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fiziks, Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics, IIT-JAM 2021, SECTION – A, Multiple Choice Questions (MCQ), Q1 – Q10 Carry One Mark Each, Q1., , The function ecos x is Taylor expanded about x  0 . The coefficient of x 2 is, (a) , , 1, 2, , (b) , , e, 2, , (c), , e, 2, , (d) Zero, , Ans. : (b), Solution: f  x   ecos x, , Taylor expansion about x  x0 is, f  x   f  x0    x  x0  f   x0 , ,  x  x0 , , , 2, , 2!, , f   x0   ..., ,  Taylor expansion about x  0, f  x   f  0   xf   0   x 2, , Coefficient of x 2 is, , f   0 , , 2!, , , , f   0 , , 2!, , f  x   ecos x  f   x    sin xecos x, , , ,  , , , , f   x    cos xecos x  sin x  sin xecos x    cos x  sin 2 x ecos x,  f   0    1  0  e1  e, ,  coefficients of x 2 , Q2., , f   0 , , 2!, , , , e, 2, , Let M be a 2  2 matrix. Its trace is 6 and its determinant has value 8 . Its eigenvalues, are, (b) 3 and 3, , (a) 2 and 4, , (c) 2 and 6, , (d) 2 and 3, , Ans. : (a), Solution: M is a 2  2 matrix, We know trace  sum of eigenvalues, and determinant  product of eigenvalues, H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016, Phone: 011-26865455/+91-9871145498, Website: www.physicsbyfiziks.com | Email: [email protected], , 1

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fiziks, Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics, , Let eigenvalues are 1 and 2 .,  1  2  6 and 12  8, , 1  2 , 2  4, Q3., , A planet is in a highly eccentric orbit about a star. The distance of its closest approach is, 300 times smaller than its farthest distance from the star. If the corresponding speeds are, , vc and v f , then, (a), , vc, is, vf, , 1, 300, , (b), , 1, 300, , (c), , 300, , (d) 300, , Ans. : (d), Solution: Using conservation of angular momentum, mvc rc  mv f rf , Q4., , vc rf 300rc,  ,  300, v f rc, rc, , An object of density  is floating in a liquid with 75% of its volume submerged. The, density of the liquid is, (a), , 4, , 3, , (b), , 3, , 2, , (c), , 8, , 5, , (d) 2 , , Ans. : (a), Solution: Weight of object equal to buoyancy force, , 3, 4, V  g  Vdg  d  , 4, 3, Q5., , An experiment with a Michelson interferometer is performed in vacuum using a laser of, wavelength 610 nm . One of the beams of the interferometer passes through a small glass, cavity 1.3cm long. After the cavity is completely filled with a medium of refractive, index n, 472 dark fringes are counted to move past a reference line. Given that the speed, of light is 3 108 m / s , the value of n is, (a) 1.01, , (b) 1.04, , (c) 1.06, , (d) 1.10, , Ans. : (a), Solution: Change in optical path x  m  2  n  1 t  m, H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016, Phone: 011-26865455/+91-9871145498, Website: www.physicsbyfiziks.com | Email: [email protected], , 2

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fiziks, Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics, ,  n 1 , Q6., , n 472  610 109,  n  1  0  011  n  1  011, , 2t, 2 1 3 102, , For a semiconductor material, the conventional flat band energy diagram is shown in the, conduction band, , figure. The variables Y , X respectively, are, Y, , (a) Energy, Momentum, (b) Energy, Distance, (c) Distance, Energy, (d) Momentum, Energy, valence band, X, , Ans. : (b), , Solution: Along the y-axis energy varies while along x-axis distance is variable., Q7., , For the given circuit, VD is the threshold voltage of the diode. The graph that best, 1k , deposits variation of V0 with Vi is, 1k , , Vi, , , , V0, , , , V0, , V0, , (a), , (b), , 0, , Vi, , 0, , V0, , VD, , Vi, , V0, , (c), , (d), , 0, , Vi, , 0, , Vi, , H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016, Phone: 011-26865455/+91-9871145498, Website: www.physicsbyfiziks.com | Email: [email protected], , 3

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fiziks, Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics, , Ans. : (a), Solution:, Postive half Cycle, s.c.,  , , , , Vi, , 1k , , Negative half Cycle, , 1k , , o.c.,  , , , , , , , , , o.c.,  , , Vi, , V0  0, , 1k , , 1k , , , , , , , V0  , , Q8., , s.c.,  , , V0, , 1k, Vi  Vi, 1k, , Arrange the following telescopes, where D is the telescope diameter and  is the, wavelength, in order of decreasing resolving power:, I. D  100 m,   21 cm, , II. D  2 m,   500 nm, , III. D  1 m,   100 nm, , IV. D  2 m,   10 mm, , (a) III, II, IV, I, , (b) II, III, I, IV, , (c) IV, III, II, I, , (d) III, II, I, IV, , Ans. : (d), Solution: d  1.22, , , D, , d I  2.56 103 ;, , d II  3.05 107, , d III  1.22  107 ;, , d IV  6.1 103, , Resolving power , , 1, d, , RPIII  RPII  RPI  RPIV, Q9., , Metallic lithium has bcc crystal structure. Each unit cell is a cube of side a . The number, of atoms per unit volume is, (a), , 1, a3, , (b), , 2, 2a 3, , (c), , 2, a3, , (d), , 4, a3, , H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016, Phone: 011-26865455/+91-9871145498, Website: www.physicsbyfiziks.com | Email: [email protected], , 4

