1. Which of the following cannot be the eigenvalues of a real 3 × 3 matrix

(a) 2i, 0, –2i

(b) 1, 1, 1

(c)

(d) i, 1, 0

Ans. (d)

Sol. If the matrix is real then the complex eigen values always occurs with its complex conjugate. In option (d) if i is an eigen value then –i must also be an eigen value. But –i is not given in option, hence option (d) is incorrect.

2. Let u(x, y) = e^ax cos (by) be the real part of a function f(z) = u(x, y) + iv(x, y) of the complex variable z = x + iy, where a, b are real constants and . The function f(z) is complex analytic everywhere in the complex plane if and only if

(a) b = 0

(b) b = +a

(c)

(d)

Ans. (b)

Sol. The function f(z) will be analytic everywhere in the complex plane if and only if it satisfies the Cauchy Riemann equation in that region.

From equation (i)

Differentiating partially with x gives

From equation (iii) and (iv)

3. The integral along the closed contour shown in the figure is

(a) 0

(b)

(c)

(d)

Ans. (c)

Sol.

4. The function y(x) satisfies the differential equation . If y(1) = 1, the value of y(2) is

(a)

(b) 1

(c) 1/2

(d) 1/4

Ans. (d)

Sol. The given differential equation can be written as

This is a linear differential equation with Integrating factor

5. The random variable is distributed according to the normal distribution . The probability density of the random variable y = x² is

(a)

(b)

(c)

(d)

Ans. (a)

Sol.

6. The Hamiltonian for a system described by the generalized coordinate x and generalised momentum p is

where and are constants. The corresponding Lagrangian is

(a)

(b)

(c)

(d)

Ans. (a)

Sol.

We know that

We know that

Eq. (1), (2) put in Eq. (4)

Eq. (3) put in Eq. (5)

7. An inertial observer sees two events E₁ and E₂ happening at the same location but 6 µs apart in time. Another observer moving with a constant velocity v (with respect to the first one) sees the same events to be 9 µs apart. The spatial distance between the events, as measured by the second observer, is approximately

(a) 300 m

(b) 1000 m

(c) 2000 m

(d) 2700 m

Ans. (c)

Sol. The position of two events with respect to S frame are (x₁, y₁, z₁, t₁) and (x₁, y₁, z₁, t₂)

The position of two events with respect to S' frame are

8. A ball weighing 100 gm, released from a height of 5 m, bounces perfectly elastically off a plate. The collision time between the ball and the plate is 0.5 s. The average force on the plate is approximately

(a) 3 N

(b) 2 N

(c) 5 N

(d) 4 N

Ans. (c)

Sol. The velocity of ball before collision with plate is

The change in momentum due to collision is = 10 × 0.1 – (–10 × 0.1) = 2 kg-m/sec

So, the average force on the plate is

= Force due to change in momentum + force due to weight

9. A solid vertical rod, of length L and cross-sectional area A, is made of a material of Young's modulus Y. The rod is loaded with a mass M, and, as a result, extends by a small amount in the equilibrium condition. The mass is then suddenly reduced to M/2. As a result the rod will undergo longitudinal oscillation with an angular frequency

(a)

(b)

(c)

(d)

Ans. (a)

Sol. According to Hooke's law,

10. If the root-mean-squared momentum of a particle in the ground state of a one-dimensional simple harmonic potential is p₀, then its root-mean-squared momentum in the first excited state is

(a)

(b)

(c)

(d)

Ans. (b)

Sol. Root-mean square momentum of nth state of one dimensional harmonic potential is given by

So, R.M.S. momentum of ground state

So, R.M.S. momentum of first excited state

11. Consider a potential barrier A of height V₀ and width b, and another potential barrier B of height 2V₀ and the same width b. The ratio T_A/T_B of tunnelling probabilities T_A and T_B, through barriers A and B respectively, for a particle of energy V₀/100, is best approximated by

(a)

(b)

(c)

(d)

Ans. (a)

Sol. Since, the energy of the particle, very less than the potential barrier. So, tunnelling probabilities can be written as

12. A constant perturbation H' is applied to a system for time leading to a transition from a state with energy E_i to another with energy E_f. If the time of application is doubled, the probability of transition will be

(a) unchanged

(b) doubled

(c) quadrupled

(d) halved

Ans. (c)

Sol. Transition probability for due to the presence of time-dependent parturbation V_P(t'), is

For a constant perturbation H' applied to a system for time

For a transition from the state with energy E_i to a state with energy constant.

