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ime : 3 Hours], , ‘in "SPECIMEN PAPER 3 (With Solution), General Instructions: Oo , (1) The question paper is ety, P fs led into four sections:, @) Section a @, No. 1 contains 10 mutipie ay, | . . No, 2 s Pt, (i) Section B :Q. No. 3 ona “a, (i Section © :Q. No. 15 © No. 26 contat, (tw) Section D :Q. No, 93 tog oi, (2) Use of log table is allowed, Use af calculator is ny CM, (3) Figure to the right indicate fill mar ks. ™, (4) For cach MCQ's comvet ane, LOMA sevens: 70D) wy, , No. 21 contains 5, , f(a), , Q.1. Select and Write the correct answer:, (i) As the electron rey,, orbit in the hy, , lo}, , Olves in the second Bohr, en atom, the corres: mn, , current is (about) 1.3 x 10“ A If the area of, the orbit is (about) 1.4 x 1, , (a) 1.82 x 103 Am? (b) 1.82 x 10-15 A-m?, (c) 2.82 x 103 Am? (4) 2.82 x 10-5 A-m?, Ans: (a) 1.82 x 103 Am?, [Hint: M = 1A, = (1.3 x 104) (1.4 x 1079), = 1.82 x 1073 Am?], Gi) In a common-base configuration the, , transistor has an emitter current of 10mA, , and collector current of 9.8 mA; the value, of the base current is, , (2)0.1mA (6)0.2mA (c)0.3mA (4)0.4mA, , Ans: (b) 0.2 mA, , [Hint: I, = 10 mA, I, = 9.8 mA, , 1, Sia I,, Therefore, the base current,, I, =i-k, , 10-9.8, = 0.2 mA], , (iii) The relation between relative permeability, and magnetic susceptibility is given by, (a)z,,=u,+ 1 (b)%,,=-H,- 2, (c) p, Sm (A) B= 1+%n, Ans: (d)p,=1+7%,,, , (iv) The internal energy of one mole of argon at, 300 Kis.......... (R= 8-314 J/mol.K), , (a)3541J (b)3741d (ce) 39415 (a)4041 5, , nark), Ans: (b) 3741Jd (1 mar, , [Hint: The internal energy of one mole of organ, at 300K is if R = 8.314 J mol K is, , (1 mark), , (1 mark), , , , (1 mark), , 3 3 300, U = SRT = 3 x8.314%, =3741J]), , O multiple choice type, , UV short answer type, 0. 14 contains 12 short anal, 8 12 short ans, , ai miner MUSE be tortiten atong 1, , , , (209), , __[Max. Marks : 70, , of questions carrying 1 mark each,, , of questions carrying 1 marke each., , wer (ype questions carrying 2 marks each,, ser type questions carrying 3 marks each., , 3 long ansiver (ype of questions carrying 4 marks each., Hallowed,, , Wh its alphabets., ete. only first attempt will be considered for evaluation, , , , The ratio of kinetic energy to the total, , energy of an electron in a Bohr orbit of the, hydrogen atom is, , (a)l:1 (b) 1-1, Ans:(b) 1: -1, , (c)2:-1 (d)1:-2, , (1 marke), (Hint: For an electron in Bohr orbit of H-atom, K.E. =-T.E., al, 1, (vi) In series LCR circuit, at resonance, phase, difference between current and emf of, , Le. 12-1], , , , source is, (a)nrad (b) Grad, (c) 3 rad (d)zero rad, , Ans: (d) zero rad, (vii) Henry is equivalent to, (a)ampere/second (b) ampere-second, (c) ohm/second (d)ohm-second, Ans: (d) ohm-second I mark}, , (viii) For polyatomic molecule having ‘F’, vibrational modes, the ratio of two specific, , , , , , , , , , heats, — is, eats, e, , Gq, , 1+F Q+F 4+F 5+F, Oorr Ost Oar OTF, Ans: (c) 34F nark, , (ix) The magnitude of centripetal force cannot, be expressed as,, 4n?mr, , — a -s(c) nwo, , (a) mw? (b) Tr, , (d) mv/a, Ans: (d) mv/o 1 mark), , () In which of the following substance, surface, tension increases with increase in, , temperature?, (a) Copper (b) Molton copper, (c) tron (d) Molton iron, , Ans: (b) Molton copper (1 mark), , Scanned with CamScanner

