XII student, came to know that her parents are planning to rent out the top floor, of their house to a mobile company she protested. She tried hard to convince her parents that this, move would be a health hazard. Ultimately her parents agreed., (a) In what way can the setting up of transmission tower by a mobile company in a residential colony, prove to be injurious to health?, (b) By objecting to this move of her parents, what value did Sunita display?, (c) Estimate the range of e.m. waves which can be transmitted by an antenna of height 20 m. (Given, radius of the earth = 6400 km)., 21. A potentiometer wire of length 1 m has a resistance of 10 W. It is connected to a 6 V battery in series, with a resistance of 5 W. Determine the emf of the primary cell which gives a balance point at 40 cm., 22. (a) Draw a labelled ray diagram showing the formation of a final image by a compound, microscope at least distance of distinct vision., (b) The total magnification produced by a compound microscope is 20. The magnification, produced by the eye piece is 5. The microscope is focussed on a certain object. The distance, between the objective and eyepiece is observed to be 14 cm. If least distance of distinct vision, is 20 cm, calculate the focal length of the objective and the eye piece., 23. (a) A mobile phone lies along the principal axis of a concave mirror. Show, with the help of a, suitable diagram, the formation of its image. Explain why magnification is not uniform., (b) Suppose the lower half of the concave mirror’s reflecting surface is covered with an opaque, material. What effect this will have on the image of the object? Explain., 24. (a) Obtain the expression for the energy stored per unit volume in a charged parallel plate capacitor., (b) The electric field inside a parallel plate capacitor is E. Find the amount of work done in moving, a charge q over a closed rectangular loop abcda., +, , +, , +, , +, , +, , +, , +, , a, , b, , d, , c, , OR
Page 511 :
510, , Xam idea Physics—XII, , (a) Derive the expression for the capacitance of a parallel plate capacitor having plate area A and, plate separation d., (b) Two charged spherical conductors of radii R1 and R2 when connected by a conducting wire, acquire charges q1 and q2 respectively. Find the ratio of their surface charge densities in terms, of their radii., 25. (a) State Ampere’s circuital law, expressing it in the integral form., (b) Two long coaxial insulated solenoids, S1 and S2 of equal lengths are wound one over the other, as shown in the figure. A steady current “I” flow through the inner solenoid S1 to the other end, B, which is connected to the outer solenoid S2 through which the same current “I” flows in the, opposite direction so as to come out at end A. If n1 and n2 are the number of turns per unit, length, find the magnitude and direction of the net magnetic field at a point (i) inside on the, axis and (ii) outside the combined system., r2, , B, , l, , A, , r1, , S1, , I, n1 turns, , S2, , n1 turns, , 26. Answer the following:, (a) Name the em waves which are suitable for radar systems used in aircraft navigation. Write the, range of frequency of these waves., (b) If the earth did not have atmosphere, would its average surface temperature be higher or lower, than what it is now? Explain., (c) An em wave exerts pressure on the surface on which it is incident. Justify., 27. (a) Deduce the expression, N = N0e - lt , for the law of radioactive decay., (b), , (i) Write symbolically the process expressing the b + decay of, nuclear process underlying this decay., (ii) Is the nucleus formed in the decay of the nucleus, , 28. (a), , 22, 11 Na,, , 22, 11 Na., , Also write the basic, , an isotope or isobar?, , (i) ‘Two independent monochromatic sources of light cannot produce a sustained interference, pattern’. Give reason., (ii) Light waves each of amplitude “a” and frequency “w”, emanating from two coherent, light sources superpose at a point. If the displacements due to these waves is given by, y1 = a cos wt and y 2 = a cos ( wt + f) where f is the phase difference between the two,, obtain the expression for the resultant intensity at the point.
Page 512 :
Examination Papers, , 511, , (b) In Young’s double slit experiment, using monochromatic light of wavelength l, the intensity of, light at a point on the screen where path difference is l, is K units. Find out the intensity of light at, a point where path difference is l/3., OR, (a) How does one demonstrate, using a suitable diagram, that unpolarised light when passed through, a polaroid gets polarised?, (b) A beam of unpolarised light is incident on a glass-air interface. Show, using a suitable ray, diagram, that light reflected from the interface is totally polarised, when m = tan i B , where m is, the refractive index of glass with respect to air and iB is the Brewster’s angle., 29. (a) Describe a simple experiment (or activity) to show that the polarity of emf induced in a coil is, always such that it tends to produce a current which opposes the change of magnetic flux that, produce it., (b) The current flowing through an inductor of self inductance L is continuously increasing. Plot a, graph showing the variation of, (i) Magnetic flux versus the current, (ii) Induced emf versus dI/dt, (iii) Magnetic potential energy stored versus the current., OR, (a) Draw a schematic sketch of an ac generator describing its basic elements. State briefly its, working principle. Show a plot of variation of, (i) Magnetic flux and, (ii) Alternating emf versus time generated by a loop of wire rotating in a magnetic field., (b) Why is choke coil needed in the use of fluorescent tubes with ac mains?, 30. (a) State briefly the processes involved in the formation of p-n junction explaining clearly how the, depletion region is formed., (b) Using the necessary circuit diagrams, show how the V-I characteristics of a p-n junction are, obtained in, (i) Forward biasing, (ii) Reverse biasing, How are these characteristics made use of in rectification?, OR, (a) Differentiate between three segments of a transistor on the basis of their size and level of doping., (b) How is a transistor biased to be in active state?, (c) With the help of necessary circuit diagram, describe briefly how n-p-n transistor in CE, configuration amplifies a small sinusoidal input voltage. Write the expression for the ac current, gain., , SET–II (Questions Uncommon to Set–I), 1. Define the term ‘electrical conductivity’ of a metallic wire. Write its S.I. unit., 2. The carrier wave is represented by, C(t) = 5 sin (10p t) volt
Page 513 :
512, , Xam idea Physics—XII, , A modulating signal is a square wave as shown. Determine modulation index., 2, m(t) in volt, , 1, , 2, , t in second, , 3. Show variation of resistivity of Si with temperature in a graph., 10. An electric dipole of length 2 cm, when placed with its axis making an angle of 60° with a uniform, electric field, experiences a torque of 8 3 Nm. Calculate the potential energy of the dipole, if it has a, charge of ±4 nC., 15. 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 and (ii) less kinetic energy? Give reasons, to justify your answer., ®, , 16. Given a uniform electric field E = 2 × 103 i$ N/C, find the flux of this field through a square of side 20, cm, whose plane is parallel to the y-z plane. What would be the flux through the same square, if the, plane makes an angle of 30° with the x-axis?, 20. A 12.9 eV beam of electrons is used to bombard gaseous hydrogen at room temperature. Upto which, energy level the hydrogen atoms would be excited?, Calculate the wavelength of the first member of Paschen series and first member of Balmer series., 22. Answer the following:, (a) Name the em waves which are used for the treatment of certain forms of cancer. Write their, frequency range., (b) Thin ozone layer on top of stratosphere is crucial for human survival. Why?, (c) Why is the amount of the momentum transferred by the em waves incident on the surface so, small?, 24. A potentiometer wire of length 1.0 m has a resistance of 15 W. It is connected to a 5 V battery in series, with a resistance of 5 W. Determine the emf of the primary cell which gives a balance point at 60 cm., , SET–III (Questions Uncommon to Set-I and II), 1. Define the term ‘drift velocity’ of charge carriers in a conductor and write its relationship with the, current flowing through it., 2. The carrier wave of a signal is given by C(t) = 3 sin (8pt) volt. The modulating signal is a square wave, as shown. Find its modulation index., 1.5, m(t) in volt, , 1, , 2, , t in second, , 4. Plot a graph showing variation of current versus voltage for the material GaAs., 9. An electric dipole of length 1 cm, which placed with its axis making an angle of 60° with uniform, electric field, experiences a torque of 6 3 Nm. Calculate the potential energy of the dipole if it has, charge ± 2 nC.
Page 514 :
Examination Papers, , 513, , 12. A deuteron and an alpha particle are accelerated with the same accelerating potential. Which one of, the two has, (a) greater value of de-Broglie wavelength, associated with it and, (b) less kinetic energy? Explain., 20. A 12.3 eV electron beam is used to bombard gaseous hydrogen at room temperature. Upto which, energy level the hydrogen atoms would be excited?, Calculate the wavelengths of the second member of Lyman series and second member of Balmer, series., 24. Answer the following questions:, (a) Name the em waves which are produced during radioactive decay of a nucleus. Write their, frequency range., (b) Welders wear special glass goggles while working. Why? Explain., (c) Why are infrared waves often called as heat waves? Give their one application., 25. A potentiometer wire of length 1 m has a resistance of 5 W. It is connected to a 8 V battery in series, with a resistance of 15 W. Determine the emf of the primary cell which gives a balance point at 60 cm., zzz
Page 515 :
514, , Xam idea Physics—XII, , Solutions, SET–I, 1. Mobility is defined as the magnitude of the drift velocity per unit electric field., | v | ltE, lt, m = d =, =, E, mE m, where t is the average collision time for electrons., The SI unit of mobility is m 2 /Vs or m 2 V -1 s -1, Amplitude of modulated signal, 2. Modulation index m =, Amplitude of carrier waves, Am, 1m, =, = 0.5, Ac, 2m, 3. The work done in moving a charge from one point to another on an equipotential surface is zero. If, electric field is not normal to the equipotential surface, it would have non-zero component along the, surface. In that case work would be done in moving a charge on an equipotential surface., 4. Glass would reach earlier. This is because there is no effect of electromagnetic induction in glass, due, to presence of Earth’s magnetic field, unlike in the case of metallic ball., Resistivity r (10–8Wm), , 5., , 0.4, , I 0.2, , 0, 50, 100, Temperature T(K), , 150, , 6. The focal length of the lens = 20 cm, Explanation:, , Point, P, Object, , P, , As the image of this combination coincides with the object itself, the rays from the object, after, refraction from the lens should fall normally on the plane mirror, so that they retrace their path. So the, rays from the point object after refraction from the lens must form parallel beam. For clarity, mirror, has been placed at a small distance from the lens., ®, , 7. Lorentz force,, , ®, , ®, , F = q( v ´ B ), ®, , ®, , Obviously, the force on charged particle is perpendicular to both velocity v and magnetic field B .
Page 516 :
Examination Papers, , 515, , 8. X represents communication channel., Function: It connects the transmitter to the receiver., 9. A ¾ ¾, ¾® Paramagnetic (Q m r > 1), B ¾¾, ¾® Diamagnetic (Q m r < 1), Susceptibility for A ® Positive, Susceptibility for B ® Negative, ®, , 10. Here, E = 5 ´ 10 3 i$ N/C, i.e. field is along positive direction of x-axis., Surface area, A = 10 cm × 10 cm, = 0.10 m × 0.10 m = 10–2 m2, (i) When plane parallel to y - z plane, the normal to plane is along x axis. Hence, q = 0°, f = EA cos q = 5 ´ 10 3 ´ 10 -2 cos 0°, = 50NC -1 m 2, (ii) When the plane makes a 30° angle with the x-axis, the normal to its plane makes 60° angle with, x-axis. Hence, q = 60°, f = EA cos q, = 5 ´ 10 3 ´ 10 -2 cos 60°, = 25 NC–1 m2, 11. Case I: The overlapping of the contributions of the wavelets from two halves of a single slit produces, a minimum because corresponding wavelets from two halves have a path difference of l/2., Case II: The overlapping of the wavefronts from the two slits produces first maximum because these, wavefronts have the path difference of l., 12., A, , Input, B, , Output, Y = A (A + B), , 0, 1, 0, 1, , 0, 0, 1, 1, , 0, 1, 0, 1, , R : OR gate, S : AND gate, OR, P is NAND gate and Q is OR gate., Input, A, 0, 1, 0, 1, , B, 0, 0, 1, 1, , Output, X, 1, 1, 1, 1
Page 517 :
516, , Xam idea Physics—XII, , 13. Junction rule: In an electric circuit, the algebraic sum of currents at any junction is zero., At any junction, the sum of the currents entering the junction is equal to the sum of currents leaving, the junction., SI = O, Justification: This rule is based on the law of conservation of charge., Loop rule: The algebraic sum of charges in potential around any closed loop involving resistors and, cells in the loop is zero., SDV = O, Justification: This rule is based on the law of conservation of energy., 14., , (i) The reactance of the capacitor will decrease, this results in increase of the current in the circuit., So, the bulb will glow brighter., (ii) Increased resistance will decrease the current in the circuit, which will decrease glow of the, bulb., , 15. A cyclotron makes use of the principle that the energy of the charged particles or ions can be made to, increase in presence of crossed electric and magnetic fields., The magnetic field acts on the charged particle and makes them move in a circular path inside the dee., Every time the particle moves from one dee to another it is acted upon by the alternating electric field,, and is accelerated by this field, which increases the energy of the particle., 16. Torque,, , t = pE sin q, 4 3 = pE sin 60°, , 3, Þ pE = 8, 2, Now, potential energy, U = - pE cos q, = – 8 cos 60°, 1, = -8 ´ = -4 J, 2, h, 17. (i) de Broglie wavelength, l =, 2mqV, 4 3 = pE ´, , Here V is same for proton and deutron., As mass of proton < mass of deutron and q p = q d, Therefore, l p > l d for same accelerating potential., h, (ii) We know that momentum =, l, Therefore, l p > l d, So, momentum of proton will be less than that of deutron., 18., , (i) Power = nhn, where n = number of photons per second, 2.0 × 10–3 = n × 6.6 × 10–34 × 6 × 1014, n =, , 2.0 ´ 10 -3, 6.6 ´ 10 –34 ´ 6 ´ 1014
Page 518 :
Examination Papers, , =, , 2.0 ´ 10 –3, 39.6 ´ 10 –20, , = 0.050 ´ 1017, , = 5 ´ 1015 photons per second, , Photo electric current, , (ii), , Intensity of light, , 19. Here, DE = 12.5 eV, Energy of an electron in the nth orbit of hydrogen atom is E n = -, , 13.6, n2, , eV, , For ground state n = 1,, 13.6, E1 = eV = -13.6 eV, 12, For first excited state n = 2,, 13.6, E2 = eV = -3.4 eV, 22, For second excited state n = 3,, 13.6, E3 = eV = –1.51 eV, 32, Energy required to excite hydrogen atoms from ground state to the second excited state, = E final - E initial, = -151, . - ( -13.6) = 12.09 eV, Thus hydrogen atoms would be excited upto third energy level (n = 3)., For Lyman series,, æ 1, 1, 1 ö÷, = Rç, ç n2 n2 ÷, l, i ø, è f, æ 1, 1, 1 ö, ÷÷, = 1.097 ´ 107 çç, 2, l, è1, 22 ø, 1, 3, = 1.097 ´ 107 ´, l, 4, , Þ, , l = 122 ´ 10 -9 m = 122 nm, For Balmer series, æ 1, 1, 1 ö, ÷÷, = 1.097 ´ 107 çç, 2, l, è2, 32 ø, , 1, = 0.82275 ´ 107 m -1, l, , 517
Page 519 :
518, , Xam idea Physics—XII, , 1, 5, = 1.097 ´ 107 ´, l, 36, 1, = 015236, ., ´ 107, l, l = 656.3 nm, , Þ l = 6.563 × 10–7, , 20. (a) Electromagnetic radiations emitted by an antenna can cause cancer, cardiac problem and, headache., (b) Sunita displayed awareness and scientific temperament., (c) Range = 2hR, = 2 ´ 20 ´ 6400 ´ 1000, = 4 ´ 64 ´ 10 6 = 16000 m, 21. Here,, l = 1 m, R1 = 10 W, V = 6 V, R 2 = 5W, Current flowing in potentiometer wire,, V, 6, 6, I =, =, =, = 0.4 A, R1 + R 2 10 + 5 15, Potential drop across the potentiometer wire, V ¢ = IR = 0.4 ´ 10 = 4V, V¢ 4, Potential gradient, K =, = = 4 V/m, l, 1, Emf of the primary cell = KI, = 4 × 0.4 = 1.6 V, Eyepiece, , 22. (a), , uo, , ue, , vo, , D, Objective, , Eye, , B, Fo, , A", A, , Fe', , O, , A', , B', , B", , ve, , (b) Here, m = -20, m e = 5, v e = -20 cm, v, For eyepiece, m e = e, ue, Þ, , 5 =, , -20, ue, , Þ, , ue =, , -20, = - 4 cm, 5, , E, , Fe
Page 520 :
Examination Papers, , 519, , Using lens formula,, 1, 1, 1, =, ve u e, fe, -, , 1, 1, 1, + =, 20 4 f e, , -1 + 5, 1, =, 20, fe, , Þ, , f e = 5 cm, , Now, total magnification, m = me ´ m0, -20 = 5 ´ m 0, m 0 = -4, Also | v 0 | + | u e | = 14, | v 0 | + |- 4| = 14, v 0 = 14 - 4 = 10 cm, v, m0 = 1 - 0, f0, -4 = 1 -5 = -, , 10, f0, , 10, f0, , Þ, , f o = 2 cm., , 23. (a), B, , A, , C, , F, , B', A', , The position of the image of different parts of the mobile phone depends on their position with, respect to the mirror. The image of the part which is on the plane perpendicular to principal, axis will be on the same plane. It will of the same size, i.e., B ¢C = BC. The images of the other, parts of the phone are getting magnified in accordance with their object distance from the, mirror., (b) Taking the laws of reflection to be true for all points of the remaining (uncovered) part of the, mirror, the image will be that of the whole object. As the area of the reflecting surface has been, reduced, the intensity of the image will be low (in this case half)., 24. (a) When a capacitor is charged by a battery, 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., Consider a capacitor of capacitance C. Initial charge on capacitor is zero. Initial potential, difference between capacitor plates = zero. Let a charge Q be given to it in small steps. When
Page 521 :
520, , Xam idea Physics—XII, , charge is given to capacitor, the potential difference between its plates increases. Let at any, q, instant when charge on capacitor be q, the potential difference 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, which may be obtained by integration. Therefore total work, Q, Q, q, W =ò, V dq = ò, dq, 0, 0, C, =, , 1 é q2 ù, ú, ê, C êë 2 úû, , Q, , =, 0, , 1, C, , æ Q2 0 ö Q2, ç, - ÷=, ç 2, 2 ÷ø 2C, è, , If V is the final potential difference between capacitor plates, then Q = CV, (CV ) 2, 1, 1, = CV 2 = QV, W =, 2C, 2, 2, This work is stored as electrostatic potential energy of capacitor i.e.,, Q2, 1, 1, = CV 2 = QV, 2C, 2, 2, Energy density: Consider a parallel plate capacitor consisting of plates, each of area A,, separated by a distance d. If space between the plates is filled with a medium of dielectric, constant K, then, K e0 A, Capacitance of capacitor,, C =, d, If s is the surface charge density of plates, then electric field strength between the plates, s, E =, Þ s = K e0 E, K e0, U =, , Electrostatic potential energy,, , Q = sA = K e 0 E A, , Charge on each plate of capacitor, \, , Energy stored by capacitor,, , U =, , ( K e 0 EA) 2, Q, 1, = K e0 E, =, 2 ( K e 0 A / d) 2, 2C, 2, , 2, , A d, , But Ad = volume of space between capacitor plates, 1, Energy stored,, U = K e 0 E 2 Ad, \, 2, U, 1, Electrostatic Energy stored per unit volume, u e =, = Ke 0 E 2, Ad, 2, This is expression for electrostatic energy density in medium of dielectric constant K., 1, In air or free space ( K = 1), therefore energy density, u e = e 0 E 2, 2, (b) Work done in moving a charge q from a to b = 0, Work done in moving a charge q from c to d = 0, This is because the electric field is perpendicular to the displacement., Now, work done from b to c = – work done from d to a, Therefore, total work done in moving a charge q over a closed loop = 0.
Page 522 :
Examination Papers, , 521, , OR, (a), , Area A, , Surface, charge density s, , I, 1, , + + + + + + + + + + + + + +, , E, , –, , d, , –, , –, , –, , –, , –, , –, , –, , –, , –, , –, , –, , –, , 2, Surface, charge density - s, , II, , In the region between the plates the net electric field is equal to the sum of the electric fields, due to the two charged plates. Thus, the net electric field is given by, s, s, s, E =, +, =, 2e o, 2e o, eo, The electric field is constant in the region between the plates. Therefore, the potential, difference between the plates will be, sd, V = Ed =, eo, Now, capacitance, Q Qe 0, C =, =, V, sd, Surface charge density s =, C =, , Q, , where A is the area of cross-section of the plates., A, , Qe o A e o A, =, Qd, d, , (b) When two charged spherical conductors are connected by a conducting wire, they acquire the, same potential., kq1, kq, = 2, R1, R2, Or, , q1, q, = 2, R1, R2, , Þ, , q1, R, = 1, q2, R2, , Hence, the ratio of surface charge densities, q / 4pR12, q R2, s1, = 1, = 1 2, s2, q 2 / 4pR 22, q 2 R12, =, , R1 R 22, R, ´, = 2, R 2 R12, R1
Page 523 :
522, , Xam idea Physics—XII, , 25. (a) The line integral of magnetic field (B) around any closed path in vacuum is m 0 times the net, current (I) threading the area enclosed by the curve., ® ®, , ò B . dl, , = m oI, , (b) Magnetic field, B = m 0 nI, (i) Magnitude of net magnetic field inside the combined system on the axis,, B = B1 – B2, = m 0 n1 I – m 0 n 2 I, = m 0 ( n1 – n 2 ) I, (ii) Outside the combined system, net magnetic field is zero., 26. (a) Microwaves, Frequency range : 1010 Hz to 1012 Hz, (b) Average surface temperature will be lower. This is because there will be no green house effect, in absence of atmosphere., (c) An electromagnetic wave exerts pressure on the surface on which it is incident because these, waves carry both energy and momentum., 27. (a) Radioactive decay Law: The rate of decay of radioactive nuclei is directly proportional to the, number of undecayed nuclei at that time., Derivation of Formula, Suppose initially the number of atoms in radioactive element is N 0 and N the number of atoms, after time t., After time t, let dN be the number of atoms which disintegrate in a short interval dt, then rate of, dN, disintegration will be, , this is also called the activity of the substance/element., dt, According to Rutherford-Soddy law, dN, µN, dt, dN, or, ...(i), = - lN, dt, where l is a constant, called decay constant or disintegration constant of the element. Its unit is, s - 1 . Negative sign shows that the rate of disintegration decreases with increase of time. For a, given element/substance l is a constant and is different for different elements. Equation (i) may, be rewritten as, dN, = - ldt, N, Integrating log e N = - lt + C, ...(ii), where C is a constant of integration., At, t = 0, N = N 0, \, log e N 0 = 0 + C Þ C = log e N 0, \ Equation (ii) gives, log e N = - lt + log e N 0, or, log e N - log e N 0 = - lt
Page 524 :
No. of undecayed nuclei, , Examination Papers, , 523, , N0, , N, N–DN, O, , t, , t+ Dt, , Time t, , N, = - lt, N0, , or, , log e, , or, , N, = eN0, , \, , N = N0 e-, , lt, lt, , ...(iii), , No. of undecayed nuclei, , According to this equation, the number of undecayed atoms/nuclei of a given radioactive element, decreases exponentially with time (i. e. , more rapidly at first and slowly afterwards)., N, Mark of N = 0 in terms of T1/2 is shown in fig., 16, N0, , N0, 2, N0, 4, N0, 8 N, , P, 0, , 16, , (b) (i), , 22, ¾® 22, 10 Ne, 11 Na ¾ ¾, , T1, , 2T1, , 2, , 2, , Time t, , 3T1, , 4T1, , 2, , 2, , + e+ + n, , Basic nuclear process:, P ¾¾, ¾® n + e + + n, (ii), 28. (a), , Isobar, , (i) The light waves, originating from two independent monochromatic sources, will not have, a constant phase difference. Therefore, these sources will not be coherent and therefore, would not produce a sustained interference pattern., (ii) The resultant displacement will be given by, y = y1 + y 2, = a cos wt + a cos( wt + f), = a[cos wt + cos( wt + f)]
Page 525 :
524, , Xam idea Physics—XII, , (b) A, \, A, \, , = 2a cos(f /2) cos(wt + f /2), The amplitude of the resultant displacement is 2a cos(f /2), The intensity of light is directly proportional to the square of amplitude of the wave. The, resultant intensity will be given by, f, I = 4a 2 cos 2 ., 2, path difference of l, corresponds to a phase difference of 2p, I, Intensity,, or, I = 4a 2, a2 =, 4, l, 2p, path difference of , corresponds to a phase difference of, 3, 3, I, I, 2 2p, Intensity = 4 ´ . cos, =, 4, 3, 4, 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. So, when an unpolarised, light falls on a polaroid, it lets only those of its electric vectors that are oscillating along a, direction perpendicular to its aligned molecules to pass through it. The incident light thus gets, linearly polarised., , D, B, , A, , Unpolarised light Polaroid Plane polarised, light, , Incident, ray, , ip r', , Reflected ray, 90°, , O, , Whenever unpolarised light is incident on the boundary, r, Refracted ray, between two transparent media, the reflected light gets, partially or completely polarised. When reflected light is, perpendicular to the refracted light, the reflected light is a, C, E, completely polarised light., D, (b) Condition : The reflected ray is totally plane polarised,, B, A, when reflected and refracted rays are perpendicular to, each other., Incident, Reflected ray, ÐBOC = 90°, ip r', ray, 90°, If i p is angle of incidence, r ¢ is angle of reflection and r, O, the angle of refraction, then according to law of reflection, r, Refracted ray, ip = r¢, and from fig. r ¢ + 90° + r = 180°, …(i), Þ i p + r = 90°, C, E, …(ii), Þ, r = ( 90° - i p ), From Snell’s law, refractive index of second medium relative to first medium (air) say.
Page 526 :
Examination Papers, , n =, Þ, \, , sin i p, sin r, , =, , sin i p, , =, , sin ( 90° - i p ), , 525, , sin i p, cos i p, , n = tan i p, , Angle of incidence, i p = tan -1 ( n)., , 29. (a) When the North pole of a bar magnet moves towards the closed coil, the magnetic flux through, the coil increases. This produces an induced emf which produces (or tend to produce if the coil, is open) an induced current in the anti-clockwise sense. The anti-clockwise sense corresponds, to the generation of North pole which opposes the motion of the approaching N pole of the, magnet. The face of the coil, facing the approaching magnet, then has the same polarity as that, of the approaching pole of the magnet. The induced current, therefore, is seen to oppose the, change of magnetic flux that produces it., , N, , N, , Magnetic Flux, , (b), , When a North pole of a magnet is moved away from the coil, the current (I) flows in the, clock-wise sense which corresponds to the generation of South pole. The induced South, pole opposes the motion of the receding North pole., (i) Magnetic flux versus the current, , Current
Page 527 :
Xam idea Physics—XII, , (ii) Induced emf versus dI/dt, dI, dt, , dI, dt, , Induced, emf, , or, , Induced, emf, , When current is increasing, at constant rate, , (iii) Magnetic energy stored, Magnetic energy, stored, , 526, , Current, OR, (a) Working: When the armature coil is rotated in the strong magnetic field, the magnetic flux, linked with the coil changes and the current is induced in the coil, its direction being given by, Fleming’s right hand rule. Considering the armature to be in vertical position and as it rotates in, anticlockwise direction, the wire ab moves upward and cd downward, so that the direction of, induced current is shown in fig. In the external circuit, the current flows along B1 R L B 2 . The, direction of current remains unchanged during the first half turn of armature. During the second, half revolution, the wire ab moves downward and cd upward, so the direction of current is, reversed and in external circuit it flows along B 2 R L B1 . Thus the direction of induced emf and, current changes in the external circuit after each half revolution., w, Expression for Induced emf: If N is the number, Armature coil, of turns in coil, f the frequency of rotation, A area, b, of coil and B the magnetic induction, then induced, c, emf, df d, e==, {NBA (cos 2p f t )}, dt, dt, N B, S, Field magnet, = 2p NBA f sin 2p f t, a, Obviously, the emf produced is alternating and, d, hence the current is also alternating., B1, R1, Current produced by an ac generator cannot be, measured by moving coil ammeter; because the, Brushes Load, Slip rings, RL, average value of ac over full cycle is zero., R, The source of energy generation is the mechanical, 2, B2, energy of rotation of armature coil.
Page 528 :
Examination Papers, , 527, , (i) Plot of variation of magnetic flux with time., j = NBA cos wt, , Magnetic, Flux j, , T, , O, , 3T, 4, , 0, , T, , 4, T, , 2, , Time, , (ii) Plot of variation of alternating emf with time., e = NABw sin wt, eo = sin wt, , Alternating, emf, O 0, , T, T, , T, , 2, 3T, , 4, , 4, , Time, , (b) Choke coil reduces the voltage across the fluorescent tube without wastage of power., 30. (a), , Electron drift, , Electron diffusion, _ _, _ _, , p, , _ _, _ _, _ _, , Hole diffusion, , + +, + +, + +, , n, , + +, + +, Hole drift, , Depletion region, , Two processes occur during the formation of a p-n junction are diffusion and drift. Due to the, concentration gradient across p and n-sides of the junction, holes diffuse from p-side to n-side, (p ® n) and electrons diffuse from n-side to p-side (n ® p). This movement of charge carriers, leaves behind ionised acceptors (negative charge f - immobile) on the p-side and donors, (positive charge immobile) on the n-side of the junction. This space charge region on either, side of the junction together is known as depletion region., (b) The circuit arrangement for studying the V–I characteristics of a diode are shown in Fig. (a), and (b). For different values of voltages the value of current is noted. A graph between V and I, is obtained as in Figure (c).
Page 529 :
528, , Xam idea Physics—XII, , From the V-I characteristic of a junction diode it is clear that it allows current to pass only, when it is forward biased. So if an alternating voltage is applied across a diode the current, flows only in that part of the cycle when the diode is forward biased. This property is used to, rectify alternating voltages., Voltmeter(V), , p, , Voltmeter(V), , n, , p, , n, , Milliameter, (mA), , Milliameter, (mA), Switch, , Switch, , + –, , + –, (a), , (b), , I(mA), 100, 80, 60, 40, 20, 100 80 60 40 20, 0.2 0.4 0.6 0.8 1.0, , Vbr, , V(V), , 10, 20, 30, , I(mA), , (c), , OR, (a) Emitter: It is of moderate size and heavily doped., Base: It is very thin and lightly doped., Collector: The collector side is moderately doped and larger in size as compared to the emitter., (b) Transistor is said to be in active state when its emitter-base junction is suitably forward biased, and base-collector junction is suitably reverse biased., (c), IC, C1, , RB, , C, , IB, , C2, RL, , B, Vi, , +, –, , E, , VBB, , IE, , +, VCC, –, , V0, Output, waveform, , If a small sinusoidal voltage with amplitude v s is superposed on the dc base bias by connecting, the source of that signal in series with the VBB supply, then the base current will have
Page 530 :
Examination Papers, , 529, , sinusoidal variations superimposed on the value of IB. As a consequence the collector current, also will have sinusoidal variations superimposed on the value of IC producing in turn, corresponding change in the value of V0 ., æ DI ö, AC current gain B ac = çç C ÷÷ ., è DI B øV, cE, , SET–II (Questions Uncommon to Set–I), 1. The reciprocal of the resistivity of a material is called its conductivity and is denoted by s., 1, Conductivity =, Resistivity, The SI unit of conductivity is ohm–1 m–1 or S m–1., A, 2, 2. Modulation index = m = = 0.4, Ac 5, 3., r, , T, , Resistivity of Si decreases with increasing temperatures., 10. Torque,, , t = pE sin q, 8 3 = pE sin 60°, 8 3 = pE ´, , 3, 2, , pE = 16, Potential energy, u = - pE cos q, = -16 ´ cos 60°, 1, = -16 ´ = -8 J, 2, 15., , (i) de Broglie wavelength, h, h, l= =, p, 2mqV, For same V , l µ, \, , lp, lµ, , =, , 1, mq, , mµ qµ, =, mpqp, , 4m p 2e, = 8 =2 2, ., mp e
Page 531 :
530, , Xam idea Physics—XII, , Clearly, l p > l µ ., Hence, proton has a greater de-Broglie wavelength., (ii) Kinetic energy, K = qV, For same V , K µ q, Kp, qp, e, 1, =, =, =, Kµ, qµ, 2e 2, Clearly, K p < K µ ., Hence, proton has less kinetic energy., ®, , 16. Here, E = 2 ´ 10 3 i$ N/C, i.e., field is along positive direction of x-axis., Surface area, A = 20 cm × 20 cm, = 0.20 m × 0.20 m = 4 × 10–2 m, (i) When plane parallel to y - z plane, the normal to plane is along x-axis. Hence q = 0°., f = EA cos q, = 2 ´ 10 3 ´ 4 ´ 10 -2 cos 0° = 80NC -1 m 2, (ii) When the plane makes a 30° angle with the x-axis, the normal to its plane makes 60° angle with, x-axis., Hence, q = 60°, f = EA cos q, = 2 ´ 10 3 ´ 4 ´ 10 -2 cos 60° = 40 NC–1 m2., 13.6, 20. Energy of an electron, E n = eV, n2, 13 . 6, For n = 1, E1 = = -13 . 6 eV, 12, 13 . 6, For n = 2, E 2 = = -3 . 4 eV, 22, 13 . 6, For n = 3, E 3 = = - 1.51 eV, 32, 13 . 6, For n = 4, E 4 = = - 0.85 eV, 42, Energy required to excite hydrogen atoms from ground state to excited state = E f - Ei, = - 0.85 – (–13.6) = 12.75 eV, Thus, hydrogen atom would be excited upto level n = 4., For Paschen series, é1, 1, 1 ù, ú, = Rê, l, êë n f2 ni2 úû, é1, 1, 1 ù, = 1. 097 ´ 107 ê, ú, 2, l, ë3, 42 û, 1, é1 1 ù, = 1. 097 ´ 107 ê - ú, l, ë 9 16 û
Page 532 :
Examination Papers, , 531, , 1, 7, = 1. 097 ´ 107 ´, l, 144, 144, l=, = 18 .75 ´ 10 -7 m = 1875 nm, 7 . 679 ´ 107, For Balmer Series, é1, 1, 1 ù, = Rê, ú, 2, l, ë2, 32 û, 1, é1 1 ù, = 1. 097 ´ 107 ´ ê - ú, l, ë4 9 û, 1, 5, = 1. 097 ´ 107 ´, l, 36, 36, = 6 .56 ´ 10 -7 m = 656 nm, l=, 7, 5 . 485 ´ 10, 22. (a) X rays or g rays, Range: 1018 Hz to 10 22 Hz., (b) Ozone layer absorbs the ultraviolet radiations from the sun and prevents it from reaching the, earth’s surface., (c) Momentum transferred,, u, p =, c, where u = energy transferred, c = speed of light, Due to the large value of speed of light (c), the amount of momentum transferred by the em, waves incident on the surface is small., V, 24. Current I =, R1 + R 2, 5, = 0 . 25 A, 15 + 5, Potential drop across the potentiometer wire, V = IR = 0 . 25 ´ 15 = 3 .75 V, Potential gradient,, V 3 .75, k =, =, = 3 .75 V/m, l, 1. 0, =, , \, , Unknown emf of the cell = KI, = 3.75 × 0.6 = 2.25 V., , SET–III (Questions Uncommon to Set-I and II), 1. Drift velocity is defined as the average velocity acquired by the free electrons in a conductor under the, influence of an electric field applied across the conductor. It is denoted by v d ., Current, I = neA. v d
Page 533 :
532, , Xam idea Physics—XII, , 2. Modulation index =, =, , Am, Ac, 1.5, = 0.5, 3. 0, , Current I (mA), , 4. There is more than one value of V for the same current I. A material exhibiting such behaviour is, GaAs., Non-Linear Negative, region, resistance, region, , Voltage V(V), , 9. Torque,, , t = pE sin q, 6 3 = pE sin 60°, 6 3 = pE ´, , 3, 2, , pE = 12, Potential energy, U = - pE cos q, = -12 cos 60°, 1, = -12 ´ = - 6 J, 2, 12. (a) de Broglie wavelength, h, h, l= =, p, 2mqV, For same V, l µ, , 1, mq, , \, , ld, =, lµ, , mµ qµ, =, md qd, , 2m d ´ 2q d, 2, =, md qd, 1, , \, ld > lµ, (b) Kinetic energy, K = qV, So, q µ > q d, For same V, we have, Kµ > Kd, 20. The energy of electron in the nth orbit of hydrogen atom is, 13 . 6, eV, En = n2, when the incident beam of energy 12.3 eV is absorbed by hydrogen atom. Let the electron jump from, n = 1 to n = n level.
Page 535 :
534, , Xam idea Physics—XII, , = 0.4 × 5 = 2 V, Potential gradient, V, K=, l, 2, = = 2 Vm -1, 1, Unknown emf of the cell = k l¢, \, = 2 × 0.6 V = 1.2 V, zzz
Page 536 :
CBSE, Examination Paper, All India-2014, Time allowed : 3 hours, , Maximum marks: 70, , General Instructions: Same as CBSE Examination Paper Delhi-2014., , SET–I, 1. Using the concept of force between two infinitely long parallel current carrying conductors, define, one ampere of current., 2. To which part of the electromagnetic spectrum does a wave of frequency 5 × 1019 Hz belong?, 3. Two equal balls having equal positive charge ‘q’ coulombs are suspended by two insulating strings of, equal length. What would be the effect on the force when a plastic sheet is inserted between the two?, 4. Define intensity of radiation on the basis of photon picture of light. Write its SI unit., 5. The electric current flowing in a wire in the direction from B to A is decreasing. Find out the direction, of the induced current in the metallic loop kept above the wire as shown., , A, , B, , 6. Why is it found experimentally difficult to detect neutrinos in nuclear b-decay?, 7. Why is the use of AC voltage preferred over DC voltage? Give two reasons., 8. A biconvex lens made of a transparent material of refractive index 1.25 is immersed in water of, refractive index 1.33. Will the lens behave as a converging lens? Give reason., 9. Using Rutherford model of the atom, derive the expression for the total energy of the electron in, hydrogen atom. What is the significance of total negative energy possessed by the electron?, OR, Using Bohr’s postulates of the atomic model, derive the expression for radius of nth electron orbit., Hence obtain the expression for Bohr’s radius., 10. A parallel plate capacitor of capacitance C is charged to a potential V. It is then connected to another, uncharged capacitor having the same capacitance. Find out the ratio of the energy stored in the, combined system to that stored initially in the single capacitor., 11. Considering the case of a parallel plate capacitor being charged, show how one is required to, generalise Ampere’s circuital law to include the term due to displacement current.
Page 537 :
536, , Xam idea Physics—XII, , 12. A cell of emf ‘E’ and internal resistance ‘r’ is connected across a variable resistor ‘R’. Plot a graph, showing variation of terminal voltage ‘V’ of the cell versus the current ‘I’. Using the plot, show how, the emf of the cell and its internal resistance can be determined., 13. Explain, with the help of a circuit diagram, the working of a p-n junction diode as a half-wave, rectifier., 14. Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area 1.0 ×, 10–7 m2 carrying a current of 1.5 A. Assume the density of conduction electrons to be 9 ´ 10 28 m–3., 15. Two monochromatic rays of light are incident normally on the face AB of an isosceles right-angled, prism ABC. The refractive indices of the glass prism for the two rays ‘1’ and ‘2’ are respectively 1.35, and 1.45. Trace the path of these rays after entering through the prism., A, 45°, '1', '2', 45°, B, , C, , 16. Write the functions of the following in communication systems:, (i) Transducer, (ii) Repeater, 17. Show diagrammatically the behaviour of magnetic field lines in the presence of (i) paramagnetic and, (ii) diamagnetic substances. How does one explain this distinguishing feature?, 18. Draw a circuit diagram of n-p-n transistor amplifier in CE configuration. Under what condition does, the transistor act as an amplifier?, 19. (a) Using the phenomenon of polarisation, show how transverse nature of light can be demonstrated., (b) Two polaroids P1 and P2 are placed with their pass axes perpendicular to each other., Unpolarised light of intensity I 0 is indident on P1 . A third polaroid P3 is kept in between P1 and, P2 such that its pass axis makes an angle of 30° with that of P1 . Determine the intensity of light, transmitted through P1 , P2 and P3 ., 20. Define the term ‘mutual inductance’ between the two coils., Obtain the expression for mutual inductance of a pair of long coaxial solenoids each of length l and, radii r1 and r 2 (r 2 >> r1 ). Total number of turns in the two solenoids are N1 and N 2 respectively., 21. Answer the following:, (a) Why are the connections between the resistors in a meter bridge made of thick copper strips?, (b) Why is it generally preferred to obtain the balance point in the middle of the meter bridge wire?, (c) Which material is used for the meter bridge wire and why?, OR, A resistance of R W draws current from a potentiometer as shown, in the figure. The potentiometer has a total resistance R o W. A, voltage V is supplied to the potentiometer. Derive an expression, for the voltage across R when the sliding contact is in the middle, of the potentiometer., , V, Ro, A, , C, , R
Page 538 :
Examination Papers, , 537, , 22. A convex lens of focal length 20 cm is placed coaxially with a convex mirror of radius of curvature, 20 cm. The two are kept at 15 cm from each other. A point object lies 60 cm in front of the convex, lens. Draw a ray diagram to show the formation of the image by the combination. Determine the, nature and position of the image formed., 23. A voltage V = V0 sin wt is applied to a series LCR circuit. Derive the expression for the average, power dissipated over a cycle., Under what condition is (i) no power dissipated even though the current flows through the circuit,, (ii) maximum power dissipated in the circuit?, 24. Write any two distinguishing features between conductors, semiconductors and insulators on the, basis of energy band diagrams., 25. For the past some time, Aarti had been observing some erratic body movement, unsteadiness and lack, of coordination in the activities of her sister Radha, who also used to complain of severe headache, occasionally. Aarti suggested to her parents to get a medical check-up of Radha. The doctor, thoroughly examined Radha and diagnosed that she has a brain tumour., (a) What, according to you, are the values displayed by Aarti?, (b) How can radioisotopes help a doctor to diagnose brain tumour?, 26. Write two basic modes of communication. Explain the process of amplitude modulation. Draw a, schematic sketch showing how amplitude modulated signal is obtained by superposing a modulating, signal over a sinusoidal carrier wave., 27. An electron microscope uses electrons accelerated by a voltage of 50 kV. Determine the de-Broglie, wavelength associated with the electrons. Taking other factors, such as numerical aperture etc. to be, same, how does the resolving power of an electron microscope compare with that of an optical, microscope which uses yellow light?, 28. Draw a labelled diagram of Van de Graaff generator. State its working principle to show how by, introducing a small charged sphere into a larger sphere, a large amount of charge can be transferred to, the outer sphere. State the use of this machine and also point out its limitations., OR, (a) Deduce the expression for the torque acting on a dipole of dipole, ®, , ®, , moment p in the presence of a uniform electric field E ., (b) Consider two hollow concentric spheres, S1 and S 2 , enclosing, charges 2Q and 4Q respectively as shown in the figure. (i) Find, out the ratio of the electric flux through them. (ii) How will the, electric flux through the sphere S1 change if a medium of, dielectric constant ‘e r ’ is introduced in the space inside S1 in, place of air? Deduce the necessary expression., , 4Q, , 2Q, , S2, , S1, , 29. (a) In Young’s double slit experiment, describe briefly how bright and dark fringes are obtained on, the screen kept in front of a double slit. Hence obtain the expression for the fringe width., (b) The ratio of the intensities at minima to the maxima in the Young’s double slit experiment is, 9 : 25. Find the ratio of the widths of the two slits., OR, (a) Describe briefly how a diffraction pattern is obtained on a screen due to a single narrow slit, illuminated by a monochromatic source of light. Hence obtain the conditions for the angular, width of secondary maxima and secondary minima.
Page 539 :
538, , Xam idea Physics—XII, , (b) Two wavelengths of sodium light of 590 nm and 596 nm are used in turn to study the, diffraction taking place at a single slit of aperture 2 × 10–6 m. The distance between the slit and, the screen is 1.5 m. Calculate the separation between the positions of first maxima of the, diffraction pattern obtained in the two cases., 30. (a) Deduce an expression for the frequency of revolution of a charged particle in a magnetic field, and show that it is independent of velocity or energy of the particle., (b) Draw a schematic sketch of a cyclotron. Explain, giving the essential details of its construction,, how it is used to accelerate the charged particles., OR, (a) Draw a labelled diagram of a moving coil galvanometer. Describe briefly its principle and, working., (b) Answer the following:, (i) Why is it necessary to introduce a cylindrical soft iron core inside the coil of a, galvanometer?, (ii) Increasing the current sensitivity of a galvanometer may not necessarily increase its, voltage sensitivity. Explain, giving reason., , SET–II (Questions Uncommon to Set–I), 1. A conducting loop is held above a current carrying wire ‘PQ’ as shown in the figure. Depict the, direction of the current induced in the loop when the current in the wire PQ is constantly increasing., , P, , Q, , 4. Why do the electrostatic field lines not form closed loops?, 5. A biconvex lens made of a transparent material of refractive index 1.5 is immersed in water of, refractive index 1.33. Will the lens behave as a converging or a diverging lens? Give reason., 6. The graph shows the variation of stopping, potential with frequency of incident radiation, for two photosensitive metals A and B. Which, one of the two has higher value of, work-function? Justify your answer., , Metal B, , Metal A, , Stopping, potential, (V0 ), O, , – We0, W', , – e0, , n0, , n'0, Frequency of incident, radiation (n)
Page 540 :
Examination Papers, , 539, , 7. To which part of the electromagnetic spectrum does a wave of frequency 3 ´ 1013 Hz belong?, 9. Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area 2.5 ×, 10–7 m2 carrying a current of 1.8 A. Assume the density of conduction electrons to be 9 × 1028 m–3., 12. Two monochromatic rays of light are incident normally on the face AB of an isosceles right-angled, prism ABC. The refractive indices of the glass prism for the two rays ‘1’ and ‘2’ are respectively 1.3, and 1.5. Trace the path of these rays after entering through the prism., A, 45°, '2', '1', 45°, B, , C, , 13. Write the functions of the following in communication systems:, (i) Transmitter, (ii) Modulator, 21. (a) Show, with the help of a diagram, how unpolarised sunlight gets polarised due to scattering., (b) Two polaroids P1 and P2 are placed with their pass axes perpendicular to each other., Unpolarised light of intensity I 0 is incident on P1 . A third polaroid P3 is kept in between P1 and, P2 such that its pass axis makes an angle of 45° with that of P1 . Determine the intensity of light, transmitted through P1 , P2 and P3 ., 22. Define the term self-inductance of a solenoid. Obtain the expression for the magnetic energy stored in, an inductor of self-inductance L to build up a current I through it., 24. A convex lens of focal length 20 cm is placed coaxially with a concave mirror of focal length 10 cm at, a distance of 50 cm apart from each other. A beam of light coming parallel to the principal axis is, incident on the convex lens. Find the position of the final image formed by this combination. Draw the, ray diagram showing the formation of the image., , SET–III (Questions Uncommon to Set–I and II), 2. The graph shows variation of stopping potential, V0 versus frequency of incident radiation v for, two photosensitive metals A and B. Which of, the two metals has higher threshold frequency, and why?, , Metal B, , Metal A, , Stopping, potential, (V0 ), O, , – We0, W', , – e0, , n0, , n'0, Frequency of incident, radiation (n)
Page 541 :
540, , Xam idea Physics—XII, , 5. Why do the electric field lines never cross each other?, 6. To which part of the electromagnetic spectrum does a wave of frequency 5 ´ 1011 Hz belong?, 10. Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area, 2.5 × 10 –7 m 2 carrying a current of 2.7 A. Assume the density of conduction electrons to be, 9 × 1028 m–3., 18. Write the functions of the following in communication systems:, (i) Receiver, (ii) Demodulator, 19. A convex lens of focal length 20 cm is placed coaxially with a convex mirror of radius of curvature, 20 cm. The two are kept 15 cm apart. A point object is placed 40 cm in front of the convex lens. Find, the position of the image formed by this combination. Draw the ray diagram showing the image, formation., 25. (a) A rod of length l is moved horizontally with a uniform velocity ‘v’ in a direction perpendicular, to its length through a region in which a uniform magnetic field is acting vertically downward., Derive the expression for the emf induced across the ends of the rod., (b) How does one understand this motional emf by invoking the Lorentz force acting on the free, charge carriers of the conductor? Explain., , zzz
Page 542 :
Examination Papers, , 541, , Solutions, SET–I, 1. One ampere is the value of steady current which when maintained in each of the two very long,, straight, parallel conductors of negligible cross-section and placed one metre apart in vacuum would, exert a force of 2 ´ 10 -7 N on 1 metre length of either wire., 2. X-rays or g-rays., 3. Force will decrease., Reason: Force between two charges each ‘q’ in vacuum is, q2, 1, ., 4pe 0 r 2, On inserting a plastic sheet (a dielectric K > 1), F0 =, , F, q2, 1, i.e.,, Force F = 0, ., 2, 4pe 0 K r, K, The force between charged balls will decrease., 4. The amount of light energy or photon energy incident per metre square per second is called intensity, of radiation., W, or J/s-m2, SI unit :, 2, m, 5. The current in the wire produces a magnetic field vertically, downward in the vicinity of the coil. When the current in wire BA, decreases, according to Lenz's law, the current induced in the coil, opposes this decrease; so the current in the coil will be in clockwise, direction., Then F =, , A, , B, , 6. Neutrinos are chargeless (neutral) and almost massless particles that hardly interact with matter., 7. (i) The generation of AC is more economical than DC., (ii) Alternating voltage can be stepped up or stepped down as per requirement during transmission, from power generating station to the consumer., (iii) Alternating current in a circuit can be controlled by using wattless devices like the choke coil., (iv) Alternating voltages can be transmitted from one place to another, with much lower energy loss, in the transmission line., 8. As a diverging lens, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, , –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, , –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, , –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, , –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, , –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, , –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, , –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, , –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, , –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, , – – – – – –, – – – – – –, – – – – – –, – – –ng –=1.25, – –, – – – – – –, – – – – – –, – – – – – –, – – – – – –, – – – – – –, – – – – – –, – – – – – –, – – – – – –, – – – n– =1.33, – –, medium, – – – –w – –, – – – – – –, – – – – – –, – – – – – –, – – – – – –, – – – – – –, – – – – – –, – – – – – –, – – – – – –, , –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, , –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, –, , As the light travels from rarer to denser, it diverges from its path.
