PYQ 2007 - VedPrep

GATE PHYSICS 2007
Previous Year Question Paper with Solution.

1. The eigenvalues of a matrix are i, –2i and 3i. The matrix is

(a) unitary

(b) anti-unitary

(d) anti-Hermitian

Ans. (d)

Sol. Eigenvalues of an anti-Hermitian matrix are either zero or purely imaginary

2. A space station moving in a circular orbit around the Earth goes into a new bound orbit by firing its engine radially outwards. This orbit is

(a) a larger circle

(b) a smaller circle

(d) a parabola

Ans. (c)

Sol. When engine is fired radially outward, its velocity is not perpendicular to radial direction. Therefore, new bound orbit cannot be circle. So new bound orbit must be an ellipse.

3. A power amplifier gives 150W output for an input of 1.5W. The gain, in dB, is

(a) 10

(b) 20

(d) 100

Ans. (b)

Sol. P₀– 150W; P_i = 1.5 W

Power gain = = 100, Gain in dB = 10 log₁₀, 100 = 20 dB

Note: For power we take 10 log P

4. Four point charges are placed in a plane at the following positions: +Q at (1, 0), –Q at (–1, 0), +Q at (0, 1) and –Q at (0, –1). At large distances the electrostatic potential due to this charge distribution will be dominated by the

(a) monopole moment

(b) dipole moment

(d) octopole moment

Ans. (b)

Sol. Since, total charge is zero. So, monopole moment is zero.

Also,quadrupole moment,

Electrostatic potential will be due to charge distribution of dipole moment.

5. A charged capacitor (C) is connected in series with an inductor (L). When the displacement current reduces to zero, the energy of the LC circuit is

(a) stored entirely in its magnetic field

(b) stored entirely in its electric field

(d) radiated out of the circuit

Ans. (b)

Sol. When displacement current reduces to zero, the current in the circuit will be zero. So, total energy will be stored entirely in its electric field.

6. Match the following

P. Frank-Hertz experiment 1. electronic excitation of

molecules

Q. Hartree-Fock method 2. wave function of atoms

R. Stern-Gerlach experiment 3. spin angular momentum of

atoms

S. Franck-Condon principle 4. energy levels in atoms

(a) (b) (c) (d)

P-4 P-1 P-3 P-4

Q-2 Q-4 Q-2 Q-1

R-3 R-3 R-4 R-3

S-1 S-2 S-1 S-2

Ans. (a)

Sol. P : Frank-Hertz experiment was performed to identify energy levels in atoms.

Q: Hartree-Fock method is used for the wave functions of atoms.

R: Stern-Gerlach experiment was performed to check the space quantisation of "spin angular momentum of atoms".

S: Frank-Condon principle was given for the electronic excitation of molecules.

7. The wavefunction of a particle, moving in a one-dimensional time-independent potential V(x) is given by , where a and b are constants. This means that the potential V(x) is of the form.

(a)

(b)

(d)

Ans. (c)

Sol.

This is wave function of free particle: So, V(x) = 0

8. The D₁ and D₂ lines of Na (3²P_1/2 3²S_1/2, 3²P_3/2 3²S_1/2) will split on the application of a weak magnetic field into

(a) 4 and 6 lines respectively

(b) 3 lines each

(d) 6 lines each

Ans. (a)

Sol. In weak magnetic field (Anomalous Zeeman effect), selection rules are

. Thus, D₁ and D₂ will give 4 and 6 transitions.

9. In a He-Ne laser, the laser transition takes place in

(a) He only

(b) Ne only

(d) He first, then in Ne

Ans. (b)

Sol. In a He-Ne laser (Four level laser), the population inversion is achieved by electric discharge He atoms help in achieving a population inversion in the Ne atoms. Thus, the laser transition takes place in Ne only.

10. The partition function of a single gas molecule is . The partition function of N such non-interacting gas molecules is then given by

(a)

(b)

(c)

(d)

Ans. (a, b)

Sol. If we take classical approach, the gas molecules are taken to be distinguishable. Hence,

If semi-classical approach is taken, the gas molecules are taken to be indistinguishable.