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210, , Q.2. Answer the following: [08], , (i) A metal rod of resistance of 15Q is moved, to the right at a constant speed 60 cm/s, along two parallel conducting rails 2 cm, apart and shorted at one end. A magnetic, field of mangitude 0.35 T points into the, page. Calculate the induced emf., , Ans: R= 15Q, v=0.6 m/s, l= 0.25 m, B=0.35T, -. Induced emf =e = Blu, , = (0.35) (0.25) (0.6), = 0.0525 V, =52.5V (1 inark), , (ii) The maximum velocity of a particle, , performing SHM is 6.28 cm/s. If the length, of its path is 8 cm, calculate the period., , Ans: V,. =6.28 cm/s = 2m cm/s, Time period = T =?, , (i) V,,,=Ao (i) o= 2, A= Path length of SHM = 8 24cm, 2 2, From formula (i) and (ii),, ‘Qn, ‘+ Vinax = AUT, on, 2n = 4(2), T = 4 sec (mark), , i.e., Time period of oscillation is 4 sec., , SECTION - B, , Attempt any EIGHT of the following: [16], Q.3. Explain Thomson's model of atom. (2), Ans: Thomson's atomic model:, (i) Thomson proposed his model of an atom, in 1903., , (ii) According to this model an atom is a, sphere having a uniform positive charge, in which electrons are embeded. (% mark), , (iii) This model is referred to as plum-pudding, model. The total positive charge is equal, to the total negative charge of electrons, in the atom, rendering in electrically, neutral. ( mark, , (iv) As the whole solid sphere is uniformly, positively charged, the positive charge can, not come out and only the negatively, charged electrons which are samll, can, , be emitted. (2 mark), , , , , , UTTAM SX Phy sp, TS, , , , , , , , , , , , , , , , has following truth *, (iii) A eee of this Hite, table, Stat, thy |, g |¥ }, o |o |o |, 1 Oo Oo, 0 1 0, 1 1 1, Ans: AND gate Ue, , What is Lorentz force?, , Lorentz force:When a charged Particle mop,, through a region in which both electric., magnetic fields are present, then the net fo, experienced by that charged particle is a, of electrostatic force and magnetic force ‘id, is called as Lorentz force. fig, , (v) What is mean by fluid?, , Ans: Any substance that can flow is a fluid. may, , (vi) How is the heat defined?, , Ans: Heat is the form of energy that is transferrej, between the system and its environment due, to temperature difference that exists between, the two. (1 mary, , (vii) Define the diathermanous substance., , Ans: A substance through which heat radiations, , can pass is known as a diathermanous, , substance. (I mart}, , (viii) In relation I = MK?, K stands for what?, , (iv), Ans:, , Ans: K = Radius of gyration. (1 mart), (v) The model also explained the fora, (am, , of ions and ionic compounds., Q.4. State Faraday’s laws of electromagn®, induction. i, Ans: Faraday's laws of electromagnet!, induction: :, (a) First law: Whenever there is 4 chang, , the magnetic flux associated Wi, , secult:, circuit, an emf is induced in the it, (yma, , tic, (2, , e, , (b) Second law: The magnitude atop, induced emfis directly proportion?“ jys, time rate of change of magnet’ nil, through the circuit. , go, , pe, , NU, , , , A, , Scanned with CamScanner

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men Paper-3 2tt, 5, Adrop of water of radius 6 mm breaks into, 9° Sumber of droplets each of radius 1 mm. How ihe ay (fa.rtiark), many droplets will be formed? (2) 6.63, ans: Given, Radius of big drop = R= 6 mm = (6.63x10-*)(10'4) (% mark), Radius of smaller drop=r=1mm 1, : Number of droplets (n) = ? * 107, Formula: Von = 102 (1 mark), Number of droplets = Vole meat big drop _ Q.9. Compare the rate of emission of heat by a, ¢ of each small drop black body at 327°C with the rate of, 4 (4 mark) emission of heat of same body at 27°C. (2), =nR® R, n=3 Ans: T, = 327 +273=600K, R- =?, 45,3 (% mari), 3 T, =27+273=300K, = (2 e _(6P .. Rate of emmission is,, n=(7) =|7] =216 (1 mark) RsoT* (4 mark), 216 droplets will be formed, XR ff, Q6. The peak value of AC through a resistor of R, = TF, 100 Q is 2 A. If the frequency of AC is 50 Hz, 4 4, find the heat produced in the resistor in one ( = (602. Vo, cycle. (2) =\a,) = (300 ism), Ans: R = 100 Q, i, =2A, f= 50 Hz = (2), He Be Ri _ 16, Of (1 mark) R, ~ 1, 100 (2)? 