## IIT JAM PHYSICS 2014

Previous Year Question Paper with Solution.

1. For vectors and , the vector product is

(a) in the same direction as

(b) in the direction opposite to

(c) in the same direction as

(d) in the direction opposite to

Ans. (a)

Sol.

2. A particle of mass m carrying charge q is moving in a circle in a magnetic field B. According to Bohr's model, the energy of the particle in the n^{th} level is

(a)

(b)

(c)

(d)

Ans. (d)

Sol. Radius of circular orbit,

Angular momentum of particle = mvr =

According to Bohr's model,

Angular momentum =

Therefore, energy of particle =

3. A conducting slab of copper PQRS is kept on the xy plane in a uniform magnetic field along x-axis as indicated in the figure. A steady current I flows through the cross section of the slab along the y-axis. The direction of the electric field inside the slab, arising due to the applied magnetic field is along the

(a) negative Y direction

(b) positive Y direction

(c) negative Z direction

(d) positive Z direction

Ans. (d)

Sol. Force on positive charge is in the direction . Therefore, the positive charge will accomulate on lower surface. Therefore, direction of induced electric field will be from lower plate to upper plate. That is in positive z-direction.

4. A collimated beam of light of diameter 1mm is propagating along the x-axis. The beam is to be expanded to a collimated beam of diameter 10 mm using a combination of two convex lenses. A lens of focal length of 50 mm and another lens with focal length F are to be kept at a distance d between them. The values of F and d respectively, are

(a) 450 mm and 10 mm

(b) 400 mm and 500 mm

(c) 550 mm and 600 mm

(d) 500 mm and 550 mm

Ans. (d)

Sol.

d = 500 + 50 = 550 mm

5. Octal equivalent of decimal number 478_{10} is

(a) 736_{8}

(b) 673_{8}

(c) 637_{8}

(d) 367_{8}

Ans. (a)

Sol. Correct answer is (a)

6. A spherical ball of ice has radius R_{0} and is rotating with an angular speed about an axis passing through its centre. At time t = 0, it starts acquiring mass because the moisture (at rest) around it starts to freeze on it uniformly. As a result its radius increases as R(t) = R_{0} + t, where is a constant. The curve which best describes its angular speed with time is

(a)

(b)

(c)

(d)

Ans. (b)

Sol.

Angular momentum of system remains constant.

Therefore, as radius increases decreases non-linearly. Therefore, correct answer is (b)

7. In 1-dimension, an ensemble of N classical particles has energy of the form . The average internal energy of the system at temperature T is

(a)

(b)

(c) 3Nk_{B}T

(d) Nk_{B}T

Ans. (d)

Sol. For one particle, the energy,

The average internal energy of one particle

or, According to equipartitian theorem,

["The contribution of each quadratic term in Hamiltonian to average thermal energy is "]

Average energy of N-particle system

U = NU_{1} = NkT

8. In a photoelectric effect experiment, ultraviolet light of wavelength 320 nm falls on the photocathode with work function of 2.1 eV. The stopping potential should be close to

(a) 1.8 V

(b) 1.6 V

(c) 2.2 V

(d) 2.4 V

Ans. (a)

Sol. Einstein's photoelectric equation,

(E_{k})_{max} = h – W_{0}

_{ }where, (E_{k})_{max} = K.E. of the emitted photo electrons _{ }

= frequency of the incident light.

W_{0} = work function of the metal.

9. In an ideal operational amplifier depicted below, the potential at node A is

(a) 1 V

(b) 0 V

(c) 5 V

(d) 25 V

Ans. (b)

Sol. Due to virtual ground concept potential at node A = 0V. Since non-inverting terminal is at 0V

10. To operate a npn transistor in active region, its emitter-base and collector-base junction respectively, should be

(a) forward biased and reversed biased

(b) forward biased and forward biased

(c) reversed biased and forward biased

(d) reversed biased and reversed biased

Ans. (b)

Sol. For a transistor to be operated in active region.

Emitter base junction = Forward biased

Collector base junction = Reverse biased.

11.

(a)

(b)

(c)

(d)

Ans. (a)

Sol.

Both are infinite G.P. Series and the sum of a infinite G.P. Series is , where a is the first term and r is the common ratio of G.P. series.

12. In a parallel plate capacitor the distance between the plates is 10 cm. Two dielectric slabs of thickness 5cm each and dielectric constants K_{1} = 2 and K_{2} = 4 respectively, are inserted between the plates. A potential of 100V is applied across the capacitor as shown in the figure. The value of the net bound surface charge density at the interface of the two dielectrics is

(a)

(b)

(c)

(d)

Ans. (a)

Sol. Let be the surface change density on the capacitor's plate.

Therefore, charge density at the interface will be

13.

