PYQ 2006 - VedPrep

GATE PHYSICS 2006
Previous Year Question Paper with Solution.

1. The trace of a 3 × 3 matrix is 2. Two of its eigenvalues are 1 and 2. The third eigenvalue is

(a) –1

(b) 0

(d) 2

Ans. (a)

Sol. Let are eigenvalues of 3 × 3 matrix.

Trace of the matrix = sum of eigenvalues of the matrix

2. The value of along a square loop of side L in a uniform field is

(a) 0

(b) 2LA

(d) L²A

Ans. (a)

Sol. According to Stokes theorem,

3. A particle of charge q, mass m and linear momentum enters an electromagnetic field of vector potential and scalar potential . The Hamiltonian of the particle is

(a)

(b)

(c)

(d)

Ans. (b)

Sol.

4. A particle is moving in an inverse square force field. If the total energy of the particle is positive, then trajectory of the particle is

(a) circular

(b) elliptical

(d) hyperbolic

Ans. (d)

Sol. For positive energy eccentricity becomes greater than one. Therefore path is hyperbolic

5. In an electromagnetic field, which one of the following remains invariant under Lorentz transformation ?

(a)

(b) E² – c²B²

(d) E²

Ans. (b)

Sol.

6. A sphere of radius R has uniform volume charge density. The electric potential at a point r(r < R) is

(a) due to the charge inside a sphere of radius r only

(b) due to the entire charge of the sphere

(d) independent of r

Ans. (b)

Sol.

depends on total charge. Therefore, V(r) also depends on total charge.

7. A free particle is moving in +x-direction with a linear momentum p. The wavefunction of the particle normalised in a length L is

(a)

(b)

(c)

(d)

Ans. (d)

Sol. Since, particle is moving along +x direction. The normalized wave function within length L is

[Since, wave function is just a plane wave , where normalized over length L]

8. Which one of the following relations is true for Pauli matrices ?

(a)

(b)

(c)

(d)

Ans. (c)

Sol.

9. The free energy of a photon gas enclosed in a volume V is given by , where a is a constant and T is the temperature of the gas. The chemical potential of the photon gas is

(a) 0

(b)

(c)

(d) aVT^–4

Ans. (a)

Sol.

10. The wavefunctions of two identical particles in stated n and s are given by , respectively. The particles obey Maxwell-Boltzmann statistics. The state of the combined two particle system is expressed as

(a)

(b)

(c)

(d)

Ans. (d)

Sol. The particles are distinguishable in Maxwell-Boltzmann statistics and hence their wave functions do not overlap. So, the state of the combined two particle system, will be

11. The target of an X-ray tube is subjected to an excitation voltage V. The wavelength of the emitted X-rays is proportional to

(a)

(b)

(c)

(d) V

Ans. (c)

Sol.

12. The principal series of spectral lines of lithium is obtained by transitions between

(a) nS and 2P, n > 2

(b) nD and 2P, n > 2

(d) nF and 3D, n > 3

Ans. (c)

Sol. The principal series of spectral lines of lithium is obtained by transition between nP to 2S transitions (for n > 1).

Hence, no option is correct. For sodium option (c) is correct.

13. Which one of the following is NOT a correct statement about semiconductors ?

(a) The electrons and holes have different mobilities in a semiconductor

(b) In an n-type semiconductor, the Fermi level lies closer to the conduction band edge

(d) Silicon has diamond structure

Ans. (c)

Sol. Silicon is an indirect band gap semiconductor.

14. Which one of the following axes of rotational symmetry is NOT permissible in single crystals ?

(a) two-fold axis

(b) three-fold axis

(d) five-fold axis

Ans. (d)

Sol.

And the number of possible fold axis,

Since,

So, possible are 360º, 180º, 120º, 90º, 60º

Corresponding the number of fold axis are 1, 2, 3, 4, 6

So, five-fold axis is not possible.

15. Weak nuclear forces act on

(a) both hadrons and leptons

(b) hadrons only

(d) all charged particle

Ans. (c)

Sol. Correct option is (c)

16. Which one of the following disintegration series of the heavy elements will give ²⁰⁹Bi as a stable nucleus ?

(a) Thorium series

(b) Neptunium series

(d) Actinium series

Ans. (b)

Sol. Since, Thorium series is 4n series.

