CSIR NET PHYSICS (JUNE 2015)
Previous Year Question Paper with Solution.

1. Consider the periodic function f(t) with time period T as shown in the figure below:

The spikes, located at , where n = 0, +1, +2, ...., are Dirac-delta functions of strength +1. The amplitudes an in the Fourier expansion

are given by

(a) (–1)n

(b)

(c)

(d)

Ans. (c)

Sol. Given function is periodic in nature having time period T = 2

The function can be written as

The fourier expansion of f (t) can be written as

Where,

2. A particle moves in two dimensions on the ellipse x2 + 4y2 = 8. At a particular instant it is at the point (x, y) = (2, 1) and the x-component of its velocity is 6 (in suitable units). Then the y-component of its velocity is

(a) –3

(b) –2

(c) 1

(d) 4

Ans. (a)

Sol. We have, x2 + 4y2 = 8

At point, (2, 1), x – component of velocity

Therefore, y – component of velocity will be

3. Consider the differential equation . If x = 0 at t = 0 and x = 1 at t = 1, the value of x at t = 2 is

(a) e2 + 1

(b) e2 + e

(c) e + 2

(d) 2e

Ans. (b)

Sol. Given differential equation

Let x = emt be the trial solution and putting these in the given equation,

We get

So, the general solutions, x = A e2t + B et

Now, at

and at

Therefore, the solution will be

At

4. The value of the integral is

(a)

(b)

(c)

(d)

Ans. (a)

Sol.

Condition of sigularity : 1 + z4 = 0

Poles of f (z) will be and will be order 1.

Out of them, lies on the upper half of complex plane

5. The Laplace transform of 6t3 + 3 sin 4t is

(a)

(b)

(c)

(d)

Ans. (a)

Sol. L[f(t)] = L[6t3] + L[3 sin 4t] = 6L[t3] + 3[L(sin 4t)]

Using

We get,

6. If the Lagrangian of a dynamical system is two dimensions is , then its Hamiltonian is

(a)

(b)

(c)

(d)

Ans. (c)

Sol.

We know that

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

Substracting (1) and (2)

From Eq. (2)

Eq. (5), (6) put in Eq. (4)

7. A particle of mass m moves in the one-dimensional potential where . One of the equilibrium points is x = 0. The angular frequency of small oscillations about the other equilibrium point is

(a)

(b)

(c)

(d)

Ans. (b)

Sol.

For equilibrium point

8. A particle of unit mass moves in the xy-plane in such a way that and . We can conclude that it is in a conservative force-field which can be derived from the potential

(a)

(b)

(c) x + y

(d) x – y

Ans. (a)

Sol. We have,

Therefore, potential,

9. Consider three inertial frames of reference A, B, and C. The frame B moves with a velocity c/2 with respect to A and C moves with a velocity c/10 with respect to B in the same direction. The velocity of C as measured in A is

(a)

(b)

(c)

(d)

Ans. (b)

Sol.

The velocity of B frame with respect to A frame is,

The velocity of C frame with respect to B frame is,

Therefore, the velocity of C frame witll respect to A frame is

10. A plane electromagnetic wave is travelling along the positive z-direction. The maximum electric field along the x-direction is 10 V/m. The approximate maximum values of the power per unit area and the magnetic induction B, respectively, are

(a) 3.3 × 10–7 watts/m2 and 10 tesla

(b) 3.3 × 10–7 watts/m2 and 3.3 × 10–8 tesla

(c) 0.265 watts/m2 and 10 tesla

(d) 0.265 watts/m2 and 3.3 × 10–8 tesla

Ans. (d)

Sol.

11. Suppose the yz-plane forms a chargeless boundary between two media of permittivities and where . If the uniform electric field on the left is (where c is a constant), then the electric field on the right is

(a)

(b)

(c)

(d)

Ans. (c)

Sol. Normal component of D is continuous and tangential component of E field is continuous

12. A proton moves with a speed of 300 m/s is a circular orbit in the xy-plane in a magnetic field 1 tesla along the positive z-direction. When an electric field of 1 V/m is applied along the positive y-direction, the center of the circular orbit

(a) remains stationary

(b) moves at 1 m/s along the negative x-direction

(c) moves at 1 m/s along the positive z-direction

(d) moves at 1 m/s along the positive x-direction

Ans. (d)

Sol. Velocity of centre of circle

Direction of shift =

13. Which of the following transformations of the electrostatic potential V and the vector potential is a gauge transformation?

