Paper-with-solution-physical-sciences.pdf

CSIR-UGC-NET JUNE 2018

PAPER WITH SOLUTION

DEC. 2017

PHYSICAL SCIENCES Previous Years’ Question TIFR | IIT-JAM | BHU | HCU

w w w. c a re e re n d e a v o u r. c o m

CSIR-UGC-NET/JRF Dec. 2017

1

CSIR-UGC–NET/JRF– DEC - 2017 PHYSICAL SCIENCES BOOKLET - [A]

PART - B 21.

Consider the differential equation

dy  ay  e bt with the initial condition y (0)  0 . Then the Laplace dx

transform Y ( s ) of the solution y (t ) is (a) 22.

1 ( s  a ) ( s  b)

1 b(s  a )

(b)

(c)

1 a ( s  b)

e a  e b ba

(d)

Consider the matrix equation

1 1 1   x   0      1 2 3   y    0  2 b 2c   z   0       The condition for existence of a non-trivial solution, and the corresponding normalised solution (upto a sign) is (a) b  2c and ( x, y, z ) 

1 (1,  2, 1) 6

(c) c  b  1 and ( x, y , z )  23.

Consider the real function f ( x)  (a) (b) (c) (d)

24.

25.

1 (2,  1,  1) 6

(b) c  2b and ( x, y, z ) 

1 (1, 1,  2) 6

(d) b  c  1 and ( x, y , z ) 

1 (1,  2, 1) 6

1 . The Taylor expansion of f ( x ) about x  0 converges ges ( x  4) 2

for all values of x for all values of x except x   2 in the region 2  x  2 for x  2 and x   2

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   Let A be a non-singular 3 × 3 matrix, the columns of which are denoted by the vectors a , b and c ,    respectively. Similarly, u , v and w denote the vectors that form the corresponding columns of ( AT ) 1. Which of the following is true ?   (a) u  a  0, u  b  0, u  c  1 (b) u  a  0, u  b  1, u  c  0   (c) u  a  1, u  b  0, u  c  0 (d) u  a  0, u  b  0, u  c  0

The number of linearly independent power series solutions, around x  0 , of the second order linear

d 2 y dy differential equation x 2   xy  0 is dx dx (a) 0 (this equation does not have a power series solution) (b) 1 (c) 2 (d) 3

CSIR-UGC-NET/JRF Dec. 2017

2

26.

A disc of mass m is free to rotate in a plane parallel to the xy-plane with an angular velocity  zˆ about a massless rigid rod suspended from the roof of a stationary car (as shown in the figure below). The rod is free to orient itself along any direction. z

y

x ax –z

The car accelerates in the positive x-direction with an acceleration a  0 . Which of the following statements is true for the coordinates of the centre of mass of the disc in the reference frame of the car ? (a) only the x and the z coordinates change (b) only the y and the z coordinates change (c) only the x and the y coordinates change (d) all the three coordinates change 27.

A cyclist, weighing a total of 80 kg with the bicycle, pedals at a speed of 10 m/s. She stops pedalling at an instant which is taken to be t  0 . Due to the velocity dependent frictional force, her velocity is found to vary as 10 v (t )  m/s, t   1    30  where t is measured in seconds. When the velocity drops to 8 m/s, she starts pedalling again to maintain a constant speed. The energy expended by her in one minute at this (new) speed, is (a) 4 kJ (b) 8 kJ (c) 16 kJ (d) 32 kJ

28.

A light signal travels from a point A to a point B, both within a glass slab that is moving with uniform velocity (in the same direction as the light) with speed 0.3 c with respect to an external observer. If the refractive index of the slab is 1.5, then the observer will measure the speed of the signal as (a) 0.67 c (b) 0.81 c (c) 0.97 c (d) c

29.

A monoatomic gas of volume V is in equilibrium in a uniform vertical cylinder, the lower end of which is closed by a rigid wall and the order by a frictionless piston. The piston is pressed lightly and released. Assume that the gas is a poor conductor of heat and the cylinder and piston are perfectly insulating. If the cross-sectional area of the cylinder is A, the angular frequency of small oscillations of the piston about the point of equilibrium, is

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5 gA (3V )

(a) 30.

4 gA (3V )

(b)

5 gA 3 V

(c)

7 gA (5V )

(d)

The normalized wavefunction of a particle in three dimensions is given by 1 (r , , )  e  r /2a 3 8a where a  0 is a constant. The ratio of the most probable distance from the origin to the mean 

distance from the origin, is [You may use

 dx x

n

e  x  n ! ].

0

(a)

1 3

(b)

1 2

(c)

3 2

(d)

2 3

CSIR-UGC-NET/JRF Dec. 2017

31.

3

The state vector of a one-dimensional simple harmonic oscillator of angular frequency , at time t  0 , is given by  (0) 

1  0  2  , where 0 and 2 are the normalized ground state and 2

the second excited state, respectively. The minimum time t after which the state vector  (t ) is orthogonal to (0) , is  2  4 (b) (c) (d) 2    The normalized wavefunction in the momentum space of a particle in one dimension is

(a) 32.

( p ) 

    (c) (d) 3    2 Let x denote the position operator and p the canonically conjugate momentum operator of a particle. The commutator  1 2 2 1 2 2  2m p   x , m p  x  ,   where  and  are constants, is zero if

(a) 33.

 , where  and  are real constants. The uncertainty x in measuring its position is p  2 2



 2

(a)    34.

36.

(b)   2

(c)   2

(d) 2  

1 d2 1 3d  Q (d) Q 4  0 z 3 4  0 z 2 A rectangular piece of dielectric material is inserted partially into the (air) gap between the plates of a parallel plate capacitor. The dielectric piece will (a) remain stationary where it is placed. (b) be pushed out from the gap between the plates. (c) be drawn inside the gap between the plates and its velocity does not change sign. (d) execute an oscillatory motion in the region between the plates.