Uranium series is (4n + 2) series.

Actinium series is (4n + 3) series and Neptunium series is (4n + 1) series.

17. The order of magnitude of the binding energy per nucleon in a nucleus is

(a) 10^–5 MeV

(b) 10^–3 MeV

(d) 10 MeV

Ans. (d)

Sol. Since average binding energy per nucleon i.e. B/A for all the nuclei 8 MeV. (i.e. of order 10 MeV)

18. The interaction potential between two quarks, separated by a distance r inside a nucleon, can be described by (a, b and are positive constants)

(a)

(b)

(c)

(d)

Ans. (c)

Sol. Interaction potential can be given as

because nuclear forces are short range and decreases with decreasing distance,

because nuclear forces will be of repulsion type at long range.

19. The high input impedance of field effect transistor (FET) amplifier is due to

(a) the pinch-off voltage

(b) its very low gate current

(d) the geometry of the FET

Ans. (b)

Sol.

The larger input resistance of JFET is due to

(1) Reverse biasing gate to source junction

(2) Negligible gate current I_G(nA)

• Input junction of FET is always reverse bias

• I_g (nA) practically

20. The circuit shown in the figure function as

(a) an OR gate

(b) an AND gate

(d) a NAND gate

Ans. (a)

Sol. BJT as Digital switch.

Then, Y = A + B. So, given circuit is OR gate.

Note: If switch is closed, current will flow and y will have output.

y = 0, when both switches are open.

21. A linear transformation T, defined as , transform a vector for a 3-dimensional real space to a 2-dimensional real space. The transformation matrix T is

(a)

(b)

(c)

(d)

Ans. (a)

Sol.

22. The value of where is the position vector and S is a closed surface enclosing the origin, is

(a) 0

(b)

(c)

(d)

Ans. (c)

Sol.

Since the closed surface encloses the origin i.e.

23. The value of , where C is a circle defined by |z| = 3, is

(a)

(b)

(c)

(d)

Ans. (a)

Sol.

Residue of f(z) at z = –1

24. The kth Fourier component of is

(a) 1

(b) 0

(c)

(d)

Ans. (c)

Sol.

25. An atom with net magnetic moment and net angular momentum is kept in a uniform magnetic induction . The magnetic moment is

(a)

(b)

(c)

(d)

Ans. (c)

Sol.

We know, torque = rate of change of angular momentum,

26. A particle is moving in a spherically symmetric potential is a positive constant. In a stationary state, the expectation value of the kinetic energy of the particle is

(a)

(b)

(c)

(d)

Ans. (a)

Sol.

We know from Virial theorem

27. A particle of mass 2 kg is moving such that at time t, its position, in metre, is given by . The angular momentum of the particle at t = 2 s about the origin, in kg m² s^–1, is

(a)

(b)

(c)

(d)

Ans. (b)

Sol.

28. A system of four particles is in xy-plane. Of these, two particles each of mass m are located at (1, 1) and (–1, –1). The remaining two particles each of mass 2m are located at (1, 1) and (–1, –1). The xy-component of the moment of inertia tensor of this system of particles is

(a) 10m

(b) –10m

(d) –2m

Ans. (c)

Sol.

29. For the given transformations (i) Q = p, P = –q and (ii) Q = p, P = q, where p and q are canonically conjugate variables, which one of the following statement is true ?

(a) Both (i) and (ii) are canonical

(b) Only (i) is canonical

(d) Neither (i) nor (ii) is canonical

Ans. (b)

Sol. For canonical transformation,

Poisson Bracket [Q, P]_{q, p} = 1

If Q = p, P = –q

It Q = p, P = q

only (i) is canonical transformation

30. The mass m of a moving particle is is its rest mass. The linear momentum of the particle is

(a) 2m₀c

(b)

(d)

Ans. (d)

Sol.