Page 544 :
Examination Papers, , 543, , On solving equation (i) and (ii), 2, , 1, æ nh ö, . Ze 2, mç, ÷ r =, 4pe 0, è 2pmr ø, Þ, Þ, , n2h2, 2, , 4p mr, r =, , =, , 1, . Ze 2, 4pe 0, , 4pe 0 n 2 h 2, , =, , e0 n2 h2, , pmZe 2, 4p 2 mZe 2, For Bohr. radius, n = 1 i.e., for K shell, rB =, , e0 h2, pZme 2, , ., , 10. The charge on the capacitor q = CV and initial energy stored in the capacitor, 1 q2, 1, ...(i), = CV 2, 2 C, 2, (a) If another uncharged capacitor is connected in series then the same amount of the charge will, transfer as shown in figure., U1 =, , +, +, +, +, +, +, +, , -, , +, +, +, +, +, +, +, , q=CV, , -, , +, +, +, +, +, +, +, , q=CV, , -, , q=CV, , Keeping charge constant, and final voltage v¢ = 2v, q2, 1 q2, 1 q2, ...(ii), Uf =, +, =, 2 C, 2 C, C, q2, q2, :, = 2 :1, U f : Ui =, C, 2C, Alternately, 1, 1, U f : Ui = CV 2 : C S . V ' 2, 2, 2, 1, 1 C, 1, = CV 2 : ´ ´ ( 2V ) 2 = CV 2 : CV 2 = 1 : 2, 2, 2 2, 2, 11., , A, , I(t), , +, , -, , +, , -, , +, , -, , +, , -, , +, , -, , +, , -, , +, , -, , During charging C capacitor, a time varying current I(t) flows through the conducting wire, so on, applying Ampere’s circuital law (for loop A), , ® ®, , ò B . dl = m o I (t), , … (i)
Page 545 :
544, , Xam idea Physics—XII, B, A, , I(t), , +, +, +, +, +, +, +, , -, , Now we consider a pot like surface enclosing the positively charged plate and nowhere touches the, conducting wire,, ® ®, , ò B . dl = 0, , … (ii), , From equation (i) and (ii), we have a contradiction, B, A, , I(t), , -, , +, +, +, +, +, +, +, , ®, , If surfaces A and B forms a tiffin box, and electric field E is passing through the surface (B);, constitute an electric flux, Q, Q, s, … (iii), f = | E| | A| =, | A| =, | A| =, e0, Ae 0, e0, If the charge on the plate in the tiffin box is changing with time, there must be a current between the plates., From equation (iii), dQ, df, d, I =, =, ( e 0 f) = e 0, dt, dt, dt, This is the missing term in Ampere’s circuital law., The inconsistency may disappear if displacement current is included between the plates., So generalised Ampere’s circuital law can be given as, ® ®, df, ò B . dt = m 0 I(t ) + m 0 Id = m 0 I(t ) + m 0 e 0 dt, 12., (E,r), , R
Page 546 :
Examination Papers, , Suppose a current I flows through the circuit and using loop rule, E - IR - Ir = 0, [V = IR], Þ, E - Ir = V, Þ, V = E - Ir, If terminal voltage V is the function of current I, Reason – Equation of, A, straight line, y = -mx + c = c - mx, Then,, V, Using the graph, For point A, I = 0 and on using equation (i), V = E -0´r = E, Hence voltage intercept (intercept on the vertical axis) measures emf of the cell., For point B,, V = 0, from equation (i), O = E - Ir, E, r =, Þ, I, i.e., negative of the slope if V - I graph measures the internal resistance r., 13., , … (i), , B, , I, , VP, , VP1, P1, , O, , 545, , A, R2, , T, t, P2, , B, 0, , T, , T/2, , t, , Step down Transformer, , Working, (i) During positive half cycle of input alternating voltage, the diode is forward biased and a current, flows through the load resistor R L , and we get an output voltage., (ii) During other negative half cycle of the input alternating voltage, the diode is reverse biased and, it does not conduct (under break down region)., Hence AC voltage can be rectified in the pulsating and unidirectional voltage., 14. Flow of current in the conductor due to drift velocity of the free electrons is given by, I = neAv d, 15, ., I, vd =, =, neA 9 ´ 10 28 ´ 1.6 ´ 10 -19 ´ 1.0 ´ 10 -7, = 1.048 ´ 10 -3 m/s ~ 1 mm/s, 15. As we know that the critical angle depends on refractive index, m as, 1, sin i c =, m, If m =, , 2 = 1.414 then Ðc = 45°, , 1 ö, If m = 1.35, Ðc = sin -1 æç, ÷ > 45°, 135, ., è, ø, , 45°, '1', '2', , 45°, , 45°, 45°
Page 547 :
546, , Xam idea Physics—XII, , So the ray 1 refracted out., 1 ö, If m = 1.45, Ðc = sin -1 æç, ÷ < 45°, è 1.45 ø, So the ray 2, totally reflect back., 16., , (i) Function of Transducer: Any device that converts one form of energy into another., (ii) Function of Repeater: A repeater, picks up the signals from the transmitter, amplifies and, retransmits it to the next receiver sometimes with a change in carrier frequency., , 17., S, , N, , S, , Paramagnetic, substance, (i), , N, , Diamagnetic, substance, (ii), , A paramagnetic material tends to move from weaker field to stronger field regions of the magnetic field., So, the number of lines of magnetic field increases when passing through it., Magnetic dipole moments are induced in the direction of magnetic field., Paramagnetic materials has a small positive susceptibility., A diamagnetic material tends to move from stronger field to weaker field region of the magnetic field., So, the number of lines of magnetic field passing through it decreases., Magnetic dipole moments are induced in the opposite direction of the applied magnetic field., Diamagnetic materials has a negative susceptibility in the range (–1 £ x < 0)., 18. A circuit diagram of n–p–n transistor as an amplifier is shown in figure., C2, , C, , C1, , RB, IB B, , Rc, E, , Vi, , V0, , IE, VBB, , VCC, , Condition:, The base-emitter junction of the transistor must be forward biased and the collector-emitter junction, must be reverse biased., 19. (a) Light from a source S is allowed to fall normally on the flat surface of a thin plate of a, tourmaline crystal, cut parallel to its axis. Only a part of this light is transmitted through A., If now the plate A is rotated, the character of transmitted light remains unchanged. Now another, similar plate B is placed at some distance from A such that the axis of B is parallel to that of A., If the light transmitted through A is passed through B, the light is almost completely transmitted, through B and no change is observed in the light coming out of B.
Page 548 :
Examination Papers, , 547, , If now the crystal A is kept fixed and B is gradually rotated in its own plane, the intensity of, light emerging out of B decreases and becomes zero when the axis of B is perpendicular to that, of A. If B is further rotated, the intensity begins to increase and becomes maximum when the, axes of A and B are again parallel., Thus, 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., From this experiment, it is obvious that light waves are transverse and not longitudinal; because, if, they were longitudinal, the rotation of crystal B would not produce any change in the intensity of, light., Polarised, light, , Unpolarised, light, , Polarised, light, , S, (a), , A, Unpolarised, light, , B, , Polarised, light, No light, , S, (b), , (b) If light of intensity I 0 passes through the first polaroid., I, The intensity of light transmitted through P1 = I 1 = 0, 2, If axis of polaroids P1 and P3 are at 30°., So intensity of light transmitted through P3 is given by, I 3 = I 1 cos 2 30°, 2, , æ 3ö, 3I, ÷ = 0, ç, ç 2 ÷, 8, ø, è, Light transmitted through P3 is allowed to pass through P2 ., I, = 0, 2, , So intensity of light transmitted through P2 = I 2 = I 3 cos 2 60°, =, , 20., r2, , l, , rns, n 1 tu, , rns, n 2 tu, , r1, S1, S2, , 3I 0 æ 1 ö 2 3I 0, ´ç ÷ =, 8, 32, è2ø
Page 549 :
548, , Xam idea Physics—XII, , From relation f2 = MI 1 or f1 = MI 2, Mutual inductance between a pair of coils is equal to the magnetic flux linked with one of the coils, due to unit current flow in the another coil., dI, dI, From relation E 2 = -M 21 1 or E1 = -M 12 2, dt, dt, The mutual inductance for a pair of coils is equal to the magnitude of the induced emf in one of the, coils, when the current in the other coil changes at rate 1 Amp/sec., When a current I 2 is set up through outer coil S 2 , it in turn sets up a magnetic flux through coil S1 , so, magnetic flux through S1 is given by, … (i), N1 f1 = M 12 I 2, The magnetic field due to the current I 2 in S 2 is m 0 n 2 I 2 ., This magnetic field, on passing through the coil S1 produces a magnetic flux, N1 f1 = (m 0 n 2 I 2 )( n1l) pr1 2, = m 0 n1 n 2 pr12 l. I 2, , … (ii), , where n, l is the total number of turns in solenoid S1 ., From equation (i) and (ii), M 12 I 2 = m 0 n1 n 2 l. pr12 . I 2, Þ, , M 12 = m 0 n1 n 2 ( pr12 )l, , 21. (a) A thick copper strip offers a negligible resistance, so does not alter the value of resistances used, in the meter bridge., (b) If the balance point is taken in the middle, it is done to minimise the percentage error in, calculating the value of unknown resistance., (c) Generally alloys magnin/constantan/nichrome are used in meter bridge, because these materials, have low temperature coefficient of resistivity., OR, Net resistance between points A and C, R, R net = R p + o, 2, R0, ´R, R, R0 R, R0, 2, =, +, = 0 +, R, 2, R 0 + 2R, 2, 0, +R, 2, R ( R + 2R) + 2R 0 R R 0 ( R 0 + 4R), = 0 0, =, 2( R 0 + 2R), 2( R 0 + 2R), Net current in between points A and C, V ´ 2( R 0 + 2R), I net =, R 0 ( R 0 + 4R), Current in Resistance R, æR ö, V p = I net R = I 1 ç 0 ÷ = I 2 ( R), è 2 ø
Page 551 :
550, , Xam idea Physics—XII, L, , M, , I', I, , O, , 15 cm, 60 cm, , For convex lens, , 30 cm, , 30 cm, , 1 1 1, = f, v u, , f = + 20 cm, u = - 60 cm, 1, 1, 1, = \, +20 v - 60, 1, 1, 1, 1, =, =, Þ v = + 30 cm, v 20 60 30, In the absence of the mirror, the lens would have formed the image of I ¢ which acts as a virtual, object for the convex mirror., u 2 = distance of virtual object I ¢ from mirror, = + 30 cm – 15 cm = + 15 cm, R + 20, f = =, = +10 cm, 2, 2, 1, 1, 1, From mirror formula, =, +, fm, v2 u2, 1, 1, 1, =, +, I0, v2, + 15, Þ, , 1, 1, 1, 1, =, =, Þ, v 2 10 15 30, , v 2 = +30 cm, , Hence the final image I is a virtual image formed at a distance of 30 cm to the right of convex mirror, or 45 cm from the convex lens., 23., , R, , L, , C, , The voltage V = V0 sin wt is applied across the series L – CR circuit. However due to impedance of the, circuit, either current lags or leads the voltage by phase opposite so the current in the circuit is given by, I = I 0 sin( wt - f)
Page 553 :
552, , Xam idea Physics—XII, , Distinguishing features, (a) In conductors: Valence band and conduction band overlap on each other., In semiconductors: Valence band and conduction band are separated by a small energy gap., In insulators: They are separated by a large energy gap., (b) In conductors: Large number of free electrons are available in conduction band., In semiconductors: A very small number of electrons are available for electrical conduction., In insulators: Conduction band is almost empty i.e., no electron is available for conduction., 25. (a) Values displayed by Aarti:, Keen observer, helpful, responsible, concerned to family, respectful towards elders in the, family., (b) Radio isotopes help the doctors in observing the difference between the movement of an, appropriate radio-isotope through a normal brain and the brain having tumour in it., 26. Basic modes of communication are, (a) Point-to-point –, (Telephony is an example), (b) Broadcast, –, (Radio and Television), In Amplitude modulation, the amplitude of the carrier wave is made to vary with time in accordance, with the modulating signal varies with time., +2, , cm(t) for AM 0, , t, , –2, 0, , 0.5, , 1, , 0.5, , 2, , 2.5, , 3, , 27. Wavelength associated with the particle (electron) is, h, h, h, 12.27, l= =, =, =, Å, p, 2mk, 2mev, V, l=, , 6.63 ´ 10 -34, 2 ´ 91, . ´ 10 -31 ´ 1.6 ´ 10 -19 ´ 50 ´ 10 3, , = 5 .34 ´ 10 -12 m, The resolving power of an electron microscope is given by, 2m sin v, 1, R.P. =, =, d min, 1.22l, For constant numerical aperture, 1, R. P. µ, l, Wavelength of yellow light is in between 700 nm to 400 nm (say 600 nm) or 6000 Å), The wavelength of the accelerating electrons is much smaller than yellow light, hence resolving, power of electron microscope is much better than optical microscope., 28. Van de Graaff generator is excluded from syllabus., OR, (a) Expression for torque
Page 554 :
Examination Papers, +q A, , 553, , ®, , +qE, , a, E, a, q, , ®, , – qE, , C, , B, , ®, , An electric dipole having charges ± q, and of size 2a is placed in uniform electric field E as, ®, , ®, , shown in figure. The forces, acting on the charges are + q E and - q E ., ®, , ®, , ®, , The net force on the dipole is F = + q E + ( -q E ) = 0, Both forces provides an equivalent torque with magnitude, t = | qE | ´ Perpendicular distance (AC), = q | E| .2a sin q, = | P | | E | sin q, The direction of the torque can be given by, ®, , ®, , ®, , t = p´ E, , (b) Using Gauss’s Theorem, , ® ®, , ò E . ds =, , q(T ), e0, , Electric flux through sphere S1 , f1 =, Electric flux through sphere S 2 , f =, , 2( q), e0, , ( 2Q + 4Q) 6Q, =, e0, e0, , 2Q, f, e, 1, Ratio 1 = 0 =, 6Q 3, f, e0, If a medium of dielectric constant K( = e r ) is filled in the sphere S1, electric flux through, 2Q, 2Q, sphere, f¢1 =, =, er e0, Ke 0, 29. (a) Condition for formation of bright and dark fringes., Suppose a narrow slit S is illuminated by monochromatic light of wavelength l.
Page 556 :
Examination Papers, , 555, , dx, where n = 0, 1, 2, 3 …, = nl, D, nDl, x =, d, For n = 0,, Central bright fringe, x0 = 0, Dl, For n = 1,, 1st bright fringe, x1 =, d, 2Dl, 2nd bright fringe, For n = 2,, x2 =, d, nDl, For n = n,, nth bright fringe, xn =, d, The distance between two consecutive bright fringes is, b = x n - x n -1, nDl ( n - 1) Dl Dl, =, =, d, d, d, For destructive interference (dark fringes), dx, l, Path difference, = ( 2n - 1), D, 2, Dl, where n = 1, 2, 3, …, x = ( 2n - 1), 2d, Dl, For n = 1, x 1¢ =, for 1st dark fringe, 2d, 3Dl, For n = 2, x ¢2 =, for 2nd dark fringe, 2d, Dl, for nth dark fringe., For n = n, x ¢n = ( 2n - 1), 2d, The distance between two consecutive dark fringe is, Dl, Dl Dl, b¢ = ( 2n - 1), - {2( n - 1) - 1}, =, 2d, 2d, d, The distance between two consecutive bright or dark fringes is called fringe width (w)., Dl, Fringe width =, \, d, The expression for fringe width is free from n. Hence the width of all fringes of red light are, broader than the fringes of blue light., (b) Intensity of light (using classical theory) is given as, I µ (Width of the slit), µ (Amplitude)2, Path difference,, , I max, (a + a2 ) 2, 25, = 1, =, 2, 9, I min, ( a1 - a 2 ), a1 + a 2, 5, =, a1 - a 2, 3, Intensity ratio, , Þ, , a1, 4, =, a2, 1
Page 557 :
556, , Xam idea Physics—XII, , I1, w, a2, = 1 = 1, I2, w2, a 22, I 1 æ 4 ö 2 16, =ç ÷ =, I2 è 1 ø, 1, OR, (a) Diffraction of light at a single slit : When monochromatic light is made incident on a single slit,, we get diffraction pattern on a screen placed behind the slit. The diffraction pattern contains, bright and dark bands, the intensity of central band is maximum and goes on decreasing on both, sides., Explanation : Let AB be a slit of width ‘a’ and a parallel beam of monochromatic light is, incident on it. According to Fresnel the diffraction pattern is the result of superposition of a large, number of waves, starting from different points of illuminated slit., 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 path difference between rays diffracted at points A and B,, D = BP - AP = BN, In D ANB , ÐANB = 90°, \ and ÐBAN = q, BN, or BN = AB sin q, sin q =, \, AB, As AB = width of slit = a, \ Path difference,, ....(i), D = a sin q, To find the effect of all coherent waves at P, we have to sum up their contribution, each with a, different phase. This was done by Fresnel by rigorous calculations, but the main features may be, explained by simple arguments given below :, At the central point C of the, screen, the angle q is zero., P, Hence the waves starting from, all points of slit arrive in the, A, q, same phase. This gives Light from, q, M1, maximum intensity at the, O, C, M2 90 N, central point C., source, q, B, If point P on screen is such that, the path difference between, rays starting from edges A and, B is l , 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, each of width . Now for, 2, a, every point, M 1 in AO, there is a corresponding point M 2 in OB, such that M 1 M 2 = ; Then, 2, o
Page 558 :
Examination Papers, , 557, , a, l, sin q = ., 2, 2, This means that the contributions from the two halves of slit AO and OB are opposite in phase, and so cancel each other. Thus equation (2) gives the angle of diffraction at which intensity falls, nl, to zero. Similarly it may be shown that the intensity is zero for sin q =, , with n as integer., a, Thus the general condition of minima is, ...(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, path difference between waves arriving at P and starting from M 1 and M 2 will be, , I0, , –4l, a, , –3l, a, , –2l, a, , –l, a, , 0, , l, a, , 2l, a, , 3l, a, , 4l, a, , Let us now divide the slit into three parts. If we take the first two of parts of slit, 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. The, 2, remaining third part will contribute to the intensity at a point between two minima. Clearly there, will be a maxima between first two minima, but this maxima will be of much weaker intensity, than central maximum. This is called first secondary maxima. In a similar manner we can show, that there are secondary maxima between any two consecutive minima; and the intensity of, maxima will go on decreasing with increase of order of maxima. In general the position of nth, maxima will be given by, 1 ö, æ, ...(iv), a sin q = ç n +, [ n = 1, 2, 3, 4, . . . . ], ÷ l,, 2 ø, è, The intensity of secondary maxima decrease with increase of order n because with increasing n,, the contribution of slit decreases., For n = 2, it is one-fifth, for n = 3, it is one-seventh and so on.
Page 560 :
Examination Papers, , 559, , (b) Cyclotron: The cyclotron, devised by Lawrence and Livingston, is a device for accelerating ions, to high speed by the repeated application of accelerating potentials., Construction: The cyclotron consists of two flat semi-circular metal boxes called ‘dees’ and are, arranged with a small gap between them. A source of ions is, located near the mid-point of the gap between the dees (fig.)., Magnetic Pole, The dees are connected to the terminals of a radio frequency, N, oscillator, so that a high frequency alternating potential of, S, several million cycles per second exists between the dees., Dee, Dee, Thus dees act as electrodes. The dees are enclosed in an, insulated metal box containing gas at low pressure. The whole, Magnetic Pole, S, apparatus is placed between the poles of a strong, electromagnet which provides a magnetic field perpendicular, to the plane of the dees., Working: The principle of action of the, apparatus is shown in fig. The positive, Magnetic field, Deflecting plate, out of paper, ions produced from a source S at the, centre are accelerated by a dee which is, at negative potential at that moment., Exit port, Due to the presence of perpendicular, magnetic field the ion will move in a, Charged, particle, circular path inside the dees. 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 D, 1, their polarity (positive becoming, negative and vice-versa) and the ion is, D2, further accelerated and moves with, higher velocity along a circular path of, greater radius. The phenomenon is, continued till the ion reaches at the, Electric Oscillator, periphery of the dees where an auxiliary, negative electrode (deflecting plate), deflects the accelerated ion on the target to be bombarded., Role of electric field, (i) Electric field accelerates the charge particle passing through the gap with the help of electric, oscillator., (ii) Electric oscillator imparts the energy to charged particle till it comes out from the exit slit., Role of magnetic field, As the accelerated charge particle enters normally to the uniform magnetic field, it exerts a, magnetic force in the form of centripetal force and charge particle moves on a semicircular path, of increasing radii in each dee ( D1 or D2 )., Expression for Period of Revolution and Frequency:, Suppose the positive ion with charge q moves in a dee with a velocity v, then,, mv 2, mv, ...(i), or r =, r, qB, 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., qvB =
Page 561 :
560, , Xam idea Physics—XII, , The angular velocity w of the ion is given by,, v qB, (from eq. i), ...(ii), w= =, r, m, The time taken by the ion in describing a semi-circle, i.e., in turning through an angle p is,, p pm, ...(iii), t = =, w Bq, Thus the time is independent of the speed of the ion i.e., although the speed of the ion goes on, increasing with increase in the radius (from eq. i) when it moves from one dee to the other, yet it, takes the same time in each dee., m, being known, B can be calculated for, From eq. (iii) it is clear that for a particular ion,, q, producing resonance with the high frequency alternating potential., OR, (a) Diagram of moving coil galvanometer, Scale, , Permanent magnet, , Pointer, Coil, , Sp, N, , S, Pivot, Soft-iron, core, Uniform radial, magnetic field, , Principle: A current carrying coil, placed in a uniform magnetic field, experiences a torque, t = NIBA sin q, ®, , ®, , =m´ B, ®, , where m is the magnetic moment of the coil., Working: If plane of the coil is parallel to the magnetic field lines, the coil experiences a, maximum torque, and it tends to rotate the coil., Spring attached with the coil produces a counter torque, which balances the magnetic torque, and, coil deflects by angle f. So in equilibrium, Kf = NIBA, [sin q = 1], where K is torsional constant., Kf, I =, \, NBA, It means the deflection of the coil is proportional to the current flowing through the coil.
Page 562 :
Examination Papers, , (b), , (i) The cylindrical, soft iron core makes the (i) field radial and (ii) increases the strength of, the magnetic field, i.e., the magnitude of the torque., f NBA, (ii) Current sensitivity Si = =, I, K, f, f, NBA, and voltage sensitivity S v = =, =, V, IR, KR, Since the resistance of the coil may vary, it means an increase in current sensitivity may, not necessarily increase the voltage sensitivity., , SET–II (Questions Uncommon to Set–I), 1. Clockwise, , P, , Q, , 4. The electrostatic field lines start from positive charge and end on negative charge., , –q, , +q, , 5. As the light travels from denser to rarer medium, so the lens behaves as converging lens., Alternative method, From lens makers formula, æn, öæ 1, 1, 1 ö, ÷÷, = ç r - 1÷÷çç, f w çè n w, R, R, 2 ø, øè 1, On using sign convention, R1 = +ve and R 2 = -ve, Also n g = 1.5 and n w = 1.33, ö, æ 1.5, 1, =ç, - 1÷÷, f w çè 1.33, ø, 1, 1, +, = + ve, R1 R 2, ö, æ 1.5, - 1÷÷ = + ve, çç, ø, è 1.33, Hence,, f w = + ve, So, the lens behaves as a converging lens., , and, , 561
Page 563 :
562, , Xam idea Physics—XII, , 6. Metal A, Since work function W = hn 0, and n¢0 > n 0 so work function of metal A is more., Aliter:, On stopping potential axis -, , w¢0, e, , >-, , w0, ., e, , Hence work function w¢0 of metal A is more., 7. Infrared radiation, 9. Flow of current in the conductor due to drift velocity of the free electrons is given by, I = neAv d, 1.8 A, I, =, vd =, 28, neA 9 ´ 10 ´ 1.6 ´ 10 -19 ´ 2.5 ´ 10 -7, = 5 × 10–4 ms–1, 12. Explanation is same as Q. 15 Set–I., , 45°, ‘2', ‘1', , 45°, , 45°, 45°, , 13., , (i) Transmitter: A transmitter processes the incoming messages/information signals so as to make, it at suitable frequency for transmission through a channel., (ii) Modulator: It is a device in which amplitude of a high frequency carrier wave is made to, change in accordance with the amplitude of message/information signal during superposition., , 21. (a) The incident sunlight is unpolarised. The dots stand for polarisation perpendicular to the plane, of the figure. The double arrows show polarisation in the plane of the figure. Under the, influence of the electric field of incident radiation, the acceleration of the charges in the, scattering molecules can be in two mutually perpendicular directions., The observer, however, receives the scattered light corresponding to only one of the two sets of, accelerated charges i.e., electrons oscillating perpendicular to the direction of propagation.
Page 564 :
Examination Papers, , 563, , Incident Sunlight, (Unpolarised), , Scattered Light, (Polarised), , To Observer, , (b) If light of intensity I 0 passes through the first polaroid, the intensity of light transmitted, I, through P1 = I 1 = 0 ., 2, Since axis of polaroids P1 and P3 are at 30°, so intensity of light transmitted through P3 is, given by, I 3 = I 1 cos 2 45°, 2, , I, I0 æ 1 ö, ç, ÷ = 0, 4, 2 è 2ø, Light transmitted through P3 is allowed to pass through P2 , so intensity of light transmitted, through P2 = I 2 = I 3 cos 2 45°, =, , 2, , =, , I0 æ 1 ö, I, ´ç, ÷ = 0, 8, 4 è 2ø, , 22. Self inductance – Using formula f = LI, if I = 1 Ampere then L = f, Self inductance of the coil is equal to the magnitude of the magnetic flux linked with the solenoid coil,, when a unit current flows through it., or, dI, Using formula | - e | = L, dt, dI, If, = 1 A/s then L = | - e |, dt, Self inductance of the coil is equal to the magnitude of induced emf produced in the solenoid coil,, when the current varies at rate 1 A/s., Expression for magnetic energy, -e, B or j, , Increasing, current, , When a time varying current flows through the coil, back emf (-e) produces, which opposes the, growth of the current flow. It means some work needs to be done against back emf in establishing a, current I. This work done will be stored as magnetic potential energy.
Page 565 :
564, , Xam idea Physics—XII, , For the current I at any instant, the rate of work done is, dW, = ( -e) I, dt, Only for inductive effect of the coil | - e | = L, \, , dI, dt, , dW, æ dI ö, = Lç ÷I, dt, è dt ø, dW = L I dI, , From work-energy theorem, dU = L I dI, I, , \, , U =, , ò LIdI, , =, , 0, , 24., , 1 2, LI, 2, , L, , M, , 20 cm, , F, O, , IL, , 20 cm, , Cm, , Im, , 30 cm, 15 cm, , For the convex lens,, u = ¥, f = 20 cm, 1 1 1, = f, v u, \, , v = 20 cm, , For the concave mirror, the image formed by the lens acts as the object., Hence, u = - (50 - 20) cm = – 30 cm and f = -10 cm, Using mirror formula, we get, 1 1 1, + =, v u f, 1, 1, 1, +, =, v -30 -10, 1, 1, 1, 1, =, =v 30 10, 15, v = -15 cm, , P
Page 566 :
Examination Papers, , 565, , The lens-mirror combination, therefore, forms a real image I 2 at a distance of 15 cm to the left of the, concave mirror or at a distance of 35 cm to the right of the convex lens., , SET–III (Questions Uncommon to Set-I and II), 2. From the graph n¢ > n 0, 0, So, the minimum frequency at which the photoemission starts is more for metal A. Hence metal A has, higher threshold frequency., 5. If the field lines cross each other, then at the point of intersection, there will be two directions for the, same electric field which is not possible., 6. Microwaves or short radiowaves., 10. We know that, I = neAv d, I, vd =, neA, =, , 2 .7, 9 ´ 10, , 28, , ´ 1. 6 ´ 10 –19 ´ 2 .5 ´ 10 –7, , = 7 .5 ´ 10 – 4 m/s, 18., , (i) Receiver: It extracts the desired message signals from the received signals at the end of, channel., (ii) Demodulator: It is a device used to retrieve information/message signals from the carrier wave, at the receiver (or just after the rectifier)., , 19. For convex lens,, u = – 40 cm, f = 20 cm, L, , M, , I, , O, , 15cm, 40cm, , 1 1 1, = f v u, 1 1, 1, = 20 v – 40, 1 1, 1, =, v 20 40, v = + 40 cm, , 50/3 cm, 40cm, , I'
Page 567 :
566, , Xam idea Physics—XII, , This image acts as a virtual object for the convex mirror., \, , u = 40 – 15 = 25 cm, 20, = + 10 cm, f=, 2, , Using mirror formula,, 1 1 1, = +, f v u, , 1 1 1, = +, 10 v 25, , Þ, , 1 1, 1, =, Þ, v 10 25, , v=, , 15, cm ~ 16.67 cm, 3, , Hence, the final image is a virtual image formed at a distance of 16.67 cm., 25. (a), , V, , l, , x, , Suppose a rod of length ‘l’ moves with velocity V inward in the region having uniform, magnetic field B., Initial magnetic flux enclosed in the rectangular space is f = | B | lx, dx, As the rod moves with velocity -V =, dt, Using Lenz’s law, df, d, æ dx ö, e== - ( Blx ) = Bl ç ÷, dt, dt, è dt ø, \, e = Blv, (b) Suppose any arbitrary charge ‘q’ in the conductor of length ‘l’ moving inward in the field as, shown in figure, the charge q also moves with velocity V in the magnetic field B., The Lorentz force on the charge ‘q’ is F = qvB and its direction is downwards., So, work done in moving the charge ‘q’ along the conductor of length l, W = F. l, W = qvBl, Since emf is the work done per unit charge, W, \, e=, = Blv, q, This equation gives emf induced across the rod., zzz
Page 568 :
CBSE Examination Paper,, Delhi Region-2015 (Code No. 55/1/1/D), Time allowed: 3 hours, , Maximum marks: 70, , General Instructions:, (i) All questions are compulsory. There are 26 questions in all., (ii) This question paper has five sections : Section A, Section B, Section C, Section D and Section E., (iii) Section A contains five questions of one mark each, Section B contains five questions of two marks, each, Section C contains twelve questions of three marks each, Section D contains one value based, questions of four marks and Section E contains three questions of five marks each., (iv) 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 the three questions of five marks weightage. You have to, attempt only one of the choices in such questions., (v) Use log tables, if necessary. Use of calculators is not allowed., c = 3 × 108 m/s, h = 6.63 × 10–34 Js, e = 1.6 × 10–19 C, e o = 8.854 ´ 10 -12 C 2 N-1 m -2, m e = 9.1 ´ 10 -31 kg, , Mass of proton = 1.673 ×10–27 kg, Boltzmann constant = 1.38 ×10–23 JK–1, , µ0 = 4p ´ 10 -7 T m A -1, 1, = 9 ´ 10 9 N m 2 C –2, 4pe o, Mass of neutron = 1.675 ×10–27 kg, Avogadro’s number = 6.023 ×1023 per gram mole, , SECTION–A, 1., 2., 3., , Current I, , 4., 5., , Define capacitor reactance. Write its S.I. units.?, What is the electric flux through a cube of side 1 cm which encloses an electric dipole?, A concave lens of refractive index 1.5 is immersed in a medium of refractive index 1.65. What is the nature, of the lens?, How are side bands produced?, Graph showing the variation of current versus voltage for a material GaAs is shown in the figure. Identify the, region of, D, C, (i) negative resistance, (ii) where Ohm’s law is obeyed., , A, , E, , B, Voltage V, , SECTION–B, 6., 7., 8., , A proton and an a-particle have the same de-Broglie wavelength. Determine the ratio of (i) their accelerating, potentials (ii) their speeds., Show that the radius of the orbit in hydrogen atom varies as n2, where n is the principal quantum number of, the atom., Distinguish between ‘intrinsic’ and ‘extrinsic’ semiconductors.
Page 569 :
596 Xam idea Physics–XII, Use the mirror equation to show that an object placed between f and 2f of a concave mirror produces a real, image beyond 2f., OR, Find an expression for intensity of transmitted light when a polaroid sheet is rotated between two crossed, polaroids. In which position of the polaroid sheet will the transmitted intensity be maximum?, 10. Use Kirchhoff’s rules to obtain conditions for the balance condition in a Wheatstone bridge., , 9., , SECTION–C, 11. Name the parts of the electromagnetic spectrum which is, (a) suitable for radar systems used in aircraft navigation., (b) used to treat muscular strain., (c) used as a diagnostic tool in medicine., Write in brief, how these waves can be produced., 12. (i) A giant refracting telescope has an objective lens of focal length 15 m. If an eye piece of focal length, 1.0 cm is used, what is the angular magnification of the telescope?, (ii) If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the, objective lens? The diameter of the moon is 3.48 × 106 m and the radius of lunar orbit is 3.8 × 108 m., 13. Write Einstein’s photoelectric equation and mention which important features in photoelectric effect can be, explained with the help of this equation., The maximum kinetic energy of the photoelectrons gets doubled when the wavelength of light incident on, the surface changes from l1 to l2. Derive the expressions for the threshold wavelength l0 and work function, for the metal surface., 14. In the study of Geiger-Marsdon experiment on scattering of a-particles by a thin foil of gold, draw the, trajectory of a-particles in the coulomb field of target nucleus. Explain briefly how one gets the information, on the size of the nucleus from this study., From the relation R = R0 A1/3, where R0 is constant and A is the mass number of the nucleus, show that, nuclear matter density is independent of A., OR, Distinguish between nuclear fission and fusion. Show how in both these processes energy is released., Calculate the energy release in MeV in the deuterium-tritium fusion reaction:, 2, 1H, , +, , 3, 1H, , ¾¾, ¾®, , 3, 1 He, , + n, , Using the data:, m( 21 H) = 2.014102 u, , 15., 16., , 17., , 18., , m( 31 H) = 3.016049 u, , m( 31 He) = 4.002603 u, , mn = 1.008665 u, 1u = 931.5 MeV/c2, Draw a block diagram of a detector for AM signal and show, using necessary processes and the waveforms,, how the original message signal is detected from the input AM wave., A cell of emf ‘E’ and internal resistance ‘r’ is connected across a variable load resistor R. Draw the plots of, the terminal voltage V versus (i) R and (ii) the current I., It is found that when R = 4 W, the current is 1 A and when R is increased to 9 W, the current reduces to 0.5 A., Find the values of the emf E and internal resistance r., Two capacitors of unknown capacitances C1 and C2 are connected first in series and then in parallel across a, battery of 100 V. If the energy stored in the two combinations is 0.045 J and 0.25 J respectively, determine, the value of C1 and C2. Also calculate the charge on each capacitor in parallel combination., State the principle of working of a galvanometer., A galvanometer of resistance G is converted into a voltmeter to measure upto V volts by connecting a, resistance R1 in series with the coil. If a resistance R2 is connected in series with it, then it can measure upto, V/2 volts. Find the resistance, in terms of R1 and R2, required to be connected to convert it into a voltmeter, that can read upto 2 V. Also find the resistance G of the galvanometer in terms of R1 and R2.
Page 570 :
CBSE Examination Papers 597, 19. With what considerations in view, a photodiode is fabricated? State its working with the help of a suitable, diagram., Even though the current in the forward bias is known to be more than in the reverse bias, yet the photodiode, works in reverse bias. What is the reason?, 20. Draw a circuit diagram of a transistor amplifier in CE configuration., Define the terms: (i) Input resistance and (ii) Current amplification factor. How are these determined using, typical input and output characteristics?, 21. Answer the following questions:, (a) In a double slit experiment using light of wavelength 600 nm, the angular width of the fringe formed on a, distant screen is 0.1°. Find the spacing between the two slits., (b) Light of wavelength 5000 Å propagating in air gets partly reflected from the surface of water. How will, the wavelengths and frequencies of the reflected and refracted light be affected?, 22. An inductor L of inductance XL is connected in series with a bulb B and an ac source., How would brightness of the bulb change when (i) number of turn in the inductor is reduced, (ii) an iron rod, is inserted in the inductor and (iii) a capacitor of reactance XC = XL is inserted in series in the circuit. Justify, your answer in each case., , SECTION–D, 23. A group of students while coming from the school noticed a box marked “Danger H.T. 2200 V” at a, substation in the main street. They did not understand the utility of a such a high voltage, while they argued,, the supply was only 220 V. They asked their teacher this question the next day. The teacher thought it to be, an important question and therefore explained to the whole class., Answer the following questions:, (i) What device is used to bring the high voltage down to low voltage of a.c. current and what is the, principle of its working?, (ii) Is it possible to use this device for bringing down the high dc voltage to the low voltage? Explain., (iii) Write the values displayed by the students and the teacher., , SECTION–E, 24. (a) State Ampere’s circuital law. Use this law to obtain the expression for the magnetic field inside an air, cored toroid of average radius ‘r’, having ‘n’ turns per unit length and carrying a steady current I., (b) An observer to the left of a solenoid of N turns each of cross section area ‘A’ observes that a steady, current I in it flows in the clockwise direction. Depict the magnetic field lines due to the solenoid, specifying its polarity and show that it acts as a bar magnet of magnetic moment m = NIA., A, , N, , OR, (a) Define mutual inductance and write its S.I. units., (b) Derive an expression for the mutual inductance of two long co-axial solenoids of same length wound one, over the other., (c) In an experiment, two coils c1 and c2 are placed close to each other. Find out the expression for the emf, induced in the coil c1 due to a change in the current through the coil c2., 25. (a) Using Huygens’s construction of secondary wavelets explain how a diffraction pattern is obtained on a, screen due to a narrow slit on which a monochromatic beam of light is incident normally., (b) Show that the angular width of the first diffraction fringe is half that of the central fringe.
Page 571 :
598 Xam idea Physics–XII, 1 l, (c) Explain why the maxima at q = æç n + ö÷ become weaker and weaker with increasing n., 2ø a, è, , OR, (a) A point object ‘O’ is kept in a medium of refractive index n1 in front of a convex spherical surface of, radius of curvature R which separates the second medium of refractive index n2 from the first one, as, shown in the figure., Draw the ray diagram showing the image formation and deduce the relationship between the object, distance and the image distance in terms of n1, n2 and R., n1, , n2, C, , O, , u, , R, , (b) When the image formed above acts as a virtual object for a concave spherical surface separating the, medium n2 from n1 (n2 > n1), draw this ray diagram and write the similar (similar to (a)) relation. Hence, obtain the expression for the lens maker’s formula., ®, 26. (a) An electric dipole of dipole moment p consists of point charges +q and –q separated by a distance 2a, ®, apart. Deduce the expression for the electric field E due to the dipole at a distance x from the centre of, ®, the dipole on its axial line in terms of the dipole moment p . Hence show that in the limit x >> a,, ®, E ®, , ®, , 2 p, , ( 4pe 0 x 3 ), , ., , ®, (b) Given the electric field in the region E = 2xi$, find the net electric flux through the cube and the charge, enclosed by it., y, , x, a, z, , OR, (a) Explain, using suitable diagrams, the difference in the behaviour of a (i) conductor and (ii) dielectric in, the presence of external electric field. Define the term polarization of a dielectric and write its relation, with susceptibility., (b) A thin metallic spherical shell of radius R carries a charge Q on its surface. A, Q, point charge, is placed at its centre C and an other charge +2Q is placed, 2, outside the shell at a distance x from the centre as shown in the figure. Find, (i) the force on the charge at the centre of shell and at the point A, (ii) the, electric flux through the shell., , A, 2Q, , Q, , Q C, 2, , zzz
Page 572 :
CBSE Examination Papers 599, , Solutions, SECTION–A, 1. The imaginary/virtual resistance offered by a capacitor to the flow of an alternating current is called capacitor, 1, reactance, X C =, . Its SI unit is ohm., wC, 2. Net electric flux is zero., Reason : (i) Independent to the shape and size., (ii) Net charge of the electric dipole is zero., 3. Concave lens, in medium of high refractive index, behaves as a convex lens (or a converging lens)., ö, æmg, 1, 1, 1, Reason :, = çç, - 1÷÷ æç - ö÷, fm èm m, R, R, è, ø, ø, 1, Since, = +ve value, mm > mg Þ, fm, So, f m > 0 . Hence acts a convex lens., 4. Side bands (Lower side band and upper side band) are produced due to superposition of low frequency, modulating signals on high frequency carrier waves., Reason: The frequency of lower side band is wc - wm and of upper side band is wc + wm ., 5., , DV, < 0., DI, (ii) The region BC approximately passes through the origin, (or current also increases with the increase of, DV, voltage). Hence, it follows Ohm’s law and in this region, > 0., DI, (i) In region DE, material GaAs (Gallium Arsenide) offers negative resistance, because slop, , SECTION–B, 6. de Broglie wavelength l =, , h, =, p, , h, 2mqV, , where, m = mass of charge particle, q = charge of particle, V = potential difference, h2, (i), l2 =, 2mqV, V=, \, \, (ii), , Vp, Va, , =, , h2, 2mql2, 2m a q a 2 ´ 4m2q 8, =, =, 2m p q p, 2mq, 1, , Vp : Va = 8 : 1, h, h, h, , lp =, , la =, l=, mv, m pv p, m a Va, l p = la, h, h, =, m pv p ma va, vp, va, , =, , ma 4, = =4:1, mp 1
Page 574 :
CBSE Examination Papers 601, P1, , I2 = I 0 cos 2 q . cos 2 ( 90 - q), I0, ( 2 sin q cos q) 2, 4, I, = 0 sin 2 ( 2q), 4, =, , P3, , q, , P2, , For maximum value of I 2 ,, sin 2q = ± 1, 2q = 90°, , Þ, , Þ, q = 45°, It is possible only when polaroid P3 is placed at angle 45° from each polaroid P1 (or P2 )., B, 10. Condition of balance of a Wheatstone bridge:, The circuit diagram of Wheatstone bridge is shown in fig., P, K2, P, Q, R and S are four resistance forming a closed bridge, called Wheatstone, bridge. A battery is connected across A and C, while a galvanometer is, I1, G, connected between B and D. At balance, there is no current in galvanometer. A, I2, Derivation of Formula: Let the current given by battery in the balanced, position be I. This current on reaching point A is divided into two parts, R, I 1 and I 2 . As there is no current in galvanometer in balanced state, current in, I, resistances P and Q is I 1 and in resistances R and S it is I 2 ., D, Applying Kirchhoff’s I law at point A, + –, ...(i), I - I 1 - I 2 = 0 or I = I 1 + I 2, E, Applying Kirchhoff’s II law to closed mesh ABDA, ...(ii), - I 1 P + I 2 R = 0 or I 1 P = I 2 R, Applying Kirchhoff’s II law to mesh BCDB, ...(iii), - I 1 Q + I 2 S = 0 or I 1 Q = I 2 S, Dividing equation (ii) by (iii), we get, I1P, I R, P, R, or, ...(iv), = 2, =, I1Q, I 2S, Q S, , I1, Q, , C, , S, I2, , K1, , This is the condition of balance of Wheatstone bridge., , SECTION–C, 11. (a) Microwave, (b) Infrared, (c) X-rays, Microwave are produced by special vacuum tubes, like klystorms, magnetrons and gunn diodes., Infrared are produced by the vibrating molecules and atoms in hot bodies., X-rays are produced by the bombardment of high energy electrons on a metal target of high atomic, weight (like tungston)., 12. (i) Given f 0 = 15 m, f e = 1.0 cm = 1.0 × 10–2 m, Objective lens, Moon, Angular magnification of telescope,, f, m =- 0, fe, =-, , a, r, , 15, 1.0 ´ 10, , D, , -2, , Image of, moon, a, , = – 1500, , Negative sign shows that the final image is inverted., , d, , f0, , (ii) Let D be diameter of moon, d diameter of image of moon formed by objective and r the distance of moon, from objective lens, then from Fig.
