## CSIR NET PHYSICS (DEC 2012)

Previous Year Question Paper with Solution.

1. A 2 × 2 matrix A has eigenvalues and . The smallest value of 'n' such that A^{n} = 1 is

(a) 20

(b) 30

(c) 60

(d) 120

Ans. (c)

Sol. A has eigenvalues

A^{n} has eigenvalues

So, |A^{n}| = product of eigenvalues

Since, N^{n} = I

So, the smallest value of 'n' is 60.

2. The graph of the function f(x) as shown below is best described by

(a) The bessel function J_{0}(x)

(b) cos x

(c) e^{–x} cos x

(d)

Ans. (a)

Sol. At x = 0, Bessel function J_{0}(x) = 1

J_{0}(x) has zeroes at x = 2.4048, 5.5201, 8.6537 ................... etc.

So, the graph is best described by J_{0}(x)

The other functions are following plots.

3. In a series of five cricket matches, one of the captains calls "Heads" every time when the toss is taken. The probability that he will win 3 times and lose 2 times is

(a) 1/8

(b) 5/8

(c) 3/16

(d) 5/16

Ans. (d)

Sol. Number of trials i.e. tosses 'n' = 5

Number of success i.e. getting heads 'r' = 3

Probability of success i.e. getting head in one trial 'p' = 1/2

Probability of failure i.e. getting tail in one trial 'q' = 1/2

Required probability:

4. The unit normal vector at the point on the surface of the ellipsoid , is

(a)

(b)

(c)

(d)

Ans. (a)

Sol. Given surface of ellipsoid:

Normal to the surface at point :

Unit normal vector to the surface at :

Option (a) is correct but it is wrongly printed in the question.

5. A solid cylinder of height H, radius R and density , floats vertically on the surface of a liquid of density . The cylinder will be set into oscillatory motion when a small instantaneous downward force is applied. The frequency of oscillation is

(a)

(b)

(c)

(d)

Ans. (d)

Sol.

Cylinder is partially submerged and in equilibrium

(for equilibrium)

Now we applying some force in downward direction then

F_{net} = F_{mg} – F_{B}

_{ }

_{ }

6. Three particles of equal mass 'm' are connected by two identical massless springs of stiffness constant 'k' as shown in the figure.

If x_{1}, x_{2} and x_{3} denote the displacements of the masses from their respective equilibrium positions, the potential energy of the system is:

(a)

(b)

(c)

(d)

Ans. (d)

Sol. Potential energy for first spring

Potential energy for second spring

Potential energy of the system,

7. Let v, p and E denotes the speed, the magnitude of the momentum, and the energy of a free particle of rest mass 'm'. Then

(a)

(b) p = mv

(c)

(d) E = mc^{2}

Ans. (c)

Sol. The energy of a particle of rest mass 'm' is given by

The velocity of particle, v = v_{g} = group velocity

8. A binary star system consists of two stars S_{1} and S_{2}, with masses m and 2m respectively separated by a distance 'r'. If both S_{1} and S_{2} individually follow circular orbits around the centre of the mass with intantaneous speeds v_{1} asnd v_{2} respectively, the ratio of speeds v_{1}/v_{2} is:

(a)

(b) 1

(c) 1/2

(d) 2

Ans. (d)

Sol.

9. Three charges are located on the circumference of a circle of radius 'R' as shown in the figure below. The two charges Q subtend an angle 90º at the centre of the circle. The charge 'q' is symmetrically placed with respect to the charges Q. If the electric field at the centre of the circle is zero, what is the magnitude of Q?

(a)

(b)

(c) 2q

(d) 4q

Ans. (a)

Sol.

According Lami's theorem

10. Consider a hollow charged shell of inner radius 'a' and outer radius 'b'. The volume charge density is (where k is a constant) in the region a < r < b. The magnitude of the electric field produced at distance r > a is:

(a)

(b)

(c)

(d)

Ans. (c)

Sol.