(a)

(b)

(c)

(d)

Ans. (d)

Sol. Gauge transformation is the one which gives same and

14. The ratio of the energy of the first excited state E1, to that of the ground state E0 of a particle in a three-dimensional rectangular box of sides L, L and L/2 is

(a) 3:2

(b) 2:1

(c) 4:1

(d) 4:3

Ans. (a)

Sol. We know that energy of the nth level of three dimensional rectangular box having sides, L, L, L/2

Ground state energy

First excited state energy,

Therefore, the ratio of first excited energy to ground state energy = 3 : 2

15. The waveform of a particle in one-dimension is denoted by in the coordinate representation and by in the momentum representation. If the action of an operator on is given by , where a is a constant, then is given by

(a)

(b)

(c)

(d)

Ans. (c)

Sol.

16. If Li are the components of the angular momentum operator , then the operator equals

(a)

(b)

(c)

(d)

Ans. (b)

Sol. First Method:

Second Method:

Similarly,

17. A particle moves in one dimension in the potential , where k(t) is a time dependent parameter. Then , the rate of change of the expression value of the potential energy, is

(a)

(b)

(c)

(d)

Ans. (a)

Sol. We have,

18. A system of N distinguishable particles, each of which can be in one of the two energy levels 0 and , has a total energy n, where n is an integer. The entropy of the system is proportional to

(a)

(b)

(c)

(d)

Ans. (d)

Sol. Since there have two state 0 and , and total number of particle N and total energy n. Therefore, n particle in the state of energy and (N-n) particle are in the 0 state.

Therefore, entropy

19. A system of N non-intersecting classical particles, each of mass m is in a two-dimensional harmonic potential of the form where is a positive constant. The canonical partition function of the system at temperature T is .

(a)

(b)

(c)

(d)

Ans. (d)

Sol. We have,

Partition function

20. In a two-state system, the transition rate of a particle from state 1 to state 2 is t12, and the transition rate from state 2 to state 1 is t21. In the steady state, the probability of finding the particle in state 1 is

(a)

(b)

(c)

(d)

Ans. (a)

Sol.

Time during which the particle will be in state 1 and 2 are proportional to and respectively.

So, probability of finding to particle in state 1

21. The condition for the liquid and vapour phases of a fluid to be in equilibrium is given by the approximate equation (Clausius-Clayperon equation), where vvap is the volume per particle in the vapour phase, and Ql is the latent heat, which may be taken to be a constant. If the vapour obeys ideal gas law, which of the following polots is correct?

(a)

(b)

(c)

(d)

Ans. (c)

Sol.

For ideal gas PV = RT

Where P0 and T0 are pressure and temeprature of vapour in liquid vapour equilibrium.

22. Consider the circuits shown in Figure (a) and (b) below.

If the transistors in Figures (a) and (b) have current gain of 100 and 10 respectively, then they operate in the

(a) active region and saturation region respectively

(b) saturation region and active region respectively

(c) saturation region in both cases

(d) active region in both cases

Ans. (b)

Sol. Since both the resistors are n-p-n. VCE is positive means active, VCE is negative means saturation.

Case-I: Transistor saturates

Case-II: Active region

23. The viscosity of a liquid is given by Poiseuille's formula . Assume that l and V acan be measured very accurately, but the pressure P has an rms error of 1% and the radius a has an independent rms error of 3%. The rms error of the viscosity is closest to

(a) 2%

(b) 4%

(c) 12%

(d) 13%

Ans. (c)

Sol.

24. Which of the following circuits behaves as a controlled inverter?

(a)

(b)

(c)

(d)

Ans. (b)

Sol. Option (b) : It produces O/P expression

When any I/P in 1. O/P is the complement of another.