Page 576 :
CBSE Examination Papers 603, 14., , 2, , 3, , 1, , 1, 2, 3, 4, , (i) For large impact parameter, almost all alpha particles go nearly undeviated and have small deflection., This shows that the mass of the atom is concentrated in a small volume in the form of nucleus and gives, an idea of size of nucleus., (ii) An alpha particle having small impact parameter suffers large scattering and in case of, head-on-collision, the alpha particle rebounds back., The radius (size) R of nucleus is related to its mass number (A) as, R = R0 A1/ 3, , \, , If m is the average mass of a nucleon, then mass of nucleus = mA, where A is mass number, 4, 4, 4, Volume of nucleus = pR 3 = p ( R 0 A 1 / 3 ) 3 = pR 03 A, 3, 3, 3, mass, mA, m, 3m, Density of nucleus, r N =, =, =, =, 4, 4, volume, 3, 3, 4pR 03, pR 0 A, pR 0, 3, 3, Clearly nuclear density r N is independent of mass number A., OR, Nuclear fission, , Nuclear fusion, , (i) Heavy nucleus splits into two smaller nuclei., , (i) Two lighter nuclei combine to form a heavy nucleus., , (ii) It takes place at ordinary temperature., , (ii) It requires extremely high temperature of the order of, 107 K., , In both the process, mass of nucleus participating in the reaction is greater than mass of the product nuclei., This difference in mass (Dm) is converted into energy and released., Mass defect = Dm, Energy released = Dm × 931 MeV, 2, 1H, , +, , 3, 1H, , ¾¾, ¾®, , 4, 2 He, , + n, , Dm = (+2.014102 + 3.016049 – 4.002603 – 1.008665) u, = 0.018883 u, Energy released = Dm × 931.5 MeV, = 0.018883 × 931 MeV, = 17.59 MeV, 15., AM wave, , Envelope, detector, , Rectifier, , (a), , (b), , Time, AM input wave, , m(t), , Output, , (c), , Time, Rectified wave, , Time, Output (without, RF component
Page 577 :
604 Xam idea Physics–XII, The AM signals are allowed to pass through a rectifier which gives a series of positive half cycles of radio, frequency pulse (rectified wave). These signals are allowed to pass through envelope detector (consisting of, L-C circuit) which separates envelope (message signal) from rectified wave., r, , E, , 16., I, , R, , Þ, , V = E – Ir, , V=, , ER, R +r, , V=e, , V=e, , V, , V, I, , R, , R1 = 4 W,, , R2 = 9 W, , I1 = 1 A,, , I2 = 0.5 A, , E = IR + Ir, E = I (R + r), Substituting the values, E = 1 (4 + r), , ...(i), , E = (9 + r), , ...(ii), , 4 + r = (9 + r), , Þ 8r + 2r = 9 + r, , r= 1W, E = 1 (4 + 1) = 5 V, Þ, r = 1 W, E = 5 V, 1, 17. Energy stored in a capacitor,, E = CV 2, 2, 1, In parallel,, 0.25 = (C1 + C2) (100)2, 2, Substituting in (i),, , In series,, From (i), , From (ii), , 0.045 =, , 1 æ C1C 2, ç, 2 çè C1 + C 2, , ö, ÷÷ (100) 2, ø, , ...(i), ...(ii), , C1 + C2 = 0.25 × 2 × 10–4, C1 + C2 = 5 × 10–5, C1C 2, = 0.045 × 2 × 10–4, C1 + C 2, , ...(iii), , C1C 2, = 0.09 × 10–4 = 9 × 10–6, C1 + C 2, From (iii), , C1 C2 =, , 2 ´ 0.045 ´ 5 ´ 10 –5, 10 4, , = 4.5 × 10–10, , C1 – C2 = (C1 + C 2 ) 2 – 4C1 C 2, C1 – C2 = 2.64 × 10–5, , ...(iv)
Page 578 :
CBSE Examination Papers 605, Solving (iii) and (iv), , C1 = 38.2 µF, C2 = 11.8 µF, Q1 = C1 V = 38.2 × 10–6 × 100 = 38.2 × 10–4 C, , In parallel, , Q2 = C2 V = 11.8 × 10–6 × 100 = 11.8 × 10–4 C, 18. When current flows through a coil kept in uniform radial magnetic field, it experiences a torque which tends, to rotate the coil and produces an angular deflection. The deflection of the coil is directly proportional to the, current., NAB, q=, .I, K, Let Ig be the current through galvanometer at full deflection, To measure V volts,, V = Ig (G + R1), ...(i), V, V, volts,, = Ig (G + R2), ...(ii), 2, 2, 2 V volts,, 2 V = Ig (G + R3), ...(iii), To measure for conversion of range dividing (i) by (ii),, G + R1, 2=, G + R2, G = R1 – 2R2, Putting the value of G in (i), we have, V, Ig =, R1 – 2R 2 + R1, Ig =, , V, 2R1 – 2R 2, , Substituting the value of G and Ig in equation (iii), we have, V, 2V=, (R1 – 2R2 + R3), 2R1 - 2R 2, 4R1 – 4R2 = R1 – 2R2 + R3, R3 = 3R1 – 2R2, 19. A photodiode (for special purpose) is fabricated with a transparent window to, allow light to fall on the junction of the diode., When the junction of photodiode is reverse bias it is illuminated with light of, energy hn > E g of the semiconductor, the electron-hole pairs are generated, due to absorption of photon., , p, , n, , Due to electric field of the function, electron are collected in n side and holes are, collected on p-side and produces an emf. When an external load is connected, an, electric current flows in the external circuit. The magnitude of the electric current depends on the intensity, of incident light., Reason of Reverse Biasing :, If there is no illumination, majority electrons are more than minority holes in n region of the diode. On, illumination the junction, some electrons-hole pairs are generated. Let the excess electrons and holes, generated are Dn and Dp respectively. So, at a particular illumination,, n ¢ = n + Dn, and, , p ¢ = p + Dp, , Since, Dn = Dp and n > p, Hence,, , Dn, DP, <<, n, P
Page 579 :
606 Xam idea Physics–XII, It means the minority carriers dominated reverse bias current is more easily measurable, than the fractional, change in the forward bias current. Hence, photodiodes are preferably used in the reverse bias for measuring, light intensity., 20., IC, C1, , RB, , C, , IB, , C2, RL, , B, , +, , E, , Vi, , +, –, , V0 Output, waveform, , VCC, , VBB, , IE, , –, , Input resistance (Ri): This is defined as the ratio of change in base-emitter voltage (DVBE) to the resulting, change in base current (DIB) at constant emitter voltage., æ DV ö, VCE, Ri = çç BE ÷÷, IB, D, I, B øV, è, CE, , Input resistance is determined from this slope of IB versus VBE plot at constant VCE., Current amplification factor: This is defined as the ratio of the change in, collector current to the change in base current at a constant collector-emitter voltage, (VCE)., æ DI ö, b = çç C ÷÷, è DI B øV, CE, , It is obtained from this slop of collector Ic versus, VCE plot using different values of IB ., DVBE, Ri =, DI B, b=, , IC, IC, , 2, , IC, , 1, , I B2 – I B1, p, p, =, 180 180°, , d =?, Angular width of fringes, l, where d = separation between two slits, b0 = ,, d, 600 ´ 10 - 9, p, =, 180°, d, Þ, , d =, , IB, , VCB, , IC 2 – IC 1, , 21. (a) l = 600 nm, b 0 = 0.1° = 0.1 ´, , IB, , 600 ´ 10 - 9 ´ 1800, = 3.43 × 10–4 m = 0 .34 m, p, , (b) l = 5000 Å = 5000 × 10–10 = 5 × 10–7 m, Reflected ray : No change in wavelength and frequency., Refracted ray : Frequency remains same, wavelength decreases, l, Wavelength l¢ =, m, , 2, , 1, , DIB, DVBE, VBE
Page 580 :
CBSE Examination Papers 607, 2, 22. Brightness of the bulb depends on square of the I rms (i.e., I rms, ), , Impedance of the circuit, Z =, , R 2 + ( w L) 2, , V, Z, (i) XL = wL. If number of turns in the inductor coil is reduced, then impedance of the combination will, decrease. It results in increase in the current. Hence, brightness of the bulb increases., (ii) When iron (being a ferromagnetic substance) rod is inserted in the coil, its inductance increases and in, turn, impedance of the circuit increases. As a result, a larger fraction of the applied ac voltage appears, across the inductor, leaving less voltage across the bulb. Hence, brightness of the bulb decreases., (iii) On inserting a capacitor, the impedance of the circuit becomes, and, , Current in the circuit, I =, , Z¢ =, , R 2 + (X L - X C ) 2 ,, , i.e., Z ¢ < Z, , (a) If X L = X C then Z ¢ = R. It results in increase in current resulting in increase in brightness of the bulb., , SECTION–D, (b) On inserting capacitor with X C = X L , voltage drop across inductor and capacitor becomes zero, and, maximum voltage appears across the bulb. Hence, brightness of the bulb will increase., 23., , (i) Step down transformer, Mutual induction: Whenever an alternative voltage is applied in the primary windings, an emf is induced, in the secondary coil., (ii) No, transformer cannot bring down dc voltage since no mutual induction will take place because flux, will not change in dc current., (iii) Student: Curiosity, scientific temper, Teacher: Friendly, easy to approach, honesty, , SECTION–E, 24., , (a) Ampere’s circuital law: The line integral of magnetic field around any closed path, in vacuum/air is equal to µ0 times the total current enclosed by that path., ®®, ò B . dl = µ0 I, Consider an air cored toroid of average radius ‘r’ having ‘n’ turns per unit length and, carrying a steady current I., , r, , I, , Consider an ampere loop of radius r as shown. Apply Ampere’s circuital law., ®®, ò B . dl = µ0 2pr n I, where 2pr · n is the number of times the current is threading., ®, B .2pr = µ0 2pr n I, ®, B = µ0 nI, , I, B, , dl
Page 581 :
608 Xam idea Physics–XII, (b), Observer S, , N, , For observer, current is flowing in clockwise direction hence we will see magnetic field lines going, towards south pole., The solenoid can be regarded as a combination of circular loops, placed side by side, each behaving like a magnet of magnetic, moment IA, where I is the current and A area of the loop., S, N S N, S N, These magnets neutralise each other except at the ends where, south and north poles appear., S, N, Magnetic moment of bar magnet, = NIA, OR, (a) When current flowing in one of two nearby coils is changed, the magnetic flux linked with the other, coil changes; due to which an emf is induced in it (other coil). This phenomenon of electromagnetic, induction is called mutual induction. The coil, in which current is changed is called the primary coil and, the coil in which emf is induced is called the secondary coil., The SI unit of mutual inductance is henry., (b) Suppose there are two coils C1 and C2. The current I1 is flowing in primary coil C1; due to which an, effective magnetic flux f 2 is linked with secondary coil C2. By experiments, f 2 µ I1 or f 2 = MI1, , ...(i), , where M is a constant, and is called the coefficient of mutual induction or mutual inductance., From (i), M=, , f2, I1, , If I1 = 1 ampere, M = f2, i.e., the mutual inductance between two coils is, numerically equal to the effective flux linkage with, secondary coil, when current flowing in primary coil is, 1 ampere., , N1, , r1, , r2, , N2, , I, Mutual Inductance of Two Co-axial Solenoids:, Consider two long co-axial solenoid each of length l with number of turns N1 and N2 wound one over the, other. Number of turns per unit length in outer (primary) solenoid, n = If I1 is the current flowing in, primary solenoid, the magnetic field produced within this solenoid., m N I, ...(ii), B1 = 0 1 1, l, The flux linked with each turn of inner solenoid coil is f2 = B1A2, where A2 is the cross-sectional area of, inner solenoid. The total flux linkage with inner coil of N2-turns., m N N, æm N N ö, F 2 = N2f 2 = N2B1A2 = N 2 ç 0 1 1 ÷ A2 = 0 1 2 A 2 I 1, l, l, ø, è, f2 m 0 N 1 N 2 A 2, =, By definition Mutual Inductance, M21 =, I1, l, , If n1 is number of turns per unit length of outer solenoid and r2 is radius of inner solenoid,, then M = µ0 n1 N2 pr22.
Page 582 :
CBSE Examination Papers 609, (c) When the current in coil C2 changes, the flux linked with C1 changes. This change in flux linked with C1, induces emf in C1., C1, , Flux linked with C1 = flux of C2, m I, f12 = B·A = 0 . A, 2r, m A, f12 = 0 . I, 2r, , C2, , I, , r, , df, d m 0 AI, emf in C1 = 12 =, dt, dt 2r, 25. (a) When a plane wavefront from a distance source illuminates a slit,, each point within the slit becomes a source of secondary spherical, wavelets. These spherical wavelets go on increasing in size, and, superpose on each other, in the region between slit and screen., (i) If all path differences are zero, then all the parts of the slit, contribute in phase. This gives maximum intensity at point P., (ii) If all path differences are non-zero, these secondary wavelets, may either superpose in phase or out of phase, resulting in either, occurrence of intensity of light or in zero occurrence., These variation in the light intensity on the screen is termed as, diffraction of light., , A = area of coil, To P1, A, , q, q, , From S, , P, , q, B, , (b) The angular width of central maxima is the angular separation between the first minima on the two sides, of the central maximum., Position of first minima sin q = nl, l, q1 = ±, d, 2l, Similarly, q2 = ±, d, Angular width of first diffraction fringe, l, bq =, d, Angular width of central maxima, +l, –l, to, d, d, 2l, b¢q =, d, Hence angular width of the first diffraction fringe is half of the central fringe., (c) The path difference between the extreme rays can be given as, BC = BP1 - AP1 = a sin q, where ‘a’ is the size of the slit., l, 2, will contribute no intensity at the point. However, the remaining one-third of the slit contributes to the, intensity at the point between two minima. Hence, the intensity of light will be much weaker than the, central maxima., If the slit is divided into three equal parts, first two-third of the slit which have a path difference of, , So, path difference, where n = 2, 3, 4, etc., , 1, a sin q = æç n + ö÷ l, 2ø, è
Page 585 :
612 Xam idea Physics–XII, OR, (a), Conductor, –, , Dielectric, E, , +, , –, , E=0, , –, , +, –, , –+, , –+, , +, , E, K, , +, , where k is dielectric constant, 1. No electric field lines travel inside conductor. 1. Alignment of atoms takes place due to electric field., 2. Electric field inside a conductor is zero., , 2. This results in a small electric field inside dielectric in, opposite direction., E, Net field inside the dielectric is ., K, , Polarization of a dielectric: The induced dipole moment developed per unit volume in an dielectric is, called polarization (P)., P = ceE, c ® susceptibility;, , e ® relative permeability;, , E ® electric field, , (b), Q, , 2Q, Q, C, 2, , x, , Q, at C = 0 because there is no electric field inside the shell., 2, Force on 2Q charge., Charge can be considered to be at C for a conducting sphere., , (i) Force on, , x, , C, , 2Q, , 3Q, 2, , A, , 3Q, 2, ®, 2 = 3KQ along CA, FA =, x2, x2, q, Q, =, f=, e 0 2e 0, K 2Q, , (ii), , Q, is the flux coming out of the shell., 2e 0, zzz
Page 586 :
CBSE Examination Paper,, Ajmer Region-2015 (Code No. 55/1/A), Time allowed: 3 hours, , Maximum marks: 70, , General Instructions: Same as CBSE Examination paper, Delhi–2015., , SECTION–A, 1., , Define dielectric constant of a medium. What is its SI unit?, , 2., , V– I graph for a metallic wire at two different temperatures T1 and T2 is as shown in the figure. Which of the, two temperatures is higher and why?, , T1, V, T2, , I, , 3., , A planar loop of rectangular shape is moved within the region of a uniform magnetic field acting, perpendicular to its plane. What is the direction and magnitude of the current induced in it?, , 4., , When light travels from an optically denser medium to a rarer medium, why does the critical angle of, incidence depend on the colour of light?, , 5., , What is the function of a band pass filter used in a modulator for obtaining AM signal?, , 6., , A variable resistor R is connected across a cell of emf e and internal resistance r as shown in the figure. Draw, a plot showing the variation of (i) terminal voltage V and (ii) the current I, as a function of R., , SECTION–B, , R, , +, , –, , e,r, , 7., , When the electron orbiting in hydrogen atom in its ground state moves to the third excited state, show how, the de Broglie wavelength associated with it would be affected., , 8., , Define a wavefront. Using Huygens’ principles, draw the shape of a refracted wavefront, when a plane wave, is incident on a convex lens., OR, (a) When a wave is propagating from a rarer to a denser medium, which characteristic of the wave does not, change and why?, (b) What is the ratio of the velocity of the wave in the two media of refractive indices m 1 and m 2 ?
Page 587 :
614 Xam idea Physics–XII, 9., , Differentiate between amplitude modulated (AM) and frequency modulated (FM) waves by drawing suitable, diagrams. Why is FM signal preferred over AM signal?, , 10., , Determine the distance of closest approach when an alpha particle of kinetic energy 4.5 MeV strikes a, nucleus of Z = 80, stops and reverses its direction., , SECTION–C, 11., , ®, , ®, , Deduce the expression for the torque t acting on a planar loop of area A and carrying current I placed in a, ®, , uniform magnetic field B ., If the loop is free to rotate, what would be its orientation in stable equilibrium?, 12., , How are electromagnetic waves produced? What is the source of the energy carried by a propagating, electromagnetic wave?, Identify the electromagnetic radiations used, (i) in remote switches of household electronic devices; and, (ii) as diagnostic tool in medicine., , 13., , (a) A ray of light is incident normally on the face AB of a right-angled glass prism of refractive index a m g = 1.5., The prism is partly immersed in a liquid of unknown refractive index. Find the value of refractive index, of the liquid so that the ray grazes along the face BC after refraction through the prism., A, , B, 60°, , C, , (b) Trace the path of the rays if it were incident normally on the face AC., 14., , The outputs of two NOT gates are fed to a NOR gate. Draw the logic circuit of the combination of gates., Write its truth table. Identify the gate equivalent to this circuit., OR, You are given two circuits (a) and (b) as shown in the figures, which consist of NAND gates. Identify the, logic operation carried out by the two. Write the truth tables for each. Identify the gates equivalent to the two, circuits., A, Y, B, , (a), A, , C, , Y, B, , D, , (b)
Page 588 :
CBSE Examination Papers 615, 15., , With the help of a circuit diagram, explain the working of a junction diode as a full wave rectifier. Draw its, input and output waveforms. Which characteristic property makes the junction diode suitable for, rectification?, , 16., , A potential difference V is applied across a conductor of length L and diameter D. How is the drift velocity,, vd, of charge carriers in the conductor affected when (i) V is halved, (ii) L is doubled and (iii) D is halved?, Justify your answer in each case., , 17., , A cyclotron’s oscillator frequency is 10 MHz. What should be the operating magnetic field for accelerating, protons? If the radius of its ‘dees’ is 60 cm, calculate the kinetic energy (in MeV) of the proton beam produced by, the accelerator., , 18., , In Young’s double slit experiment, the two slits are separated by a distance of 1.5 mm and the screen is, placed 1 m away from the plane of the slits. A beam of light consisting of two wavelengths 650 nm and 520, nm is used to obtain interference fringes. Find., (a) the distance of the third bright fringe for l = 520 nm on the screen from the central maximum., (b) the least distance from the central maximum where the bright fringes due to both the wavelengths, coincide., , 19., , (a) The figure shows the plot of binding energy (BE) per nucleon as a function of mass number A. The letters, A, B, C, D and E represent the positions of typical nuclei on the curve. Point out, giving reasons, the two, processes (in terms of A, B, C, D and E), one of which can occur due to nuclear fission and the other due, to nuclear fusion., , D, , C, , BE, A, , E, , B, A, , Mass Number A, , (b) Identify the nature of the radioactive radiations emitted in each step of the decay process given below:, A, Z, , 20., , X ¾ ¾®, , A–4, Z – 2Y, , ¾ ¾®, , A -4, Z - 1W, , Name the three different modes of propagation in a communication system., State briefly why do the electromagnetic waves with frequency range from a few MHz upto 30 MHz can, reflect back to the earth. What happens when the frequency range exceeds this limit?, , 21., , Define the terms ‘stopping potential’ and ‘threshold frequency’ in relation to photoelectric effect. How does, one determine these physical quantities using Einstein’s equation?, , 22., , Three circuits, each consisting of a switch ‘S’ and two capacitors, are initially charged, as shown in the, figure. After the switch has been closed, in which circuit will the charge on the left-hand capacitor (i), increase, (ii) decrease and (iii) remain same? Give reasons., S, , 6Q, , S, , +, , +, C, , 2C, , (a), , 3Q, , 6Q, , S, +, , +, C, , C, , (b), , +, , +, , 3Q, 6Q, , 3C, , C, , (c), , 3Q
Page 589 :
616 Xam idea Physics–XII, , SECTION–D, 23., , Sunita and her friends visited an exhibition. The policeman asked them to pass through a metal detector., Sunita’s friends were initially scared of it. Sunita, however, explained to them the purpose and working of the, metal detector., Answer the following questions:, (a) On what principle does a metal detector work?, (b) Why does the detector emit sound when a person carrying any metallic object walks through it?, (c) State any two qualities which Sunita displayed while explaining the purpose of walking through the, detector., , SECTION–E, 24., , (a) A point-object is placed on the principal axis of a convex spherical surface of radius of curvature R,, which separates the two media of refractive indices n1 and n2 (n2 > n1). Draw the ray diagram and deduce, the relation between the distance of the object (u), distance of the image (v) and the radius of curvature, (R) for refraction to take place at the convex spherical surface from rarer to denser medium., (b) Use the above relation to obtain the condition on the position of the object and the radius of curvature in, terms of n1 and n2 when the real image is formed., OR, (a) Draw a labelled ray diagram showing the formation of image by a compound microscope in normal, adjustment. Derive the expression for its magnifying power., (b) How does the resolving power of a microscope change when, (i) the diameter of the objective lens is decreased,, (ii) the wavelength of the incident light is increased?, Justify your answer in each case., , 25., , (a) State Faraday’s law of electromagnetic induction., (b) Explain, with the help of a suitable example, how we can show that Lenz’s law is a consequence of the, principle of conservation of energy., (c) Use of expression for Lorentz force acting on the charge carriers of a conductor to obtain the expression, for the induced emf across the conductor of length l moving with velocity v through a magnetic field B, acting perpendicular to its length., OR, (a) Using phasor diagram, derive the expression for the current flowing in an ideal inductor connected to, an ac source of voltage, v = v0 sin wt. Hence plot graphs showing variation of, (i) applied voltage and, (ii) the current as a function of wt., (b) Derive an expression for the average power dissipated in a series LCR circuit., , 26., , (a) State Gauss’s law in electrostatics. Show, with the help of a suitable example along with the figure, that, the outward flux due to a point charge ‘q’, in vacuum within a closed surface, is independent of its size or, shape and is given by q e o ., (b) Two parallel uniformly charged infinite plane sheets, ‘l’ and ‘2’, have charge densities + s and -2s, respectively. Give the magnitude and direction of the net electric field at a point., (i) in between the two sheets and, (ii) outside near the sheet ‘1’.
Page 590 :
CBSE Examination Papers 617, OR, (a) Define electrostatic potential at a point. Write its SI unit., Three point charges q1, q2 and q3 are kept respectively at points A, B and C as shown in the figure. Derive, the expression for the electrostatic potential energy of the system., A, r12, , B, , r13, , r23, , C, , (b) Depict the equipotential surfaces due to, (i) an electric dipole,, (ii) two identical positive charges separated by a distance., zzz
Page 591 :
618 Xam idea Physics–XII, , Solutions, SECTION–A, 1. It is the ratio of capacitance of a capacitor with dielectric medium to that without dielectric medium i.e.,, C, K = medium, C0, It has no unit., V, = R , i.e., smaller the slope, larger the resistance. As resistance of a metal, I, increases with the increase of temperature then T1 > T2 ., 3. If planer loop moves within the region of uniform magnetic field, no current will be induced in the loop., Hence no direction., b, 4. The refractive index is different for different colour wavelength as m = a +, . Hence, critical angle, l2, 1, sin iC = would also be different for different colour of light., m, , 2. The slope of V – I graph is, , 5. A bandpass filter blocks lower and higher frequencies and allows only a band of frequencies to pass through., , SECTION–B, 6. Current in the circuit I =, , E, =, R +r, , and terminal voltage V = IR =, , E, r, R æç1 + ö÷, Rø, è, E, , æ1 + r ö, ç, ÷, Rø, è, , I0, , E, , I, , V, , R, , 7. We know,, de Broglie wavelength,, Þ, Also, \, , l=, , h, h, =, p mv, , 1, ,, v, 1, vµ, n, lµ, , lµn, , \ de Broglie wavelength will increase., , R
Page 592 :
CBSE Examination Papers 619, 8. A wavefront is defined as the continuous locus of all the particles which are vibrating in the same phase., Transmitted, spherical wavefront, Incident, plane, wavefront, , F, , f, , OR, (a) Frequency remains constant, because number of waves entering at the point of two mediums per second, remains same., (b) Velocity in different media can be given as, c, c, c1 = 0 and c 2 = 0, m1, m2, \, , c1 m 2, =, c2 m 1, , i.e., Speed of light is inversely proportional to the refractive index of the media., 9., , EC, , (a) Carrier wave ec = Ec sin wct, (a) Carrier wave, , Em, B, , C, (b) Modulating wave (em), , Envelope, , E, , A, D, (b) Modulating signal, Phase increased, , Em, Ec, , (c) Modulated wave, , (c) Modulated wave e(t), , In frequency modulation (FM) the noise is minimised, hence FM is preferred., 10. Let r be the centre to centre distance between the alpha particle and the nucleus (Z = 80). When the alpha, particle is at the stopping point, then, 1 ( Ze) ( 2e), K=, r, 4p e 0, or, , r=, =, =, , 1, 2Ze 2, ., k, 4p e 0, 9 ´ 10 9 ´ 2 ´ 80 e 2, 4.5 MeV, 9 ´ 10 9 ´ 2 ´ 80 ´ 1.6 ´ 10 - 19, 4.5 ´ 10 6
Page 593 :
620 Xam idea Physics–XII, =, , 9 ´ 160 ´ 1.6, ´ 10 - 16, 4.5, , = 512 ´ 10 - 16 m, = 5.12 × 10–14 m, , SECTION–C, 11. 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, , N, , angle ( 90° - q) with the direction of magnetic field., ®, , ®, , ®, , q, , Q, , Therefore each of the forces F2 and F4 acting on these sides, has same magnitude F¢ = Blb sin ( 90° - q) = Blb cos q., , Axis of loop, or normal to loop, , F2, , B, R, , F1, I, , ®, , According to Fleming’s left hand rule the forces F2 and F4, , F3, P, , are equal and opposite but their line of action is same., Therefore these forces cancel each other i.e. the resultant of, , S, F4, , ®, , ®, , F2 and F4 is zero., The sides PQ and RS of current loop are perpendicular to the, , F1=IlB, , and F3 is, F = IlB sin 90° = IlB., ®, , loo, , q, , N, b, , q, , B, , ®, , 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, ®, , of, , I, (Upward), , ax, is, , ®, , p, , ®, , magnetic field, therefore the magnitude of each of forces F1, , I (Downward), N', , ®, , F1 and F3 is zero, but they form a couple called the, , b sin q, F3=IlB, , 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., \ Moment of couple or Torque,, t = (Magnitude of one force F) × perpendicular distance = ( BIl ) × ( b sin q) = I (lb) B sin q, lb = area of loop = A (say), Torque, t = IAB sin q, \, If the loop contains N-turns, then t = NI AB sin q, But, , ®, , ®, , ®, , In vector form t = NI A ´ B., The magnetic dipole moment of rectangular current loop = M = NIA, \, , ®, , ®, , ®, , t = M ´ B
Page 594 :
CBSE Examination Papers 621, Direction of torque is perpendicular to direction of area of loop as well as the direction of magnetic field, ®, , ®, , i.e., along I A ´ B ., The current loop would be in stable equilibrium, if magnetic dipole moment is in the direction of the, ®, , magnetic field ( B) ., 12. Electromagnetic waves are produced by oscillating charges which produce oscillating electric field and, magnetic field. Source of the energy carried by a propagating electromagnetic wave is the energy of the, accelerated charge., The electromagnetic radiations used in, (i) Infra-red radiations, (ii) X-rays, 13. (a) From Snell’s law, am g, , sin i c = a m l sin 90°, , A, , 1.5 × sin 60° = a m l, \, , am l, , = 1.5 ´, , B, 60°, , 3, 2, , IC, , = 1.3, , 90°, , (b) The ray strikes at an angle of 30° < i c . So, the ray of light deviates, apart from the normal., A, , B, C, , 60°, 30°, , r, , C, , 14. Output of gate G1 = A, , A, , Output of gate G 2 = B, Gate G 3 is NOR gate, its inputs are A and B,, , Y, B, , So, output of gate G 3 is Y = A + B = A . B = AB., , G2, , This is Boolean Expression for AND gate., Truth table, A, , B, , A, , B, , A + B, , 0, , 0, , 1, , 1, , 0, , 0, , 1, , 1, , 0, , 0, , 1, , 0, , 0, , 1, , 0, , 1, , 1, , 0, , 0, , 1, , OR, Circuit (a):, , C, G1, , Output of first gate C = A × B, Output Y = C × C = C = A × B = AB, , D, , G3
Page 595 :
622 Xam idea Physics–XII, Truth table, A, , B, , Y, , 0, , 0, , 0, , 1, , 0, , 0, , 0, , 1, , 0, , 1, , 1, , 1, , This is Boolean expression for AND gate., Output C = A , Output D = B, , Circuit (b):, , Output Y = A × B = A + B = A + B, Truth table, A, , B, , Y, , 0, , 0, , 0, , 1, , 0, , 1, , 0, , 1, , 1, , 1, , 1, , 1, , This is Boolean expression for OR gate., 15., , VP, , VP1, P1, , O, , A, R2, , T, t, P2, , B, 0, , T/2, , T, , t, , Step down Transformer, , Working, (i) During positive half cycle of input alternating voltage, the diode is forward biased and a current flows, through the load resistor R L , and we get an output voltage., (ii) During other negative half cycle of the input alternating voltage, the diode is reverse biased and it does, not conduct (under break down region)., Hence, AC voltage can be rectified in the pulsating and unidirectional voltage., The junction diode allows the current to pass only when it is forward biased. This makes the diode, suitable for rectification., V R, I, V, V, 16. Drift velocity, =, vd =, =, =, r, l, rl, neA, neA, ne, æ, ö, neA ç, ÷, è A ø, (i) As v d µ V , when V is doubled the drift velocity is halved., 1, (ii) As v d µ , when l is doubled the drift velocity is halved., l, (iii) As v d is independent of D, when D is doubled drift velocity remains unchanged., 17. The oscillator frequency should be same as proton cyclotron frequency, then, Magnetic field, B =, , 2p mv 2 ´ 3.14 ´ 1. 67 ´ 10 - 27 ´ 107, =, q, 1.6 ´ 10 - 19, , = 0.66 T, v = r w = r ´ 2pv, = 0.6 ´ 2 ´ 3.14 ´ 107 = 3.78 ´ 107 m/s
Page 596 :
CBSE Examination Papers 623, 1, mv 2, 2, 1, = ´ 1.67 ´ 10 - 27 ´ (3.78 ´ 107 ) 2 J, 2, , So,, , Kinetic energy, K =, , =, , 1 1.67 ´ 10 - 27 ´ 14.3 ´ 1014, MeV, ´, 2, 1.6 ´ 10 - 19 ´ 10 6, , = 7.4 MeV, 18. Given l1 = 650 nm = 650 ´ 10 (i) y n =, , nDy1, Þ, d, , y3 =, , m, l 2 = 520 nm = 520 ´ 10 -, , 9, , 3 ´ 1 ´ 520 ´ 10 1. 5 ´ 10, , - 3, , 9, , = 1 × 04 ´ 10 -, , 9, 3, , m, m ~ 1 mm, , (ii) For least distance of coincidence of fringes, there must be a difference of 1 in order of l1 and l 2 ., As l1 > l 2 ,, n1 < n 2, If n1 = n, 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 =, =, =4, Þ nl1 = ( n + 1) l 2, Þ n=, ( 650 - 520) nm, l1 - l 2, 130, \ Least distance, 19., , y min =, , (a) If a heavy nuclei of low, , nDl1, 4 ´ 1 ´ 650 ´ 10 =, d, 15, . ´ 10 - 3, , 9, , = 1 × 733 ´ 10 -, , 3, , m ~ 17, . mm, , BE, BE, splits up into two fragments, then, of the product nuclei increases and, A, A, , becomes stable. So,, E ®C + D, BE, BE, fuse together, the, of the product nuclei increases and becomes stable. So,, If two nuclei of low, A, A, A + B ®C, (b) If atomic number decreases by 2 and mass number decreases by 4 an alpha particle is emitted out. So,, A, Z, , X, , a, , ¾®, , A-4, Z - 2Y, , If a b - is emitted out, the atomic number increases by 1, while mass number remains unchanged. So,, A-4, Z - 2Y, , b-, , ¾®, , A-4, Z -1W, , 20. Three modes of propagation are, (i) Ground waves (ii) Sky waves (iii) Space waves, Ionosphere acts as reflector for the range of frequencies from few MHz to 30 MHz, hence it bends the radio, waves back to the Earth. Waves of frequencies greater then 30 MHz penetrate the ionosphere and escape., 21. Threshold Frequency: The minimum frequency of incident light which is just capable of ejecting electrons, from a metal is called the threshold frequency. It is denoted by v 0 ., 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 V0 (or VS )., By Einstein’s equation, eV0 = hn – f0, For any given frequency n > n 0 , V0 can be determined.
Page 597 :
624 Xam idea Physics–XII, f, h, Stopping Potential, V0 = æç ö÷ n – 0, e, è eø, f0 = hv 0, f, Threshold frequency, V0 = 0, h, 22. From law of conservation of charge., as, , Let q1 and q 2 be the charges on the left and right capacitors,, ( q1 + q 2 ) before = ( q1¢ + q¢2 ) after, For circuit (a), 6Q + 3Q = 2CV + CV, 9Q 3Q, =, \ Common potential difference, V =, 3C, C, Charge on the left capacitor, after closing the key S., q1¢ = 2CV, = 2C ´, , 3Q, C, , = 6Q, Charge on the left capacitor in the circuit (a) remains same (or unchanged)., For circuit (b), 6Q + 3Q = CV + CV, 9Q, Common potential difference, V =, 2C, Charge on the left capacitor, after closing the switch S, q1¢ = CV, =C ´, , 9Q, 2C, , = 4.5Q, i.e., the charge on the left capacitor in the circuit (b) will decrease., For circuit (c), 6Q + 3Q = 3CV + CV, 9Q, Common potential difference, V =, 4C, and the charge on the left capacitor becomes, 9Q 27QC, =, = 6.75 Q, 4C, 4C, i.e., the charge on the left capacitor in circuit c will increase., q1¢ = 3C ´, , SECTION–D, 23., , (a) Metal detector works on the principle of resonance in ac circuits., (b) When a person carrying any metallic object walks through the gate of a metal detector, the impedance of, the circuit changes, resulting in significant change in current in the circuit that causes a sound to be, emitted as an alarm., (c), , (i) Knowledge, (ii) Scientific temperament
Page 598 :
CBSE Examination Papers 625, , SECTION–E, 24., , N, (a) Relation of object and image distances of a, S, convex spherical surface : Let SPS¢ be the, n2, n1 Rarer medium, i A, convex, spherical, refracting, surface,, B Denser medium, `1', r, h, '2', separating the two media of refractive indices, I, g, C b, a, O, P, n1 and n 2 respectively ( n1 < n 2 ) i.e., M, Q, I, u, R, medium ‘1’ is rarer and medium ‘2’ is denser., v, Let P be the pole, C the centre of curvature, and PC the principal axis of convex refracting, S', surface. O is a distant point object on the, principal axis. The ray OA starting from O is incident on point A of the spherical surface, CAN is normal, at point A of the surface. Due to going from rarer to denser medium the ray OA deviates along the normal, CAN and is refracted along the direction AB. The another ray OP starting from O is incident normally on, the spherical surface and passes undeviated after refraction along PQ. Both the rays AB and PQ meet at, point I on the principal axis, i.e., I is the real image of point object O., , Let i be the angle of incidence of ray OA and r the angle of refraction in the denser medium i.e., Ð OAN = i and Ð CAI = r . Let Ð AOP = a , Ð AIP = b and Ð ACP = g, i=g + a, , In triangle OAC,, , g =b+r, n, sin i, = 2, sin r, n1, , In triangle AIC,, From Snell’s law, , or, , ...(i), r =g -b, , ...(ii), ...(iii), , If point A is very close to P, then angles i, r , a, b and g will be very small, therefore, sin i = i, , and, , sin r = r, , From equation (iii),, n, i, = 2, r, n1, Substituting values of i and r from (i) and (ii), we get, g + a n2, =, g -b, n1, or, , n1 ( g + a) = n 2 ( g - b), , ...(iv), , Let h be the height of perpendicular drawn from A on principal axis i.e. AM = h. As a , b and g are very, small angles., \, , tan a = a , tan b = b, , and, , tan g = g, , Substituting these values in (iv), n1 (tan g + tan a) = n 2 (tan g - tan b), As point A is very close to point P, point M is coincident with P., AM, h, From figure, tan a =, =, OM OP, AM, h, tan b =, =, MI, PI, AM, h, tan g =, =, MC, PC, Substituting these values in (v), we get, h, h ö, æ h - h ö, n1 æç, +, ÷ = n2 ç, ÷, OP ø, PI ø, è PC, è PC, , ...(v)
Page 599 :
626 Xam idea Physics–XII, 1, 1 ö, æ 1 - 1 ö, n1 æç, +, ÷ = n2 ç, ÷, OP ø, PI ø, è PC, è PC, , or, , ...(vi), , If the distances of object O, image I, centre of curvature C from the pole be u, v and R respectively, then, by sign convention PO is negative while PC and PI are positive. Thus,, u = - PO,, , v = + PI ,, , R = + PC, , Substituting these values in (vi), we get, 1 1, 1 1, n1 æç - ö÷ = n 2 æç - ö÷, R, u, R, vø, è, ø, è, , n1, n, n, n, - 1 = 2 - 2, R, u, R, v, , or, , n2, n, n - n1, - 1 = 2, v, u, R, , \, , The focal length of a convex lens is given by, æ 1, 1, 1 ö, = ( n - 1) ç, ÷, f, R2 ø, è R1, According to Cauchy’s formula, n =a+, , b, 2, , l, , +, , c, l4, , + ..., , Then n inversely as l, When wavelength increases, the refractive index n decreases; so focal length of lens increases with, increase of wavelength., n 2 n1, n - n1, (b), = 2, v, u, R, For real image, v is positive, n2, =0, \, u, n1 n 2 – n1, Therefore,, <, | u|, R, n1 R, | u|, R, >, Þ, | u| >, Þ, n1 n 2 – n1, n 2 – n1, which is the required condition., OR, (a) If image A ¢ B¢ is exactly at the focus of the eyepiece, then image A ¢¢ B¢¢ is formed at infinity., u0, , F0, B, , ue=fe, , v0, , B', , Fe, , Fe¢, , F¢, , 0, , 8, , Objective, , If the object AB is very close to the focus of the objective lens of focal length fo, then magnification Mo by, the objective lens
Page 600 :
CBSE Examination Papers 627, Me =, , L, fo, , where L is tube length (or distance between lenses Lo and Le), Magnification Me by the eyepiece, D, Me =, fe, where D = Least distance of distinct vision, Total magnification, m = MoMe, æ L öæ D ö, = ç ÷ç ÷, è f o øè f e ø, (b) The resolving power of the microscope µ, , n sin q, ., l, , where q is the semi-angle of the cone., (i) If the diameter of the objective lens decreases, then the value of resolving power will decrease as, resolving power is directly proportional to the diameter., 1, (ii) Resolving power µ , therefore, the resolving power of the microscope will decrease with the, l, increase of wavelength of light used., 25. (a) Faraday’s Laws of Electromagnetic Induction, (i) Whenever there is a change in magnetic flux linked with a coil, an emf is induced in the coil. The, induced emf is proportional to the rate of change of magnetic flux linked with the coil., Df, i.e.,, eµ, Dt, (ii) emf induced in the coil opposes the change in flux, i.e.,, Df, Df, eµÞ e=-k, Dt, Dt, where k is a constant of proportionality., Negative sign represents opposition to change in flux., Df, In SI system f is in weber, t in second, e in volt, when k = 1, e = Dt, Df, If the coil has N-turns, then e = - N, Dt, (b) 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.”, Example: When the north pole of a coil is brought near a closed coil, the direction, of current induced in the coil is such as to oppose the approach of north pole. For, this the nearer face of coil behaves as north pole. This necessitates an, anticlockwise current in the coil, when seen from the magnet side [fig. (a)], Similarly when north pole of the magnet is moved away from the coil, the, direction of current in the coil will be such as to attract the magnet. For this the, nearer face of coil behaves as south pole. This necessitates a clockwise current in, the coil, when seen from the magnet side [fig. (b)]., , S, , N, , N, , (a), , S, , N, (b), , S, , Conservation of Energy in Lenz’s Law: Thus, in each case whenever there is a, relative motion between a coil and the magnet, a force begins to act which opposes the relative motion., Therefore to maintain the relative motion, a mechanical work must be done. This work appears in the, form of electric energy of coil. Thus Lenz’s law is based on principle of conservation of energy.
Page 601 :
628 Xam idea Physics–XII, (c) Expression for Induced emf: We know that if a charge q moves, ®, , ®, , C, , with velocity v in a magnetic field of strength B , making an, angle q , then magnetic Lorentz force, , P, , F = q v B sin q, ®, , D, , ++, ++, , B, , Fe, , ®, , If v and B are mutually perpendicular, then q = 90° ,, , –, , l, , v, , F = qvB sin 90° = qvB, ®, , ®, , Fm, , The direction of this force is perpendicular to both v and B and is, given by Fleming’s left hand rule., Suppose a thin conducting rod PQ is placed on two parallel, , –– ––, Q, , M, , ®, , N, , metallic rails CD and MN in a magnetic field of strength B. The, ®, , ®, , direction of magnetic field B is perpendicular to the plane of paper, downward. In fig B is represented by, ®, , cross (´) marks. Suppose the rod is moving with velocity v , perpendicular to its own length, towards the, right. We know that metallic conductors contain free electrons, which can move within the metal. As, charge on electron, q = - e, therefore, each electron experiences a magnetic Lorentz force, Fm = evB ,, whose direction, according to Fleming’s left hand rule, will be from P to Q. Thus the electrons are, displaced from end P towards end Q. Consequently the end P of rod becomes positively charged and end, Q negatively charged. Thus a potential difference is produced between the ends of the conductor. This is, the induced emf., Due to induced emf, an electric field is produced in the conducting rod. The strength of this electric field, V, ...(i), E =, l, and its direction is from (+) to (–) charge, i.e., from P to Q., The force on a free electron due to this electric field, Fe = eE, , ...(ii), , The direction of this force is from Q to P which is opposite to that of electric field. Thus the emf produced, opposes the motion of electrons caused due to Lorentz force. This is in accordance with Lenz’s law. As, the number of electrons at end Q becomes more and more, the magnitude of electric force Fe goes on, ®, , ®, , increasing, and a stage comes when electric force F e and magnetic force F m become equal and, opposite. In this situation the potential difference produced across the ends of rod becomes constant. In, this condition, Fe = Fm, eE = evB or, , E = Bv, , ...(iii), , \ The potential difference produced,, V = El = B v l volt, Bvl, V, Also the induced current I =, =, ampere, R, R, OR, (a) Inductive Reactance : The opposition offered by an inductor to the flow of, alternating current through it is called the inductive reactance. It is denoted by, X L . Its value is X L = wL = 2pfL, where L is inductance and f is the frequency of the applied voltage., Obviously, , XL, , XL µ f, , That the graph between X L and frequency ( f ) is linear (as shown in fig.)., , f
Page 602 :
CBSE Examination Papers 629, Phase Difference between Current and Applied Voltage in Purely Inductive circuit :, AC circuit containing pure inductance : Consider a coil of self-inductance L and negligible ohmic, resistance. An alternating potential difference is applied across its ends. The magnitude and direction of AC, changes periodically, due to which there is a continual change in magnetic flux linked with the coil., Therefore according to Faraday’s law, an induced emf is produced in the coil, which opposes the applied, voltage. As a result the current in the circuit is reduced. That is inductance acts like a resistance in ac, circuit. The instantaneous value of alternating voltage applied, V = V0 sin wt, , ...(i), di, If i is the instantaneous current in the circuit and, , the rate of change of current in the circuit at that, dt, instant, then instantaneous induced emf, di, e=-L, L, dt, According to Kirchhoff’s loop rule, V + e=0 Þ V -L, , di, =0, dt, , di, di V, or, =, dt, dt, L, V sin w t, di V0 sin wt, or di = 0, =, dt, dt, L, L, , V=V0 sin wt, (a), , V =L, , or, or, , Integrating with respect to time ‘t’,, i=, , V0, L, , ò, , sin wt dt =, , V0 ì, cos wt ü, í ý, L î, w, þ, , =or, , i=, , =-, , V0, cos wt, wL, , V0, p, sin æç, - wt ö÷, wL, 2, è, ø, , V0, p ö, sin æç wt ÷, wL, 2 ø, è, , …(ii), , This is required expression for current, or, , p ö, i = i 0 sin æç wt ÷, 2, è, ø, , where, , i0 =, , ...(iii), , V0, wL, , ...(iv), , is the peak value of alternating current, Also comparing (i) and (iii), we note that current lags behind the applied voltage by an angle, , p, (Fig. b)., 2, , Phasor diagram : The phasor diagram of circuit containing pure inductance is shown in Fig. (b)., V0, , O, , wt, p/2, , Reference axis, , iO, , (b) Phasor diagram
Page 603 :
630 Xam idea Physics–XII, Graphs of V and I versus wt for this circuit is shown in fig. (c)., V0, , V, , i0, i, V, or O, i, , p, , p/2, , 2p, , 3p, wt, , (c), , (b) V = Vm sin wt, i = i m ( wt + f), P = Vi, , and instantaneous power,, , = Vm sin wt . i 0 sin ( wt + f) = Vm i m sin wt sin ( wt + f), 1, = Vmi m 2 sin wt . sin ( wt + f), 2, From trigonometric formula, 2 sin A sin B = cos ( A - B) - cos ( A + B), 1, \ Instantaneous power,, P = Vmi m [cos ( wt + f - wt) - cos ( wt + f + wt)], 2, 1, = Vm i m [cos f - cos ( 2wt + f)], 2, , … (i), , Average power for complete cycle, ®, , P =, , 1, Vm i m [cos f - cos ( 2wt + f) ], 2, , where cos ( 2wt + f) is the mean value of cos ( 2wt + f) over complete cycle. But for a complete cycle,, cos ( 2wt + f) = 0, \ Average power,, ®, , P =, , V i, 1, Vmi m cos f = 0 0 cos f, 2, 2 2, , ®, , P = Vrms i rms cos f, , 26. (a) Statement: The net-outward normal electric flux through any closed surface of any shape is equal to, 1 / e 0 times the total charge contained within that surface, i.e.,, , òS, where ò, , S, , ®, , ®, , E • dS, , =, , 1, Sq, e0, , indicates the surface integral over the whole of the closed surface, Sq is the algebraic sum of, , all the charges (i.e., net charge in coulombs) enclosed by surface S., , E, DS, , Proof:, Electric field at any point on S is given by, q, 1, E=, ×, 4p e 0 r 2, , r, , Let a point charge + q be placed at centre O of a sphere S. Then S is a Gaussian surface., , S, , q
Page 606 :
CBSE Examination Paper,, Allahabad Region-2015 (Code No. 55/1/RU), Time allowed: 3 hours, , Maximum marks: 70, , General Instructions: Same as CBSE Examination paper, Delhi–2015., , SECTION–A, 1., , Define the term ‘self-inductance’ of a coil. Write its SI unit., , 2., , Why does bluish colour predominate in a clear sky?, , 3., , I – V graph for a metallic wire at two different temperatures, T1 and T2 is as shown in the figure. Which of the, two temperatures is lower and why?, T1, , I, T2, , V, , 4., , Which basic mode of communication is used for telephonic communication?, , 5., , Why do the electrostatic field lines not form closed loops?, , SECTION–B, 6., , When an electron in hydrogen atom jumps from the third excited state to the ground state, how would the de, Broglie wavelength associated with the electron change? Justify your answer., , 7., , Write two factors which justify the need of modulating a low frequency signal into high frequencies before, transmission., , 8., , Use Kirchhoff’s rules to determine the potential difference between the points A and D when no current, flows in the arm BE of the electric network shown in the figure., F, , 3W, , E, , 1V, , D, , R1, R, , 2W, 3V, A, 6V, , B, , C, 4V
Page 607 :
634 Xam idea Physics–XII, 9., , You are given two converging lenses of focal lengths 1.25 cm and 5 cm to design a compound microscope. If, it is desired to have a magnification of 30, find out the separation between the objective and the eyepiece., OR, A small telescope has an objective lens of focal length 150 cm and eyepiece of focal length 5 cm. What is the, magnifying power of the telescope for viewing distant objects in normal adjustment?, If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the image of the tower, formed by the objective lens?, , 10., , Calculate the shortest wavelength in the Balmer series of hydrogen atom. In which region (infra-red, visible,, ultraviolet) of hydrogen spectrum does this wavelength lie?, , SECTION–C, 11., , Calculate the potential difference and the energy stored in the capacitor C2 in the circuit shown in the figure., Given potential at A is 90 V, C1 = 20 mF, C2 = 30 mF, C3 = 15 mF., A, C1, , C2, , C3, , 12., , Find the relation between drift velocity and relaxation time of charge carriers in a conductor., A conductor of length L is connected to a d.c. source of emf ‘E’. If the length of the conductor is tripled by, stretching it, keeping ‘E’ constant, explain how its drift velocity would be affected., , 13., , State clearly how an unpolarised light gets linearly polarised when passed through a polaroid., (i) Unpolarised light of intensity I0 is incident on a polaroid P1 which is kept near another polaroid P2, whose pass axis is parallel to that of P1. How will the intensities of light, I1 and I2, transmitted by the, polaroids P1 and P2 respectively, change on rotating P1 without disturbing P2?, (ii) Write the relation between the intensities I1 and I2., , 14., , Define modulation index. Why is its value kept, in practice, less than one?, A carrier wave of frequency 1.5 MHz and amplitude 50 V is modulated by a sinusoidal wave of frequency, 10 kHz producing 50% amplitude modulation. Calculate the amplitude of the AM wave and frequencies of, the side bands produced., ®, A uniform magnetic field B is set up along the positive x-axis. A particle of charge ‘q’ and mass ‘m’ moving, ®, with a velocity v enters the field at the origin in X-Y plane such that it has velocity components both along, ®, and perpendicular to the magnetic field B . Trace, giving reason, the trajectory followed by the particle. Find, out the expression for the distance moved by the particle along the magnetic field in one rotation., , 15., , 16., , (a) Determine the value of phase difference between the current and the voltage in the given series LCR, circuit., R = 400 W, , V = V0 sin (1000t + f), , C = 2 mF, , L = 100 mH, , (b) Calculate the value of the additional capacitor which may be joined suitably to the capacitor C that, would make the power factor of the circuit unity., 17., , Write the expression for the generalised form of Ampere’s circuital law. Discuss its significance and describe, briefly how the concept of displacement current is explained through charging/discharging of a capacitor in, an electric circuit.