31. Three point charges q, q and –2q are located at (0, –a, a), (0, a, a) and (0, 0, –a), respectively. The net dipole moment of this charge distribution is

(a)

(b)

(c)

(d)

Ans. (a)

Sol. The total charge,

So, the dipole moment of this charge distribution does not depend on choice of origin

Therefore, net dipole moment,

32. A long cylindrical kept along z-axis carries a current density is a constant and r is the radial distance form the axis of the cylinder. The magnetic induction inside the conductor at a distance d from the axis of the cylinder is

(a)

(b)

(c)

(d)

Ans. (c)

Sol. According to Ampere's law,

33. The vector potential in a region is given as . The associated magnetic induction is is

(a)

(b)

(c)

(d)

Ans. (b)

Sol.

34. At the interface between two linear dielectrics (with dielectric constants ), the electric field lines bend, as shown in the figure. Assume that there are no free charges at the interface. The ratio is

(a)

(b)

(c)

(d)

Ans. (a)

Sol. We know the boundary condition,

Therefore, (ii) divided by (i), we get

35. Which one of the following sets of Maxwell's equations for time-independent charge density and current density is correct ?

(a)

(b)

(c)

(d)

Ans. (b)

Sol. The Maxwell's equations are

Therefore, the Maxwell's equation for time independent charge density and current density are

36. A classical charged particle moving with frequency in a circular orbit of radius a, centred at the origin in the xy-plane, electromagnetic radiation. At points (b, 0, 0) and (0, 0, b), where b >> a, the electromagnetic waves are

(a) circularly polarized and elliptically polarized, respectively

(b) plane polarized and elliptically polarized, respectively

(d) circularly polarized and plane polarized, respectively

Ans. (c)

Sol. Observer will sec linearly polarized light when he will look from x-direction. But if observer looks from z-axis then he will see electric field vary as a circularly polarized light with time.

37. A particle of mass m is represented by the wavefunction , where k is the wavevector and A is a constant. The magnitude of the probability current density of the particle is

(a)

(b)

(c)

(d)

Ans. (a)

Sol. Probability current density,

38. A one-dimensional harmonic oscillator is in the state , where , and are the ground, first excited and second excited states, respectively. The probability of finding the oscillator in the ground state is

(a) 0

(b)

(c)

(d) 1

Ans. (c)

Sol.

The probability of finding the oscillator in ground state, is

39. The wavefunction of a particle in a one-dimensional potential at time t = 0 is

where and are the ground and the first excited states of the particle with corresponding energies E₀ and E₁. The wavefunction of the particle at a time t is

(a)

(b)

(c)

(d)

Ans. (d)

Sol.

Therefore, wave function at time t is

40. The commutator [L_x, y], where L_x is the x-component of the angular momentum operator and y is the y-component of the position operator, is equal to

(a) 0

(b)

(c)

(d)

Ans. (d)

Sol. [L_x, y] = [(yp_z – zp_y), y] = [y, y]p_z + y[p_z, y] – [z, y]p_y – z[p_y, y]

41. In hydro genic states, the probability of finding an electron at r = 0 is

(a)

(b)

(c)

(d)

Ans. (a, c, d)

Sol. The probability of finding the electron at r = 0 is zero for 1s, 2s, 2p

42. Each of the two isolated vessels, A and B of fixed volumes, contains N molecules of a perfect monatomic gas at a pressure P. The temperatures of A and B are T₁ and T₂, respectively. The two vessels are brought into thermal contact. At equilibrium, the change in entropy is

(a)

(b)

(c)

(d) 2Nk_B

Ans. (c)

Sol.

The final temperature would be as both the verssels have all same parameters.

Now change in entropy is

43. The internal energy of n moles of a gas is given , where V is the volume of the gas at temperature T and a is a positive constant. One mole of the gas in state (T₁, V₁) is allowed to expand adiabatically into vacuum to a final state (T₂, V₂). The temperature T₂ is

(a)

(b)

(c)

(d)

Ans. (c)

Sol. Since the gas is expanded adiabatically into vaccum, it is a case of free expansion. Hence, internal energy does not change

44. The mean internal of a one-dimensional classical harmonic oscillator in equilibrium with a heat bath of temperature T is

(a)

(b) k_BT

(c)

(d) 3k_BT

Ans. (b)

Sol. The Hamiltonian for one-dimensional harmonic oscillator is

Since, the Hamiltonian has two quadratic terms and as per as law of equipartition of energy, each contributes .

