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PG Exam 2017 : Physics (Banaras Hindu University (BHU), Varanasi )

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Brijmohan
Banaras Hindu University (BHU), Varanasi
PhD Condensed matter Physics

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fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES MATEMATICAL PHYSICS SOLUTIONS NET/JRF (JUNE-2011) Q1. The value of the integral dz z 2 e z , where C is an open contour in the complex z -plane as C lm z shown in the figure below, is: (a) (b) e (c) Ans: 5 e e 5 e e (d) 5 e 0,1 C 5 e e 1,0 (c) 1,0 Re z Solution: If we complete the contour, then by Cauchy integral theorem 2 z 2 z 2 z 2 z 2 z z z dzz e dzz e 0 dzz e dzz e z e 2 ze 2e 1 1 1 Q2. 1 C C 1 1 Which of the following matrices is an element of the group SU 2 ? 1 1 (a) 0 1 1 i (b) 3 1 3 1 3 1 i 3 i 2 i (c) 3 1 i Ans: 5 e e (d) 3 2 1 2 1 2 3 2 (b) Solution: SU 2 is a group defined as following: SU 2 clearly (b) hold the property of SU 2 . 1 i 3 , 1 3 2 : , C; and 1 i 3 , 1 3 2 1 . Note: SU 2 has wide applications in electroweak interaction covered in standard model of particle physics. Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 1 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Q3. Let a and b be two distinct three dimensional vectors. Then the component of b that is perpendicular to a is given by a b a a b b b a a b a b (a) (b) (c) (d) a2 b2 a2 b2 Ans: (a) where n is perpendicular to plane containing Solution: a b ab sin n a and b and pointing upwards. b a 2b sin k a a b ab sin a n a a b a b a b sin k b sin k . a2 a2 Q4. b sin k a Let p n x (where n 0,1, 2, ......) be a polynomial of degree n with real coefficients, 4 defined in the interval 2 n 4 . If pn x pm x dx nm , then 2 (a) p0 x (c) p0 x Ans: 1 2 and p1 x 1 and p1 x 2 3 3 x 2 3 3 x 2 (b) p0 x (d) p0 x 1 and p1 x 3 3 x 2 1 2 and p1 x 3 3 x 2 (d) Solution: For n not equal to m kroneker delta become zero. One positive and one negative term can make integral zero. So answer may be (c) or (d). Now take n m 0 so p0 x 1 2 and then integrate. (d) is correct option because it satisfies the equation Check by integration and by orthogonal property of Legendre polynomial also. Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 2 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Which of the following is an analytic function of the complex variable z x iy in the Q5. domain z 2 ? (a) 3 x iy (b) 1 x iy 7 x iy 7 4 (c) 1 x iy 7 x iy 4 Ans: 3 3 (d) x iy 1 1/ 2 (b) Solution: Put z x iy . If z x iy appears in any of the expressions then that expression is non-analytic. For option (d) we have a branch point singularity as the power is 1 which 2 is fractional. Hence only option (b) is analytic. 1 1 1 Consider the matrix M 1 1 1 1 1 1 Q6. A. The eigenvalues of M are (a) 0, 1, 2 Ans: (b) 0, 0, 3 (d) 1, 1, 3 (c) 1, 1, 1 (b) 1 1 1 1 1 Solution: For eigen values 1 0 1 1 1 1 1 2 1 1 1 1 1 1 0 1 1 2 2 1 0 2 2 3 2 2 2 0 3 3 2 0 2 3 0 0, 0, 3 For any n n matrix having all elements unity eigenvalues are 0, 0, 0,..., n . B. The exponential of M simplifies to (I is the 3 3 identity matrix) e3 1 (a) e I 3 M (b) e M I M (c) e M I 33 M (d) e M e 1 M M Ans: M2 2! (a) Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 3 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Solution: For e M let us try to diagonalize matrix M using similarity transformation. 