PREVIOUS YEAR (2020) COMMON P.G ENTRANCE PHYSICS QUESTION SOLUTION -2

 

 *FOR QUESTION 1-35 ; CLICK THE LINK  https://biswatutorialphysics.blogspot.com/2021/08/previous-year-2020-common-pg-entrance.html

Q/36.  The Poynting vector S of an electromagnetic wave is ?

 (A) 𝑆 = 𝐸⃗ × 𝐻⃗          (B) 𝑆 = 𝐸⃗ × 𝐵⃗           (C) 𝑆 = 𝐸⃗/𝐵⃗        (D) 𝑆 = 𝐸⃗/𝐻⃗

 Ans:  Poynting vector 𝑆 = (1/𝜇)(𝐸⃗ × 𝐵⃗ )

       If the medium is linear, 𝐻⃗ = 𝐵⃗/𝜇

       So,  𝑆 = (1/𝜇)(𝐸⃗ × 𝜇𝐻⃗ ) = 𝐸⃗ × 𝐻⃗

 

Q/ 37. For good conductors, skin depth (𝛿) varies with frequency (𝜔) as?

 (A) 1/𝜔            (B) 1/√𝜔                 (C) 𝜔                 (D) √𝜔

Ans : The skin depth is that distance below the surface of a conductor where the current density has reduced  to (1/𝑒) of its value at the surface.

   skin depth, 𝛿 = √(2/𝜇𝜎𝜔)

   Skin depth (𝛿) varies with frequency (𝜔) as 𝛿 ∝ 1/√𝜔

 

Q/38. The thickness of half wave plate of quartz for a wavelength of 5000 A⁰ (given refractive index µ_{Extraordinary} = 1.553 and refractive index µ_Ordinary=1.544) is ?

 (A) 2.78×10^-3 cm              (B) 2.78×10^-5 cm

(C) 2.78×10^-7 cm                (D) 3.78×10^-3 cm

 

 Ans : A Wavelength, 𝜆 = 5000𝐴⁰

          Refractive index, 𝜇_e = 1.553 ;          𝜇₀ = 1.544

    Thickness of half wave plate, 𝑡_1/2 = 𝜆/{2(𝜇_e−𝜇₀)}

 𝑡_1/2 = 5000×10^-10m/{2(1.553−1.544)}

          = 5×10^-7/2×0.009 =  2.78 × 10^-5𝑚 = 2.78 × 10^-3𝑐𝑚

 

Q/39. In a micro-canonical ensemble, a system A of fixed volume is in contact with a reservoir B. Then?

 (A) A can exchange only energy with B

 (B) A can exchange only particles with B

 (C) A can exchange neither energy nor particles with h B

 (D) A can exchange both energy and particles with B

 

Ans : The microcanonical ensemble is a collection of independent systems, having the same number of particles N, volume V and an energy between E and E+δE.

 So, in microcanonical ensemble, neither energy nor particle is exchanged.

 

Q/40. The Fermi-Dirac distribution function is given by

                          𝑓_𝐹𝐷(𝜀) = 1/ {exp( 𝜀−𝜀_F/𝐾𝑇 )+1}

 where 𝜀_F is the Fermi energy. The value of 𝑓_𝐹𝐷(𝜀_𝐹) at the absolute temperature is?

 (A) 0                 (B) 1                   (C) ½          (D) Infinity

Ans :   At absolute temperature, T=0K,

                                                    𝑓(𝜀) = { 1, 𝑓𝑜𝑟 𝜀 < 𝜀_F

                                                                   0, 𝑓𝑜𝑟 𝜀 > 𝜀_F

            Thus, at T= 0K the highest occupied energy level is Fermi energy. So, the value of 𝑓_𝐹𝐷(𝜀_𝐹) at the absolute temperature is 1

 

q/ 41. Two bodies have their moments of inertia I and 2I respectively about their axis of rotation. If their kinetic energies of rotation are equal, their angular momenta will be in the ratio of  ?