For small

Therefore, transition probability

Since, , therefore, if the time of application is doubled, then the transition probability will be quadrupled i.e. 4 times.

13. The two vectors are orthonormal if

(a)

(b)

(c)

(d)

Ans. (c)

Sol. The given two vectors are

According to normalized condition,

14. Two long hollow co-axial conducting cylinders of radii R₁ and R₂ (R₁ < R₂) are placed in vaccum as shown in the figure below.

The inner cylinder carries a charge per unit length and the outer cylinder carries a charge per unit length. The electrostatic energy per unit length of this system is

(a)

(b)

(c)

(d)

Ans. (c)

Sol. Electrostatic energy of a capacitor

15. A set of N concentric circular loops of wire, each carrying a steady current I in the same direction, is arranged in a plane. The radius of the first loop is r₁ = a and the radius of the n^th loop is given by r_n = n r_{n – 1}. The magnitude B of the magnetic field at the centre of the circles in the limit N → , is

(a)

(b)

(c)

(d)

Ans. (d)

Sol. Magnetic field at the center is given by

16. An electromagnetic wave (of wavelength in free space) travels through an absorbing medium with dielectric permittivity given by where . If the skin depth is , the ratio of the amplitude of electric field E to that of the magnetic field B, in the medium (in ohms) is

(a)

(b) 377

(c)

(d)

Ans. (d)

Sol.

We know that,

k = nk₀

We know that,

Again, we know that,

17. The vector potential (where a and k are constants) corresponding to an electromagnetic field is changed to . This will be a gauge transformation if the corresponding change in the scalar potential is

(a) akr²e^–at

(b) 2akr²e^–at

(c) –akr²e^–at

(d) –2akr²e^–at

Ans. (c)

Sol. According to gauge transformation

18. A thermodynamic function

G(T, P, N) = U – TS + PV

is given in terms of the internal energy U, temperature T, entropy S, pressure P, volume V and the number of particles N. Which of the following relations is true? (In the following µ is the chemical potential.)

(a)

(b)

(c)

(d)

Ans. (a)

Sol. Given: G(T, P, N) = U – TS + PV

DG = dU – TdS – SdT + PdV + VdP

where, dU = TdS – PdV + µdN

dG = µdN – SdT + VdP

19. A box, separated by a movable wall, has two compartments filled by a monoatomic gas of . Initially the volumes of the two compartments are equal, but the pressures are 3P₀ and P₀, respectively. When the wall is allowed to move, the final pressures in the two compartments become equal. The final pressure is

(a)

(b)

(c)

(d)

Ans. (c)

Sol.

Using adiabatic equation for compartment I, we have

Similarly, for compartment II, we have

From the above figure,

V₁ + V₂ = 2V₀

Now adding (1) and (2), we get

20. A gas of photons inside a cavity of volume V is in equilibrium at temperature T. If the temperature of the cavity is changed to 2T, the radiation pressure will change by a factor of

(a) 2

(b) 16

(c) 8

(d) 4

Ans. (b)

Sol. For a gas of photons inside a cavity,

21. In a thermodynamic system in equilibrium, each molecule can exist in three possible states with probabilities 1/2, 1/3 and 1/6 respectively. The entropy per molecule is

(a) k_B ln 3

(b)

(c)

(d)

Ans. (c)

Sol. The entropy per molecule is defined as

22. In the n-channel JFET shown in figure below, V_i = –2V, C = 10 pF, V_DD = +16V, and .

If the drain D-source S saturation current I_DSS is 10 mA and the pinch-off voltage V_i, is –8V, then the voltage across points D and S is

(a) 11.125 V

(b) 10.375 V

(c) 5.75 V

(d) 4.75 V

Ans. (d)

Sol. V_i = –2 volt

V_DD = 16 volt

I_DSS = maximum drain current = 10 mA

V_P = –8 (pinch off voltage)

To - calculate voltage (OR) Resistance value we apply DC - Analysis:

DC - Analysis means f = 0,

Open-circuit capacitor

Drawing circuit, making capacitor open circuit.

V_GS = V_G – V_S

V_GS = –2 volt

V_S = 0

I_D = 5.625 mA

KVL at output, 16 = 2×I_D + V_DS

Put I_D, V_DS = 16 – 2 × 5.625

V_DS = 4.75 volt

23. The gain of the circuit given below is .

The modification in the circuit required to introduce a dc feedback is to add a resistor

(a) between a and b

(b) between positive terminal of the op-amp and ground

(c) in series with C

(d) parallel to C

Ans. (d)

Sol. Gain of circuit

"–" sign due to in verting op-amp.