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fiziks, Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics, , Ans. : (c), 1, Solution: The effective number of atoms in BCC is neff   8  1  2, 8, The volume of the bcc unit cell of side a is a 3 ., Thus, the number of atoms per unit volume is , , 2, ., a3, , Therefore, the correct option is (c)., Q10., , The moment of inertia of a solid sphere (radius R and mass M ) about the axis which is, at a distance of, (a), , R, from the centre is, 2, , 3, MR 2, 20, , (b), , 1, MR 2, 2, , (c), , 13, MR 2, 20, , (d), , 9, MR 2, 10, , Ans. : (c), Solution:, Using parallel axis theorem, I 0  I C .M  Md 2 ; d , , R, 2, , I C .M  MR 2, 5, 2, , 2, 2, MR 2  8  5 , 13, R, 2, I 0  MR 2  M    MR 2 , , MR 2,  MR , 5, 2, 5, 4, 20, 20,  , , , 2, , Q11 – Q30. carry two marks each, , Q11., , Let  x, y  denote the coordinates in a rectangular Cartesian coordinate system C . Let, ,  x, y, , denote the coordinates in another coordinate system C  defined by, , x  2 x  3 y, y   3 x  4 y, The area element in C  , is, (a), , 1, dxdy, 17, , (b) 12dxdy, , (c) dxdy, , (d) xdxdy, , Ans. : (a), Solution: C  x, y   C   x, y , , dxdy  Jdxdy, H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016, Phone: 011-26865455/+91-9871145498, Website: www.physicsbyfiziks.com | Email: [email protected], , 5

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fiziks, Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics, , Ans. : (a), Solution: y, , dy, dy, y 2 3x 2, , c,  3x  0  y,  3x   ydy   3 xdx   0 , 2, 2, dx, dx, , 02 3 1, 3, c c, Finding the value of c , at x  1, y  0 ;  , 2, 2, 2, 2, , , , y 2 3x 2 3, y2, x2,  x2  1  2 , ,   y 2  3x 2  3 , 3, 2, 2, 2, 1, , y2, ,  3, , 2, , 1, , Which is equation of an ellipse., Q14., , In the figure below, point A is the object and point B is the image formed by the lens., Let l1 , l2 and l3 denote the optical path lengths of the three rays 1, 2 and 3 , respectively., Identify the correct statement., 1, 2, , A, , B, , 3, , (a) l1  l2  l3, , (b) l1  l2  l3, , (c) l1  l3  l2, , (d) l1  l3  l2, , Ans. : (a), Solution: Fermat’s principle: A ray of light in traveling between two points requires either a, minimum or a maximum time., Q15., , A particle initially at the origin in an inertial frame S , has a constant velocity Viˆ . Frame, S  is rotating about the z - axis with angular velocity  (anticlockwise). The coordinate, , axes of S  coincide with those of S at t  0 . The velocity of the particle Vx, Vy  in the, S  frame, at t , ,  V, , (a)  , , V ,  2, , , , is, 2, (b)  V , V , ,  V, , (c) , , V ,  2, , ,  3V , , , V , (d) ,  2, , , H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016, Phone: 011-26865455/+91-9871145498, Website: www.physicsbyfiziks.com | Email: [email protected], , 7

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fiziks, Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics, , Ans. : (d), , A, , Solution: Y1  A  0  A0  A0  A, , Y2  A  A  AA  AA  0, , Y1, , Y, , Y2, , Thus Y  A  0  A0  A0  A, Q17., , The total charge contained within the cube (see figure), in which the electric field is given, , by E  K 4 x 2iˆ  3 yjˆ , where  0 is the permittivity of free space, is, , , , , , y, , 01 0, , 10 0, , 0 0 0, , x, ,  001, z, (a) 7 K  0, , (b) 5K  0, , (c) 33K  0, , (d) Zero, , Ans. : (a), , ,  , Solution: (i) x  1, d a  dydzxˆ , E  d a  4 Kx 2 dydz  4 Kdydz so, ,  , (ii) x  0, d a  dydzxˆ , E  d a  4 Kx 2 dydz  0 so, , ,  , (v) z  1, d a  dxdyzˆ, E  d a  0 so, , 0, , 0, ,  ,  E da  0 ., , ,  , (iii) y  1, d a  dxdzyˆ , E  d a  3Kydxdz  3Kdxdz so, , ,  , (iv) y  0, d a  dxdzyˆ , E  d a  3Kydxdz  0 so, ,  , 1, 1,  E  d a  4 K  dy  dz  4 K ., ,  ,  E da  0 ., ,  , 1, 1, E, , d, a, , 3, K, dx, ,   dz  3K ., 0, , 0, ,  ,  E da  0 ., , ,  , (vi) z  0, d a  dxdyzˆ, E  d a  0 so, , , , , ,  E da  0 ., ,  , Thus Qenc   0  E  d a  4 K  0  0  3K  0  0  0  0  7 K  0, , H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016, Phone: 011-26865455/+91-9871145498, Website: www.physicsbyfiziks.com | Email: [email protected], , 9