11. A solid superconductor is placed in an external magnetic field and then cooled below its critical temperature. The superconductor

(a) retains its magnetic flux because the surface current supports it

(b) expels out its magnetic flux because it behaves like a paramagnetic materials

(d) expels out its magnetic flux because the surface current induces a field in the direction opposite to the applied magnetic field

Ans. (d)

Sol. According to Meissner effect, the superconductor expels out its magnetic flux because the surface current induces a field in the direction opposite to the applied magnetic field and it behaves like a perfect diamagnet

12. A particle with energy E is in a time-independent double well potential as shown in the figure.

Which of the following statements about the particle is NOT correct?

(a) The particle will always be in a bound state

(b) The probability of finding the particle in one well will be time-dependent

(d) The particle can tunnel from one well to the other, and back

Ans. (b)

Sol. Since, potential is time independent, therefore the probability of finding the particle in one well will be time independent.

13. It is necessary to apply quantum statistics to a system of particles if

(a) there is substantial overlap between the wavefunctions of the particles

(b) the mean free path of the particles is comparable to the inter-particle separation

(d) the particles are interacting

Ans. (a)

Sol. The quantum statistics applies when the de–Broglic wavelength of the particles (or associated with the particles) is comparable to the size of the particles. Under this condition, their wave-functions start overlapping.

14. When liquid oxygen is poured down close to a strong bar magnet, the oxygen stream is

(a) repelled towards the lower field because it is diamagnetic

(b) attracted towards the higher field because it is diamagnetic

(d) attracted towards the higher field because it is paramagnetic

Ans. (d)

Sol. Electron configuration of oxygen,

So, it has two free electron in the outer shell.

So, it will be attracted towards the higher field because it is paramagnetic.

15. Fission fragments are generally radioactive as

(a) they have excess of neutrons

(b) they have excess of protons

(d) their total kinetic energy is of the order of 200 MeV

Ans. (a)

Sol. Since number of neutron increases in comparison to number of protons with atomic number i.e. (N-Z) increases with Z. For the radioactive elements like ₉₂U²³⁵ and ⁹²U²³⁸, (N-Z) values are 51 and 54 respectively. When a uranium nucleus, undergoes fission fragments (Kr and Bal generally contain many neutrons.

16. In a typical npn transistor the doping concentrations in emitter, base and collector regions are C_E, C_B and C_C respectively. These satisfy the relation

(a) C_E > C_C > C_B

(b) C_E > C_B > C_C

(d) C_E = C_C > C_B

Ans. (a)

Sol. N_E > N_C > N_B

As per manufacturing of BJT doping of emitter > collector > base.

17. The allowed states for He (2p²) configuration are

(a) ¹S₀, ³S₁, ¹P₁, ³P_0,1,2, ¹D₂ and ³D_1,2,3

(b) ¹S₀, ³P_0,1,2 and ¹D₂

(d) ¹S₀ and ¹P₁

Ans. (b)

Sol. For He (2p²) configuration, we have two equivalent electrons in 2d² state.

Thus, For one electron,

For other electron,

Thus, possible values of s and l are

Multiplicity (2s +1) = 1, 3 and

l = |l₁ – l₂|, |l₁– l₂| + 1,..., (l₁ + l₂) – |1 – 1|, |1 – 1| + 1,...,(1 + 1) = 0, l, 2 (S, P, D, states).

Thus, in total we have six terms, three singlet and three triplet terms.

All these terms are even because the configuration 2d² is even ( = 1 + 1 = 2). We can write these terms as ¹S, ¹P, ¹D, ³S, ³P, ³D

To take into account spin–orbit interaction, let us combine L and S to form J.

Now, J = |L – S|,... |L + S|.

For the equivalent electrons (according Hund's rule), terms ¹P₁, ²S₁ ³D_{1, 2, 3} will not be present and terms ¹S₀, ¹D₂, ³P₀,_1,2 will be present.