Page 608 :
CBSE Examination Papers 635, 18., , Use Huygens’ principle to show how a plane wavefront propagates from a denser to rarer medium. Hence,, verify Snell’s law of refraction., , 19., , Identify the gates P and Q shown in the figure. Write the truth table for the combination of the gates shown., Name the equivalent gate representing this circuit and write its logic symbol., A, P, , X, , B, , Q, , Y, , 20., , Draw a circuit diagram of a C.E. transistor amplifier. Briefly explain its working and write the expression for, (i) current gain, (ii) voltage gain of the amplifier., , 21., , (a) Write three characteristic properties of nuclear force., (b) Draw a plot of potential energy of a pair of nucleons as a function of their separation. Write two, important conclusions that can be drawn from the graph., , 22., , (a) Describe briefly three experimentally observed features in the phenomenon of photoelectric effect., (b) Discuss briefly how wave theory of light cannot explain these features., OR, (a) Write the important properties of photons which are used to establish Einstein’s photoelectric equation., (b) Use this equation to explain the concept of (i) threshold frequency and (ii) stopping potential., , SECTION–D, 23., , One morning an old man walked bare-foot to replace the fuse wire in kit kat fitted with the power supply, mains for his house. Suddenly he screamed and collapsed on the floor. His wife cried loudly for help. His, neighbour’s son Anil heard the cries and rushed to the place with shoes on. He took a wooden baton and used, it to switch off the main supply., Answer the following questions:, (i) What is the voltage and frequency of mains supply in India?, (ii) These days most of the electrical devices we use require a.c. voltage. Why?, (iii) Can a transformer be used to step up d.c. voltage?, (iv) Write two qualities displayed by Anil by his action., , SECTION–E, 24. (a) Define electric flux. Write its SI unit., “Gauss’s law in electrostatics is true for any closed surface, no matter what its shape or size is.” Justify, this statement with the help of a suitable example., (b) Use Gauss’s law to prove that the electric field inside a uniformly charged spherical shell is zero., OR, (a) Derive the expression for the energy stored in a parallel plate capacitor. Hence obtain the expression for, the energy density of the electric field., (b) A fully charged parallel plate capacitor is connected across an uncharged identical capacitor. Show that, the energy stored in the combination is less than that stored initially in the single capacitor., 25., , Explain, using a labelled diagram, the principle and working of a moving coil galvanometer. What is the, function of (i) uniform radial magnetic field, (ii) soft iron core?, Define the terms (i) current sensitivity and (ii) voltage sensitivity of a galvanometer. Why does increasing the, current sensitivity not necessarily increase voltage sensitivity?, OR, ®, ®, (a) Write using Biot – Savart law, the expression for the magnetic field B due to an element dl carrying, ®, current I at a distance r from it in a vector form.
Page 609 :
636 Xam idea Physics–XII, Hence derive the expression for the magnetic field due to a current carrying loop of radius R at a point P, distant x from its centre along the axis of the loop., (b) Explain how Biot-Savart law enables one to express the Ampere’s circuital law in the integral form, viz.,, ® ®, , ò B × dl = m 0 I, where I is the total current passing through the surface., 26. (a) Consider two coherent sources S1 and S2 producing monochromatic waves to produce interference, pattern. Let the displacement of the wave produced by S1 be given by, Y1 = a cos wt, and the displacement by S2 be, Y2 = a cos ( wt + f), Find out the expression for the amplitude of the resultant displacement at a point and show that the, intensity at that point will be, f, I = 4a2 cos 2, 2, Hence establish the conditions for constructive and destructive interference., (b) What is the effect on the interference fringes in Young’s double slit experiment when (i) the width of the, source slit is increased; (ii) the monochromatic source is replaced by a source of white light?, OR, (a) A ray ‘PQ’ of light is incident on the face AB of a glass prism ABC (as shown in the figure) and, emerges out of the face AC. Trace the path of the ray. Show that, Ði + Ðe = ÐA + Ðd, where d and e denote the angle of deviation and angle of emergence respectively., A, , i, Q, P, B, , C, , Plot a graph showing the variation of the angle of deviation as a function of angle of incidence. State the, condition under which Ðd is minimum., (b) Find out the relation between the refractive index (m) of the glass prism and Ð A for the case when the, angle of prism (A) is equal to the angle of minimum deviation (dm). Hence obtain the value of the, refractive index for angle of prism A = 60°., zzz
Page 610 :
CBSE Examination Papers 637, , Solutions, SECTION–A, 1. The self-inductance of the coil is defined as the magnetic flux linked with the coil changed when unit current, flows through it., The SI unit of self-inductance is henry (H)., 2. At the time of sunrise or sunset, the sun is just above the horizon; sunlight travels a longer distance through, the atmospheric particles. The amount of scattering is inversely proportional to the fourth power of the, wavelength, i.e.,, 1, Iµ, l4, Since l B << l R . Hence, the bluish colour predominates in the clear sky., 3. If a constant current I flows through the conductor, resistance at, T1, temperature T1 and T2 is, T2, V, I, R1 = 1, I, V2, and, R2 =, I, V, V1, V2, Since, V2 > V1 Þ R 2 > R1, The resistance of the wire increases with rise of temperature. Hence, T1 is lower than T2 ., 4. In point to point communication mode, the communication takes place over a link between a single, transmitter and a receiver., 5. Electric field lines start from positive charge and terminate at negative charge. If there is a single positive, charge, the field lines start from the charge and terminate at infinity. So, the electric field lines do not form, closed loops., , SECTION–B, 6. de Broglie wavelength associated with a moving charge particle having a KE ‘K’ can be given as, é, p2 ù, h, h, 1, …(1) êK = mv 2 =, l= =, ú, p, 2, 2m û, 2mK, ë, The kinetic energy of the electron in any orbit of hydrogen atom can be given as, æ 13.6, ö 13.6, K= - E = -ç, …(2), eV ÷ =, eV, 2, è n, ø, n2, Let K1 and K 4 be the KE of the electron in ground state and third excited state, where n1 = 1 shows ground, state and n 2 = 4 shows third excited state., Using the concept of equation (1) & (2), we have, l1, =, l4, , K4, =, K1, , l1, =, l4, , 12, 4, , 2, , l4, 4, i.e., the wavelength in the ground state will decrease., , Þ, , l1 =, , =, , n12, n 22, 1, 4
Page 611 :
638 Xam idea Physics–XII, l, . [For transmitting, 4, a base signal of frequency 20 kHz or wavelength 15 km, the size of the antenna should be 3.75 km., Hence, an antenna of size 3.75 km is not possible to construct and operate.], , 7. (a) For transmitting a signal of wavelength l, the size of the antenna should have a size, , 2, , l, (b) The power radiated by an antenna P µ æç ö÷ . It implies that for the same length of the antenna, the, è lø, power radiated for the base band signals is very low, and increases with the decrease in wavelength of, em waves., (c) If many transmitters are radiating signals in nearly base band mode, these signals will get mixed up, and, cannot be distinguished., I3 E, 3, I1, 8. According to Kirchhoff’s junction rule at E or B, F, D, I 3 = I 2 + I1, I2, 1V, Since I 2 = 0 in the arm BE as given in the question, R, R1, I3, Þ, I 3 = I1, Using loop rule in the loop AFEBA, 3V, I1, 2, I2, +6V – 2I 3 + 1V – 3I 3 – I 2 R1 + 3V = 0, A, B, C, Þ, 2I 3 + 3I 3 + I 2 R1 = 10 V, I1, I3, 6V, 4V, Since I 2 = 0, so, 5I 3 = 10 V, Þ, I3 = 2 A, \, I 3 = I1 = 2 A, The potential difference between A and D, along the branch AFED of the closed circuit., VA - 2 I 3 + 1 V - 3 I 3 - VD = 0, Þ, VA – VD = 2 I 3 - 1 V + 3 I 3, =2×2–1+3×2, =9V, 9. The magnification due to the objective lens, vo, mo =, (- uo ), If the object is close to focus of the objective lens then, u o = f o and v = L (L = distance between two lenses), L, mo =, fo, If the final image is at the near point, then magnification due to eye lens is, æ, Dö, m e = çç1 +, ÷÷, f, è, e ø, \, , M = mo ´ me =, , Use f o = 1.25 cm and f e = 5 cm, , L, f0, , æ, Dö, ÷÷, çç1 +, f, è, e ø, , L æ, 25 ö, ç1 +, ÷, 1.25 è, 5 ø, L, 30 =, ´ 6, 1.25, 30 ´ 1.25, L=, = 6.25 cm, 6, Separation between two lenses is 6.25 cm., 30 =
Page 612 :
CBSE Examination Papers 639, OR, If the telescope is in normal adjustment, i.e., the final image is at infinity., f, M= o, fe, Since f o = 150 cm, f e = 5 cm, , 150, = 30, 5, If tall tower is at distance 3 km from the objective lens of focal length 150 cm. It will form its image at, distance v o . So,, 1, 1, 1, =, f o vo, uo, , \, , M=, , 1, 1, 1, =, ( - 3 km), 150 cm v o, 1, 1, 1, =, v o 1.5 m 3000 m, 3000 ´ 1.5, 4500, =, = 1.5 m, 3000 – 1.5 2998 .5, h, v, I, Magnification, mo = = i = o, O ho, uo, vo =, , hi, 1.5 m 1 .5, =, =, 100 m 3 km 3000, 1.5 ´ 100 1, =, m, hi =, 3000, 20, hi = 0.05 m, 10. If electron jumps from ni = 3, 4, . . . , ¥ to n f = 2 state. It is known as Balmer series., From Rydberg formula,, æ 1, 1 ö÷, 1, =R ç, ç n2, l, ni2 ÷ø, è f, For shortest wavelength n f = 2 and ni = ¥, 1, 1 ö, æ 1, =R ç, ÷, 2, ls, è2, ¥2 ø, ls =, , 4, R, , Since R = 1.09 ´ 107 m - 1, \, , ls =, , 4, 1.09 ´ 10, , 7, , =, , 4000, ´ 10 - 10 m, 1.09, , = 3669.7 ´ 10 - 10 m = 3669.7 Å, Since range of visible light is 4000 Å to 7500 Å. So, it must be in ultraviolet-region., , SECTION–C, 11. Capacitors C1 , C 2 and C 3 are in series. So, its net capacitance is, 1, 1, 1, 1, =, +, +, C s C1, C2, C3
Page 613 :
640 Xam idea Physics–XII, 1, 1, 1, +, +, 20 30 15, 20, Cs =, mF, 3, Net charge on the capacitors, C1 , C 2 and C 3 remain same., q = C s (VA - VE ), 20, =, mF ´ ( 90 - 0), 3, = 600 µC, The p.d across C 2 due to charge 600 mC is, q, V2 =, C2, =, , =, Energy stored in the capacitor C 2 ,, , 600, = 20 V, 30, , 1 q2 æ 1, 2ö, ç or C 2 V2 ÷, 2 C2 è 2, ø, 1, = ´ 30 mF ´ ( 20) 2, 2, , U2 =, , = 6000 mJ = 6 ´ 10 - 3 J, 12. Inside the conductor, all free electrons suffer collisions with the heavy fixed ions, and after collisions these, electrons again emerge with the same speed, but in random directions. So, at given time, net velocity of the, electrons is zero, i.e.,, ®, 1 n ®, (Ui ) av =, å Ui = 0, N i=1, If the electron field established inside the conductor, electrons get accelerated, so, eE, a=–, m, Let t i be the time elapsed after the last collision of the ith electron. So, the average velocity of all electrons, i = 1, 2, . . . , n, ®, eE, (t i ) av, v d = (vi ) av = ( vi ) av m, eE, t, vd = 0 m, where t = average time known as relaxation time., If emf of d.c. source is e, along the conductor of length L., e æeö, vd = ç ÷t, m èLø, If emf e is constant, then, 1, …(1), vd µ, L, On stretching the length L ¢ = 3L, 1, …(2), v¢d µ, 3L
Page 614 :
CBSE Examination Papers 641, From equations (1) and (2), we have, 1 1, : =1:3, 3L L, v, So,, v¢d = d, 3, 13. In unpolarised light, electric vectors oscillate in all possible directions and strike on a polaroid. It is observed, that (i) oscillations in a direction perpendicular to that of the alignment of the molecules in a polaroid are able, to pass through it, while the components of light along the alignment gets blocked. Hence, the light gets, linearly polarised. (ii) Half components of all vibrating electric field vectors pass through the polaroid,, results that the intensity of light gets reduced to half., I, (i) If unpolarised light of intensity I 0 incident on a poloroid P1 , its intensity gets reduced 0, 2, (a) If polaroid P2 is placed close to P1 , and pass axis is parallel to that of P, then, I, I1 = 0, 2, I0, and, I2 =, 2, (b) If polaroid P1 is rotated, keeping pass axis of P2 remain at rest., I, I 1 = 0 remain same, 2, but I ¢2 < I 2, Þ, I2 < I0, (c) If polaroid P1 is rotated by 90°, then, I, I1 = 0, 2, But I ¢2 = 0, (ii) Let q be the angle between pass axis of P1 and P2 , then, v¢d : v d =, , I 2 = I 1 cos 2 q, , I2 =, , I0, cos 2 q, 2, , It is also known as Malus law., 14. The ratio of amplitude of modulating signal to the amplitude of carrier wave is called modulation index., A, µ= m, Ac, Am < A c, (a) If A m = A c then m = 1, (b) If A m > A c , then m > 1, It occurs in the carrier waves distortion. So the value of modulation index should be m £ 1., 1, µ = 50% =, 2, Am, Since, µ=, Ac, A, 1, = m, 2 50 V, Þ, , A m = 25 V, Frequency of LSB = m c - m m, = 1.5 MHz – 10 KHz
Page 615 :
642 Xam idea Physics–XII, = 1.5 MHz – 0.01 MHz, = 1.49 MHz, Frequency of USB = Dc + v m, = 1.5 MHz – 0.01 MHz, = 1.51 MHz, y, 15. If component v x of the velocity vector is along the magnetic, field, and remain constant, the charge particle will follow a, helical trajectory; as shown in fig., O, If the velocity component v y is perpendicular to the magnetic, field B, the magnetic force acts like a centripetal force qv y B., So, , q vy B =, , vy, , vx, , +q, , mv 2y, , x, , z, , r, qBr, vy =, m, , B, , Since tangent velocity v y = rw, qBr, m, qB, Þ, w=, m, 2p 2pm, Time taken for one revolution, T =, =, w, qB, , Þ, , rw=, , and the distance moved along the magnetic field in the helical path is, x = vx . T, 2pm, = vx ., qB, R = 400 W, , 16. (a), , V = V0 sin (1000t + f), , C = 2 mF, , L = 100 mH, , From phasor diagrams,, Veff, , Z, VL – VC, , f', , OR, Ieff, , VR, , Phase angle, tan f=, , XL – XC, , f', R, , X L - XC, R, , X L = w L = 1000 ´ 100 ´ 10 - 3 = 100 ohm, 1, 1, 1000, =, =, = 500 ohm, wC 1000 ´ 2 ´ 10 - 6, 2, 500 - 100, =1, \, tan f¢ =, 400, \, f¢ = 45°, p, The phase angle between the current and applied voltage is 45° æç = ö÷ ., è 4ø, XC =, , V
Page 616 :
CBSE Examination Papers 643, (b) If power factor of the circuit is unity. It means the series LCR would be in resonance. It is possible, if, another capacitor C ¢ is used in the circuit., X C¢ = X L, 1, = wL, wC¢, , So,, , C¢ =, , Þ, , =, , 1, w2 L, 1, (1000), , 2, , ´ 100 ´ 10 - 3, , = 10 - 5 F, C ¢ = 10 mF, Since C ¢ > C, so an additional capacitor of 8 mF would be connected in parallel to the capacitor of, C = 2 mF., 17. The generalisation in Ampere’s circuital law is made by Maxwell, as, , ò, , ®, , ®, , B × dl = m 0 ( I c + I d ), = m 0 Ic + m 0 Id, = m 0 I c + m 0 e0, , dfe, dt, , dfE, is displacement current., dt, Significance : This expression signifies that the source of magnetic field is not just due to the conduction, current in the metallic conductors, but also due to the time rate of change of electric flux called displacement, current., During charging and discharging of a capacitor, electric field between the plates will change. Hence there, will be a change in electric flux, called displacement current, between the plates., 18. We assume a plane wavefront AB propagating in denser, Incident, B, medium incident on the interface PP ¢ at angle i$ as shown in, wavefront, Medium, I, Fig. Let t be the time taken by the wave front to travel a, i, (Denser), distance BC. If v1 is the speed of the light in medium I., i, P, P', A, r C, So,, BC = v1 t, r, Medium II, In order to find the shape of the refracted wavefront, we, (Denser), Refracted, draw a sphere of radius AE = v 2 t , where v 2 is the speed of, wavefront, light in medium II (rarer medium). The tangent plane CE, E, represents the refracted wave front, vt, BC, In DABC,, = 1, sin i =, AC, AC, AE v 2 t, and in DACE,, sin r =, =, AC, AC, v1, sin i BC v1t, =, …(1), \, =, sin r AE v 2 t v 2, where I d = e 0, , Let c be the speed of light in vacuum, So,, , µ1 =, , C, v1, , and, , µ2 =, , C, v2
Page 617 :
644 Xam idea Physics–XII, m2, v, = 1, m 1 v2, , …(2), , From equations (1) and (2), we have, sin i m 2, =, sin r m 1, m 1 sin i = m 2 sin r, It is known as Snell’s law., 19. Gate P is AND gate., Gate Q is NOT gate., Truth Table, A, , B, , X, , Y, , 0, , 0, , 0, , 1, , 0, , 1, , 0, , 1, , 1, , 0, , 0, , 1, , 1, , 1, , 1, , 0, , Since Y = A × B. The equivalent gate is NAND, Logic symbol is, A, Y, B, C2, , 20., C1, , IC, , C, , Rb, , B, , Rc, , Ib, E, Vi, , VBB, , IE, , VCC, , Vo, , When an ac input signal v i (to be amplified) is superimposed on the bias VBB , the output, which is measured, between collector and ground, increases., æ DI ö, Current gain, b ac = çç c ÷÷, è DI b ø, Voltage gain of the amplifier, A v =, , DVo, DI c R c, =DVi, DI b . r, , Rc, r, Negative sign shows that the output voltage Vo is in opposite phase with the input voltage Vi ., 21. (a) Characteristics of Nuclear Forces:, (i) Nuclear force is much stronger than the coulomb force acting between charges., (ii) The nuclear force between two nucleons falls rapidly to zero as their distance is more than a few, femtometer. This leads to saturation of forces., (iii) The binding force exists at a distance of few femtometers. So, it is a short range force., (iv) The nuclear force exists between proton-proton, proton-neutron and neutron-neutron. So, it is, charge independent., Av = - b .
Page 618 :
CBSE Examination Papers 645, (b) (1) The potential energy is minimum at a distance about r 0 = 0.8 fm., , A, +100, , MeV, , Repulsive, B, , 0, D, Attractive, , –100, , C, , 1, , 2, , 3, , 4, , r (fm), , (2) If the distance r 0 > 0.8 fm, the nuclear force is attractive., (3) If the distance between the nucleons r < 0.8 fm. The nuclear force is repulsive., 22. (a) (i) Keeping the frequency of the incident radiations and accelerating potential constant, the, photoelectric current is directly proportional to the intensity of radiations., (ii) The maximum kinetic energy of the photoelectrons varies linearly with the frequency of incident, radiations, but is independent of its intensity., (iii) For frequency of incident radiations lower than the cut-off frequency n 0 , no photoelectric emission, is possible even if the intensity is large., (iv) The photoelectric emission is an instantaneous process without any time lag (i.e., less than 10 - 9 )., (b) According to classical wave theory of light, the greater the intensity of light, the greater is the amplitude, of electric and magnetic fields. When the radiations fall on the metal surface, the electrons absorb greater, energy from the radiations of greater intensity. It means the maximum kinetic energy of the, photoelectrons is then expected to increase with increase intensity., No matter what the frequency of radiations, this expectation directly contradict the experimental, observed phenomenon., When radiations of suitable freqeuncy and radiation fall on the metal surface, the absorption of energy by, electrons take place continuously over the entire wave front of radiations. Since, the large number of, electrons absorb energy, the energy absorbed per electron per second turns out to be small. It is estimated, that the electron will take hours or more to pick up sufficient energy to over come the work function and, come out from the metal surface. It again contradicts the observation, in which photoelectrons come out, within time less than 10 - 9 second., OR, (a) Important properties of Photon :, (i) In interaction of radiation with matter, radiation behaves as it is made up of particles called photon., h, (ii) Each photon has energy E ( = hn) and momentum P æç = ö÷ , if energy of photon is greater than the, è dø, work function of the metal, electron emission occurs., When a photon strikes on a single electron, the energy needed for the electron to escape from the, metal surface should be greater than the work function of the metal. So,, K max = hn - w, 1, 2, (b) From relation K max = mv = hn - w, 2, (i) K max depends linearly on frequency m and independent of intensity of radiation., Since K max = zero (or non-negative, the radiations of frequency m 0 then, 0 = hn 0 - W, Þ, W = hn 0
Page 619 :
646 Xam idea Physics–XII, Work function of the metal is defined as a certain minimum amount of energy required by an, electron to just pull it out from the metal surface., (ii) For a particular frequency of incident radiations, the minimum negative potential, for which the, photocurrent stops (or becomes zero) is called the cut off or stopping potential., , SECTION–D, 23., , (a) In most generators, the coil is held stationary and electromagnets are rotated at 50 rev/sec, so the, frequency of the current is 50 Hz and voltage supply to the consumer is at 220 V., (b), , (i) AC supply can be stopped up/stopped down., (ii) It can be converted into dc by using full wave rectifiers., (iii) Power loss in transmission line is minimum., , (c) No, because induced emf in the secondary coil of the transformer is only due to time rate of change of, magnetic flux., (d) Anil is brave/kind/knowledgeable about ac/dc insulators/conductors, and has awareness about safety, measures., , SECTION–E, 24. (a) The total number of electric field lines crossing (or diverging) a surface normally is called electric flux., SI unit of electric flux is volt metre N. m 2 c -1 ., According to Gauss theorem, the electric flux through a closed surface depends only on the net charge, enclosed by the surface and not upon the shape or size of the surface., For any closed arbitrary slope of the surface enclosing a charge the outward flux is the same as that due to, a spherical Gaussian surface enclosing the same charge., Justification: This is due to the fact that, (i), , electric field is radial and, 1, (ii) the electric field E µ, R2, , (b) Electric field inside the shell (hollow charged conducting sphere): The, charge resides on the surface of a conductor. Thus a hollow charged, conductor is equivalent to a charged spherical shell. To find the electric, field inside the shell, we consider a spherical Gaussian surface of radius, , Ei, r, , ®, , r ( < R), concentric with the given shell. If E is the electric field inside the, shell, then by symmetry electric field strength has the same magnitude Ei, on the Gaussian surface and is directed radially outward. Also the, , ®, , dS, R, , ®, , directions of normal at each point is radially outward, so angle between E i and d S is zero at each point., Hence, electric flux through Gaussian surface, ®, , =, , òS, , =, , ò, , ®, , Ei . d S, , Ei dS cos 0 = Ei . 4pr 2, , Now, Gaussian surface is inside the given charged shell, so charge enclosed by Gaussian surface is zero., Hence, by Gauss’s theorem, ®, Ei, S, , ò, Þ, , Ei 4pr 2 =, , ®, , . dS, , 1, =, ´ charge enclosed, e0, , 1, ´ 0, e0, , Þ Ei = 0, , E, , R, , r, , Thus, electric field at each point inside a charged thin spherical shell is zero. The graph is shown in fig.
Page 620 :
CBSE Examination Papers 647, OR, (a) When a capacitor is charged by a battery, 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., , A, , Consider a capacitor of capacitance C. Initial charge on capacitor is, zero. Initial potential difference between capacitor plates is zero., Let a charge Q be given to it in small steps. When charge is given to, capacitor, the potential difference between its plates increases. Let, at any instant when charge on capacitor be q, the potential, q, difference between its plates V = ., C, , B, , + +Q, +, +, +, +, +, +, +, , VA, , –Q –, –, –, –, –, –, –, –, , VB, , VAB = V, , 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,, which may be obtained by integration. Therefore total work, Q, Q, q, W =ò, V dq = ò, dq, 0, 0, C, 1 é q2 ù, =, ê, ú, C ë 2 û, , Q, , =, 0, , æ Q2 0 ö Q2, ç, - ÷=, ç 2, 2 ÷ø 2C, è, , 1, C, , If V is the final potential difference between capacitor plates, then Q = CV, \, , W =, , (CV ) 2, 1, = CV, 2C, 2, , 2, , =, , 1, QV, 2, , This work is stored as electrostatic potential energy of capacitor i.e.,, Electrostatic potential energy, U =, , Q2, 1, = CV, 2C, 2, , 2, , =, , 1, QV, 2, , Energy density: Consider a parallel plate capacitor consisting of plates, each of area A, separated by a, distance d. If space between the plates is filled with a medium of dielectric constant K, then, K e0 A, Capacitance of capacitor,, C =, d, If s is the surface charge density of plates, then electric field strength between the plates, s, E =, Þ s = K e0 E, K e0, Charge on each plate of capacitor, Q = sA = K e 0 E A, \, , Energy stored by capacitor, U =, , ( K e 0 EA) 2, Q2, 1, = K e0 E, =, 2C 2 ( K e 0 A / d) 2, , But Ad = volume of space between capacitor plates, 1, Energy stored, U = K e 0 E 2 Ad, \, 2, U, 1, Electrostatic Energy stored per unit volume, u e =, = Ke 0 E, Ad, 2, , 2, , A d, , 2, , This is expression for electrostatic energy density in medium of dielectric constant K., 1, In air or free space ( K = 1), therefore energy density, u e = e 0 E 2 ., 2
Page 621 :
648 Xam idea Physics–XII, (b) Suppose a capacitor C1 is charged to potential V and another capacitor C1 is uncharged., 1, The energy stored in the capacitor C1 is C1V 2 = U., 2, After connection, the charge redistributes in such a way that the potential difference across the, combination becomes V., From law of conservation of charge, Q = Q¢ + Q¢, C1V = C1V ¢ + C1V ¢, So, common potential after connection, V, V¢ =, 2, Final energy stored in the combination, 1, 1, U ¢ = C1V ¢ 2 + C1V ¢ 2, 2, 2, 1, = C1 ( 2V ¢ 2 ), 2, 2, , 1, 1, V, C1 ´ 2æç ö÷ = C1V 2, 4, 2, è2ø, 1, Hence U ¢ < U , i. e. , the energy stored in the combination is the energy stored initially., 2, 25. 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., A soft iron core in cylindrical form is placed between the coil., One end of coil is attached to suspension wire which also serves as one terminal (T1 ) of galvanometer. The, other end of coil is connected to a loosely coiled strip, which serves as the other terminal (T2 ). The other end, of the suspension is attached to a torsion head which can be rotated to set the coil in zero position. 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. The levelling screws are also provided at the base of the instrument., The pole pieces of the permanent magnet are cylindrical so that the magnetic field is radial at any position of, the coil., Principle and working: When current ( I ) is passed in the coil, 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, N is the number, of turns in a coil., When the magnetic field is radial, as in the case of cylindrical pole pieces and soft iron core, then in every, position of coil the plane of the coil, is parallel to the magnetic field lines, so that q = 90° and sin 90° = 1 ., The coil experiences a uniform coupler., Deflecting torque,, t = NIAB, If C is the torsional rigidity of the wire and q is the twist of suspension strip, then restoring torque = C q, For equilibrium, deflecting torque = restoring torque, i.e., NIAB = C q, NAB, ...(i), \, q=, I, C, i.e., qµI, deflection of coil is directly proportional to current flowing in the coil and hence we can construct a linear, scale., Importance (or function) of uniform radial magnetic field: Torque as current carrying coil in a magnetic, field is t = NIAB sin q
Page 622 :
CBSE Examination Papers 649, In radial magnetic field sin q = 1, so torque is t = NIAB., This makes the deflection ( q) proportional to current. In other words, the radial magnetic field makes the, scale linear., (i) The cylindrical, soft iron core makes the (i) field radial and (ii) increases the strength of the magnetic, field, i.e., the magnitude of the torque., NAB, q, ... (i), (ii) Current sensitivity, S I = æç ö÷ =, C, èIø, NAB, q, Voltage sensitivity, SV = æç ö÷ =, GC, èV ø, Dividing (ii) by (i), SV, 1, 1, SI, =, Þ SV =, G, SI, G, , ... (ii), , Clearly the voltage sensitivity depends on current sensitivity and the resistance of galvanometer. If we, increase current sensitivity and resistance G is larger, then it is not certain that voltage sensitivity will be, increased. Thus, the increase of current sensitivity does not imply the increase of voltage sensitivity., Sensitivity of galvanometer : Current sensitivity : It is defined as the deflection of coil per unit current, flowing in it., NAB, q, Sensitivity, S I = æç ö÷ =, I, C, è ø, Voltage sensitivity: It is defined on the deflection of coil per unit potential difference across its ends, q, NAB, i.e.,, SV =, =, ,, V, Rg . C, where Rg is resistance of galvanometer., Clearly for greater sensitivity number of turns N, area A and magnetic field strength B should be large and, torsional rigidity C of suspension should be small., Dividing (ii) by (i), SV, 1, 1, Þ SV =, =, SI, SI, G, G, Clearly the voltage sensitivity depends on current sensitivity and the resistance of galvanometer. If we, increase current sensitivity and resistance G is larger, then it is not certain that voltage sensitivity will be, increased. Thus, the increase of current sensitivity does not imply the increase of voltage sensitivity., OR, (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, ®, , dB =, , ®, , dl, , q, , P, r, , ®, , m 0 I dl ´ r, 4p, r3, , I, , where m 0 is permeability of free space. Its value is, m 0 = 4p ´ 10 -7 Wb/A-m., 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 as shown in the figure.
Page 623 :
650 Xam idea Physics–XII, ®, , ®, , ®, , The direction of magnetic field dB is perpendicular to the plane containing I dl and r ., Magnetic field at the axis of a circular loop: Consider a circular loop of radius R carrying current I,, 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. Consider a small element of length dl of the coil at point A., ®, , The magnitude of the magnetic induction dB at point P due to this element is given by, ®, , dB =, , m 0 I dl sin a, 4p, r2, , ...(i), , ®, , ®, , ®, , The direction of dB is perpendicular to the plane containing dl and r and is given by right hand screw, ®, , ®, , ®, , rule. As the angle between I dl and r is 90°, the magnitude of the magnetic induction dB is given by,, ®, , dB =, , m 0 I dl sin 90° m 0 I dl, =, ×, 4p, r2, 4p r 2, , ...(ii), , If we consider the magnetic induction produced by the whole of the circular coil, then by symmetry the, components of magnetic induction perpendicular to the axis will be cancelled out, while those parallel to, ®, , the axis will be added up. 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,, ®, , dB x =, But sin a =, , m 0 I dI, , sin a, , 4p r 2, , ...(iii), , R, and r = ( R 2 + x 2 ) 1 / 2, r, ®, , \, , dB x =, , m 0 I dl, 4p r, , 2, , ×, , m 0 IR, R m 0 IR, =, dl =, dl, 3, r, 4p r, 4p ( R 2 + x 2 ) 3 / 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, 2, 4p ( R + x ), 4p ( R + x 2 ) 3 / 2, But, , ò dl = length, , Therefore,, , of the loop = 2pR, B =, , ...(v), , m 0 IR, 4p ( R, , 2, , ®, B = B x i$ =, , ( 2 p R), , + x 2)3/ 2, m 0 IR 2, , 2 (R 2 + x 2 ) 3 / 2, , i$., , If the coil contains N turns, then, B =, , m 0 NIR 2, 2 (R 2 + x 2 ) 3 / 2, , tesla., , ...(vi), , ®, (b) Ampere’s circuital law gives a relationship between the line integral of a magnetic field B and the total, current I which produces this field.
Page 624 :
CBSE Examination Papers 651, Biot-Savart law can be expressed as Ampere’s circuital law by considering the surface to be made up a, large number of loops. The sum of the tangential components of the magnetic field multiplied by the, length of all such elements, gives the result., ®®, ò B . dl = m o I, 26., , (a) When two monochromatic waves, represented by y1 = a cos wt and y 2 = a cos ( wt + f) superpose on, each other, the resultant wave is, y = y1 + y 2, = a cos wt + a cos ( wt + f), = a [cos wt + cos ( wt + f)], f, fö, æ, = 2a cos cos ç wt + ÷, 2, 2ø, è, f, The amplitude of the resultant is, A = 2a cos, 2, Since the intensity is directly proportional to square of the amplitude,, f, Intensity,, A2 = 4a 2 cos 2, 2, Conditions for constructive interference:, If f = 0, ± 2p, ± 4p, . . . the intensity at the points will be, I = 4a 2, Phase difference, f= 2np ( n = 0, 1, 2, 3), l, Path difference =, ´ 2np = nl, 2p, Conditions for destructive interference:, If f = ± p, ± 3p, ± 5p, . . . . there is a destructive interference leading to zero intensity., f= ( 2n – 1) p , n = 1, 2, 3, l, Path difference = ( 2n + 1) , n = 1, 2, 3, 2, (b) The size of the fringe width interference pattern is given by, lD, b=, d, (i) Let s be the size of the source and S be its distance from the plane of the slits. To observe the, s, l, interference fringes, < must be satisfied., S, d, s, l, If width of the source slit is increased, the condition £ is not satisfied, hence interference, S, d, pattern will become less and less sharp., (ii) When monochromatic source is replaced by a white light, the interference pattern due to different, colours of white light overlap on each other incoherently., (1) For central bright fringes for different colours remain at the same position., l, (2) For other points S 2 P - S1 P = b , the blue component will be absent and the fringe will appear red., 2, l, (3) At another point Q, S 2 Q - S1 Q = l b = r the red colour predominates the blue colour., 2, Hence, fringe close to the central bright is red, the far point will appear blue and few-fringes will not, appear clear and no clear fringe pattern is seen., OR
Page 625 :
652 Xam idea Physics–XII, (a) Graph of deviation in d with variation in angle of incidence i :, The homogeneous transparent medium (such as glass) enclosed, by two plane refracting surfaces is called a prism. The angle, between the refracting surfaces is called the refracting angle, or angle of prism. The section cut by a plane perpendicular to, the refracting surfaces is called the principal section of the, prism., Let PQR be the principal section of the prism. The refracting, angle of the prism is A., , P, A, , N1, , O, , (i1–r1), i1, F, , r1, , q, , d, r2, , N2, (i2–r2), G, , i2, , N, , E, Q, , H, R, , A ray of monochromatic light EF is incident on face PQ at angle of incidence i1 . The refractive index of, material of prism for this ray is n. This ray enters from rarer to denser medium and so is deviated towards, the normal FN and gets refracted along the direction FG. The angle of refraction for this face is r1 . The, refracted ray FG becomes incident on face PR and is refracted away from the normal GN 2 and emerges, in the direction GH. The angle of incidence on this face is r 2 (into prism) and angle of refraction (into air), is i 2 . The incident ray EF and emergent ray GH when produced meet at O. The angle between these two, rays is called angle of deviation ‘d‘., Ð OFG = i1 - r1, , and, , Ð OGF = i 2 - r 2, , In D FOG, d is exterior angle, \, , d = Ð OFG + Ð OGF = (i1 - r1 ) + (i 2 - r 2 ), = (i1 + i 2 ) - (r1 + r 2 ), , ...(i), , The normals FN1 and GN 2 on faces PQ and PR respectively, when produced meet at N. Let Ð FNG = q, In D FGN,, , r1 + r 2 + q = 180°, , ...(ii), , In quadrilateral PFNG, Ð PFN = 90° , Ð PGN = 90°, \, , A + 90° + q + 90° = 360° or, , A + q = 180°, , ...(iii), , r1 + r 2 = A, , Comparing (ii) and (iii),, , ...(iv), , Substituting this value in (i), we get, or, From Snell’s law, , d = i1 + i 2 - A, , ...(v), , i1 + i 2 = A + d, sin i1, sin i 2, n =, =, sin r1, sin r 2, , ...(vi), , Minimum Deviation : From equation (v), it is clear that the, angle of deviation depends upon the angle of incidence i1 . As, the path of light is reversible, therefore if angle of incidence, be i 2 , then angle of emergence will be i1 . Thus for two angles, of incidence i1 and i 2 , there will be one angle of deviation., , ...(vii), , d, , dm, If we determine experimentally, the angles of deviation, corresponding to different angles of incidence and then plot i d, (on X-axis) and d (on Y-axis), we get a curve as shown in, figure. Clearly if angle of incidence is gradually increased,, from a small value, the angle of deviation first decreases,, x, i1, i, i2, becomes minimum for a particular angle of incidence and, i, then begins to increase. Obviously for one angle of deviation, ( d), there are two angles of incidences i1 and i 2 , but for one and only one particular value of angle of, incidence (i), the angle of deviation is the minimum. This minimum angle of deviation is represented by, d m . For minimum deviation i1 and i 2 become coincident, i.e.,i1 = i 2 = i (say), , So from (vii), , r1 = r 2 = r (say)
Page 626 :
CBSE Examination Papers 653, Hence from (iv) and (vi), 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,, n =, =, A ö, sin r, sin æç, ÷, è 2 ø, Ðd will be minimum for i = e., (b), , æ A + d min, sin ç, 2, è, µ=, A, sin, 2, sin A, =, A, sin, 2, A, A, 2 sin . cos, 2, 2, =, A, sin, 2, , ö, ÷, ø, , (Q A = dm), , = 2 cos, , A, 2, , If A = 60°,, 60, = 2 cos 30, 2, 3, = 3, =2´, 2, , m = 2 cos, , zzz
Page 627 :
CBSE Examination Paper,, Bhubaneshwer Region-2015 (Code No. 55/3/B), Time allowed: 3 hours, , Maximum marks: 70, , General Instructions: Same as CBSE Examination Paper, Delhi–2015., , SECTION–A, 1., 2., 3., , For the same angle of incidence, the angle of refraction in two media A and B are 25° and 35° respectively. In, which medium is the speed of light less?, Define the term ‘quality factor’ of resonance in series LCR circuit. What is its SI unit?, Two metallic resistors are connected first in series and then in parallel across a d c supply. Plot of I – V graph, is shown for the two cases. Which one represents a parallel combination of the resistors and why?, A, I, , B, , V, , 4., 5., , A point charge +Q is placed in the vicinity of a conducting surface. Trace the field lines between the charge, and the conducting surface., Write the function of a transducer in communication system., , SECTION–B, 6., , In the potentiometer circuit shown, the null point is at X. State with reason, where the balance point will be, shifted when., (a) resistance R is increased, keeping all other parameters unchanged;, (b) resistance S is increased, keeping R constant., P, , R, , X, , A, , S, , 7., 8., , 9., , B, , G, , Distinguish between ‘sky wave’ and ‘space wave’ modes of propagation. Why is the sky wave mode of, propagation restricted to frequencies upto 40 MHz?, Determine the value of the de Broglie wavelength associated with the electron orbiting in the ground state of, hydrogen atom (Given En = – (13.6/n2) eV and Bohr radius r0 = 0.53 Å). How will the de Broglie wavelength, change when it is in the first excited state?, Given the value of the ground state energy of hydrogen atom as – 13.6 eV, find out its kinetic and potential, energy in the ground and second excited states?
Page 628 :
CBSE Examination Papers 655, 10., , A ray of light incident on an equilateral glass prism propagates parallel to the base line of the prism inside it., Find the angle of incidence of this ray. Given refractive index of material of glass prism is 3 ., OR, A biconvex lens of glass of refractive index 1.5 having focal length 20 cm is placed in a medium of refractive, index 1.65. Find its focal length. What should be the value of the refractive index of the medium in which the, lens should be placed so that it acts as a plane sheet of glass?, , SECTION–C, 11., , 12., 13., 14., 15., , 16., , 17., , 18., , (a) Which one of the following electromagnetic radiations has least frequency:, UV radiations, X-rays, Microwaves?, (b) How do you show that electromagnetic waves carry energy and momentum?, (c) Write the expression for the energy density of an electromagnetic wave propagating in free space., Draw the magnetic field lines distinguishing between diamagnetic and paramagnetic materials., Give a simple explanation to account for the difference in the magnetic behaviour of these materials., Explain, with the help of suitable diagram, the two important processes that occur during the formation of p-n, junction. Hence define the terms: depletion region and barrier potential., How is a light emitting diode fabricated? Briefly state its working. Write any two important advantages of, LEDs over the conventional incandescent low power lamps., Two capacitors of capacitance 10 mF and 20 µF are connected in series with a 6 V battery. After the, capacitors are fully charged, a slab of dielectric constant (K) is inserted between the plates of the two, capacitors. How will the following be affected after the slab is introduced:, (a) the electric field energy stored in the capacitors?, (b) the charges on the two capacitors?, (c) the potential difference between the plates of the capacitors?, Justify your answer., Define the activity of a radioactive sample. Write its SI unit., A radioactive sample has activity of 10,000 disintegrations per second (dps) after 20 hours. After next 10 hours, its activity reduces to 5,000 dps. Find out its half life and initial activity., Answer the following:, (a) In what way is diffraction from each slit related to the interference pattern in a double slit experiment?, (b) When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at, the centre of the shadow of the obstacle. Explain, why?, (c) How does the resolving power of a microscope depend on (i) the wavelength of the light used and (ii) the, medium used between the object and the objective lens?, Define the term ‘intensity of radiation’ in photon picture of light., Ultraviolet light of wavelength 2270 Å from 100 W mercury source irradiates a photo cell made of a given, metal. If the stopping potential is –1.3 V, estimate the work function of the metal. How would the photocell, respond to a high intensity (~ 105 Wm–2) red light of wavelength 6300 Å produced by a laser?, OR, , 19., , Set up Einstein’s photoelectric equation using the photon picture of electromagnetic radiation. Explain, briefly how this equation accounts for all the observations in the photoelectric effect., (a) Show that the average power consumed in an inductor L connected to an a.c. source is zero., (b) In a series LR circuit, XL = R and the power factor of the circuit is P1. When a capacitor with capacitance, C such that XC = XL is put in series, the power factor becomes P2. Find out P1/P2., , 20., , Define the electric resistivity of a conductor., Plot a graph showing the variation of resistivity with temperature in the case of a (a) conductor,, (b) semiconductor.
Page 629 :
656 Xam idea Physics–XII, Briefly explain, how the difference in the behaviour of the two can be explained in terms of number density, of charge carriers and relaxation time., 21., , Draw a block diagram of a simple modulator for obtaining amplitude modulated signal., A carrier wave of peak voltage 12 V is used to transmit a message signal. What should be the peak voltage of, the modulating signal in order to have a modulation index of 75%?, , 22., , (a) The light from a clear blue portion of the sky shows a rise and fall of intensity when viewed through a, polaroid which is rotated. Describe, with the help of a suitable diagram, the basic phenomenon/process, which occurs to explain this observation., (b) Show how light reflected from a transparent medium gets polarised. Hence deduce Brewster’s law., , SECTION–D, 23., , Asha’s uncle was advised by his doctor to have an MRI (magnetic resonance imaging) scan of his brain. Her, uncle felt that it was too expensive and wanted to postpone it., When Asha learnt about this, she took the help of her family and when she approached the doctor, he also, offered a substantial discount. She thus convinced her uncle to undergo the test to enable the doctor to know, the condition of his brain. The resulting information greatly helped his doctor to treat him properly., Based on the above paragraph, answer the following questions:, (a) What according to you are the values displayed by Asha, her family and the doctor?, (b) What in your view could be the reason for MRI test to be so expensive?, (c) Assuming that MRI test was performed using a magnetic field of 0.1 T, find the maximum and minimum, values of the force that the magnetic field could exert on a proton (charge = 1.6 × 10–19 C) that was, moving with a speed of 104 m/s., , SECTION–E, 24., , (a) Describe, with the help of a suitable diagram, how one can demonstrate that emf can be induced in a coil, due to the change of magnetic flux. Hence state Faraday’s law of electromagnetic induction., (b) Two loops, one rectangular of dimensions 10 cm × 2.5 cm and second of square shape of side 5 cm are, ®, moved out of a uniform magnetic field B perpendicular to the planes of the loops with equal velocity v as, is shown in the figure., (i) In which case will the emf induced be more?, (ii) In which case will the current flowing through the two loops be less?, Justify your answer., 10 cm, 2.5 cm, , 5 cm, 5 cm, , OR, (a) State the principle of an a.c. generator., (b) Explain briefly, with the help of labelled diagram, its working and obtain the expression for the emf, generated in the coil., (c) Draw a schematic diagram showing the nature of the alternating emf generated by the rotating coil in the, magnetic field during one cycle.
Page 630 :
CBSE Examination Papers 657, , 25., , ®, (a) Derive the expression for the potential energy of an electric dipole of dipole moment P placed in a, ®, uniform electric field E ., Find out the orientation of the dipole when it is in (i) stable equilibrium, (ii) unstable equilibrium., (b) Figure shows a configuration of the charge array of two dipoles., a, +q, , (a), (b), (c), 26., , (a), (b), , (c), , (a), (b), , a, –2q, , +q, r, , P, , Obtain the expression for the dependence of potential on r for r >> a for a point P on the axis of this array, of charges., OR, Define electric flux. Write its SI unit., Using Gauss’s law, obtain the electric flux due to a point charge ‘q’ enclosed in a cube of side ‘a’., Show that the electric field due to a uniformly charged infinite plane sheet at any point distant x from it, is, independent of x., Define a wavefront., Using Huygens’ principle, draw the diagrams to show the nature of the wavefronts when an incident, plane wavefront gets, (i) reflected from a concave mirror, (ii) refracted from a convex lens., Draw a diagram showing the propagation of a plane wavefront from denser to a rarer medium and verify, Snell’s law of refraction., OR, A concave mirror produces a real and magnified image of an object kept in front of it. Draw a ray, diagram to show the image formation and use it to derive the mirror equation., A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12 cm, from P. At what point does the beam converge if the lens is, (i) a convex lens of focal length 20 cm ?, (ii) a concave lens of focal length 16 cm ?, zzz
Page 631 :
658 Xam idea Physics–XII, , Solutions, 1. In medium A, Reason : The refractive index of the medium is, c, sin i, mm =, = 0, sin r, cm, cm =, , Þ, , c 0 ´ sin r, sin i, , rA < rB, , Since, , ( cm ) A < ( cm ) B, 2. The quality factor (Q) of a series LCR circuit is given by the ratio of resonant frequency to frequency band, width of the resonant curve, wr, w L, i.e.,, Q =, = r, w2 - w1, R, Clearly, smaller the value of R, larger is the quality factor and sharper the resonance. Thus quality factor, determines the nature of sharpness of resonance., SI Unit – No unit, 3. Line A represents the parallel combination., , A, , Reason : At a given potential difference V, current in the combination A is, more than in the combination B., , I, , B, , IA > IB, , i.e.,, V, V, and R B =, IA, IB, , Since R A =, , V, , Þ, , RA < RB, , 4., +Q, , Metal plate, , –Q, , 5. Transducer is a device which converts energy from one form to another form., P, , 6., , A, , R, , l, , X, , S, , B, , G, , Let l be the balance length of the segment AX on the potentiometer wire for given resistance R and S.
Page 632 :
CBSE Examination Papers 659, (a) If resistance R is increased, the current flow in the main circuit (or wire AB) will decrease. From relation, rI, the potential gradient along the wire AB will decrease. To balance the emf of the cell, the point X, K =, L, will shift toward the point B, i.e.,, e = kl = k ¢l¢, If k ¢ < k, so, l¢ > l, (b) For the given resistance R, the potential gradient along the wire remain same. Balance length ‘l’ remain, constant., e = kl, and no current flows in the resistance S, if resistance S is increased/decreased. There is no change in the, balance length., 7., S.No., , Sky Wave, , Space Wave, , (i), , In this mode the waves are achieved by the In this mode, the waves travels in straight line from, ionospheric reflection., transmitting antenna to the receiving antenna., , (ii), , Frequency range from few MHz upto 30 to 40 MHz., , The same frequency (or any frequency) are used for, LOS communication., , (iii), , Electromagnetic waves of frequencies higher than, 30 MHz penetrate the ionosphere and escape., , Electromagnetic waves above 40 MHz are used in, satellite communication., , (iv), , Electromagnetic waves follows total, reflection at any layer of ionosphere., , internal In satellite communication the em waves do not, follows total internal reflection at any stage in the, space., , The electromagnetic waves of frequnecy greater than 40 MHz penetrate the ionosphere and escape. So, sky wave mode is restricted upto 40 MHz., 8. In ground state, the kinetic energy of the electron is, + 13.6 eV, K =-E =, 12, \, , K = 13.6 ´ 1.6 ´ 10 - 19 J, = 2.18 ´ 10 - 18 J, de Broglie wavelength, l =, l1 =, , h, =, p, , h, 2m K, 6.63 ´ 10 - 34, , 2 ´ 91, . ´ 10 - 31 ´ 218, . ´ 10 - 18, , = 9.33 ´ 10 - 9 = 0.33 nm, Kinetic energy in the first excited state ( n = 2), 13.6, K = - E=+, eV = + 3.4 eV, 22, = 3.4 ´ 1.6 ´ 10 - 19 J = 0.54 × 10 - 18 J, de Broglie wavelength, l 2 =, =, , h, 2mK, 6.63 ´ 10 - 34, 2 ´ 91, . ´ 10 - 31 ´ 0.544 ´ 10 - 18, , = 2 × 0.33 nm = 0.66 nm, i.e., de Broglie wavelength will increase (or double).
