IIT JAM PHYSICS 2006
Previous Year Question Paper with Solution.

1. In a crystalline solid, the energy band structure (E-k relation) for an electron of mass M is given by . The effective mass of electron in the crystal is

(a) m

(b)

(c)

(d) 2m

Ans. (c)

Sol.

2. Two electric dipole P1 and P2 are placed at (0, 0, 0) and (1, 0, 0) respectively, with both of them pointing in the +z-direction. Without changing the orientations of the dipoles, P2 is moved to (0, 2, 0). The ratio of the electrostatic potential energy of the dipoles after moving to that before moving is:

(a)

(b)

(c)

(d)

Ans. (d)

Sol. The electric field at the centre at the dipole 2 due to dipole 1 is

The potential energy of the dipole 2 in this field is

3. The truth table for the given circuit is:

(a)

(b)

(c)

(d)

Ans. (c)

Sol.

4. At a given point in space the total light wave is composed of three phasors P1 = a, P2 = and . The intensity of light at this point is

(a)

(b)

(c)

(d)

Ans. (b)

Sol. Resultant phasors,

The intensity of light at this point is I = P2 = 4a2 cos4

5. A small magnetic dipole is kept at the origin in the x-y plane. One wire L1 is located at z = –a in the x-z plane with a current I flowing in the positive x-direction. Another wire L2 is at z = +a in y-z plane with the same current I as in L1, flowing in the positive y-direction. The angle made by the magnetic dipole with respect to the positive x-axis is:

(a) 225º

(b) 120º

(c) 45º

(d) 270º

Ans. (a)

Sol. The magnetic field at O due to L1 is in negatively y direction whereas magnetic field due to L2 is in negative x-direction and both are of some magnitude.

So, resultant magnetic field will make an angle of 225º with x axis and angle hence, θ made by magnetic dipole w.r.t. axis = 225º.

6. The ratio of the inner radii of two glass tubes of same length is . A fluid of viscosity 8.0 cP flows through the first tube, and another fluid of viscosity 0.8 cP flows through the second one when equal pressure difference is applied across both of them. The ratio of flow rate in the first tube to that in the second tube is:

(a) 1.6

(b)

(c)

(d) 0.4

Ans. (d)

Sol.

7. The relation between angular frequency and wave number k for given type of waves is . The wave number k0 for which the phase velocity equals the group velocity is:

(a)

(b)

(c)

(d)

Ans. (c)

Sol.

8. A neutron of mass mn = 10–27 kg is moving inside a nucleus. Assume the nucleus to be a cubical box of size 10–14 m with impenetrable walls. Take Js and 1 MeV 10–13 J. An estimate of energy in MeV of the neutron is:

(a) 80 MeV

(b)

(c) 8 MeV

(d)

Ans. (b)

Sol. Using Heisenbeg's uncertainty principle we get,

Matching the order of value we get,

9. A spring-mass system has undamped natural angular frequency = 100 rad s–1. The solution x(t) at critical damping is given by x(t) = x0(1 + t)exp(–t), where x0 is a constant. The system experiences the maximum damping force at time

(a) 0.01 s

(b) 0.1 s

(c)

(d)

Ans. (a)

Sol. x = x0(1 + w0t)exp (–w0t)

Damping force is maximum when speed is maximum.

10. In an intrinstic semiconductor, the free carrier concentration n (in cm–3) varies with temperature T (in Kelvin) as shown in the figure below. The band gap of the semiconductor is (use Boltzmann constant kB = 8.625 × 10–5 eVK–1).

(a) 1.44 eV

(b) 0.72 eV

(c) 1.38 eV

(d) 0.69 eV

Ans. (d)

Sol. Carrier concentration

Comparing with slope

From graph, we get,

 

11. represents an electromagnetic wave. Possible directions of the fast axis of a quarter wave plate which converts this wave into a circularly polarized wave are

(a)

(b)

(c)

(d)

Ans. (a)

Sol. Directions of vibration of wave lie in direction of unit vector in this direction is .

In order to convert the plane light to circularly polarized fast axis of quarter wave plate must make 45º angle with .

12. A particle of rest mass m0 is moving uniformly in a straight line with relativistic velocity c, where c is the velocity of light in vacuum and 0 < < 1. The phase velocity of the de-Broglie wave associated with the particle is:

(a)

(b)

(c) c

(d)

Ans. (b)

Sol.

13. Electrons of energy E coming in from x = impinge upon a potential barrier of width 2a and height V0 centred at the origin with V0 > E, as shown in the figure below. Let . In the region –a < x < a, the wave function for the electrons is a linear combination of

(a) ekx and e–kx

(b) eikx and ekx

(c) eikx and eikx

(d) eikx and ekx

Ans. (a)

Sol. Schrodinger's equation

14. A solid melts into a liquid via order phase transition. The relationship between the pressure P and the temperature T of the phase transition is P = –2T + P0, where P0 is a constant. The entropy change associated with the phase transition is 1.0 J mole–1 K–1. The Clausius-Clapeyron equation for the latent heat is L = T. Here = vliquid – vsolid is the change in molar volume at the phase transition. The correct statement relating the values of the volumes is:

(a) vliquid = vsolid

(b) vliquid = vsolid – 1

(c)

(d) vliquid = vsolid + 2

Ans. (c)

Sol.

15. The symmetrical part of is:

(a)

(b)

(c)

(d)

Ans. (d)

Sol.