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fiziks, Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics, , Q18., , Four charges are placed very closed to each other, as shown. The separation between the, two charges on the y -axis is a . The separation between the two charges on the x -axis, is also a . The leading order (non-vanishing) form of the electrostatic potential, at point, , P , at a distance r from the origin  r  a  , is, , P, , r, q, , a, , 600, q, , (a), , qa, 4 0 2r 2, , (c), , 1 qa, 4 0 r 2, , 1, , , , , , , , O q, q, a, , 3 1, , (b), , 1 2qa, 4 0 r 2, , , , (d), , 1 qa, 1 3, 4 0 r 2, , 5 1, , , , , , Ans. : (a), Solution: Qmono  q  q  q  q  0, , , a, p  q, 2, , Vdip, , ,  a   a  a , yˆ   q   xˆ   q   yˆ   q  xˆ   qa   xˆ  yˆ , ,  2   2  2 , , 1 p.rˆ, 1 qa, , ,   xˆ.rˆ  yˆ.rˆ , 2, 4 0 r, 4 0 r 2, ,  Vdip ,  Vdip , , 1 qa, 1 qa  1, 3,  cos 600  sin 600  , , , , 2 , 2 , 4 0 r, 4 0 r  2 2 , 1, , qa, 4 0 2r 2, , , , , , 3 1, , H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016, Phone: 011-26865455/+91-9871145498, Website: www.physicsbyfiziks.com | Email: [email protected], , 10

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fiziks, Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics, , Q19., , At t  0, N 0 number of a radioactive nuclei A start decaying into B with a decay, constant a . The daughter nuclei B decay into nuclei C with a decay constant b . Then,, the number of nuclei B at small time t (to the leading order) is, (a) a N 0t, , (b)  a  b  N 0t, , (c)  a  b  N 0t, , (d) b N 0 t, , Ans. : (a), a, b, A ,  B , C, , Solution:, NB , , N 0 a, N, N, e  at  e  bt   0 a 1  a t  1  bt   0 a  b  a  t, b  a, b  a, b  a, , N B  N 0 a t, Q20., , , The electric field of an electromagnetic wave has the form E  E0 cos t  kz  iˆ . At, , t  0 , a test particle of charge q is at z  0 , and has velocity v  0.5ckˆ , where c is the, , speed of light. The total instantaneous force on the particle is, (a), , qE0 ˆ, i, 2, , (b), , , , qE0 ˆ ˆ, ij, 2, , , , (c), , , , qE0 ˆ ˆ, i k, 2, , , , (d) Zero, , Ans. : (a), ,  E, , Solution: E  E0 cos t  kz  iˆ and corresponding B  0 cos t  kz  ˆj, c, ,  , , Force on charge particle q is F  qE  q v  B, , , , , , , E, c, , F  qE0 cos t  kz  iˆ  q  kˆ  0 cos t  kz  ˆj , c, 2, , , qE, qE, F  qE0 cos t  kz  iˆ  0 cos t  kz  iˆ  0 cos t  kz  iˆ, 2, 2, Thus the total instantaneous force on the particle is, , qE0 ˆ, i., 2, , H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016, Phone: 011-26865455/+91-9871145498, Website: www.physicsbyfiziks.com | Email: [email protected], , 11

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fiziks, Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics, , Q21., , The rms velocity of molecules of oxygen gas is given by v at some temperature T . The, molecules of another gas have the same rms velocity at temperature, , T, . The second gas, 16, , is, (a) Hydrogen, , (b) Helium, , (c) Nitrogen, , (d) Neon, , Ans. : (a), Solution: vlinear , , 3k BT, 3RT, , M is molecular mass., , m, M, , For oxygen, molecule, M  2  16  32, 2, 2, vrms,  oxygen   vrms,  unknown gas  , , Q22., , 3RT 3R  T , ,    M  2 i.e., Hydrogen gas., 32, M  16 , , A system undergoes a thermodynamic transformation from state S1 to state S2 via two, different paths 1 and 2 . The heat absorbed and work done along path 1 are 50 J and, , 30 J , respectively. If the heat absorbed along path 2 is 30 J , the work done along path, 2 is, (b) 10 J, , (a) Zero, , (c) 20 J, , (d) 30 J, , Ans. : (b), Solution: QI  50 J , WI  30 J, , I, , QII  30 J , WII  ?, S1, , Q  U  W  U  Q  W, , S2, , U I  U II  QI  WI  QII  WII, , II, ,  50  30  30  WII  WII  10 J, Q23., , The condition for maxima in the interference of two waves, Ae, , k, i 0,  2, , , , , , , 3x  y  t , , , and Ae, , k, , i  0  x  y   t ,  2, , , is given in terms of the wavelength  and m , an integer, by, , , (c) , , (a), ,   , 2  x  1  2  y  m, , 3  2 x  1  2 y  2m, 3, , , (d) , , (b), ,   , 2  x  1  2  y   2m  1 , , 3  2 x  1  2 y  2m, 3, , H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016, Phone: 011-26865455/+91-9871145498, Website: www.physicsbyfiziks.com | Email: [email protected], , 12

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fiziks, Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics, , Ans. : (a), Solution: 1 , , k0, 2, , , , k0,  x  y   t, 2, , 2 , ,   1  2 , , k0 , k, 3 x  y  2 x  2 y   0 , , 2, 2 , , For maxima:, ,   2m, , k0 , 2 , , , Q24., , , , 3 x  y  t ,, , , ,  , , , ,  , , , , 3  2 x  1 2 y, , , , , 3  2 x  1  2 y   2m, , ,  , , , , 3  2 x  1 2 y , , 4m,  2m, 2 / , , A semiconductor pn junction at thermal equilibrium has the space charge density   x , profile as shown in the figure. The figure that best depicts the variation of the electric, field E with x is ( W denotes the width of the depletion layer), , , 0, , W, , 2, , p-type, , E, , (a), , n-type, , (b), , 0, W, , 2, , W, 2, , W, 2, , x, , 0, , W, 2, , x, , E, , 0, , W, , 2, , E, , W, , 2, , E, , (c), , x, , W, 2, , x, , (d), , 0, , W, , 2, , W, 2, , x, , H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016, Phone: 011-26865455/+91-9871145498, Website: www.physicsbyfiziks.com | Email: [email protected], , 13

