CSIR NET PHYSICS (JUNE 2011)
Previous Year Question Paper with Solution.
1. A particle of unit mass moves in a potential , where a and b are positive constants. The angular frequency of small oscillations about the minimum of the potential is:
(a)
(b)
(c)
(d)
Ans. (b)
Sol.
2ax4 + (–2)b = 0
We know that
2. A signal of frequency 10 kHz is being digitized by an A/D converter. A possible sampling time which can be used is:
(a) 100 µs
(b) 40 µs
(c) 60 µs
(d) 200 µs
Ans. (c)
Sol. Signal frequency = 10 kHz = 104 Hz
According to sampling theorem,
Minimum frequency for sampling = 2 × (signal frequency) = 2 × 104 Hz
Sampling time
Minimum sampling time = 50 µs which is closest to 60 µs.
3. The electrostatic potential V(x, y) in free space in a region where the charge density is zero is given by V(x, y) = 4e2x + f`(x) – 3y2. Given that the x-component of the electric field, Ex, and V are zero at the origin, f(x) is:
(a) 3x2 – 4e2x + 8x
(b) 3x2 – 4e2x + 16x
(c) 4e2x – 8
(d) 3x2 – 4e2x
Ans. (d)
Sol. In a region where charge density is zero, the electrostatic potential V satisfies Laplace equation.
Using in equation (i), we get
Using, Ex (x = 0) = 8 + f '(0) = 0, we get f '(0) = –8
Using, V(x, y) = 4e2x + f(x) – 3y2 is zero at origin (0, 0). So, f(0) = –4
On integrating f "(x) from equation (ii), we get
So, f (x) = 3x2 – 4e2x
4. Consider the transition of liquid water to steam as water boils at a temperature of 100ºC under a pressure of 1 atmosphere. Which one of the following quantities does not change discontinuously at the transition?
(a) The Gibbs free energy
(b) The internal energy
(c) The entropy
(d) The specific volume
Ans. (a)
Sol. Liquid water to steam is first order phase transition. So, G is continuous and 1st order derivative of G is discontinuous.
U = G + TS – PV is also discontinuous.
5. The value of the integral , where C is an open contour in the complex z-plane as shown in the figure below, is:
(a)
(b)
(c)
(d)
Ans. (c)
Sol.
Since f(z) = z2ez is an analytic function, then
[Since, along CR, y = 0 and dy = 0]
Then, value of open contour of integral
6. Which of the following matrices is an element of the group SU(2)?
(a)
(b)
(c)
(d)
Ans. (b)
Sol. A matrix , with det(U) = 1 is said to be an element of the special unitary group of order 2 i.e. SU(2).
For the matrix,
7. For constant uniform electric and magnetic fields and
, it is possible to choose a gauge such that the scalar potential
and vector potential
are given by
(a)
(b)
(c)
(d)
Ans. (b)
Sol. First Method:
Second Method:
For any constant vector field ,
Since, and
are constant uniform fields.
8. Let and
be two distinct three-dimensional vectors. Then the component of
that is perpendicular to
is given by
(a)
(b)
(c)
(d)
Ans. (a)
Sol. Component of that is perpendicular to
9. The wavefunction of a particle is given by , where
and
are the normalized eigenfunctions with energies E0 and E1 corresponding to the ground state and first excited state, respectively. The expectation value of the Hamiltonian inthe state
is:
(a)
(b)
(c)
(d)
Ans. (d)
Sol.
10. A particle is confined to the region x > 0 by a potential which increases linearly as u(x) = u0x. The mean position of the particle at temperature T is:
(a)
(b) (kBT)2/u0
(c)
(d) u0kBT
Ans. (a)
Sol.
11. Circularly polarized light with intensity I0 is incident normally on a glass prism as shown in the figure. The index of refraction of glass is 1.5. The intensity I of light emerging from the prism is:
(a) I0
(b) 0.96 I0
(c) 0.92 I0
(d) 0.88 I0
Ans. (b)
Sol. The right angle prism behave as a total internal reflector.