A becomes only for one I/P terminal is 1. {i.e. controlled inversion operation option (d) produces .}

When any I/P is zero produces complement operation

{is controlled inverter}

25. The concentration of electrons, n and holes, p for an intrinsic semiconductor at a temperature T can be expressed as , where Eg is the band gap and A is a constant. If the mobility of both types of carries is proportional to T–3/2, then the log of the conductivity is a linear function of T–1 with slope

(a)

(b)

(c)

(d)

Ans. (c)

Sol. We have,

26. Three real variables a, b and c are each randomly chosen from a uniform probability distribution in the interval [0, 1]. The probability that a + b > 2c is

(a)

(b)

(c)

(d)

Ans. (c)

Sol. Three random variables a, b, c are chosen randomly from an uniform distribution in the interval [0, 1]

(a + b – 2c)max = 2

(a + b – 2c)max = – 2

So, sample space for the experiment will be

–2 < (a + b – 2c) < 2 ...(1)

We have to calculate the probability that

0 < (a + b – 2c) < 2 ...(2)

Assuming a + b = X and 2c = Y

Conditions (1) and (2) and be written as

–2 < X – Y < 2 and 0 < X – Y < 2

For a uniform probability distribution

27. The rank-2 tensor xixj, where xi are the Cartesian coordinates of the position vector in three dimensions, has 6 independent elements. Under rotation, these 6 elements decompose into irreducible sets (that is the elements of each set transform only into linear combinations of elements in that set) containing

(a) 4 and 2 elements

(b) 5 and 1 elements

(c) 3, 2 and 1 elements

(d) 4, 1 and 1 elements

Ans. (b)

Sol. The rank – 2 tensor Tij = xixj can be expressed as a 3 × 3 matrix.

Tij can be decomposed into irreducible representation in the following way:

Here, is rank 0 tensor and transform under rotation like a scalar and it will have only one independent component.

Here, is rank 1 tensor and transform under rotation like a vector and represented by

i.e. no independent component.

Here, is a rank 2 tensor and represented by

Since, is symmetric and traceless, there are 5 independent elements.

28. Consider the differential equation with the initial condition y = 2 at x = 0. Let y(1) and y(1/2) be the solutions at x = 1 obtained using Euler's forward algorithm with step size 1 and 1/2 respectively.

The value of (y(1) – y(1/2))/y(1/2) is

(a)

(b) –1

(c)

(d) 1

Ans. (b)

Sol. with initial conditions y (x = 0) = 2

Step Size: h = 1

Step Size: h = ½:

So,

29. Let f(x, t) be a solution of the wave equation in 1-dimension. If at t = 0, and for all x, then f(x, t) for all future times t > 0 is described by

(a)

(b)

(c)

(d)

Ans. (d)

Sol.

Since, f (x, t) is the solution this equation.

Only option (d) satisfies this differential equation

30. Let q and p be the canonical coordinate and momentum of a dynamical system. Which of the following transformations is canonical?

(a) neither A nor B

(b) both A and B

(c) only A

(d) only B

Ans. (d)

Sol.

31. The differential cross-section for scattering by a target is given by . If N is the flux of the incoming particles, the number of particles scattered per unit time is

(a)

(b)

(c)

(d)

Ans. (d)

Sol. Total scattering cross section

Therefore, total number of particles scattered per unit time

32. Which of the following figures is a schematic representation of the phase space trajectories (i.e., contours of constant energy) of a particle moving in a one-dimensional potential ?

(a)

(b)

(c)

(d)

Ans. (a)

Sol.

For equilibrium point

so, V'(x) = 0 gives, x = 0, x = +1

V"(x) = –1 + 3x2 so, V"(x = 0) = –1 < 0 corresponds to maxima

V"(x = +1) = 2 > 0 corresponds to minima

33. Consider a rectangular wave guide with transverse dimensions 2m × 1m driven with an angular frequency = 109 rad/s. Which transverse electric (TE) modes will propagate in this wave guide?

(a) TE10, TE01 and TE20

(b) TE10, TE11 and TE20

(c) TE01, TE10 and TE11

(d) TE01, TE10 and TE22

Ans. (a)

Sol. Cutoff frequency for wave guide is

So, TE22, TE11 will not propagate.

Therefore, only TE01, E10, TE20 mode will porpagate.

34. A rod of length L carries a total charge Q distributed uniformly. If this is observed in a frame moving with a speed v along the rod, the charge per unit length (as measured by the moving observer) is

(a)

(b)

(c)

(d)

Ans. (c)

Sol. Charge is invariant quantity, and length undergoes contraction by

Therefore, charge per unit length

35. The electric and magnetic fields in the charge free region z > 0 are given by

where , k1 and k2 are positive constants. The average energy flow in the x-direction is

(a)

(b)

(c)

(d)

Ans. (a)

Sol.