18. The energy levels of a particle of mass m in a potential of the form

are given, in terms of quantum number n = 0, 1, 2, 3, ...., by

(a)

(b)

(c)

(d)

Ans. (c)

Sol. For the given potential, 0 for x > 0 and = 0 for v < 0.

For the sake of continuity at x = 0

should be zero at x = 0

The normalized wave function of a particle moving under a linear harmonic oscillator potential is given by

(x = 0) will be non-zero for n = 0, 2, 3, 4, 6... and will be zero 1, 3, 5, 7........So, possible value of 'n' is odd.

The energy eigenvalues of the particle will be

19. The electromagnetic field due to a point charge must be described by Lienard Weichert potentials when

(a) the point charge is highly accelerated

(b) the electric and magnetic fields are not perpendicular

(d) the calculation is done for the radiation zone, i.e. far away from the charge

Ans. (c)

Sol. Correct option is (c)

20. The strangeness quantum number is conserved in

(a) strong, weak and electromagnetic interactions

(b) weak and electromagnetic interactions only

(d) strong and electromagnetic interactions only

Ans. (d)

Sol. Strangeness is conserved in strong and electromagnetic both interactions.

21. The eigenvalues and eigenvectors of the matrix are

(a)

(b)

(c)

(d)

Ans. (a)

Sol.

22. A vector field is defined everywhere as . The net flux of associated with a cube of side L, with one vertex at the origin and sides along the positive X, Y, and Z axes, is

(a) 2L³

(b) 4L³

(d) 10L³

Ans. (a)

Sol.

The net flux of F associated with the cube

23.

(a)

(b)

(c)

(d)

Ans. (c)

Sol.

24. Consider a vector in the coordinate system . The axes are rotated anti-clockwise about the Y axis by an angle of 60º. The vector in the rotated coordinate system is

(a)

(b)

(c)

(d)

Ans. (a)

Sol. The rotation matrix corresponding to the anti-clockwise rotation about y-axis by an angle will

25. The contour integral is to be evaluated on a circle of radius 2a centered at the origin. It will have contributions only from the points

(a)

(b) ia and –ia

(c)

(d)

Ans. (d)

Sol.

26. Inverse Laplace transform of is

(a)

(b)

(c)

(d)

Ans. (b)

Sol.

27. The points, where the series solution of the Legendre differential equation

will diverge, are located at

(a) 0 and 1

(b) 0 and –1

(d)

Ans. (c)

Sol. The solution of Legendre differential equation will be y = A P_n(x) + BQ_n (x)

where P_n(x) is Legendre polynomial of order n and Q_n(X) is Legendre function of second kind and given by

This shows that, Q_n(x) will diverge at x = ± 1.

28. Solution of the differential equation with the boundary condition that y = 1 at x = 1 is

(a) y = 5x4 – 4

(b)

(c)

(d)

Ans. (d)

Sol.

29. Match the following

P. rest mass 1. time like vector

Q. charge 2. Lorentz invariant

R. four-momentum 3. tensor of rank 2

S. electromagnetic field 4. conserved and Lorentz invariant

(a) (b) (c) (d)

P-2 P-4 P-2 P-4

Q-4 Q-2 Q-4 Q-2

R-3 R-1 R-1 R-3

S-1 S-3 S-3 S-1

Ans. (c)

Sol. (P) rest mass (2) Lorentz invariant

(Q) charge (4) conserved and Lorentz invariant

(R) Four momentum (1) time like vector

(S) Electromagnetic field (3) tensor of rank 2

30. The moment of inertia of a uniform sphere of radius r about an axis passing through its centre is given by . A rigid sphere of uniform mass density and radius R has two smaller spheres of radius hollowed out of it, as shown in the figure. The moment of inertia of the resulting body about the Y axis is

(a)

(b)

(c)

(d)

Ans. (b)

Sol.

31. The Lagrangian of a particle of mass m is

where V, W and are constants.