1 d2 Q 4  0 z 3

(b)

1 2d Q 4 0 z 2

(c)

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An  electromagnetic wave is travelling in free space (of permittivity 0 ) with electric field ˆ cos q ( x  ct ). The average power (per unit area) crossing planes parallel to 4 x  3 y  0 will E  kE 0 be (a)

37.



Two point charges  3Q and Q are placed at (0, 0, d ) and (0, 0, 2d ) respectively, above an infinite grounded conducting sheet kept in the xy-plane. At a point (0, 0, z ) , where z  d , the electrostatic potential of this charge configuration would approximately be (a)

35.

(b)

4 0c E02 5

(b) 0 cE02

(c)

1 0 c E02 2

(d)

16 0 c E02 25

A plane electromagnetic wave from within a dielectric medium (with   4 0 and   0 ) is incident  on its boundary with air, at z  0 . The magnetic field in the medium is H  ˆjH 0 cos t  kx  k 3 z , where  and k are positive constants. The angles of reflection and refraction are, respectively, (a) 45º and 60º (b) 30º and 90º (c) 30º and 60º (d) 60º and 90º





CSIR-UGC-NET/JRF Dec. 2017

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38.

 The dispersion relation of a gas of spin - 12 fermions in two dimensions is E  v k , where E is the  energy, k is the wave vector and v is a constant with the dimension of velocity. If the Fermi energy

at zero temperature is  F , the number of particles per unit area is  2F 3/2 3F F (c) (d) (2v 2 2 ) (3v 3 3 ) (62 v3 2 ) The relation between the internal energy U, entropy S, temperature T, pressure p, volume V, chemical potential µ and number of particles N of a thermodynamic system is dU  TdS  pdV  dN . That U is an exact differential implies that

(a) 39.

F (4v)

(a) 

(c) p 40.

(b)

p T  S V , N V U T

 S,N

(b) p S,N

1 U T V

(d) S ,

U T

U V

S S,N

p T  S V , N V

S ,

S ,N

The number of microstates of a gas of N particles in a volume V and of internal energy U, is given by 3 N /2

 aU  (U , V , N )  (V  Nb)  ,   N  (where a and b are positive constants). Its pressure P, volume V and temperature T, are related by N

aN  (a)  P  V 

aN  (b)  P  2 V 

  (V  Nb)  N k BT 

(c) PV  N k BT 41.

  (V  Nb)  N k BT 

(d) P(V  Nb)  N k BT

Consider a system of identical atoms in equilibrium with blackbody radiation in a cavity at temperature T. The equilibrium probabilities for each atom being in the ground state 0 and an excited state 1 are P0 and P1 , respectively. Let n be the average number of photons in a mode in the cavity that causes transition between the two states. Let W0 1 and W1 0 denote, respectively, the squares of the matrix elements corresponding to the atomic transitions 0  1 and 1  0 . Which of the following equations hold in equilibrium ? (a) P0 nW0 1  P1 W1  0 (b) P0 W0 1  P1 nW1  0 (c) P0 nW0  1  P1 W1 0  P1 nW1  0 (d) P0 nW0  1  P1 W1 0  P1 nW1  0

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42.

In the circuit below the voltages VBB and VCC are kept fixed, the voltage measured at B is a constant, but that measured at A fluctuates between a few V to a few mV. VCC +9V RC B VBB +3V

RB A

From these measurements it may be inferred that the (a) base is open internally (b) emitter is open internally (c) collector resistor is open (d) base resistor is open

CSIR-UGC-NET/JRF Dec. 2017

43.

The full scale voltage of an n-bit Digital-to-Analog Converter is V. The resolution that can be achieved in it is (a)

44.

V (2  1)

(b)

n

V (2  1)

V 22 n

(c)

n

(d)

V n

The spring constant k, of a spring of a mass ms , is determined experimentally by loading the spring with mass M and recording the time period T, for a single oscillation. If the experiment is carried out for different masses, then the graph that correctly represents the result is

T2

T2

(a)

T2

(b) (0, 0)

45.

5

(c) (0, 0)

M

T2

(d) (0, 0)

M

(0, 0)

M

M

A Zener diode with an operating voltage of 10 V at 25 ºC has a positive temperature co-efficient of 0.07% per ºC of the operating voltage. The operating voltage of this Zener diode at 125 ºC is (a) 12.0 V (b) 11.7 V (c) 10.7 V (d) 9.3 V PART - C

46.

Consider an element U () of the group SU(2), where  is any one of the parameters of the group. Under an infinitesimal change      , it changes as U ()  U ()  U ()  (1  X ()) U () . To order  , the matrix X () should always be (a) positive definite (b) real symmetric (c) hermitian (d) anti-hermitian

47.

The differential equation

dy ( x )  x 2 , with the initial condition y (0)  0 , is solved using Euler’s ’s dx

method. If y E ( x ) is the exact solution and y N ( x ) the numerical solution obtained using n steps of

( y N ( x)  yE ( x)) is proportional to yE ( x )

equal length, then the relative error

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(a)

1 n2

(b)

1 n3

1 n4

(c)

(d)

1 n 1

48.

The interval [0, 1] is divided into n parts of equal length to calculate the integral  ei 2 x dx using the 0

trapezoidal rule. The minimum value of n for which the result is exact, is (a) 2 (b) 3 (c) 4 (d)  49.

The generating function G (t , x) for the Legendre polynomials Pn (t ) is

1

G (t , x ) 





 x n Pn (t ), for

1  2 xt  x 2 n  0 If the function f ( x) is defined by the integral equation x

 f ( x) dx  xG (1, x), 0

x  1.

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it can be expressed as 

(a)

1  x n  m Pn (1) Pm  2  n, m  0

n, m  0



x

nm

(d)

Pn (1) Pm (1)

A particle moves in one dimension in a potential V ( x )   k 2 x 4  2 x 2 , where k and  are constants. Which of the following curves best describes the trajectories of this system in phase space ? p

p

x

(a)

x

(b)

p

p

x

(c)

51.