Page 633 :
660 Xam idea Physics–XII, 9. Total energy of the electron in any state is, 13.6, E =eV, n2, For ground state n = 1, \, , E = – 13.6 eV, , Since, , K =–E, = – (– 13.6 eV), = + 13.6 eV, , and, , Potential energy, U = 2E, = 2 (– 13.6 eV), = – 27.2 eV, , For second excited state, n = 3, K =-E =, , \, , + 13 .6, eV, 9, , = 1.51 eV, 2 ´ (– 13 .6 eV), U = 2E =, 9, , and, , = – 3.02 eV, 10. From the figure, we see, r = 30°, We know, sin i, n 21 =, Þ, sin r, , 60°, i, , sin i, 3 =, sin 30°, , Þ, , Þ, , sin i = 3 sin 30° = 3 ´, , Þ, , i = 60°, , 1, 2, , r 30°, 60°, , OR, mg = 1.5, , mm = 1.65, , From lens formula, when lens in a medium, æmg, öæ 1, 1, 1 ö, = çç, - 1÷÷ ç, ÷, fm èm m, R2 ø, ø è R1, , …(i), , When lens in air, 1, = (m g - 1), fa, From equation (i) and (ii), we get, , æ 1, 1 ö, ç, ÷, R, R, è 1, 2 ø, , …(ii)
Page 635 :
662 Xam idea Physics–XII, 11. (a) Microwave, (b) When a charge oscillates with some frequency. It produces an oscillating electric field and magnetic field, in space. So, an electromagnetic wave is produced., The frequency of the em wave is equal to the frequency of oscillation of the charge., Hence energy associated with the em wave comes at the expense of the energy of the source., If the em wave of energy U strikes on a surface and gets completely absorbed, total momentum delivery, U, to the surface is p = ., E, Hence em wave also carry momentum., (c) The em wave consists of oscillating electric and magnetic fields, So net energy density of em wave is, U = UE + UB, U =, , 1, 1 B2, e0 E 2 +, 2, 2 m0, , 12., S, , N, , Paramagnetic, substance, (i), , S, , N, , Diamagnetic, substance, (ii), , A paramagnetic material tends to move from weaker field to stronger field regions of the magnetic field., So, the number of lines of magnetic field increases when passing through it., Magnetic dipole moments are induced in the direction of magnetic field., Paramagnetic materials has a small positive susceptibility., A diamagnetic material tends to move from stronger field to weaker field region of the magnetic field., So, the number of lines of magnetic field passing through it decreases., Magnetic dipole moments are induced in the opposite direction of the applied magnetic field., Diamagnetic materials has a negative susceptibility in the range (–1 £ x < 0)., 13. Two important processes occurring during the formation of a p-n junction are (i) diffusion and (ii) drift., (i) Diffusion: In n-type semiconductor, the concentration of, Electron diffusion, Electron drift, electrons is much greater as compared to concentration of, – +, holes; while in p-type semiconductor, the concentration of, p, n, holes is much greater than the concentration of electrons. When, – +, a p-n junction is formed, then due to concentration gradient, the, Electron, holes diffuse from p side to n side ( p ® n) and electrons, – +, diffuse from n side to p-side ( n ® p). This motion of charge, Hole diffusion, Hole, carriers gives rise to diffusion current across the junction., Hole drift, (ii) Drift: The drift of charge carriers occurs due to electric field., Vo Potential barrier, Due to built in potential barrier, an electric field directed from, n-region to p-region is developed across the junction. This field, + +, causes motion of electrons on p-side of the junction to n-side and, – +, motion of holes on n-side of junction to p-side. Thus a drift P, N, – +, current starts. This current is opposite to the direction of, w, diffusion current., Depletion, Depletion region : The space charge region on either side of the, layer, junction is known as depletion region., Barrier potential : When the p-n junction is in equilibrium, the potential difference across the junction is, called a barrier potential.
Page 636 :
CBSE Examination Papers 663, 14. A light emitting diode is fabricated as, (a) p-region and n-regions are heavily dropped and, (b) the diode must be encapsulated in the transparent cover, so that light can come out., Working, A light emitting diode is simply a forward biased p-n junction which emits spontaneous light radiations. At, the junction, energy is released in the form of photons due to the recombination of the excess minority charge, carrier with the majority charge carrier., Advantages, (i) Low operational voltage and less power consumption., (ii) Fast action and no warm up time required., (iii) Long life and ruggedness., C1C 2, 15. Let Q be the charge on each capacitor. So, Q =, V., C1 + C 2, Initial electric field energy in each capacitor becomes, U1 =, , 1 Q2, 2 C1, , and U 2 =, , 1 Q2, 2 C2, , Initial charge on each capacitor, Q = C1V1 , Q = C 2 V2, , and Q =, , C1C 2, .V, C1 + C 2, , where V1 and V2 are p.d across the capacitors, On inserting the dielectric slab the capacitance of each capacitor becomes, C1¢ = KC1, , and C ¢2 = KC 2, , and equivalent capacitance becomes, KC1 ´ KC 2, C1C 2, C ¢eq =, =K, KC1 + KC 2, C1 + C 2, New charge on the capacitor becomes, æ C1C 2 ö V, Q¢ = C ¢eq V ¢ = K ç, ÷´, è C1 + C 2 ø K, Q¢ =, , C1C 2, .V, C1 + C 2, Q¢ = Q, , (a) New electric field energy becomes, U1¢ =, , 1 Q¢ 2, 1 Q2, =, 2 KC1, 2 KC1, , U 2¢ =, , 1 Q¢ 2, 1 Q2, =, 2 KC 2, 2 KC 2, , i.e., electric field energy decreases in each capacitor., (b) Charge on each capacitor remain same, i.e.,, Q¢ = Q
Page 637 :
664 Xam idea Physics–XII, (c) New p.d. across each capacitor, , and, , V1¢ =, , V1, Q, =, K, KC1, , V2¢ =, , V2, Q, =, K, KC 2, , i.e., p.d across each capacitor decreases., 16. The activity of a radioactive element at any instant is equal to its rate of, decay at that instant. SI unit of activity is becqueral., Let R 0 be initial activity of the sample, and its activity at any, instant ‘t’ is, , R0, , R, , R = R0 e- l t, If t = 20 h, then R = 10000., So,, , 10000 = R 0 e - l, , ´ 20, , t, , …(i), , t, , After next 10 h, i.e.,, at time t ¢ = 30 h, \, , R ¢ = 5000, 5000 = R 0 e - l, , ´ 30, , …(ii), , Dividing (i) by (ii), we get, 10000 e - 20 l, =, = e10 l, 30, l, 5000, e, On taking log on both side, 10 l = log e 2, As we know that, l T1 / 2 = log e 2, \, T1 / 2 = 10 h, From initial time t = 0 to t = 20 h, there are two half lives., 2, , So,, , R, 1, = æç ö÷, R0 è 2 ø, , or, , 10, 000 1, =, R0, 4, , Initial activity at t = 0 is, R 0 = 4 ´ 10000, 17., , = 40000 dps, (a) When a plane wavefront of monochromatic light, illuminates, the slit LN, each point in the slit LN becomes, source of secondary wavelets., The secondary wavelets originating from different points, superpose on each, while travelling towards point C and, point P; at an angle q. However, the superposition of the, secondary wavelets produces a diffraction pattern of, varying intensity, as shown in fig., , To P, L, , From S, , M, , N, , q, , To C
Page 638 :
CBSE Examination Papers 665, Intensity, Io, , O, Angle q, , (b) When a plane wavefront coming from a distant source strikes on a circular obstacle; each point on the, edge becomes a source of secondary wavelets, and diffracted wavefronts interfere constructively at the, centre of the shadow, and produces a bright spot., (c) Resolving Power of a Microscope, The resolving power of a microscope is defined as its ability to form separate images of two close objects, placed near the microscope (e.g., two lines drawn on a sheet)., The minimum distance between close objects for which microscope can just form separate images of the, objects is called the limit of resolution of microscope. Smaller the limit, larger the resolving power., 1.22l, The angular resolving limit of microscope is dq =, ,, 2n sin q, where n is the refractive index of medium between object and objective., n sin q, Resolving power µ, l, (i) Resolving power µ n; therefore resolving power of a compound microscope increases when, refractive index (n) between the object and objective lens increases., 1, (ii) Resolving power µ ; therefore, resolving power of a compound microscope decreases with the, l, increase of wavelength of light used., 18. The intensity of light of certain frequency (or wavelength) is defined as the number if photons passing, through unit area in unit time., For a given wavelength, (l) of light, hc, =W + K, l, (where Vs is stopping potential), = W + eVs, 6.63 ´ 10 - 34 ´ 3 ´ 108, 2270 ´ 10 - 10, \, , = W + 1.6 ´ 10 - 19 ´ ( - 1.3 eV), , æ 6.63 ´ 10 - 34 ´ 3 ´ 108, ö, W =ç, . ÷ eV, - 13, ç 2270 ´ 10 - 10 ´ 1.6 ´ 10 - 19, ÷, è, ø, W = 4.2 eV, , The wavelength of red light 6300 Å >> 2270 Å. So, the energy of red light must be, hc, in eV, E = hn =, el, =, , 6.63 ´ 10 - 34 ´ 3 ´ 108, 1.6 ´ 10 - 19 ´ 6300 ´ 10 - 10, , 198.9, 6.63 ´ 3, eV = 1.973 eV, ´ 10 =, 1.6 ´ 63, 1.6 ´ 63, The energy of red light is very less than its work function, even intensity is very high. Hence no emission of, electron is possible., =
Page 639 :
666 Xam idea Physics–XII, OR, The energy of each photon is hn, where h is plank’s constant and n the frequency of light. Each photon, interacts with one electron., Incident photon energy (hv) is used up in two ways:, (1) A part of this energy is used in liberating the electron from the metal surface which is equal to the work, function (Wo) of the metal., (2) Remaining part of the energy imparts KE to the emitted electrons., hv = Wo + KE, 1, hv – Wo = mv 2 = eVo, 2, Einstein's photoelectric equation is E k = hn - W for a single photon ejecting a single electron., (i) Explanation of frequency law: When frequency of incident photon (n), increases, the kinetic energy of, emitted electron increases. Intensity has no effect on kinetic energy of photoelectrons., (ii) Explanation of intensity law: When intensity of incident light increases, the number of incident, photons increases, as one photon ejects one electron; the increase in intensity will increase the number of, ejected electrons. In other words, photocurrent will increase with increase of intensity. Frequency has no, effect on photocurrent., (iii) Explanation of no time lag law: When the energy of incident photon is greater than work function, the, photoelectron is immediately ejected. Thus, there is no time lag between incidence of light and emission, of photoelectrons., 19. (a) The power, P = Vrms i rms cos f, where, , R, R, ; For ideal inductor R = 0,, \ cos f =, =0, Z, Z, cos f = 0, i. e., power dissipated by an ideal inductor in ac circuit is zero., , cos f =, , \ P = Vrms i rms, (b) For L-R circuit, Power factor, , P1 = cos f =, , R, z, , Z, , R, , =, R, , 2, , XL, , f, , + X L2, , R, , Since X L = R, P1 =, , \, , 1, 2, , For series LCR circuit, Power factor, P2 =, , R, =, 2, , R, R 2 + (X L + X C ) 2, , Since X L = X C, \, , P2 = 1, 1, Ratio of power factors,, , P1, 1, = 2 =, P2, 1, 2, , 20. The resistivity of the material of a conductor is defined as the resistance offered by a conductor of length 1 m, and area of cross-section 1 m 2 . Its S.I. unit is ohm ´ metre (Wm).
Page 640 :
CBSE Examination Papers 667, , Resistivity r (10–8Wm), , (i), 0.4, , I 0.2, , 0, 50, 100, Temperature T(K), , 150, , Resistivity rT of metallic conductor as a function of temperature T ., , (ii), r, , Silicon, , T, , The resistivity of a material is given by, r=, , m, , ne 2 t, (a) In conductors (metal), the electron density n does not depend on the temperature to any appreciable, extent but relaxation time decreases with the rise of temperature., (b) In semiconductors, however electron density increases with the rise of temperature but relaxation time, nearly remain same hence resistivity of the semiconductors decreases with the rise of temperature., 21., , m(t), , x(t), , Am sin wm(t), (Modulating signal), , y(t), , c(t), , AC sin wc(t), , (carrier wave), , Modulation index, µ =, , Bx(t) + cx2(t), , Am, Ac, , 3, Since A c = 12 V, and m = 75% = 0.75 or æç ö÷, è4ø, A, 0.75 = m, \, 12, Am = 0.75 × 12 = 9 V, Þ, 22. (a) Sun emits unpolarised light, and represented as, dots and double arrow. The dots stand for, polarisation perpendicular to the plane and double, arrow in the polarisation of plane., When the unpolarised light strikes on the, atmospheric molecules, the electrons in the, molecules acquire components of motion in both, directions. The charge accelerating parallel to, Polaroid, To observer, double arrow do not radiate energy towards the, observer, so the component of electric field represented by dots radiate towards the observer., If the scattered radiations represented by dots is viewed through an artificial polaroid. It shows the, variation in its intensity with the rotation of the polaroid.
Page 641 :
668 Xam idea Physics–XII, (b) The reflected ray is totally plane polarised, when reflected and, D, refracted rays are perpendicular to each other., A, ÐBOC = 90°, When reflected wave is perpendicular to the refracted wave, the Incident, ip r', reflected wave is a totally polarised wave. The angle of incidence ray, in this case is called Brewster’s angle and is denoted by i B ., O, r, If r ¢ is angle of reflection and r the angle of refraction, then, according to law of reflection, iB = r ¢, and from fig. r ¢ + 90° + r = 180°, E, …(i), Þ, i B + r = 90°, …(ii), Þ, r = ( 90° - i B ), From Snell’s law, refractive index of second medium relative to first medium (air) say., sin i B, sin i B, sin i B, n =, =, =, sin r, sin ( 90° - i B ) cos i B, Þ, n = tan i B, This is known as Brewtser law., \, , Angle of incidence, i B = tan -1 ( n)., , 23. (a) Values displayed by, (i) Asha : Scientific awareness, (ii) Her family : Helpfulness, Empathy, (iii) Doctor : Generous, helping nature., (b) A strong magnetic field is required by using an expensive setup., ®, , (c) Magnetic force on moving charge particle in uniform magnetic field B can be given as, ®, , ®, , ®, , F = q( v ´ B), , F = qvB sin q, ( i) Maximum force at q = 90º, or, , F = qvB, = 1.6 × 10–19 × 104 × 0.1, = 1.6 × 10–16N, (ii) Minimum force at q = 0º and 180º, F=0, i.e., charge particles either move parallel or antiparallel to the magnetic field lines., 24. (a), C1, , N, , I, G, , S, , B, Reflected ray, 90°, Refracted ray, , C
Page 642 :
CBSE Examination Papers 669, (i) A coil connected to a galvanometer is shown in fig. When N pole of a bar magnet is pushed towards, the coil, the pointer in the galvanometer deflects indicating the presence of the electric current in the, coil., (ii) When the N-pole of the bar magnet is pulled away from the coil, the galvanometer shows deflection, in the opposite direction., This experiment shows the generation of electric current in the coil., Faraday’s Laws of Electromagnetic Induction, (i) Whenever there is a change in magnetic flux linked with a coil, an emf is induced in the coil. The, induced emf is proportional to the rate of change of magnetic flux linked with the coil., Df, i.e.,, eµ, Dt, (ii) emf induced in the coil opposes the change in flux, i.e.,, Df, Df, eµÞ e=-k, Dt, Dt, where k is a constant of proportionality., Negative sign represents opposition to change in flux., Df, In SI system f is in weber, t in second, e in volt, when k = 1, e = Dt, Df, If the coil has N-turns, then e = - N, Dt, (b), 10 cm, , 2.5 cm, , 5 cm, 5 cm, , (i) The induced emf e = Blv. If the wire has a cut at any point., Since lsq > lrect . Hence, e sq > e react ., Emf will be more in case of square loop., (ii) The L rect > L sq , i. e. , rectangular wire has more resistance. So, current will be loss in the rectangle., OR, (a) AC generator: A dynamo or generator is a device which converts mechanical energy into electrical, energy., w, Principle: It works on the principle of electromagnetic, Armature coil, induction. When a coil rotates continuously in a magnetic, b, field, the effective area of the coil linked normally with the, c, magnetic field lines, changes continuously with time. This, variation of magnetic flux with time results in the, production of an alternating emf in the coil., N B, S, Field magnet, (b) Working of AC generator: When the armature coil is, a, rotated in the strong magnetic field, the magnetic flux, d, linked with the coil changes and the current is induced in, B1, the coil, its direction being given by Fleming’s right hand, R1, rule. Considering the armature to be in vertical position, Brushes Load, Slip rings, RL, and as it rotates in anticlockwise direction, the wire ab, moves upward and cd downward, so that the direction of, R2, B2, induced current is shown in fig. In the external circuit, the
Page 643 :
670 Xam idea Physics–XII, current flows along B1 R L B 2 . The direction of current remains unchanged during the first half turn of, armature. During the second half revolution, the wire ab moves downward and cd upward, so the, direction of current is reversed and in external circuit it flows along B 2 R L B1 . Thus the direction of, induced emf and current changes in the external circuit after each half revolution., Expression for Induced emf: If N is the number of turns in coil, f the frequency of rotation, A area of, coil and B the magnetic induction, 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., Current produced by an ac generator cannot be measured by moving coil ammeter; because the average, value of ac over full cycle is zero., The source of energy generation is the mechanical energy of rotation of armature coil., (c), , e0, e, T/4, , T/2, , t, , e = e0 sin wt, , 25. (a) The potential energy of an electric dipole in 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., The work done in this process may be supposed to be done in two parts., (i) The work done (W1 ) in bringing the dipole perpendicular to electric field from infinity., (ii) Work done (W2 ) in rotating the dipole such that it finally makes an angle q from the direction of, electric field., Let us suppose that the electric dipole is brought, +q, qE, from infinity in the region of a uniform electric, ®, , field such that its dipole moment p always, 2l, From, remains perpendicular to electric field. The, infinity, electric forces an charges +q and - q are qE and, –q, qE, qE, along the field direction and opposite to field, direction respectively. As charges + q and - q, traverse equal distance under equal and opposite, forces; therefore, net work done in bringing the dipole in the region, +q, qE, of electric field perpendicular to field-direction will be zero, i.e.,, W1 = 0., p, Now the dipole is rotated and brought to orientation making an, angle q with the field direction (i.e., q1 = 90° and q 2 = q° ), qE, –q, therefore, work done, W2 = pE (cos q1 - cos q 2 ), = pE (cos 90° - cos q) = - pE cos q, \ Total work done in bringing the electric dipole from infinity, i. e. , Electric potential energy of electric, dipole., U = W1 + W2 = 0 - pE cos q = - pE cos q, In vector form, , ® ®, , U =- p × E, , For stable equilibrium, q = 0°, For unstable equilibrium, q = 180° .
Page 644 :
CBSE Examination Papers 671, OR, (a) Electric flux: It is defined as the total number of electric field lines passing through an area normal to its, surface., ®, , ®, , Also, f = ò E . d s, , The SI unit is Nm2/C or volt-metre., q, independent to the side length ‘a’., e0, , (b) Electric flux through the cube, f =, , (c) Let electric charge be uniformly distributed over the surface of a thin, non-conducting infinite sheet. Let, the surface charge density (i.e., charge per unit surface area) be s. We need to calculate the electric field, strength at any point distance x from the sheet of charge., q=90o, dS3, o, S1 E 90 S3, , S2, , dS1, E, q=0o, , dS2, E, q=0o, B, , A, Sheet, , To calculate the electric field strength near the 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). 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., Total electric flux, ®, , ò SE, or, , ®, , ò SE, , ®, , =, , • d S, , ®, , ò SE, 1, , ®, , • dS, , =, , òS, , =E, , 1, , ò, , ®, , • d S1 +, , ®, , ò SE, 2, , E dS1 cos 0° +, dS1 + E, , ò, , ®, , • d S2, , òS, , 2, , +, , ®, , ò SE, 3, , ®, , • d S3, , E dS 2 cos 0° +, , òS, , 3, , E dS 3 cos 90°, , dS 2 = E a + E a = 2E a, , \ Total electric flux = 2E a ., As s is charge per unit area of sheet and a is the intersecting area, the charge enclosed by Gaussian, surface = s a, According to Gauss’s theorem,, 1, Total electric flux =, ´ (total charge enclosed by the surface), e0, i.e.,, , 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. If the surface charge density s is negative the, electric field is directed towards the surface charge., 26. (a) The sources of light emitting waves having a constant initial phase difference are called coherent sources., The sources of light emitting waves with a random phase difference are called incoherent sources.
Page 645 :
672 Xam idea Physics–XII, (b), Spherical wavefront at radius R/2, Concave mirror, at radius R, , Incident, plane wave, , F, , Convex lens, , F, Spherical wavefront, at radius f, , (c) Figure shows the refraction from denser medium ( n1 ) into rarer medium ( n 2 < n1 ). The angle of, refraction r > i and from Snell’s law, ...(i), n1 sin i = n 2 sin r, At critical angle, i = i c, \, , n1 sin i c = n 2 sin 90° = n 2, n, sin i c = 2, n1, , ...(ii), , From equation (i), we have, n2, sin i, =, = ratio of the speed of light, n1, sin r, Let v1 and v 2 be the speed of wavefronts in medium n1 and n 2 ., v2, sin i, Hence,, =, sin r, v1, , ...(iii), , OR, (a) Draw DN^A ¢P, In DABN and DA ¢B ¢N, AB, AN, =, ¢, ¢, AB, A ¢N, In DDNF and DB ¢A ¢F, DN, FN, =, A ¢B ¢ A ¢F, But AB = DN, AN, FN, FN, =, =, \, A ¢N A ¢N A ¢N – FN, If aperture of mirror is very small, the point N will, be very near to P,, So, FN = FP, , B, A', , D, , C, A, , N, , F, f, u, , B', , v, , P
Page 646 :
CBSE Examination Papers 673, \, , AP, FP, FP, =, =, A ¢P A ¢P A ¢P – FP, –f, –u, =, – v –( v – f ), , uv – uf = vf, Dividing both sides by uvf , we get, 1 1 1, – =, f, v u, 1 1 1, + =, v u f, (b), , (i) Here, the object is virtual image is real., From lens formula,, 1 1 1, = f, v u, , f = + 20 cm, , Since f = + 20 cm, and u = + 12 cm from point P, 1, 1, 1, = \, 20 v 12, 1, 1, 1, =, +, v 20 12, 20 ´ 12 240 30, v =, =, =, = 7.5 cm, 20 + 12, 32, 4, , P', , P, , v, 12 cm, , Image would be at distance +7.5 cm from optical centre of the lens., (ii), , P, , P', , 12 cm, f = – 16 cm, , From lens formula,, 1 1 1, = f, v u, 1, 1, 1, = - 16 v 12, 1, 1, 1, =+, v, 16 12, 16 ´ 12, 16 ´ 12, v =, =, = 48 cm, - 12 + 16, + 4, Image would be at distance + 48 cm from the optical centre of the lens., zzz
Page 647 :
CBSE Examination Paper,, Chennai Region-2015 (Code No. 55/1/MT), Time allowed: 3 hours, , Maximum marks: 70, , General Instructions: Same as CBSE Examination Paper Delhi-2015., , SECTION–A, 1., , Depict the electric field lines due to two positive charges kept a certain distance apart., , 2., , The line AB in the ray diagram represents a lens. State whether the lens is convex or concave., A, , B, , 3., , Define the term ‘wattless current’., , 4., , Why are repeaters used in communication system?, , 5., , I – V graph for two identical conductors of different materials A and B is shown in the figure. Which one of, the two has higher resistivity?, A, , B, , I, , O, , V, , SECTION–B, 6., , Deuterons and a-particles are accelerated through the same potential. Find the ratio of the associated de, Broglie wavelengths of the two.
Page 648 :
CBSE Examination Papers 675, 7., , The energy levels of an atom are given below in the diagram., 0 eV, –1 eV, , A, , B, , D, , C, , E, , –3 eV, , –10 eV, , Which of the transitions belong to Lyman and Balmer series? Calculate the ratio of the shortest wavelengths, of the Lyman and the Balmer series of the spectra., 8. For an amplitude modulated wave, the maximum amplitude is found to be 10 V while the minimum, amplitude is 2 V. Determine the value of modulation index m., What would be the value of m if the minimum amplitude is zero volt? Why is m generally kept less than 1?, 9. A ray of light incident on one of the faces of a glass prism of angle ‘A’ has angle of incidence 2A. The, refracted ray in the prism strikes the opposite face which is silvered, the reflected ray from it retracing its, path. Trace the ray diagram and find the relation between the refractive index of the material of the prism and, the angle of the prism., OR, An object is placed 40 cm from a convex lens of focal length 30 cm. If a concave lens of focal length 50 cm is, introduced between the convex lens and the image formed such that it is 20 cm from the convex lens, find the, change in the position of the image., 10. In the circuit diagram shown, AB is a uniform wire of resistance 15 W and length 1 m. It is connected to a cell, E1 of emf 2 V and negligible internal resistance and a resistance R. The balance point with another cell E2 of, emf 75 mV is found at 30 cm from end A. Calculate the value of the resistance R., R, , E1, , X, , A, , E2, , B, G, , SECTION–C, 11. A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to, another uncharged 300 pF capacitor. Calculate how much electrostatic energy is lost in the process. What is, the source of energy loss?, 12. How are the following types of electromagnetic waves produced:, (i) Microwaves, (ii) Infra-red waves, Write two important uses of each of the above electromagnetic radiations.
Page 649 :
676 Xam idea Physics–XII, 13. Draw the circuit arrangement for studying the V – I characteristics of a p-n junction diode (i) in forward bias, and (ii) in reverse bias. Draw the typical V – I characteristics of a silicon diode., Describe briefly the following terms:, (i) “minority carrier injection” in forward bias, (ii) “breakdown voltage” in reverse bias., 14. (a) Give a brief description of the basic elementary process invovled in the photoelectric emission in, Einstein’s picture., (b) When a photosensitive material is irradiated with the light of frequency v, the maximum speed of, electrons is given by Vmax. A plot of V2max is found to vary with frequency v as shown in the figure., , V2 max, , 0, , n, , v, , l, , Use Einstein’s photoelectric equation to find the expressions for, (i) Planck’s constant and (ii) work function of the given photosensitive material, in terms of the, parameters l, n and mass m of the electron., 15. Draw a circuit diagram to study the input and output characteristics of an n-p-n transistor in common emitter, configuration. Explain briefly how this arrangement is used to obtain the typical input/output characteristics, of a transistor. Draw the graphs showing the nature of input/output curves., OR, , 16., 17., , 18., , 19., , Draw a labelled circuit diagram of a full wave rectifier. Explain its underlying principle and working. Depict, the input and output waveforms., Describe briefly the role of a capacitor in filtering., Draw a schematic sketch of a cyclotron. Explain its working principle. Obtain the necessary mathematical, expression to show how this machine is used to accelerate charged particles., Write the expression for the resultant intensity at a point due to the superposition of two monochromatic waves, y1 = a cos wt, y 2 = a cos ( wt + f), Where f is the phase difference between the two waves and a and w denote the amplitude and angular, frequency., In Young’s double slit experiment using monochromatic light of wavelength l, the intensity of light at a point, on the screen where path difference is l is k units. Find the intensity at a point on the screen where path, difference is l 4., Suppose you wish to transmit an electronic signal in the audio frequency range over a long distance directly., Write briefly the three important factors which prevent you from doing so and explain how you overcome, these factors., (a) Plot a graph showing the variation of binding energy per nucleon as a function of mass number., Which property of nuclear force explains the approximate constancy of binding energy in the range, 30 < A < 170?
Page 650 :
CBSE Examination Papers 677, (b) A radioactive nucleus ‘A’ undergoes series of decays shown in the following scheme:, A, , a, , b, , A1, , g, , A2, , A3, , If the mass number and atomic number of A3 are 176 and 69 respectively, find the mass number and atomic, number of A., 20. Discuss briefly, with the help of a suitable diagram, what happens when unpolarised light passes through two, identical polaroids where the orientation of one polaroid is fixed and the second is rotated with respect to the, one. Draw a graph showing the dependence of intensity of transmitted light on the angle between the, polariser and analyser. Explain clearly how one understands this variabtion using Malus’ law., 21. A resistor of 100 W and a capacitor of 100 p mF are connected in series to a 220 V, 50 Hz a.c. supply., (a) Calculate the current in the circuit., (b) Calculate the (rms) voltage across the resistor and the capacitor. Do you find the algebraic sum of these, voltages more than the source voltage? If yes, how do you resolve the paradox?, 22. When a metallic conductor is subjected to a certain potential V across its ends, discuss briefly how the, phenomenon of drift occurs. Hence define the term ‘drift velocity’ of charge carriers and show that the, ®, , ®, , current density j is related to the applied electric field E by the relation, ®, , ®, , j =sE, , where s defines the conductivity of the material., , SECTION–D, 23. Deepika and Ruchika were asked by their teacher to perform an experiment using a galvanometer. Before, doing the experiment they were very keen to know the different parts of the galvanometer which was given to, them in the form of a small box. They approached the teacher and asked for the permission. The teacher, thought it would be a good idea if the galvanometer be opened before the whole class and explained its, construction and working to all of them., Based on the above paragraph, answer the following questions:, (a) What, in your opinion, were the qualities displayed by Deepika, Ruchika and the teacher?, (b) State briefly the working principle of the galvanometer., (c) What is the shape of the magnets used and why is it so designed?, , SECTION–E, 24. A right circular cylinder of length ‘a’ and radius ‘r’ has its centre at the origin and its axis along the x-axis so, that one face is at x = + a/2 and the other at x = – a/2, as shown in the figure. A uniform electric field is acting, ®, , ®, , parallel to the x-axis such that E = E 0 i$ for x > 0 and E = - E 0 i$ for x < 0., y-axis, , ®, E, , ®, E, , x-axis, , O, , x = –a/2, , x = a/2
Page 651 :
678 Xam idea Physics–XII, Find out the flux (i) through the flat faces, (ii) through the curved surface of the cylinder. What is the net, outward flux through the cylinder and the net charge inside the cylinder?, OR, (a) Show, using Gauss’s law, that for a parallel plate capacitor consisting of two large plane parallel, conductors having surface charge densities + s and – s, separated by a small distance in vacuum, the, electric field., (i) in the outer regions of both the plates is zero., (ii) is s e 0 in the inner region between the charged plates., Hence obtain the expression for the capacitance of a parallel plate capacitor., (b) Explain what is the effect of inserting a dielectric slab of dielectric constant k in the intervening space, inside the plates on, (i) the electric field,, (ii) the capacitance of the capacitor., 25. (a) State two main considerations taken into account while choosing the objective in optical telescopes with, large diameters., (b) Draw a ray diagram for the formation of image by a reflecting type telescope. What is its magnifying, power?, (c) What are the advantages of a reflecting type telescope over the refracting type?, OR, (a) Draw a ray diagram for the formation of image by a compound microscope in normal adjustment., (b) Obtain the expression for the minimum separation between the two points seen as distinct in a, microscope. What is its relation with the resolving power? Mention the factors by which the resolving, power of a microscope can be increased., 26. (a) Explain the meaning of mutual inductance of a solenoid S1 with respect to the solenoid S2., (b) Consider two concentric circular coils, one of radius r1 and the other of larger radius r2 (r1 << r2), placed, coaxially with centres coinciding with each other. Obtain the expression for the mutual inductance of the, arrangement., (c) In an experimental arrangement of two coils C1 and C2 placed coaxially parallel to each other, find out, the expression for the emf induced in the coil C1 (of N1 turns) corresponding to the change of current I2 in, the coil C2 (of N2 turns)., OR, (a) State Ampere’s circuital law. Show that the magnetic field B at a distance r outside the straight infinite, wire carrying current I is tangential and is given by B = m 0 I (2pr )., (b) Consider a long straight cylindrical wire of circular cross-section of radius a, as shown in the figure. The, current I is uniformly distributed across this cross-section. Calculate the magnetic field B in the region, r < a and r > a. Plot a graph of B versus r from the centre of the wire., I, a, , zzz
Page 652 :
CBSE Examination Papers 679, , Solutions, SECTION–A, 1., N, , q1, , d, , q2, , 2. It is a convex or converging lens., Reason: The refracted ray is bending towards the principal axis., 3. Wattless current is that component of alternating current which does not contribute to power consumption in, the main circuit where the power is transferred from the source., Or, When pure inductor and/or pure capacitor is connected to ac source, the current flows in the circuit, but with, no power loss, such a current is called the wattless current., 4. Repeaters extend the range of communication system, because a repeater picks up a signal, amplifies it, and, retransmits it to the next repeater., 5. The resistivitiy of material B is higher., Reason: If the same amount of the current flows through them, then VB > VA , and from Ohm’s law, R B > R A . Hence the resistivity of the material B is higher., , SECTION–B, 6. Kinetic energy gained by the charged particles at a given potential can be given by, p2, 1, mv 2 =, = qV, 2, 2m, p =, , \, , 2mqV, , de Broglie wavelength associated with a moving particle is, h, l=, p, \ de Broglie wavelengths associated with deutron and a-particles can be given as, h, ld =, 2m d q d V, la =, ld : la =, , h, 2m a q a V, ma qa :, , q d = e , q a = 2e, m d = 2m p, \, , md qd, and m a = 4m p, , ld : la = 2 : 1, , 7. Transition C and E belong to Lyman series., Reason: In Lyman series, the electron jumps to lowest energy level from any higher energy levels., Transition B and D belong to Balmer series., Reason: The electron jumps from any higher energy level to the level just above the ground energy level.
Page 653 :
680 Xam idea Physics–XII, The wavelength associated with the transition is given by, hc, l=, DE, Ratio of the shortest wavelength, lL : lB =, =, , hc, hc, :, DE L DE B, 1, 1, :, = 3 : 10, 0 - ( - 10) 0 - ( - 3), , 8. Amax = Ac + Am = 10 V, Amin = Ac – Am = 2 V, Modulation index =, , A max – A min 10 – 2 8, 2, =, =, =, A max + A min 10 + 2 12 3, , If the value of minimum amplitude, Ac – Am = 0, \, , Ac = Am = 5 V, A, \, m= m =1, Ac, Modulation index is generally kept less than one, to transmit the distortion less information to the distant, receivers., 9. From Snell’s law, , x, , m =, , sin i, sin 2A, =, sin r, sin r, , …(i), , A, Q, , In DXQR,, , i = 2A, , 90°, , R, , r, , ( 90° - r ) + A + 90° = 180°, r =A, , or, , …(ii), , y, , From Eq. (i) and (ii), we get, m =, \, , sin 2A, = 2 cos A, sin A, , A = cos - 1 (m / 2), OR, , For the convex lens, f 1 = + 30 cm and object distance u1 = - 40 cm, therefore,, 1, 1, 1, =, f1, v1 u1, 1, 1, 1, =, + 30 v1, - 40, 1, 1, 1, 1, =, =, v1 30 40 120, Þ, , P, , v1 = + 120 cm, a real image is formed., , On introducing a concave lens, f 2 = - 50 cm, and u 2 = 120 - 20 = + 100 cm from the concave lens., 1, 1, 1, =, f2, v2, u2, , z
Page 654 :
CBSE Examination Papers 681, 1, 1, 1, =, - 50 v 2, + 100, 1, 1, 1, 1, =+, =v2, 50 100, 100, , \, , v 2 = - 100 cm, A virtual image is formed at the distance of 100 cm from the concave lens., The change in position between the real image and the virtual image is 100 cm + 100 cm = + 200 cm, to the left of its origional position., 10. Current drawn from the cell, E1 = 2 V, I =, , E1, 2, =, 15 + R 15 + R, , Potential drop across the wire AB, VAB = I ´ 15 =, , 2 ´ 15, 30, =, 15 + R 15 + R, , Since wire length is 1 m or 100 cm., So, potential gradient along the wire,, VAB, 30, K =, =, 100 cm 100 (15 + R), At the balance point, E 2 = k l2, 75 mV =, , 30, ´ 30 cm, 100 (15 + R), , 75 ´ 10 - 3 ´ 100 (15 + R) = 900, 15 + R =, \, , 9000, 75, , R = 120 - 15 = 105 ohm., , SECTION–C, 11. Before connecting another capacitor, charge on the capacitor, C1 is, q1 = C1V1, , –, , +, , C1 = 600 pF, , V1 = 200 V, = 600 ´ 200 V = 120000 pC, On connecting another capacitor C 2 charge distribution occurs, till the system comes to a common potential, q1 = q1¢ + q¢2, C1 –, , \, , +, , 120000 pC = C1V + C 2 V, = (600 pF + 300 pF) V, 120000 400, volt, V=, =, 900, 3, , Amount of energy loss, , +, , DU = U f - Ui, 1, 1, = (C1 + C 2 ) V 2 - C1V12, 2, 2, =, , 1, æ 400 ö, ´ 900 pF ´ ç, ÷, 2, è 3 ø, , 2, , -, , 1, ´ 600 pF ´ ( 200) 2 = - 4 ´ 10 6 pJ, 2, , This energy loss appears in the form of spark light and sound., , C2, , –
Page 655 :
682 Xam idea Physics–XII, 12. (i) Microwaves are produced by special Vacuum tubes, namely; klystrons, magnetrons and gunn diodes., Uses: Radar system, ovens, air craft navigation., (ii) Infrared are produced by vibrations of atoms and hot bodies., Uses: Physical therapy, detectors in earth satellites., Voltmeter(V), , 13., , p, , Voltmeter(V), , n, , p, , n, , Milliammeter, (mA), , Microammeter, (mA), Switch, , Switch, + –, , – +, (a), , (b), , I(mA), 100, 80, 60, 40, 20, 100 80 60 40 20, , Vbr, , 0.2 0.4 0.6 0.8 1.0, , V(V), , 10, 20, 30, , I(mA), , (c), , (i) Minority Carrier Injection: Due to the applied voltage, electrons from n-side cross the depletion, region and reach p-side (where they are minority carriers). Similarly, holes from p-side cross this, junction and reach the n-side (where they are minority carriers). This process under forward bias is, known as minority carrier injection., (ii) Breakdown Voltage: It is a critical reverse bias voltage at which current is independent of applied, voltage., , le c, , hn, , toe, , t, en, cid n, In hoto, P, , Einstein explained photoelectric effect on the basis of quantum theory. The, main points are, , tro, n, , 14. (a) Einstein’s Explanation of Photoelectric Effect: Einstein’s Photoelectric Equation, , Ph, o, , EK, 1. Light is propagated in the form of bundles of energy. Each bundle of, energy is called a quantum or photon and has energy hn, where, h = Planck’s constant and n = frequency of light., w, Metal, 2. The photoelectric effect is due to collision of a photon of incident light and, a bound electron of the metallic cathode., 3. When a photon of incident light falls on the metallic surface, it is completely absorbed. Before being, absorbed it penetrates through a distance of nearly 10 -8 m (or 100 Å). The absorbed photon, transfers its whole energy to a single electron. The energy of photon goes in two parts: a part of, energy is used in releasing the electron from the metal surface (i. e. , in overcoming work function), and the remaining part appears in the form of kinetic energy of the same electron.
Page 657 :
CE, , =, 10, , VCE = 5V, , IB, , 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 obtained by, plotting collector current IC 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, , 684 Xam idea Physics–XII, , V, , VBE, , The characteristic curves show:, When collector-emitter voltage VCE is increased from zero, 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 value, of VCE upto which collector current IC changes is called the knee voltage Vknee ., Definition of Input resistance:, These characteristic curves are obtained by plotting the base current I B versus base-emitter voltage VBE for, fixed collector-emitter voltage VCE . The observations of I B versus VBE are repeated for fixed values of VCE ., Definition of Amplification Factor:, Common Base Current Amplification factor (or ac current gain) a is defined as the ratio of small change in, collector current to the small change in emitter current for constant value of collector voltage in common, base configuration, i.e.,, æ DI C ö, ÷÷, a = çç, è DI E øVC = constant, Practical value of a ranges from 0.9 to 0.99 for junction transistor., Common emitter current amplification factor (or ac 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, b = çç, è DI B, , ö, ÷÷, øVC, , = constant, , Current gain, , æ DI, b = çç C, è DI B, , ö, ÷÷, øVCE, , We take the active region of output characteristics i, the, region where collector current ( I C ) is almost, independent of VCE ., Now we choose any two characteristic curves for given, values of I B and find the two corresponding values of, IC ., æ DI, b = çç C, è DI B, , Then, , ö (IC ) 2 - (IC )1, ÷÷ =, ø (I B ) 2 - (I B )1, , From graph ( I C ) 1 = 5 × 2 mA, ( I C ) 2 = 7 × 3 mA, ( I B ) 1 = 30 mA,, b=, , ( I B ) 2 = 40 mA, , (7 × 3 - 5 × 2) mA, 2 × 1 ´ 10 - 3, =, = 210, ( 40 - 30) mA, 10 ´ 10 - 6, , Collector current in (IC) mA, , Determination of Current Gain, , 12.5, Base current ( IB), , 10, , 60 µA, 50 µA, 40 µA, , 7.5, , 30 µA, , 5, , 20 µA, 10 µA, , 2.5, 0.5, , 1, , 1.5, , 2, , 2.5, , 3, , 3.5, , Collector to emitter voltage (VCE) in volts, , 4
Page 658 :
CBSE Examination Papers 685, OR, , Working: The AC input voltage across secondary, s1 and s 2 changes polarity after each half cycle., Suppose during the first half cycle of input AC signal,, the terminal s1 is positive relative to centre tap O and s 2, is negative relative to O. Then diode D1 is forward, biased and diode D2 is reverse biased. Therefore, diode, D1 conducts while diode D2 does not. The direction of, current (i1 ) due to diode D1 in load resistance R L is, directed from A to B. In next half cycle, the terminal s1 is, negative and s 2 is positive relative to centre tap O. The, diode D1 is reverse biased and diode D2 is forward, biased. Therefore, diode D2 conducts while D1 does, not. The direction of current (i 2 ) due to diode D2 in, load resistance R L is still from A to B. Thus the current, in load resistance R L is in the same direction for both, half cycles of input AC voltage. Thus for input AC, signal the output current is a continuous series of, unidirectional pulses., In a full wave rectifier, if input frequency is f hertz,, then output frequency will be 2f hertz because for, each cycle of input, two positive half cycles of output, are obtained., Capacitor Filtering: A capacitor of high value,, filters out the ac ripples and give a pure dc voltage., , P1 N1, , p1, Input AC signal, to be rectified, , Rectification: Rectification means conversion of ac, into dc. A p-n diode acts as a rectifier because an ac, changes polarity periodically and a p-n diode conducts, only when it is forward biased; it does not conduct when, reverse biased., , i1, , s1, D1, Centre tap, , +, A, , s2, P2 N2, , i2, , Output, , RL, , p2, B, , D2, –, , Waveform, at P1, , Waveform, at P2, , O, , T, , T, 2, , 3, T, 2, , t, 2T, , O, , Due to, D1, Output, waveform O, (across RL), , 3, T, 2, , T, , T, 2, , Due to, D1, , Due to, D2, , T, , T, 2, , 2T, , t, , Due to, D2, , 3, T, 2, , 2T, , t, , In absence of external load, the capacitor remains charged to the peak voltage of the rectified output, voltage. When a load is connected in parallel to the capacitor it gets discharged through the load and the, voltage across the capacitor begins to fall. In the next half cycle of the rectified output voltages. The, capacitor again gets charged to the peak voltage. It means, the output voltage obtained by using capacitor, input filter is nearer to the peak voltage of the rectified voltage., 16. Cyclotron: The cyclotron, devised by Lawrence and Livingston, is a device for accelerating charge particles, to high speed by the repeated application of accelerating potentials., Construction: The cyclotron consists of two flat semi-circular metal, boxes called ‘dees’ and are arranged with a small gap between them. A, source of ions is located near the mid-point of the gap between the dees, (fig.). The dees are connected to the terminals of a radio frequency, oscillator, so that a high frequency alternating potential of several million, cycles per second exists between the dees. Thus dees act as electrodes., The dees are enclosed in an insulated metal box containing gas at low, pressure. The whole apparatus is placed between the poles of a strong, electromagnet which provides a magnetic field perpendicular to the plane, of the dees., , Magnetic Pole, N, S, Dee, , Dee, , Magnetic Pole, S
Page 659 :
686 Xam idea Physics–XII, Working: The principle of action of the, apparatus is shown in figure. The, positive ions produced from a source S, at the centre are accelerated by a dee, which is at negative potential at that, moment. Due to the presence of, perpendicular magnetic field the ion, will move in a circular path inside the, dees. 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., , Magnetic field, out of paper, , Deflecting plate, , Exit port, Charged, particle, , D1, D2, , Electric Oscillator, , Role of electric field, Electric field accelerates the charge particle passing through the gap., Role of magnetic field, As the accelerated charge particle enters normally to the uniform magnetic field, it exerts a magnetic force in, the form of centripetal force and charge particle moves on a semicircular path of increasing radii in each dee, ( D1 or D2 ) alternatively., Expression for period of revolution and frequency:, Suppose the positive ion with charge q moves in a dee with a velocity v, then,, qvB =, , mv 2, r, , or r =, , mv, qB, , ...(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., The angular velocity w of the ion is given by,, v qB, (from eq. i), ...(ii), w= =, r, m, The time taken by the ion in describing a semi-circle, i.e., in turning through an angle p is,, p pm, ...(iii), t = =, w Bq, Thus the time is independent of the speed of the ion i.e., although the speed of the ion goes on increasing with, increase in the radius (from eq. i) when it moves from one dee to the other, yet it takes the same time in each dee., m, From eq. (iii) it is clear that for a particular ion, being known, B can be calculated for producing resonance, q, with the high frequency alternating potential., 17. Suppose two coherent waves superpose, as given, at a point the resultant displacement at the point will be, y = y1 + y 2, = a cos wt + a cos ( wt + f), fö, æ fö, æ, a[cos w t + cos (w t + f)] = 2a cos ç ÷ cos ç wt + ÷, 2ø, è2ø, è
Page 660 :
CBSE Examination Papers 687, æ fö, The amplitude of the resultant displacement = 2a cos ç ÷, è2ø, Since the intensity is proportional to the square of the amplitude, the intensity at that point will be, æ fö, I = 4I 0 cos 2 ç ÷, è2ø, where I0 represents the intensity produced by each one of the individual sources; I0 in proportional to a2., f, Intensity, I = 4I 0 cos 2, 2, When path difference is l, phase difference is 2p, I = 4I 0 cos 2 p = 4I 0 = k, (given), …(i), \, l, When path difference, D = , the phase difference, 4, 2p, 2p l p, D=, ´ =, f¢ =, l, l, 4 2, f¢, 2, (since k = 4I 0 ), \, I ¢ = 4I 0 cos, 2, p k, = k cos 2, =, 4 2, 18. Three factors that prevent us from sending the signals directly are:, (i) Size of the antenna or aerial, (ii) Effective power radiated by the antenna, (iii) Intermixing of signals, To overcome these factors, (i) Size of antenna should be comparable to wavelength (at least l 4 in dimension)., 1, (ii) Power radiated is proportional to, . This implies that for the same antenna length power radiated, l2, increases with decreasing wavelength (l) ., (iii) Message signal should be modulate a high frequency carrier wave so that a band of frequencies can be, allotted to each message signal for its transmission., 19. (a) The graph shown in figure., , Binding Energy per Nucleon (in MeV), , 9.0, O16, 8.0C12, F18, He4 N14, 7.0, , Fe56, U238, , 6.0, Li7, , 5.0, 4.0, 3.0, 2.0, , H2, , 1.0, 0.0, 0, , 20, , 40, , 60, , 80 100 120 140 160 180 200 220 240, Mass Number, , Nuclear forces are short ranged and saturated., a, , (b) 70 A 180 ¾®, , b, , –, , 176, ¾®, 68 A1, , 176, 69 A 2, , g, , ¾®, , 176, 69 A 3, , Mass number of A is 180 and atomic number of A is 70.