1 x1 0 2 1 For 3 , 1 2 1 x 2 0 1 2 1 x3 0 2 x1 x2 x 0 , x1 2 x 2 x3 0 , x1 x 2 2 x3 0 3 3 x 2 3 x3 0 or x 2 x3 x1 x 2 x3 k . 1 Eigen vector is 1 3 1 where k 1 . 1 For 0 , 1 1 1 x1 0 1 1 1 x 0 x x x 0 1 2 3 2 1 1 1 x3 0 Let k1 1 x1 k1 , x2 k 2 and x3 k1 k 2 . Eigen vector is k 2 1 / 2 1 k 1 k 2 1 where k1 k 2 1 . Let x1 k1 , x2 k 2 and x3 k1 k 2 . Other Eigen vector 1 1/ 2 0 1 where k1 1, k 2 1. 0 1 1 1 2 1 1 1 1 S 1 0 1 S 2 1 1 D S MS , M SDS . 1 1 1 1 1 1 eM 1 0 0 3 e M 1 e 1 M Se D S 1 e D 0 1 0 3 3 0 0 e Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 4 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES NET/JRF (DEC-2011) Q7. An unbiased dice is thrown three times successively. The probability that the numbers of dots on the uppermost surface add up to 16 is (a) Ans: 1 16 1 36 (b) (c) 1 108 (d) 1 216 (b) Solution: We can get sum of dice as 16 in total six ways i.e. three ways (6, 5, 5) and three ways (6, 6, 4). Total number of ways for 3 dice having six faces 6 6 6 6 1 6 6 6 36 Q8. The generating function F x, t Pn x t n for the Legendre polynomials n 0 is F x, t 1 2 xt t 2 (a) 5 / 2 Ans: (b) 3 / 2 2 . The value of P3 1 is (c) 1 (d) 1 (d) Solution: P3 Q9. 1 1 1 1 3 5 x 3 3x P3 1 5 1 3 1 5 3 1 2 2 2 The equation of the plane that is tangent to the surface xyz 8 at the point 1, 2, 4 is (a) x 2 y 4 z 12 (b) 4 x 2 y z 12 (c) x 4 y 2 0 Ans: Pn x (d) x y z 7 (b) Solution: To get a normal at the surface lets take the gradient xy 8i zx 4 xyx yzi j k j 2k 4 j 2k 8i 64 0. 16 4 We want a plane perpendicular to this so: r r0 x 1 i y 2 j z 4 k 8i 4 j 2k 0 4x 2 y z 12 . Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 5 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Q10. A 3 3 matrix M has Tr M 6, Tr M 2 26 and Tr M 3 90 . Which of the following can be a possible set of eigenvalues of M ? 1,1, 4 (a) Ans: (b) 1, 0, 7 (c) 1, 3, 4 (d) 2, 2, 2 (c) Solution: Tr M 2 1 3 4 also Tr M 3 1 3 4 90 . Q11. 2 2 2 3 3 3 Let x1 t and x2 t be two linearly independent solutions of the differential equation d 2x dx dx t dx t 2 f t x 0 and let w t x1 t 2 x2 t 1 . If w 0 1, then w 1 is 2 dt dt dt dt given by (b) e 2 (a) 1 Ans: (d) 1 / e 2 (c) 1 / e (d) Solution: W t is Wronskian of D.E. Pdt W e e 2t W 1 e 2 since P 2 . Q12. for 2n x 2n 1 1 The graph of the function f x 0 for 2n 1 x 2n 2 ~ where n 0,1, 2,...... is shown below. Its Laplace transform f s is f x 1 e s 1 e s (a) (b) s s 1 (c) Ans: 1 s 1 e s (d) 1 s 1 e s 0 1 2 3 4 5 x (c) Solution: L f x e sx 1 f x dx e 0 1 sx 2 1dx e 0 3 sx 3 0dx e sx 1dx ...... 1 2 e sx e sx 1 s 1 3 s 0 e 1 e e 2 s ...... ...... s s s 0 s 2 Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 6 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Since S Q13. 1 1 1 e s e 2 s e 3s ........ 1 e s e 2 s e 3s .... s s 1 1 a where r e s and a 1 S . 1 r s 1 e s The first few terms in the Taylor series expansion of the function f x sin x around x 4 are: 1 (a) 2 2 3 1 1 1 x x x ..... 4 2! 4 3! 4 2 3 1 1 1 (b) 1 x x x ..... 4 2! 4 3! 4 2 3 1 (c) x x ..... 4 3! 4 (d) Ans: 1 x 2 x3 1 x ..... 2! 3! 2 (c) Solution: f x sin x 1 f 2 4 1 f cos 4 2 4 1 f sin 4 2 4 So Taylor s series is given by 2 3 1 1 1 1 x x x ..... 4 2! 4 3! 4 2 Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 7 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES NET/JRF (JUNE-2012) Q14. A vector perpendicular to any vector that lies on the plane defined by x y z 5 , is j (a) i Ans: j k (c) i j k (b) 3 j 5k (d) 2i (c) Solution: Let x y z 5 i x j y k z x y z 5 i j k . Q15. 