Converting circuit to a practical integrator, connecting "Resistor" parallel to capacitor

Concept: Given circuit is idela LPF/Integrator

if we connect "R" in parallel to "Capacitor" it becomes practical integrator with gain sign due to inverting OPAMP.

24. A 2 × 4 decoder with an enable input can function as a

(a) 4 × 1 multiplexer

(b) 1 × 4 demultiplexer

(c) 4 × 2 encoder

(d) 4 × 2 priority encoder

Ans. (b)

Sol. 2 × 4, Decoder

Enable – input means inverter/NOT gate

Decoder = n × 2ⁿ

If we are considering input as selection line

Format DEMUX/demultiplexer

Above - CKT is 1 × 4 Demux

25. The experimentally measured values of the variables x and y are 2.00 +0.05 and 3.00 +0.02, respectively. What is the error in the calculated value of z = 3y – 2x from the measurements?

(a) 0.12

(b) 0.05

(c) 0.03

(d) 0.07

Ans. (a)

Sol. z = 3y – 2x

26. The Green's function satisfying with the boundary conditions g(–L, x₀) = 0 = g(L, x₀), is

(a)

(b)

(c)

(d)

Ans. (a)

Sol.

boundary conditions:

g(–L, x₀) = 0 = g(L, x₀)

The homogeneous equation for Green's function is

Solution of above equation is

Applying boundary condition

From continuity of Green's function at x = x₀ we have

A(x₀ + L) = C(x₀ – L)

From discontinuity of derivative of Green's function

We have

Thus, the required solution of Green's function is given by

27. Let be the Pauli matrices and

Then the coordinates are related as follows

(a)

(b)

(c)

(d)

Ans. (b)

Sol.

28. The interval [0, 1] is divided into 2n parts of equal length to calculate the integral using Simpson's rule. What is the minimum value of n for the result to be exact?

(a)

(b) 2

(c) 3

(d) 4

Ans. (b)

Sol. Exact Method:

Simpson's 1/3^rd rul: (Taking n = 2, 2n = 4)

Length of sub-intervals,

29. Which of the following sets of 3 × 3 matrices (in which a and b are real numbers) form a group under matrix multiplication?

(a)

(b)

(c)

(d)

Ans. (c)

Sol. In order to form the group the required matrix must satisfy the following conditions.

(a) For any three matrices A(BC) = (AB)C

(b) There must exist an identity element I such that AI = IA = A

(c) There must exist an inverse element for each element belonging to G.

(d) For any two matrices , the matrix AB G (closer property).

All the given matrices satisfy property (a), if we put a = 0 and b = 0, we see that property (b) also holds for all the given matrices.

If we put a = 1 and b = 1, in option (a), two rows becomes identical and matrix in option (a) is non-invertible. If we put a = 1 and b = 1, in option (d), two rows becomes identical and the matrix is non-invertible.

Now only option (b) and (c) remains. For option (b) take two matrices and multiply

we see that the resulting matrix does not satisfy closure property. For option (3) take two matrices and multiply

we see that the resulting matrix satisfy the closure property. Hence the correct option is (c)

30. The Lagrangian of a free relativistic particle (in one dimension) of mass m is given by where . If such a particle is acted upon by a constant force in the direction of its motion, the phase space trajectories obtained from the corresponding Hamiltonian are

(a) ellipses

(b) cycloids

(c) hyperbolas

(d) parabolas

Ans. (c)

Sol.

Corresponding Hamiltonian is

This is equation of hyperbola in x-p_x plane.

31. A Hamiltonian system is described by the canonical coordinate q and canonical momentum p. A new coordinate Q is defined as , where t is the time and is a constant, that is, the new coordinate is a combination of the old coordinate and momentum at a shifted time. The new canonical momentum P(t) can be expressed as

(a)

(b)

(c)

(d)

Ans. (d)

Sol. In canonical transformation coordinate and momenta at one instant of time are transformed into new coordinate and momenta Further, if q, p Q, P be a cannonical transformation then Poisson bracket

This condition is satisfied only if we take

32. The energy of a one-dimensional system, governed by the Lagrangian

where k and n are two positive constants, is E₀. The time period of oscillation satisfies

(a)

(b)

(c)

(d)

Ans. (b)

Sol.