45. A monatomic crystalline solid comprises of N atms, out of which n atoms are in interstitial positions. If the available interstitial sites are N', then number of possible microstates is

(a)

(b)

(c)

(d)

Ans. (d)

Sol. Number of ways of choosing n-atoms out of N-atoms = ^NC_n

and number of ways for putting n-atoms in N'-sites = ^N'C_n.

Therefore, total number of microstates,

46. A system of N localized, non-interacting spin-½ ions of magnetic moment µ each is kept in an external magnetic field H. If the system is in equilibrium at temperature T, then Helmholtz free energy of the system is

(a)

(b)

(c)

(d)

Ans. (b)

Sol. The energy of a ion is

The Helmholtz free energy is

47. The phase diagram of a free particle of mass m kinetic energy E, moving in one-dimensional box with perfectly elastic walls at x = 0 and x = L, is given by

(a)

(b)

(c)

(d)

Ans. (a)

Sol.

There is no force on the particle except due to collision.

Therefore momentum of particle remains constant and changes its direction due to collision.

48. In the microwave spectrum of identical rigid diatomic molecules, the separation between the spectral lines is recorded to be 0.7143 cm^–1. The moment of inertia of the molecule, in kg m², is

(a) 2.3 × 10^–36

(b) 2.3 × 10^–40

(d) 7.8 × 10^–46

Ans. (d)

Sol. Microwave region (10^–4 m – 10^–2 m) of the electromagnetic spectrum corresponds to the pure rotational spectra.

Energy of the level (in terms of wave number)

E(J) = BJ(J + 1), J = 0, 1, 2, ...

Selection rules

= ±1,

Therefore, wave number of emitted or absorbed radiation is

v = 2B(J + 1)

Putting J = 0, 1, 2, 3

v = 2B, 4B, 6B is the position of spectral lines.

Separation between spectral lines

Δv = 4B – 2B = 2B = constant

Given:

2B = 0.7143 cm^–1 = 0.7143 × 10² m^–1

49. Which one of the following electronic transitions in Neon is NOT responsible for LASER action in a helium-neon laser ?

(a)

(b)

(c)

(d)

Ans. (a)

Sol. For He-Ne laser, the laser transition takes place in Ne only.

According to above diagram, the electronic transition 6s 5p is not responsible for LASER action in He-Ne laser.

50. In the linear Stark effect, the application of an electric field

(a) completely lifts the degeneracy of n = 2 level on hydrogen atom and splits n = 2 level into four levels

(b) partially lifts the degeneracy of n = 2 level on hydrogen atom and splits n = 2 level into three levels

(d) does not affect the n = 2 levels

Ans. (b)

Sol. For n = 2 level, the allowed values of n, n₁, n₂ and m_l are given in following table:

The Stark shift of hydrogen n = 2 energy level is shown below:

From above figure, it is clear that two levels corresponding to m_l = –1 and m_l = +1 falls together i.e. application of electric field partially lifts the degeneracy of n = 2 level and splits n = 2 into three levels.

51. In hyperfine interaction, there is coupling between the electron angular momentum and nuclear angular momentum , forming resultant angular momentum . The selection rules for the corresponding quantum number F in hyperfine transitions are

(a)

(b)

(c)

(d)

Ans. (c)

Sol. The selection rule for the quantum number F in hyperfine transitions : = 0, ± 1 but .

52. A vibrational electronic spectrum of homonuclear binary molecules, involving electronic ground state and excited , exhibits a continuum at . If the total energy of the dissociated atoms in the excited state exceeds the total energy of the dissociated atoms in the ground state by E_ex cm^–1, then dissociation energy of the molecule in the ground state is

(a)

(b)

(c)

(d)

Ans. (c)

Sol.