The conserved quantities are

(a) energy and z-component of linear momentum only

(b) energy and z-component of angular momentum only

(d) energy and z-components of both linear and angular momenta

Ans. (c)

Sol. Lagrangian L explicitly depends on time t, so energy is not conserved,

z is cyclic coordinate, so p_z is conserved

L is invariant under rotation about z axis, so L_z is conserved.

32. Three particles of mass m each situated at x1(t), x2(t) and x3(t) respectively are connected by two springs of spring constant k and un-stretched length l. The system is free to oscillate only in one dimension along the straight line joining all the three particles. The Lagrangian of the system is

(a)

(b)

(c)

(d)

Ans. (d)

Sol. Kinetic energy

Potential energy

Therefore, the Lagrangian of the system is.

33. The Hamiltonian of a particle is where q is the generalized coordinate and p is the corresponding canonical momentum. The Lagrangian is

(a)

(b)

(c)

(d)

Ans. (b)

Sol.

p = m (q – q)

Putting the value of p from above

34. A toroidal coil has N closely-wound turns. Assume the current through the coil to be I and the toroid is filled with a magnetic material of relative permittivity µr. The magnitude of magnetic induction inside the toroid, at a radial distance r from the axis, is given by

(a) µrµ0NIr

(b)

(c)

(d)

Ans. (c)

Sol.

35. An electromagnetic wave with is travelling in free space and crosses a disc of radius 2m placed perpendicular to the z-axis. If E₀ = 60 Vm^–1, the average power, in Watt, crossing the disc along the z-direction is

(a) 30

(b) 60

(d) 270

Ans. (c)

Sol. We have, I = l50 Wm^–2

36. Can the following scalar and vector potentials describe an electromagnetic field?

where is a constant

(a) Yes, in the Coulomb gauge

(b) Yes, in the Lorentz gauge

(d) No

Ans. (d)

Sol.

37. For a particle of mass m in a one-dimensional harmonic oscillator potential of the form , the first excited energy eigenstate is . The value of a is

(a)

(b)

(c)

(d)

Ans. (c)

Sol. Schrodinger equation,

Comparing the coefficient of x² from both sides,

38. If the value of [x³, p] is

(a)

(b)

(c)

(d)

Ans. (c)

Sol.

39. There are only three bound states for a particle of mass m in a one-dimensional potential well of the form shown in the figure. The depth V₀ of the potential satisfies

(a)

(b)

(c)

(d)

Ans. (a)

Sol. There will be only three bound state if

40. An atomic state of hydrogen is represented by the following wavefunction:

where a₀ is a constant. The quantum numbers of the state are

(a) l = 0, m = 0, n = 1

(b) l = 1, m = 1, n = 2

(d) l = 2, m = 0, n = 3

Ans. (c)

Sol.

Since, there is only one cosine term so = 1.

Since, there are no term containing , so m = 0.

So, the state is 2P.

41. Three operators X, Y and Z satisfy the commutation relations

The set of all possible eigenvalues of the operator Z, in units of , is

(a)

(b)

(c)

(d)

Ans. (c)

Sol.

Since, component of total angular momentum follow the above relation

So, z operator is behaving like J_z and it has eigen values ± where

42. A heat pump working on the Carnot cycle maintains the inside temperature of a house at 22ºC by supplying 450 kJ s^–1. If the outside temperature is 0ºC, the heat taken, in kJ s^–1, from the outside air is approximately

(a) 487

(b) 470

(d) 417

Ans. (d)

Sol. Given: T₁ = 0 °C = 273 K, Q₁ = ? and T₂ = 22 °C = 295 K, Q₂ = 450 kJs^–1

43. The vapour pressure p (in mm of Hg) of a solid, at temperature T, is expressed by ln p = 23 – 3863/T and that of its liquid phase by ln p = 19 – 3063/T. The triple point (in Kelvin) of the material is

(a) 185

(b) 190

(d) 200

Ans. (d)

Sol. At triple point, vapour pressure of a solid phase gets equal to that of liquid phase.