16. Consider a Body Centered Cubic (BCC) crystal with lattice constant a. Determine

(a) The Miller indices for the (1, 0, 0) plane,

(b) The number of atoms per unit area in the (1, 1, 1) plane.

Sol. (a) The plane (1, 0, 0) cut the x, y and z axis at 1, 0, 0 respectively.

Therefore, The Miller indices of the plane are

(b) Atoms in (111) planes are shown in the figure below

The number of atoms inside the triangle =

Area of the triangle is

Therefore, number of atoms per unit area in the (1, 1, 1) plane

17. The equation of state of a gas is , while during an adiabatic process the gas obeys = K, where a and K are positive constants. All other symbols have their usual meaning. Find the worked done by the gas when it is expanded first isothermally from (P, V) to (P1, 2V). And then adiabatically from (P1, 2V) to where P1 < P.

Sol. Equation of state of gas is given by

In isothermal expansion, gas expanded from (P, V) to (P, 2V) at T constant.

Work done in isothermal expansion

Now gas is expanded adiabatically from (P, 2V) to

Word done in adiabatic expansion

Total work done, W = w1 + w2

18. A conducting sphere of radius RA has a charge Q. It is surrounded by a dielectric spherical shell of inner radius RA and outer radius RB (as shown in the figure below) having electrical permitivity

(a) Find the surface bound charge density at r = RA.

(b) Find the total electrostatic energy stored in the dielectric (region B).

Sol.

19. For the transistor circuit shown below, evaluate VE, RB and RC, given IC = 1 mA, VCE = 3.8 V, VBE = 0.7 V and VCC = 10 V. Use the approximation IC IE.

Sol. By Kirchoff's voltage law.

VCC = iCRC + VCE + iCRE

20. For the vector field

(a) Calculate the volume integral of the divergence of over the region defined by –a < x < a, –b < y < b and 0 < z < c.

(b) Calculate the flux of out of region through the surface at z = c. Hence deduce the net flux through the rest of the boundary of the region.

Sol.

(b) Flux of out the region through the surface z = c

Flux of the vector field across the closed surface of the region defined by

So, flux through the rest of the boundary region = 0

21. The spherical surface of a plano-convex lens of radius of curvature R = 1 m is gently placed on a flat plate. The space between them is filled with a transparent liquid of refractive indices of the lens and the flat plate are 1.5 and 1.6 respectively. The radius of the sixteenth dark Newton's ring in the reflected light of wavelength is found to be mm.

(a) Determine the wavelength (in microns) of the light.

(b) Now, the transparent liquid is completely removed from the space between the lens and the flat plate. Find the radius (in mm) of the twentieth dark ring in the reflected light after the change.

Sol. (a) Optical path difference = 2 t, where is refractive index of liquid.

For destructive interference, and t is thickness of the film at A.

Substituting (2) into (1)

16th dark ring will correspond to n = 15.

(b) When transparent liquid is removed, only ray 2 suffers a phase change of , and path difference .

22. A resistor of 1 and an inductor of 5 mH are connected in series with a battery of emf 4 V through a switch. The switch is closed at time t = 0. In the following, you may use e3 20.

(a) Find the current flowing in the circuit at t = 15 micro-second.

(b) Find the heat dissipated through the resistor during the first 15 micro-second.

Sol. At t = 0, switch is closer

The induced emf developed is

Applying Kirchoff's voltage law,

Integrating factor IF = eRt/L

At t = 0, i = 0

Current flowing the circuit at t = 15 µs.

(b) Heat disipated through resistor during first 15µs.

23. A photon of energy Eph collides with an electron at rest and gets scattered at an angle 60º with respect to the direction of the incident photon. The ratio of the relativistic kinetic energy T of the recoiled electron and the incident phont energy Eph is 0.05.

(a) Determine the wavelength of the incident photon in terms os the Compton wavelength , where h, me, c are Planck's constant, electron rest mass and velocity of light respectively.

(b) What is the total energy Ee of the recoiled electron in units of its rest mass?

Sol. (a) Photon of energy Eph collides with an electron at rest and gets seattered at = 60º with respect to the direction of the incident photon.

According to compton effect.

where wavelength of scattered photon, kinetic energy transferred to electron,

Substituting value of in (1)

Now using relation on (2) and substituting the value of T

(b) Total energy of recoiled electron Ee = Rest mass energy + K.E. of electron

24. A particle moves in a plane with velocity such that . The time dependence of the magnitude of the velocity = 5t. It is given that r = 1, = 0 and vr > 0 at t = 0. (In the following, you may use e3 20).

(a) Determine the trajectory r of the particle.

(b) At what time will become 4 radian?

Sol.

25. A body of mass 1 kg moves under the influence of a central force, with a potential energy function Joule, where r is in meters. It is found to move in a circular orbit of radius r = 2m. (In the following, you may use e3 20).

(a) Find its angular momentum L and total energy E.

(b) A piece of mass m1 = 0.5 kg breaks off suddenly from the body and begins to fall radially inwards with velocity v = 10 cm s–1. What are the values of angular momentum L2 and the total energy E2 of the remaining piece, assuming that the potential energy function remains the same?

Sol.

(b) m1 = 0.5 kg, v = 10 cm/second

Laws of conservation, L1 + L3 = L, L1 = 0

So angular momentum of L2 =

Kinetic energy of mass m1, kE1 =

 

So, conservation of energy E = E1 + E2