(a) a bright fringe will be obtained for cos = –1

(b) a bright fringe intensity is given by (E_{10})^{2} + (E_{20})^{2}

(c) a dark fringe will be obtained for cos = 1

(d) a dark fringe intensity is given by (E_{10} – E_{20})^{2}

Ans. (d)

Sol. It is the problem of beats bright fringe is obtained when

From interference we know that maximum density = (E_{01} + E_{02})^{2} and min = (E_{01} – E_{02})^{2}

14. A solid metallic cube of heat capacity S is at temperature 300 K. It is brought in contact with a reservoir at 600 K. If the heat transfer takes place only between the reservoir and the cube, the entropy change of the universe after reaching the thermal equilibrium is

(a) 0.69 S

(b) 0.54 S

(c) 0.27 S

(d) 0.19 S

Ans. (d)

Sol.

15. If the surface integral of the field over the closed surface of an arbitrary unit sphere is to be zero, then the relationship between is

(a)

(b)

(c)

(d)

Ans. (b)

Sol. According to Divergence Theorem,

16. The moment of inertia of a disc about one of its diameters is I_{M}. The mass per unit area of the disc is proportional to the distance from its centre. If the radius of the disc is R and its mass is M, the value of I_{M} is

(a)

(b)

(c)

(d)

Ans. (c)

Sol. Moment of inertia about diameter

Correct answer is (c)

17. A rigid uniform horizontal wire PQ of mass M, pivoted at P, carries a constant current I. It rotates with a constant angular speed in a uniform vertical magnetic field B. If the current were switched off, the angular acceleration of the wire, in terms of B, M and I would be

(a) 0

(b)

(c)

(d)

Ans. (c)

Sol. here magnetic torque is being balanced by some other torque

When magnetic field is switched off the other torque remains present. Therefore, angular acceleration is produced by the other torque.

18. Two points N and S are located in the northern and southern hemisphere, respectively, on the same longitude. Projectiles P and Q are fired from N and S, respectively, towards each other. Which of the following options is correct for the projectiles as they approach the equator?

(a) Both P and Q will move towards the east

(b) Both P and Q will move towards the west

(c) P will move towards the east and Q towards the west

(d) P will move towards the west and Q towards the east

Ans. (b)

Sol. Projectiles will deviate in the direction of coriolis of force. Since, direction of coriolis. Force on the both the projectiles is towards west (as determined by ). So, both will deviate towards west.

19. Two particles A and B of mass m and one particle C of mass M are kept on the x axis in the order ABC. Particle A is given a velocity . Consequently there are two collisions, both of which are completely inelastic. If the net energy loss because of these collisions is of the initial energy, the value of M is (ignore frictionless losses)

(a) 8m

(b) 6m

(c) 4m

(d) 2m

Ans. (b)

Sol.

20. The line integral of a vector field , where r^{2} = x^{2} + y^{2}, is taken around a square (see figure) of side unit length and centered at (x_{0}, y_{0}) with |x_{0}| > and |y_{0}| > . If the value of the integral is L, then

(a) L depends on (x_{0}, y_{0})

(b) L is independent of (x_{0}, y_{0}) and its value is – 1

(c) L is independent of (x_{0}, y_{0}) and its value is 0

(d) L is independent of (x_{0}, y_{0}) and its value is 2

Ans. (c)

Sol.

the field in defined everwhere within and on the square of side unit length and centered at (x_{0}, y_{0}) with

21. Diamond lattice can be considered as a combination of two fcc lattice displaced along the body diagonal by one quarter of its length. There are eight atoms per unit cell. The packing fraction of the diamond structure is

(a) 0.48

(b) 0.74

(c) 0.34

(d) 0.68

Ans. (c)

Sol.

22. Thermal neutrons (energy = 300 k_{B} = 0.025 eV) are sometimes used for structural determination of materials. The typical lattice spacing of a material for which these can be used is

(a) 0.01 nm

(b) 0.05 nm

(c) 0.1 nm

(d) 0.15 nm

Ans. (c)

Sol. Energy of thermal neutrons = k_{B}T_{}= 300 k_{B} = 0.025 eV

From Bragg's law: 2d sin

23. What is the maximum height above the dashed line attained by the water stream coming out at B from a thin tube of the water tank assembly shown in the figure? Assume h = 10 m, L = 2m, and = 30º.

(a) 10 m

(b) 2 m

(c) 1.2 m

(d) 3.2 m

Ans. (d)

Sol. Maximum height above dashed line

H_{max} = + max height of water above B

On substituting the values, we get

H_{max} = 3.2 m

Therefore, correct answer is (d)

24. A steady current in a straight conducting wire produces a surface charge on it. Let E_{out} and E_{in} be the magnitudes of the electric field just outside and just inside the wire, respectively. Which of the following statements is true for these fields?