For normal incidence:
n1 = 1.5, n2 = 1.0
Intensity of emerging light = I = (0.96)I0
12. The acceleration due to gravity (g) on the surface of Earth is approximately 2.6 times that on the surface of Mars. Given that the radius of Mars is about one half the radius of Earth, the ratio of the escape velocity on Earth to that on Mars is approximately:
(a) 1.1
(b) 1.3
(c) 2.3
(d) 5.2
Ans. (c)
Sol.
Equation (2) put in equation (1)
13. A plane electromagnetic wave is propagating in a lossless dielectric. The electric field is given by , where c is the speed of light in vacuum, E0, A and k0 are constants and
and
are unit vectors along the x- and z-axes. The relative dielectric constant of the medium,
and the constant A are
(a)
(b)
(c)
(d)
Ans. (a)
Sol.
We know that
Comparing (1) and (2)
Relative dielectric constant for the medium
We know that
14. Consider the digital circuit shown below in which the input C is always high (I).
The entries in the Z column (vertically) are
(a) 1010
(b) 0100
(c) 1111
(d) 1011
Ans. (d)
Sol.
Truth Table:
15. The energy levels of the non-relativistic electron in a hydrogen atom (i.e. in a Coulomb potential are given by
, where n is the principal quantum number, and the corresponding wave functions are given by
, where
is the orbital angular momentum quantum number and m is the magnetic quantum number. The spin of the electron is not considered. Which of the following is a correct statement?
(a) There are exactly different wave functions
, for each
.
(b) There are different wave functions
, for each
.
(c) does not depend on and m for the Coulomb potential.
(d) There is a unique wave function and
.
Ans. (c)
Sol.
The energy En depends only on n and not on and m. The degeneracy of energy levels is n2 (neglecting spin) that arises due to the spherical symmetry of the Columb potential which depends only on radial distance r of the electron from the nucleus which is considered as the origin. There is no angular dependence
in V(r).
16. The Hamiltonian of an electron in a constant magnetic field is given by
where µ is a positive constant and
denotes the Pauli matrices. Let
and I be the 2 × 2 unit matix. Then the operator
simplifies to
(a)
(b)
(c)
(d)
Ans. (b)
Sol. We know that,
17. The Hamiltonian of a system with n degrees of freedom is given by
H(q1, ................., qn; p1, .................., pn; t),
with an explicit dependence on the time t. Which of the following is correct?
(a) Different phase trajectories cannot intersect each other.
(b) H always represents the total energy of the system and is a constant of the motion.
(c) The equations are not vali since H has explicit time dependence.
(d) Any initial volume element in phase space remains unchanged in magnitude under time evolution.
Ans. (d)
Sol. According to Liouville's theorem, the phase volume occupied by a collection of system evolve according to Hamilton's equation of motion, and will be preserved in time.
18. If the perturbation H' = ax, where a is a constant; is added to the infinite square well potential
The first order correction to ground state energy is:
(a)
(b)
(c)
(d)
Ans. (a)
Sol. For a square well potential of infinite height at 0 and .
Ground State wave function will be,
First-order correction to the ground state is:
19. Let pn(x) (where n = 0, 1, 2 .........) be a polynomial of degree n with real coefficients, defined in the interval 2 < n < 4. If , then
(a)
(b)
(c)
(d)
Ans. (d)
Sol. According to the given orthonormality condition,
Only option (d) satisfies all the above equation.
20. A cavity contains blackbody radiation in equilibrium at temperature T. The specific heat per unit volume of the photon gas in the cavity is of the form where
is a constant. The cavity is expanded to twice its original volume and then allowed to equilibrate at the same temperature T. The new internal energy per unit volume is:
(a)
(b)
(c)
(d)
Ans. (d)
Sol. For Black-body radiation, where,
is Stefan-Boltzman constant.