53. The NMR spectrum of ethanol (CH₃CH₂OH) comprises of three bunches of spectral lines. The number of spectral lines in the bunch corresponding to CH₂ group is

(a) 1

(b) 2

(d) 4

Ans. (d)

Sol. In the NMR spectroscopy of CH₂CH₂OH three bunches are observed

Three type of proton exist in this molecule so multiplicity (2N1 + 1) for every proton

For –CH₃: 3H^b (2N1 + 1)

where, N is number of near by proton (H)

I = spin of proton (H)

For-OH group: ^aH alcoholic hydrogen broad signal

So, Due to –CH₂ group quartet are obtain four time are observed

54. The energy of electrons of wavevector in a solid is given by , where A and B are constants. The effective mass of the electron at is

(a)

(b)

(c)

(d)

Ans. (c)

Sol.

From equation (ii), equation (i) implies m^*

55. Which one of the following statements is NOT correct about the Brillouin zones (BZ) of a square lattices with constant a ?

(a) The first BZ is a square of side in k_x – k_y plane

(b) The areas of the first BZ and third BZ are the same

(d) The area of the second BZ is twice that of the first BZ

Ans. (d)

Sol. The first BZ is a square of side and second BZ is a square of side

Therefore, area of second BZ = area of first BZ.

56. In a crystal of N primitive cells, each cell contains two monovalent atoms. The highest occupied energy band of the crystal is

(a) one-fourth filled

(b) one-third filled

(d) completely filled

Ans. (c)

Sol. The highest occupied energy band of the crystal is half filled.

57. If the number density of a free electron gas changes from 10²⁸ to 10²⁶ electrons/m³, the value of plasma frequency (in Hz) changes from 5.7 × 10¹⁵ to

(a) 5.7 × 10¹³

(b) 5.7 × 10¹⁴

(d) 5.7 × 10¹⁷

Ans. (b)

Sol. The plasma frequency () is directly proportional to density of free electrons

Initially the electron density, N₁ = 10²⁸ electrons/m³

Finally the electron density, N₂ = 10²⁶ electrons/m³ and initial plasma frequency ()₁ = 5.7 × 10¹⁵ Hz. Let the final plasma frequency ()₂ = x

58. Which one of the following statements about superconductors is NOT true ?

(a) A type I superconductor is completely diamagnetic

(b) A type II superconductor exhibits Meissner effect upto the second critical magnetic field

(d) Both type I and type II superconductors exhibits sharp fall in resistance at the superconducting transition temperature

Ans. (b)

Sol. Type-I superconductors exhibit complete Meissner effect and are completely. Type-II superconductors exhibit complete Meissner effect upto H_C1 only. Above H_C1, the flux begins to penetrate the specimen and for H = H_C2, the complete penetration occurs and material becomes completely normal conductor. In the region between the field H_C4 and H_C2, the Meissner effect is not strictly followed.

Therefore, the statement in option (b) is not true.

59. Two dielectric materials A and B exhibit both ionic and orientational polarizabilities. The variation of their susceptibilities with temperature T is shown in the figure, where is the relative dielectric constant. It can be inferred from the figure that

(a) A is more polar and it has a smaller value of ionic polarizability than that of B

(b) A is more polar and it has a higher value of ionic polarizability than that of B

(d) B is more polar and it has a smaller value of ionic polarizability than that of A

Ans. (d)

Sol.

Since, slope of B slope A, therefore, B is more polar.

Since intercept of A is larger than B, therefore ionic polarizability of B is lesser in comparison to A.

60. The experimentally measured spin g factors of proton and a neutron indicate that

(a) Both proton and neutron are elementary point particles

(b) Both proton and neutron are not elementary point particles

(d) While neutron is an elementary point particle, proton is not

Ans. (b)

Sol. The experimentally measured spin g-factors of proton and neutron indicate that both proton and neutron are not elementary point particles.

61. By capturing an electron, transform into releasing

(a) a neutrino

(b) an antineutrino

(d) a positron

Ans. (a)

Sol.

For the conservation of lepton number in the above reaction (equation), we need to add a lepton member on the right side. i.e. either electron or electron neutrino or or moun or moun neutrino.

Since, on left side there is–an electron so We need a neutrino on right side.