2 RR = 16:1 (1 mark), = 960) = 4J (mari) Q.10. Find the angular speed of revolution of, . ‘th required so that the body on it, This is the required quantity. ean one, surface, at equator would feel i, Q.7. Define end correction. State any two (R = 6400 ae = 9.8 m/s? no weight, limitations of end correction. (2) ‘Ages Radi f me = 2:8 mis) (2), Ans. End correction: The distance of the antinode s ee = 6400 tea, from the open end of the pipe is called end &=9.8 m/s, correction. (1 mark) At equator,, Limitations of end correction: Centripetal force = Weight of the body, (i) Inner diameter of the tube must be mro? = mg (ema, uniform. wei ‘, (ii) Effects of air flow and temperature outside r, the tube are ignored. oO =, (iii) The prongs of the tuning fork should be, perpendicular to the air coloumn in the 7, tube, with their tips at the centre of the “V64x (4% mark), tube and a small distance above ve ni = 1.237 x 10°, a of the tube. ; datienddipenuenty n @ =1.237 x 10° rad/s (1 mark), .8. ion, aleve oa. o jealate the auuaber of | Q-11-Draw a neat labelled diagram to determine, iz is 6. ’ (2) the resistance of a galvanometer by using, photons in the radiation. a meter bridge. (2), Ans: E=nhv, Ans: Labellings:, where hv is the energy of a photon in a AC - uniform wire, , radiation of frequency v and, nis the number of photons in the radiation., , , , E - Battery ellimanator, K_ - Key, , I, , Scanned with CamScanner

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212, , UTTAM'S Xil Physics Pay 18 Solution, , G- Galvanometer, , R- Resistance box, - Rheostat, , J - Jockey, , , , , , , , , , Fig. Resistance of galvanometer by using a meter, bridge (Kelvin method), @iagram - 1 mark, Labellings - 1 mark), Q.12.Magnetic field lines can be entirely, confined within the core ofa toroid, but not, within a straight solenoid. Why? (2), , (i) A toroid is a solenoid of finite length bent, into a hollow circular tube. The magnetic, field lines are concentric circles in the, toroid. (1 mark), , (ii) Magnetic field lines are entirely confined, within the core of a toroid because the, toroid has no starting and finishing ends., , (@1 mark), , Ans:, , Q.13. When water boils, why does its temperatur,, remains constant? (2), When water boils, the heat energy used 4,, Increase the interatomic distances of the, atoms, hence expansion takes place. Work is, done against the atmospheric pressure, The, change in internal energy is Zero. Heng,, temperature is constant. (1 mark, According to first law of thermodynamics,, , Ans:, , dg =dU+dw, ' 4 dQ =dW (1 mark), Q.14.State Gauss’s law. What is a Gaussian, surface? (2), , Gauss’s law: The electric flux over a closeq, , Ans: ‘, . surface of any size and any shape is equal to, , 3 times total charge enclosed by that, 0, , surface., Gausian surface: The surface over which, Gauss’s theorem is applied is called as, Gaussian surface. (1 mark), , (1 mark}, , SECTION - C, , Q.15. What is transformer? Explain step up and, , step down transformer. (3), , Ans: Transformer: Transformer is a