21. Consider a system of N non-interacting spins, each of which has classical magnetic moment of magnitude µ. The Hamiltonian of this system in an external magnetic field is
, where
is the magnetic moment of the ith spin. The magnetization per spin at temperature T is:
(a)
(b)
(c)
(d)
Ans. (b)
Sol. Interaction energy
22. Which of the following is an analytic function of the complex variable z = x + iy in the domain |z| < 2?
(a) (3 + x – iy)7
(b) (1 + x + iy)4(7 – x – iy)3
(c) (1 – 2x – iy)4(3 – x – iy)3
(d) (x + iy – 1)1/2
Ans. (b)
Sol. Any complex function which contains only 'z', will be analytic in nature.
But any complex function which contains , will not be analytic in nature.
Option-a:
Option-b: (1 + x + iy)4 (7 – x – iy)3 = (1 + z)4 (7 – z)3
Option-c: (1 – 2x – iy)4 (3 – x – iy)3 = (1 – Re.z – z)4 (3 – z)3
Option-d: (x + iy – 1)1/2 = (z – 1)1/2 has a branch point at z = 1
23. A particle in one dimension moves under the influence of a potential V(x) = ax6, where a is a real constant. For lasrge n the quantized energy level En depends on n as:
(a) En ~ n3
(b) En ~ n4/3
(c) En ~ n6/5
(d) En ~ n3/2
Ans. (d)
Sol. V(x) = ax6
For classical turning points, E = V(x) = ax6
Using quantization condition of WKB approximation,
Using dimensional analysis in terms of energy.
24. The Lagrangian of a particle of charge e and mass m in applied electric and magnetic fields is given by , where
and
are the vector and scalar potentials corresponding to the magnetic and electric fields, respectively. Which of the following statements is correct?
(a) The canonically conjugate momentum of the particle is given by
(b) The Hamiltonian of the particle is given by
(c) L remains unchanged under a gauge transformation of the potentials.
(d) Under a gauge transformation of the potentials, L changes by the total time derivative of a function of and t.
Ans. (d)
Sol.
We know that and
fields are in variant under gauge transformation.
where is an arbitrary scalar function.
The expression in bracket is just the total derivative of .
If we add a total time derivative of a function of and t to the Lagrangian, the equations of motion do not change.
25. A static, spherically symmetric charge distribution is given by where A and k are positive constants. The electrostatic potential corresponding to this charge distribution varies with r as
(a) r e–kr
(b)
(c)
(d)
Ans. (b)
Sol.
26. Consider two independently diffusing non-interacting particles in 3-dimensional space, both placed at the origin at time t = 0. These particles have different diffusion constants D1 and D2. The quantity where
and
are the positions of the particles at time t, behaves as:
(a) 6t(D1 + D2)
(b) 6t(D1 – D2)
(c)
(d)
Ans. (a)
Sol. Diffusion equation in one-dimension for one particle is
with as solution by assuming x = 0 at t = 0.
For this Gausian distribution,
27. A resistance is measured by passing current through it and measuring the resulting voltage drop. If the voltmeter and the ammeter have uncertainties of 3% and 4%, respectively, then
(A) The uncertainty in the value of resistance is:
(a) 7.0%
(b) 3.5%
(c) 5.0%
(d) 12.0%
Ans. (c)
Sol.
Uncertainties in V and I are independent. So, they must add in quadrature
(B) The uncertainty in the computed value of the power dissipated in resistance is
(a) 7%
(b) 5%
(c) 11%
(d) 9%
Ans. (b)
Sol. Power Dissipated P = VI
28. In the absence of an applied torque a rigid body with three distinct principal moments of inertia given by I1, I2 and I3 is rotating freely about a fixed point inside the body. The Euler equations for the components of its angular velocity are
(A) The equilibrium points in space are
(a) (1, –1, 0), (–1, 0, 1) and (0, –1, 1)
(b) (1, 1, 0), (1, 0, 1) and (0, 1, 1)
(c) (1, 0, 0), (0, 1, 0) and (0, 0, 1)
(d) (1, 1, 1), (–1, –1, –1) and (0, 0, 0)
Ans. (c)
Sol. Euler equations:
The general solution is for any value of a, b and c.