(a) E_{out} is always greater than E_{in}

(b) E_{out }is always smaller than E_{in}

(c) E_{out} could be greater or smaller than E_{in}

(d) E_{out} is equal to E_{in}

Ans. (b)

Sol. Using boundary condition we get

25. A small charged spherical shell of radius 0.01 m is at a potential of 30V. The electrostatic energy of the shell is

(a) 10^{–10} J

(b) 5 × 10^{–10} J

(c) 5 × 10^{–9} J

(d) 10^{–9} J

Ans. (b)

Sol.

26. At an instant shown, three point masses m, 2m and 3m rest on a horizontal surface, and are at the vertices of an equilateral triangle of unit side length. Assuming that G is the gravitational constant, the magnitude and direction of the torque on the mass 3m, about the point O, at that instant is

(a) Zero

(b)

(c)

(d)

Ans. (d)

Sol.

27. A sine wave of 5V amplitude is applied at the input of the circuit shown in the figure. Which of the following waveforms represents the output most closely?

(a)

(b)

(c)

(d)

Ans. (d)

Sol.

28. 1011 binary input have been applied at X_{3}X_{2}X_{1}X_{0} input in the shown logic circuit made of XOR gates. The binary output Y_{3}Y_{2}Y_{1}Y_{0} of the circuit will be

(a) 1101

(b) 1010

(c) 1111

(d) 0101

Ans. (a)

Sol.

29. A ring of radius R carries a linear charge density It is rotating with angular speed The magnetic field at its centre is

(a)

(b)

(c)

(d)

Ans. (b)

Sol. Magnetic field at centre

30. A stationary source (see figure below) emits sound waves of frequency *f* towards a wall. If an observer moving with speed u in a direction perpendicular to the wall, measures a frequency at the instant shown, then u is related to the speed of sound V_{S} as

(a) V_{S}

(b) V_{S}/2

(c) V_{S}/4

(d) V_{S}/8

Ans. (c)

Sol. The point on the wall from where sound gets reflected acts as the source.

The approach velocity of the observer toward the source u cos 60º =

31. A real gas has specific volume v at temperature T. Its coefficient of volume expansion and isothermal compressibility are and k_{T}, respectively. Its molar specific heat at constant pressure C_{p} and molar specific heat at constant volume C_{v} are related as

(a) C_{p} = C_{v} + R

(b)

(c)

(d) C_{p} = C_{v}

Ans. (c)

Sol.

Let us consider that entropy, S = S (T, V). Then the change in entropy

By using Maxwell's thermodynamic relation.

From equation (1) and (2)

Let P = P(T, V)

From equation (3) and (4)

32. Two frames, O and O´, are in relative motion as shown. O´ is moving with speed c/2, where c is the speed of light. In frame O, two separate events occur at (x_{1}, t_{1}) and (x_{2}, t_{2}). In frame O´, these events occur simultaneously. The value of (x_{2} – x_{1}) / (t_{2} – t_{1}) is

(a) c/4

(b) c/2

(c) 2c

(d) c

Ans. (c)

Sol. Suppose, two events occur in the frame 'O' occurs at

According to Lorentz transformation,

In frame O´, two events will be simultaneous is

33. White light is incident on a grating G_{1} with groove density 600 lines/nm and width 50 mm. A small portion of the diffracted light is incident on another grating G_{2} with groove density 1800 lines/mm and width 15 mm. The resolving power of the combined system is

(a) 3 × 10^{3}

(b) 57 × 10^{3}

(c) 81 × 10^{7}

(d) 108 × 10^{5}

Ans. (c)

Sol. After passing from first grating

The wavelengths get resolved, where

= 600 × 50 (in the first order, resolving power = number of lines)

After passing from the second grating, wavelengths get resolved

34. Four particles of mass m each are inside a two dimensional square box of side L. If each state obtained from the solution of the Schrodinger equation is occupied by only one particle, the minimum energy of the system in units of is

(a) 2

(b)

(c)

(d)

Ans. (b)

Sol. Energy eigenvalue of a particle of mass 'm' inside a two dimensional square box of length 'L' is

Ground state corresponds to (1, 1) i.e. E_{11} =

First excited state corresponding to (1, 2), (2, 1)

Second excited state corresponding to (2, 2) and so on.

Since, each state is occupied by only one particle, then minimum energy will be

35. At atmospheric pressure (= 10^{5} Pa), aluminium melts at 550 K. As it melts, its density decreases from 3 × 10^{3} kg/m^{3} to 2.9 × 10^{3} kg/m^{3}. Latent heat of fusion of aluminium is 24 × 10^{3} J/kg. The melting point of aluminium at a pressure of 10^{7} Pa is closest to

(a) 551.3 K

(b) 552.6 K

(c) 558.7 K

(d) 547.4 K

Ans. (b)

Sol. Using Clausius-Clapeyron equation

We get, melting point = 552.6 K

36. Find the solution of the differential equation with the boundary condition y(0) = 2 and = 2, giving all steps clearly. Find the value of x where y = 0.