1 1 (  1) and   (  1) 2 2

(b)  

1 1 (  1) and   (  1) 2 2

1 1 1 1 (  1) (  1) and   (  1) (d)   (  1) and   2 2 2 2 Consider a set of particles which interact by a pair potential V  ar 6 , where r is the inter-particle separation and a  0 is a constant. If a system of such particles has reached virial equilibrium, the ratio of the kinetic to the total energy of the system is

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(c)  

(a) 53.

x

(d)

Let ( x, p) be the generalized coordinate and momentum of a Hamiltonian system. If new variables   ( X , P ) are defined by X  x sinh ( p) and P  x cosh ( p) , where ,  and  are constants, then the conditions for it to be a canonical transformation, are (a)  

52.

x n  m Pn (0) Pm (1)

 n, m  0

n, m  0

50.

x n  m Pn (1) Pm (1)

 



(c)



(b)

1 2

(b)

1 3

(c)

3 4

(d)

2 3

 In an intertial frame S, the magnetic vector potential in a region of space is given by A  az iˆ (where a is a constant) and the scalar potential is zero. The electric and magnetic fields seen by an inertial observer moving with a velocity viˆ with respect to S, are respectively. [In the following



1 1  v2 c 2

(a) 0 and ajˆ

]. (b)  vakˆ and aiˆ

(c) vakˆ and va ˆj

(d) vakˆ and a ˆj

CSIR-UGC-NET/JRF Dec. 2017

54.

7

In the rest frame S1 of a point particle with electric charge q1 , another point particle with electric charge q2 moves with a speed v parallel to the x-axis at a perpendicular distance l. The magnitude of the electromagnetic force felt by q1 due to q2 when the distance between them is minimum, is [In the following  

(a)

v2 1 2 c

].

1 q1 q2 4 0 l 2

(b)

1  q1 q2 4  0 l 2

 v2  1 q1 q2  v 2  1  (d) 1 2   2  2  4   c  l 0    c  A circular current carrying loop of radius a carries a steady current. Aconstant electric charge is kept  at the centre of the loop. The electric and magnetic fields, E and B respectively, at a distance d vertically above the centre of the loop satisfy          (a) E  B (b) E  0 (c)  E  B  0 (d)   E  B  0 (c)

55.

1

1  q1 q2 4 0 l 2









56.

A phase shift of 30º is observed when a beam of particles of energy 0.1 MeV is scattered by a target. When the beam energy is changed, the observed phase shift is 60º. Assuming that only s-wave scattering is relevant and that the cross-section does not change with energy, the beam energy is (a) 0.4 MeV (b) 0.3 MeV (c) 0.2 MeV (d) 0.15 MeV

57.

The Hamiltonian of a two-level quantum system is H 

58.

 cos 4   cos 8   cos 2     (a)  (b)  (c)     sin sin sin  4   8   2  Consider a one-dimensional infinite square well

1 1 1    . A possible initial state in 2  1 1  which the probability of the system being in that quantum state does not change with time, is  cos 6   (d)   sin  6 

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 0 for 0  x  a, V ( x)   otherwise  If a perturbation

V V ( x)   0 0

for 0  x  a3 , otherwise

is applied, then the correction to the energy of the first excited state, to first order in V , is nearest to (a) V0 59.

(b) 0.16 V0

(c) 0.2 V0

(d) 0.33 V0

The energy eigenvalues En of a quantum system in the potential V  cx 6 (where c  0 is a constant), for large values of the quantum number n, varies as (a) n 4/3 (b) n3/2 (c) n5/4 (d) n6/5

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60.

Consider a quantum system of non-interacting bosons in contact with a particle bath. The probability of finding no particle in a given single particle quantum state is 10–6. The average number of particles in that state is of the order of (a) 103 (b) 106 (c) 109 (d) 1012

61.

A closed system having three non-degenerate energy levels with energies E  0,   , is at temperature T. For   2 k BT , the probability of finding the system in the state with energy E  0 , is 1 1 1 1 (b) (c) cosh 2 (d) cosh 2 (2 cosh 2) 2 (1  2cosh 2) Two non-degenerate energy levels with energies 0 and  are occupied by N non-interacting particles at a temperatures T. Using classical statistics, the average internal energy of the system is

(a) 62.

(a) 63.

N

1  e

 / k BT



N

(b)

1  e

 / k BT

(c) N  e   / kBT



(d)

3 N k BT 2

In the circuit below, D1 and D2 are two silicon diodes with the same characteristics. If the forward voltage drop of a silicon diode is 0.7 V, then the value of the current I1  I D1 , is I1 1 k ID1 D1

10 V

(a) 18.6 mA 64.

(b) 9.3 mA

ID2 D2

Vo

(c) 13.95 mA

(d) 14.65 mA

The circuit below comprises of D-flip flops. The output is taken from Q3 , Q2 , Q1 and Q0 , as shown in the figure. LSB D

MSB Q Q Q Q www.careerendeavour.com 0

Q

1

D

Q

2

D

Q

3

D

Q Q

CLR

CLR

CLR

CLR

CLK RST

The binary number given by the string Q3 Q2 Q1 Q0 changes for every clock pulse that is applied to the CLK input. If the output is initialized at 0000, then the corresponding sequence of decimal numbers that repeats itself, is (a) 3, 2, 1, 0 (b) 1, 3, 7, 14, 12, 8 (c) 1, 3, 7, 15, 12, 14, 0 (d) 1, 3, 7, 15, 14, 12, 8, 0

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65.

9

Two physical quantities T and M are related by the equation

2 M  b , a 2 where a and b are constant parameters. The variation of T as a function of M was recorded in an experiment to determine the value of a graphically. Let m be the slope of the straight line when T 2 is plotted vs. M, and m be the uncertainty in determining it. The uncertainty in determining a is T

66.

a  m  b  m  2  m   m  (b) a  (c) (d)        2 m  2a  m  a  m   m  The sensitivity of a hot cathode pressure gauge is 10 mbar–1. If the ratio between the numbers of the impinging charged particles to emitted electrons is 1:10, then the pressure is (a) 10 mbar (b) 10–1 mbar (c) 10–2 mbar (d) 102 mbar

67.