62. Which one of the following nuclear processes is forbidden ?

(a)

(b)

(c)

(d)

Ans. (b)

Sol.

63. To explain the observed magnetic moment of deuteron (0.8574 µ_N), its ground state wavefunction is taken to be an admixture of S and D states. The expectation values of the z-component of the magnetic moment in pure S and pure D states are 0.8797 µ_N and 0.3101 µ_N respectively. The contribution of the D state to the mixed ground state is approximately

(a) 40%

(b) 4%

(d) 0.04%

Ans. (b)

Sol. Deuteron is a single two nucleon (one p and one n) bound system which is found in nature. The important experimental detrimenta properties about deuteron are given below:

(i) The B.E. of deuterium/nucleon is very small compare to other nuclei i.e. it is a weakly system.

(ii) The ground state spin of deuteron l_d = 1

(iii) The ground state parity of deuteron = even (+)

(iv) The quadrupole momento of deuteron (Q_d) 0

(v) The magnetic moment of deutron is slightly different from the sum of intrinsic mag. moments of neutron + proton i.e. (µ_n + µ_p) – µ_d → 0

These factor represents that the ground state of Deuteron is a mixture of states in which ³S contribution is 96% and 3D contribution is only 4%.

64. A sinusoidal input voltage v_in of frequency is fed to the circuit shown in the figure, where C₁ >> C₂. If v_m is the peak value of the input voltage, then output voltage (v_out) is

(a) 2v_m

(b)

(c)

(d)

Ans. (a)

Sol. Considering positive half cycle

D₁, (ON) and D₂ (OFF)

Considering (–)ve half cycle:

D₁ (OFF), D₂(ON)

Vm + V_C1 + V_C2 = 0

Vm + Vm + V_C2 = 0

V_C2 + V_out = – 2V_m

V_out + 2V_m

Note: This circuit is also called voltage doubler circuit

Consider magnitude. So, change V_out (Polarity)

65. The low-pass active filter shown in the figure has a cut-off frequency of 2 kHz and a pass band gain of 1.5. The values of the resistors are

(a)

(b)

(c)

(d)

Ans. (d)

Sol. Given:

f_cutoff = 2KHz

f_H= 2KHz [LPF]

Gain = 1.5

Applying virtual ground concept at inverting input, we get

66. In order to obtain a solution of the differential equation , involving voltages v(t) and v₁, an operational amplifier (Op-Amp) circuit would require at least

(a) two Op-Amp integrators and one Op-Amp adder

(b) two Op-Amp differentiators and one Op-Amp adder

(d) one Op-Amp integrator, one Op-Amp differentiator and one Op-Amp adder

Ans. (a)

Sol. Consider the differential equation,

This is a second order differential equation.

If we can integrate it two times, we can easily get the solution of the above equation.

(Since, because it is involving second order derivatives)

So, two op-amp integrator must be there.

Since, there will be two solution of differential equation v(t) and vt.

So, we also need an adder to add the outputs.

67. In the given digital logic circuit, A and B form the input. The output Y is

(a)

(b)

(c)

(d)

Ans. (d)

Sol. Output Y = ?

68. The largest analog output voltage from a 6-bit digital to analog converter (DAC) which produces 1.0 V output for a digital input of 010100, is

(a) 1.6 V

(b) 2.9 V

(d) 5.0 V

Ans. (c)

Sol. 1 volt analog output for

69. A ripple counter designed with JK flip-flops provided with CLEAR (CL) input is shown in the figure. In order that this circuit functions as a MOD-12 counter, the NAND gate input (X₁ and X₂) should be

(a) A and C

(b) A and D

(d) C and D

Ans. (d)

Sol. CLOCK signal is applied to Flip-Flop "A" so it becomes "LSB" of Flip-Flop.

(i) Here clock is applied at negative edge trigger and output is taken at Q so it is up-counter.

(ii) It is MOD 12 counter and here Q_A is LSB so output must be 1100 (DCBA) to reset the counter.

