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im, , oe, , 1., , a, , (ii), , (iii), , liv), , ba, , ¢ : 3 Hours], eral Instructions:, tion paper is dj, ii) The quest iwided into, (y Section A :Q, No, 1 contains, ; Q. No. 2 contair, (i Section B 19. No. 3 to 2 No., (ii) Section C :Q, No, 2, (iv) Section D:Q. No. 27 ¢,, (g) Use of log table is allowed, Us, (a) Figure to the right indicate fiat, (4) For each MCQ's correct ansux, , £6.) ne MB) sees HO) onan Wg ee, , Jour sections:, , , , Select and Write the Correct answer:, , Asimple pendulum is Suspended from t!, , roof of a train. If the train is moving or, an acceleration 49 em/s?, Then the angle, of inclination of the String about the, , [1o], , vertical will be, (a) 20° (b) zero (c) 30° (@)3°, Ans: (d) 3° (1 marie), [Hint: f=mgsin® => a=gsing (+ f= ma), 49 _ a, a0 =sine => sin @=0.05, @ = 39, , The Boolean expression for an exclusive OR, gate is, , @A+B (b)A@B, Ans: (b) A®@B, What is the angular momentum of the, , electron in the third Bohr orbit in the, hydrogen atom?, , (J) A+B @A-B, , (1 mark), , it = 1.055 x 10° kg m?/s], {a)3.165 x 10° kg m2/s, (b) 3.165 x 10% kg m?/s, () 3.165 x 10 kg m2/s, (4) 3.165 x 1074 kg m?/s, Ans: (a) 3.165 x 10-4 kg m?/s, , (mark), , nh, (Hint: .. Angular momentum, L = mur= 97, , Forn=3,L= (3) = 3x (1.055 x 10), = 3.165 x 10° kgm?/s], , For which of the following does the centre, , of mass lie outside the body?, , (8) Peneil (b) Ashotput, , © Adise (d) A bangle, , Ans: (d) A bangle, , 'Hint: Out of the four given bodi, Mass of a bangle lies outside it wie, other bodies it lies within the body., , (1 mark), , les the centre of, as in all, , 10. multiple choice, 8 very short answe:, 14 contains 12 sh, , written along, , ype of, , ort anse, , , , T type of, , ____[Max. Marks : 70, , questions carrying 1 marke each,, , Questions carrying 1 mark each., , ver Wipe questions carrying 2 marks each., wer, , S 5 long answer, , ‘ype questions carrying 3 marks each., type of questions carrying 4 marks each., , with ts alphabets,, , D oossee ete. only fi, “Ste, only first attempt will be considered for evaluation., LI cece, , (v), , (vi), , (vii), , (viii, , , , (195), , Ans: (a) the first overtone, , aay, , An ideal transformer has 100 turns in the, Primary and 250 turns in the secondary. The, peak value of the ac is 28 V. The rms, secondary voltage is nearest to, , (a)100V_(b) 70V (c) 50V (d)40V, Ans: (c) 50V (1 mark), e, ny, [Hint: e, = ny, e, = 20 x28=70V, es, ms = IQ, , 70 x 0.707 = 49.49 V = 50 V], , The magnetic dipole moment of current loop, is independent of, , (a) number of turns, (b) area of loop, (c) current in the loop, (d) magnetic field in which it is lying, Ans: (d) magnetic field in which it is lying, (1 mark), , The null point of a potentiometer wire will, shift beyond the potentiometer wire if, (a) emf of driving cell is low, , (b) emf of accumulator is high, , (c) length of wire is small, , (d) length of wire is large, , Ans: (a) emf of driving cell is low ( mark), , ) Asonometer wire vibrates with three nodes, , and two antinodes. The corresponding mode, of vibration is, , (a) the first overtone, (b) the second overtone, (c) the third overtone, (@) the fourth overtone, , (1 mark), , , , Scanned with CamScanner