44. The free energy for a photon gas is given by , where a is a constant. The entropy S and the pressure P of the photon gas are

(a)

(b)

(c)

(d)

Ans. (a)

Sol.

45. A system has energy levels E₀, 2E₀, 3E₀, ..., where the excited states are triply degenerate. Four non-interacting bosons are placed in this system. If the total energy of these bosons is 5E₀, the number of microstates is

(a) 2

(b) 3

(d) 5

Ans. (b)

Sol. Given: Every excited state 3-fold degenerate

The possible states where the total energy is 5F₀ are as follows

So number of possible states are three.

46. In accordance with the selection rules for electric dipole transitions, the 4³P₁ state of helium can decay by photon emission to the states

(a) 2¹S₀, 2¹P₁ and 3¹D₂

(b) 3¹P₁, 3¹D₂ and 3¹S₀

(d) 2³S₁, 3³D₂ and 3³D₁

Ans. (d)

Sol. For Helium: 4 ³P₁

For one electron transition – ± 1 and = 0, ± 1.

For an atom as a whole,

= 0, = ± 1 (for one electron transition) and = 0, ± 1 (for more than one electron transition),

= 0, ± 1 and = any integer.

Out of these four options, only option (d) satisfies the above conditions.

47. If an atom is in the ³D₃ state, the angle between its orbital and spin angular momentum vectors ( and ) is

(a)

(b)

(c)

(d)

Ans. (a)

Sol. For ³D₃ state, L = 2, J = 3, 2S + 1 = 3 S = 1

The angle between the orbital and spin angular momentum can be calculated using formula,

48. The hyperfine structure of Na(3²P_3/2) with nuclear spin I = 3/2 has

(a) 1 state

(b) 2 states

(d) 4 states

Ans. (d)

Sol. For Na(3²P_3/2) with nuclear spin 1 = , we have, J = , 2S+1 = 2 S = and L = 1

For hyperfine structure,

Total angular momentum F = total angular momentum of electron + total nuclear spin

= |J + I|,....,|J–I

Thus the hyperfine structure of Na (3²P_1/2) with nuclear spin has 4 states as shown below.

49. The allowed rotational energy levels of a rigid hetero-nuclear diatomic molecule are expressed as where B is the rotational constant and J is a rotational quantum number.

In a system of such diatomic molecules of reduced mass µ; some of the atoms of one element are replaced by a heavier isotope, such that the reduced mass is changed to 1.05µ. In the rotational spectrum of the system, the shift in the spectral line, corresponding to a transition is

(a) 0.475 B

(b) 0.50 B

(d) 1.0 B

Ans. (c)

Sol. The allowed rotational energy

E_J = BJ(J + 1) ...(i)

The wave number () of the emitted or absorbed radiation under allowed transition is given by

= 2B(J + l) ...(ii)

where J is rotational quantum number and is rotational constant.

For the reduced mass µ the wave number corresponding to the transition J = 4 5 is given by

[We take lower value of J i.e. J= 4, here]

For the reduced mass µ' = 1.05 µ and J = 4 5 transition the wave number is given by

From equations (iii) and (iv), we get

50. The number of fundamental vibrational modes of CO₂ molecule is

(a) four : 2 are Raman active and 2 are infrared active

(b) four : 1 is Raman active and 3 are infrared active

(d) three : 2 are Raman active and 1 is infrared active

Ans. (c)

Sol. The number of fundamental vibrational modes of CO₂ molecule is three. Out of these, two are infrared active and one is Raman active.

51. A piece of paraffin is placed in a uniform magnetic field H₀. The sample contains hydrogen nuclei of mass m_p, which interact only with external magnetic field. An additional oscillating magnetic field is applied to observe resonance absorption. If gi is the g-factor of the hydrogen nucleus, the frequency, at which resonance absorption takes place, is given by

(a)

(b)

(c)

(d)

Ans. (c)

Sol. The magnetic moment of the nucleus

The energy absorbed at resonance frequency

If v₀ is the resonance frequency, then

52. The solid phase of an element follows van der Waals bonding with inter-atomic potential , where P and Q are constants. The bond length can be expressed as

(a)

(b)

(c)

(d)

Ans. (a)

Sol.