Page 661 :
688 Xam idea Physics–XII, Or, a, , 180, ¾®, 70 A, , b, , +, , 176, ¾®, 70 A1, , 176, 69 A 2, , g, , ¾®, , 176, 69 A 3, , Mass number of A is 180 and atomic number of A is 72., 20. Light from a source S is allowed to fall normally on the flat surface of a thin plate of a tourmaline crystal, cut, parallel to its axis. Only a part of this light is transmitted through A., If now the plate A is rotated, the character of, transmitted light remains unchanged. Now, another similar plate B is placed at some, distance from A such that the axis of B is, parallel to that of A. If the light transmitted, through A is passed through B, the light is, almost completely transmitted through B and, no change is observed in the light coming, out of B., , Polarised, light, , Unpolarised, light, , Polarised, light, , S, (a), , A, Unpolarised, light, , B, , Polarised, light, No light, , S, , If now the crystal A is kept fixed and B is, gradually rotated in its own plane, the, (b), intensity of light emerging out of B, decreases and becomes zero when the axis of B is perpendicular to that of A. If B is further rotated, the, intensity begins to increase and becomes maximum when the axes of A and B are again parallel., Thus, 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., From this experiment, it is obvious that light waves are transverse and not longitudinal; because, if they were, longitudinal, the rotation of crystal B would not produce any change in the intensity of light., Graph showing the variation of intensity of transmitted light through analyser is shown in fig., I, , I = I0 cos2 q, , I0, , O, , p/2, , p, , 3p/2, , 2p, , q, , Explanation: If the pass axis of analyser makes an angle q with the pass axis of polariser, the component, E cos q of the polarised light passes through the pass axis of analyser. As we rotate the analyser, keeping, polariser fixed, the intensity will vary as, I = I 0 cos 2 q, where I 0 is the intensity of the polarised light. This is known as Malus’ law., 1, 1, 21. (a) Capacitive reactance, XC =, =, wC, 2 p nC, , R = 100 W, , C=, , 100 mF, p, , 1, , =, , 2p ´ 50 ´, , 100, mF, p, , = 100 W, Impedance of the circuit,, , Z =, , R 2 + X C2, , = (100) 2 + (100) 2 = 100, , Erms = 220 V, n = 50 Hz, , 2
Page 662 :
CBSE Examination Papers 689, Current in the circuit, , I rms =, =, , (b) Voltage across resistor,, , E rms, Z, 220, 100 2, , = 1.55 A, , VR = I rms R, = 155, . ´ 100 = 155 V, , Voltage across capacitor, VC = I rms ´ C = 155, . ´ 100 = 155 V, The algebraic sum of voltages across the combination is, Vrms = VR + VC, = 155 V + 155 V, = 310 V, While Vrms of the source is 220 V. Yes, the voltages across the combination is more than the voltage of, the source. The voltage across the resistor and capacitor are not in phase., This paradox can be resolved as when the current passes through the capacitor, it leads the voltage VC by, p, phase . So, voltage of the source can be given as, 2, Vrms = VR2 + VC2, = (155) 2 + (155) 2 = 155 2 = 220 V, 22. When a metallic conductor is subjected to a certain potential V across its ends the electrons get accelerated, due to electric field. Each electron experiences acceleration for an average time, t, called the relaxation time., It then undergoes a collision and velocity of electron is completely random. The average velocity of all the, electrons contributes to the flow of current., The average velocity of free electrons along the direction of positive potential, acquired through their, eE, acceleration for average time t is called drift velocity. Vd =, t, m, The current flowing through the conductor due to drifting of the electrons can be given as, I = ne A v d, …(i), and drift velocity of the electron, | e| E, …(ii), vd =, .t, m, From relation (i) and (ii), we get, I = neA, I=, So,, Þ, Since,, \, , current density, J =, , | e| E, .t, m, , ne 2 A t, .E, m, neAVd, I, ne 2 t, .E, =, =, A, A, m, , ne 2 t ®, . E, m, m, 1, r= =, s ne 2 t, , ®, , J=, , ®, , ®, , J=s E, , (in vector form)
Page 663 :
690 Xam idea Physics–XII, , SECTION–D, 23. (a) Qualities displayed by Deepika, Ruchika are their inquisitiveness for practical knowledge, Teacher displayed concern for the students., (b) When a current passes through a coil, placed in a uniform magnetic field, it experiences a torque. It does, not experience a net force, t = NIAB sin q, where q is the angle between the normal to plane of coil and the magnetic field of strength B, N is the, number of turns in a coil., When the magnetic field is radial, as in the case of cylindrical pole pieces and soft iron core, then in, every position of coil the plane of the coil, is parallel to the magnetic field lines, so that q = 90° and, sin 90° = 1, Deflecting torque,, t = NIAB, If C is the torsional rigidity of the wire and q is the twist of suspension strip, then restoring torque = C q, For equilibrium, deflecting torque = restoring torque, i.e., NIAB = C q, NAB, q=, I, \, C, i.e., qµI, deflection of coil is directly proportional to current flowing in the coil and hence we can construct a, linear scale., (c) The shape of the magnets used is concave., It is so designed to produce a radial magnetic field., , SECTION–E, ^, , 24., , j, , –i, , r, , ^, , i, , a/2, , a/2, , (i) Flux through the flat faces (both), f1 = – E 0 i$ . pr 2 i$ = | E 0 | p r 2, (ii) Flux through the curved surface, f2 = E 0 i$ . ( 2p ra) $j, [Q i$ . j$ = 0], , =0, , (Field and area vector are perpendicular to each other), Net outward flux through the cylinder,, fnet = 2f1 + f2, = 2E 0 p r 2, , According to Gauss’s theorem, fnet =, , Q, e0, , \ Change inside the cylinder, Q = 2p e 0 r 2 E 0, , [Q i$. i$ = 1]
Page 664 :
CBSE Examination Papers 691, OR, (a) The electric field at point P1 and P3 in the outer region can be given as, ®, E P1, , =, , ®, , E P3 =, , s $, s, ( -i$ ), i +, 2e 0, 2e 0, s, s, ( -i$) +, ( +i$ ), 2e 0, 2e 0, , =0, Electric field intensity at point P2, ®, E P2, , \, , –s, , +s, , P1, , s $, s, s, i, =, ( -i$) +, ( -i$) = e0, 2e 0, 2e 0, , | E P2 | =, , P2, , P3, , s, e0, , Electric potential differences between the plates due to existing field, , s, can be given as, e0, , V0 = ( E P2 ) . d, where d is the separation between the plates, each of area A., q, So, capacitance of the capacitor,, C =, V0, e A, s. A, = 0, d, æ s ö, çç, ÷÷ d, e, è 0ø, (b) (i) If a dielectric slab of thickness ‘d’ is inserted between the plates of the capacitor,, E, s, . This is because the oppositely directed field due to, the Electric field decreases to, E = 0 =, k, e0 k, the polarised dielectric., k e0 A, (ii) Capacitance of the capacitor becomes C = k e 0 =, d, Thus capacitor with dielectric increases by a factor k because the electric field decreases by a factor k., 25. (a) Two main considerations are (i) light gathering power and (ii) resolving power., Light gathering power depends on the area of the objective lens, i.e., fainter objects can be observed., Resolving power means the ability to observe two objects, which are close to each other in the same, direction, distinctly., C =, , (b), , Secondary, convex mirror, , Eyepiece, Objective, concave mirror, , In normal adjustment, magnifying power of the telescope, M =, , F0, ., Fe
Page 665 :
692 Xam idea Physics–XII, (c) Advantages:, (i) Parabolic mirror is used to minimise the spherical aberration., (ii) No chromatic aberration occurs., (iii) Light mechanical support is required, because mirror weighs much less than a lens of equivalent, optical quality., OR, (a), , u0, , ue=fe, , v0, , F0, B, , B', , Fe, , Fe¢, , F¢, , 0, , 8, , Objective, , æ 1.22l ö, Size of image of the point object in the image plane = S = v ç, ÷, è D ø, Two objects whose images are closer this distance will not be resolved, they will be seen as one., The minimum separation between two distinctly seen points in the object plane,, 1.22fl, S, S, =, =, =, Magnifying Power D f, D, Resolving power of the microscope can be increased., (i) By choosing a medium of high refraction index, (ii) By choosing light of short wavelength., (iii) By increasing the aperture of this objective., 26. (a) Mutual Induction: When two solenoids are placed nearby and the current in one solenoid (often called, primary solenoid) is changed, the magnetic flux linked with the neighbouring solenoid (often called, secondary solenoid) changes; due to which an emf is induced in the neighbouring solenoid. This effect is, called the mutual induction. If M is mutual inductance of two solenoids, then f2 µ I 1 or f2 = MI 1, Definition of mutual inductance :, , M =, , f2, ., I1, , The mutual inductance of two solenoids is defined as the magnetic flux linked with the secondary, solenoid when the current in primary solenoid is 1 ampere., DI 1, e2, Also induced emf in secondary solenoid e 2 = - M, Þ M =, ., Dt, DI 1 / Dt, The mutual inductance of two solenoids is defined as the emf induced in the secondary solenoid when the, rate of change of current in the primary solenoid is 1 A/s., The S.I. unit of mutual inductance is also henry (H). The mutual inductance of two solenoids does not, depend on the fact which solenoid carries the current and in which solenoid emf is induced i.e.,, M 12 = M 21 = M, This is also called reciprocity theorem of mutual inductance., If L1 and L 2 are self-inductances of two solenoids with 100% flux linkage between them, then, otherwise, M = L1 L 2 ,, M = k L1 L 2, where k is coefficient of flux linkage between the solenoids.
Page 666 :
CBSE Examination Papers 693, Mutual Inductance of solenoid-solenoid system, m N N A, M = 0 1 2, l, where A is area of solenoid, l is length of solenoid, N1 is number of turns in solenoid and N 2 is number of, turns in solenoid., (b) Let a current I 2 flow through the outer coil C 2 . The magnetic field at the centre of the coil is, N m I, …(i), B2 = 2 0 2, 2r 2, If the magnetic field B 2 is considered constant over the cross-sectional area of, the coil C1 ., Hence,, , magnetic flux f1 = N1 B 2 p r12, , …(ii), , C1, r1, , From equation (i) and (ii), we get, , r2, I2, , æN m I ö, f1 = N1 çç 2 0 2 ÷÷ p r12, 2r 2, è, ø, Since, f1 = M 12 I 2, \, , Mutual inductance, M12 =, , C2, , m 0 N1 N 2 p r12, 2 r2, , (c) Let f 1 be the flux through coil C1 (of N1 turns) when current in coil C2 is I2. Then, we have, N1 f1 = MI2, , ...(i), , For current varying with time,, d( N1 f1 ) d( MI 2 ), =, dt, dt, Since induced emf in coil C1 is given by, d( N1 f1 ), e1 = –, dt, dI, e1 = – m 2, dt, , [from (i)], , It shows that varying current in a coil induces emf in the neighbouring coil., OR, (a) Ampere’s Circuital Law: 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.e.,, ®, , ®, , ò B . dl = m 0 I, Let us consider an Amperian loop of radius r around the infinite long conductor. If, magnetic field is tangential to the circumference of the circle., By Ampere’s circuital law, ®®, ò B . dl = m 0 I, , \, , B, r, , I, , B . ( 2p r ) = m 0 I, m I, B = 0, 2p r, , (b) Magnetic field due to a straight thick wire of uniform cross-section: Consider an infinitely long, cylinderical wire of radius a, carrying current I. Suppose that the current is uniformly distributed over
Page 667 :
694 Xam idea Physics–XII, whole cross-section of the wire. The cross-section of wire is, circular. Current per unit cross-sectional area., I, …(i), i=, pa 2, , Circular, cross-section, of cylinder, , Magnetic field at external points (r > a): We consider a circular, path of radius r ( > a) passing through external point P concentric, with circular cross-section of 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. So, , a, , ®, , line integral of magnetic field B around the circular path, , ò, , ®, , ®, , B × dl =, , I, , 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, m 0I, 2pr, This expression is same as the magnetic field due to a long current carrying straight wire., , Þ, , B 2pr = m 0 ´ I Þ, , B=, , This shows that for external points the current flowing in wire may be supposed to be concerned at, the axis of cylinder., Magnetic Field at Internal Points ( r < a): Consider a circular path of, radius r ( < a), passing through internal point Q, concentric with circular, cross-section of the wire. In this case the assumed circular path encloses, only a path of current carrying circular cross-section of the wire., , a, , \ Current enclosed by path, = current per unit cross-section ´ cross section of assumed circular path, æ I, = i ´ pr 2 = çç, è pa 2, , r, , O, , P, , ö, Ir 2, ÷÷ ´ pr 2 =, ø, a2, , \ By Ampere’s circuital law, ®, , ®, , ò B × dl = m 0, , Q, , ´ current closed by path, , Þ, , B × 2pr = m 0 ´, , Þ, , B =, , Ir, , r, , 2, , R, , a, , O, , a2, , a, , m 0 Ir, 2pa 2, , Clearly, magnetic field strength inside the current carrying wire is, directly proportional to distance of the point from the axis of, wire., At surface of cylinder r = a, 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., , Bs, Bµr, , B, , O, , B µ1, r, , a, , r, , zzz
Page 668 :
CBSE Examination Paper,, Guwahati -2015 (Code No. 55/2/G), Time allowed: 3 hours, , Maximum marks: 70, , General Instructions: Same as CBSE Examination Paper, Delhi–2015., , SECTION–A, 1., , The figure shows the field lines of a positive point charge. What will be the sign of the potential energy, difference of a small negative charge between the points Q and P ? Justify your answer., , Q, P, , +, , 2., , 3., 4., , The radii of curvature of both the surfaces of a lens are equal. If one of the surfaces is made plane by, grinding, how will the focal length and power of the lens change?, 1, Two wires, one of copper and the other of manganin, have same resistance and equal thickness. Which wire, is longer? Justify your answer., Name the boxes X and Y shown in the block diagram of a generalized communication system:, Message signal, , X, , Y, , Receiver, , Message signal, , 5., , In an ac circuit, the instantaneous voltage and current are V = 200 sin 300 t volt and I = 8 cos 300 t ampere, respectively. Is the nature of the circuit capacitive or inductive? Give reason., , 6., , A screen is placed 90 cm away from an object. The image of the object on the screen is formed by a convex, lens at two different locations separated by 20 cm. Determine the focal length of the lens., Define the term ‘power loss’ in a conductor of resistance R carrying a current I. In what form does this power, loss appear? Show that to minimise the power loss in the transmission cables connecting the power stations to, homes, it is necessary to have the connecting wires carrying current at enormous high values of voltage., (a) In the following nuclear reaction, , SECTION–B, , 7., , 8., , n+, , 9., , 235, 92 U, , ¾¾, ¾®, , 144, Z Ba, , +, , A, 36 X, , + 3n,, , assign the values of Z and A., (b) If both the number of protons and the number of neutrons are conserved in each nuclear reaction, in what, way is the mass converted into energy? Explain., Explain briefly how communication takes place using mobile telephony., OR, Which basic mode of communication is used in satellite communication? Which type of wave propagation is, used in this mode? Write the expression for the maximum line of sight distance d between two antennas, having heights h1 and h2.
Page 669 :
696 Xam idea Physics–XII, 10. The wavelength of light from the spectral emission line of sodium is 589 nm. Find the kinetic energy of the, electron for which it would have the same de Broglie wagelength., , SECTION–C, 11. Draw a block diagram of a simple modulator to explain how the AM wave is produced. Can the modulated, signal be transmitted as such? Explain., 12. (a) Monochromatic light of frequency 6 × 1014 Hz is produced by a laser. The power emitted is 2.0 × 10–3 W., How many photons per second on an average are emitted by the source?, (b) Figure shows variation of stopping potential (V0) vs. frequency (v) of incident radiation for two metals X, and Y. Which metal will emit electrons of larger kinetic energy for same wavelength of incident, radiation? Explain., V0, , X, , 0.5, , 1.0, , Y, , v, , 13. (a) Assume that the light of wavelength 6000 Å is coming from a star. Find the limit of resolution of a, telescope whose objective has a diameter of 250 cm., (b) Two slits are made 1 mm apart and the screen is placed 1 m away. What should be the width of each slit to, obtain 10 maxima of the double slit pattern within the central maximum of the single slit pattern?, 14. A metallic rod of length ‘l’ is rotated with a frequency ‘v’, with one end hinged at the centre and the other end, at the circumference of a circular metallic ring, about an axis passing through the centre and perpendicular to, the plane of the ring. A constant and uniform magnetic field B parallel to the axis is present everywhere., (a) Obtain the expression for the emf induced between the centre and the ring., (b) Given that the rod has resistance ‘R’, then how much power will be generated?, × × × × × × ×, × × × × × × ×, Metallic ring, × × × × l× × ×, × × × × × × ×, × × × × × × ×, × × × × × × ×, × × × × × × ×, 15. Write the expression for the generalized Ampere’s circuital law. Through a suitable example, explain the, significance of time-dependent term., 16. Describe briefly using the necessary circuit diagram, the three basic processes which take place to generate, the emf in a solar cell when light falls on it. Draw the I – V characteristics of a solar cell., Write two important criteria required for the selection of a material for solar cell fabrication., OR, Describe briefly how light emitting diode is fabricated and explain its working. Write three important, advantages of LEDs over conventional incandescent lamps., 17. Compare and explain three distinguishing features observed in Young’s double slit interference pattern with those, seen for a coherently illuminated single slit producing diffraction pattern., 18. Use Bohr’s postulates of hydrogen atom to deduce the expression for the kinetic energy (K.E.) of the electron, e2, , where rn is the radius of the nth orbit. How is the, revolving in the nth orbit and show that K.E. =, 8 p e o rn, potential energy in the nth orbit related to the orbital radius rn?
Page 670 :
CBSE Examination Papers 697, 19. Using truth tables of AND gate and NOT gate show that NAND gate is an AND gate followed by a NOT, gate. Hence write the truth table of NAND gate., Why are NAND gates called ‘Universal Gates’?, 20. A network of four 10 µF capacitors is connected to a 500 V supply as shown in the figure. Determine the, (a) equivalent capacitance of the network and, (b) charge on each capacitor., +, +, +, , –, –, –, , C2, –– –, ++ +, , ++ +, , C1, , C3 – – –, C4, +, +, +, , –, –, –, , C2, , +, , –, , 500 V, , 21. A 16 W resistance wire is bent to form a square. A source of emf 9 V is connected across one of its sides as, shown. Calculate the current drawn from the source. Find the potential difference between the ends C and D., If now the wire is stretched uniformly to double the length and once again the same cell is connected in the, same way, across one side of the square formed, what will now be the potential difference across one of its, diagonals?, R, , D, , R, , C, , R, , R, A, , B, , 9V, –4, , 22. A uniform magnetic field of 6.5 × 10 T is maintained in a chamber. An electron enters into the field with a, speed of 4.8 × 106 m/s normal to the field. Explain why the path of the electron is a circle. Determine its, frequency of revolution in the circular orbit. Does the frequency depend on the speed of the electron?, Explain., SECTION–D, 23. Hari is a student of Class X in a school near his village. His uncle gifted him a bicycle with a dynamo fitted in, it. He was thrilled to find that while cycling during night, he could light the bulb and see the objects on the, road clearly. He, however, did not know how this device works. He asked this question to his teacher. The, teacher considered it an opportunity and explained the working of a dynamo to the whole class., Answer the following questions:, (a) State the principle and working of a dynamo., (b) Write two values each displayed by Hari and his Science teacher., , SECTION–E, 24. (a) State the essential conditions for the phenomenon of total internal reflection to take place., (b) Draw a ray diagram to show how a right isosceles prism made of crown glass can be used to obtain the, inverted image.
Page 671 :
698 Xam idea Physics–XII, (c) Explain briefly with the help of a necessary diagram, how the phenomenon of total internal reflection is, used in optical fibres. Illustrate giving an example how optical fibres can be employed for transmission, of optical signals., OR, (a) Draw a suitable diagram to demonstrate that the given shape of a wavefront at t = 0, its shape at a later, time t1 can be obtained using Huygens’ geometrical construction., (b) Consider the propagation of a plane wavefront from a rarer to a denser medium and verify Snell’s law of, refraction. Show that when a wave gets refracted into a denser medium, the wavelength and speed of, propagation but the frequency remains the same., 25. (a) Define the term ‘electric flux’. Write its SI unit., (b) Given the components of an electric field as E x = ax , E y = 0 and Ez = 0, where a is a dimensional, constant. Calculate the flux through each face of the cube of side ‘a’, as shown in the figure, and the, effective charge inside the cube., y, , a, n, , n, a, , x, a, , a, , z, , (a), (b), (c), 26. (a), (b), , (c), (a), (b), (c), , OR, Define equipotential surface. Why is the electric field at any point on the equipotential surface directed, normal to the surface?, Draw the equipotential surfaces for an electric dipole. Why does the separation between successive, equipotential surfaces get wider as the distance from the charges increases?, For this dipole, draw a plot showing the variation of potential V versus x, where x (x >> 2a), is the, distance from the point charge – q along the line joining the two charges., Write the principle of working of a transformer. Show, with the help of suitable diagrams, how the, windings of a step-up transformer are done., Assuming the transformer to be an ideal one, deduce the expression for the ratio of (i) output voltage to, input voltage and (ii) output current to input current in terms of the number of turns in the primary and, secondary coils., What are the main sources of energy loss in actual transformers and how are these reduced?, OR, Draw a labelled diagram of a moving coil galvanometer. State its working principle. What is the function, of a cylindrical soft irone core used in it?, Define the terms (i) current sensitivity and (ii) voltage sensitivity., Explain the underlying principle used in converting a galvanometer into a (i) voltmeter and (ii) ammeter., zzz
Page 672 :
CBSE Examination Papers 699, , Solutions, 1. The sign of the potential energy difference of a small negative charge will be positive. This is because, negative charge moves a point at a lower potential energy to a point at a higher potential energy., 1, 1 ö, æ 1, 2. Focal length of lens = ( n - 1) ç, +, ÷, f, R ø, è R, R, f =, 2 ( n - 1), When one surface is made plane,, \ f ¢ =, As P =, , 1, = ( n - 1), f, , 1ö, æ1, ç + ÷, R, ¥, è, ø, , R, = 2f . That is the focal length will be doubled., ( n - 1), , 1, , so power will be halved., f, , (By grounding will be replaced by – by grinding.), 3. Copper, Reason : Let l1 and l2 be lengths of copper and manganin wires having same resistance R and thickness i.e., area of cross-section (A)., r l, Resistance of copper wire, R = 1 1, A, r l, Resistance of manganin wire R = 2 2, A, Þ, , r 1l1 = r 2 l2, , (As rl = constant), , Since r 1 <<< r 2, l1 >>> l2, , So,, i.e., copper wire would be longer., 4. X ® Transmitter Y ® channel, , 5. Capacitive: As the current leads the applied voltage by phase angle, , p, ., 2, , V = 200 sin 300t, , Reason :, , I = 8 cos 300t, , and, , pö, æ, = 8 sin ç 300 t + ÷, 2ø, è, , p öù, é, æ, êQ cos q = sin ç q + 2 ÷ú, è, øû, ë, , 6. Given separation between object and screen, D = 90 cm, Separation between two positions of lens, x = 20 cm, \, , ( 90) 2 - ( 20) 2, 8100 - 400, D2 - x 2, =, =, 4D, 4 ´ 90, 4 ´ 90, 7700, =, = 21.4 cm, 4 ´ 90, , Focal length of lens, f =, , 7. The energy gained by the charge carriers dissipated in the form of heat (or emission of heat and light waves).
Page 673 :
700 Xam idea Physics–XII, Consider a device of resistance R to which a power P is to be delivered is connected to the transmission cable,, which has a resistance R c from power station to homes., If V is the voltage across R and a current I flows through it, then, P = VI, , …(1), , and power dissipated in the transmission line of resistance R c is, Pc = I 2 R c, , …(2), , From equation (1) and (2), we get, 2, , æPö, Pc = ç ÷ R c, èV ø, Pc =, \, , P2, V2, , Pc µ, , . Rc, 1, , V2, To reduce the power loss in the transmission cable, the wire must carry current at high voltage., , 8. (a), , n+, , 235, 92 U, , ¾®, , 144, Z Ba, , +, , A, 36 X, , + 3n, , From law of conservation of atomic number, 0 + 92 = Z + 36, Þ, Z = 92 - 36 = 56, From law of conservation of mass number,, 1 + 235 = 144 + A + 3 ´ 1, A = 236 - 147 = 89, (b) (i) BE of, , 235, 92 U, , < BE of ( 144, Ba +, 56, , 89, 36 X ) and, , due to difference in BE of the nuclides. A large amount, , of the energy will released in the fission of, , 235, 92 U., , (ii) Mass number of the reactant and product nuclides are same but there is an actual mass defect. This, difference in the total mass of the nuclei on both sides gets converted into energy, i.e., DE = Dmc 2 ., 9., , (i) A given physical area is divided into smaller parts called cells or cell zones, usually of a hexagonal, design., (ii) A radio antenna is installed to receive and send radio signals to all the mobile phones present inside that, cell., (iii) Handling of all outgoing and incoming calls is managed through a central control called Mobile, Telephone Switching Office (MTSO). The working range of the cell antenna varies from a minimum of, 1.5 to 2 km to about 50 to 60 km., (iv) MTSO records the location and identifies the cell of the mobile phone., OR, Line of sight mode is used in the satellite communication., Space wave propagation is used., Expression for the maximum line of sight as distance between two antennas having heights h1 and h2, d = 2Rh1 + 2Rh2, , where R is the radius of the earth., 10. de Broglie wavelength,, l=, , h, =, p, , h, 2mK, , é, p2 ù, ú, êSince K =, 2m úû, êë
Page 674 :
CBSE Examination Papers 701, \, , l2 =, , or, , K =, =, , h2, 2m K, h2, 2ml2, ( 6.63 ´ 10 - 34 ) 2, 2 ´ 91, . ´ 10 - 31 ´ (589 ´ 10 - 9 ) 2, , = 6.95 ´ 10 - 25 J, 11. Block diagram of simple modulator, x(t), , m(t), +, Am sin wmt, c(t), (Modulating signal ), AC sin wct, (carrier wave), , Square, law device, , y(t), , Bandpass, Filter, centred at wc, , AM wave, , B x(t) + cx2(t), , No, AM wave cannot be transmitted as such because, (a) Modulator is to be followed by a power amplifier, which will provide the necessary power., (b) Size of the antenna should be of such order that will radiate the same power of AM wave, which is to be, transmitted in the space., nhn, 12. (a) Power of radiation, P =, = Nhn, t, where N is number of photons per sec., P, or, N =, hn, =, , 2.0 ´ 10 - 3, 6.63 ´ 10 - 34 ´ 6 ´ 1014, , = 5 ´ 1015 photons per second, (b) From the graph of stopping potential versus frequency work function, f y > fx, From Eienstein’s equation, hc, K =, -f, l, Since wavelength l is constant, so, hc, Kx =, - fx, l, hc, and, Ky =, - fy, l, Þ, Kx > K y, From metal X, the electrons emitted out with larger kinetic energy., 13. (a) The limit of resolution of the objective lens in the telescope is, 1.22 l, Dq =, D, Since D = 250 cm = 2.5 cm and l2 = 6000 Å = 6 ´ 10 - 7 m, 1.22 ´ 6 ´ 10 - 7, = 2.9 ´ 10 - 7 radian, 2.5, (b) If a is the size of single slit for diffraction pattern then. For first maxima, l, q=, ( n = 1), a, \, , Dq =
Page 675 :
702 Xam idea Physics–XII, , 14., , and angular separation of central maxima in the diffraction pattern, 2l, …(1), q¢ = 2 q =, a, The angular size of the fringe in the interference pattern, b l, a =, D d, If there are 10 maxima within the central maxima of the diffraction pattern, then, 10 a = q¢, æ lö 2 l, 10 ç ÷ =, a, èdø, d, a =, Þ, 5, The distance between two slits is 1 mm., 1, \ Size of the single slit a = mm = 0.2 mm, 5, (a) As the rod rotated, the free electrons in the rod move towards the outer, end, and causes a negative charge near the rim and a positive charge at, the centre., The magnitude of emf in the element of size dr, at distance r from the, centre, | d e | = B v dr, and net emf at the ends of the rod, E =, , l, , ò0, , r=l, r, , B v dr, , For rotating rod at frequency, n, v = rw = 2pnr, \, , | E| =, , l, , ò0, , B ´ 2pvr dr, , = 2p v B, , l, , ò0, , æ l2, = 2p v B ç, ç 2, è, , r dr, ö, ÷, ÷, ø, , | E | = pn l2B, (b) Power generated due to induced emf, ( pnl 2 B) 2, e2, p2n 2l 4 B 2, =, =, R, R, R, ® ®, df E, 15. f B× dl = m 0 I C + m 0 e 0, dt, dfE ö, æ, = m 0 ç IC + e 0, ÷ = m 0 (IC + I D ), dt ø, è, df E, represents the displacement current., The time dependent term i.e., e 0, dt, P =, , ®, , It exists in the region in which the electric flux ( j 0 ) i.e., the electric field (E ) changes with time., Example: During charging or discharging of a capacitor, the current in the wire connecting the capacitor, plates to the source is conduction current whereas in between the plates it is displacement current due to the, change of electric field between the plates which makes the circuit complete., The conduction current is always equal to the displacement current.
Page 676 :
CBSE Examination Papers 703, 16., IL, , n, , p, B, , A, Depletion, region, , Basic Processes :, (a) Generation of electron and hole pairs due to the light falling at the function, energy of the light must be, hn > E g ., (b) Separation of electrons and holes due to the electric field of the depletion region inside the junction., (c) The electrons reaching the n side, produces a negative potential at terminal A, and holes reaching the, p-side produces a positive potential at terminal B., I-V Characteristic of solar cell:, I, VOC, , V, , ISC, , where VOC = Open circuit voltage, and I SC = Short circuit current., Criteria : (i) Band gap E g must be of range 1.0 to 1.8 eV., (ii) High optical absorption, about 10 4 cm - 1, (iii) Strong electrical conductivity, (iv) Availability of the raw material, (v) Material must be low cost., OR, (i) An LED is fabricated from heavily doped p-n junction having a band gap of 1.8 eV approximate., (ii) The diode is encapsulated with a transparent cover so emitted light can come out., Working : When the LED is forward biased, the electrons move from n side to p-side and holes start to move, from p-side to n-side. Thus, at the junction boundary on either side of the junction, excess minority carries, are there which recombine with the majority carrier near the junction., On recombination, the energy is released in the form of photons., (i) When forward current of the diode is small, the intensity of emitted light is small., (ii) As the forward current increases, intensity of light also increases and reachers a maximum value., (iii) Further increase in current results decrease in the intensity., Advantages :, (i) Low operational voltage and less power., (ii) Fast action and no warm up time required., (iii) The band width of emitted light is 100 Å to 500 Å., (iv) Long life and ruggedness., (v) Fast on-off switching capability.
Page 677 :
704 Xam idea Physics–XII, 17. Difference between interference and diffraction, Interference, , Diffraction, , (i) It is due to the superposition of two waves, coming from two coherent sources., , (i) It is due to the superposition of secondary wavelets, originating from different parts of the same wavefront., , (ii) The width of the interference bands is equal., , (ii) The width of the diffraction bands is not the same., , (iii) The intensity of all maxima (fringes) is same., , (iii) The intensity of central maximum is maximum and, goes on decreasing rapidly with increase in order of, maxima., , 18. 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., ( Z e) ( e), mv 2, 1, [from Rutherford model], … ( i), =, r, 4p e 0, r2, Z e2, 1, 4p e 0, r, 1, So, Kinetic energy [ K ] = mv 2, 2, mv 2 =, , or,, , Z e2, 1, 4p e 0 2 r, ( Z e) ( -e), 1, Potential energy =, 4p e 0, r, K =, , 1, Ze 2, 4p e 0 r, , =Total energy,, , E = KE + PE =, E =-, , æ, Z e2, Z e2, 1, 1, + çç 4p e 0, 4p e 0 2r, r, è, , ö, ÷, ÷, ø, , 1 Z e2, 4p e 0 2r, , For nth orbit, E can be written as E n, so,, , En = -, , Z e2, 1, 4p e 0 2 rn, , ...(ii), , Negative sign indicates that the electron remains bound with the nucleus (or electron-nucleus form an, attractive system), From Bohr's postulate for quantization of angular momentum., n h, nh, mvr =, Þ v =, 2p, 2p mr, Substituting this value of v in equation (i), we get, m, r, , é n h ù, ê 2pmr ú, ë, û, , 2, , =, , or,, , r =, , or,, , rn =, , Ze 2, 1, 4p e 0 r 2, e0 h2 n2, p m Ze 2, e0 h2 n2, p m Ze 2, , …(iii)
Page 679 :
706 Xam idea Physics–XII, The capacitor C 4 is the parallel to capacitor C s ., Equivalent capacitance C eq of the network, 10, 40, C eq = C 4 + C s = 10 mF +, mF =, mF, 3, 3, (b) The charge Q on C1 , C 2 , C 3 is same being in series. So,, 10, Q = C sV =, mF ´ 500 V, 3, 5000, =, mC = 1.67 ´ 10 3 mC = 1.67 ´ 10 - 3 C, 3, and charge on C 4 at V = 500 V is, Q4 = C 4 V, = 10 mF ´ 500 V, = 5000 mC = 5.0 ´ 10 - 3 C, 16, 21. Resistance in each branch =, =4W, 4, , C, , D, R, , Resistances in branch AD, DC and CB are in series. So,, Rs = R + R + R = 3 R, , R, , R, , =3´ 4, , R, , = 12 W, , A, , Resistance in branch AB is in parallel of series combination. So, equivalent, resistance, 1, 1, 1, =, +, R eq, Rs, RA, , B, , 9V, , 1, 1, +, 12, 4, 48, R eq =, =3W, 16, V, 9, Current drawn from the source I =, = = 3 Amp, R eq, 3, =, , P.d. between points C and D, , VCD = I s ´ RCD, V, VCD =, ´ RCD, Rs, =, , 9 V, ´ 4 W = 3 volt, 12 W, , A, . Results that the resistance of each branch, 2, becomes 4 times of its original value. Hence, resistance of each branch, 16 W, becomes 16 W., C, D, , If length of the wire is doubled, the l¢ = 2l and area A ¢ =, , PD across A and C, , æ V ö, ÷÷ ´ ( R AD + R AC ), VAC = çç, è RS ø, =, , 9, 9, ´ (16 + 16) =, ´ 32, 48, 48, , =6 V, , 16 W, , 16 W, 16 W, , A, , B
Page 680 :
CBSE Examination Papers 707, PD across B and D, VA - VB = 9 V (due to parallel combination), V, ´ R AD, VA - VD =, RS, =, , 9, ´ 16 = 3.0 V, 48, , (VA - VB ) - (VA - VD ) = (9 – 3) V, Þ, , VD - VB = 6.0 V, , ®, , ®, , 22. Electron enters normal to the magnetic field lines. The Lorentz-force is normal to B and v . The electron, follows a circular trajectory. This is because the Lorentz force acting on electron provides the necessarily, centripetal force., The frequency of the electron is its circular path of radius r is, qB, w=, m, qB, or, (independent to the radius and velocity), n=, 2p m, =, , 1.6 ´ 10 -, , 19, , ´ 6.5 ´ 10 - 4, , 2 ´ 314, . ´ 91, . ´ 10 - 31, , = 18, . ´ 107 Hz, No, the frequency is independent of speed, but depends on the strength of the magnetic field., 23. (a) Principle : It is based on Faraday law of electromagnetic induction. The changing magnetic flux through, the coil provides the necessarily induced emf across the ends of the coil., Working: The coil (called armature) is mechanically rotated in the uniform magnetic field by some, external means the rotation of the coil causes the magnetic flux through it to change, so an emf is induced, in the coil. The ends of the coil are connected to an external circuit by means of slip rings and brushes., ®, , When a coil of N turns, each of area A rotates in uniform magnetic field B, the magnitude of induced emf, can be given as, E = NBAw sin wt, (b) Values displayed by Hari are :, (i) Curiosity, (ii) Learning attitude, (iii) Scientific temperament, Values displayed by Science teacher are :, (i) Responsive to the students, (ii) Carring and concerned, (iii) Encourging the scientific temperament, 24. (a) Conditions for total internal reflection are :, (i) Light must travel from denser to rarer medium., (ii) Angle of incidence in denser medium must be greater than critical angle (C)., (b), B, , 45°, , 90°, , A, A', B', , 45°
Page 681 :
708 Xam idea Physics–XII, (c) Optical Fibre: An optical fibre is a device based, Coating n = 1.5, on total internal reflection by which a light, signal may be transmitted from one place to, another with a negligible loss of energy. It is a, n = 1.7, very long and thin pipe of quartz ( n = 1 × 7) of, - 4, thickness nearly » 10, m coated all around, with a material of refractive index 1 × 5. A large, number of such fibres held together form a light, pipe and are used for communication of light, A, signals. When a light ray is incident on one end, B, at a small angle of incidence, 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. 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., 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., Application : Optical fibre is used to transmit light signal to distant places., OR, (a), A, , A1, , A2, , A2, 1, , A1, 1, , 2, , 2, , 3, , 3, , S, , 4, , 4, B2, B, (a) B1, , A, , 5, , 5, B2, , B B1, (b), , AB is a wavefront of point source O at time t = 0., A1 B1 and A 2 B 2 are two envelops on the secondary wavelets at another time t ¹ 0 . Forward envelop, A1 B1 exists but the backward envelope A 2 B 2 does not exist., (b) 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, it is deviated from its, B, path. This phenomenon is called refraction. In transversing from, 90o, first medium to another medium, the frequency of wave remains, i, i, B', unchanged but its speed and the wavelength both are changed. Let X, Y, r, A, r, XY be a surface separating the two media ‘1’ and ‘2’. Let, 90o, v1 and v 2 be the speeds of waves in these media., A', 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., Clearly BB¢ = v1t. As the wavefront AB advances, it strikes the points between A and B¢ of boundary, surface. According to Huygen’s principle, secondary spherical wavelets originate from these points, which, travel with speed v1 in the first medium and speed v 2 in the second medium.
Page 682 :
CBSE Examination Papers 709, First of all secondary wavelet starts from A, which traverses a distance AA¢ ( = v 2 t) in second medium in, time t. In the same time-interval t, the point of wavefront traverses a distance BB¢ ( = v1t) in first medium, and reaches B¢ , from, where the secondary wavelet now starts. Clearly BB¢ = v1t and AA¢ = v 2 t., 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 incident wavefront AB advances, the secondary wavelets start from points between A and, B¢ , one after the other and will touch A¢ B¢ simultaneously. According to Huygen’s principle A¢ B¢ is the, new position of wavefront AB in the second medium. Hence A¢ B¢ will be the refracted wavefront., First law: As AB, A¢ B ¢ and surface XY are in the plane of paper, therefore the perpendicular drawn on, them will be in the same plane. As the lines drawn normal to wavefront denote the rays, therefore we may, say that the incident ray, refracted ray and the normal at the point of incidence all lie in the same plane., This is the first law of refraction., Second law: Let the incident wavefront AB and refracted wavefront A¢ B¢ make angles i and r, respectively with refracting surface XY., In right-angled triangle AB¢ B, Ð ABB¢ = 90°, vt, BB¢, ...(i), \, sin i = sin Ð BAB¢ =, = 1, AB¢, AB¢, Similarly in right-angled triangle AA¢ B ¢ , Ð AA¢ B ¢ = 90°, v t, AA¢, ...(ii), \, sin r = sin Ð AB¢ A¢ =, = 2, AB¢ AB¢, Dividing equation (i) by (ii), we get, v, sin i, ...(iii), = 1 = constant, sin r, v2, As the rays are always normal to the wavefront, therefore the incident and refracted rays make angles, i and r with the normal drawn on the surface XY i.e. i and r are the angle of incidence and angle of, refraction respectively. According to equation (iii):, The ratio of sine of angle of incidence and the sine of angle of refraction for a given pair of media is a, constant and is equal to the ratio of velocities of waves in the two media. This is the second law of, refraction, and is called the Snell’s law., 25. (a) Reason: Let l1 and l 2 are the wavelengths of light in medium 1 and medium 2. If BB ¢ = l1 and, AA ¢ = l 2 ,, l1, BB ¢ v1, =, =, l2, AA ¢ v 2, v1, v, = 2, l1, l2, , Þ, , Thus, when a wave gets refracted into a denser medium, its wavelength and speed decrease but, vö, æ, frequency ç n = ÷ remaining same., lø, è, (b), L, , y, , R, a, , n, , n, , O, a, a, z, , Since E x = a x exists, so flux through the faces normal to X-axis is non-zero and flux through the faces, normal to Y and Z axes are zero.
Page 683 :
710 Xam idea Physics–XII, Flux through face R at distance n = 2a., fR = a x ( a) 2, = a ( 2a) a, , ( q = 0° ), 2, , = 2a a 3, Flux through the face L at distance x = a, fL = - a x . ( a) 2, = - aa. a, , ( q = 180° ), , 2, , = - a a3, Net flux through the cube is,, , fnet = fL + fR, , = - a a 3 + 2a a 3 = a a 3, , Net charge q if exists inside the cube, so from Gauss’s law, q, fnet =, e0, Charge q = e 0 fnet = a e 0 a 3, OR, (a) A surface around a point charge (or group of charges) with a constant value of potential at all points on, the surface is called an equipotential surface., If the electric field is not normal to the equipotential surface, then to move a charge q 0 against the, direction of the component of the field, some work would have to be done. But this work done is in, contradiction to the definition of an equipotential surface., If there is no potential difference, there is no work done., Hence, electric field is directed normally to the equipotential surface., (b), , +Q, , –Q, , The electric field intensity decreases as the distance from the charge increases, i.e., | E | µ, and we know that, or, , 1, r2, , dV, dr, dV, dr = | E|, , | E|=-, , As the value of E decreases, it results increase in the separation of equipotential surfaces from each other., (c) For an ideal electric dipole, assuming x > > 2a the distance either, –x, O, X, measured from the mid-point of the dipole (or from any charge + q and, - q), potential at a point can be given as, 1, P, V =±, V, V µ x12, 4p e 0 | x | 2, Y
Page 684 :
CBSE Examination Papers 711, 26. (a) Transformer: Transformer is a device by which an alternating voltage may be decreased or increased. It, is based on the principle of mutual-induction., Construction: It consists of laminated core of soft iron, on which two coils of insulated copper wire are, separately wound. These coils are kept insulated from each other and from the iron-core, but are coupled, through mutual induction. 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. 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. Transformers are of two types:, 1. 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. (i. e. , N S > N p )., 2. 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. e. , N S < N p ), (A.C. mains), Primary, , laminated, iron core, , Primary, , Primary, , Secondary, , Secondary, , (a) Step up, , (b) Step down, , Core, , Secondary, Transformer, , Working: When alternating current source is connected to the ends of primary coil, the current changes, continuously in the 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., (b) Let N p be the number of turns in primary coil, NS the number of turns in secondary coil and f the, magnetic flux linked with each turn. 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. According to Faraday’s laws the emf, induced in the primary coil, Df, ...(i), ep = - Np, Dt, and emf induced in the secondary coil, Df, ...(ii), eS = - NS, Dt, From (i) and (ii), eS, N, ...(iii), = S, ep, Np, If the resistance of primary coil is negligible, the emf ( e p ) induced in the primary coil, will be equal to, the applied potential difference (V p ) across its ends. Similarly if the secondary circuit is open, then the, potential difference VS across its ends will be equal to the emf ( e S ) induced in it; therefore, VS, e, N, ...(iv), = S = S = r (say), Vp, ep, Np
Page 685 :
712 Xam idea Physics–XII, where r =, , NS, is called the transformation ratio. If i p and i s are the instantaneous currents in primary, Np, , and secondary coils and there is no loss of energy; then, For about 100% efficiency, Power in primary = Power in secondary, V p i p = VS i S, Vp, Np, iS, 1, \, =, =, =, ip, VS, NS, r, Ns > N p, , In step up transformer,, , ...(v), , ® r >1;, , So, VS > V p and i S < i p, i.e. step up transformer increases the voltage., In step down transformer, N S < N p ® r < 1, so, VS < V p and i S > i p, i.e. step down transformer decreases the voltage, but increases the current., (c) (i) Joule Heating: Energy is lost in resistance of primary and secondary windings as heat ( I 2 Rt). It can, be reduced by using thick wires., (ii) Flux Leakage: Energy is lost due to coupling of primary and secondary coils not being perfect, i.e.,, whole of magnetic flux generated in primary coil is not linked with the secondary coil. It can be, reduced by winding the primary and secondary coils one over the other., OR, (a) 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., A soft iron core in cylindrical form is placed between the coil., One end of coil is attached to suspension wire which also serves as one terminal (T1 ) of galvanometer., The other end of coil is connected to a loosely coiled strip, which serves as the other terminal (T2 ). The, other end of the suspension is attached to a torsion head which can be rotated to set the coil in zero, position. 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. The levelling screws are also provided at the base of the, instrument., The pole pieces of the permanent magnet are cylindrical so that the magnetic field is radial at any position, of the coil., H, , Suspension wire, M, , T1, T2, , Coil, , NIBl, b, , N, , S, , N, , S, , N, , S, , NIBl, Coiled strip, , (a), , (b), Magnetic lines of, force of radial magnetic field, , (c), , Principle and working: When current ( I ) is passed in the coil, torque t acts on the coil, given by, t = NIAB sin q
Page 686 :
CBSE Examination Papers 713, where q is the angle between the normal to plane of coil and the magnetic field of strength B, N is the, number of turns in a coil., When the magnetic field is radial, as in the case of cylindrical pole pieces and soft iron core, then in every, position of coil the plane of the coil, is parallel to the magnetic field lines, so that q = 90° and sin 90° = 1, . The coil experiences a uniform coupler., Deflecting torque,, t = NIAB, If C is the torsional rigidity of the wire and q is the twist of suspension strip, then restoring torque = C q, For equilibrium, deflecting torque = restoring torque, i.e., NIAB = C q, NAB, ...(i), \, q=, I, C, i.e., qµI, deflection of coil is directly proportional to current flowing in the coil and hence we can construct a, linear scale., Importance (or function) of uniform radial magnetic field: Torque as current carrying coil in a, magnetic field is t = NIAB sin q, In radial magnetic field sin q = 1, so torque is t = NIAB., This makes the deflection ( q) proportional to current. In other words, the radial magnetic field makes the, scale linear., (b) Sensitivity of a galvanometer : Current sensitivity : It is defined as the deflection of coil per unit, current flowing in it., æ q ö NAB, Sensitivity S I = ç ÷ =, C, èIø, Voltage sensitivity: It is defined on the deflection of coil per unit potential difference across its ends, q, NAB, i.e.,, SV =, =, ,, V, Rg . C, where Rg is resistance of galvanometer., Clearly for greater sensitivity, number of turns N, area A and magnetic field strength B should be large, and torsional rigidity C of suspension should be small., (c) A galvanometer can be changed into voltmeter by connecting a high resistance in series, so that a, negligible current flows through it., A galvanometer can be converted into an ammeter by connecting a shunt resistance in parallel with the, coil. So, the ammeter does not reduce the current in the circuit., zzz
Page 687 :
CBSE Examination Paper, Panchkula Region -2015 (Code No. 55/1/C), Time allowed : 3 hours, , Maximum marks: 70, , General Instructions: Same as CBSE Examination paper, Delhi–2015., , SECTION–A, 1., 2., , In a series LCR circuit, VL = VC ¹ VR. What is the value of power factor?, What happens when a forward bias is applied to a p-n junction?, , 3., , Write a relation for polarisation P of a dielectric material in the presence of an external electric field E ., Define the term ‘mobility’ of charge carriers. Write its SI unit., The focal length of an equiconvex lens is equal to the radius of curvature of either face. What is the refractive, index of the material of the lens?, , 4., 5., , ®, , ®, , SECTION–B, 6., , Write the important characteristic features by which the interference can be distinguished from the observed, diffraction pattern., OR, , Explain the basic differences between the construction and working of a telescope and a microscope., State Bohr’s postulate of hydrogen atom which successfully explains the emission lines in the spectrum of, hydrogen atom., Use Rydberg formula to determine the wavelength of Ha line., [Given: Rydberg constant R = 1.03 × 107 m–1], 8. State the two Kirchhoff’s rules used in electric networks. How are these rules justified?, 9. Given the ground state energy E0 = –13.6 eV and Bohr radius a0 = 0.53Å. Find out how the de Broglie, wavelength associated with the electron orbiting in the ground state would change when it jumps into the first, excited state., 10. (a) Distinguish between ‘Analog’ and ‘Digital’ forms of communication., (b) Explain briefly two commonly used applications of the ‘Internet’., 7., , SECTION–C, 11. Plot a graph showing the variation of current density (j) versus the electric field (E) for two conductors of, different materials. What information from this plot regarding the properties of the conducting material, can, be obtained which can be used to select suitable materials for use in making (i) standard resistance and (ii), connecting wires in electric circuits?, Electron drift speed is estimated to be of the order of mms–1. Yet large current of the order of few amperes, can be set up in the wire. Explain briefly., 12. State Biot–Savart law. Deduce the expression for the magnetic field at a point on the axis of a current, carrying circular loop of radius ‘R’, distant ‘x’ from the centre. Hence write the magnetic field at the centre, of a loop., 13. What does a polaroid consist of? Show, using a simple polaroid, that light waves are transverse in nature., Intensity of light coming out of a polaroid does not change irrespective of the orientation of the pass axis of, the polaroid. Explain why?