(A) 1:2              (B) 2:1          (C) 1:√2             (D) √2: 1

Ans :  Let I₁=I and I₂= 2I

      Kinetic energy of rotation, 𝐾 = 𝐼𝜔²/2 = 𝐿²/2𝐼                  (𝐿 = 𝐼𝜔= angular momentum

                     ω=angular velocity and

  If their kinetic energies of rotation are equal,  𝐾₁ = 𝐾₂

                   ⇒ 𝐿₁²/2𝐼₁ = 𝐿₂²/2𝐼₂

                  ⇒ 𝐿₁²/𝐿₂² =𝐼/2𝐼 = 1/2

                  ⇒ 𝐿₁/𝐿₂  = √(1/2) = 1/√2

 

Q/42. A particle executing simple harmonic motion of amplitude 5 cm has a maximum speed of 31.4 cm/s. The frequency of its oscillation is ?

 (A) 4 Hz            (B) 3 Hz               (C) 2 Hz             (D) 1 Hz

 Ans :     Amplitude, A=5 cm

          Maximum speed, v_max=31.4 cm/s

         Velocity of a particle executing simple harmonic motion, 𝑣 = ±𝜔√(𝐴² – 𝑥²)

       At the mean position 𝑥 = 0

                                𝑣 = 𝜔𝐴  ;     𝜔 = 2𝜋𝑓,  

     𝜔 is angular frequency            𝑓is frequency.

                    𝑣 = 2𝜋𝑓𝐴 ⇒ 𝑓 = 𝑣/2𝜋𝐴 = 31.4/(2×3.14×5) = 31.4/31.4 = 1Hz

 

Q/43. In a reversible cycle, the value of the integral ∮ (𝑑𝑄/𝑇) is ?

        (A) ∮𝑑𝑄/𝑇 > 0                         (B) ∮ 𝑑𝑄/𝑇 < 0

        (C) ∮ 𝑑𝑄/𝑇 = 0                        (D) ∮ 𝑑𝑄/𝑇 =constant

 Ans :  In reversible cycle, ∮ 𝑑𝑄/𝑇 = 0. This is known as Clausius  theorem.

 

 

Q/44. What is the correct expression for the phase angle in an RLC series circuit?

 (A) 𝜑 = tan^-1{(𝑋_𝐿−𝑋_𝐶)/𝑅}                      (B) 𝜑 = tan^-1{(𝑋_𝐿+𝑋_𝐶)/𝑅}                      

(C) 𝜑 = tan{(𝑋_𝐿−𝑋_𝐶)/𝑅}                      (D) 𝜑 = tan^-1(𝑋_𝐿−𝑋_𝐶)                    

Answer:  tan𝜑 = (𝑋_𝐿−𝑋_𝐶)/𝑅                     

               ⇒ 𝜑 = tan^-1{(𝑋_𝐿−𝑋_𝐶)/𝑅}                      

 

Q/45. The time constant of an R-C circuit is ?

       (A) RC               (B) R/C               (C) R                 (D) C

Ans :  ꚍ = RC

Q/46. The value of the time constant in the R-L circuit is ?

 (A) L/R                   (B) R/L                  (C) R                (D) L

 Ans :  ꚍ = L/R                   

 

Q/47. In Newton’s rings experiment, the diameter of the 15th ring was found to be 0.590 cm and that of the 5th ring was 0.336 cm. If the radius of the plano-convex lens is 100 cm, the wavelength of light used ?