53. Consider the atomic packing factor (APF) of the following crystal structures:

P. Simple Cubic

Q. Body Centred Cubic

R. Face Centred Cubic

S. Diamond

T. Hexagonal Close Packed

Which two of the above structures have equal APF?

(a) P and Q

(b) S and T

(d) R and T

Ans. (d)

Sol.

From above table, it is clear that, the APF of face-centered cubic (R) and hexagonal close packed (T) are same.

54. In a powder diffraction pattern recorded from a face-centred cubic sample using x-rays, the first peak appears at 30º. The second peak will appear at

(a) 32.8º

(b) 33.7º

(d) 35.3º

Ans. (d)

Sol. For face-centered cubic sample,

For first peak, (hkl) = (11)

For second peak, (hkl) = (200)

Therefore, from Bragg's law: 2d sin

55. Variation of electrical resistivity with temperature T of three solids is sketched (on different scales) in the figure, as curves P, Q and R.

Which one of the following statements describes the variations most appropriately?

(a) P is for a superconductor, and R for a semiconductor

(b) Q is for a superconductor, and P for a conductor

(d) R is for a superconductor, and P for a conductor

Ans. (b)

Sol.

56. An extrinsic semiconductor sample of cross-section A and length L is doped in such a way that the doping concentration varies as where N₀ is a constant. Assume that the mobility µ of the majority carriers remains constant. The resistance R of the sample is given by

(a)

(b)

(c)

(d)

Ans. (a)

Sol.

Consider an elementary length dx.

Therefore, the resistance of the elementary length

57. A ferromagnetic mixture of iron and copper having 75% atoms of Fe exhibits a saturation magnetization of 1.3 × 10⁶ Am^–1. Assume that the total number of atoms per unit volume is 8 × 10²⁸ m^–3. The magnetic moment of an iron atom, in terms of the Bohr Magneton, is

(a) 1.7

(b) 2.3

(d) 3.8

Ans. (b)

Sol. Ferromagnetic mixture of Iron (Fe) and Copper (Cu)

Amount (%) of Fe = 75%; % of Cu = 25%

Total number of atoms, = n = 8 × 10²⁸ m^–3

75% of n contributes for saturation magnetisation

So, n_eff = 0.75 × 8 × 10²⁸ m³

58. Half life of a radio-isotope is 4 × 10⁸ years. If there are 10³ radioactive nuclei in a sample today, the number of such nuclei in the sample 4 × 10⁹ years ago were

(a) 128 × 10³

(b) 256 × 10³

(d) 1024 × 10³

Ans. (d)

Sol. Given that half life T_1/2 – = 4 × 10⁸ years

Total time = 4 × 10⁹ years.

Therefore, number of half life time

Therefore, remaining number of particles after 10 half life time

59. In the deuterium + tritium (d + t) fusion more energy is released as compared to deuterium + deuterium (d + d) fusion because

(a) tritium is radioactive

(b) more nucleons participate in fusion

(d) the reaction product ⁴He is more tightly bound

Ans. (a)

Sol. Since, tritium is radioactive, therefore more energy releases in (deuterium + tritium) fusion.

60. According to the shell model the ground state spin of the ¹⁷O nucleus is

(a)

(b)

(c)

(d)

Ans. (b)

Sol. For ₈O¹⁷, Z = 8, N = 9

9N = (1s_1/2)² (1p_3/2)⁴ (1p_1/2)² (1d_5/2)¹

Therefore, parity = (–1)²= +1(even)

Therefore, spin-parity =

61. A relativistic particle travels a length of 3 × 10^–3 m in air before decaying. The decay process of the particle is dominated by

(a) strong interactions

(b) electromagnetic interactions

(d) gravitational interactions

Ans. (c)

Sol. Travelling distance = 3 × 10^–3 m

The time taken by the particle to travel the given distance

Therefore, interaction is weak.