The Zeeman shift of the energy of a state with quantum numbers L, S, J and mJ is

(a)

HZ 

mJ  B B   L  J   gS  S  J   J ( J  1)

where B is the applied magnetic field, g S is the g-factor for the spin and  B h  1.4 MHz-G–1, where h is the Planck constant. The approximate frequency shift of the S  0, L  1 and mJ  1 state, at a magnetic field of 1 G, is (a) 10 MHz (b) 1.4 MHz (c) 5 MHz (d) 2.8 MHz 68.

The separations between the adjacent levels of a normal multiplet are found to be 22 cm–1 and 33 cm– 1 . Assume that the multiplet is described well by the L-S coupling scheme and the Lande’s interval rule, namely E ( J )  E ( J  1)  AJ , where A is a constant. The term notations for this multiplet is (a) 3P0, 1, 2 (b) 3F2, 3, 4 (c) 3G3, 4, 5 (d) 3D1, 2, 3

69.

If the fine structure splitting between the 2 2P3/2 and 2 2P1/2 levels in the hydrogen atom is 0.4 cm–1, the corresponding splitting in Li2+ will approximately be (a) 1.2 cm–1 (b) 10.8 cm–1 (c) 32.4 cm–1 (d) 36.8 cm–1

70.

A crystal of MnO has NaCl structure. It has a paramagnetic to anti-ferromagnetic transition at 120 K. Below 120 K, the spins within a single [111] plane are parallel but the spins in adjacent [111] planes are anti-parallel. If neutron scattering is used to determine the lattice constants, respectively, d and d  , below and above the transition temperature of MnO then

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d d (b) d  (c) d  2d  (d) d  2d  2 2 A metallic nanowire of length l is approximated as a one-dimensional lattice of N atoms with lattice

(a) d  71.

spacing a. If the dispersion of electrons in the lattice is given as E (k )  E0  2t cos ka, where E0 and t are constants, then the density of states inside the nanowire depends on E as (a) N 3

t2 E  E0

2

(b)

 E  E0   2t   1  

(c) N 3

E  E0 t2

(d)

N (2t ) 2  ( E  E0 )2

CSIR-UGC-NET/JRF Dec. 2017

10

72.

Consider a two-dimensional material of length l and width w subjected to a constant magnetic field B applied perpendicular to it. The number of charge carriers per unit area may be expressed as kqB n , (2) where k is a positive real number and q is the carrier charge. Then the Hall resistivity  xy is (a)

73.

2k q2

l w

(b)

2 w kq 2 l

(c)

2 kq 2

(d)

2k q2

The spin-parity assignments for the ground and first excited states of the isotrope

57 28 Ni ,

in the single

particle shell model, are 

1 3 (a)   and   2 2

74.





5 7 (b)   and   2 2





3 5 (c)   and   2 2

The first excited state of the rotational spectrum of the nucleus



238 92 U

ground state. The energy of the second excited state (in keV), is (a) 150 (b) 120 (c) 90 75.



3 5 (d)   and   2 2



has an energy 45 keV above the (d) 60

Which of the following processes is not allowed by the strong interaction but is allowed by the weak interaction? (a) K 0  0  K 0     

(b) p  n  d  p  p

(c)    K 0  p  n

(d) p     n   

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CSIR-UGC-NET/JRF Dec. 2017

CSIR-UGC-NET/JRF DEC 2017 PHYSICAL SCIENCES BOOKLET-[A]

PART - B 21.

dy  ay  e  bt with y  t  0   0 dt Taking Laplace Transform on both sides,

Given:



s  Y  s   y t  0  a  Y  s  



Y  s s  a 



Y  s 

1 s  b

1 s  b 1

 s  a  s  b 

Correct option is (a) 22.

Given :  1 1 1  x   0        1 2 3  y    0   2 b 2c  z   0       For the existance of non-trivial solution of x, y, z 1 1

1

1 2

3 0

2 b 2c

[Apply c '2  c2  c1, c '3  c3  c1 , we get]



1

0

1

1

www.careerendeavour.com 2 0 0

2 b  2 2c  2 



 2c  2   2  b  2   0 b  c 1

From the given matrix equation, x y z 0 x  2 y  3z  0

... (1) ... (2)

From equations (1), (2),  y  2 z  y  2 z From equation (1),  x  z 1 2 and y   6 6 Correct option is (d)

Therefore, x  z 

11

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12

23.

Given : f  x  

1 x 4 2

Consider the corresponding complex function to be f  z  

1 z 4 2

Here, z  2i are the singular points of f  z  Im(z) Largest possible circle centred about z = 0, within which f(z) is analytic.

z = 2i

Re(z) z = –2i

Radius of convergence of the Taylor series expansion of f  z  about z = 0, is R = 2 and corresponding region of convergence is z  2 Therefore, the Taylor series expansion of f  x  about x = 0 converges in the region

x  2 i.e.  2  x  2 Correct option is (c) 24.

   Assume, a  a1iˆ  a2 ˆj  a3kˆ ; b  b1iˆ  b2 ˆj  b3kˆ , c  c1iˆ  c2 ˆj  c3kˆ

 a1 A   a2  a3

A  T

1

b1

c1     c2   Det.  A   a  b  c c2 



b2 b3

T

 

 A1





T 1  Co-factor matrix T   Det   A 



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1  Co-factor matrix  Det   A 

 comp. of 1        a  b  c   b  c    b c Therefore, u     a  b c



 







      So, u  a  1, u  b  0, u  c  0 Correct option is (c)

  

comp. of        c  a  

comp.of    a  b





  

CSIR-UGC-NET/JRF Dec. 2017

25.

Given : x

13

d 2 y dy   xy  0 dx 2 dx

d2y dy  x  x2 y  0 2 dx dx Similar to the form Bessel D.E. (n = 0) 

x2

d2y dy  x  x 2  n2 y  0 2 dx dx The given differential equation will have two linearly independent power series solution i.e.



x2

J0  x   1 



x2 x4 x6    ................ 22 22  42 22  62  62

Y0  x   J 0  x  

1 x  J 0  x  

2

dx

Correct option is (c) 26.