device used to, , convert low alternating voltage at higher current, , into high alternating voltage at lower current and, , vice versa. In other words, a transformer is an, , electrical device used to increase or decrease, , alternating voltage. (1 mark), , e Types of transformer: There are two types of, transformer, (i) Step-up transformer: The transformer, which converts low alternating voltage at, higher current into a high alternating, voltage at lower current is called step up, transformer. In other words a step up, transformer gives increased alternating, voltage output. Symbol of step up, transformer is as shown below in figure., , “a:, , Fig. Step-up transformer (1 mark), , | (ii) Step-down transformer: The transformer, which converts high alternating voltage, at lower current into a low alternating, voltage at higher current is called step, down transformer. In other words, a step, donw transformer gives decreased, alternating output voltage. Symbol of, stepdown transformer is as shown, following in figure., , Fig. Step-down transformer (1 martd, , 9.16. Find the shortest wavelength in Pasche™, , series if, the longest wavelength in Balme!, Series is 6563 A. (3), 4, =?2, = 6563 A, , For shortest wavelength of Paschen series:, P=3andn=c, , Ans:, , , , Scanned with CamScanner

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pecan Paper a, hg n[ iS, nr D n, 1 1 |, her([}-2], 9 ”, A= R oe WG mario, For longest wavelength of Balmer series, , p=2andn=3, , 2 oll, % “R479, , _5R, ~ 36, = 38, » = BR s+ Gi) (1 mark), Dividing eqn (i) by eqn (ii) we have,, A 9 SR _ 45, 4. R” 36 36, 45), A= 36, _ 45x 6563, ~ 36, _ 295335, ~ 36, 2. =8203 A (1 mario), , 1, Q.17.Consider the cyclic process ABCA ona, sample of 2.0 mol of an ideal gas as shown, in figure, the temperatures of the gas at A, and B are 300 K and 500 K respectively. A, total of 1200 J heat is withdrawn from the, sample in the process. Find the workdone, by the gas in part BC. Take R = 8.3 J/mol-K., , (3), , , , : Given: n= 2.0, T, = 300K. T, = 500K, Q= 1200J, R=8.3d/mol-K, Workdown = Wy. =?, , Formulae:, faQ= AU+W, (b) W = P(V,-V), {c) PAV = nRAT, , Calculation: The change in internal energy, , , , during the cyclic process is pe ole, formula (a) the heat supplied to, , equal to the work done by it., , 213, , , , , , * = Wonca, = Way + Wire + W, ‘A, =-1200J (4 mark), As, the volume remains constant during the, path CA, W,, = 0, 4 Way +t Wie = - 1200 5 vee (I) (4 mark), From formula, (a) and (c), Way = ARC, -,), = 2.0 x 8.3 x (500 — 300), = 400 x 8.3, = 33203 (4 mark), From equation (i),, 3320 + W,,, =-1200J, Wy = 745205 (1 mark), Negative sign indicates that the work is done, on the gas. (4 mark), Thus, Work of amount -4520 J is done by the, gas in part BC., , Q.18.A circular loop of radius 9.7 cm carries a, current 2.3 A. Obtain the magnetic field, (a) at the centre of the loop and, (b) at a distance of 9.7 cm from the centre, , of the loop but on the axis. (3), Ans: r=9.7cm,1=2.3A,B=?,B.=?, Hol, ~BEQr (42 mark), , 4nx10~’ x2.3 _ 28.88, , -5, 19.4 x10, , ~ 2x9.7x10, B = 1.488 x 10°T, , , , (4 mark), B _ Hol r?, ms 9 2 + x2 f2, 4nx 1077 x2.3x(9.7x 107)?, 2\9.7x102 +(9.7x10°2)2]?, 4nx1077 x2.3x(9.7x10°)?, abe7zxi022/, 4nx2.3x107 x, ~ 9x2x1.414x(9.7x10, , _ nX2.3x 107, *1,414x9.7x10 7, B,,, =0-5270 x 10°T, , (4 mark), , @mark), , , , I, , , , , , , , , , x9.7x107, , log 3.142, , , , , , , , , -log 2.3, (1 mark) |7°8, -log 1.414, log 9.7, , , , = 1.7218, AL( 1.7218) = 0.5270, , , , Scanned with CamScanner