62. The strange baryon has the quark structure

(a) uds

(b) uud

(d)

Ans. (c)

Sol. For baryon, quark structure is 'uus'

63. A neutron scatters elastically from a heavy nucleus. The initial and final states of the neutron have the

(a) same energy

(b) same energy and linear momentum

(d) same linear and angular momenta

Ans. (a)

Sol. Since there is no loss of energy in elastic scattering. Therefore, the initial and final states of the neutron have the same energy.

64. The circuit shown is based on ideal operational amplifiers. It acts as a

(a) subtractor

(b) buffer amplifier

(d) divider

Ans. (b)

Sol.

Ideal op-amp has infinite input resistance so there will no current pass through op-amp and V_A and V_B equal to zero due to virtual ground concept.

V₃ = V₁

Output is same as input. So, it works as Buffer amplifier.

65. Identify the function F generated by the logic network shown

(a) F = (X + Y)Z

(b)

(d) F = XYZ

Ans. (d)

Sol.

66. In the circuit shown, the ports Q₁ and Q₂ are in the state Q₁ = 1, Q₂ = 0. The circuit is now subjected to two complete clock pulses. The state of these ports now becomes

(a) Q₂ = 1, Q₁ = 0

(b) Q₂ = 0, Q₁ = 1

(d) Q₂ = 0, Q₁ = 0

Ans. (c)

Sol. Given Q₁ = 1, Q₂ = 0

For J = K = 1 output is complemented (Toggle), and Q₂ will be toggle when Q₁ become from 0 1 (as it is clock pulse of Q₂)

So, after two clock pulse output will be 11.

67. The registers Q_D, Q_C, Q_B and Q_A shown in the figure are initially in the state 1010 respectively. An input sequence SI = 0101 is applied. After two clock pulses, the state of the shift registers (in the same sequence Q_D Q_C Q_B Q_A) is

(a) 1001

(b) 0100

(d) 1010

Ans. (a)

Sol. Serial input is S₁ =0101

Parallel output of 2nd clock = 10 0 1

68. For the circuit shown, the potential difference (in Volts) across R_L is

(a) 48

(b) 52

(d) 65

Ans. (a)

Sol. To find potential difference (in volts) across R_L.

Nodal-Analysis at (a)

V_a = 48 Volt

69. In the circuit shown, the voltage at test point P is 12V and the voltage between gate and source is –2V. The value of R (in ) is

(a) 42

(b) 48

(d) 70

Ans. (d)

Sol.

V_GS = – 2 Volt

KVL (1)

+V_GS + 1_D4 = 42 I₁

–2 + I_D × 4 = 42 I₁

Pul 1_D = 2mA; 6 = 421₁; 1₁ = mA

KVL as per dashed line shown

16 = 1₁ × R–2 + 8 Put I₁ = mA

16 = I₁ × R + 6

10 = 1₁ × R Put 1, value

R = 70

70. When an input voltage Vi, of the form shown, is applied to the circuit given below, the output voltage V₀ is of the form

(a)

(b)

(c)

(d)

Ans. (c)

Sol.

Equivalent circuit.

Case - (1) Case - (2)

0 < V_in < 2.3 V 2.3V < V_i < 12 V

Diode – ON Diode – OFF

Case – (3)

V_in < 0

Diode – ON

Note: When the diode is upward direction the signal will transmitted above the reference voltage minus cut in voltage of diode (V = 0.7) (Plus minus depend on polarity of reference voltage).

Common Data for Questions 71, 72, 73:

A particle of mass m is confined in the ground state of a one-dimensional box, extending from x = –2L to x = +2L. The wavefunction of the particle in this state is where is a constant

71. The normalization factor of this wavefunction is

(a)

(b)

(c)

(d)

Ans. (c)

Sol.

