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

 

Q/1. The directional derivative of the scalar function 𝜑 = 𝑥²𝑦𝑧 + 4𝑥𝑧² at the point (1, -2, -1) in the direction 2𝑖̂− 𝑗̂− 2𝑘̂ ?

(A) 3/27                    (B) 27/3                     (C) 35                         (D) 20

 

 Ans:      Scalar function 𝜑 = 𝑥²𝑦𝑧 + 4𝑥𝑧²

                    ∇𝜑   = grad 𝜑 = 𝑖̂ (𝜕𝜑/𝜕𝑥) + 𝑗̂ (𝜕𝜑/𝜕𝑦) + 𝑘̂ (𝜕𝜑/𝜕𝑧)

               = 𝑖̂ 𝜕(𝑥²𝑦𝑧 + 4𝑥𝑧²)/𝜕x+ 𝑗̂ 𝜕(𝑥²𝑦𝑧 + 4𝑥𝑧²)/𝜕y + 𝑘̂ 𝜕(𝑥²𝑦𝑧 + 4𝑥𝑧²)/𝜕z

          = 𝑖̂(2𝑥𝑦𝑧 + 4𝑧²) + 𝑗̂(𝑥²𝑧) + 𝑘̂(𝑥²𝑦 + 8𝑥𝑧)

       ∇𝜑 at the point (1 ,-2,-1)  =  𝑖̂(4 + 4) + 𝑗̂(−1) + 𝑘̂(−2 − 8) = 8𝑖̂− 𝑗̂− 10𝑘̂

Let 𝑎 = 2𝑖̂− 𝑗̂− 2𝑘̂   ;     Unit vector 𝑎̂ = 𝑎⃗/|𝑎| = (2𝑖̂−𝑗̂−2𝑘̂)/3

Directional derivative at the point (1,-2,-1) is  𝑑𝜑/𝑑𝑠 = ∇𝜑∙𝑎̂

            = (8𝑖̂− 𝑗̂− 10𝑘̂)∙{( 2𝑖̂−𝑗̂−2𝑘̂)/3} = 37/3    (All option are wrong)

 

Q/2. The value of div ( 𝑟⃗/𝑟³) ?

           (A) 0                  (B) 1                    (C) 3               (D) ∞

 

Ans:   div ( 𝑟⃗/𝑟³) = ∇⃗∙(𝑟^-3 𝑟⃗)

As We know  ∇⃗∙(f . 𝑟⃗) = (∇⃗f) . 𝑟⃗ +f(∇⃗∙ 𝑟⃗)   ;  f=constant

        ∇⃗∙( 𝑟^-3. 𝑟⃗) = (∇⃗𝑟^-3) . 𝑟⃗ + 𝑟^-3 (∇⃗∙ 𝑟⃗)  

                         =-3 𝑟^-2𝑟̂. (𝑟̂/𝑟) + 𝑟^-3.3                             ∇⃗∙ 𝑟⃗=3

  =-3 𝑟^-3 + 𝑟^-3.3 =0  

 

 

Q/3. The value of  ^∞∫_-∞ 𝑓(𝑥)𝛿(𝑥 − 2) 𝑑𝑥 is?

 (A) 𝑓(0)                        (B) 𝑓(1)                (C) 𝑓(2)                        (D) 𝑓(∞)

 

 Ans: As we know for one dimensional Dirac delta function  

                                   ^∞∫_-∞ 𝑓(𝑥)𝛿(𝑥 − 𝑎)𝑑𝑥 = 𝑓(𝑎)

            So   ^∞∫_-∞ 𝑓(𝑥)𝛿(𝑥 − 2) 𝑑𝑥  = 𝑓(2)

 

Q/4. Moment of inertia of a sphere of mass M and radius R about one of its diameter is?

   (A) (2/5)𝑀𝑅²              (B) (2/3)𝑀𝑅²                   (C)(1/2)𝑀𝑅²             (D)𝑀𝑅²

 

 Ans: A Moment of inertia of a sphere of mass M and radius R about one of its diameters is (2/5)𝑀𝑅²

 

Q/5. Which of the following is true for the relation between modulus of rigidity (𝜂), Young’s modulus (𝑌) and Poisson’s ratio (𝜎) ?

(A) 𝜂 = 𝑌/2(1+𝜎)         (B) 𝜂 = 𝑌/3(1+𝜎)       (C) 𝜂 = 𝑌/2(1−𝜎)     (D) 𝜂 = 𝑌/ 3(1−𝜎)

 

Ans:  𝜂 = 𝑌/2(1 + 𝜎)

 

Q/6. At what speed will the mass of a body be 1.25 times it’s rest mass?