Any point on the co-ordinate axes is an equilibrium point; choose, a = b = c = 1.
(B) The constants of motion are
(a)
(b)
(c)
(d)
Ans. (b)
Sol.
L is a constant of motion.
29. In a system consisting of two spin-½ particles labeled 1 and 2, let and
denote the corresponding spin operators. Here
and
are the three Pauli matrices.
(A) In the standard basis the matrices for the operators and
respectively,
(a)
(b)
(c)
(d)
Ans. (c)
Sol. Pauli Spin matrices:
(B) These two operators satisfy the relation
(a)
(b)
(c)
(d)
Ans. (d)
Sol.
30. Consider the matrix
(A) The eigenvalues of M are
(a) 0, 1, 2
(b) 0, 0, 3
(c) 1, 1, 1
(d) –1, 1, 3
Ans. (b)
Sol.
Eigenvalue equation
For any square matrix M of order n having all elements equal to 1, have eigenvalues n, 0, 0, 0, ...........
Therefore, eigen values of M are: 0, 0, 3
(B) The exponential of M simplifies to (I is the 3 × 3 identity matrix)
(a)
(b)
(c) eM = I + 33 M
(d) eM = (e – 1) M
Ans. (a)
Sol.
M3 = M2.M = 3M.M = 32M and so on.
31. The radius of a nucleus is measured to be 4.8 × 10–13 cm.
(A) The radius of a nucleus can be estimated to be
(a) 2.86 × 10–13 cm
(b) 5.2 × 10–13 cm
(c) 3.6 × 10–13 cm
(d) 8.6 × 10–13 cm
Ans. (c)
Sol. Since, R = R0(A)1/3
(B) The root-mean-square (rms) energy of a nucleon in a nucleus of atomic number A in its ground state varies as
(a) A4/3
(b) A1/3
(c) A–1/3
(d) A–2/3
Ans. (c)
Sol. The ground state energy of a nucleon can be given as
(where, m0c2 = rest mass energy)
32. The character table of C3, the group of symmetries of an equilateral triangle is given below
In the above C1, C2, C3 denotes the three classes of C3v, containing 1, 3 and 2 elements respectively, and and
are the characters of the three irreducible representations
and
of C3v.
(A) The entries a, b, c and d in this table are, respectively
(a) 2, 1, –1, 0
(b) –1, 2, 0, –1
(c) –1, 1, 0, –1
(d) –1, 1, 1, –1
Ans. (b)
Sol. The correct character table for C3V point group is:
Order of group = Number of classes = 1 + 3 + 2 = 6 = R
The entries satisfy great orthogonality theorem.
1(1)2 + 3(1)2 + 2(1)2 = 6; 1 (+1)2 + 3(–1)2 + 2(1)2 = 6; 1(2)2 + 3(0)2 + 2(–1)2 = 6
1(1)(1) + 3(1)(–1) + 2(1)(1) = 0; (1)(2) + 3(1)(0) + 2(1)(–1) = 0;
1(1)(2) + 3(–1)(0) + (2(1)(–1) = 0
(B) The reducible representation of C3v with character
= (4, 0, 1) decomposes into its irreducible representations
as
(a)
(b)
(c)
(d)
Ans. (c)
Sol.
33. Light of wavelength 660 nm and power of 1 mW is incident on a semiconductor photodiode with an absorbing layer of thickness of (ln 4) µm.
(A) If the absorption coefficient at this wavelength is 104 cm–1 and if 1% power is lost on reflection at the surface, the power absorbed will be
(a) 750 µW
(b) 675 µW
(c) 250 µW
(d) 225 µW
Ans. (c)
Sol. Power of incident light = 1000 µW.