**Electric field for a < r < b:**

According Gauss's law

**Electric field for r > b:**

According to Gauss's theorem

Total charge enclosed inside the Gaussian surface,

11. Consider the interference of two coherent electromagnetic wave whose electric field vectors are given by and where is the phase difference. The intensity of the resulting wave is given by , where is the time average of E^{2}. The total intensity is

(a) 0

(b)

(c)

(d)

Ans. (b)

Sol.

No effect of phase on average value of total electric field as electric fields are perpendicular to each other.

Hence,

12. Four charge (two +q and two –q) are kept fixed at the four vertices of a square of side 'a' as shown

At the point P which is at a distance R from the centre (R > > a), the potential is proportional to

(a)

(b)

(c)

(d)

Ans. (c)

Sol.

Q monopole moment = = +q–q+q–q = 0

Dipole moment does not depend upon choice of origin.

13. A point charge 'q' of mass 'm' is kept at a distance 'd' below a grounded infinite conducting sheet which lies in the xy-plane. What is the value of 'd' for which the charge remains stationary?

(a)

(b)

(c) There is no finite value of 'd'

(d)

Ans. (a)

Sol. (acting upward direction) ... (1)

(acting downward direction) ... (2)

From equation (1) and (2)

F_{e} = mg (net force on charge q is zero because electrostatics and gravitational force equal in magnitude and opposite in direction)

14. The wave function of a state of the hydrogen atom is given by

where denotes the normalized eigen function of the state with quantum numbers n, *l* and m in the usual notation. The expectation value of L_{z} in the state Ψ is:

(a)

(b)

(c)

(d)

Ans. (d)

Sol. The given wavefunction is

The expectation value of L_{z} is given by

15. The energy eigenvalues of a particle in the potential are

(a)

(b)

(c)

(d)

Ans. (a)

Sol. The potential

The Schrodinger equation,

For S.E., ... (1)

The energy is given by

On comparing equation (2) by equation (1), the energy corresponding to equation (2) are given by

16. If a particle is represented by the normalized wave function

the uncertainty in its momentum is

(a)

(b)

(c)

(d)

Ans. (d)

Sol. The given normalized wavefunction is

17. Given the usual canonical commutation relations, the commutator [A, B] of A = i(xp_{y} – yp_{x}) and B = (yp_{z} + zp_{y}) is:

(a)

(b)

(c)

(d)

Ans. (c)

Sol. A = i (xp_{y} – up_{x}) & B = (yp_{z} + zp_{y})

[A, B] = [i(xp_{y} – yp_{x}), (yp_{z} + zp_{y})]

According to distributive properties,

[AB, C] = A[B, C] + [A, C]B & [A, BC] = B[A, C] + [A, B] C

And associative properties [A, B + C] = [A, B] + [A, C]

We can simplify

[A, B] = i [(xp_{y} – yp_{x}), yp_{z}] + i [(xp_{y} – yp_{x}), zp_{y}]

= i [xp_{y}, yp_{z}] – i [yp_{x}, yp_{z}] + i [xp_{y}, zp_{y}] – i [yp_{x}, zp_{y}]

[* out of 4 commutation relations 2^{nd} and 3^{rd} will be zero as there will be no canonically conjugate pair term].

[A, B] = i [xp_{y}, yp_{z}] – i [yp_{x}, zp_{y}]

= ix [p_{y}, yp_{z}] + i [x, yp_{z}]p_{y} – iy [p_{x}, zp_{y}] – i [y, zp_{y}]p_{x}

[2^{nd} and 3^{rd} will be also zero, reason will be same as above]

18. The entropy of a system, S, is related to the accessible phase space volume by where E, N and V are the energy, number of particles and volume respectively. From this one can conclude that

(a) does not change during evolution to equilibrium

(b) Oscillates during evolution to equilibrium

(c) Is a maximum in equilibrium

(d) Is a minimum in equilibrium

Ans. (c)

Sol. In the micrononical ensemble, the phase volume is maximized to obtain the characteristic thermodynamics quantity which is entropy

19. Let be the work done in a quasistatic reversible thermodynamics process. Which of the following statements about is correct ?