 (A) 0.1 c                   (B) 0.3 c                (C) 0.4 c               (D) 0.6 c

 

 Ans:   As we know  m =  𝛾m₀                where 𝛾 =1/√(1 – 𝑣²/𝑐²)

            𝑚 = relativistic mass,                          𝑚₀ = rest mass or proper mass.

                     𝑚 = 1.25 𝑚₀

1.25𝑚₀ = 𝑚₀ /√(1 – 𝑣²/𝑐²)

⟹√(1 – 𝑣²/𝑐²)  =4/5

⟹ (1 – 𝑣²/𝑐²) = 16/25

⟹𝑣²/𝑐² = 1- 16/25 = 9/25

      ⟹ 𝑣/𝑐  =  3/5 ⟹ 𝑣 = (3/5)𝑐 = 0.6c

 

Q/7. The electric field intensity 𝐸⃗ due to an infinitely charged plane sheet at a distance 𝑟 from the sheet is related as?

 (A) 𝐸⃗ ∝ 𝑟            (B) 𝐸⃗∝ 𝑟^-1         (C) 𝐸⃗∝ 𝑟²         (D) 𝐸⃗ is independent of 𝑟

 

 Answer:  The electric field intensity 𝐸⃗ due to an infinitely charged plane sheet is           

                 𝐸⃗ = 𝜎𝑛̂/2𝜖₀

 So 𝐸⃗ is independent of distance 𝑟.

 

Q/8. The magnetic field due to long straight current carrying conductor of radius 𝑅, when 𝑟 > 𝑅 (𝑟 is the distance between the point and the axis of wire) proportional to?

 (A) 𝑟              (B) 𝑟^-1           (C) 𝑟²          (D) 𝑟^-2

 

Ans:  The direction of magnetic field is circumferential, circling around the wire. The magnitude of magnetic field is constant around an Amperian loop of radius r, centred on the wire.

 So, Ampere’s law gives         ∮ 𝐵⃗∙𝑑𝑙 = 𝜇₀𝐼

                                           𝐵 ∮𝑑𝑙 = 𝐵 2𝜋𝑟 = 𝜇₀𝐼

                             ⇒ 𝐵 = 𝜇₀𝐼/2𝜋𝑟

                               ⇒ 𝐵 ∝ r^-1

 

Q/9.  The self-inductance of a coil with turns 50, flux 3 units and a current of 0.5 A is:

 (A) 75          (B) 150                  (C) 300                      (D) 450

 

 Ans:  Number of turns N=50               Flux 𝜑= 3 units             Current I=0.5A

       Self-inductance L  = 𝑁𝜑/I  = 50∗3/0.5 = 300

 

Q/10. The dielectric constant for a material with electric susceptibility of 5 is  ?

(A) 6            (B) 4             (C) 3             (D) 0         

 

Ans:  Electric susceptibility 𝜒_e =5

 Dielectric constant or relative permittivity 𝜖_r = 𝜖/𝜖₀ = 1 + 𝜒_e

𝜖 is permittivity of the material                  𝜖₀ is permittivity of free space.

Dielectric constant 𝜖_r = 1 + 𝜒_e = 1 + 5 = 6

 

Q/11. Two thin convex lenes having focal length 2 cm and 5 cm are coaxial and separated by a distance of 3cm. The equivalent focal length is ?

 (A) 0.5 cm           (B) 2.5 cm                     (C) 1.5 cm          (D) 3.5 cm

 

Ans:  Focal lengths of convex lenses are f₁ = 2 𝑐𝑚,           f₂ = 5 𝑐𝑚

               Separation between them is 𝑑 = 3 𝑐𝑚

 The equivalent focal length is f_eq

 (1/f_eq) = (1/f₁) +  (1/f₂) −  𝑑/𝑓₁𝑓₂

⇒  (1/f_eq)  = 1/2 + 1/5 – 3/10 = (5 + 2 – 3)/10 = 4/10 = 2/ 5

  ⟹ 𝑓_eq = 5/2 = 2.5cm

 

Q/12. In Young’s double slit experiment, the separation of slits is 1.9 mm and fringe spacing is 0.31 mm at a distance of 1 meter from the slits. The wavelength of the slit is:

 (A) 4890 A°             (B) 5890 A°             (C) 6890 A°               (D) 7890 A°   

 

Ans :  Separation of slits, 𝑑 = 1.9 × 10^-3 𝑚

        Fringe width, 𝛽 = 0.31 × 10^-3𝑚         Screen distance, 𝐷 = 1𝑚

                   𝛽 = 𝜆𝐷/𝑑

        ⇒ 𝜆 = 𝛽𝑑/𝐷 = 0.31 × 10^-3 × 1.9 × 10^-3 = 0.589 × 10^-6 m = 5890 A°

 

Q/13.  𝑓(𝑡) is a periodic function with period 𝑇. The average value is?