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fiziks, Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics, ,  n x ,  n x , g  x   a0   an cos ,   bn sin , ,  L ,  L ,  Fourier series of a function t   1/ 2,1/ 2, , , n t, n t , , g  t   a0    an cos, , a, ,  bn sin,  an cos t  bn sin nt , , 0, , T /2, T /2, n 1 , n 1, , where an , , 2 T /2, 2n t, g  t  cos, dt ,, , T T / 2, T, , bn , , 2 T /2, 2n t, g  t  sin, dt, , T T / 2, T, , bn , , 2 T / 2 f1  t , 2n t, 0, sin, , , T, /, 2, T, , T, even, , , For f1  t   4t 2  3, , an , , 2 T / 2 f1  t , 2n t, 0;, cos, , , T, /, 2, T, , T, even, , even, , odd, , For f 2  t   6t 3  7t (odd function), an , , 2 T / 2 f2 t , 2n t, cos, dt  0 ;, , , T, /, 2, T, , T, odd, , , bn , , even, ,  For f1  t  : an  0 and bn  0, Q29., , 2 T / 2 f2 t , 2n t, sin, dt  0, , , T, /, 2, T, , T, odd, , odd, , and, , For f 2  t  : an  0 and bn  0, , A thin circular disc lying in the xy -plane has a surface mass density  , given by,  , r2 , , 1, , if r  R, ,   r    0  R 2 , , 0, if r  R, , where r is the distance from its center. Its moment of inertia about the z -axis, passing, through its center is, (a), ,  0 R4, , (b), , 4, ,  0 R 4, 6, , (c)  0 R 4, , (d) 2 0 R 4, , Ans. : (b),  2, , Solution: I z  , ,  r dm, 2, , where   r  , , 0 0, ,  2, , dm,  dm   rdrd, rdrd, , ,  R 4 R 4   0 R 4, r2 , I z    r  rdrd  2 0  r 1  2  dr  2 0 , , , 6 , 6,  R ,  4, 0 0, 0, R, , 2, , 3, , H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016, Phone: 011-26865455/+91-9871145498, Website: www.physicsbyfiziks.com | Email: [email protected], , 16

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fiziks, Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics, SECTION – B, MULTIPLE SELECT QUESTIONS (MSQ), Q31 – Q40 Carry Two Marks Each, , Q31., , A gaseous system, enclosed in an adiabatic container, is in equilibrium at pressure P1 and, volume V1 . Work is done on the system in a quasi-static manner due to which the, pressure and volume change to P2 and V2 , respectively, in the final equilibrium state. At, every instant, the pressure and volume obey the condition PV   C , where  , , CP, and, CV, , C is a constant. If the work done is zero, then identify the correct statement(s), , (a) PV, 2 2  PV, 1 1, , (b) PV, 2 2   PV, 1 1, , (c) PV, 2 2     1 PV, 1 1, , (d) PV, 2 2     1 PV, 1 1, , Ans. : (a), Solution:, ,  P1 ,V1 , , adiabatic, process, ,  P2 ,V2 , , Q  0 [ the process is adiabatic], W  0 [given], Q  U  W  U  0  T  0, i.e. the process is isothermal too, ,  PV, 1 1  PV, 2 2 must satisfy., Alternate,, Since the process is adiabatic [i.e Q  0 ], And work done, W  0 [given], We know, for adiabatic process, Work , 0, , 1,  PV  PV ,  1 1 1 2 2, , 1,  PV  PV   PV, 1 1  PV, 2 2,  1 1 1 2 2, , H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016, Phone: 011-26865455/+91-9871145498, Website: www.physicsbyfiziks.com | Email: [email protected], , 18

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fiziks, Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics, , Q32., , An isolated ideal gas is kept at pressure P1 and volume V1 . The gas undergoes free, expansion and attains a pressure P2 and volume V2 . Identify the correct statements(s), , CP ,  , , CV , , , (b) PV, 1 1  PV, 2 2, , (a) This is an adiabatic process, , , (c) PV, 1 1  PV, 2 2, , (d) This is an isobaric process, , Ans. : (a), (b), Solution:, , Vaccum, ,  P1 ,V1 , ,  P2 ,V2 , , Free expansion, Free expansion of ideal gas is irreversible adiabatic process. (Since the entropy increase), Work done in free expansion is zero. (Ideal gas), , W, , 1,  PV  PV   0  PV, 1 1  PV, 2 2,  1 2 2 1 1, , is true for a reversible adiabatic process but since here the process is irreversible, it does, not apply here., , H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016, Phone: 011-26865455/+91-9871145498, Website: www.physicsbyfiziks.com | Email: [email protected], , 19