(a) is a perfect differential if the process is isothermal

(b) is a perfect differential if the process is adiabatic

(c) is always a perfect differential

(d) cannot be a perfect differential.

Ans. (b)

Sol. According to the first law of thermodynamics

dQ = dU + dW

where dQ is te amount of heat absorbed by a system, dU is the change in the internal energy of the system which depends only on the temperature, and dW is the amount of work done. If , then dW is not a perfect differential as it depends upon the path chosen. However for an adiabatic process

dQ = 0, dU = –dW

Since dU is a perfect differential, therefore dW is also a perfect differential.

20. Consider a system of three spins S_{1}, S_{2} and S_{3} each of which can take values +1 and –1. The energy of the system is given by E = –J[S_{1}S_{2} + S_{2}S_{3} + S_{3}S_{1}], where J is a positive constant. The minimum energy and the corresponding number of spin configurations are, respectively,

(a) J and 1

(b) –3J and 1

(c) –3J and 2

(d) –6J and 2

Ans. (c)

Sol. The energy of the system is given by

E = – J [S_{1}S_{2} + S_{2}S_{3} + S_{3}S_{1}], where J is positive constant and each among S_{1}, S_{2} and S_{3} can take +1 and –1 values.

Suppose S_{1}, S_{2}, S_{3} all have value +1, then the energy of the system is

E = – J (1 × 1 + 1 × 1 + 1 × 1) = – 3J

Suppose S_{1}, S_{2}, S_{3} all have value –1, then the energy of the system is

E = – J [(–1) × (–1) + (–1) × (–1) + (–1) × (–1)] = – 3J

If S_{1}, S_{2}, S_{3} have different values then energy will be greater than –3J.

21. The minimum energy of a collection of 6 non-interacting electrons of spin -½ placed in a one dimensional infinite square well potential of width L is

(a)

(b)

(c)

(d)

Ans. (a)

Sol. The energy of a particle of mass 'm' in a one-dimension square well potential of width L is given by

Only two electron with spin ½ can fill in the same state.

The minimum energy of the system is

22. A live music broadcast consists of a radio-wave of frequency 7 MHz, amplitude-modulated by a microphone output consisting of signals with a maximum frequency of 10 KHz. The spectrum of modulated output will be zero outside the frequency band

(a) 7.00 MHz to 7.01 MHz

(b) 6.99 MHz to 7.01 MHz

(c) 6.99 MHz to 7.00 MHz

(d) 6.995 MHz to 7.005 MHz

Ans. (b)

Sol. Carrier frequency f_{c} = 7 MHz

Modulation frequency f_{m} = 10 KHz = 0.01 MHz

Modulated signal will have frequency components (f_{c} + f_{m}) = 7.01 MHz and (f_{c} – f_{m}) = 6.99 MHz

23. In the op-amp circuit shown in the figure, V_{i} is a sinusoidal input signal of frequency 10 Hz and V_{0} is the output signal.

The magnitude of the gain and the phase shift, respectively, are close to the values

(a)

(b)

(c) 10 and zero

(d) 10 and

Ans. (d)

Sol.

Phase difference is 180º =

24. The logic circuit shown in the figure below.

implements the Boolean expression

(a)

(b)

(c) y = A.B

(d) y = A + B

Ans. (a)

Sol.

High means 1.

Output of first EXOR gate is

Output of second EXOR gate is

Output of OR gate is

As per De Morgan's theorem

25. A diode D as shown in the circuit as an I-V relation which can be approximated by

The value of v_{D} in the circuit is:

(a)

(b) 8V

(c) 5V

(d) 2V

Ans. (d)

Sol.