 (A) ₀∫^T𝑓(𝑡) 𝑑𝑡      (B) (1/𝑇) ₀∫^T𝑓(𝑡) 𝑑𝑡    (C) (2/𝑇) ₀∫^T𝑓(𝑡) 𝑑𝑡       (D) (1/2𝑇) ₀∫^T𝑓(𝑡)𝑑𝑡

 

Ans:  The average value of 𝑓(𝑡) is   (1/𝑇) ₀∫^T𝑓(𝑡) 𝑑𝑡   

 

Q/14.  If _-1∫⁺¹ 𝑃_n (𝑥) 𝑑𝑥 = 2  ; then 𝑛 is ?

       (A) 0                   (B) 1                  (C) -1          (D) None of these

 

Ans:  Legendre polynomial

 

               𝑃₀(𝑥) = 1   ;         𝑛 = 0,

_-1∫⁺¹ 𝑃_n (𝑥) 𝑑𝑥= _-1∫⁺¹ 𝑃₀(𝑥) 𝑑𝑥 =_-1∫⁺¹ 1 𝑑𝑥

                     = [x]_-1⁺¹  =  2

 

Q/15.  If Γ(𝑛) = Γ(𝑛+1)/𝑛 , then Γ(−𝑛) is ?

 (A) 0          (B) 1           (C) ∞                  (D) None of these

Ans:     Γ(n+1) = n!

            Γ(𝑛) = Γ(𝑛+1)/𝑛  → ∞ as 𝑛 → 0

 If n=0 ;        Γ(0) = Γ(0+1)/0 = Γ(1)/0 = 1/0 → ∞

 n=-1 ;      Γ(−1) = Γ(−1+1) /−1 = Γ(0)/−1 → ∞

n=-2 ;      Γ(−1) = Γ(−2+1) /−2 = Γ(-1)/−2 → ∞

So Γ(𝑛) becomes infinite not only at zero but also at all negative integers. Then Γ(−𝑛) is  ∞.

 

 Q/16.  The efficiency of Carnot’s engine working between the steam point and ice point is?

 (A) 24.31%       (B) 25.21%         (C) 23.52%            (D) 26.80%

Ans:   Efficiency of Carnot engine in terms of the absolute temperatures of its two heat reservoirs is 𝜂 = 1 – 𝑇_L 𝑇_H

T_L is temperature of cooler reservoir      T_H is temperature of hotter reservoir.

 T_L =absolute temperature of ice point=273K

T_H =absolute temperature of steam point=373K

 Efficiency 𝜂 = 1 – 𝑇_L /𝑇_H = 1 – (273/373) = 1 − 0.7319 = 0.268

                  ⇒ 𝜂 = 0.268 * 100= 26.80%

 

Q/17. Two ends of a rod are kept at 127℃ and 227℃ . When 2000 Cal of heat flows in this rod, then change in entropy is ?

 (A) 1 Cal/K          (B) 20 Cal/K             (C) 6.9 Cal/K                 (D) 0.7 Cal/K

 Ans :  Temperatures of two ends of the rod are

                        𝑇₁ = 127℃ = (127 + 273)𝐾 = 400𝐾

                       𝑇₂ = 227℃ = (227 + 273)𝐾 = 500𝐾

Heat flowing through two ends of the rod is ∆𝑄 = 2000 𝐶𝑎𝑙

Change in entropy ∆𝑆 = ∆𝑄/∆𝑇 = ∆𝑄/(𝑇₂−𝑇₁) = 2000/100 = 20 Cal/K

 

Q/18 . In a gas, the relative velocity of the most probable speed (𝑉_p), the average speed (𝑉̅) and root mean square speed (𝑉_rms) of the molecule are?