  Initial torque on the disc = r  Fpseudo  kˆ   maiˆ  maˆj





Since, the disc has angular momentum, it will move in y-direction. So, that change in angular momentum occurs in y-direction due to initial torque. After a little later,      r  F   cos   kˆ  sin  ˆj   ma iˆ  mg  kˆ   



 



 

 

 m  a cos  ˆj  a sin  kˆ  g sin  iˆ  Since, torque is in all directions, therefore disc will move in all directions. So, all coordinates will change. Correct option is (d) 27.

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m  80kg , vi  10 m /s v t  

10 dv 1 1    2 t dt 3  t  1 1    30  30 

When velocity is 8 m/s 10 1 dv 1 2   0.8     0.8 t  t  dt 3   1   1    30   30  Force at the moment, when velocity is 8 m/s 8

dv 1 51.2 2  80    0.8    dt 3 3 To maintain constant speed, the cyclist must apply the this much force F m

CSIR-UGC-NET/JRF Dec. 2017

14

Therefore, power applied = fv 

51.2 409.6 8  3 3

Energy expended in one minute =

409.6  60  8192 Joule  8.192 kJ 3

Correct option is (b) 28.

Velocity of light in glass, V 'x 

c c  n 1.5

Velocity of glass  S ' w.r.t.s  V  0.3c Velocity of light in ground frame,

S

1  0.3c V 'x  V 1.45 1.5 Vx    c  0.81c V ' x V 0.3 1.8 1 1 2 1.5 c

29.

For equilibrium of piston, mg  PA

Ground

... (1)

For adiabatic process PV   constant Differentiate to get dPV    PV  1dV  0 If x be displacement of piston from equilibrium position, dV  Ax



dP  

 PAx V

Restroing force,

 PA2 x F  dPA    S .H .M . V Force constant, k 

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 PA2  mgA  k V V

Angular frequency of oscillation,   For monoatomic gas,   5 gA 3V Correct option is (a) 



5 3

k  gA  m V

S light

v

CSIR-UGC-NET/JRF Dec. 2017

30.

1

Given :   r ,  ,   

8 a 3

15

e r /2a  a  0 

Method-1: 







r   *r 4 r 2 dr  0



r 

1 4  r / a 3 1 e  r dr  3 L r 3 3  8 a 0 2a

  s

1 a

1 3!  3a 3 2a 1/ a 4

Radial probability density, 2

P  r   Rn  r  r 2 

1 e r / a r 2 3 8 a

At most probable radial distance rmp ,

dP 1    0  e  r / a  2r  r 2  dr r rmp a  r r 

mp



rmp  2a

2 3 Correct option is (d)

Therefore, rmp /

31.

r



Given :   t  0  

1 1 0  2 2 2

Method-1:

 

 t    t 

1 1 0 e iE0t /   2 e iE2t /  2 2

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1 1  0 e  it /2  2 e  i 5t /2 2 2

Since,   t  should be orthogonal to   t  0  , therefore,

  t  |  t  0   0  



1 it /2 1 i 5t /2 e  e 0  2 2 t

 2n  1  2

  for n  0  2 Correct option is (a) tmin 

1 1  1 it /2   1  0 2 ei 5t /2   0  2  0  2e 2 2     2 ei 4t /2  1  e 

 n  0, 1, 2, .......

i 2 n 1

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16

32.

We can easily solve this question using dimension analysis. Given : Momentum space wave function,

 P 

 p 2 2

Dimension of p = Dimension of   Dimension of p = Dimension of 

   Dimension of x = Dimension of p = Dimension of  Correct option is (c) 33.

 1 2 2 1 2 2  2m p   x , m p   x   



1   2 2   2 2  p2 , p2   p , x    x , p     x 2 , x 2   0 2    2m m 2m



  2 2   2 2 p ,x  p ,x  0 2m  m

      2 2 Correct option is (b) The electrostatic potential at (0, 0, z) If

34.

  0, 0,3 

(0, 0, z)

1  Q 3Q 3Q Q       4 0   z  2d   z  d   z  d   z  2d  

–Q 3Q

Q  2d 3d 3d 2d  Q  1  3 3 1    2d   4 0 z  z z z z  4 0 z 2 35.

–3Q

Correct option is (b) The capacitor has width a and length l. The combient capacitance of the capacitor is

Q

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C  C1  C2  

ax a    x   0  d d

a  x     x   0  d

Therefore, the electrostatic potential energy of the system is

d x

(l–x)

1 a u  CV 2   x     x   0  V 2 2 2d Therefore, the force on the dielectric slab is dU a     0 V     0  dx 2d Therefore, the force on the slab will be inwards and when it reach at the other end the force on it will be opposite direction. So, slab will execute oscillations. Correct option is (d) F 

CSIR-UGC-NET/JRF Dec. 2017

36.

17

Given electric field is  ˆ cos q  x  ct  E  kE 0

 E Therefore, the magnetic field, B   yˆ 0 cos q  x  ct  c

     E2 Now, the pointing vector, S  E  H  1 E  B  xˆ 0 cos 2 q  x  ct  µ0 cµ0

Therefore, the average power per unit area crossing planes parallel to 4 x  3 y  0 is  E2 3 4 P  S  nˆ  xˆ 0 cos 2 q  x  ct    iˆ  cµ0 5 5

4 E02 1 2    0cE02 5 µ0c 2 5

ˆj  

1  2  cos q  x  ct    2 

Correct option is (a) 37.

The x-component of propagating vector is k x  k and the z-component of propagation vector k z  3k

Therefore, tan i 

kx k 1   kz 3k 3 i x.

1   i  tan    30º  3 According to Snell’s law, 1 

x r z

n1 sin 1  n2 sin  2 

2sin 30º  1sin  2



n1   n  4  2    90º Correct option is (b)  p  Given: E  v k , where k  



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38.