 (A)4880 A⁰         (B) 5880 A⁰           (C) 6680 A⁰             (D) 7680 A⁰

Ans :  In Newton’s ring experiment, wavelength of light

                                                 𝜆 = (𝐷_𝑛+𝑚 ² – 𝐷_n²)/4𝑚𝑅

 𝐷_𝑛+𝑚 = 0.590 𝑐𝑚 = 0.590 × 10^-2𝑚,        𝐷_n = 0.336 𝑐𝑚 = 0.336 × 10^-2𝑚

                  𝑅 = 100 𝑐𝑚 = 1 𝑚     𝑛 = 5,     𝑛 + 𝑚 = 15  ;  𝑚 = 10

                  ⇒ 𝜆 = {(0.590)²× 10^-4 − (0.336)²× 10^-4}/4 × 10 × 1

                           = 5.88 × 10^-7𝑚 =  5880𝐴⁰

 

Q/48. If a charged particle of mass m is accelerated through a potential difference V Volts, the de-Broglie wavelength is proportional to ?

 (A) V               (B V^-1/2                    ) (C) V²         (D) V^1/2

Ans :  If a charged particle of mass m and charge q is accelerated through a potential difference V Volts, Kinetic energy 𝐾 = 𝑞𝑉

            Momentum 𝑝 = √(2𝑚𝐾) = √(2𝑚𝑞𝑉)

  The de-Broglie wavelength 𝜆 = ℎ/𝑝 = ℎ/√(2𝑚𝑞𝑉)

        𝜆 α 1/√V = V^-1/2                    

 

Q/49. The lowest energy possible for a particle in a potential box is 2eV. The next highest energy of the particle is ?

(A)4 eV           (B)16 eV            (C) 32 eV        (D) 8 eV

Ans :  The energy eigenvalues constituting the energy levels for a particle in a box,

 𝐸_n = 𝑛²𝜋²ℏ² /2𝑚𝐿²   ;  𝑛 = 1,2,3, …

The lowest energy possible for a particle,

for 𝑛 = 1  ;    𝐸₁ = 𝜋²ℏ²/2𝑚𝐿²  = 2 𝑒𝑉

The next highest energy of the particle, for 𝑛 = 2

   𝐸₂ = (2)²𝜋²ℏ²/2𝑚𝐿²  = 4*(𝜋²ℏ²/2𝑚𝐿²) = 4 × 2 𝑒𝑉 = 8 𝑒𝑉

 

Q/50. A rod has length of 1 meter. If the rod is placed inside a satellite moving with a velocity 0.8 c relative to the laboratory, the length of the rod by the observer in laboratory is ?

 (A) 0.5 meter         (B) 0.6 meter        (C) 0.7 meter          (D) 0.8 meter

Ans :  𝐿 = 𝐿₀√(1 – 𝑣²/𝑐²)

             𝐿₀ = 1 𝑚,     𝑣 = 0.8𝑐

        So, 𝐿 = 𝐿₀√(1 – 𝑣²/𝑐²) = √(1− 0.64) = √0.36 = 0.6 𝑚

 

Q/51. As an object approaches the speed of light, its mass becomes ?

 (A) Zero            (B) Double         (C) Remains same           (D) Infinite

Ans :  Relativistic mass of an object, m = m₀√(1 – 𝑣²/𝑐²)

 As an object approaches the speed of light, 𝑣 → 𝑐 .

 So, 𝑚 = 𝑚₀√(1− 𝑐²/𝑐²) → ∞ .

Q/52. Nuclear forces are :

 (A) Gravitational attractive             

 (B) Electrostatic repulsive  

(C) Long range and strong attractive

 (D) Short range and strong attractive

Ans :  The short-range attractive forces between nucleons arise from the strong interaction.

Q/53. The binding energy per nucleon is maximum for the nucleus ?

 (A) ⁵⁶Fe         (B)⁴He         (C)²⁰⁸Pb           (D) ¹⁰¹Mp

Ans :  The binding energy per nucleon is maximum for the ⁵⁶Fe nucleus.

Q/54. The mean life time of one of the atoms of a radioactive sample with disintegration constant λ is ?

 (A) 1⁄𝜆        (B) ln 2⁄𝜆           (C) 𝜆 ln 2            (D) ln 𝜆⁄2

Ans :  𝑇̅ = 1/𝜆

Q/55. If a generalized coordinate has the dimensions of momentum, the generalized velocity will have the dimension of ?