 (A) 𝑉𝑟𝑚𝑠 > 𝑉̅ > 𝑉𝑝                                        (B) 𝑉̅ > 𝑉𝑟𝑚𝑠 > 𝑉𝑝

 (C) 𝑉𝑝 > 𝑉̅ > 𝑉𝑟𝑚𝑠                                       (D) 𝑉𝑝 > 𝑉𝑟𝑚𝑠 > 𝑉̅

Ans:  Root mean square speed 𝑉_rms  = √(3𝐾𝑇/𝑚) =1.732√(𝐾𝑇/𝑚)

           Average speed 𝑉̅ = √(8𝐾𝑇/𝜋𝑚) = 1.595√(𝐾𝑇/𝑚)

          Most probable speed 𝑉𝑝 = √(2𝐾𝑇/𝑚) =1.414√(𝐾𝑇/𝑚)

           So 𝑉𝑟𝑚𝑠 > 𝑉̅ > 𝑉𝑝

 

Q/19 . The residue of 𝑧/(𝑧−𝑎)(𝑧−𝑏) at infinity is ?

 (A) 1                     (B) -1                  (C) 0                     (D) ∞

Ans:  𝑓(𝑧) = 𝑧/(𝑧 − 𝑎)(𝑧 − 𝑏)

        Residue of 𝑓(𝑧) at infinity is Res_𝑧→∞ 𝑓(𝑧)

          = lim_𝑧→∞ [−𝑧𝑓(𝑧)] = lim_𝑧→∞ [ −𝑧²/ (𝑧−𝑎)(𝑧−𝑏) ]

              = lim_𝑧→∞ [ −1/(1− 𝑎/𝑧 )(1− 𝑏/𝑧 )] = −1

 

Q/20 . The Fourier transform of the function 𝑓(𝑥) is 𝐹(𝑘) = ∫ 𝑒^ikx𝑓(𝑥)𝑑𝑥. The Fourier transform of 𝑑𝑓(𝑥)/𝑑𝑥 is ?

 (A) 𝑑𝐹(𝑘)/𝑑𝑘           (B)  ∫𝐹(𝑘)𝑑𝑘           (C) −𝑖𝑘𝐹(𝑘)          (D) 𝑖𝑘𝐹(𝑘)

 Ans:  Fourier transform of the function 𝑓(𝑥) is 𝐹[𝑓(𝑥)] = 𝐹(𝑘) = ∫ 𝑒^𝑖𝑘𝑥𝑓(𝑥)𝑑𝑥

           Fourier transform of 𝑑ⁿ𝑓(𝑥)/𝑑𝑥ⁿ = 𝐹 [𝑑ⁿ𝑓(𝑥)/𝑑𝑥ⁿ] = (−𝑖𝑘)^𝑛  𝐹[𝑓(𝑥)]

           So, Fourier transform of 𝑑𝑓(𝑥)/𝑑𝑥 = 𝐹[𝑑𝑓/𝑑𝑥] = (−𝑖𝑘)𝐹[𝑓(𝑥)] = −𝑖𝑘𝐹(𝑘)

 

Q/21. If 𝑓(𝑠) =₀∫^∞  e^−𝑠𝑡𝐹(𝑡)𝑑𝑡 is the Laplace transform of the function 𝐹(𝑡), then Laplace transform of 𝑘𝑡 is ?

    (A) 1/𝑠²            (B) 𝑘/𝑠²            (C) 𝑘/𝑠                (D) 𝑠²/𝑘

 

 Ans:  Laplace transform of 𝐹(𝑡) is

                            𝐿[𝐹(𝑡)] =₀∫^∞  e^−𝑠𝑡𝐹(𝑡)𝑑𝑡

         So Laplace transform of 𝑘𝑡 is  ;

                        𝐿[𝑘𝑡] = ₀∫^∞  e^−𝑠𝑡 k𝑡𝑑𝑡  = k₀∫^∞  e^−𝑠𝑡 𝑡𝑑𝑡

                                    = 𝑘/𝑠²

 

 

22. When an electron jumps from fourth orbit to second orbit, one can get?

     (A) First line of Pfund series                (B) Second line of Lyman series

    (C) Second line of Paschen series        (D) Second line of Balmer series

 Ans :  If the quantum number of the initial higher energy state is n_i =3,4,5, …  and quantum number of final lower energy state is n_f =2, then for Balmer series

 Electron transition from n_i=3 to n_f=2 corresponds to first line of Balmer series.

 Electron transition from n_i=4 to n_f=2 corresponds to second line of Balmer series.

Q/23. Davisson and Germer experiments relate to ?