Poyinting vector in x – direction.

36. A uniform magnetic field in the positive z-direction passes through a circular wire loop of radius 1 cm and resistance 1 lying in the xy-plane. The field strength is reduced from 10 tesla to 9 tesla in 1 s. The charge transferred across any point in the wire is approximately

(a) 3.1 × 10–4 coulomb

(b) 3.4 × 10–4 coulomb

(c) 4.2 × 10–4 coulomb

(d) 5.2 × 10–4 coulomb

Ans. (d)

Sol. Charge flown

37. The Dirac Hamiltonian for a free electron corresponds to the classical relation . The classical energy-momentum relation of a particle of charge q in a electromagnetic potential is . Therefore, the Dirac Hamiltonian for an electron in an electromagnetic field is

(a)

(b)

(c)

(d)

Ans. (d)

Sol. Dirac Hamiltonian for free particle

Linear momentum of a particle of charge q in electromagnetic field with potential

Dirac Hamiltonian in electromagnetic field with potential

For electron, q = –e

Dirac Hamiltonian in electromagnetic field with potential

38. A particle of mass m is in a potential , where is a constant. Let . In the Heisenberg picture is given by

(a)

(b)

(c)

(d)

Ans. (b)

Sol. We have,

39. A particle of energy E scatters off a repulsive spherical potential

where V0 and a are positive constants. In the low energy limit, the total scattering, cross-section is , where . In the limit the ratio of to the classical scattering cross-section off a sphere of radius a is

(a) 4

(b) 3

(c) 1

(d) 1/2

Ans. (a)

Sol. We have, where,

As

So,

and the classical scattering cross – section of sphere of radiua a is

Therefore,

40. Two different sets of orthogonal basis vectors and are given for a two-dimensional real vector space. The matrix representation of a linear operator in these bases are related by a unitary transformation. The unitary matrix may be chosen to be

(a)

(b)

(c)

(d)

Ans. (c)

Sol.

Change of basis from basis 1 to basis 2, will correspond to the diagonalization of operator .

Diagonalization of is done using similarity transformation i.e.

Where is the required unitary matrix formed by eigen vectors of in basis 1.

Therefore,

41. A large unber N of Brownian particles in one-dimension start their diffusive motion from the origin at time t = 0. The diffusion coefficients is D. The number of particles crossing a point at a distance L from the origin, per unit time, depends on L and time t as

(a)

(b)

(c)

(d)

Ans. (a)

Sol. The equation of Brownian motion,

Where, = density of particles.

The equation (1) is also known as diffusion equation. The solution of equation (1) in 3 – D.

Number of particles per unit length, per unit time

Dimension is not mentioned in the question but if we consider it as 3 – D problem then solution (2) takes form.

And if we use equation (2) or (4) to find

Mean square position,

And if we consider the problem as 1–D then solution given by equation (3) or in option (1) then

42. Consider three Ising spins at the vertices of a triangle which interact with each other with a ferromagnetic Ising interaction of strength J. The partition function of the system at temperature T is given by :

(a)

(b)

(c)

(d)

Ans. (a)

Sol. Partition function =

The energy of two particles degeneracy

For lsing model of three – spin total combination

J1 = J2 = J

E = J and degeneracy 6

Therefore, partition function

43. An ideal Bose gas in d-dimensions obeys the dispersion relation , where A and s are constants. For Bose-Einstein condensation to occur, the occupancy of excited states

where c is a constant, should remain finite even for µ = 0. This can happen if

(a)

(b)

(c)

(d)

Ans. (c)

Sol. For B.E. condensation to take place Ne = finite i.e. integral converge,

44. For the circuit and the input sinusoidal waveform shown in the figures below, which is the correct waveform at the output?

(The time scales in all the plots are the same).

(a)

(b)

(c)

(d)

Ans. (b)

Sol. FET have acting as a switch for negative have cycle O/P is positive option (a) and (c) neglected. For negative half cycle FET forward bias then 10K and 5 K are parallel. Equivalent is 3.33 K.

(for input negative half cycle)

45. For the logic circuit given below, the decimal count sequence and the modulus of the circuit corresponding to A B C D are

(a)

(b)

(c)

(d)

Ans. (b)

Sol.

The above representation is equivalent representation of a ring counter.