1 2 3 The eigen values of the matrix A 2 4 6 are 3 6 9 (a) 1, 4, 9 Ans: (b) 0, 7, 7 (c) 0,1,13 (d) 0, 0,14 (d) 1 Solution: For eigenvalues A I 0 2 3 3 4 6 0 6 9 2 1 4 9 36 2 2 9 18 3 12 3 4 0 1 4 9 36 1 4 9 36 9 0 3 14 2 0 2 14 0 0, 0, 14 . Q16. The first few terms in the Laurent series for 1 z 1 z 2 in the region 1 z 2 and around z 1 is (a) 1 z z 2 z3 1 z z 2 .... 1 .... 2 2 4 8 (c) 1 z2 1 1 2 4 1 2 .... 1 2 .... z z z z (b) 1 2 3 z 1 z 1 z .... 1 z (d) 2 z 1 5 z 1 7 z 1 .... 2 3 Ans: Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 8 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Solution: 1 z 1 z 2 1 1 1 1 1 1 1 1 z z 2 z 1 1 z z 1 1 1 z Q17. 1 2 1 z 2 1 2 3 1 z 3 ... 1 1 1 z 1 z 21 31 1 2 3 z 1 z 1 z .... 1 z Let u x, y x 1 2 x y 2 be the real part of analytic function f z of the complex 2 variable z x i y . The imaginary part of f z is (a) y xy Ans: (b) xy (c) y (d) y 2 x 2 (a) Solution: u x, y x Check 1 2 x y 2 , v x, y ? 2 u v u v . and x y y x u v , x y v 1 x y v y xy f x u v v y y x x v yx f y y xy f x yx f y f x 0 f y y V xy y Q18. Let y x be a continuous real function in the range 0 and 2 , satisfying the inhomogeneous differential equation: sin x d2y dy cos x x 2 dx dx 2 The value of dyldx at the point x / 2 (a) is continuous (b) has a discontinuity of 3 (c) has a discontinuity of 1/3 (d) has a discontinuity of 1 Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 9 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Ans: (d) Solution: After dividing by sin x , Integrating both sides, d2y dy cot x cosec 2 x 2 dx dx x 2 dy dy cot x dx cosec x x dx dx 2 dx dy cot x y cosec x ydx 1 dx Using Dirac delta property: f x x x f x (it lies with the limit). 0 0 dy cos x y y sin 2 xdx 1 , at x ; sin x 0 . So this is point of discontinuity. dx sin x Q19. A ball is picked at random from one of two boxes that contain 2 black and 3 white and 3 black and 4white balls respectively. What is the probability that it is white? (a) 34 / 70 Ans: (b) 41/ 70 (c) 36 / 70 (d) 29 / 70 (b) Solution: Probability of picking white ball 2 B 3W 3B 4W 3 4 From box I and from box II 5 7 Probability of picking a white ball from either of the two boxes is Q20. 1 3 4 41 2 5 7 70 The eigenvalues of the antisymmetric matrix, 0 A n3 n2 n3 0 n1 n2 n1 0 where n1 , n 2 and n3 are the components of a unit vector, are (a) 0, i, i (c) 0,1 i, 1, i Ans: (b) 0,1, 1 (d) 0, 0, 0 (a) Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 10 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES 0 Solution: A n3 n 2 n3 0 n 3 n2 n3 1 0 but so, 0 n1 0 n1 n2 0 T n1 A n3 0 n2 n3 0 n1 n2 n1 0 n2 n1 0 2 2 2 n12 n2 n3 2 2 3 n12 n2 n3 2 2 n12 n2 n3 1 1 0 , 2 L , 3 L A AT (Antisymmetric). Eigenvalues are either zero or purely imaginary. Q21. Which of the following limits exists? N 1 (a) lim ln N N m 1 m N 1 (c) lim ln N N m 1 m N 1 (b) lim ln N N m 1 m N 1 (d) lim N m 1 m Ans: (b) Q22. A bag contains many balls, each with a number painted on it. There are exactly n balls which have the number n (namely one ball with 1, two balls with 2, and so on until N on them). An experiment consists of choosing a ball at random, noting the number on it and returning it to the bag. If the experiment is repeated a large number of times, the average value the number will tend to (a) 2N 1 3 (b) N 2 (c) N 1 2 (d) N N 1 2 Ans: (a) Solution: Total number of balls 1 2 3 4 ..... N N N 1 2 The probability for choosing a k th ball at random k N N 1 2 Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 11 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Average of it is given by k k P Q23. 