Page 688 :
CBSE Examination Papers 715, 14. How is Zener diode fabricated? What causes the setting up of high electric field even for small reverse bias, voltage across the diode?, Describe with the help of a circuit diagram, the working of Zener diode as a voltage regulator., OR, (a) Explain with the help of a diagram, how depletion region and potential barrier are formed in a junction, diode., (b) If a small voltage is applied to a p-n junction diode how will the barrier potential b affected when it is, (i) forward biased, and (ii) reverse biased?, 15. Light of intensity ‘I’ and frequency ‘v’is incident on a photosensitive surface and causes photoelectric, emission. What will be the effect on anode current when (i) the intensity of light is gradually increased, (ii), the frequency of incident radiation is increased, and (iii) the anode potential is increased? In each case, all, other factors remain the same., Explain, giving justification in each case., 16. When is a transistor said to be in active state? Draw a circuit diagram of a p-n-p transistor and explain how it, works as a transistor amplifier. Write clearly, why in the case of a transistor (i) the base is thin and lightly, doped and (ii) the emitter is heavily doped., 17. (a) State three important factors showing the need for translating a low frequency signal into a high, frequency wave before transmission., (b) Draw a sketch of a sinusoidal carrier wave along with a modulating signal and show how these are, superimposed to obtain the resultant amplitude modulated wave., 18. You are given three circuit elements X, Y and Z. When the element X is connected across an a.c. source of a, given voltage, the current and the voltage are in the same phase. When the element Y is connected in series, with X across the source, voltage is ahead of the current in phase by p/4. But the current is ahead of the, voltage in phase by p/4 when Z is connected in series with X across the source. Identify the circuit elements, X, Y and Z., When all the three elements are connected in series across the same source, determine the impedance of the, circuit., Draw a plot of the current versus the frequency of applied source and mention the significance of this plot., 11, 11, +, 19. Write symbolically the nuclear b decay process of C. Is the decayed product X an isotope or isobar of C?, 6, 6, 11, Given the mass values m ( C) = 11.011434 u and m (X) = 11.009305 u. Estimate the Q-value in this process., 6, 20. An object is placed 15 cm in front of a convex lens of focal length 10 cm. Find the nature and position of the, image formed. Where should a concave mirror of radius of curvature 20 cm be placed so that the final image, is formed at the position of the object itself?, 21. Arrange the following electromagnetic waves in the order of their increasing wavelength:, (a) g-rays, , (b) Microwaves, , (c) X-rays, , (d) Radio waves, , How are infra-red waves produced? What role does infra-red radiation play in (i) maintaining the Earth’s, warmth and (ii) physical therapy?, 22. Explain briefly the process of charging a parallel plate capacitor when it is connected across a d.c. battery., A capacitor of capacitance ‘C’ is charged to ‘V’ volts by a battery. After some time the battery is, disconnected and the distance between the plates is doubled. Now a slab of dielectric constant,, 1 < k < 2, is introduced to fill the space between the plates. How will the following be affected:, (a) The electric field between the plates of the capacitor, (b) The energy stored in the capacitor, Justify your answer by writing the necessary expressions.
Page 689 :
716 Xam idea Physics–XII, , SECTION–D, 23. Ajit had a high tension tower erected on his farm land. He kept complaining to the authorities to remove it as, it was occupying a large portion of his land. His uncle, who was a teacher, explained to him the need for, erecting these towers for efficient transmission of power. As Ajit realised its significance, he stopped, complaining., Answer the following questions:, (a) Why is it necessary to transport power at high voltage?, (b) A low power factor implies large power loss. Explain., (c) Write two values each displayed by Ajit and his uncle., , SECTION–E, 24. (a) Define self-inductance of a coil. Obtain an expression for the energy stored in a solenoid of, self-inductance ‘L’ when the current through it grows from zero to ‘I’., (b) A square loop MNOP of side 20 cm is placed horizontally in a uniform magnetic field acting vertically, downwards as shown in the figure. The loop is pulled with a constant velocity of 20 cms–1 till it goes out, of the field., , M, , 20 cm, , N, v, , P, , O, 1m, , (i) Depict the direction of the induced current in the loop as it goes out of the field. For how long would, the current in the loop persist?, (ii) Plot a graph showing the variation of magnetic flux and induced emf as a function of time., OR, ®, , (a) Draw the magnetic field lines due to a circular loop of area A carrying current I. Show that it acts as a bar, ®, , ®, , magnet of magnetic moment m = IA., (b) Derive the expression for the magnetic field due to a solenoid of length ‘2l’, radius ‘a’ having ‘n’ number, of turns per unit length and carrying a steady current ‘I’ at a point on the axial line, distant ‘r’ from the, centre of the solenoid. How does this expression compare with the axial magnetic field due to a bar, magnet of magnetic moment ‘m’?, 25. (a) In Young’s double slit experiment, deduce the conditions for obtaining constructive and destructive, interference fringes. Hence deduce the expression for the fringe width., (b) Show that the fringe pattern on the screen is actually a superposition of single slit diffraction from each, slit., (c) What should be the width of each slit to obtain 10 maxima of the double slit pattern within the central, maximum of the single slit pattern, for green light of wavelength 500 nm, if the separation between two, slits is 1 mm?, OR, (a) Two thin convex lenses L1 and L2 of focal lengths f1 and f2 respectively, are placed coaxially in contact., An object is placed at a point beyond the focus of lens L1. Draw a ray diagram to show the image, formation by the combination and hence derive the expression for the focal length of the combined, system.
Page 690 :
CBSE Examination Papers 717, (b) A ray PQ incident on the face AB of a prism ABC, as shown in the figure, emerges from the face AC such, that AQ = AR., A, , Q, , R, , P, , B, , C, , Draw the ray diagram showing the passage of the ray through the prism. If the angle of the prism is 60°, and refractive index of the material of the prism is 3, determine the values of angle of incidence and, angle of deviation., ®, , ®, , 26. (a) Deduce the expression for the potential energy of a system of two charges q1 and q2 located at r1 and r 2, respectively in an external electric field., , (b) Three point charges, + Q, + 2Q and – 3Q are placed at the vertices of an equilateral triangle ABC of side l., If these charges are displaced to the mid-points A1, B1 and C1 respectively, find the amount of the work, done in shifting the charges to the new locations., A (+Q), , A1, , B (+2Q), , C1, , B1, , C (–3Q), , OR, Define electric flux. Write its SI unit., State and explain Gauss’s law. Find out the outward flux due to a point charge + q placed at the centre of a, cube of side ‘a’. Why is it found to be independent of the size and shape of the surface enclosing it? Explain., zzz
Page 691 :
718 Xam idea Physics–XII, , Solutions, SECTION–A, 1. Power factor,, , Eeff, , cos f =, Since VL = VC ,, , VR, VR2, , + (VL - VC ) 2, , VL – VC, , V, cos f = R = 1, VR, , f, , Ieff, VR = Ieff .R, The value of power factor is 1., 2. The direction of the applied voltage (V) is opposite to the built-in potential V0. As a result, depletion layer, width decreases and the barrier height is reduced to V0 – V ., ®, ®, 3. P = e 0 x E, , P – polarisation density, x ® electric susceptibility, e 0 ® relative permittivity., ®, vd, 4. Mobility: Mobility of charge carrier is defined as drift velocity per unit electric field. m =, | vd |, E, m / s m2, =, V / m Vs, R1 = R2 = f, 1 æ m 2 – m1 öæ 2 ö, ÷ç ÷, =ç, ÷ç f ÷, f çè m 1, øè ø, m2, 1, 3, = + 1 = = 1.5, m1, 2, 2, Refractive index of material = 1.5., SI unit =, , 5., , Þ, , 1 m2, =, –1, 2 m1, , SECTION–B, 6., Interference, (i) It is due to the superposition of two waves coming, from two coherent sources., (ii) The width of the interference bands is equal., (iii) The intensity of all maxima (fringes) is same., , Diffraction, (i) It is due to the superposition of secondary wavelets, originating from different parts of the same, wavefront., (ii) The width of the diffraction bands is not the same., (iii) The intensity of central maximum is maximum and, goes on decreasing rapidly with increase of order of, maxima., , OR, Telescope, , Microscope, , Construction, , Objective has large focal length and large Objective has very short focal length and short, aperture but eyepiece of short focal length and aperture and eye piece has short focal length and, short aperture, large aperture (fe > fo), , Working, , It forms magnified image of distant object, , It forms magnified image of a small nearby object, , 7. The electron does not radiate energy when in a stationary orbit. The quantum of energy (or photon) is emitted, or absorbed when an electron jumps from one stationary orbit to the other. The frequency of emitted or, absorbed photon is given by, hv = |Ei – Ef |
Page 692 :
CBSE Examination Papers 719, First line of Balmer is called H a ., é1, 1, 1 ù 5R, =R ê, - ú =, 2, l, ë2, 12 û 36, 1, 5, =, ´ 1.03 ´ 107, l 36, 36, l=, = 6990 Å, 5 ´ 1.03 ´ 107, 8. Junction law: The algebraic sum of currents meeting at any junction is zero., åI = 0, This law is based on law of conservation of charge., Loop law: The algebraic sum of potential differences of different circuit elements of a closed circuit is zero., åV = 0, This law is based on law of conservation of energy., h, h, 9. de Broglie wavelength, l = =, p, 2mK, In ground state,, , …(1), , E0 = - 13.6 eV, , So KE, , K1 = + 13.6 eV, (Independent to radius), 13.6, 13.6, In first excited state E = eV = = - 3.4 eV, 2, n, 22, , So K2, , ...(2), , Since K = - E and U = 2E, , = + 3.4 eV, , ...(3), , From Eq. (1), (2) and (3), we get, l1, =, l2, , K2, K1, , =, , 3.4, 1, =, 13.6, 2, , Þ, , l 2 = 2l1, , de Broglie wavelength will become double., 10. (a) Aanalog – An analog message signal is in the form of a continuous wave e.g., sine wave., Digital – A digital message signal is discrete and in the form of a pulse coded by 0 and 1., 1, V, , V, O, , t, , t, Digital, , Analog, , (b) Uses of Internet, e-mail, e-banking, e-shopping, e-ticketing, chatting, surfing, file transfer etc., , SECTION–C, 11. As we know that the drift velocity of the free electrons in a conducting material is given by, – eE, vd =, t, m, and the flow of the current due to drifting electrons, I = neAv d, From relation (1) and (2), we get, 2, æ eE ö ne A, I = neA ç, t.E, t÷ =, m, è m ø, , …(1), , …(2)
Page 693 :
720 Xam idea Physics–XII, Current density, | J | =, , Þ, , I, A, , | J |=, , n e2 A, E, m, , | J |=, , 1, E, r, , é, m ù, êwhere, r = 2 ú, ë, ne P û, , Þ, , Current density, r µ E, 1, where represents the slope of J-E graph., r, , B, , Let r A the resistivity of the standard resistance A and r B is the resistivity of, the connective wire, i.e., r B < r A . So the graph of J vs E can be given as, , J, , A, , Information regarding current flow:, E, (i) When the circuit is closed, an electric field gets established through the, conductor with the speed of light., (ii) Localised electrons at all places inside the conductor begins to drift instantly and current flow occurs, immediately., (iii) In the relation i = neAv d , the free electron number density is of order of 10 29 m - 3 , even drift velocity is, , of order of few mm/sec or 10 - 3 mm/sec, the current flow becomes of order of few ampere., I = neAv d ,, Here, drift velocity Vd is very small (in the order of mm s–1), but electron number density (n) is very, large. Hence, large current can be set up in the wire., 12. 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, 0, dB =, 4p r 3, , dl, I, , q, ®, r, , P, , where µ0 is permeability of free space. Its value is, µ0 = 4p × 10–7 Wb/A-m., (i) Magnetic field at the axis of a circular loop: Consider a A dl, circular loop of radius R carrying current I, with its plane, r, Q, N, R, perpendicular to the plane of paper. Let P be a point of, I, a, observation on the axis of this circular loop at a distance x, a, x, from its centre O. Consider a small element of length dl of O, M, P, the coil at point A. The magnitude of the magnetic induction, ®, N¢, Q¢, dB at point P due to this element is given by, A¢ dl, ® m I dl sin a, 0, ...(i), dB =, 4p, r2, ®, ®, ®, This direction of dB is perpendicular to the plane containing dl and r and is given by right hand screw, ®, ®, ®, rule. As the angle between I dl and r is 90°, the magnitude of the magnetic induction dB is given by,, ® m I dl sin 90° m I dl, ...(ii), = 0, dB = 0, 4p, r2, 4p r 2, If we consider the magnetic induction produced by the whole of the circular coil, then by symmetry the, components of magnetic induction perpendicular to the axis will be cancelled out, while those parallel to
Page 694 :
CBSE Examination Papers 721, the axis will be added up. Thus the resultant magnetic induction at axial point P is along the axis and may, be evaluated as follows:, ®, The component of dB along the axis,, ® m I dl, sin a, ...(iii), dB x = 0, 4p r 2, R \, r = (R2 + x2)1/2, But sin a = and, r, ® m I dl R m IR, . = 0, \, dB x = 0, dl, 4p r 2 r 4p r 3, m 0 IR, , ...(iv), dl, 4p( R + x 2 ) 3 / 2, 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, 2, 4p( R + x ), 4p( R + x 2 ) 3 / 2, =, , But, Therefore,, , 2, , ò dl = length of the loop = 2pR, ®, B=, , m 0 IR, 4p( R, , 2, , ®, B = B x i$ =, , + x 2)3/ 2, , ...(v), , ( 2pR), , m 0 IR 2, 2( R 2 + x 2 ) 3 / 2, , i$., , At centre,, , x =0, ® m I, \, B= 0, 2R, 13. A polaroid consists of long chain molecules aligned in a particular direction., , Unpolarised light, , Polarised light, Polaroid, , The electric vectors (associated with the propagating light wave) along the direction of the aligned molecules get, absorbed. Thus, 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. This shows, that light wave is transverse in nature, i.e., electric field is oscillating perpendicular to the direction of propogation., On rotating the pass axis of the polaroid there is no change in the intensity of the transmitted light because in, unpolarised light electric field vector oscillates in all direction so when light is incident the electric field, vector in the pass axis will be absorbed and transmitted light will be linearly polarised. It does not depend on, this orientation of the polaroid., 14. Zener diode is fabricated by heavily doping both p- and n- side of the junction. However, due to diffusion of, maximum number of electrons and holes, the depletion region of order < 10 - 6 m is formed. If a reverse bias, voltage of 5 V is subjected across the junction, an extremely high electric field of order 5 ´ 10 6 V/m gets, established across the junction.
Page 695 :
722 Xam idea Physics–XII, Zener diode as voltage regulator :, The current diagram of zener diode as DC voltage regulator is shown in fig., A, Unregulated, voltage, , RS, , IL, , B, , C, RL Regulated, voltage, D, , The unregulated d.c. voltage, which is filtered output of a rectifier, is connected across the terminals A and B, of the series combination of resistance R s and zener diode such that zener diode remain reverse biased., (i) If input voltage increases, the current through resistance R s and zener diode also increases. This increase, in current increases the voltage drop across resistance R s without any change in the voltage drop across, the zener diode even through the current through zener diode increases., (ii) If input voltage decreases, the current through the resistance R s and zener diode also decreases. The, voltage drop across R s decreases without any change in the voltage across the zener diode., (iii) If voltage drop becomes less than the break down voltage of the junction of zener diode. It does not work., Since the voltage across the zener diode remain constant and works as a d.c. voltage regulator., OR, Electron drift, , Electron diffusion, , (a) Two processes occur during the, formation of a p-n junction are diffusion, – – + +, and drift. Due to the concentration, – – + +, gradient across p and n-sides of the, p, n, – – + +, junction, holes diffuse from p-side to, n-side (p ® n) and electrons diffuse from, – – + +, n-side to p-side (n ® p). This movement, – – + +, of charge carriers leaves behind ionised, Depletion region, Hole diffusion, Hole drift, acceptors (negative charge immobile) on, the p-side and donors (positive charge, immobile) on the n-side of the junction. This potential difference created is called potential barrier. This, space charge region on either side of the junction together is known as depletion region. It is the region, without charge carriers., (b), , (i) Forward Bias: When connected in forward bias holes from P and electrons from n are pushed, towards depletion layer hence barrier potential decreases., (ii) Reverse Bias: When connected in reverse bias holes from P and electrons from n are pulled away, from the depletion layer hence the barrier potential increases., 15. (i) When intensity of light is gradually increased more photons are incident hence more electrons will come, out. Thus anode current will increase with increase of intensity., (ii) When frequency of incident radiation is increased electrons will come out with more kinetic energy but, number of electrons will not change hence there is no change in current., Frequency of light affects the maximum kinetic energy of the emitted photoelectrons., , I, , I, , n0, , n, , n0, , Intensity
Page 696 :
CBSE Examination Papers 723, (iii) Anode potential increased, on increasing anode potential, less number of, electrons having maximum kinetic energy reaches to the negatively, charged plate, and current flow in the circuit decreases. However, at, maximum negative potential V0 of the plate, no electron reaches to the, plate, and hence no current flows in the circuit., , I, , V0, Anode potential, , 16. Active state: In a transistor when change in input voltage causes change in output voltage then it is said to be, in active state., , C1, , p-n-p transistor as amplifier, p-n-p, C, IC, IB, , C2, , B, Vi, , RL, , E, Ri, , IE, , VEE, –, , Output, voltage, V0, , +, , –, , +, VCC, , 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 RL 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., Current Gain. The ratio of change in collector current (DIc) to the change in base current (DI b ) is defined as, the alternating current gain denoted by b . Thus,, DI, b (ac) = c, DI b, b has positive values and is generally greater than 20., Voltage Gain. The voltage gain of common-emitter transistor amplifier is given by, DVout, R DI c æ DI c ö R L, ÷., = L, =ç, Av =, DVin, Ri DI b çè DI b ÷ø Ri, R, Þ, Av = b L, Ri, (i) Base is thin and lightly doped so that maximum charge carriers emitted from emitter reach base and very, few of them recombine to give a small base current (IB), I, b= C, IB, smaller base current give large current gain., (ii) Emitter is heavily doped because the total charge carriers that move in transistor are emitted by emitter, only. Larger is the emitter current more will be collector current hence large current gain., 17. The modulation is needed due to, lö, æ, (i) transmission of audio frequency electrical signals need long impracticable antenna ç h ~ ÷ ., 4ø, è, æ, 1 ö, ÷÷ ., (ii) the power radiated at audio frequency is quite small, hence transmission is quite lossy çç P µ, è, l2 ø, (iii) The various information signals transmitted at low frequency get mixed and hence cannot be distinguished.
Page 698 :
CBSE Examination Papers 725, | E eff | 2 = v R2 + (VL - VC ) 2, E eff = I eff, , R 2 + (X L - X C ) 2, , …(1), , Since, E eff = I eff Z, The impedance of the circuit can be given, Z =, , …(2), , R 2 + (X L - X C ) 2, , The flow of the current in series LCR circuit varies as a function of, frequency of a.c source is shown in figure., Current amplitude is maximum at resonant frequency w0 ; and can be, Veff, æ, 1 ö, ÷÷., given as I max =, . It is possible at Z = R çç or w0 =, R, LC ø, è, Significance :, The resonance phenomenon is exhibit by the circuit only when both, L and C are present in the circuit. When voltages across L and C, cancel each other (on being out of phase). The current in the circuit, Veff, reaches to its maximum value and can be given as, . So, we can, R, not have resonance in a R.L. or R.C. circuits., 11, 11, b + decay, 19. 6 C ¾ ¾ ¾ ¾¾® 5 X + 0 b + 0 v, 0, +1, X is isobar of, , 11, 6 C, 11, Dm = m ( 11, 6 C) - m( 5 X ), , Mass defect,, , = (11.011434 – 11.009305), = 0.002129 u, \, , Q = Dm ´ 9315, . MeV, = 0.002129 × 931.5 MeV, , = 1.98 MeV, 20. u = –15 cm, f = + 10 cm (convex lens), v = ?, 1 1 1, Lens formula,, – =, v u f, 1, 1, 1, 1, 1, 1, =, Þ, =, –, v (–15) 10, v 10 15, 150, v=, = 30 cm, 5, Image formed is real and magnified., A¢¢, A, B¢¢¢, B, , B¢¢, , F, , B¢, , A¢¢¢, A¢, 30 cm, , A¢B¢ ® Image formed by lens, , 20 cm, , I, , w0, , w (m rod/s)
Page 699 :
726 Xam idea Physics–XII, A²B² ® Image formed by mirror, A¢²B¢² ® Image of A²B² formed by lens, For the image to be formed at object the image formed by the mirror should be at object for the mirror. Hence, the image formed by the lens should lie at centre of curvature of the mirror. So that its image can be formed at, object by the lens hence mirror must get placed 50 cm from the lens., \ D = (30 + R) cm = (30 + 20) cm = 50 cm, 21. g rays < X rays < Microwaves < Radio waves, Infra red rays are produced by the vibration of atoms and molecules., (i) Maintaining Earth’s Warmth: Infrared rays are absorbed by the earth’s surface and reradiated as, longer wave length infrared rays. These radiations are trapped by green house gases such as CO2 and, maintain earth’s warmth., (ii) Physical Therapy: Infrared rays are easily absorbed by water molecules present in body. After, absorption, their thermal motion increases causing heating which is used as physical therapy., +Q – Q, 22. When capacitor plates are connected to battery, free electrons present in the, plate are attracted by + ve terminal of the battery giving + Q charge to the, ¯e, e, plate. The electrons from the – ve terminal are accumulated on the other plate, to produce charging on the second plate by induction from first plate. Due to, this transfer of electrons and induction both plates attain equal and opposite, charge., e, When the battery is disconnected the charge on the plates of the capacitor ¯e, e®, e®, does not change., On increasing the distance between the plates patential difference gets doubled., (a) Electric field decreases, s, 1 s, and E f = ,, ,, This is because Ei =, e0, K e0, E =, , Ei, K, , (b) Energy stored increases, New capacitance, , æe Aö, C = çç 0 ÷÷ k, è 2d ø, =, , Initially Energy =, , K, C0 ,, 2, , \C < C 0, , Q2, Q2 2, and Energy =, as 1 < K < 2, ., 2C, C K, , SECTION–D, 23. (a) Power loss in the form of heating through wires is given by I2R., If the power is transmitted at high voltage then low current flows through wires, and hence there is less power loss P = VI, If V increases I decreases., (b) In ac ,, , Pav = EvIv cos f, , where cos f is power factor and f is the phase between EV and IV., Smaller value of cos f will result in higher value of current which will result in, more power loss. Hence low power factor implies large power loss., (c) Values displayed by: Ajit: Scientific temper, keen to learn, social awareness., Uncle: Intelligent, aware, professional honesty., , IV, , f, , EV
Page 700 :
CBSE Examination Papers 727, , SECTION–E, 24. (a) Self inductance of coil: The property of a coil by virtue of which it opposes any change in the amount of, current flowing through it is called self inductance of the coil., Energy stored in a solenoid, When a current grows through an inductor, the induced emf opposes its growth and as such it has to do, some work. When the current in the circuit has attained its maximum value, the induced emf vanishes, and as such the current has to do no work now. The total amount of work done by the current in attaining, its maximum value is stored in the inductor in the form of its magnetic energy. When the circuit is broken, and current stops flowing in it, this stored up energy is released and it maintains the flow of current for, more time., Let current flowing through the circuit at any instant t be i., di, Rate of growth of current at that time =, dt, Induced emf set up in the circuit = e, Clearly, induced emf set up in the inductor, i.e.,, di, (neglecting –ve sign), e= L, dt, Work done by the current in time dt, i.e.,, æ di ö, dW = eidt = Lç ÷i dt, or dW = Li di, è dt ø, Total work done by the current in increasing from 0 to its maximum value (I), i.e.,, I, , I, , 0, , 0, , W = ò dW = ò, , I, , I, , 0, , 0, , i2, Li di = L ò i di = L, 2, , æI2, ö 1, 1, = Lç, – 0 ÷ = LI 2 or W = LI 2, ç 2, ÷ 2, 2, è, ø, , It is this amount of work which is stored in the circuit in the form of its magnetic energy (UB), i.e.,, 1, UB = LI 2, 2, (b) (i) Clockwise NMOP., v = 20 cm/s; d = 20 cm, Time taken by the loop to move out of magnetic field., d, 20, t= =, =1s, v, 20, Induced current will last for 1 second till the length 20 cm moves out of the field., (ii), f, , e=–, , df, dt, , 4, , 5, , t, , 4, , 5, , t
Page 701 :
728 Xam idea Physics–XII, OR, (a) When current flows through a circular coil of area A, magnetic field lines, enter from one face and leaves of other., One face of coil becomes north pole and other south pole it is similar to a, bar magnet., , N, , Magnetic field due to circular loop on its axis for far off points, m 2IA, B= 0, 4p x 3, , C, S, , Magnetic field due to bar magnet at an axial point, m 2m, B= 0, 4p x 3, ®, By comparison, magnetic moment of coil = IA, ®, Hence coil will behave like a bar magnet with magnetic moment IA, (b) Let the solenoid consists of n turns per unit, length. Let its length be 2l and radius a. We can, evaluate the axial field at a point P, at a distance, r from the centre O of the solenoid. To do this,, consider a circular element of thickness dx of the, solenoid at a distance x from its centre. It, consists of n dx turns. Let I be the current in the, solenoid., The magnitude of the field at point P due to the, circular element is, dB =, , x, dx, a, P, O, , r, , l, 2l, , m 0 n dx I a 2, 2[(r – x ) 2 + a 2 ] 3 / 2, , The magnitude of the total field is obtained by summing over all the elements — in other words by, integrating from x = – l to x = + l. Thus,, B=, , m 0 nIa 2, 2, , dx, , l, , ò– l [(r – x ) 2, , + a 2 ]3 / 2, , Consider the far axial field of the solenoid, i.e., r >> a and r >> l. Then the denominator is approximated, by [(r – x)2 + a2]3/2 » r3, and, , B=, , m 0 n I a2, 2r 3, , l, , ò– l dx, , =, , m 0 n I 2la 2, 2, r3, , Note that the magnitude of the magnetic moment of the solenoid is, m = n (2l) I (pa2) – (total number of, turns × current × cross-sectional area). Thus,, m 2m, B= 0, 4p r 3, This is also the far axial magnetic field of a bar magnet which one may obtain experimentally. Thus, a bar, magnet and a solenoid produce similar magnetic fields. The magnetic moment of a bar magnet is thus, equal to the magnetic moment of an equivalent solenoid that produces the same magnetic field., 25., , (a) The intensity of light at a point due to two coherent sources, can be given as, f, I = 4I 0 cos 2, 2
Page 702 :
CBSE Examination Papers 729, (i) If f = 0, ± 2p, ± 4p, . . . the constructive interference leading to maximum intensity, I = 4I 0, (ii) If f = ± p, ± 3p, ± 5p, . . . the destructive interference leads to the zero intensity., X, P, A, S*, , Q, d/2, , d, , O, , d/2, , R, , B, , Y, , D, , PQ = (x – d/2) and PR = (x + d/2), 2, , BP – AP2 = [D2 + (x + d/2)2] – [D2 + (x – d/2)2], , We have, or, , (BP + AP) (BP – AP) = 2xd, 2xd, 2xd, xd, (BP – AP) =, =, =, BP + AP 2D, D, , (as BP » AP » D), , D = (BP – AP) = xd/D, xd, Condition for maxima,, Þ, = nl, n = 0, 1, 2, D = nl, D, nl D, x =, d, l, n = 0, 1, 2, Condition for minima, D = (2n + 1) ,, 2, ( 2n + 1) lD, xd, l, = (2n + 1), Þ, x=, D, 2, 2d, Fringe width (b) is the distance between two successive bright or dark fringe., nlD, lD, lD, Þ, – ( n – 1), b=, b = xn – xn – 1 =, d, d, d, Thus, path difference,, , (b) In the double-slit experiment the pattern on the screen is, superposition of single-slit diffraction from each slit, and the double-slit interference pattern., In the broad diffraction peak there appears several, fringes of smaller width due to double slit interference., 10l 2l, d, (c), =, Þ, a = = 0.2 mm, d, a, s, OR, L, , (a) O – object, , 1, , L, , 2, , I¢ ® Image formed by L1/virtual object, for L2, I ® Image formed by L2, Lens formula,, , 1, 1 1, + =, –u v f, , For lens L1, , 1, 1, 1, +, =, –u v¢ f 1, , O, , u, , I, , v, , I¢, , v¢, , ...(i)
Page 704 :
CBSE Examination Papers 731, 4kQ 2 6kQ 2 12kQ 2, 14kQ 2, –, –, =–, l, l, l, l, Work done in shifting the charge is equal to difference in potential energy., 14 kQ 2 æç - 7kQ 2, W = U A 1B 1C 1 – UABC = ç, l, l, è, =, , ö, ÷, ÷, ø, , - 7kQ 2, l, OR, Electric flux: The electric flux through a given area in an electric field represents the total number of electric, field lines crossing the area normally., SI unit – Nm2 C–1., 1, Gauss’s law: Gauss’s law states that the total electric flux passing through a closed surface is equal to, e0, times the net charge enclosed by the closed surface., ® ®, q, ò E . ds = e 0, If the net charge enclosed is zero then the flux is also zero. Flux through, q, . Flux only depends on the net charge enclosed hence, the cube is, q, e0, shape and size of surface is independent. Cube of any size or closed, surface of any shape will have same flux., =, , Let q be the charge inside the cube, and the cube inside enclosed, surface. Total electric lines emanating from the charge passes, through the cube and surface of any shape. Hence,, q, f = fcube = fs =, e0, Hence, f if independent to the shape and size of the surfaces., zzz
Page 705 :
CBSE Examination Paper, Patna Region-2015 (Code No. 55/1/P), Time allowed : 3 hours, , Maximum marks: 70, , General Instructions: Same as CBSE Examination paper, Delhi–2015., , SECTION–A, 1., 2., , Distinguish between emf and terminal voltage of a cell., The field lines of a negative point charge are as shown in the figure. Does the kinetic energy of a small, negative charge increase or decrease in going from B to A?, , B, A, , –q, , 3., 4., , What is the function of a ‘Repeater’ used in communication system?, The line AB in the ray diagram represents a lens. State whether the lens is convex or concave., A, , B, , 5., , Draw a graph to show variation of capacitive-reactance with frequency in an ac circuit., , SECTION–B, 6., , In a meter bridge shown in the figure, the balance point is found to be 40 cm from end A. If a resistance of, 10 W is connected in series with R, balance point is obtained 60 cm from A. Calculate the value of R and S., R, , S, , G, A, , B, E2
Page 706 :
CBSE Examination Papers 733, 7., , A ray of light passes through an equilateral glass prism such that the angle of incidence is equal to the angle, of emergence and each of these angles is equal to 3/4 of angle of prism. Find the angle of deviation., OR, , Calculate the speed of light in a medium whose critical angle is 45°., Does critical angle for a given pair of media depend on the wavelength of incident light? Give reason., 8. How does one explain, using de Broglie hypothesis, Bohr’s second postulate of quantization of orbital, angular momentum?, 9. What is ground wave communication? Explain why this mode cannot be used for long distance, communication using high frequencies., 10. The equivalent wavelength of a moving electron has the same value as that of a photon of energy 6 × 10–17J., Calculate the momentum of the electron., , SECTION–C, 11. Find the equivalent capacitance of the network shown in the figure, when each capacitor is of 1 µF. When the, ends X and Y are connected to a 6 V battery, find out (i) the charge and (ii) the energy stored in the network., , X, , Y, , 12. State the underlying principle of a potentiometer. Write two factors by which current sensitivity of a, potentiometer can be increased. Why is a potentiometer preferred over a voltmeter for measuring the emf of a, cell?, 13. (a) Define the term ‘intensity of radiation’ in terms of photon picture of light., (b) Two monochromatic beams, one red and the other blue, have the same intensity. In which case (i) the, number of photons per unit area per second is larger, (ii) the maximum kinetic energy of the, photoelectrons is more? Justify your answer., 14. (a) Give two reasons to explain why reflecting telescopes are preferred over refracting type., (b) Use mirror equation to show that convex mirror always produces a virtual image independent of the, location of the object., 15. State Lenz’s law. Illustrate, by giving an example, how this law helps in predicting the direction of the, current in a loop in the presence of a changing magnetic flux., In a given coil of self-inductance of 5 mH, current changes from 4 A to 1 A in 30 ms. Calculate the emf, induced in the coil., OR, In what way is Gauss’s law in magnetism different from that used in electrostatics? Explain briefly., The Earth’s magnetic field at the Equator is approximately 0.4 G. Estimate the Earth’s magnetic dipole, moment. Given : Radius of the Earth = 6400 km., 16. How are electromagnetic waves produced? What is the source of energy of these waves?, Draw a schematic sketch of the electromagnetic waves propagating along the + x-axis. Indicate the directions, of the electric and magnetic fields. Write the relation between the velocity of propagation and the magnitudes, of electric and magnetic fields., 17. Obtain the relation between the decay constant and half life of a radioactive sample., The half life of a certain radioactive material against a-decay is 100 days. After how much time, will the, undecayed fraction of the material be 6.25% ?, 18. Write two important considerations used while fabricating a Zener diode. Explain, with the help of a circuit, diagram, the principle and working of a Zener diode as voltage regulator., 19. Draw the necessary energy band diagrams to distinguish between conductors, semiconductors and insulators., How does the change in temperature affect the behaviour of these materials? Explain briefly.
Page 707 :
734 Xam idea Physics–XII, 20. (a) What are the three basic units in communication systems? Write briefly the function of each of these., (b) Write any three applications of the internet used in communication systems., 21. (a) Write the necessary conditions to obtain sustained interference fringes., (b) In Young’s double slit experiment, plot a graph showing the variation of fringe width versus the distance, of the screen from the plane of the slits keeping other parameters same. What information can one obtain, from the slope of the curve?, (c) What is the effect on the fringe width if the distance between the slits is reduced keeping other parameters, same?, 22. A series LCR circuit is connected across an ac source of variable angular frequency ‘w’. Plot a graph showing, variation of current ‘i’ as a function of ‘w’ for two resistances R1 and R2 (R1 > R2)., Answer the following questions using this graph:, (a) In which case is the resonance sharper and why?, (b) In which case is the power dissipation more and why?, , SECTION–D, 23. During a thunderstorm the ‘live’ wire of the transmission line fell down on the ground from the poles in the, street. A group of boys, who passed through, noticed it and some of them wanted to place the wire by the, side. As they were approaching the wire and trying to lift the cable, Anuj noticed it and immediately pushed, them away, thus preventing them from touching the live wire. During pushing some of them got hurt. Anuj, took them to a doctor to get them medical aid., Based on the above paragraph, answer the following questions:, (a) Write the two values which Anuj displayed during the incident., (b) Why is it that a bird can sit on a suspended ‘live’ wire without any harm whereas touching it on the, ground can give a fatal shock?, (c) The electric power from a power plant is set up to a very high voltage before transmitting it to distant, consumers. Explain, why., , SECTION–E, 24. Two infinitely long straight parallel wires, ‘1’ and ‘2’, carrying steady currents I1 and I2 in the same direction, ®, are separated by a distance d. Obtain the expression for the magnetic field B due to the wire ‘1’ acting on, wire ‘2’. Hence find out, with the help of a suitable diagram, the magnitude and direction of this force per, unit length on wire ‘2’ due to wire ‘1’. How does the nature of this force change if the currents are in opposite, direction? Use this expression to define the SI unit of current., OR, Draw a necessary arrangement for winding of primary and secondary coils in a step-up transformer. State its, underlying principle and derive the relation between the primary and secondary voltages in terms of number, of primary and secondary turns. Mention the two basic assumptions used in obtaining the above relation., State any two causes of energy loss in actual transformers., 25. (a) State Kirchhoff’s rules and explain on what basis they are justified., (b) Two cells of emfs E1 and E2 and internal resistances r1 and r2 are connected in parallel. Derive the, expression for the (i) emf and (ii) internal resistance of a single equivalent cell which can replace this, combination., OR, (a) ‘‘The outward electric flux due to charge + Q is independent of the shape and size of the surface which, encloses it.’’ Give two reasons to justify this statement., (b) Two identical circular loops ‘1’ and ‘2’ of radius R each have linear charge densities – l and + l C/m, respectively. The loops are placed coaxially with their centres R 3 distance apart. Find the magnitude, and direction of the net electric field at the centre of loop ‘1’.
Page 708 :
CBSE Examination Papers 735, 26. (a) Use Huygens’ principle to show the propagation of a plane wavefront from a denser medium to a rarer, medium. Hence find the ratio of the speeds of wavefronts in the two media., (b), , (i) Why does an unpolarised light incident on a polaroid get linearly polarised?, (ii) Derive the expression of Brewster’s law when unpolarised light passing from a rarer to a denser, medium gets polarised on reflection at the interface., OR, , A biconvex lens with its two faces of equal radius of curvature R is made of a transparent medium of, refractive index µ1. It is kept in contact with a medium of refractive index µ2 as shown in the figure., , R, , R, , µ1, , µ2, , (a) Find the equivalent focal length of the combination., (b) Obtain the condition when this combination acts as a diverging lens., (c) Draw the ray diagram for the case µ1 > (µ2 + 1) /2, when the object is kept far away from the lens. Point, out the nature of the image formed by the system., zzz
Page 709 :
736 Xam idea Physics–XII, , Solutions, SECTION–A, (i) The emf of a cell is equal to the terminal voltage, when the circuit is open., The emf of a cell is less than the terminal voltage, when the cell is being charged, i.e.,, V = E + ir, 2. The kinetic energy of a negative charge decreases while going from point B to point A, against the movement, of force of repulsion., 3. A repeater picks up a signal, amplifies it, and retransmit it, thereby extending the range of a communication, system., 4. Concave lens., XC, 5., Capacitive Reactance, , 1., , w/n, , Frequency, , SECTION–B, 6., , R 40, 2S, =, Þ 3R = 2S Þ R =, S, 60, 3, R + 10 60, =, Þ 2R + 20 = 3S, S, 40, From equation (i) and (ii), we get, 2S, 2´, + 20 = 3S, 3, Þ, , S = 12 ohm, , From equation (i), we get, R =, Þ, , R = 8 ohm, i + e =A + D, , 7. In prism, and i = e =, , 2 ´ 12, 3, , 3, A, 4, , (given), , 3, 3, A + A, 4, 4, 3A, A, D =, Þ, -A =, 2, 2, Since A = 60° (being an equilateral glass prism), So,, , A + D =, , ...(i), ...(ii)
Page 710 :
CBSE Examination Papers 737, So,, , D =, , 60°, = 30°, 2, OR, , Critical angle in the medium, iC = 45°, So,, , refractive index, µ =, , Þ, , 1, 1, =, sin iC, sin 45°, , µ = 2, Refractive index, µ =, 2 =, , Þ, , cm =, , c0, cm, 3 ´ 108, cm, 3 ´ 108, 2, , = 2.1 ´ 108 m/s, , Yes, critical angle for a pair of media depends on wavelength, because m = a +, , b, l2, , , where a and b are, , constants of the media., 8. For a stationary orbit,, 2pr = n l, , ...(i), , From de Broglie hypothesis wavelength of the electron,, h, ...(ii), l=, p, From equations (i) and (ii), we get, æhö, 2pr = n ç ÷, è pø, Þ, , rp =, , nh, 2p, , angular momentum, l = r ´ tangent momentum P, nh, l =, \, 2p, It is known as Bohr’s second postulate of quantization of angular momentum., 9. When an electromagnetic wave glides over the earth’s surface, its changing magnetic field induces an electric, current on the surface, which in turn produces a changing magnetic field. It is known as ground wave, communication., At higher frequency, the rate of variation of magnetic field is larger inducing a larger current on the surface., So, energy dissipation of the signal increases. So, the higher is the frequency, more attenuation occur, hence, this mode cannot be used for long distance communication., 10. From de Broglie hypothesis, the wavelength of the electron, h, ...(i), le =, p, Since,, , and, Since, , photon energy, hn =, , hc, l photon, , l photon = l e, , From equations (i) and (ii), we get, , = 6 ´ 10 - 17 J, , ...(ii)
Page 711 :
738 Xam idea Physics–XII, hc, 6 ´ 10, , - 17, , h, p, , =, , 6 ´ 10 - 17, 6 ´ 10 - 17, =, c, 3 ´ 108, , Þ, , p =, , Þ, , p = 2 ´ 10 - 25 kg m/s, , SECTION–C, 11. The given circuit can be rearranged as, A, , C, , C, X, , C/2, , C, , C, Y, , C, , X, , Y, , C, , C/2, C, , C, , C, , C, , Þ, , Þ, , Þ, , B, , It is known as wheatstone bridge of the capacitor., Since VA = VB , so the bridge capacitor between points A and B can be removed., (i) The equivalent capacitor of the network, C eq =, =, , C´C, C´C, +, C +C, C +C, C, C, +, 2, 2, , = C = 1 mF, Charge in the network,, , Q = C eq V, =C ´V, = 1 mF ´ 6 V = 6 mC, , (ii) Energy stored in the capacitor,, 1, C eq V 2, 2, 1, = ´ 1 mF ´ ( 6) 2, 2, , U =, , = 18 mJ, 12. Principle: The potential drop across a part of the potentiometer wire is directly proportional to the length of, that part of the wire of uniform cross section., V =kl, where k is potential gradient., Current sensitivity of potentiometer wire is also known as potential gradient, and it can be increased., (i) By increasing the total length of the wire, keeping terminal voltage constant., (ii) By connecting a suitable extra resistance R in series with the potentiometer. So, less amount of the, current flows through the potentiometer wire., Reasons: At the balance point, there is no net current drawn from the cell, and cell is in open circuit, condition. Voltmeter has some resistance, when connected across the cell. Some current is drawn, as a result, emf of the cell decreases. Hence, emf of the cell cannot be measured by the voltmeter.
Page 712 :
CBSE Examination Papers 739, 13., , (a) The number of photons incident normally per unit area per unit time is determined the intensity of radiations., (b), , 14., , (i) Red light, because the energy of red light is less than that of blue light, ( hn) R < ( hn) B, (ii) Blue light, because the energy of blue light is greater than that of red light, ( hn) B > ( hn) R, (a) Advantages: (i) Parabolic mirror is used to remove the spherical aberration., (ii) No chromatic aberration in mirror., (iii) Light mechanical support is required, because mirror weighs much less than a lens of equivalent, optical quality., 1 1 1, (b) From equation, = +, f v u, u < 0, i. e. , always negative for the object position in front of convex/concave mirror, f > 0, i. e. , focal, length is always positive for a convex mirror., 1 1 1, = v f, u, Þ, On using sign convention,, , v =, , fu, u -f, , v =, , ( + f ) ( - u), ( -u) - ( + f ), , v = +ve, v >0, Hence, image is behind the mirror and convex mirror always forms a virtual image., 15. When the North pole of a bar magnet moves towards the closed, coil, the magnetic flux through the coil increases. This produces an, induced emf which produces (or tend to produce if the coil is open), an induced current in the anti-clockwise sense. The anti-clockwise, sense corresponds to the generation of North pole which opposes, the motion of the approaching N pole of the magnet. The face of, N, the coil, facing the approaching magnet, then has the same polarity, as that of the approaching pole of the magnet. The induced current,, therefore, is seen to oppose the change of magnetic flux that, produces it., When a North pole of a magnet is moved away from the coil, the, current (I) flows in the clock-wise sense which corresponds to the, generation of South pole. The induced South pole opposes the, N, motion of the receding North pole., It states that the polarity of induced emf is such that it tends to, produce a current which opposes the change in magnetic flux that produced it., Magnitude of the induced emf in an inductor coil is given by, dI, | e| =L, dt, =, , 5 ´ 10 - 3 ´ ( 4 - 1), 30 ´ 10 - 3, , = 0.5 V
Page 713 :
740 Xam idea Physics–XII, OR, As we know that, ®, , ®, , Isolated positive or negative charge exists freely. So, Gauss’s law states that ò E . d S =, , 1, [ q], e0, , Isolated magnetic poles do not exist. So, Gauss’s law states that, , ò, , ®, , ®, , B . dS = 0, , Magnetic field intensity at the equator is, m, m, B = 0 ., 4p R 3, m, = 10 - 7, R3, m = 107 . B | R | 3, , \, , = 107 ´ 0.4 ´ 10 - 4 ´ ( 6400 ´ 10 3 ) 3, = 1.1 ´ 10 23 Am 2, 16. Electromagnetic waves are produced by the accelerated charge. The battery or an electric field that, accelerates the charge particles is the source of energy of electromagnetic waves., x, , E, , B, , z, B, y, , ®, Directions of E along y axis/ z axis., ®, Directions of B Along z axis/ y axis., | E0 |, The speed of em wave is | C | =, | B0 |, 17. Relation between Half-life and Decay Constant : The radioactive decay equation is, N = N 0 e - lt, N, when t = T , N = 0, 2, N0, = N 0 e - lT, \, 2, 1, or, e - lT =, 2, Taking log of both sides, -lT log e e = log e 1 - log e 2, or, lT = log e 2, log e 2, T=, \, l, 2 .3026 log10 2 2 .3026 ´ 0 .3010, =, =, l, l, 0 . 6931, or, T=, l, , …(i), , ...(ii), , ...(iii)
Page 714 :
CBSE Examination Papers 741, We know that, N æ1ö, =ç ÷, Nv è 2 ø, , n, , 6.25 æ 1 ö, =ç ÷, 100 è 2 ø, , n, , 1 æ1ö, =ç ÷, 16 è 2 ø, , n, , 4, , æ1ö, æ1ö, ç ÷ =ç ÷, è2ø, è2ø, , n, , Electron energy, , n=4, Þ, Required time, t = n ´ T1 / 2 = 4 ´ 100 = 400 days., 18. (i) Heavily doping both p and n sides of the junction., (ii) Appropriate break down voltage under reverse bias., A zener diode is considered as a special purpose semiconductor diode, because it is designed to operate under reverse bias in the breakdown, region., We know that reverse current is due to the flow of electrons (minority, carriers) from p ® n and holes from n ® p. As the reverse bias voltage, is increased, the electric field at the junction becomes significant. When, the reverse bias voltage V = Vz, 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. These electrons causes high current at, breakdown., 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. Now, any Unregulated, voltage, increase/decrease in the input voltage results in, (VL), increase/decrease of the voltage drop across R s without, any change in voltage across the zener diode. Thus, the, zener diode acts as a voltage regulator., 19., , I (mA), , Forward bias, Reverse bias, Vz, V(V), , I (µA), , Rs, , IL, Load, RL, , Conduction band, Ev, Eg @ 0, , Ec, , Valence band, , Conduction band, Ec, E g < 3 eV, Ev, , Electron energy, , Electron energy, , Conductor, Ec, E g > 3 eV, Ev, , Valence band, Semiconductor, , Insulator, , Regulated, voltage, (Vz)
Page 715 :
742 Xam idea Physics–XII, Distinguishing features, (a) In conductors: Valence band and conduction band overlap each other., In semiconductors: Valence band and conduction band are separated by a small energy gap., In insulators: They are separated by a large energy gap., (b) In conductors: Large number of free electrons are available in conduction band., In semiconductors: A very small number of electrons are available for electrical conduction., In insulators: Conduction band is almost empty i.e., no electron is available for conduction., Effect of Temperature:, (i) In conductors: At high temperature, the collision of electrons become more frequent with the, atoms/molecules at lattice site in the metals as a result the conductivity decreases (or resistivity, increases)., (ii) In semiconductors: As the temperature of the semiconducting material increases, more electron hole, pairs becomes available in the conduction band and valance band, and hence the conductivity increases, or the resistivity decreases., (iii) In insulators: The energy band between conduction band and valance band is very large, so it is, unsurpassable for small temperature rise. So, there is no change in their behaviour., 20. The three basic units in communication system are (i) transmitter (ii) communication channel and, (iii) Receiver., Function:, (i) Transmitter: It processes the incoming message signal on suitable carrier waves to makes it suitable for, transmission., (ii) Communication Channel: It is a link for propagating the signals from transmittor to receiver., (iii) Receiver: It extracts the message signal from the signal received at the end of the channel., Applications of Internet:, (i) Internet surfing, (ii) E-banking, (iii) E-booking, (iv) E-shopping, (v) E-mails, (vi) Social networking etc., 21. (a) For conditions of sustained interference fringes :, (i) The interfering sources must be coherent i.e., sources must have same frequency and constant initial, phase., (ii) Interfering waves must have same or nearly same amplitude, so that there may be contrast between, maxima and minima., Dl, Fringe width b =, d, where D = distance between slits and screen., d = separation between slits., l = wavelength of light., (b) Information from the slope :, wavelength,, , (c) Effect: From relation,, Fringe width,, , l = Slope ´ d, = d . tan q, lD, b=, d, 1, bµ, d, , Fringe, width, , (b), , If distance d between the slits is reduced, the size of fringe width will increase., , q, D
Page 716 :
CBSE Examination Papers 743, w0 L, R, 1, Sharpness of resonance µ, R, , 22. (a) Sharpness of resonance =, Þ, , Io, , Since R 2 < R1 , hence resonance is sharp at R 2 ., , R2, , I, , V2, R, 1, (for constant V), P µ, R, , (b) Power dissipation, P =, Þ, , R1 > R2, , Since R 2 < R1 . Hence, power dissipation is more at R 2 ., , w, , wo, , SECTION–D, 23. (a) Values displayed by Anuj are:, Presence of mind, helpful, awareness, (b) When a bird perches on a bare high power line; its points of contact are at the same potential, so no, transfer of charge takes place. In other words no current flows through the bird; so nothing happens to, the bird; on the other hand if a man standing on ground touches the same line, there is established a, potential difference between the (touching part) and feet of man; so there is instantaneous flow of, sufficient quantity of charge from power line to earth through the conducting human body; hence the, man gets a fatal shock., (c) As we know that power loss in the current carrying wire is P = I 2 R. When the power is transmitted in the, wire by a step up transformer, the current decreases to a very low level. Hence, I2R loss is cut down or, minimised., , SECTION–E, 24. 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. It has been observed, experimentally that when the currents in the, wire are in the same direction, they experience, an attractive force (fig. a) and when they carry, currents in opposite directions, they experience a, repulsive force (fig. b)., Let the conductors PQ and RS carry currents, I 1 and I 2 in same direction and placed at, separation d., Consider a current–element ‘ab’ of length DL of, wire RS. The magnetic field produced by, current-carrying conductor PQ at the location of, other wire RS, m I, B1 = 0 1, 2p d, , P, , R, , P, , R, , b, , I2, DL, , DF, , a, , I1, Q, , B, , B, I2, , I1, , S, , Q, , a, DL, b, , DF, , S, , d, Fig. (a), , Fig. (b), , ...(i), , According to Maxwell’s right hand rule or right hand palm rule number 1, the direction of B1 will be, perpendicular to the plane of paper and directed downward. Due to this magnetic field, each element of other, wire experiences a force. The direction of current element is perpendicular to the magnetic field; 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 d
Page 717 :
744 Xam idea Physics–XII, \, , 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 d, 2p d, , \, , Force acting per unit length of conductor, F m 0 I1I 2, f =, =, N/ m, L, 2p d, , ...(ii), , According to Fleming’s left hand rule, the direction of magnetic force will be towards PQ i.e. the force will, be attractive., On the other hand if the currents I 1 and I 2 in wires are in opposite directions, the force will be repulsive., The magnitude of force in each case remains the same., Definition of SI unit of Current (ampere): In SI system of fundamental unit of current ‘ampere' is defined, assuming the force between the two current carrying wires as standard., The force between two parallel current carrying conductors of separation d is, F m 0 I1I 2, f =, =, N/ m, L, 2pd, If I 1 = I 2 = 1 A, d = 1 m, then, m, f = 0 = 2 ´ 10 - 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., OR, Transformer: Transformer is a device by which an alternating voltage may be decreased or increased. It is, based on the principle of mutual-induction., Construction: It consists of laminated core of soft iron, on which two coils of insulated copper wire are, separately wound. These coils are kept insulated from each other and from the iron-core, but are coupled, through mutual induction. 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. 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. Transformers are of two types:, 1. 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. (i. e. , N S > N p )., 2. 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. e. , N S < N p ), (A.C. mains), Primary, , laminated, iron core, , Secondary, Transformer, , Primary, , Primary, , Secondary, , Secondary, , (a) Step up, , (b) Step down, , Core
Page 718 :
CBSE Examination Papers 745, Working: When alternating current source is connected to the ends of primary coil, the current changes, continuously in the 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., Let N p be the number of turns in primary coil, NS the number of turns in secondary coil and f the magnetic, flux linked with each turn. 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. According to Faraday’s laws the emf induced in the primary, coil, Df, ...(i), ep = - Np, Dt, and emf induced in the secondary coil, Df, ...(ii), eS = - NS, Dt, From (i) and (ii), N, eS, ...(iii), = S, Np, ep, If the resistance of primary coil is negligible, the emf ( e p ) induced in the primary coil, will be equal to the, applied potential difference (V p ) across its ends. Similarly if the secondary circuit is open, then the, potential difference VS across its ends will be equal to the emf ( e S ) induced in it; therefore, VS, e, N, ...(iv), = S = S = r (say), Vp, ep, Np, where r =, , NS, is called the transformation ratio. If i p and i s are the instantaneous currents in primary and, Np, , secondary coils and there is no loss of energy; then, For about 100% efficiency, Power in primary = Power in secondary, V p i p = VS i S, Vp, Np, iS, 1, ...(v), \, =, =, =, ip, VS, NS, r, In step up transformer, N s > N p, , ® r >1;, , So, VS > V p and i S < i p, i.e. step up transformer increases the voltage., In step down transformer, N S < N p ® r < 1, so, VS < V p and i S > i p, i.e. step down transformer decreases the voltage, but increases the current., Assumptions:, (a) Flux linked with each turn of primary and secondary coils must be constant., (b) Induced emf in the primary coil must be equal to applied alternating voltage across it., (c) The primary resistance and current must be small., Causes of energy loss in actual transformers:, (i) Joule Heating: Energy is lost in resistance of primary and secondary windings as heat ( I 2 Rt)., (ii) Flux Leakage: Energy is lost due to coupling of primary and secondary coils not being perfect, i.e.,, whole of magnetic flux generated in primary coil is not linked with the secondary coil., 25. (a) Kirchhoff’s Laws, (i) First law (or junction law): The algebraic sum of currents meeting at any junction is zero,, i.e.,, SI = 0, This law is based on conservation of charge.