Apply KVL

26. The Taylor expansion of the function , where 'x' is real, about the point x = 0 starts with the following terms:

(a)

(b)

(c)

(d)

Ans. (b)

Sol. Taylor series expansion of f(x) about x = x_{0} is:

Now, in question x_{0} = 0, f (x) = ln (cosh x)

f "(x) = sech^{2 }x;

f "'(x) = 2sech x (–sech x.tanh x) = –2sech^{2 }x.tanh x

f ""(x) = – 2 (2sech x) (–sech x.tanh x) tanh x – 2sech^{2} x.sech^{2}x = 4sech^{2}x.tanh^{2}x – 2sech^{4}x

Required Taylor series expansion:

27. Given a 2 × 2 unitary matrix U satisfying U'U = UU' = 1 with det U = , one can construct a unitary matrix V(V'V = VV' = 1) with det V = 1 from it by

(a) Multiplying U by

(b) Multiplying any single element of U by

(c) Multiplying any row or column of U by

(d) Multiplying U by .

Ans. (a)

Sol.

We have to multiply U with to get V with determinant = 1

28. The vlaue of the integral , where C is a closed contour defined by the equation 2|z| – 5 = 0, traversed in the anti-clockwise direction, is:

(a)

(b)

(c)

(d)

Ans. (a)

Sol.

Only z = 2 i.e. point (2, 0) is within 'c'

29. A function f(x) obeys the differential equation and satisfies the conditions f(0) = 1 and as . The value of is:

(a)

(b)

(c)

(d)

Ans. (c)

Sol. Given differential equation :

Let, f = c.e^{mx} be trial solution.

Substituting in the given equation, we get

30. A planet of mass 'm' moves in the gravitational field of the Sun (mass M). If the semi-major and semi-minor axes of the orbit are 'a' and 'b' respectively, the angular momentum of the planet is:

(a)

(b)

(c)

(d)

Ans.

Sol. None of the option is correct.

31. The Hamiltonian of a simple pendulum consisting of a mass 'm' attached to a massless string of length *l* is . If L denotes the Lagrangian, the value of is:

(a)

(b)

(c)

(d)

Ans. (a)

Sol.

We know that

Eq. (1), (2) put in Eq. (3)

Eq. (5) put in Eq. (4)

32. Which of the following set of phase-space trajectories which one is not possible for a particle obeying Hamilton's equations of motion (for a time-independent Hamiltonian)?

(a)

(b)

(c)

(d)

Ans. (c)

Sol. According to Louville's theorem, phase space trajectories cannot cross.

33. Two bodies of equal mass 'm' are connected by a massless rigid rod of length '*l*' lying in the xy-plane with the centre of the rod at the origin. If this system is rotating about the z-axis with a frequency , its angular momentum is

(a)

(b)

(c)

(d)

Ans. (b)

Sol.

L = angular momentum = I, where 'I' is moment of inertia of the rod about an axis z with centre at O.

34. An infinite solenoid with its axis of symmetry along the z-direction carries a steady current I.

The vector potential at a distance R from the axis.

(a) Is constnat inside and varies as R outside the solenoid.

(b) Varies as R inside and is constant outside the solenoid.

(c) Varies as 1/R inside and as R outside the solenoid.

(d) Varies as R inside and as 1/R outside the solenoid.

Ans. (d)

Sol. Infinite solenoid carrying a current I. The vector potential at a distance R from the axis, then

We know that,

Let radius of solenoid is r.

35. Consider an infinite conducting sheet in the xy-plane with a time dependent current density , where K is a constant. The vector potential at (x, y, z) is given by

The magnetic field is:

(a)

(b)

(c)

(d)

Ans. (d)

Sol.

We know

36. When a charged particle emits electromagnetic radiation, the electric field and the Poynting vector at a large distance 'r' from the emitter vary as and respectively. Which of the following choices for 'n' and 'm' are correct?

(a) n = 1 and m = 1

(b) n = 2 and m = 2

(c) n = 1 and m = 2

(d) n = 2 and m = 4

Ans. (c)

Sol.