Power of reflected light = 10 µW
Power of light entering the semiconductor = 990 µW
Amount existing in the semiconductor = (990 µW)e–µx
where
Therefore, amount exiting in the semiconductor
Therefore, amount of power transmitted through semiconductor
So, power absorbed in the semiconductor = (1000 µW – 750 µW) = 250 µW
Correct answer = 250 µW
[Note: If 1% light is not reflected only then the exact answer will be 750 µW.
(B) The generated photo-current for a quantum efficiency of unity will be
(a) 360 µA
(b) 400 µA
(c) 133 µA
(d) 120 µA
Ans. (c)
Sol.
34. The magnetic field of the TE11 mode of a rectangular waveguide of dimensions a × b as shown in the figure is given by , where x and y are in cm.
(A) The dimensions of the waveguide are
(a) a = 3.33 cm, b = 2.50 cm
(b) a = 0.40 cm, b = 0.30 cm
(c) a = 0.80 cm, b = 0.60 cm
(d) a = 1.66 cm, b = 1.25 cm
Ans. (a)
Sol. Hz = H0 cos kxx cos kyy (TEnm mode)
(B) The entire range of frequencies f for which the TE11 mode will propagate is:
(a) 6.0 GHz < f < 7.5 GHz
(b) 7.5 GHz < f < 9.0 GHz
(c) 7.5 GHz < f < 12.0 GHz
(d) 7.5 GHz < f
Ans. (d)
Sol. Cut-off frequency
All the frequencies greater than cut-off frequency (fc)11 will posible for TE11.
35. Consider the energy level diagram (as shown in the figure below) of a typical three level ruby laser system with 1.6 × 1019 Chromium ions per cubic centimeter. All the atoms excited by the 0.4 µm radiation decay rapidly to level E2 which has a lifetime = 3ms
(A) Assuming that there is no radiation of wavelength 0.7 µm present in the pumping cycle and that the pumping rate is R atoms per cm3, the population density in the level N2 builds up as:
(a)
(b)
(c)
(d) N2(t) = Rt
Ans. (b)
Sol.
(B) The minimum pump power required (per cubic centimeter) to bring the system to transparency, i.e. zero gain, is
(a) 1.52 kW
(b) 2.64 kW
(c) 0.76 kW
(d) 1.32 kW
Ans. (d)
Sol.
We get, (Power)Threshold = 1.32 × 103 W = 1.32 kW
36. A flux quantum (fluxoid) is approximately equal to 2 × 10–7 gauss-cm2. A type II superconductor is placed in a small magnetic field, which is then slowly increased till the field starts penetrating the superconductor. The strength of the field at this point is gauss
(A) The penetration depth of this superconductor is
(a) 100Å
(b) 10Å
(c) 1000Å
(d) 314Å
Ans. (a)
Sol.
(B) The applied field is further increased till superconductivity is completely destroyed. The strength of the field is now gauss. The correlation length of the superconductor is:
(a) 20Å
(b) 200Å
(c) 628Å
(d) 2000Å
Ans. (a)
Sol.
[Does not match with any option, however we can still eliminate the wrong options, has different expression in different approximations]
For a type II semiconductor.
Only option (a) satisfies this.
37. A beam of pions is incident on a proton target, giving rise to the process
(A) Assuming that the decay proceeds through strong interactions, the total isospin I and its third component I3 for the decay products, are
(a)
(b)
(c)
(d)
Ans. (c)
Sol.
I = 5/2, I3 = 3/2.
Note: Reaction is forbidden . For strong interaction but is allowed by EM interaction.
(B) Using isospin symmetry, the cross-section for the above process can be related to that of the process
(a)
(b)
(c)
(d)
Ans. (b)
Sol. In isospin symmetry, the particles are replaced by the corresponding antiparticles
Reaction of particles :
Reaction of antiparticle :
38. The two dimensional lattice of graphene is an arrangement of Carbon atoms forming a honeycomb lattice of lattice spacing a, as shown below. The carbon atoms occupy the vertices.
(A) The Wigner-Seitz cell has an area of
(a) 2a2
(b)
(c)
(d)
Ans. (d)
Sol. The Wigner Seitz cell has the area of a hexagon.
