## IIT JAM PHYSICS 2012

Previous Year Question Paper with Solution.

1. Given a function f(x, t) of both position x and time t, the value of (where ) is:

(a)

(b)

(c)

(d)

Ans. (b)

Sol.

2. If is a constant vector and is the position vector then would be

(a)

(b)

(c)

(d)

Ans. (b)

Sol.

3. Three masses m, 2m and 3m are moving in x-y plane with speeds 3u, 2u and u, respectively, as shown in the figure. The three masses collide at the same time at P and stick together. The velocity of the resulting mass would be

(a)

(b)

(c)

(d)

Ans. (d)

Sol. Using conservation of momentum

We get, 6m V_{x} = 3mu – 3mu cos 60 – 4 mu cos 60 = –0.5 mu; V_{x} =

And, 6m V_{y} = 3mu sin 60º – 4 mu sin 60

4. The figure shows a thin square sheet of metal of uniform density along with possible choices for a set of principal axes (indicated by dashed lines) of the moment of inertia, lying in the plane of the sheet. The correct choice(s) for the principal axes would be

(a) p, q and r

(b) p and r

(c) p and q

(d) p only

Ans. (c)

Sol. Axes shown in p and q have minimum value of M.I and no external torque is required to maintain uniform motion about these axes. In symmetric objects, Principal axes coincide with axes of symmetry.

Therefore axes shown in 'p' and 'q' represent principal axes.

5. A lightly damped harmonic oscillator loses energy at the rate of 1% per minute. The decrease in amplitude of the oscillator per minute will be closest to

(a) 0.5%

(b) 1%

(c) 1.5%

(d) 2%

Ans. (a)

Sol. Time dependence of energy of a lightly damped H.O. is given by

Since, E amplitude^{2} therefore power loss Amplitude^{2} thus we conclude that if power loss is 1% per minute amplitude will decrease by 0.5% per minute.

6. A parallel plate air-gap capacitor is made up of two plates of area 10 cm^{2} each kept at a distance of 0.88 mm. A sine wave of amplitude of 10V and frequency 50 Hz is applied across the capacitor as shown in the figure. The amplitude of the displacement current density (in mA/m^{2}) between the plates will be closest to

(a) 0.03

(b) 0.30

(c) 3.00

(d) 30.00

Ans. (a)

Sol. Given, v(t) = 10 sin

Therefore, electric field between the plates,

Displacement current density

Therefore, a Amplitude of

7. A tiny particle of mass 1.4 × 10^{–11} kg is floating in air at 300 K. Ignoring gravity, its r.m.s. speed (in µm/s) due to random collisions with air molecules will be closest to

(a) 0.3

(b) 3

(c) 30

(d) 300

Ans. (c)

Sol.

8. When the temperature of a blackbody is doubled, the maximum value of its spectral energy density, with respect to that at initial temperature, would become

(a) 1/16 times

(b) 8 times

(c) 16 times

(d) 32 times

Ans. (c)

Sol. Energy density (u) of black body radiation is proportional to T^{4} i.e., .

9. Light takes 4 hours to cover the distance from Sun to Neptune. If you travel in a spaceship at a speed 0.99c (where 'c' is the speed of light in vacuum), the time (in minutes) required to cover the same distance measured with a clock on the spaceship will be approximately

(a) 34

(b) 56

(c) 85

(d) 144

Ans. (a)

Sol. Distance between sun and neptune, d_{0} = 4 × 60 × 60 × c meters.

Apparent distance for the person in space ship,

Therefore, time taken,

10.
is a radioactive nucleus of half-life 2*l*n 2 × 10^{8} s. The activity of 10 g of in disintegration per second is

(a)

(b) 2

(c)

(d) 5 × 10^{14}

Ans. (d)

Sol.

11. An X-ray beam of wavelength 1.54 Å is diffracted from the (110) planes of a solid with a cubic lattice of lattice constant 3.08 Å. The first order Bragg diffraction occurs at

(a)

(b)

(c)

(d)

Ans. (b)

Sol. For first order diffraction.

12. The Boolean expression , where P and Q are the inputs to a circuit, represents the following logic gate

(a) AND

(b) NAND

(c) NOT

(d) OR

Ans. (a)

Sol.

13. Group-I contains x- and y-components of the electric field and Group-II contains the type of polarization of light

The correct set of matches is

(a)

(b)

(c)

(d)

Ans. (c)

Sol.

Amplitudes are unequal and phase difference is /2.

Therefore, resultant will be elliptically polarized.

Amplitudes are equal and phase difference is / 2.

Therefore, resultant will be circularly polarized.

Phase difference is zero.

Therefore resultant will be linearly polarized

14. For a liquid to vapour phase transition at T_{tr}, which of the plots between specific Gibbs free energy 'g' and temperature 'T' is correct?

(a)

(b)

(c)

(d)

Ans. (c)

Sol. Actually, variation of g with T is determined by the entropy. Because the entropy of gaseous phase of substance is greater than that of liquid phase. Therefore g for gas changes more steeply than that for liquid.

15. A segment of a circular wire of radius R, extending from θ = 0 to , carries a constant linear charge density, . The electric field at origin 'O' is

(a)

(b)

(c)

(d) 0

Ans. (a)

Sol.