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fiziks, Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics, , Q33., , A beam of light travelling horizontally consists of an unpolarized component with, intensity I 0 and a polarized component with intensity I p . The plane of polarization is, oriented at an angle  with respect to the vertical. The figure shows the total intensity, , I total after the light passes through a polarizer as a function of the angle  , that the axis, of the polarizer makes with respect to the vertical. Identify the correct statements(s), I total (W / m 2 ), 25, 20, 15, 10, 5, 0, , 50, , 100, , 150, , 200, , (a)   1250, , (b) I p  5 W / m 2, , (c) I 0  17.5 W / m 2, , (d) I 0  10W / m 2 ; I p  20 W / m 2, ,  (degrees), , Ans. : (d), Solution: I total , , I max , , I, I,  I p cos 2  , I min    5W / m 2  I   10W / m 2, 2, 2, , I,  I p  25  I p  20W / m 2, 2, , H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016, Phone: 011-26865455/+91-9871145498, Website: www.physicsbyfiziks.com | Email: [email protected], , 20

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fiziks, Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics, , Q34., , Consider the following differential equation that describes the oscillations of a physical, system:, , d2y, dy,  2    y  0, dt, dt, If  and  are held fixed, and  is increased, then,, (a) The frequency of oscillations increases, (b) The oscillations decay faster, (c) The frequency of oscillations decreases, (d) The oscillations decay slower, Ans. : (a), Solution:, , d 2 y  dy , ,  y0, dt 2  dt , , 2 , , , , ,  , , 0 , 2, , , , (a)   02   2 , ,  2, 1, 1,  2 , 4   2   , 4   2, 2,  4, 2, , So, option (a) is correct and option (c) is wrong, (b) A  A0 e  t, , , , ,  constant, 2, , So, option (b) and (d) are is wrong., Q35., , For the given circuit, identify the correct statement(s), , 1k, , (a) I 0  1 mA, (b) V0  3V, , 1k, , (c) If RL is doubled, I 0 will change to 0.5 mA, , , V0, , , , (d) If RL is doubled, V0 will change to 6V, 1 k, , 1 k, , 1V, , I0, , RL  1.5 k , , H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016, Phone: 011-26865455/+91-9871145498, Website: www.physicsbyfiziks.com | Email: [email protected], , 21

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fiziks, Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics, , , , TH  150 K, , QH, 2,  1.5 , 3, QL, , QH, , Change in entropy of engine in one cycle, , Sengine , , QH  QL  750  500 , , , ,  55  0, TH, TL, 150, 100, , engine, , work, , i.e change in entropy of engine in 0.5 seconds is zero., Change in entropy of engine and hot bath, ,  Sengine  S hot bath  0 , Q37., , QL, ,  QH   5 J / K, , TL  100 K, , TH, , , , A time independent conservative force F has the form, F  3 yiˆ  f  x, y  ˆj . Its, magnitude at x  y  0 is 8 . The allowed form(s) of f  x, y  is (are), (a) 3 x  8, , (b) 2 x  8  y  1, , (c) 3x  8e  y, , 2, , (d) 2 x  8cos y, , 2, , Ans. : (a), (c), , , Solution: For conservative force F ;, , xˆ,  ,   F   / x, 3y, , yˆ,  / y, f  x, y , , zˆ,  / z  0, 0, , f,  f, ,  xˆ  0  0   yˆ  0  0   zˆ   3   0 , 3, x,  x, , , H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016, Phone: 011-26865455/+91-9871145498, Website: www.physicsbyfiziks.com | Email: [email protected], , 23

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fiziks, Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics, , Q38., , The figure shows the cross-section of hollow cylindrical tank, 2.2 m in diameter which is, half filled with water (refractive index of 1.33). The space above the water is filled with a, gas of unknown refractive index. A small laser moves along the bottom surface and aims, a light beam towards the center (see figure). When the laser moves a distance of, , S  1.09 m or beyond from the lowest point in the water, no light enters the gas. Identify, the correct statement(s) (speed of light is 3  108 m / s )., , Gas, , Water surface, , S, , Laser, , (a) The refractive index of the gas is 1.05, (b) The time taken for the light beam to travel from the laser to the rim of the tank when, S  1.09 m is 8.9 ns, (c) The time taken for the light beam to travel from the laser to the rim of the tank when, S  1.09 m is 9.7 ns, (d) The critical angle for the water-gas interface is 56.77 0, Ans. : (b), (c), (d), Solution:  c , , S 1.09, ,  0.991 rad, R 1.1, ,  c  56.80  56.770, , R, , c, , So, option (d) is correct., (a) Apply Snell’s law, n sin  c  ng sin 90, , 1.09m, , 0, , , , , , ng  1.33  sin 56.77 0  1.11 or ng  1.33  sin  56.80   1.11, Option (a) is wrong., H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016, Phone: 011-26865455/+91-9871145498, Website: www.physicsbyfiziks.com | Email: [email protected], , 24

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fiziks, Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics, , (b) If S  1.09m , then time taken by laser light to reach the rim will be, t, , R R R, 1.1, 11 2.44, 1.33  1.11 , ns  8.94ns,      a  , 8 , v va c, 3  10, 3, , Option (b) is correct., (c) If S  1.09m , then time taken by laser light to reach the rim will be, t, , 2R, , , , , , 2 1.1, 22  1.33, ns  t  9.75ns, 1.33 , 8, 3  10, 3, , Option (c) is correct., Q39., , Identify the correct statement(s) regarding nuclei, (a) The uncertainly in the momentum of a proton in a nucleus is roughly 105 times the, uncertainly in the momentum of the electron in the ground state of Hydrogen atom, (b) The volume of a nucleus grows linearly with the number of nucleons in it, (c) The energy of  rays due to de-excitation of nucleus can be of the order of MeV, (d), , 56, , Fe is the most stable nucleus, , Ans. : (a), (b), (c), (d), Solution: (a) p p R p  pe Re , p p, pe, , , , , 2, , Re 1010 m, ,  105, R p 1015 m, , 4, 4, (b) V   R 3   R03 A, 3, 3, (c) The energy of different energy levels and sublevels is of the order of MeV, (d) Binding energy per nuclei is maximum for, Q40., , 56, , Fe, , A particle of mass m is in an infinite square well potential of length L . It is in a, superposed, ,   x , , state, , of, , the, , first, , two, , energy, , eigenstates,, , as, , given, , by, , 1, 2,  n 1  x    n  2  x  . Identify the correct statement(s). h is Planck’s, 3, 3, , constant., (a) p  0, , (b) p , , 3h, 2L, , (c) E , , 3h 2, 8mL2, , (d) x  0, , H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016, Phone: 011-26865455/+91-9871145498, Website: www.physicsbyfiziks.com | Email: [email protected], , 25