 (A) Velocity            (B) Acceleration              (C) Force            (D) Torque

 Ans :  If a generalized coordinate has the dimensions of momentum, the generalized velocity will have the dimension of force.

 

Q/56. Hamilton’s canonical equations of motion are ?

 (A) 𝑞_i ̇ = 𝜕𝐻/𝜕𝑝_𝑖 and 𝑝_i ̇ = 𝜕𝐻/𝜕𝑞_i

(B) 𝑞_i ̇ = 𝜕𝐻/𝜕𝑝_𝑖 and 𝑝_i ̇ = -𝜕𝐻/𝜕𝑞_i

(C) 𝑞_i ̇= 𝜕𝐻/𝜕𝑝̇𝑖 and 𝑝_i ̇ = 𝜕𝐻/𝜕𝑞_i ̇

 (D) 𝑞_i ̇= 𝜕𝐻/𝜕𝑝̇𝑖 and 𝑝_i ̇ =-𝜕𝐻/𝜕𝑞_i ̇

Ans :  Hamilton’s canonical equations of motion are

𝑞_i ̇ = 𝜕𝐻/𝜕𝑝_𝑖 and 𝑝_i ̇ = -𝜕𝐻/𝜕𝑞_i

Q/57. The generalized velocity co-ordinate 𝑞_𝑘 of a classical system with Lagrangian L is said to be cyclic if ?

 (A) 𝜕𝐿/𝜕𝑞_k  = 𝑞_k ̇               (B) 𝜕𝐿/𝜕𝑞_k = 0         

(C) 𝜕𝐿/𝜕𝑞_k ̇ = 0                 (D) None of these

Ans : The Lagrange’s equation is (𝑑/𝑑𝑡)[𝜕𝐿/𝜕𝑞_k ̇] − 𝜕𝐿/𝜕𝑞_k = 0

    The generalized momentum corresponding to the generalized coordinate 𝑞_k is 𝑝_k = 𝜕𝐿/𝜕𝑞_k ̇

So, Lagrange’s equation is 𝑑𝑝_k/𝑑𝑡 = 𝜕𝐿/𝜕𝑞_k

    ⟹ 𝑝_k ̇ = 𝜕𝐿/𝜕𝑞_k

 When the Lagrangian function does not contain a coordinate 𝑞𝑘 explicitly, the coordinate 𝑞𝑘 is called cyclic or ignorable coordinate.

    𝜕𝐿/𝜕𝑞_k = 0

So, 𝑝_k ̇= 0          ⟹ 𝑝_k  = 𝜕𝐿/𝜕𝑞_k ̇=constant

Thus, whenever the Lagrangian function does not contain a coordinate 𝑞_k  explicitly, the generalized momentum 𝑝_k  is a constant of motion. The coordinate 𝑞_k   is called cyclic or ignorable coordinate.

 

Q/58. A particle moves in circular orbit about the origin under the action of a central force 𝐹 = − 𝑘𝑟̂/𝑟³ . If the potential energy is zero at infinity, the total energy of the particle is ?

 (A) −𝑘/𝑟²           (B) −𝑘/2𝑟²          (C) 0        (D) 𝑘𝑟²

 Ans :   force 𝐹 = − 𝑘𝑟̂/𝑟³  

For circular motion, centripetal force   𝐹 = − 𝑚𝑣²𝑟̂/r

    ⟹ − 𝑚𝑣²𝑟̂/r = − 𝑘𝑟̂/𝑟³  

     ⟹ 𝑚𝑣²  = 𝑘/𝑟²

  Kinetic energy of the particle, 𝐾 = 𝑚𝑣²/2 = 𝑘/2𝑟²

    Potential energy of the particle, 𝑉 = − _r∫^∞ 𝐹.𝑑𝑟 =  = − 𝑘/2𝑟²

 Total energy of the particle, 𝐸 = 𝐾 + 𝑉 = 𝑘/2𝑟²− 𝑘/2𝑟² = 0

Q/59. The law at given temperature, the ratio of spectral emissive and absorptive powers of a body is called ?