We know that

From (1), (2) and (3)

37. The energies in the ground state and first excited state of a particle of mass in a potential V(x) are –4 and –1, respectively, (in units in which ). If the corresponding wavefunctions are related by sinh x, then the ground state eigenfunction is

(a)

(b)

(c)

(d)

Ans. (c)

Sol. The time independent schrodinger equation

Substituting the value of from equation (1) into equation (2)

38. The perturbation

acts on a particle of mass 'm' confined in an infinite square well potential

The first order correction to the ground state energy of the particle is

(a)

(b)

(c) 2ba

(d) ba

Ans. (d)

Sol. The ground state wavefunction for the given potential

There perturbation,

The first order correction in energy,

39. Let and denote the normalized eigenstates corresponding to the ground and the first excited states of a one-dimensional harmonic oscillator. The uncertainty in the state is:

(a)

(b)

(c)

(d)

Ans. (a)

Sol.

Now, [as the states of LHO have definite parity. So, the expectation value of odd operator will become zero]

40. What would be the ground state energy of the Hamiltonian

if vibrational principle is used to estimate it with the trial wavefunction with b as the variational parameter?

[Hint: ]

(a)

(b)

(c)

(d)

Ans. (c)

Sol. The Hamiltonian,

The trial wave function with b variational parameter.

41. The free energy difference between the superconducting and the normal states of a material is given by , where is an order parameter and and are constants such that > 0 in the normal and < 0 in the superconducting state, while > 0 always. The minimum value of in the superconducting state is

(a)

(b)

(c)

(d)

Ans. (b)

Sol.

42. A given quantity of gas is taken from the state A C reversibly, by two paths, A C directly and A B C as shown in the figure below.

During the A C the work done by the gas is 100 J and the heat absorbed is 150 J. If during the process A B C the work done by the gas is 30 J, the heat absorbed is:

(a) 20 J

(b) 80 J

(c) 220 J

(d) 280 J

Ans. (b)

Sol. dQ = dU + dW_{1}, dW_{1} = 100J

For path AC: 150 J = dU + 100 J

dU = 150 J – 100J = 50 J

For path ABC: dQ = dU + dW_{2} = 50J + 30J = 80J

43. Consider a one-dimensional Ising model with N spins, at very low temperatures when almost all the spins are aligned parallel to each other. There will be a few spin flips with each flip costing an energy 2J. In a configuration with r spin flips, the energy of the system is E = –NJ + 2rJ and the number of configuration is ^{N}C_{r}; r varies from 0 to N. The partition function is

(a)

(b)

(c)

(d)

Ans. (d)

Sol. Let us consider only three energy levels, E_{r} = –2J + 2rJ

E_{0} = –2J, E_{1} = 0, E_{2} = 2J

44. A magnetic field sensor based on the Hall effect is to be fabricated by implanting. As into a Si film of thickness 1 µm. The specifications require a magnetic field sensitivity of 500 mV/Tesla at an excitation current of 1 mA. The implanation dose is to be adjusted such that the average carrier density, after activation, is

(a) 1.25 × 10^{26} m^{–3}

(b) 1.25 × 10^{22} m^{–3}

(c) 4.1 × 10^{21} m^{–3}

(d) 4.1 × 10^{20} m^{–3}

Ans. (b)

Sol. Hall voltage to magnetic field ratio

Current, I = 1mA = 10^{–3} Amp

Thickness of sample, t = 1 × 10^{–6} m

We know that Hall voltage,

45. Band-pass and band-reject filters can be implemented combining a low pass and a high pass filter in series and in parallel, respectively. If the cut-off frequencies of the low pass and high pass filters are and , respectively, the condition required to implement the band-pass and band-reject filters are respectively.

(a)

(b)

(c)

(d)

Ans. (b)

Sol. Low pass cut off frequency

High pass out off frequency

Low pass response High pass responses

For band pass filter responses is: For band reject filter response is

46. The output characteristics of a solar panel at a certain level of irradiance is shown in the figure below.

If the solar cell is to power a load of 5 , the power drawn by the load is:

(a) 97 W

(b) 73 W

(c) 50 W

(d) 45 W

Ans. (d)

Sol.