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fiziks, Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics, , Q43., , A small conducting square loop of side l is placed inside a concentric large conducting, square loop of side L  L  l  . The value of mutual inductance of the system is expressed, as, , n 0 l 2, . The value of n is ______ (Round off to two decimal places), L, , Ans. : 2.828, Solution: Assume smaller loop (loop-1) is at the centre of bigger loop(loop-2). Let current I is, flowing in bigger loop., Magnetic field at the centre of a square loop is B  4 ,  B  4, , 0 I, 4 0 I, sin 450  sin  450  , 4 L / 2,  2L, , , , , , Flux through loop-1 is 1  MI  B  l 2  MI , M , , Q44., , 0 I,  sin  2  sin 1 , 4 d, , 4 0 I,  l 2  MI,  2L, ,  l2, 4 0l 2,  2.828 0, L, 2 L, , Consider N1 number of ideal gas particles enclosed in a volume V1 . If the volume is, changed to V2 and the number of particles is reduced by half, the mean free path becomes, four times of its initial value. The ratio, , V1, is ________ (Round off to one decimal place)., V2, , Ans. : 0.5, Solution:  , , 1, V, V, ,     V  N, 2, 2, N, 2n d, 2 N d, , V1 1 N1, , , V2 2 N 2, , 1 N1, N,  41   1 ,  2 , ,  0.5, , Q45. A particle is moving with a velocity 0.8cjˆ ( c is the speed of light) in an inertial frame S1 ., Frame S2 is moving with a velocity 0.8ciˆ with respect to S1 . Let E1 and E2 be the, respective energies of the particle in the two frames. Then,, , E2, is ________ (Round off, E1, , to two decimal places)., H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016, Phone: 011-26865455/+91-9871145498, Website: www.physicsbyfiziks.com | Email: [email protected], , 28

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fiziks, Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics, , Ans. : 1.66, Solution: From S1 frame the velocity of particle is 0.8cjˆ so energy is, m0 c 2, 1, , v p , S1, , m0 c 2, 10,  m0 c 2  E1  1.66m0 c 2, 1  0.64 6, , , , 2, , c2, , ux  0 , uy  0.8c , v  0.8c, v2, u  v, c 2  0.8c 1  0.64  0.8c  0.6  0.48c, ux  x,  0.8c , u y , u v, u v, 1  x2, 1  x2, c, c, uy 1, ,  0.8    0.48, , u p , S2 , E2 , , Q46., , 2, , m0 c 2, 1  0.87, , , , 2, , c  0.64  0.23c  0.87c  0.93c, , m0 c 2,  2.78m0 c 2, 0.36, , , , E2 2.78, ,  1.66, E1 1.67, , At some temperature T , two metals A and B , have Fermi energies A and B ,, respectively. The free electron density of A is 64 times that of B . The ratio, , A, is ____., B, , Ans. : 16, Solution: The Fermi energy for metal A and B is written as, A , , , , 2, 3 2 n A, 2m, , , , 2/3, , and, , , , , ,   64 , ,  16, , B , , 2, 3 2 nB, 2m, , 2/3, , The ratio of the Fermi energy is, A  nA ,  , B  nB , , 2/3, ,  64nB , , ,  nB , , 2/3, 2/3, , 0, , Q47., , 0, , A crystal has monoclinic structure, with lattice parameters, a  5.14 A , b  5.20 A ,, o, , c  5.30 A and angle   990 . It undergoes a phase transition to tetragonal structure with, 0, , 0, , lattice parameters, a  5.09 A and c  5.27 A . The fractional change in the volume, , V, V, , of the crystal due to this transition is __________ (Round off to two decimal places)., , H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016, Phone: 011-26865455/+91-9871145498, Website: www.physicsbyfiziks.com | Email: [email protected], , 29

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fiziks, Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics, , Ans. : 0.024, Solution: Volume of the monoclinic unit cell is, Vmono  abc sin     5.14  5.20  5.30  sin  99  1030 m3  139.91 1030 m3, , Volume of the tetragonal unit cell is, Vtetr  a 2 c  5.09  5.09  5.27 1030 m3  136.54 1030 m3, The fractional change in the volume, , V, is, V, , 139.91  136.54, V, ,  0.024, V, 139.91, , Q48., , A laser beam shines along a block of transparent material of length 2.5 m . Part of the, beam goes to the detector D1 while the other part travels through the block and then hits, the detector D2 . The time delay between the arrivals of the two light beams is inferred to, be 6.25 ns . The speed of light c  3  108 m / s . The refractive index of the block, , D2, , Laser, , D1, , is____________ (Round off to two decimal places)., , Ans. : 1.73 to 1.77, Solution: Time delay t , ,    1 , Q49., ,    1 x, Extra path travelled,  t , c, c, , tc 6.25 109  3 108, ,  0.75, x, 2.5, ,    1.75, , An ideal blackbody at temperature T , emits radiation of energy density u . The, corresponding value for a material at temperature, , T, u, is, ., 2, 256, , Its emissivity is______________(Round off to three decimal places)., , H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016, Phone: 011-26865455/+91-9871145498, Website: www.physicsbyfiziks.com | Email: [email protected], , 30