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(x), , @, , Ans:, , (ii), , Ans:, , (iii), , Q.3., , - N= 4000, mean radius of toroid =, , Visible light comprises of wavelengths in the, range..., (a) 100 - 400 nm, , (c) 300 - 700 nm, Ans: (d) 400 - 700 nm (1 marld, , The substance which allows heat radiations, to pass through it is, (a) dry air, , {c) iron, , , , {b) 200 - 800 nm, (a) 400 = 700 nm, , (b) water vapour, (a) wood, Ans: (a) dry air, , Answer the following:, , What is meant by a surface film?, Surface film: The layer of the liquid surface, , of thickness equal to the range of molecular, attraction is called surface film. (1 mark), , A 100 mH inductor, a 25 pF capacitor and, 15 Q resistor are connected in series to a, , 120 V, 50 Hz AC source. Calculate current, at resonance., , Given,, , L = 100mH=107H,C=25uF=25x 10°F, R = 15Q9,e,.,=120V, f=50 Hz, , At resonance, Z=R=15Q, , (1 mark), [08], , (1 mark), , A gas receives an amount of heat equal to, 110 J and performs 40 J of work. What is, the change in internal energy of the gas?, , dQ = dU+aw, , dU = dQ-dw, , = 110-40, dU = 70J (1 mark), fe, , Attempt any EIGHT of the following: [16], , A toroid of 4000 turns has outer radius of, 26 cm and inner radius of 25 cm. If the, current in the wire is 10 A, calculate the, magnetic field of the toroid. (2), , (25 + 26), 2, , or =25.50em=25.5 x 10*m, Length of toroid = [= 2nr, , , , Ni 2 s 4000" __., = TT * on x 25.510 *, B= bt nl (Y mark), ; B= Ho, 2 ,, sj Saeiee eaeenaa |, , 2nx 255x107, , a, , ——UTTAM'S XII Physics Papers 3, SS Sot, , (iv) State the relation between the sy, , and weber., The relation between the SI units visi, weber, , 1 volt = 1 weber per second, , 1 volt = 1 Wb/s be, State the drawback of Thomson's toda, atom. of, Drawback of Thomson's model of atom., model known as the plumpuding moq, to account for the observed sce, o-particles and spectra of variou, , UNits Volt, , Ans:, and, , (v), , Ans:, ons, el, Failed, ‘attering of, S elements,, C1 mark}, (Wote: It can be shown that Thomson atom cane, be stable), What should be the order of the size of an, obstacle or aperture to produce diffraction, of light?, For pronounced diffraction, the size of an, obstacle or aperture should be of the order of, the wavelength of light or greater., Define seconds pendulum., , : A simple pendulum whose period is two, seconds is called seconds pendulum. (i mati, Capacitors are combined in which arragement, when we require a large capacitance at small, potentials?, , Capacitors are combined in parallel, arrangement when we require a large, capacitance at small potentials., , (vi), Ans:, (1 mark), , (vii), , (viii), , Ans:, , (1 mark, , , , , , , , 25.5x102 log 80 | 1085,, 80 |= log 25.5 _= 1966, = 255% 10" | anio.svea) = 8, B= 3.137% 1029 Oe, G4. The length of a straight thin wire is 2™, , ie, is uniformly charged with a positive charg, , of 3 UC. If charge density of the wire wile, 10° C/m then calculate the cleo, intensity due to the wire at a point (2), away from the centre of the wire., Charge q=3 wc =3x 10°C, , Length, , Ans:, , , , 2m, r=1.5m, Charge density = 4 = 1,5 x 10°C/m, , a, , Scanned with CamScanner