Truth table of given counter:

It is clear from truth table, that as soon as 1100 appears, input to the NAND gate becomes 11 which reset the flip-flop with clear = 1

So, X₁ and X₂ should be connected Q_C = 1 and Q_D = 1 i.e. C and D.

70. The tank circuit of a Hartley oscillator is shown in the figure. If M is the mutual inductance between the inductors, the oscillation frequency is

(a)

(b)

(c)

(d)

Ans. (a)

Sol.

L_equivalent = L₁ + L₂ + 2M (Same behaviour)

Common Data for Q.71, Q.72 and Q.73 :

An unperturbed two-level system has energy eigenvalues E₁ and E₂, and eigen functions . When perturbed, its Hamiltonian is represented by

71. The first-order correct to E₁ is

(a) 4A

(b) 2A

(d) 0

Ans. (d)

Sol. Given: perturbed Hamiltonian

Therefore, the first order correction in E₁ is

72. The second-order correction to E₁ is

(a) 0

(b) A

(c)

(d)

Ans. (d)

Sol. The second order correction in E₁, is

73. The first-order correction to the eigenfunction is

(a)

(b)

(c)

(d)

Ans. (a)

Sol. The first order correction to the eigenfunction is

Common Data for Q.74 and Q.75 :

One of the eigenvalues of the matrix is 5.

74. The other two eigenvalues are

(a) 0 and 0

(b) 1 and 1

(d) –1 and –1

Ans. (c)

Sol. Sum of the eigenvalues = Trace of the matrix = 5

75. The normalized eigenvector corresponding to the eigenvalue 5 is

(a)

(b)

(c)

(d)

Ans. (d)

Sol.

Statement for Linked Answer Q.76 and Q.77 :

The powder diffraction pattern of a body centred cubic crystal is recorded by using Cu X-rays of wavelength 1.54 Å.

76. If the (002) planes diffract at 60º, then lattice parameter is

(a) 2.67 Å

(b) 3.08 Å

(d) 5.34 Å

Ans. (c)

Sol.

77. Assuming the atomic mass of the constituent atoms to be 50.94 amu, then density of the crystal in units of kg m^–3is

(a) 3.75 × 10³

(b) 4.45 × 10³

(d) 8.89 × 10³

Ans. (a)

Sol.

= 0.3749 × 10⁴ kg/m³ = 3.75 × 10³ kg/m³

Statement for Linked Answer Q.78 and Q.79 :

A particle of mass m is constrained to move in a vertical plane along a trajectory given by , , where A is a constant.

78. The Lagrangian of the particle is

(a)

(b)

(c)

(d)

Ans. (a)

Sol. L = T – V

79. The equation of motion of the particle is

(a)

(b)

(c)

(d)

Ans. (d)

Sol. Equation of motion

Statement for Linked Answer Q.80 and Q. 81 :

A dielectric sphere of radius R carries polarization , where r is the distance from the centre and k is a constant. In the spherical polar coordinate system, are the unit vectors.

80. The bound volume charge density inside the sphere at a distance r from the centre is

(a) –4kR

(b) –4kr

(d) –4kr³

Ans. (b)

Sol.

81. The electric field inside the sphere at a distance d from the centre is

(a)

(b)

(c)

(d)

Ans. (a)

Sol.

Statement for Linked Answer Q. 82 and Q. 83 :

Consider Fermi theory of -decay.

82. The number of final states states of electrons corresponding to momenta between p and p + dp is

(a) independent of p

(b) proportional to pdp

(d) proportional to p³dp

Ans. (c)

Sol. The number of final states of electrons corresponding to momenta between p and (p + dp) is proportional to p²dp.

83. The number of emitted electrons with momentum p and energy E, in the allowed approximation, is proportional to (E₀ is the total energy given up by the nucleus).

(a) (E₀ – E)

(b) p(E₀ – E)

(d) p(E₀ – E)²

Ans. (b)

Sol. The number of emitted electrons with momentum (p) and energy (E) in the allowed approximation is proportional to p (E₀ – E).

Statement for Linked Answer Question 84 and 85 :

Consider a radiation cavity of volume V at temperature T.

84. The density of states at energy E of the quantized radiation (photons) is

(a)