16. The P-V diagram below represents an ideal monatomic gas cycle for 1 mole of a gas. In terms of the gas constant R. Calculate the temperatures at the points J, K, L and M. Also calculate the heat rejected and heat absorbed during the cycle, and the efficiency of the cycle.

Sol. At point K: PV = nRT

Total work done during the cycle = Area of loop JKLM = (160 – 80) × 10^{3} × (0.1 – 0.03) = 5600J

Heat absorbed, dQ^{+} = dQ_{1} + dQ_{2} = 31600 J

Heat rejected, dQ^{–} = dQ_{3} + dQ_{4} = –26000 J

17. 2 kg of a liquid (specific heat = 2000 J K^{–1} kg^{–1}, independent of temperature) is heated from 200 K to 400 K by either of the following two processes P_{1} and P_{2}.

P_{1} : bringing it in contact with a reservoir at 400 K.

P_{2}: bringing it first in contact with a reservoir at 300 K till equilibrium is reached, and then bringing it in contact with another reservoir at 400 K.

Calculate the change in entropy of the liquid and that of the universe in processes P_{1} and P_{2}. Neglect any change in volume of the liquid.

Sol. Change in entropy of liquid

Change in entropy of reservoir

Change in entropy of universe

18. (a) Two concentric, conducting spherical shells of radii R_{1} and R_{2} (R_{1} < R_{2}) are maintained at potentials V_{1} and V_{2}, respectively. Find the potential and electric field in the region R_{1} < r < R_{2}.

(b) A polarized dielectric cube of side *l* is kept on the x-y plane as shown. If the polarization in the cube is , where 'k' is a positive constant, then find all the bound surface charge densities and volume charge density.

Sol.

**Second method:** Poisson's equation in the region R_{1} < r < R_{2} is

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19. A water cannon starts shooting a jet of water horizontally, at r = 0, into a heavy trolley of mass 'M' placed on a horizontal ground. The nozzle diameter of the water cannon is d, the density of water is and the speed of water coming out of the nozzle is u. Find the speed of the trolley as a function of time. Assume that all the water from the jet is collected in the trolley. Neglect all frictional losses.

Sol. At time t mass of trolley = M + ρutA; A = , area of cross-section

Let v be speed of trolley at time t and speed becomes v + dv at time t + dt.

Therefore, using conservation of momentum we get,

20. A long straight solenoid of radius R and n turns per unit length carries a current I = t, where is a constant. 't' is time and remains finite. The axis of the solenoid is along the z-axis. Find the magnetic field, electric field and the poynting vector inside the solenoid. Show these vectors at some instant t_{1} at any point (i) on the axis of the solenoid, and (ii) at a distance r (< R) from the axis.

Sol. Using Lenz's law we can easily conclude that direction of electric field is opposite to the direction of current.

Using integral from of Max wells third equation,

Poynting vector

At the axis of solenoid, electric field is zero, therefore it doesn't have any direction. Magnetic field is along the axis and poynting vector is radially inward.

21. In the operational amplifier circuit shown below, input voltage, V_{1} = V and V_{2} = are applied.

(a) Determine the current flowing through a resistance R_{4} and the output voltage V_{0}.

(b) In the above circuit, if V_{1} is grounded and square pulses of peak voltage 1V and frequency 100 Hz are applied at V_{2}, determine the voltage and phase change of the output pulses.

Sol.

22. A particle of mass 'm' is confined in a potential box of sides L_{x}, L_{y} and L_{z} as shown in the figure. By solving the Schrodinger equation of the particle, find its eigenfunctions and energy eigenvalues.

Sol.

We get B_{1} = 0

And energy eigen values are given by

23. A particle of mass 'm' and charge 'q' moves in the presence of a time independent magnetic field . Set up Newton's equation of motion for the particle,

Since for a magnetic field , one can write , where is a function of position. Calculate as swen by the moving particle. Show that , where is the momentum of the particle, can be written as 'q' times the gradient of a function.

Sol.

24. Consider a periodic function f(x), with periodicity 2

where 'c' is a constant.

(a) Expand f(x) in a Fourier series.

(b) From the result obtained in (a). Show that

Sol.

25. Two orthogonally polarized beams (each of wavelength 0.5 µm and with polarization marked in the figure) are incident on a two-prism assembly and energy along x-direction, as shown. The prisms are of identical and n_{0} and n_{e} are the refractive indices of the o-ray and e-ray, respectively. Use

(a) Find the value of and n_{e}.

(b) If the right hand side prism starts sliding down with the vertical component of the velocity u_{y} = 1µm/s, what would be the minimum time after which the state of polarization of the emergent beam would repeat itself?

Sol. For ray 1 (at the interface of two crystals)

For ray 2 (At the interface of two crystals)

(b) State of polarization will repeate itself when path difference between e – ra and o–ray changes by

In time t the right crystal moves down by v_{y}t distance.

Therefore, change in width of medium in time t = v_{y}t tan 30º

Due to sliding of right prism, length of medium traversed by the two beams changes, this leads to change in optical path difference.