E  v

p  pv 

The density of state in two-dimension is g  p   

g  E  dE  g  p  dp 

2 A E pdp where, p  2 h v

2 A E dE 2 A  2 2 EdE h2 v v hv

1 4 A particles, g  E   2 2 E 2 hv The total number of particle is

For spin-



N   f  E  g  E  dE ,where f  E  for fermions is 0

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18

1  E F  exp   1  k BT  At zero temperature (T = 0K), we have f E 

1, f E   0,

E  F E  F





4 A N   f  E  g  E  dE  2 2 hv 0

F



EdE 

0

4 A 2 A F2   F 2h 2 v 2 2 v 2  2

h      2 

Therefore, the number of particles per unit area is

N  F2  A 2 v 2  2 Correct option is (d) 39.

Given: dU  TdS  pdV  µdN  U   U  p    and T     V  S , N  S V , N Since, dU is an exact differential, we have 

   U      U            S  V  S V  V  S V  S



 p   T        S V , N  V  S , N

Correct option is (a) 40.

Given :  U ,V , N   V  Nb 

N

 aU     N 

3N / 2

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The Boltzmann’s entropy is S  k B ln  

3 N /2   N  aU  S  k B ln V  Nb      N   

From combined I and II law of thermodynamics, we have TdS  dU  PdV  µdN 

 S  P T    V U , N

 1  P  k BTN    P V  Nb   Nk BT  V  Nb  Correct option is (d) 

CSIR-UGC-NET/JRF Dec. 2017

41.

19

Here the transition will 3 different types : 1. Stimulated absorbtion transition: It depends on the number of atom in state 0 , photon number inside the cavity and the square of the matrix elements. Therefore, the rate of transition of atom from 0  1 is P0 nW0 1. 2. Stimulated emission transition: It depends on the number of atom in state 1 , photon number inside the cavity and the square of the matrix elements. Therefore, the rate of transition of atom from 1  0 is P1nW1  0 . 3. Spontaneous emission transition: It does not depend on the number of photon inside the cavity. It depends on the number of atom in state 1 and the square of the matrix elements. Therefore, the rate of transition of atom from 1  0 is PW 1 1 0 . At equilibrium the rate of transition of atom to the 0  1 is equal to 1  0 . Therefore, P0 nW0  1  PW 1 1  0  P1nW1 0 Correct option is (d) VCC RC B+

RB A + VBE

VBB

42.

Variable

VCE = Constant – –

RB = constant = RC

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VCC = IC RC + VCE constant, Input KVL, VCC = ICRC + VCB + VBE constant

1. Incorrect 2. Emitter-open

VCC VCB + RB 3V

+ VBE

RC + VCE

–

–

Open circuit voltage is unknown

CSIR-UGC-NET/JRF Dec. 2017

20

3.

Collector

VCC = 9V

4.

is open VCC

RB A 3V

RC VCB A VBE

B + VCE –

+

+

B VCE = Constant

+

– –

its not finite value

Vfullscale

43.

Re solution 

44.

n = number of bits. If mass of the spring is ms and a block of mass M is attached with it. The equivalent mass of the system is equal to

 in volts 

2n  1

m  M equivalent   s  M   3  Therefore, the time period of oscillation is given by

T  2

ms M 3 k

4 2  ms  M  k  3  Correct option is (a)

T2

T2 

45.

Given, Vz  10 volt 

Vz 

M

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0.07  10  125  25   0.7 volt 100

Therefore, the final zener voltage, Vz

final

 Vz  Vz  10  0.7  volt  10.7 volt

PART - C

46.

   dU d   U   U     U    U      1   d  d     X     

CSIR-UGC-NET/JRF Dec. 2017

X     

 d i d  i ˆ    i    Lz d   d  

i X †        Lˆz 

†

 



21

  X   

Therefore, X    will always be anti-hermitian Correct option is (d) 47.

dy   x 2 with y  0   0 dx 

y

 3 x c 3

 3 x 3 Exact solution (using Taylor series): Applying y  0   0  c  0  y 

yi 1  yi  f  xi , yi   h  f '  xi , yi 



yi 1 

h2 h3  f "  xi , yi  2! 3!

h2 h3  yi"  yi"' 2! 3!   Approx. solution according  yi  y 'i h   

Error in calculation

to Euler's method

If h is small, Error in one step is of the order of h2. 2 Therefore, Error in total number of steps ‘n’, will be of the order of n  h 

Correct option is (d) 48.

Exact result: 1

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1

 e i 2 x  1 i 2 0  e dx   i 2   2i e  e   0 0 0 Trapezoidal rule: i 2 x

 a, b  0,1 ;

n  2; h 

ba 1  n 2

1 x  0  x0 ;  x1;1  x2 2 f  x   1  y0 ; 1  y1;1  y2 1

e 0

i 2 x

dx 

h 1  y0  y2   2  y1    1  1  2  1   0 2 4

Correct option is (a)

1 n

CSIR-UGC-NET/JRF Dec. 2017

22

x

49.

 f  x ' dx '  x  G 1, x   x  0

 x  1  x 

1 1  2 x  x2

1

 x  x 2  x 3  ............. f  x   1  2 x  3x 2  ......... 



f  x   1  x 

2



1 1  2 xt  x 2

  x n  Pn  t 

... (1)

n 0 

1

 x m  Pm  t 



... (2) m 0 1  2 xt  x Replacing ‘t’ by 1 and multiplying (1) and (2), we get 2

1

1  x  

2





1

1  x 

1  x 



2

2

  x m n  Pm 1 Pn 1 m  0 n 0









x m n  Pm 1  Pn 1

m ,n 0 



f  x 



x m n  Pm 1  Pn 1

m , n 0

Correct option is (b) 50.

V  x    kx 4   2 x 2

Since, there is minimum (potential well). So, motion can be bounded. Therefore, phase space trajectory will be closed loops for energies less than the height of potential well. However for higher energies will not be closed loops. Further, E  K .E.  P.E . V(x)

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E  T V  x T  E V  x

x

For x  , V  x     T    Px   These features are followed by figure Correct option is (b) 51.