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fiziks, Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics, , Ans. : 0.063, Solution: We know u   T 4 , where   emissivity, For black body   1 ., u   T 4, , (1), 4, , For other material, , u, u, T, T ,  ,     , 256, 256, 16, 2, , 4, , , , u,   T 4 ………(2), 16, , From (1) and (2), , , Q50., , 1,  0.0625  0.063 (Rounded to three decimal places), 16, , A particle with positive charge 103 C and mass 0.2 kg is thrown upwards from the, ground at an angle 450 with the horizontal with a speed of 5 m / s . The projectile moves, through a horizontal electric field of 10V / m , which is in the same direction as the, horizontal component of the initial velocity of the particle. The acceleration due to, gravity is 10 m / s 2 . The range is ___________ m . (Round off to three decimal places)., , Ans. : 2.51, 2u, 1, Solution:  y  u y t  gt 2 , Time of flight T  y, g, 2, ,  x  uxt , R  ux, , 1 qE 2, 1 qE 2, t , Range R  u xT , T, 2 m, 2 m, , 1  qE   2u y ,  , , , 2  m  g , g, , 2u y, , where u x , , 2, , 5, 5, qE 103  10 1, , uy , , g  10,, , , m, 0.2, 20, 2, 2, 2, , 5 25 1 1  1  25 , 25 1, R,   , , ,  2.5  0.0125  2.512, , 2 2 10 2  20   2  10  10 80, , H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016, Phone: 011-26865455/+91-9871145498, Website: www.physicsbyfiziks.com | Email: [email protected], , 31

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fiziks, Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics, Q51 – Q60 Carry Two Marks Each., , Q51., , Consider a hemispherical glass lens (refractive index is 1.5 ) having radius of curvature, R  12 cm for the curved surface. An incoming ray, parallel to the optical axis, is incident, , on the curved surface at a height h  1cm above the optical axis, as shown in the figure., The distance d (from the flat surface of the lens) at which the ray crosses the optical axis, is ___________ cm (Round off to two decimal places)., , h, d, R, , Ans.: 16, Solution: For the curved surface, , 2, v, , , , 1, u, , , , 2  1, , , , R, , 1.5 1.0 1.5  1.0, , , v, , 12, , d, R, ,  v  36cm, , For the plane surface, 1.0 1.5 1.0  1.5, , , d 24, ,  d  16cm, , d, , I2, , I1, , 36cm, 24cm, , H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016, Phone: 011-26865455/+91-9871145498, Website: www.physicsbyfiziks.com | Email: [email protected], , 32

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fiziks, Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics, , Q52., , Twenty non-interacting spin, , 1, particles are trapped in a three-dimensional simple, 2, , harmonic oscillator potential of frequency  . The ground state energy of the system, in, units of  , is _________., Ans. : 60, 3, 5, 7, Solution: E   2 gi Ei   2 1     3  2      2  6      3  15  42    60, 2, 2, 2, i, Q53., , A thin film of alcohol is spread over a surface. When light from a tunable source is, incident normally, the intensity of reflected light at the detector is maximum for, ,   640 nm and minimum for   512 nm . Taking the refractive index of alcohol to be, 1.36 for both the given wavelengths, the minimum thickness of the film would be, , __________ nm (Round off to two decimal places)., Ans. : 470.58, Solution:, , air  a  1, alcohol  al  1.36 , surface   L , Condition of maxima, 2 L t  n1, ,    640nm , , (1), , Condition of minima, 2  L t   2m  1 2, , n, 1, , m , 2, , , , , (2), ,  2  512nm , , 2   2, 2 512 4, 2n, , ,  , 1 640 5, 2m  1 2  2   1, , nm2, , From equation (i), 2  1.36  t  2  640nm   t , ,  2  640  nm, 2  1.36, ,  t  470.58nm, , H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016, Phone: 011-26865455/+91-9871145498, Website: www.physicsbyfiziks.com | Email: [email protected], , 33

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fiziks, Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics, , Q54., , For the Boolean expression Y  ABC  ABC  ABC  ABC , the number of combinations, for which the output Y  1 is __________., , Ans. : 4, Solution: Y  ABC  ABC  ABC  ABC,  Y   AB  AB  C   AB  AB  C,  Y  DC  DC, , A, , B, , C, , D  A B, , Y  DC, , 0, , 0, , 0, , 0, , 1, , 0, , 0, , 1, , 0, , 0, , 0, , 1, , 0, , 1, , 0, , 0, , 1, , 1, , 1, , 1, , 1, , 0, , 0, , 1, , 0, , 1, , 0, , 1, , 1, , 1, , 1, , 1, , 0, , 0, , 1, , 1, , 1, , 1, , 0, , 0, , where D  AB  AB, , Y  DC, , Q55., , An RC circuit is connected to two dc power supplies, as shown in the figure. With, switch S open, the capacitor is fully charged. S is then closed at time t  0 . The voltage, across the capacitor at t  2.4 milliseconds is __________ V (Round off to one decimal, place)., , 20 V,  , 4k , , 10  F, S, , 6k , ,  , 10V, , H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016, Phone: 011-26865455/+91-9871145498, Website: www.physicsbyfiziks.com | Email: [email protected], , 34