Page 719 :
746 Xam idea Physics–XII, (ii) Second law (or loop law): The algebraic sum of potential differences of different circuit elements, of a closed circuit (or mesh) is zero, i.e.,, SV = 0, This law is based on conservation of energy., E1, , (b), , I1, , r1, B, , A, E2, , Eq, , I = I1 + I2, , I2, , r2, , I, (b), , (a), , Let I 1 and I 2 be the currents leaving the positive, terminals of the cells, and at the point B, I = I1 + I 2, …(i), Let V be the potential difference between points A and B of the combination of the cells, so, V = E1 - I 1r1, …(ii) (across the cells), and, V = E 2 - I 2 r2, From equation (i), (ii) and (iii), we get, (E - V ) (E 2 - V ), I = 1, +, r1, r2, , …(iii), , E ö, æ1, æE, 1 ö, = çç 1 + 2 ÷÷ - V çç +, ÷÷, r, r, r, r, è 1, è 1, 2 ø, 2 ø, Fig. (b) shows the equivalent cell, so for the same potential difference, V = E eq - Ir q, E eq, V, or, I =, rq, rq, , ...(iv), , ...(v), , On comparing Eq. (iv) and (v), we get, E eq, E, E, = 1 + 2, r1, r2, rq, and, , 1, 1, 1, =, +, r q r1, r2, , Þ, , r eq =, , r1r 2, r1 + r 2, , On further solving, we have, æ1, 1, E eq çç +, r2, è r1, Þ, , E, ö E1, + 2, ÷÷ =, r2, ø r1, E r + E 2 r1, E eq = 1 2, r1 + r 2, OR, , (a) In figure, A charge + Q is enclosed in side the surfaces S1 and S 2 ., (i) For a given charge Q the same number of electric field lines, emanating from the surfaces S1 and S 2 depends on the charge Q, and independent to the shape and size of the surfaces of S1 and S 2 ., (ii) From Gauss’s law the net-outward electric flux through any closed, 1, times the charge, surface of any shape and size is equal to, e0, enclosed within that surface i.e.,, ®, ®, ®, ®, Q, S1 E . d S = S 2 E . d S =, e0, , S2, , +Q, , S1
Page 720 :
CBSE Examination Papers 747, (b), , 2 +l, , 1, R, , R, , –l, , q, O2, , q, , O1, , Z=R 3, , Electric field at the centre O1 due to loop 1 is given by, ®, , | E1 | = 0, , (As Z = 0), , Electric field at a point outside the loop 2 on the axis passing normally through O2 of loop 2 is, ®, lR, Z, ., | E2 | =, 2, 2e 0 ( R + Z 2 ) 3 / 2, Since, , Z =R 3, =, , lR, R 3, ., 2, 2e 0 ( R + 3R 2 ) 3 / 2, , l 3, towards right (As l is positive), 16e 0 R, So, net electric field at the centre of loop 1, =, , ®, , ®, , ®, , E = E1 + E 2, , =0+, =, , l 3, 16e 0 R, , l 3, 16e 0 R, , 26. (a), Spherical wavefront at radius R/2, Concave mirror, at radius R, , Incident, plane wave, , F, , Convex lens, , F, Spherical wavefront, at radius f, , Figure shows the refraction from denser medium ( n1 ) into rarer medium ( n 2 < n1 ). The angle of, refraction r > i and from Snell’s law, n1 sin i = n 2 sin r, At critical angle, i = i c, , ...(i)
Page 721 :
748 Xam idea Physics–XII, , \, , n1 sin i c = n 2 sin 90° = n 2, n, sin i c = 2, n1, , ...(ii), , From equation (i), we have, n 2 sin i, =, = ratio of the speed of light, n1 sin r, Let v1 and v 2 be the speed of wavefronts in medium n1 and n 2 ., v2, sin i, =, Hence,, v1 sin r, (b), , ...(iii), , (i) A polaroid consists of long chain molecules aligned in a particular direction. The electric vectors, along the direction of the aligned molecules get absorbed. So, when an unpolarised light falls on a, polaroid, it lets only those of its electric vectors that are oscillating along a direction perpendicular, to its aligned molecules to pass through it. The incident light thus gets linearly polarised., (ii) Condition : The reflected ray is totally plane polarised, when, D, reflected and refracted rays are perpendicular to each other., B, A, ÐBOC = 90°, When reflected wave is perpendicular to the refracted wave, Incident, Reflected ray, ip r', the reflected wave is a totally polarised wave. The angle of ray, 90°, incidence in this case is called Brewster’s angle and is, O, denoted by i B ., r, Refracted ray, If r ¢ is angle of reflection and r the angle of refraction, then, according to law of reflection, C, iB = r ¢, E, and from fig. r ¢ + 90° + r = 180°, …(i), Þ, i B + r = 90°, …(ii), Þ, r = ( 90° - i B ), From Snell’s law, refractive index of second medium relative to first medium (air) say., sin i B, sin i B, sin i B, n =, =, =, sin r, sin ( 90° - i B ) cos i B, Þ, n = tan i B, This is known as Brewtser law., \, , Angle of incidence, i B = tan -1 ( n)., OR, , (a) If refraction occurs at first surface, m 1 1 (m 1 - 1), - =, v1, u, R, , 1, , 2, , ...(i), R, , R, , If refraction occurs at second surface, and the image of the first surface acts as an, object, m2, m, m - m1, ...(ii), - 1 = 2, v, v1, -R, On adding equation (i) and (ii), we get, m2, 1 2 m1 - m 2 - 1, - =, v, u, R, If rays are coming from infinity, i.e., u = - ¥ then v = f, m2, 2 m1 - m 2 - 1, 1, =, +, f, ¥, R, , µ1, , µ2
Page 723 :
CBSE Examination Paper,, Foreign-2015 (Code No. 55/2/1/F), Time allowed: 3 hours, , Maximum marks: 70, , General Instructions: Same as CBSE Examination Paper Delhi-2015., , SECTION–A, ®, , ®, , ®, , 1., , Write the expression for the torque t acting on a dipole of dipole moment p placed in an electric field E ., , 2., 3., , Plot a graph showing the variation of resistivity of a conductor with temperature., The power factor of an AC circuit is 0.5. What is the phase difference between voltage and current in this, circuit?, Is it necessary for a transmitting antenna to be at the same height as that of receiving antenna for LOS, communication?, Why does the Sun look reddish at sunset or sunrise?, , 4., 5., , SECTION–B, 6., , 7., 8., 9., , Given the resistances of 1 W, 2 W and 3 W. How will you combine them to get an equivalent resistance of, 11, 11, (i) W and (ii) W ?, 3, 5, The kinetic energy of the electron orbiting in the first excited state of hydrogen atom is 3.4 eV. Determine the, de Broglie wavelength associated with it., (a) Write the b-decay of tritium in symbolic form., (b) Why is it experimentally found difficult to detect neutrinos in this process?, The V – I characteristic of a silicon diode is as shown in the figure. Calculate the resistance of the diode at, (i) I = 15 mA and (ii) V = – 10 V., I (mA), 30, Silicon, 20, 15, 10, , –10V, , 0, 0.5, – –1mA, , 0.8, , V (V), , 10. How does the refractive index of a transparent medium depend on the wavelength of incident light used?, Velocity of light in glass is 2 × 108 m/s and in air is 3 × 108 m/s. If the ray of light passes from glass to air,, calculate the critical angle.
Page 724 :
CBSE Examination Papers 751, OR, An equiconvex lens of focal length ‘f’ is cut into two identical plane convex lenses. How will the power of, each part be related to the focal length of the original lens?, A double convex lens of + 5D is made of glass of refractive index 1.55 with both faces of equal radii of, curvature. Find the value of its radius of curvature., SECTION–C, 11. Four charges +q, – q, + q and – q are to be arranged respectively at the four corners of a square ABCD of side ‘a’., (a) Find the work required to put together this arrangement., (b) A charge q 0 is brought to the centre of the square, the four charges being held fixed. How much extra, work is needed to do this?, OR, Three point charges +q each are kept at the vertices of an equilateral triangle of side ‘l’. Determine the, magnitude and sign of the charge to be kept at its centroid so that the charges at the vertices remain in, equilibrium., 12. (a) An electron moving horizontally with a velocity of 4 × 104 m/s enters a region of uniform magnetic field, of 10–5 T acting vertically upward as shown in the figure. Draw its trajectory and find out the time it takes, to come out of the region of magnetic field., ×, , ×, , ×, , ×, , ×B, , ×, , ×, , ×, , ×, , ×, , ×, , ×, , –e, , (b) A straight wire of mass 200 g and length 1.5 m carries a current of 2A. It is suspended in mid air by a, uniform magnetic field B. What is the magnitude of the magnetic field?, 13. A plane wavefront propagating in a medium of refractive index ‘m1’ is incident on a plane surface making the, angle of incidence ‘i’ as shown in the figure. It enters into a medium of refractive index ‘m2’ (m2 > m1). Use, Huygens’ construction of secondary wavelets to trace the propagation of the refracted wavefront. Hence, verify Snell’s law of refraction., , i, , 14. A circular coil of radius 10 cm, 500 turns and resistance 200 W is placed with its plane perpendicular to the, horizontal component of the Earth’s magnetic field. It is rotated about its vertical diameter through 180° in, 0.25 s. Estimate the magnitudes of the emf and current induced in the coil. (Horizontal component of the, Earth’s magnetic field at the place is 3.0 × 10–5 T), 15. Using Rydberg formula, calculate the longest wavelengths belonging to Lyman and Balmer series. In which, region of hydrogen spectrum do these transitions lie?, [Given R = 1.1 × 107 m–1]
Page 725 :
752 Xam idea Physics–XII, 16. Why cannot two independent monochromatic sources produce sustained interference pattern?, Deduce with the help of young’s arrangement to produce interference pattern, an expression for the fringe, width., 17. Answer the following questions:, (a) Define ‘bandwidth’ and describe briefly its importance in communicating signals., (b) Distinguish between digital and analogue signals., (c) Write the functions of transducer and repeater., 18. Write briefly the important processes that occur during the formation of p–n junction. With the help of, necessary diagrams, explain the term ‘barrier potential’., 19. A beam of monochromatic radiation is incident on a photosensitive surface. Answer the following questions, giving reasons:, (a) Do the emitted photoelectrons have the same kinetic energy?, (b) Does the kinetic energy of the emitted electrons depend on the intensity of incident radiation?, (c) On what factors does the number of emitted photoelectrons depend?, 20. (a) What is an ‘integrated circuit (I.C.)’? Distinguish between (i) linear I.C. and (ii) Digital I.C., (b) Identify the equivalent gate for the following circuit and write its truth table., A, , P, , B, , Y, , X, , 21. Why does a galvanometer when connected in series with a capacitor show a momentary deflection, when it is, being charged or discharged?, How does this observation lead to modifying the Ampere’s circuital law? Hence write the generalised, expression of Ampere’s law., 22. In the given circuit in the steady state, obtain the expression for (a) the potential drop (b) the charge and, (c) the energy stored in the capacitor, C., A, , B, , V, , R, , V, , C, , C, , E, , 2R, , 2V, , F, , D, , SECTION–D, 23. Sushil is in the habit of charging his mobile and then leaving the charger connected through the mains with, the switch on. When his sister Asha pointed it out to him, he replied there was no harm as the mobile had been, disconnected. Asha then explained to him and convinced him. How the energy was still being wasted as the, charger was continuously consuming energy., Answer the following questions:, (a) What values did Asha display in convincing her brother?, (b) What measures, in your view, should be adopted to minimize the wastage of electric energy in your, household?, (c) Imagine an electric appliance of 2W, left connected to the mains for 20 hours. Estimate the amount of, electrical energy wasted.
Page 726 :
CBSE Examination Papers 753, , SECTION–E, 24. Write any two important points of similarities and differences each between Coulomb’s law for the, electrostatic field and Biot-Savart’s law for the magnetic field., Use Biot-Savart’s law to find the expression for the magnetic field due to a circular loop of radius ‘r’ carrying, current ‘I’, at its centre., OR, What are eddy currents? How are they produced?, Describe briefly three main useful applications of eddy currents., 25. (a) A small conducting sphere of radius ‘r’ carrying a charge +q is surrounded by a large concentric, conducting shell of radius R on which a charge +Q is placed. Using Gauss’s law, derive the expressions, for the electric field at a point ‘x’., (i) between the sphere and the shell (r < x < R),, (ii) out side the spherical shell., (b) Show that if we connect the smaller and the outer sphere by a wire, the charge q on the former will always, flow to the latter, independent of how large the charge Q is., OR, ® ® ®, , ®, , (a) Consider a system of n charges q1, q2, ... qn with position vectors r1 , r 2 , r 3 , ... r n relative to some origin, ®, , ®, , ‘O’. Deduce the expression for the net electric field E at a point P with position vector r p , due to this, system of charges., (b) Find the resultant electric field due to an electric dipole of dipole moment, 2aq, (2a being the separation, between the charges ± q) at a point distant ‘x’ on its equator., 26. Draw a ray diagram showing the image formation of a distant object by a refracting telescope. Define its, magnifying power and write the two important factors considered to increase the magnifying power., Describe briefly the two main limitations and explain how far these can be minimised in a reflecting, telescope., OR, (a) Draw a ray diagram showing image formation in a compound microscope. Define the term ‘limit of, resolution’ and name the factors on which it depends. How is it related to resolving power of a, microscope?, (b) Suggest two ways by which the resolving power of a microscope can be increased., (c) “A telescope resolves whereas a microscope magnifies.” Justify this statement., zzz
Page 727 :
754 Xam idea Physics–XII, , Solutions, SECTION–A, 1. Torque acting on the dipole, placed in uniform electric field is given by, ®, , ®, , ®, , q, , Resistivity r (10–8Wm), , t =P ´ E, , 2., , +q, , P, , E, , –q, , 0.4, , I 0.2, , 0, 50, 100, Temperature T(K), , 150, , 3. Power factor between voltage and current is given by cos f , where f is phase difference, 1, cos f = 0.5 =, 2, 1, p, f = cos - 1 æç ö÷ =, è2ø 3, , \, 4. No., , 5. During sunset or sunrise, the sun is just above the horizon, the blue colour gets scattered most by the, atmospheric molecules while red light gets scattered least, hence sun appears red., 1, Reason: Scattering intensity I µ, and l B << l R ., l4, , SECTION–B, 6., , 2, 11 æ, W ç = 3 + ö÷ W the resistance of 3 W is connected in series with the, 3ø, 3 è, parallel combination of resistors of 1 W and 2 W., , (i) For obtaining the resistance of, , 1W, 3W, 2W, , 11, 6, (ii) For obtaining the resistance of W æç = 1 + ö÷ W the resistance of 1 W is connected in series with the, 5 è, 5ø, parallel combination of 2 W and 3 W ., 2W, A, , B, , 1W, 3W, , 7. Kinetic energy of the electron in the first excited state is 3.4 eV i.e.;, p2, = 3.4 eV = 3.4 × 1.6 × 10–19 J, 2m, p2 = 2 × 9.1 × 10–31 × 3.4 × 1.6 × 10–19
Page 728 :
CBSE Examination Papers 755, p = 18.2 ´ 3.4 ´ 1.6 ´ 10 - 50, 99.008 ´ 10 - 25 = 9.95 ´ 10 - 25, , =, de Broglie wavelength, l =, =, 3, 1H, , 8. (a), 9., , b, , -, , ¾®, , 3, 2 He, , +, , 0, –1 e, , h, p, 6.63 ´ 10 - 34, 9.95 ´ 10, , - 25, , = 0.67 ´ 10 - 9 = 0.67 nm, , + n+Q, , (b) It is due to very weak interaction with matter., (i) By considering straight line of I-V characteristic curve between 10 mA to 20 mA; and assuming passing, through the origin, draw horizontal and vertical lines from 10 mA and 20 mA, we have, I 1 = 10 mA,, V1 = 0.7 V, and, I 2 = 20 mA, V2 = 0.8 V, Dynamic resistance in forward biasing can be given as, ( 0.8 - 0.7) V, DV, r =, =, DI, 20 mA - 10 mA, 0.1 V, =, = 10 W, 10 mA, (ii) At – 10 V. The I-V characteristic graph is a straight line parallel to the voltage axis and not showing any, variation. So, the static resistance can be given as, V = - 10 V and I = - 1 mA, V 10 V, =, \ Static resistance, r =, I, 1 mA, = 1 ´ 107 W = 10 MW, , 10. Refractive index of a transparent medium decreases with increase in wavelength of the incident light used., Refractive index of glass with respect to air is given, Speed of light in air, m =, Speed of light in glass, =, m =, , Now,, , sin i c =, , 3 ´ 108 m s, 2 ´ 108 m s, , = 1. 5, , 1, sin i c, 1, m, , æ1ö, 1, 2, i c = sin –1 ç ÷ = sin –1 æç ö÷ = sin –1 æç ö÷ ., . ø, è 15, è3ø, èm ø, OR, Power of a lens, P =, , 1, f (in m), , After cutting, the power of each part will be half of the power of original lens., Therefore, focal length = 2f, \, , Power of each part, P ¢ =, , 1, 2f
Page 730 :
CBSE Examination Papers 757, Electric potential at the centre of the square is, VC =, , + q, + q, 1, 1 æ-qö, 1, 1 æ-qö, +, +, ç, ÷+, ç, ÷, 4p e 0 s, 4p e 0 è s ø 4p e 0 s, 4p e 0 è s ø, , =0, and electric potential at infinity is always zero., Hence, work done W = 0 ., , FR, , OR, , 60°, +q, , The charge, at any one vertex will remain in equilibrium, if the net electric, force at that point, due to the other three charges, is zero., , A, , Let Q be the required charge., ®, F1 = Force at A due to the charge at B,, ®, F1 =, , F1, , F2, , S, l, S, , ®, q2, 1, ., along BA, 4pe 0 l 2, , Q, , G, , l, , C, +q, , B, +q, l, , Force at A due to the charge at C,, ®, ®, q2, 1, ., along CA, F2 =, 4pe 0 l 2, ® ®, F1 + F2 = 3 ., Centroid at, , l, 3, , ®, q2, 1, ., along GA, 4pe 0 l 2, , from A, , Force at A due to charge at G =, , Qq(3), 1, ., 4pe 0, l2, , 3Qq = – 3q 2, , \, , Q=–, , q, 3, , 12. (a) From Flemings left hand rule, the electron deflects in anticlockwise direction., As the electron comes out the megnetic field region, it will describe a semi-circular path., Magnetic force provides a centripetal force. So,, mv 2, or, r, pr, pm, T =, =, v, eB, , evB =, Time taken,, , T =, , 3.14 ´ 9.1 ´ 10, 1. 6 ´ 10, , -19, , eB =, , mv, r, , r, × × × × × ×, × × × × × ×, –e, , × × × × × ×, , -31, , × × × × × ×, , -5, , × × × × × ×, , ´ 10, , 3.14 ´ 9.1 ´ 10 -7, =, 1. 6, = 1. 97 ´ 10 -7 second, , × × × × × ×
Page 731 :
758 Xam idea Physics–XII, (b) If Ampere’s force acts in upward direction and balances the weight, that is,, Fm = mg, , Fm, B, , BIl = mg, mg, B =, Il, , mg, , 0.2 ´ 10 2, =, =, 2 ´ 1.5 3, = 0.67, 13. 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, it is deviated from its path. This, phenomenon is called refraction. In transversing from first medium to, another medium, the frequency of wave remains unchanged but its, speed and the wavelength both are changed. Let XY be a surface, separating the two media ‘1’ and ‘2’. Let v1 and v 2 be the speeds of, waves in these media., , B, i, X, , 90o, i, , B', r, , A, , o, , 90, , Y, , r, , A', , 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. Clearly BB¢ = v1t. As the, wavefront AB advances, it strikes the points between A and B¢ of boundary surface. According to Huygen’s, principle, secondary spherical wavelets originate from these points, which travel with speed v1 in the first, medium and speed v 2 in the second medium., First of all secondary wavelet starts from A, which traverses a distance AA¢ ( = v 2 t) in second medium in time, t. In the same time-interval t, the point of wavefront traverses a distance BB¢ ( = v1t) in first medium and, reaches B¢ , from, where the secondary wavelet now starts. Clearly BB¢ = v1t and AA¢ = v 2 t., 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 incident wavefront AB advances, the secondary wavelets start from points between A and B¢ , one, after the other and will touch A¢ B¢ simultaneously. According to Huygen’s principle A¢ B¢ is the new position, of wavefront AB in the second medium. Hence A¢ B¢ will be the refracted wavefront., First law : As AB, A¢ B ¢ and surface XY are in the plane of paper, therefore the perpendicular drawn on, them will be in the same plane. As the lines drawn normal to wavefront denote the rays, therefore we may say, that the incident ray, refracted ray and the normal at the point of incidence all lie in the same plane., This is the first law of refraction., Second law : Let the incident wavefront AB and refracted wavefront A¢ B¢ make angles i and r respectively, with refracting surface XY., In right-angled triangle AB¢ B, Ð ABB¢ = 90°, vt, BB¢, sin i = sin Ð BAB¢ =, = 1, \, AB¢, AB¢, Similarly in right-angled triangle AA¢ B ¢ , Ð AA¢ B ¢ = 90°, v t, AA¢, \, sin r = sin Ð AB¢ A¢ =, = 2, AB¢ AB¢, Dividing equation (i) by (ii), we get, v, sin i, = 1 = constant, sin r, v2, , ...(i), , ...(ii), , ...(iii)
Page 732 :
CBSE Examination Papers 759, As the rays are always normal to the wavefront, therefore the incident and refracted rays make angles i and r, with the normal drawn on the surface XY i.e. i and r are the angle of incidence and angle of refraction, respectively. According to equation (iii):, The ratio of sine of angle of incidence and the sine of angle of refraction for a given pair of media is a, constant and is equal to the ratio of velocities of waves in the two media. This is the second law of, refraction, and is called the Snell’s law., 14., , ^, , n, 180°, ^, , n, , B, , B, , fi = NBA, , Initial magnetic flux,, , = 500 ´ 3 ´ 10 - 5 ´ p (0.1) 2, = 15 p ´ 10 - 5 Wb, On turning by 180°, Final flux, ff = - NBA = - 15 p ´ 10 - 5 Wb, Magnitude of induced emf, e = –, =, , df, dt, 2 ´ 15p ´ 10 - 5, = 120p ´ 10 - 5 volt, 0.25, , = 376.8 ´ 10 - 5 = 0.038 volt, Induced current,, , I =, , e, 0.038, =, = 19 ´ 10 - 5 A, R, 200, , 15. Rydberg formula for electron transition is given by, æ 1, 1, 1 ö÷, =R ç, ( ni > n f ), 2 ÷, ç n2, l, n, f, i, è, ø, For longest wavelength belonging to Lyman series, n f = 1 and ni = 2, 1, 1 ö, æ 1, = 1 .1 ´ 107 ´ ç, \, ÷, 2, l, 22 ø, è1, 1, 3.3, = 1.1 ´ 107 æç1 - ö÷ =, ´ 107, 4ø 4, è, 4, and, l=, ´ 10 - 7 = 121.2 × 10–9 = 121.2 m, 3 .3, For longest wavelength of Balmer series, n f = 2 and ni = 3, 1 ö, 1, æ 1, = 1 .1 ´ 107 ´ ç, \, ÷, 2, l, 32 ø, è2, 5, = 1.1 × 107 æç ö÷, è 36 ø
Page 733 :
760 Xam idea Physics–XII, \, , l=, , 36 ´ 10 - 7, 5 .5, , = 654.5 × 10–9 = 654.5 nm, Hence,, l Lyman = 121.2 nm, and, l Balmer = 654.5 nm, First transition lies in ultraviolet region, Second transition lies in visible region., 16. Two independent sources do not maintain constant phase difference, therefore the interference pattern will, also change with time., P, , S1, , S2, , Suppose a narrow slit S is illuminated by monochromatic light of wavelength l., The light rays from two coherent sources S1 and S 2 are reaching a point P, have a path difference, (S 2 P - S1 P)., (i) If maxima (bright fringe) occurs at point P, then, (n = 0, 1, 2, 3 …), S 2 P - S1 P = nl, (ii) If minima (dark fringe) occurs at point P, then, l, (n = 0, 1, 2, 3 …), S 2 P - S1 P = ( 2n + 1), 2, P, x, , A, , S1, , x, , d/2, S, , d/2, , T, , O, , z, z, , B, , S2, D, , y, , Light waves starting from S and fall on both slits S1 and S 2 . Then S1 and S 2 behave like two coherent sources., Spherical waves emanating from S1 and S 2 superpose on each other, and produces interference pattern on the, screen. Consider a point P at a distance x from 0, the centre of screen. The position of maxima (or minima), depends on the path difference. (S 2 T = S 2 P - S1 P)., From right angled DS 2 BP and DS1 AP,, 2, 2, é, d ù é, d ù, (S 2 P) 2 - (S1 P) 2 = êD 2 + æç x + ö÷ ú - êD 2 + æç x - ö÷ ú = 2xd, 2ø ú ê, 2ø ú, è, è, êë, û, û ë, , (S 2 P + S1 P)(S 2 P - S1 P) = 2xd
Page 734 :
CBSE Examination Papers 761, Þ S 2 P - S1 P =, , 2xd, (S 2 P + S1 P), , In practice, the point P lies very close to O, therefore, S 2 P ~ S1 P = D, 2xd, xd, S 2 P - S1 P =, =, 2D, D, , … (i), , For constructive interference (Bright fringes), dx, Path difference,, = nl where n = 0, 1, 2, 3 …, D, nDl, x =, d, For n = 0,, For n = 1,, For n = 2,, For n = n,, , x0 = 0, Dl, x1 =, d, 2Dl, x2 =, d, nDl, xn =, d, , Central bright fringe, 1st bright fringe, 2nd bright fringe, nth bright fringe, , The distance between two consecutive bright fringes is, b = x n - x n -1, nDl ( n - 1) Dl Dl, =, =, d, d, d, For destructive interference (dark fringes), dx, l, Path difference, = ( 2n - 1), D, 2, Dl, where n = 1, 2, 3, …, x = ( 2n - 1), 2d, Dl, for 1st dark fringe, For n = 1,, x 1¢ =, 2d, 3Dl, For n = 2,, for 2nd dark fringe, x ¢2 =, 2d, Dl, for nth dark fringe., For n = n,, x ¢n = ( 2n - 1), 2d, The distance between two consecutive dark fringe is, Dl, Dl Dl, b¢ = ( 2n - 1), - {2( n - 1) - 1}, =, 2d, 2d, d, The distance between two consecutive bright or dark fringes is called fringe width (w)., Dl, Fringe width =, \, d, The expression for fringe width is free from n. Hence the width of all fringes of red light are broader than the, fringes of blue light., 17. (a) The frequency range is known as bandwidth; and different bandwidths are required in transmitting, audible and video signals in communication. For example, a T.V. signal contains both voice and picture, and is usually allocated 6 MHz of band width for transmission.
Page 735 :
762 Xam idea Physics–XII, (b) A signal that varies continuously with time (e. g. , sine wave form) is called an analog signal., A signal that is discrete is called a digital signal. The presence of signal is denoted by digit 1 and absence, is denoted by digit 0., (c) Transducer: A device which converts energy from one form to another form., Repeater: It is a combination of a receiver and a transmitter., 18. Two important processes occurring during the formation of a p-n junction are (i) diffusion and (ii) drift., (i) Diffusion: In n-type semiconductor, the concentration of, Electron diffusion, Electron drift, electrons is much greater as compared to concentration of, –, +, holes; while in p-type semiconductor, the concentration of, p, n, holes is much greater than the concentration of electrons. When, – +, a p-n junction is formed, then due to concentration gradient, the, Electron, holes diffuse from p side to n side ( p ® n) and electrons, – +, diffuse from n side to p-side ( n ® p). This motion of charge, Hole diffusion, Hole, carriers gives rise to diffusion current across the junction., Hole drift, (ii) Drift: The drift of charge carriers occurs due to electric field., Vo Potential barrier, Due to built in potential barrier, an electric field directed from, n-region to p-region is developed across the junction. This field, + +, causes motion of electrons on p-side of the junction to n-side and, – +, motion of holes on n-side of junction to p-side. Thus a drift P, N, – +, current starts. This current is opposite to the direction of, w, diffusion current., Depletion, layer, , 19. In photoelectric effect, an electron absorbs a quantum of energy hn of radiation, which exceeds the work, function, an electron is emitted with maximum kinetic energy,, (a), (b), (c), 20. (a), , K max = hn - W, No, all electrons are bound with different forces in different layers of the metal. So, more tightly bound, electron will emerge with less kinetic energy. Hence, all electrons do not have same kinetic energy., No, because an electron cannot emit out if quantum energy hn is less than the work function of the metal., The K.E. depends on energy of each photon., Number of emitted photoelectrons depends on the intensity of the radiations provided the quantum, energy hn is greater than the work function of the metal., An integrated circuit (IC) is miniature, low cost electronic circuit consisting of active and passive, components fabricated together on a single crystal of silicon. The active components are transistors and, diodes and passive components are resistors and capacitors., Linear Integrated Circuit, , Digital Integrated Circuits, , Analog integrated circuits., , These circuits find use in true/false logical operations, and digital computation., , Different power supplies are required to regulate It does not make use of split power supplies., different components., Linear integrated output waves in the form of sine., , Digital IC’s output a high +5 V DC or a low 0V., , (b) The output of NOR gate is connected to both the inputs of NAND gate., The equivalent gate is OR gate., A, , B, , X, , Y, , 0, , 0, , 1, , 0, , 0, , 1, , 0, , 1, , 1, , 0, , 0, , 1, , 1, , 1, , 0, , 1
Page 736 :
CBSE Examination Papers 763, 21. During charging or discharging of the capacitor, displacement current between the plates is produced. Hence,, circuit becomes complete and galvanometer shows momentary deflection., According to Ampere’s circuital Law, ®®, ò B . dl = m 0 I, At surface P,, At surface S,, \, , ò, , ®®, B . dl = m 0 I c, , ò, , ®®, B . dl = 0, ®®, B . dl ¹, , òp, , òs, , S, +, +, +, +, +, +, +, , P, , i (t), , ®®, B . dl, , -, , M, , This contradicts Ampere’s circutial law. This law must be, missing something. Hence the law needs modification., Modified form of Ampere’s circuital law, ®®, d, ù, é, ò B . dl = m 0 êëic + Î0 dt fe úû, 22. (a) In steady state, current in the circuit ACDFA, 2V - V, V, I =, =, 2R + R 3R, Potential at point E, assuming VE = 0, VE = 2 V - 2IR, , R, A, , B, , = 2 V - 2R ´, , F, , V, , V, 4V, =, 3R, 3, , E, , V, , C, , D, 2V, , 2R, , Potential difference, VBE = VE - V, 4V, V, =, -V =, 3, 3, Hence, potential drop across the capacitor is, , V, ., 3, , V, 1, (b) Charge on the capacitor q = C æç ö÷ = CV, è3ø 3, (c) Energy stored in the capacitor U =, , 1 æV ö, Cç ÷, 2 è3ø, , 2, , =, , 1, CV 2, 18, , SECTION–D, 23. (a) Asha displayed the following values in convincing her brother., (i) Knowledgeable, (ii) thoughtful, convincing, (b) (i) Always keep the appliances in off position, there may be a short circuit, and an electric power, even, may loss in the circuit components., (ii) Use such appliances, in which a wattless current flows. e.g., a wattless curernt flows through the, choke which is used in the fluoroscent tubes., (c) Electrical energy wasted in the bulb = P ´ t, = 2 ´ 20 ´ 3600, = 144000 J
Page 737 :
764 Xam idea Physics–XII, , SECTION–E, 24. Similarities:, Both electrostatic field and magnetic field:, (i) follows the principle of superposition., (ii) depends inversely on the square of distance from source to the point of interest., Differences:, (i) Electrostatic field is produced by a scalar source ( q) and the magnetic field is produced by a vector, ®, , source ( I dl ) ., (ii) Electrostatic field is along the displacement vector between source and point of interest; while magnetic, field is perpendicular to the plane, containing the displacement vector and vector source., (iii) Electrostatic field in angle independent, while magnetic field is angle dependent between source vector, and displacement vector., Magnetic field at the axis of a circular loop: Consider a circular loop of radius R carrying current I, 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. Consider a small element of length dl of the coil at point A. The magnitude of the, ®, , magnetic induction dB at point P due to this element is given by, ®, , dB =, , m 0 I dl sin a, 4p, r2, , ®, , ...(i), ®, , ®, , The direction of dB is perpendicular to the plane containing dl and r and is given by right hand screw rule., ®, , ®, , ®, , As the angle between I dl and r is 90°, the magnitude of the magnetic induction dB is given by,, ®, , dB =, , m 0 I dl sin 90° m 0 I dl, =, ×, 4p, 4p r 2, r2, , ...(ii), , If we consider the magnetic induction produced by the whole of the circular coil, then by symmetry the, components of magnetic induction perpendicular to the axis will be cancelled out, while those parallel to the, ®, , axis will be added up. 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,, ®, , dB x =, But sin a =, , m 0 I dI, , sin a, , 4p r 2, , ...(iii), , R, and r = ( R 2 + x 2 ) 1 / 2, r, ®, , \, , dB x =, , m 0 I dl, 4p r, , 2, , ×, , m 0 IR, R m 0 IR, =, dl =, dl, 3, r, 4p r, 4p ( R 2 + x 2 ) 3 / 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, 2, 4p ( R + x ), 4p ( R + x 2 ) 3 / 2, But, , ò dl = length, , of the loop = 2pR, , ...(v)
Page 738 :
CBSE Examination Papers 765, Therefore,, , m 0 IR, , B =, , 4p ( R, , 2, , ®, B = B x i$ =, , ( 2 p R), , + x 2)3/ 2, m 0 IR 2, 2 (R, , 2, , 2 3/2, , + x ), , i$., , If the coil contains N turns, then, B =, , m 0 NIR 2, 2 (R 2 + x 2 ) 3 / 2, , ...(vi), , tesla., , OR, Eddy currents: When a thick metallic piece is placed in a time varying magnetic field, the magnetic flux, linked with the plate changes, the induced currents are set up in the conductor; these currents are called eddy, currents. These currents are sometimes so strong, that the metallic plate becomes red hot., Due to heavy eddy currents produced in the core of a transformer, large amount of energy is wasted in the, form of undesirable heat., Minimisation of Eddy Currents: Eddy currents may be minimised by using laminated core of soft iron., The resistance of the laminated core increases and the eddy currents are reduced and wastage of energy is, also reduced., Application of Eddy Currents:, 1. Induction Furnace: In induction furnance, the metal to be heated is placed in a rapidly varying magnetic, field produced by high frequency alternating current. Strong eddy currents are set up in the metal produce, so much heat that the metal melts. This process is used in extracting a metal from its ore. The, arrangement of heating the metal by means of strong induced currents is called the induction furnace., 2. Induction Motor: The eddy currents may be used to rotate the rotor. Its principle is: When a metallic, cylinder (or rotor) is placed in a rotating magnetic field, eddy currents are produced in it. According to, Lenz’s law, these currents tend to reduce to relative motion between the cylinder and the field. The, cylinder, therefore, begins to rotate in the direction of the field. This is the principle of induction motion., 25., , (a) For Gaussian surface of radius, r < x < R, , ò, Þ, , ®, , ®, , E . dS =, , | E|, , q, e0, q, , ò | dS | = e 0, q, e0, , | E | . 4p x 2 =, Hence,, , | E| =, , 1, 4px, , 2, , q, e0, , For Gaussian surface of radius x > R, , ò, , ®, , ®, , E . dS =, , ( q + Q), e0, , | E | . 4p x 2 =, Hence,, , | E| =, , ( q + Q), e0, , ( q + Q), 1, ., 4p e 0, x2, , r, , +q, , +Q, R
Page 739 :
766 Xam idea Physics–XII, (b) The electric potential of the inner sphere is, V1 =, , q, Q, 1, 1, +, 4p e 0 r, 4p e 0 R, , and electric potential of the outer shell, q, Q, 1, 1, +, ., V2 =, 4p e 0 r, 4p e 0 R, Since V1 > V2 , so charge will always flow from the smaller sphere to the larger sphere., OR, Electric field due to a system of point charges., Y, ®, , E2, ®, ® +E 2, 1, , ®, , q1, , P, ®, , ®, , r1, , q0, , r1P, , r2, , ®, , E1, , ®, , r, , ®, , E, , r2P, q2, , O, , X, ® ®, , ®, , Consider a system of N point charges q1, q2, ....., qN having position vectors r1 , r 2 , . . . , r N , with respect to, ®, , origin O. We wish to determine the electric field at point P whose position vector is r . According to, Coulomb’s law, the force on charge q 0 due to charge q1 is, ®, q q, 1, . 1 0 r$1P, F1 =, 4pe 0 r 2, 1P, where r$1P is a unit vector in the direction from q1 to P and r1P is the distance between q1 and P., Hence the electric field at point P due to charge q1 is, ®, ® F, 1 q1, E1 = 1 =, r$1P, q0, 4pe 0 r 2, 1P, , Similarly, electric field at P due to charge q2 is, ®, q, 1, . 2 r$2 P, E2 =, 4pe 0 r 2, 2P, According to the principle of superposition of electric fields, the electric field at any point due to a group of, point charges is equal to the vector sum of the electric fields produced by each charge individually at that, point, when all other charges are assumed to be absent., Hence, the electric field at point P due to the system of N charges is, ®, ® ® ®, E = E1 + E 2 +. . .+ E N, =, , ù, q, q, 1 é q1, 1, r$1P + 2 r$2 P +. . .+ N r$NP ú =, ê, 2, 2, 2, 4p Î0 êr, úû 4p Î0, r2 P, r NP, ë 1P, , N, , å, , qi, , r, 2 iP, i = 1 riP, $
Page 740 :
CBSE Examination Papers 767, At a point of equatorial line: Consider a point P on broad side on, the position of dipole formed of charges + q and - q at separation, 2l. The distance of point P from mid point (O ) of electric dipole is, r. Let, , ®, E1, , and, , ®, E2, , E1, , E = E1cosq + E2cosq, , q, , E, , be the electric field strengths due to charges, , E1sinq, , P, , + q and - q of electric dipole., From fig. AP = BP = r 2 + l, ®, , \, , E1 =, ®, E2, , =, ®, , q, 1, 2, 4p e 0 r + l, q, 1, 4p e 0 r 2 + l, , 2, , E2sinq, , E2, , r, 2, , along B to P, , A, , q, O, , –q, l, , 2, , +q, , B, , l, 2l, , along P to A, , ®, , ®, , ®, , Clearly E1 and E 2 are equal in magnitude i.e. | E1 | = | E 2 | or E1 = E 2, ®, , ®, , To find the resultant of E1 and E 2 , we resolve them into rectangular components., ®, ®, Component of E1 parallel to AB = E1 cos q , in the direction to BA, ®, , Component of E1 perpendicular to AB = E1 sin q along OP, ®, ®, Component of E 2 parallel to AB = E 2 cos q , in the direction BA, ®, , Component of E 2 perpendicular to AB = E 2 sin q along PO, ®, , ®, , Clearly, components of E1 and E 2 perpendicular to AB : E1 sin q and E 2 sin q being equal and opposite, ®, , ®, , cancel each other, while the components of E1 and E 2 parallel to AB : E1 cos q and E 2 cos q, being in the, ®, same direction add up and give the resultant electric field whose direction is parallel to BA., \ Resultant electric field at P is E = E1 cos q + E 2 cos q, But E1 = E 2 =, , q, 1, 2, 4p e 0 (r + l, , 2, , ), OB, =, PB, , l, , From the figure,, , cos q, , \, , E = 2E1 cos q = 2 ´, =, , =, , r2 + l, , 2, , =, , l, (r, , 2, , +l, , 2 1/ 2, , q, 1, l, ×, 2, 2, 2, 4p e 0 (r + l ) (r + l, , 2 1/ 2, , ), , 2ql, 1, 2, 4p e 0 (r + l 2 ) 3 / 2, , But q.2l = p = electric dipole moment, p, 1, =, \E, 4p e 0 (r 2 + l 2 ) 3 / 2, , ...(iii), , If dipole is infinitesimal and point P is far away, we have l < < r , so l, and so equation (iii) gives, , ), , 2, , may be neglected as compared to r 2
Page 741 :
768 Xam idea Physics–XII, E =, , p, p, 1, 1, =, /, 2, 3, 2, 4p e 0 (r ), 4p e 0 r 3, , i.e., electric field strength due to a short dipole at broadside on position, ®, p, 1, , in the direction parallel to BA ...(iv), E =, 4p e 0 r 3, Its direction is parallel to the axis of dipole from positive to negative charge., It may be noted clearly from equations (ii) and (iv) that electric field strength due to a short dipole at any, point is inversely proportional to the cube of its distance from the dipole and the electric field strength at, axial position is twice that at broad-side on position for the same distance., Important:, Note the important point that the electric field due to a dipole at large distances falls off as, 1, 1, and not as, as in the case of a point charge., r3, r2, 26., ue, , f0, , Distant Object, , B, , Objective, , a, A, , Eye piece, , a A", , F'e, , F0, A', b, , C1, , C2, , F'e, , B', L2, , B", , ve, , Magnifying Power : The magnifying power of a telescope is measured by the ratio of angle (b) subtended, by final image on the eye to the angle ( a) subtended by object on the eye, i.e.,, Magnifying power M =, , b, a, , Magnifying power can be increased as, (i) Focal length of the objective lens must be several times greater than the focal length of the eye piece., (ii) Distance between two lens must be less than the sum of focal lengths of both convex lenses., Advantages of a reflecting telescope., (i) A reflecting telescope is free from chromatic aberration., (ii) Spherical aberration may be reduced by using objective in the shape of a paraboloid mirror., (iii) The image is brighter than that in a refracting telescope wherein intensity is reduced due to absorption, and reflection by objective lens glass., (iv) High resolution is achieved by using a mirror of large aperture., (v) Mirror requires grinding and polishing on one side only., (vi) It is easier and more economical.
Page 742 :
CBSE Examination Papers 769, OR, (a), , Eyepiece, uo, , ue, , vo, , D, Objective, , Eye, , B, A", A, , Fo, A', , Fe', , O, , Fe, , E, , B', , B", , ve, , (b) In normal adjustment, magnifying power of the telescope M =, , F0, ., Fe, , (i) If focal length of the objective lens is large in comparison to the eyepiece, magnifying power, increases., D, ., (ii) Resolving power of the telescope RP =, 1.22l, D being the diameter of the objective., To increase the resolving power of the telescope, large aperture of the objective lens is required., (c) Compound Microscope: A compound microscope essentially consists of two co-axial convex lenses of, small focal lengths. The lens facing the object is called an objective lens while that towards eye is called, the eye lens. (eyepiece), \, , Magnifying power of microscope,, M =, , v D æ, v, b, (= mo ´ me ) = o, çç 1 + e, a, uo ve è, fe, , ö, ÷÷, ø, , Separation between lenses, d = v o + u e, Special cases: (a) When final image is formed at a distance of distinct vision, v e = D, \, , M =-, , vo, uo, , æ, D, çç 1 +, f, e, è, , ö, ÷÷, ø, , and, , d = v0 + ue, , The distance between second focal point of objective and first focal point of eye lens is called the tube, length denoted by L,then, v0, L, =, u0, f0, So,, , M=-, , L, fo, , æ, D, çç 1 +, fe, è, , ö, ÷÷, ø, , When final image is formed at infinity, v e = ¥, then, v, D, M =- o ´, uo, fe, =-, , L D, ×, fo fe, , and, , d = vo + f e
Page 743 :
770 Xam idea Physics–XII, The magnifying power of telescope is, Angle subtended by final image at eye b, =, M =, a, Angle subtended by an object on eye, = (mo ´ me ) = -, , fo, fe, , f, æ, çç 1 + e, ve, è, , ö, ÷÷, ø, , and Length of telescope L = f o + u e, where, , v e = distance of final image from eye lens, u e = distance of real image A¢ B¢ from eye lens, , Special cases (a) When final image is formed at a distance of distinct vision, then v e = D, \, , M =-, , fo æ, fe ö, ÷, ç 1+, fe è, D ø, , and, , L = f o + ue, , A telescope resolves two far objects being in nearly same direction and a microscope magnifies the, objects because light come out from all points of the same object., zzz