Given new variable, X  x sinh (  p ), P  x cosh (  p ) The transformation will be canonical if the Poisson bracket of X and P is equal to 1. Therefore,  X , P x , p  1



X P X P  1 x p p x

CSIR-UGC-NET/JRF Dec. 2017



  x    1 sinh 2 (  p )    x   1 cosh 2 (  p )  1



x   1  sinh 2 (  p )    cosh 2 (  p)   1

23

L.H.S. will be 1, if     1  0 and  sinh 2 (  p )    cosh 2 (  p)  1 . Both satisfied simultaneously. But none of the given option satisfied the above two conditions simultaneously. However, only option (c) satisfies     1 . So, it may be taken as correct option. 52.

V  r   ar 6

Using virial theorem, T 

n 6 V  T  V  T 3 V 2 2

E T  V  E T 

53.

1 T 3

T 1 3 T   3 E 4

Correct option is (c)  A  aziˆ    B    A  ajˆ    A E    0 t According to Lorentz transformation, E ' x  Ex  0

B 'x  Bx  0

  E ' z    E z  vBy   a v

B ' y   B y  vE z / c 2   a

E ' y   E y  vBz  0



54.

 V 

B 'z

E '  a vkˆ



   B

z

 vE y / c 2

 0

B '   ajˆ

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Correct option is (d) The electric field at O due to point charge q2 is



 

1  v 2 / c2 q2 E  r, t   4 0 1  v 2 sin 2  / c 2





zz 3/2



2

q2   since,   90º  4 0  2

Therefore, the force felt by the charge q1 is F

1  q1q2 4 0  2

Correct option is (b)

P y

O y

x

l x

CSIR-UGC-NET/JRF Dec. 2017

24

55.

The direction of electric field on the axis due to point charge at the centre is along the axis and also the direction of magnetic field along the axis.    EB  0 E || B     E B  0 and





Correct option is (d) 56.

According to partial wave analysis, total cross-section is given as 4  2 k



  2  1 sin 2    0

Assuming that, only s-wave scattering is relevant, we obtain, 4 4  2 2 sin   sin 2  0 0 2 2 mE k As ‘  ’ does not change with energy ‘E’, therefore,



sin 2  0  constant E



sin 2  0   2   1 sin  0  2  E1 E2



sin 2  0  sin 2 60º  2 E2  E1   0.1 MeV sin 2  0  sin 2 30º   1

3/ 4   0.1 MeV  0.3 MeV 1/ 4 Correct option is (b) The possible initial state in which probability of the system being in that quantum state does not change with time, will be an eigenstate of the Hamiltonian H i.e. H   

57.

E2 

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 cos   H  1  1 1  cos   Assuming    1 1  sin   ; 2     sin  

H 

 cos   sin   1 1      2  cos   sin   2

    sin  4       2      sin        4

 

    cos  cos     1 8 8 Only for    H   i.e. H      8 2  sin    sin       8 8   Correct option is (b)

CSIR-UGC-NET/JRF Dec. 2017

58.

Unperturbed system: 1-D infinite potential well  2 2 x sin   st  x  2  x   a a 1 E .S . 0  First order correction to energy E2    2 V  x   2  1

a /3

 0





a /3

 0

2 2 2 x sin  V0 dx a a

V0 4 x   1  cos a  dx  a

V0  a  a   0   a  3  4

otherwise

 a  x  4 

a /3

 4 x  sin  a  0 

4    sin 0    sin 3  

a 3 V0 3a  V0  0.398V0  approx      3 8  3 8    Nearest answer is 0.33 V0 Correct option is (d) According to WKB approximation, 

59.

V0 a

0  x  a 

V0 a

If V  x   x  , then En  n 2  /   2 In the given question, V  x   x 6 26 / 6  2  En  n     En  n3/2 Correct option is (b)

60.

f oc  = the probability of occupancy = the average number of particles in a state

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fun  = The probability of unoccupancy

= The probability of finding number of particle in a state = 10–6 f oc  f un  1  f oc  1  10

61.

6

10 

6

 1

106

The order of average number of particles in a state = (106–1) ~ 106 Correct option is (b) The probability of finding the system in the state with energy E = 0 is P  E  0 

e

     / k BT

e 0/ kBT 1 1   / k BT   / kBT 0/ k BT   / k BT e 1 e 1  2 cosh   /k BT  e e

For   2k BT , we have 1 1  2 cosh 2 Correct option is (a) P   0  

25

CSIR-UGC-NET/JRF Dec. 2017

26

62.

The average internal energy of the system is

E g e N g e i

 Ei / k BT

i

i

E

 Ei / kB T

i

N

 e  / kBT  0  e 0/ kBT N    / k BT 0/ k BT e e 1  e  / k BT

i

Correct option is (a) 63.

Given D1 and D2 Two identical diotes Vy each   0.7 V

I1

Equivalent figure

IK +

10  0.7 I1   9.3 mA 1

9.3 mA V0

10 V 0.7 V ID

To find I1  ID1 ?

0.7 V

1

ID1  ID 2  4.65 mA I1  ID1  9.3  4.65 mA  13.95 mA

Note: Its a 4-bit Johnson counter

LSB MSB Q0 Q1 Q2 Q3 1

0

0

0

1

1

0

0

initial data (as per majority of options)

www.careerendeavour.com Modulus is 8

1

1

1

2

V0 = 0.7 V

I1  ID2  ID2

64.

ID –

0 Sequence coming as

1

1

1

1

0

1

1

1

0

0

1

1

0

0

0

1

0

0

0

0

1

0

0

0

1 3 7 15 14 12 8 0

Repeat

CSIR-UGC-NET/JRF Dec. 2017

65.

27

We have,

T

( M  b) 2 2  T 2  2 ( M  b) 2 a

2 a



(T 2 ) 2 2  2 M a



 2 2  2 2 m   2   a2  m   a 



2a  a 1  0 m  2 m a

ln a 2  ln 2 2  ln m

a m  a     2 m 

a m    2 m  Correct option is (a) 

66.

a 

The pressure (P) of a hot cathod pressure gauge 

i ii  c

where, i+ = current due to impinging particles i– = current due to emitted electrons c = sensitivity of gauge

i  number of impinging particles 1   number of electrons 10 i P

67.