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gogomenPapers2, , :, Electric intensity,, whe, pg 7 2neor (% mart), , 1S x10°8, = 2x3.14x8.85x10" 5, , , , 15 (Vo marke, , , , , , , _ 108 "Log Caloulation:, * 6,288.85 log 108, , = 17990, , = 1.799x 10'NC!, , (1 mark), , 5. State the zeroth law of thermod:, lynamics,, Give their schematic representation. (2), , Ans: Statement: If two systems are each in thermal, equilibrium with a third system, they are also, in thermal equilibrium with each other., , , , , , (1 mark), =, Bl<-[], Fig. Schematic representation of zeroth law of, thermodynamics, , (Correct Figure - 1 mark), , Q6. Draw a neat labelled diagram of schematic, of experimental set up for photoelectric, , , , , , effect. (2), Ans:, , He S, , Quartz window, Collector, , plate, Commutator, , hotoelectric, , "9. Schematic of experimental setup for p, SL, , fect., (Correct Figure ~ 1 mark., Labellings ~ | mark), , tic ratio of electron, 1x 10° kg), (2), , 27. Calculate the gyromagne, (Given: e = 1.6 x 107°C, ™, = 9., , ‘ Bs 2 mark), Ans: Gyromagnetic ratio = 2m, (Ys mark), , , , _197, , _ 16x10", ~ 2x(9.1x1079), , , , (% mark), , , , , , Log Calculation:, 6 , log 16 =0,2041, = 18.9” 10" = log 18.2 = 1.2601, = 0.0879 % 10!2 ’ Zone, AL( 2.9440) =0,0879', = B.79 x 10! Aen, , = 8.8 x 10! Ckg! (1 mark), , Q.8. Define phasor. Draw the diagram representing, it. (2), Phasor: A rotating vector that represents a, quality varing sinusoidally with time fs called, a phasor. (1 mark), , Phasor diagrams:, o o,, , Ans:, , , , I, sin wt}, , oO I, cos wt, (a) (b), , Fig. (a) and (b) : Phasor diagrams = (1 mark’, , Q.9. A drop of radius 2 x 10° m and density, 1.2 x 10° kg/m® falls through air. The, viscosity of air is 1.8 x 10% Ns/m?., Neglecting buoyancy due to air, calculate, the terminal speed of the drop. (2), , , , , , , , , , , , , , , , , Ans: r=2x 10°m, p= 1.2 x 10° kg/m?, y= 1.8x 10° Ns/m’, v=?, 9 =9.8 m/s?, -2 reg, =5- (4 mark), 2 2, se ee, on, _ 2x(2x10°) x1.2x10°x9.8, . 9x1.8x10° ee, _ 2x 4x10"! x1.2x10° x 9.8, 9x1.8x10°, 9.6x9.8x107, = 16.2x10> Log Calculation:, log 9.6 = 0.9823, = 9:6X9.8 ) gs | +18 9.8 0.9912, = 62 -log 16.2 = 1.2095, _ 4 0.7640, = 5.808 x 10 AL(0.7640) = 5.808, v = 5.8x 10? m/s, = 5.8cm/s (1 mari), , Scanned with CamScanner

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198, , , , g.10., , Ans:, , Q.11., , Q.12., Ams:, , Q.15., , Ans:, , ; Eddy currents:, , h of maximum <n, .08 jum. If the, , of a body at 700 K is 4.0:, , feneeminr of the body is raised to 1400, , K, Calculate the maximum arsine, , emitted energy:, , X= 4.08 jun = 4,08 x 10 om, , hie? T, = 700K, 'T, = 1400 K, , my, , .The wavelengt, , mtr = Agta, Amy Th |, = (% mark), Ang T, = 4.08 x 10° x a (4 mark), AL. =2.04 pm (1 mark), , =e, Why and where are eddy currents, undesirable? How are they minimized? (2), , Eddy currents result in generation of heat, (energy loss) in the cores of transformers,, motors, induction coils etc. ( mark), To minimize the eddy currents instead of solid, metal block, cores are made of thin insulated, metal strips or laminae. (1 mark), , Write a short note on Beats. (2), , Beats: This is an interesting phenomenon, based on the principle of superposition of, waves. When there is superposition of two, sound waves having same amplitude but, slightly different frequiencies, travelling in, the same direction, the intensity of sound, varies periodically with time. This, phenomenon is known as production of beats., , (2 mark), , The occurrences of maximum intensity are, called waxing and those of minimum, intensity are called waning. One waxing and, successive waning together constitute one, beat. The number of beats heard per second, is called beat frequency. (1 mark), , SECTI, , A circular race course track has a radius of, 500 m and is banked to 10°. If the coefficient, of friction