Initial count is 1 0 0 0 i.e. 8

for next clock 0 1 0 0 i.e. 4

For next clock 0 0 1 0 i.e. 2

For next clock 1 0 0 1 i.e. 9

For next clock 0 1 0 1 i.e. 5

46. In the circuit given below, the thermistor has a resistance 3 k at 25ºC. Its resistance decreases by 150 per ºC upon heating. The output voltage of the circuit at 30ºC is

(a) –3.75 V

(b) –2.25 V

(c) 2.25 V

(d) 3.75 V

Ans. (c)

Sol. At 250C RT = 3K

For 10C rise RT decreases 150

50 rise RT decreases = 5 × 120 = 750

At 300C rise RT = 2.250 = 2.25K

Apply Model:

47. The low-energy electronic excitations in a two-dimensional sheet of graphene is given by , where v is the velocity of the excitations. The density of states is proportional to

(a) E

(b) E3/2

(c) E1/2

(d) E2

Ans. (a)

Sol. We know that density of state for 2 – D

Therefore, density of states

48. X-ray of wavelength = a is reflected from the (1 1 1) plane of a simple cubic lattice. If the lattice constant is a, the corresponding Bragg angle (in radian) is

(a)

(b)

(c)

(d)

Ans. (c)

Sol. Condition of maxima for Bragg reflection:

49. The critical magnetic fields of a super-conductor at temperatures 4 K and 8 K are 11 mA/m and 5.5 mA/m respectively. The transition temperature is approximately

(a) 8.4 K

(b) 10.6 K

(c) 12.9 K

(d) 15.0 K

Ans. (b)

Sol. We know that, where, Tc is the transition temperature.

Since, then

50. A diatomic molecule has vibrational states with energies and rotational states with energies Ej = Bj (j + 1), where v and j are non-negative integers. Consider the transitions in which both the initial and final states are restricted to v < 1 and j < 2 and subject to the selection rules and . Then the largest allowed energy of transition is

(a)

(b)

(c)

(d)

Ans. (c)

Sol. Given

Ej = Bj (j + 1) (j = 0, 1, 2)

Transition possible and

Therefore, maximum allowed energy transition will correspond to and j = 1 to j = 2

Corresponding energy,

51. Of the following term symbols of the np2 atomic configurations, 1S0, 3P0, 3P1, 3P2 and 1D1 which is the grounded state?

(a) 3P0

(b) 1S0

(c) 3P2

(d) 3P1

Ans. (a)

Sol. We have electron distribution np2 i.e.

(i) Higher the value of s, lower the energy. So, sG.S. = 3

(ii) Higher the value of s, lower the energy. So,

(iii) Since, the subshells are less than half filled. Higher value of j, higher the energy. So, JG.S.=0

Therefore, ground state 2s+1Tj = 3P0

52. A He-Ne laser operates by using two energy levels of Ne separated by 2.26 eV. Under steady state conditions of optical pumping, the equivalent temperature of the system at which the ratio of the number of atoms in the upper state to that in the lower state will be 1/20, is approximately (the Boltzmann constant kB = 8.6 × 10–5 eV/K).

(a) 1010 K

(b) 108 K

(c) 106 K

(d) 104 K

Ans. (d)

Sol. According to Boltzmann distribution N = N0e–E/kT

Number of atoms in ground state

Number of atoms in upper state

53. Let us approximate the nuclear potential in the shell model by a 3-dimensional isotropic harmonic oscillator. Since the lowest two energy levels have angular momenta l = 0 and l = 1 respectively, which of the following two nuclei have magic numbers of protons and neutrons?

(a)

(b)

(c)

(d)

Ans. (a)

Sol. l = 0, state has capacity and l = 1 state has capacity 2[2 × 1 + 1] = 6

The first magic number is 2 and second magic number is 2 + 6 = 8

54. The charm quark is assigned a charm quantum number C = 1. How should the Gellmann-Nishijima formula for electric charge be modified for four flavours of quarks?

(a)

(b)

(c)

(d)

Ans. (d)

Sol. The Gellmann – Nishijma formula is

55. The reaction cannot proceed via strong interactions because it violates the conservations of

(a) angular momentum

(b) electric charge

(c) baryon number

(d) isospin

Ans. (d)

Sol.

Isospin conservation is violated.

I = 0 I = 0 I = 0 I = I

For strong interaction, angular momentum, electric charge and baryon number remain conserved.