Total energy

E = T + V ... (2)

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

Eq. (4) put in Eq. (1)

Eq. (4), (5) put in Eq. (3)

33. An electron is decelerated at a constant rate starting from an initial velocity u (where u << c) to u/2 during which it travels a distance s. The amount of energy lost to radiation is

(a)

(b)

(c)

(d)

Ans. (d)

Sol. Radiative power due to accelerated charge particle

The fraction of energy loss due to radiation

34. The figure below describes the arrangement of slits and screens in a Young's double slit experiment. The width of the slit in S₁ is a and the slits in S₂ are of negligible width.

If the wavelength of the light is , the value of d for which the screen would be dark is

(a)

(b)

(c)

(d)

Ans. (b)

Sol. The screen will be dark if both the points A and B is dark.

35. A constant current I is flowing in a piece of wire that is bent into a loop as shown in the figure.

The magnitude of the magnetic field at the point O is

(a)

(b)

(c)

(d)

Ans. (b)

Sol.

So total magnetic field

36. Consider the potential

where are the position vectors of the vertices of a cube of length a centered at the origin and V₀ is a constant. If , the total scattering cross-section, in the low-energy limit, is

(a)

(b)

(c)

(d)

Ans. (c)

Sol.

Scattering amplitude :

37. The Coulomb potential V(r) = –e²/r of a hydrogen atom is perturbed by adding H' = bx² (where b is a constant) to the Hamiltonian. The first order correction to the ground state energy is (The ground state wavefunction is .)

(a)

(b)

(c)

(d)

Ans. (b)

Sol. The first order energy correction term of the ground state is given by

38. Using the trial function

the ground state energy of a one-dimensional harmonic oscillator is

(a)

(b)

(c)

(d)

Ans. (b)

Sol. We have trial function,

According to normalized condition

The energy is given by

Therefore, ground state energy,

39. In the usual notation for the states of a hydrogen like atom, consider the spontaneous transitions and . If t₁ and t₂ are the lifetimes of the first and the second decaying states respectively, then the ratio is proportional to

(a)

(b)

(c)

(d)

Ans. (a)

Sol. Rate of spontaneous emission (in dipole approx.)

  where, = fine structure constant

Mean lifetime of the decaying state is

Note: There is no need of solving the integration for part as values of for the initial and final state are same for both transitions

40. A random variable n obeys Poisson statistics. The probability of finding n = 0 is 10^–6. The expectation value of n is nearest to

(a) 14

(b) 10⁶

(c) e

(d) 10²

Ans. (a)

Sol. In Poisson's statistics the probability of finding the value n is given by

The mean of Poisson's statistics is µ. From the question

Talking Log of both sides,

Hence the expectation value of n is

41. The single particle energy levels of a non-interacting three-dimensional isotropic system, labelled by momentum k, are proportional to k³. The ratio of the average pressure to the energy density at a fixed temperature, is

(a) 1/3

(b) 2/3

(c) 1

(d) 3

Ans. (c)

Sol. If the energy is proportional to n^th power of momentum k, i.e.

Then, where, E is the energy and is the energy density

Since, (given), n = 3

42. The Hamiltonian for three using spins S₀, S₁ and S₂, taking value , is

.

If the system is in equilibrium at temperature T, the average energy of the system, in terms of , is

(a)

(b)

(c)

(d)

Ans. (d)

Sol. The hamiltonian is,

H = – jS₀ (S₁ + S₂); S₀, S₁, S₂ = + 1

The possible values of Hamiltonian are

The average energy is,

43. Let I₀ be the saturation current, the ideality factor and v_F and v_R the forward and reverse potentials, respectively, for a diode. The ratio R_R/R_F of its reverse and forward resistances R_R and R_F respectively, varies as (In the following k_B is the Boltzmann constant, T is the absolute temperature and q is the charge.)

(a)

(b)

(c)

(d)

Ans. (b)

Sol.

44. In the figures below, X and Y are one bit inputs. The circuit which corresponds to a one bit comparator is

(a)

(b)

(c)

(d)

Ans. (c)

Sol. Given 1-bit comparator.