1  102 mbar 10  10

Correct option is (c) In Anomalous zeeman effect E  J z  g j

www.careerendeavour.com E  g  BM (Where  B 

e eh B.M J M J B  g J  B BM J  2m 4M

J B

J

E  hv  E  hv  hv = g J  B BM J  v  g J

B BM J h

Now for S = 0, L = 1  J = 1  gJ  1 

J  J  1  S S  1  L  L  1 202  1 1 2J  J  1 4

Also for J  1  M J  1, 0  1

J=1

MJ = +1 MJ = 0

 for M J  1 given  M J  1  1  V  g J  B  B  M J  1 1.4 MH Z  6   1 Gauss  1  1.4 MH Z h Correct option is (b)

MJ = –1

eh ) 4M

CSIR-UGC-NET/JRF Dec. 2017

28

68.

By Lande Interval Rule E  J   E  J 1  AJ i;e. proportional to higher J. Seperation between adjacent levels are 22 cm–1 and 33 cm–1, i.e. their ratio = 3

D1, 2, 3 as seperation between 3D1, 3D2 and 3D3 are in the ration

This ratio is given by option (4) i.e. 2 : 3 (i.e. proportional to higher J). 69.

Fine structure splitting is proportional to Z4 i.e. Tz 2 . Now for H, Z = 1 and for Li++, Z = 3

TLi



70.

22 2  33 3

TH



 3 4 1 4

 81  TLi  81 0.4  32.4 cm 1

Correct option is (c) MnO has NaCl type structure which is paramagnetic at room temperature Tc  120 K T  Tc  Antiferromagnetic (ordered) T  Tc  Paramagnetic (disordered) [111] plane. a

Lattice constant, d 



a

3 h k  The distortion below, Tc  120 K is apparently an exchange striction which compresses the crystal 2

2

2

along 111 , increasing the nearest neighbour and decreasing the next nearest neighbour distances. In antiferromagnetic state    , the lattice parameter will be double paramagnetic state. d a 2a '    2; d ' a' a'



a  2a '

Correct option is (c) a

71.

a

a

a

d  2d '

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l

Given : E  k   E0  2t cos ka

... (i)

The density of state is one-dimensional is g  p  

2 h

where, p  k

g  k  dk  g  p  dp 

2 dk dk  h 

... (ii)

Also, the equation (i) gives, dE  2at sin kadk 

dk 

dE 2at 1  cos 2 ka



dE  E  E0  2at 1     2t 

2

dE

 a

2

 2t    E  E0 

2

CSIR-UGC-NET/JRF Dec. 2017





g  E  dE  g  k  dk 

g E 

a

 a

29

dE 2

 2t    E  E0 

where,    N  1 a

2

 N  1 a 2 2  2t    E  E0 

Since, the number of atoms would be large in the nanowire, we have N

g E 

2

 2t    E  E0 



N

g E 



2

2

 2t    E  E0 

2

Correct option is (d) Number of charge carrier per unit area, n 

72.

Hall resistivity,  xy 

kqB 2 

B B  2   2    2 ne kq 2 B kq

Correct option is (c)

73.

28

Ni 57 

N  28  even  N  29  odd 

N  29 :  1 s1/ 2 

2

f 7/2 nnnnnnnn 1 d3/2 nnn

1

3/2

1

nnnnnn

p1/2

nn

p

3/2

1

s

p3/ 2 

4



1

p1/ 2 

2

6

d  1

5/ 2

2

s1/ 2 

2

4

d  1

3/ 2

2

f

5/2

1

8

f7/ 2 

n

p3/2



1

p3/ 2   for ground state

2

odd neutron

www.careerendeavour.com 1 f nnnnnnnn

nn

d

5/2

1

1

odd neutron

2

s



2

p

3/2

2

2

7/2

1





JP 

3 2

Parity   1



1

nnnn

  1  1

 odd 

2

1

1/2

Ground state

Correct option is (d)

d

3/2

nnnn

s

1/2

nn

1

1

N = 29

empty

1



JP 

5 2

Parity   1



3

d

5/2

nnnnnn

p

1/2

nn

p

3/2

nnnn

s

1/2

nn

First excited state

  1  1

 odd 

CSIR-UGC-NET/JRF Dec. 2017

30

74.

Rotational energy, EJ 

2 J  J  1 2I

For first excited state J = 2



E2 

2 2  2 45  2   2  1  45   6   2I 2I 2I 6

4+

Second excited

2+

First excited

0+

Ground excited

Therefore, for second excited state J = 4, energy is

EJ  E4 

2 45  4  4  1   4  5  150 KeV 2I 6

Correct option is (a) 75.

For option (a), K0   0  K 0      

Q : 0  0  0  1  1  Q  0 B : 0  0  0  0  0  B  0 No lepton, S : 1  0  1  0  0  S  0 and S  2 Since strangeness is not conserved, so it is neither strong nor electromagnatic. In weak interactions, involving strange particles S  1 or S  1 or  1 . However, this does not necessarily hold in second order weak reactions, where there are mixes of K0 and K 0 mesons. Since this reaction is second order weak reaction involving mixes of K0 and K 0 , thus S = 2 is allowed and it is a weak interaction. For option(2)

For option(3)

For option (4)



  K0  P  n Q :1  0  1  0  Q  0

P     n    Q :1  1  0  2  Q  0

Q :1  0  1  1  1  Q  0

B :1  0  1  1  B  0

B :1  1  1  1  B  0

Pn  d PP 



deuteron 1 H 2

www.careerendeavour.com Since Baryon Number is not Since Baryon number is not

B :1  1  2  1  1  B  0 No Lepton

conserved, so it is forbidden conserved, so it is forbidden

S:0  0  0  0 0

reaction

1 1 1 1 I3 :   0    I3  0 2 2 2 2 1 1 1 1 I: 0 2 2 2 2   (0,1)

1 2  (0,1)

Both common  I is conserved  Strong interaction

Correct option is (a)

reaction.