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fiziks, Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics, , Ans. : 18.824, Solution: Initially capacitor will be charged upto 20 V with polarity shown in figure., 20 V, When Switch S is closed, then at t  2.4 milliseconds,  , vC  V 1  e, , 2.410, , , 6103 10103,   30 1  e, , 3, ,  t / RC, ,  vC  30 1  e, , 0.04, , , , , , , 4k , ,   30 1  0.9607   1.176 Volts, , 10  F,  , S, , So net voltage across capacitor is,   20  1.176  Volts  18.824 Volts, , 6k , ,  , 10V, Q56., , A current I is uniformly distributed across a long straight nonmagnetic wire (  r  1 ) of, circular cross-section with radius a . Two points P and Q are at distances, , a, and 9a ,, 3, , respectively, from the axis of the wire. The ratio of the magnetic fields at points P and, Q is __________., , Ans. : 3, Solution: Magnetic field inside the wire is B , Magnetic field outside the wire is B , , Thus, Q57., ,  I  a / 3, 0 Ir,  BP  0, 2, 2 a, 2 a 2, , 0 I, 0 I,  BQ , 2 r, 2  9a , , BP 0 I  a / 3 2  9a , , , 3, BQ, 0 I, 2 a 2, , A particle A of mass m is moving with a velocity viˆ , and collides elastically with a, particle B , of mass 2m , B is initially at rest. After collision, A moves with a velocity, v A ˆj . If vB is the final speed of B , then v A2  kvB2 . The value of k is __________., , H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016, Phone: 011-26865455/+91-9871145498, Website: www.physicsbyfiziks.com | Email: [email protected], , 35

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fiziks, Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics, , Ans. : 1, Solution: From conservation of momentum in x -direction, , After Collision, , Before, , mv  2mvB cos  ………(1), , Collision, , From conservation of momentum in y -direction, , A m, v A ˆj, , 0  mv A  2mvB cos   mv A  2mvB sin  ……..(2), Squaring and adding both equation, , m, , v 2  v A2  4vB2 ……. (3), , A, , 2m, , v, , B, , vB cos , , , , From conservation of kinetic energy, vB sin , , 1 2 1 2 1, mv  mv A  2mvB2  v 2  v A2  2vB2 ……. (4), 2, 2, 2, , 2m, vB, , Solving simultaneously equation (3) and (4), v A2  vB2  k  1, 0, , Q58., , In an X - ray diffraction experiment with Cu crystals having lattice parameter 3.61A ,, X - rays of wavelength of 0.090 nm are incident on the family of planes 1 1 0 . The, , highest order present in the diffraction pattern is ___________., Ans. : 5, Solution: Bragg’s law is; 2d sin    n , For the highest order sin    1 ; n , , 2a, h  k2  l2, 2, , 2a, ,  h2  k 2  l 2, , sin    n, , , , 2  3.61 1010, 0.9  1010  2, ,  5.67, , The maximum order is n  5, , H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016, Phone: 011-26865455/+91-9871145498, Website: www.physicsbyfiziks.com | Email: [email protected], , 36

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fiziks, Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics, , Q59., , A parallel plate capacitor having plate area of 50 cm 2 and separation of 0.1mm is, completely filled with a dielectric (dielectric constant K  10 ). The capacitor is, connected to a 10 k resistance and an alternating voltage v  10sin 100  t  , as shown, in the figure. The switch S is initially open and then closed at t  0 . The ratio of the, displacement current in the capacitor, to the current in the resistance, at time t , , 2, , , , seconds is ____________ (Round off to three decimal places)., , 10 k , , S, , v  10 sin 100 t , , Ans. : 0.038, Solution: Displacement current density J d  , , E, v,   10sin 100  t  ,   0 r    10 0 , , t, t  d , t  0.1 103 , , J d  106  0 100   cos 100  t , , Displacement current I d  J d  A  106  0 100   cos 100  t    50  104 m 2  Amp, At time t , , 2, , , , I d  5 105  0 cos  200  Amp  I d  5  105  3.14  8.86 1012  0.939 Amp, ,  I d  130.76 107 Amp, , Current through resistor R is, IR , , 10sin 100  t , 10sin  200 , v, , Amp , Amp, Amp  3.42  104 Amp, 3, 3, 10 10, 10  10, R, , I d 130.76  107 Amp, Thus, ,  0.038, 3.42 104 Amp, IR, , H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016, Phone: 011-26865455/+91-9871145498, Website: www.physicsbyfiziks.com | Email: [email protected], , 37

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fiziks, Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics, , Q60., , The wavelength of characteristic K X - ray photons from Mo (atomic number 42 ) is, 0, , _____________ A (Round off to one decimal place)., (Speed of light is 3 108 m / s ; Rydberg constant R  1.09 107 / m ), Ans. : 0.73, Solution:, , 1, , k, , k , , 3R,  Z  b, 4, , , , 4, 3R  Z  b , , 2, , , , 4, 3  1.09  10   42  1, 7, , 2, ,  0.73 1010 m, ,  k  0.73 A0, , ------------------------------------------------------------------------------------------------------------, , H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016, Phone: 011-26865455/+91-9871145498, Website: www.physicsbyfiziks.com | Email: [email protected], , 38