Split the hexagon into six equilateral triangles.
Use Pythagoras theorem
(B) The Bravais lattice for this array is a
(a) Rectangular lattice with basis vectors and
(b) Rectangular lattice with basis vectors and
(c) Hexagonal lattice with basis vectors and
(d) Hexagonal lattice with basis vectors and
Ans. (c)
Sol. The hexagonal Bravais lattice is defined as
39. Consider the decay process in the rest frame of the
. The masses of
, and
are
and zero respectively.
(A) The energy of is :
(a)
(b)
(c)
(d)
Ans. (b)
Sol.
Let energy of is E.
Energy conservation
Momentum conservation
Energy of
(B) The velocity is is:
(a)
(b)
(c)
(d)
Ans. (a)
Sol. Now, momentum of
40. A narrow beam of X-rays with wavelength 1.5Å is reflected from an ionic crystal with an fcc lattice structure with a density of 3.32 g cm–3. The molecular weight is 108 AMU (1 AMU = 1.66 × 10–24 g).
(A) The lattice constant is:
(a) 6.00Å
(b) 4.56Å
(c) 4.00Å
(d) 2.56Å
Ans. (a)
Sol.
N = 4 (Number of atoms in fcc lattice)
M = 108 × 1.66 × 10–24,
(B) The sine of the angle corresponding to (111) reflection is:
(a)
(b)
(c) ¼
(d) 1/8
Ans. (b)
Sol.
41. If an electron is in the ground state of the hydrogen atom, the probability that its distance from the proton is more than one Bohr radius is approximately
(a) 0.68
(b) 0.48
(c) 0.28
(d) 0.91
Ans. (a)
Sol.
Probability for electron to lie outside the first Bohr radius is:
42. A time varying signal Vm is fed to an op-amp circuit with output signal V0 as shown in the figure below.
The circuit implements a
(a) High pass filter with cutoff frequency 16 Hz.
(b) High pass filter with cutoff frequency 100 Hz
(c) Low pass filter with cutoff frequency 16 Hz
(d) Low pass filter with cutoff frequency 100 Hz.
Ans. (a)
Sol.
Apply KCL at node (2)
Apply KCL at node (1)
So, it high pass filter with cutoff frequency 16 Hz.
43. The Hamiltonian of a particle of unit mass moving in the xy-plane is given to be:
in suitable units. The initial values are given to be (x(0), y(0)) = (1, 1) and
. During the motion, the curves traced out by the particles in the xy-plane and the pxpy-plane are
(a) Both straight lines
(b) A straight line and a hyperbola respectively
(c) A hyperbola an ellipse, respectively
(d) Both hyperbolas
Ans. (d)
Sol.
Hamilton's equations:
Multiplying Eq (i) and (ii)
Again Hamiltan's equations:
We get, c1 = 0, c2 = 0
So that
Now multiplying px and py, then we get
From eq. (iii), (iv) we can say that both equestion shows hyperbolas.
44. Consider an ideal Bose gas in three dimensions with the energy-momentum relation with s > 0. The range of s for which this system may undergo a Bose-Einstein condensation at a non-zero temperature is:
(a) 1 < s < 3
(b) 0 < s < 2
(c) 0 < s < 3
(d) 0 < s <
Ans. (c)
Sol.
Number of particular N' that can be accomodated in all the excited states for Bose-Einstein condensation is
45. Two gravitating bodies A and B with masses mA and mB, respectively, are moving in circular orbit. Assume that mB >> mA and let the radius of the orbit of body A be RA. If the body A is losing mass adiabatically, its orbital radius RA is proportional to
(a) 1/mA
(b)
(c) mA
(d)
Ans. (b)
Sol. As the mass loss is adiabatic, angular momentum is conserved. [One can consider mB to be the mass of the Sun while mA to be the mass of earth, as mB >> MA].
is the new radius.
Thus, replacing VA in (1) from (2) we obtain,