between tyres of vehicle and the, road surface is 0.25. Compute, (i) the maximum speed to avoid slipping., , (ii) the optimum speed to avoid wear and, tear of the tyres. (g = 9.8 m/s?) (3), , r= 500 m, 6 = 10°, 1 = 0.25,, Vnax =? Uy=?, , , , UTTAM'S Xil Physi, $$ aS eRe Sea,, , 13. On what factors does the potentiay, 9 of wire depend? What is the st int, of, , potential gradient?, , (2), Ans: The potential gradient depends i, n | difference between the coda te, , otentia, Lie and the length of the wire. ne le, Meare), volt ‘, St unit = Tete = V/™) 1 rey, , .14, Draw neat labelled diagram of Huygen:, 9. progress of a plane wavefront and Prope:, , of a spherical wavefront. a, Ans: B B, Wavefront vt, att=0 >, Secondary, wavelets, —, after time T, A AN, , Fig. (a) Progress of a plane wavefront (|, Wavefront at time T, , Secondary, wavelets, , , , Secondary, \ 4 source, , e, ASA source BB, wavefront at time t = 0, , Fig. (b) Progress of a spherical wavefront., | mat, , , , We have,, , max,, , a h, + tan (, 1-p, tand, , 10°, 500x9.8 [ean, , 1—0.25xtan10°, , 0.25 + 0.1763 |, 1—0.25x0.1763, , 4900 [, , Scanned with CamScanner

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pacemien Paper - 2, pc, , 4900 2a, 1-0.044, , = ,|4900 [S352], , 0.956, = ¥2185, , U,4, = 46.72 m/s, \lso we know that,, , = yrgtané, , Big i, = ¥500 x9.8xtan10°, 500 x9.8x0.176, , = 29.37 m/s (1 mark), , }.16. The photoelectric work function of a metal, is 3 eV. Find the maximum kinetic energy, and maximum speed of photoelectrons when, , °, , radiation of wavelength 4000 A is incident, , on the metal surface. (3), ins: ¢ =S3eV=3x 1.6 10'%J, , ec =3x 10% m/s, , 2X =4000 A = 4000 x 10° m, , h =6.63 x 10“Js, m= 9.1 x 10%! kg, , i) According to Einstein's photoelectric equation, the maximum kinetic energy of photoelectrons, , (1 mark), , (4 mark), , Lin? = hv-4, , Qe max, , (4% mark), , 6.63x10"' 3x10" _ 3.4. gx 10719, 4000 x10"?, , 4.9725 x 104.8 x 10°", , 0.1725 x 10°”, , 1.725 x 107d, , (1 mark), , = 1.725 x 10°? J, um speed of photoelectrons, , 1. ja, Hy mwinax, , “ The maxiny, , 3, v = [21.725 x10 (4 mark), max m, 3.45x107°, Sqn, 9.1x10°°, (1 mark), , v,. = 1.947 x 10°m/s, , 9.17, What is meant by the term impedence?, , uit decreases, The total impedence of a circ, when a capacitor is added in series with L, , and R. Explain why- (3), , , , Ans:, , 9.18., , , , (i) For an LR circuit, the impedence,, , Zip= YR? +Xi°, , where X, is the reactance of the inductor., , (¥ mark), , (ii) When a capacitor of capacitance C is, added in series with L and R, the, , impedance,, , IR? +(X, -XcP (mark), (iii) Because in the case of an inductor the, current lags behind the voltage by a phase, , angle of (3) rad while in the case of a, capacitor the current leads the voltage by, , Tt, a phase angle of (3) rad. (1 mark), , (iv) The decrease in net reactance decreases, the total impedance., Zoe < Ziad, , “LCR, In Young's double slit experiment the ratio, of the intensities at the maxima and, minima in the interference pattern is, , 36 : 16. What is the ratio of the widths of, , (1 mark), , , , , , , , the two slits? (3), I, = - +. (Given), min, W,, w, =?, 2, = (1 mark), 7 > \2, 36 (= + Ey }, 16 E yo — Exo, Exo t Eo -6_3, E\o-Ey, 4 2, By componendo-dividendo, Ew 34+2, Eg = 352 = 5 (1 mark), , ". Therefore, the ratio of the slit widths,, , we t= (Ee) 2, * Eb0 = (5) =25, , W, = 1, Ww,, ~ WW, 25 (1 mark), , Scanned with CamScanner