Page 1 : 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:
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Page 2 : 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 12 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:
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Page 3 : fiziks, Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics, , n 1 , Q6., , n 472 610 109, n 1 0 011 n 1 011, , 2t, 2 1 3 102, , 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:
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Page 4 : 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 2.56 103 ;, , d II 3.05 107, , d III 1.22 107 ;, , d IV 6.1 103, , 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:
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Page 5 : 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, dxdy, 17, , (b) 12dxdy, , (c) dxdy, , (d) xdxdy, , Ans. : (a), Solution: C x, y C x, y , , dxdy Jdxdy, 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:
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Page 7 : 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:
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Page 9 : 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:
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Page 10 : 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:
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Page 11 : 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, ij, 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:
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Page 12 : 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:
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Page 13 : 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:
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Page 16 : 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 nt , , 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:
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Page 18 : 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:
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Page 19 : 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:
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Page 20 : 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:
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Page 21 : 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 , , y0, 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:
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Page 23 : 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 , , , , 55 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:
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Page 24 : 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:
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Page 25 : 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 1010 m, , 105, R p 1015 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:
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Page 28 : 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, 41 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:
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Page 29 : 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, , ux 0 , uy 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:
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Page 30 : 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 1030 m3 139.91 1030 m3, , Volume of the tetragonal unit cell is, Vtetr a 2 c 5.09 5.09 5.27 1030 m3 136.54 1030 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 6.25 109 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:
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Page 31 : 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 103 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 103 10 1, , uy , , g 10,, , , m, 0.2, 20, 2, 2, 2, , 5 25 1 1 1 25 , 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:
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Page 32 : 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:
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Page 33 : 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 n1, , 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, , nm2, , 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:
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Page 34 : 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 DC, , 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 DC, , 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:
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Page 35 : 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, , 2.410, , , 6103 10103, 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:
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Page 36 : 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 1010, 0.9 1010 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:
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Page 37 : 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 0.1 103 , , J d 106 0 100 cos 100 t , , Displacement current I d J d A 106 0 100 cos 100 t 50 104 m 2 Amp, At time t , , 2, , , , I d 5 105 0 cos 200 Amp I d 5 105 3.14 8.86 1012 0.939 Amp, , I d 130.76 107 Amp, , Current through resistor R is, IR , , 10sin 100 t , 10sin 200 , v, , Amp , Amp, Amp 3.42 104 Amp, 3, 3, 10 10, 10 10, R, , I d 130.76 107 Amp, Thus, , 0.038, 3.42 104 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:
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Page 38 : 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 1010 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:
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