Normalization condition:

72. The energy eigenvalue corresponding to this state is

(a)

(b)

(c)

(d)

Ans. (d)

Sol. The energy of the ground state will be

73. The expectation value of p² (p is the momentum operator) in this state is

(a) 0

(b)

(c)

(d)

Ans. (c)

Sol. The expectation value of

Common Data for Questions 74, 75:

The Fresnel relations between the amplitudes of incident and reflected electromagnetic waves at an interface between air and a dielectric of refractive index µ, are

The subscripts || and refer to polarization, parallel and normal to the plane of incidence respectively. Here, i and r are the angles of incidence and refraction respectively

74. The coordination for the reflected ray to be completely polarized is

(a) µcos i = cos r

(b) cos i = µcos r

(d) cos i = –µcos r

Ans. (a)

Sol. When the incident angle i is become (i_p) Brewster angle then the reflected light is completely polarized.

75. For normal incidence at an air-glass interface with µ = 1.5 the fraction of energy reflected is given by

(a) 0.40

(b) 0.20

(d) 0.04

Ans. (d)

Sol. We know that

Statement for Linked Answer Questions 76 & 77:

In the laboratory frame, a particle P of rest mass m0 is moving in the positive x direction with a speed of . It approaches an identical particle Q, moving in the negative x direction with a speed of .

76. The speed of the particle P in the rest frame of the particle Q is

(a)

(b)

(c)

(d)

Ans. (c)

Sol. Speed of P in rest frame of Q

77. The energy of the particle P in the rest frame of the particle Q is

(a)

(b)

(c)

(d)

Ans. (b)

Sol. Energy of P in rest frame of Q is

Statement for Linked Answer Questions 78 & 79

The atomic density of a solid is 5.85 × 10²⁸ m^–3. Its electrical resistivity is 1.6 × 10^–8 m. Assume that electrical conduction is described by the Drude model (classical theory), and that each atom contributes one conduction electron.

78. The drift mobility (in m² V^–1 s^–1) of the conduction electrons is

(a) 6.67 × 10^–3

(b) 6.67 × 10^–6

(d) 7.63 × 10^–6

Ans. (a)

Sol. We know that conductivity,

79. The relaxation time (mean free time), in seconds, of the conduction electrons is

(a) 3.98 × 10^–15

(b) 3.79 × 10^–14

(d) 2.64 × 10^–11

Ans. (b)

Sol.

Statement for Linked Answer Questions 80 & 81:

A sphere of radius R carries a polarization , where k is a constant and is measured from the centre of the sphere.

80. The bound surface and volume charge densities are given, respectively, by

(a)

(b)

(c)

(d)

Ans. (b)

Sol. Bound surface charge density

Bound volume charge density,

81. The electric field at a point outside the sphere is given by

(a)

(b)

(c)

(d)

Ans. (a)

Sol. The electric field at the out side of the sphere

Statement for Linked Answer Questions 82 & 83:

An ensemble of quantum harmonic oscillators is kept at a finite temperature

82. The partition function of a single oscillator with energy levels is given by

(a)

(b)

(c)

(d)

Ans. (a)

Sol.

The partition function of a single oscillator is

83. The average number of energy quanta of the oscillations is given by

(a)

(b)

(c)

(d)

Ans. (a)

Sol. The average number of energy quanta are

Statement for Linked Answer Questions 84 & 85:

A 16µA beam of alpha particles, having cross-sectional area 10^–4 m², is incident on a rhodium target of thickness 1µm. This produces neutrons through the reaction

84. The number of alpha particles hitting the target per second is

(a) 0.5 × 10¹⁴

(b) 1.0 × 10¹⁴

(d) 4.0 × 10²⁰

Ans. (a)

Sol. Given that, I = 16µA = 16 × 10^–6 A

Charge (q_a) = 2e = 2 × 1.6 × 10^–19 C

85. The neutrons are observed at the rate of 1.806 × 10⁸ s^–1. If the density of rhodium is approximated as 10⁴ kg m^–3 the cross-section for the reaction (in barns) is

(a) 0.1

(b) 0.2

(d) 0.8

Ans. (b)

Sol. For a reaction, X + x y + Y or X (x, y) Y