The circuit is satisfying above truth table

45. Both the data points and a linear fit to the current vs voltage of a resistor are shown in the graph below.

If the error in the slope is , then the value of resistance estimated from the graph is

(a)

(b)

(c)

(d)

Ans. (b)

Sol. According to Ohm's law:

Let S is the slope of given graph

46. In the following operational amplifier circuit C_in = 10 nF, R_in = 20 k, R_F = 200 k and C_F = 100 pF.

The magnitude of the gain at a input signal frequency of 16 kHz is

(a) 67

(b) 0.15

(c) 0.3

(d) 3.5

Ans. (d)

Sol.

Magnitude = 20024.72

Magnitude = 89064.199

Go for nearest value

All options are incorrect

47. An atomic spectral line is observed to split into nine components due to Zeeman shift. If the upper state of the atom is ³D₂ then the lower state will be

(a) ³F₂

(b) ³F₁

(c) ³P₁

(d) ³P₂

Ans. (c)

Sol. The upper state term is ³P₂, that is J = 2 and M_J = – 2, –1, 0, 1, 2

There are 9 Zeeman transition, accoding to the selection rule

The lower state must split into three components, i.e.,

M_J = 1, 0, –1 J = 1

the multiplicity selection rule, = 0 demands the lower state term also, to be a triplet that is, S =1.

The value J = 1 and S = 1 lead L = 1 i.e. lower state is ³P₁.

48. If the coefficient of stimulated emission for a particular transition is 2.1 × 10¹⁹ m³ W^–1s^–3 and the emitted photon is at wavelength 3000 Å, then the lifetime of the excited state is approximately

(a) 20 ns

(b) 40 ns

(c) 80 ns

(d) 100 ns

Ans. (c)

Sol. We know that,

where, A₂₁ = Einstein coefficient of spontaneous emission of radiation

B₂₁ = Einstein constant of stimulated emission radiation

Also, spontaneous life time of the upper level is

Relation between Einstein coefficient is

49. If the binding energies of the electron in the K and L shells of silver atom are 25.4 keV and 3.34 keV, respectively, then the kinetic energy of the Auger electron will be approximately

(a) 22 keV

(b) 9.3 keV

(c) 10.5 keV

(d) 18.7 keV

Ans. (d)

Sol. The kinetic energy of the Auger electron will be

K.E. = (E_K – E_L) – E_L = [(25.4 – 3.34)] KeV = 18.7 KeV

50. The energy gap and lattice constant of an indirect band gap semiconductor are 1.875 eV 0.52 nm, respectively. For simplicity take the dielectric constant of the material to be unity. When it is excited by broadband radiation, an electron initially in the valence band at k = 0 makes a transition to the conduction band. The wave vector of the electron in the conduction band, in terms of the wave vector k_max at the edge of the Brillouin zone, after the transition is closest to

(a) k_max/10

(b) k_max/100

(c) k_max/1000

(d) 0

Ans. (c)

Sol. We know that the k_max in 3D solid is given by

And the wave vector of electron,

51. The electrical conductivity of copper is approximately 95% of the electrical conductivity of silver, while the electron density in silver is approximately 70% of the electron density in copper. In Drude's model, the approximate ratio of the mean collision time in copper (τ_Cu) to the mean collision time in silver is

(a) 0.44

(b) 1.50

(c) 0.33

(d) 0.66

Ans. (d)

Sol. According to the Drude-Model, the conductivity

Therefore, the mean collision time,

= conductivity and n = electron density

n_Ag = 70% of n_Cu = 0.70 n_Cu

52. The charge distribution inside a material of conductivity and permittivity at initial time t = 0 is , a constant. At subsequent times is given by

(a)

(b)

(c)

(d)

Ans. (a)

Sol. We know equation of continuity

53. If in a spontaneous -decay of at rest, the total energy released in the reaction is Q, then the energy carried by the -particle is

(a) 57Q/58

(b) Q/57

(c) Q/58

(d) 23Q/58

Ans. (a)

Sol. In spontaneous -decay

Here, A = 232, atomic mass number

54. The range of the nuclear force between two nucleons due to the exchange of points is 1.40 fm. If the mass of the pion is 140 MeV/c² and the mass of the rho-meson is 770 MeV/c², then the range of the force due to exchange of rho-mesons is

(a) 1.40 fm

(b) 7.70 fm

(c) 0.25 fm

(d) 0.18 fm

Ans. (c)

Sol.

55. A baryon X decays by strong interaction as , where is a member of the isotriplet . The third component I₃ of the isospin of X is

(a) 0

(b) 1/2

(c) 1

(d) 3/2

Ans. (a)

Sol.