For 5 resistance power drawn is P = 3 × 15 = 45 watt

47. Consider the energy level diagram shown below, which corresponds to the molecular nitrogen laser.

If the pump rate R is 10^{20} atoms cm^{–3} s^{–1} and the decay routes are as shown with ns and , the equilibrium populations of states 2 and 1 are, respectively,

(a) 10^{14} cm^{–3} and 2 × 10^{12} cm^{–3}

(b) 2 × 10^{12} cm^{–3} and 10^{14} cm^{–3}

(c) 2 × 10^{12} cm^{–3} and 2 × 10^{6} cm^{–3}

(d) zero and 10^{20} cm^{–3}

Ans. (b)

Sol.

Rate equations for level (1) and (2) are given by

48. Consider a hydrogen atom undergoing a 2P 1S transition. The lifetime t_{sp} of the 2P state for spontaneous emission is 1.6 ns and the energy difference between the levels is 10.2 eV. Assuming that the refractive index of the medium n_{0} = 1, the ratio of the Einstein coefficients for stimulated emission B_{21}()/A_{21}() is given by

(a) 0.683 × 10^{12} m^{3} J^{–1} s^{–1}

(b) 0.146 × 10^{–12} J s m^{–3}

(c) 6.83 × 10^{12} m^{3} J^{–1} s^{–1}

(d) 1.463 × 10^{–12} J s m^{–3}

Ans. (a)

Sol.

= 0.0676 × 10^{13} = 0.676 × 10^{12}

49. Consider a He-Ne laser cavity consisting of two mirrors of reflectivities R_{1} = 1 and R_{2} = 0.98. The mirrors are separated by a distance d = 20 cm and the medium in between has a refractive index n_{0} = 1 and absorption coefficient = 0. The values of the separation between the modes and the width of each mode of the laser cavity are:

(a)

(b)

(c)

(d)

Ans. (c)

Sol. Laser mode separation,

but c = 3 × 10^{8} m/sec, d = 20 cm = 20 × 10^{–2} m

R_{1} = 1, R_{2} = 0.98

50. Non-interacting bosons undergo Bose-Einstein Condensation (BEC) when trapped in a three-dimensional isotrpic simple harmonic potential. For BEC to occur, the chemical potential must be equal to

(a)

(b)

(c)

(d) 0

Ans. (c)

Sol. For 3-D isotropic harmonic potential

is the energy of the ground state. When BEC occurs, all the non-interacting bosons go to the ground state. The chemical potential = ground state of the bosonic system

51. In a band structure calculation, the dispersion relation for electrons is found to be

where is a constant and a is the lattice constant. The effective mass at the boundary of the first Brilliouin zone is

(a)

(b)

(c)

(d)

Ans. (d)

Sol.

This is tight-binding energy expression for simple cubic solid with lattice parameter a. The expression for effective mass (m*) is given by

52. The radius of the Fermi sphere of free electrons in a monovalent metal with an fcc structure, in which the volume of the unit cell is a^{3}, is

(a)

(b)

(c)

(d)

Ans. (a)

Sol.

53. The muon has mass 105 MeV/c^{2} and mean lifetime 2.2 µs in its rest frame. The mean distance traversed by muon of energy 315 MeV/c^{2} before decaying is approximately

(a) 3 × 10^{5} km

(b) 2.2 cm

(c) 6.6 µm

(d) 1.98 km

Ans. (d)

Sol. m_{0} = 105 MeV/c^{2}, m = 315 MeV/c^{2}

^{ }

^{ }

54. Consider the following particles: the proton p, the neutron n, the neutral pion and the delta resonance . When ordered in terms of decreasing lifetime, the correct arrangement is as follows:

(a)

(b)

(c)

(d)

Ans. (c)

Sol.

55. The single particle energy difference between the p-orbitals (i.e. p_{3/2} and p_{1/2}) of the nucleus is 3 MeV. The energy difference between the states in its 1f orbital is

(a) –7 MeV

(b) 7 MeV

(c) 5 MeV

(d) –5 MeV

Ans. (b)

Sol. The single particle energy difference

; where 'C' is constant.

For p-orbitals, the energy difference,

The energy difference between the states in 1f orbital.

On substituting the value of 'c' from equation (1) into (2)