(A) Interference                       (B) Polarisation              

(C) Electron diffraction            (D) Phosphorescence

 

Ans : C Davisson and Germer experiment relate to electron diffraction 11

 

Q/24. Which is incorrect according to the shell model of the nucleus?

  (A) Magic number exist                    

 (B) Nucleons interact with their nearest neighbours only

(C) Nucleons in a nucleus interact with a general force field

 (D) Large electronic quadruple moment exists for certain nuclei.

Ans : B    →The shell model of the nucleus accounts for the existence of magic numbers

→ Certain other nuclear properties in terms of nucleon behaviour in a common force field.

→Each nucleon interacts chiefly with a general force field produced by all other nucleons.

 

Q 25. Which of the following is not used as a moderator in a nuclear reactor?

     (A) H₂O           (B) D₂O                  (C) C       (D) Al

          Ans: D Al is not used as a moderator

Q/26. Which of the following is used in VLSI technology to form integrated circuit?

 (A) Transistors                 (B) Switches             (C) Diodes           (D) Buffers

 Ans : A           Very large-scale integration (VLSI) is the process of creating an integrated circuit (IC) by combining millions of MOS transistors onto a single chip.              

 

 

 

28. Digital circuit can be made by repetitive use of ?

    (A) NOT gates                    (B) OR gates           

       (C) NAND gates               (D) AND gates

 Ans :  The repeated use of NAND gates or NOR gates alone can produce all other logic gates.

Q/29. Asynchronous counters are known as?

(A) Ripple counters                                    (B) Modulus counters

(C) Decade counters                   (D) Multiple check counters

 Ans :  Asynchronous counters are known as Ripple counters.

 

Q/30. According to Schrodinger, a particle is equivalent to a ?

    (A) Single wave              (B) Sound wave             

     (C) Light wave              (D) Wave packet

 Ans :  According to Schrodinger a particle is equivalent to a wave packet.

Q/31. Position and moment operators satisfy [𝑥̂, 𝑝̂] = 𝑖ℏ, the value of [𝑝̂,[𝑥̂, 𝑝̂]] is?

 (A) 1                  (B) 0                (C) ℏ                    (D) 𝑖ℏ

 

Ans :    [𝑥̂, 𝑝̂] = 𝑖ℏ

                   [𝑝̂,[𝑥̂, 𝑝̂]] = [𝑝̂, 𝑖ℏ] = i[𝑝̂, ℏ] = 0                        ℏ= constant

  Q/32 . The normal Zeeman effect is?

   (A) Observed only in atoms with an even number of electrons

(B) Observed only in atoms with an odd number of electrons

(C) Confirmation of space quantization

 (D) Not a confirmation of space quantization

Ans :  Both A and C

      The normal Zeeman effect is due to transitions between the singlet (S=0) states of the atom. If there are even number of electrons, they will be paired off. So, their spin will be zero.

   The Zeeman effect is a vivid confirmation of space quantization.

Q/33. For Bragg’s reflection by a crystal to occur, the X-ray wavelength 𝜆 and inter atomic distance 𝑑 must be ?

    (A) 𝜆 > 2𝑑                               (B) 𝜆 = 2𝑑                    (C) 𝜆 ≤ 2𝑑          (D) 𝜆 < 2𝑑

 Ans :  Bragg’s law: 2𝑑sin 𝜃 = 𝑛𝜆                     𝑛 represents the order of reflection.

  The highest possible order is determined by the condition that sin 𝜃 can not exceed unity.

     For 𝑛 = 1    , 2𝑑 sin 𝜃 = 𝜆 ;   sin 𝜃 ≤ 1,

                           𝜆 ≤ 2𝑑

Q/34. According to the band theory of solids, the potential energy of two types of standing waves inside the crystal differ by an amount ?

(A) Energy gap         (B) 6 eV              (C) 2 eV           (D) None of these

Ans :   The potential energy of two types of standing waves inside the crystal differ by energy gap

35. The magnetic lines of force can not penetrate the body of a superconductor. This phenomenon is known as ?

 (A) Isotopic effect                                     (B) BCS theory

  (C) Meissner effect                        (D) London theory

 

Ans :  The magnetic lines of force can not penetrate the body of a superconductor . The phenomenon of flux exclusion in a superconductor is known as Meissner effect.     

FOR QUESTION (36-70) CLICK THE LINK  https://biswatutorialphysics.blogspot.com/2021/08/previous-year-2020-common-pg-entrance_10.html