Ans.

Sol.

Let us assume that trial solution by y = C.e^{mx}

Putting the trial solution in the given differential equation. We get,

The solution of the given differential equation will be

37. The electric field in an electromagnetic (EM) wave is . What is the intensity of the EM wave and the number of photons per second falling on the unit area of a perfectly reflecting screen kept perpendicular to the direction of propagation? When a photon in this beam is reflected from the screen, what is the impulse it imparts to the screen? Use this to find the pressure exerted by the EM wave on the screen.

Ans.

Sol.

Number of photons per second falling on the unit area of a perfectly reflecting screen kept perpendicular to the direction of propagation.

Linear momentum of each photon =

Total momentum of all photons falling per second on an unit area =

As the total momentum of the photons is transferring to the surface, therefore this momentum i.e. is transferred to the unit area per second.

Since, the rate of change in momentum = exerted force.

So, impulse imparted by the photon on the screen in one second = change in momentum = (for unit area)

Pressure exerted by E.M. wave on the screen

= Exerted force per unit area

= Rate of change in momentum per unit area

38. A uniform rod of mass m and length *l* is hinged at one of its ends O and is hanging vertically. It is hit at its midpoint with a very short duration impulse J so that it starts rotating about O. Find the magnitude and direction of the horizontal impulse that O applies on the rod when it is hit.

Ans.

Sol. Let J´ be angular impulse from the support

Dividing (i) by (ii), we get

39. An easy derivation of = constant for an ideal gas undergoing an adiabatic process: Consider P and V as the basic variables of an ideal gas and write the heat exchanged dQ in terms of dV and dP. Next, using the definition of C_{P} and C_{V} in the expression for dQ, obtain a differential equation relating P and V for an adiabatic process and solve it to get the desired relationship. Derivation Should Not use the first law of thermodynamics. [For a function *f*(x, y) the differential

Ans.

Sol.

For adiabatic process dQ = 0

40. As shown in the figure below, an unpolarised beam of light of wavelength 500 nm is incident on a linear polariser at AF with vertical polarisation. The light beam then passes through a wave plate BE (half wave or quarter wave plate) of thickness 1.00125 mm and gets reflected from a mirror CD. The reflected light is indicated by the dashed line (DEF) in the diagram. The ordinary and extraordinary refractive indices for the material of the wave plate are 1.658 and 1.558, respectively. Light is incident normally on all surfaces.

(a) What is the polarisation of the beam at C?

(b) What is the polarisation of the beam at E and F?

Ans.

Sol. The path difference introduced by the plate between ordinary and extraordinary rays is (1.658 – 1.558) 1.00125 mm = 0.100 × 1.00125 mm = 0.100125 mm = 100125 nm

Now, the phase difference = (path difference) =

So, the plate is a quarter wave plate.

(a) At C the light is elliptically polarized

(b) At E the light has passed the quarter wave plate twice, so the phase difference of is introduced, so the light is linearly polarized.

41. A standing wave of light is formed between two mirrors and a beam of atoms is incident on it normally (see figure below) from the left. On the right side, atoms are detected in the direction of the beam and also at an angle as shown in the figure. This is due to material waves of atoms diffracted by the standing wave that acts like a grating; the slit width of this grating is given by the distance between two maxima of the light intensity. If the atomic beam is made of atoms of mass m moving with speed v and the light wave has wavelength find the smallest angle by using the diffraction condition.

Ans.

Sol. The standing wave acts like slit whose width is (because in stating wave, the distance between two consecutive maxima called anti-nodes is half of the wavelength).

Now, the wavelength of atoms (de Broglie wavelength)

From the diffraction condition

42. According to Wien's theory of black body radiation, the spectral energy density in a blackbody cavity at temperature T is given as

where and are constants and c is the speed of light. Further, the intensity of radiation coming out of the cavity is , where is the total energy density of radiation. Given that Stefan-Boltzmann constant = 5.67 × 10^{–8} Wm^{–2} K^{–4} and = 2.90 × 10^{–3} m.K, find the values of and . The value of integral .

Ans.

Sol. According to Wein's theory of Black body radiation, the spectral energy density in a blackbody cavity at temperature T is given by,

Total intensity of radiation coming out of the cavity

43. A horizontal rod of proper length L moves with uniform speed V > 0 along the x-axis of a coordinate frame. A ground observer measures the position coordinates of its two ends at two different times, with time difference > 0. The observer finds that the difference between the two coordinates is L. Calculate in terms of L, V and the speed of light c. If measured correctly, what would have been the length of the rod in the ground frame?

Ans.

Sol.

Using inverse Lorentz transformation we get

Since, in the rest frame of rod

If measured correctly (i.e. simultaneously) length of the rod will be