 (A) Wein’s law                 (B) Kirchhoff’s law

(C) Stefan’s law              (D) Displacement law

Ans :  Kirchhoff’s law: The ratio of emissive power to absorptive power is the same for all bodies at a given temperature and is equal to the emissive power of a blackbody at that temperature.

Q/60. A permanent memory, which helps to start-up the computer and does not erase data after power off ?

(A) Network interface card               (B) CPU

(C) RAM                                                (D) ROM

Ans :  ROM, which stands for Read-Only Memory, is a storage medium used in your computer to store data. The data stored in ROM can only be read and is near impossible to modify. Unlike RAM, ROM is non-volatile, meaning that even when you turn off your computer, the contents of ROM still remain.

Q/61. Which of the following is non-volatile storage?

(A) Backup              (B) Secondary        (C) Primary       (D) Cache

Ans : Secondary  memory is used as non-volatile storage

Q/62. A half adder is a logic circuit with?

 (A) Two inputs and two outputs

 (B) Three inputs and one output

(C) Three inputs and two outputs

 (D) Two inputs and one output

 Ans : The half adder circuit has two inputs, which add two input digits and has two output which generates a carry and a sum

 

Q/63. An oscillator differs from an amplifier because ?

 (A) It has more gain                     (B) It has less gain

(C) It requires no input signal                   (D) It requires no dc supply

Ans : An oscillator differs from an amplifier because It requires no input signal                   

An oscillator is a system consisting of active and passive circuit elements to produce a sinusoidal or other repetitive waveform at the output without the application of an external input signal. An amplifier contains one or more active devices and transforms power from the dc supply into the signal power at the output, the signal amplitude at the output being proportional to that at the input.

Q/64. In a ferromagnetic material, as the applied field gradually reduced to zero, the polarization still left is known as ?

 (A) Coercive polarization                (B) Spontaneous polarization

(C) Space charge polarization             (D) Remanent polarization

Ans :  The polarization remaining in the material when the polarization field is reduced to zero, is called the remanent polarization.

Q/65. The splitting of spectral line in presence of an electric field is called as ?

 (A) Stark effect                           (B) Zeeman effect          

(C) Paschen-Back effect            (D) Raman effect

Ans : A The splitting of spectral line in presence of an electric field is called as Stark effect

 

66. Weak forces act on ?

(A) Both hadrons and leptons

(B) Hadrons only

(C) All the charged particles

 (D) None of these.

 Ans : Weak forces act on quarks and leptons. Weak forces mediate transformation of quarks and leptons.

Q/67. Which of the following elementary particle is a lepton?

(A) Photon                          (B) µ-meson

 (C) Neutron                       (D) Proton

Ans :  µ-meson  is a lepton.

Q/68. Biot-Savarts law in magnetic field is analogous to law in electric field ?

 (A) Gauss law                       (B) Faraday law

(C) Coulombs law                 (D) Ampere law

 Ans :  Biot-Savarts law in magnetic field is analogous to Coulombs law in electric field

 

Q/69. The Ampere law is based on ?

 (A) Stoke’s theorem                 (B) Green’s theorem

(C) Gauss divergence theorem

(D) Maxwell theorem

Ans : The Ampere law is based on Stoke’s theorem                 

Q/70. The magnetic flux density of a finite length conductor of radius 12 cm and current 3A in air is?

 (A) 5×10^-6         (B) 4×10^-6          (C) 6×10^-6         (D) 7×10^-6         

Ans :  Current, I= 3A, Radius, r=12 cm

The magnetic field intensity, 𝐻 = 𝐼/2𝜋𝑟

The magnetic flux density, 𝐵 = 𝜇₀𝐻,

 𝐵 = 𝜇₀𝐻 = 𝜇₀𝐼/2𝜋𝑟 = (4𝜋 ×10^-7× 3)/( 2×𝜋×12×10^-2)=  5 ×10^-6