2N 1 3 2 N N 1 2 N 1 2 k 2 N N 1 N N 1 6 where k 2 N N 1 2 N 1 . 6 Consider a sinusoidal waveform of amplitude 1V and frequency f 0 . Starting from an arbitrary initial time, the waveform is sampled at intervals of 1 . If the corresponding 2 f0 Fourier spectrum peaks at a frequency f and an amplitude A , them (a) f 2 f 0 and A 1V (c) f 0 and A 1V Ans: (b) f 2 f 0 and 0 A 1 A (d) f f0 1 and A V 2 2 (b) Solution: y 1sin 2 f 0t . y The fourier transform is: F y IV 1 f f 0 f f 0 2 0 1 In Fourier domain f f 0 , A . 2 t T NET/JRF (DEC-2012) Q24. a b c The unit normal vector of the point , , on the surface of the ellipsoid 3 3 3 x2 y2 z 2 1 is a2 b2 c2 (a) (c) ca bci j abk (b) a 2 b2 c2 c bi j ak (d) a2 b2 c2 b ai j ck a2 b2 c2 i j k 3 Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 12 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Ans: All the options given are incorrect. Solution: Here x2 y2 z2 1. a2 b2 c2 Unit normal vector is . 2 y 2 xi x2 y 2 z 2 j 2 zk So, i j k 1 2 b2 c2 a2 x y z b2 c2 a c , , 3 3 3 a b 4 4 4 2 b2c 2 a 2c 2 a 2c 2 3a 2 3b 2 3c 2 a 2b 2 c 2 3 Q25. 2 2 2 i j k a 3 b 3 c 3 a b c , , 3 3 3 2 2 2 i j k ca bci j abk a 3 b 3 c 3 2 b 2 c 2 c 2 a 2 a 2b 2 b 2 c 2 c 2 a 2 a 2b 2 abc 3 Given a 2 2 unitary matrix U satisfying U U UU 1 with det U e i , one can construct a unitary matrix V V V VV 1 with det V 1 from it by (a) multiplying U by e i / 2 (b) multiplying any single element of U by e i (c) multiplying any row or column of U by e i / 2 (d) multiplying U by e i Ans: (a) Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 13 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES The graph of the function f x shown below is best described by 1.00 (a) The Bessel function J 0 x 0.25 0 .50 (b) cos x 0.25 (c) e x cos x 0.00 0.25 1 (d) cos x 0.50 x 0.75 (a) 1.00 0 1 2 3 4 5 6 7 8 9 10 x In a series of five Cricket matches, one of the captains calls Heads every time when the f (x) Q26. Ans: Q27. toss is taken. The probability that he will win 3 times and lose 2 times is (a) 1 / 8 Ans: (b) 5 / 8 (c) 3 / 16 (d) 5 / 16 (d) 1 Solution: P 2 3 1 1 2 5 3 2 5! 5! 1 1 . 3! 5 3 ! 8 2 3! 5 3 ! 1 5 4 3! 20 5 5 32 3! 2! 32 8 2 16 The probability of getting exactly k successes in n trials is given by probability mass function Q28. n! n k p k 1 p , k successes, n trials. k ! n k ! The Taylor expansion of the function ln cosh x , where x is real, about the point x 0 starts with the following terms: (a) (b) 1 2 1 4 x x .... 2 12 (c) Ans: 1 2 1 4 x x .... 2 12 1 2 1 4 x x .... 2 6 (d) 1 2 1 4 x x .... 2 6 (b) Solution: cosh x e x e x .Tailor s series expansion of f x about x a 2 Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 14 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES f x f a f a x a f ' ' a x a 2 f ' ' ' x a 3 ... . Here a 0 . 1! 2! 3! e x e x 1 e x e x ex ex f x tanh x 0 , f x log 0 x 0 2 e x e x e x e x 2 x 0 2 e f ' ' x x e x e x e x e x e x e x e x e x e x 2 e x e x e e 2 x e x e x 2 x 2 1 tanh 2 x At x 0, f ' ' x 1, f ' ' ' x 2 f x Q29. 1 2 1 4 x x ....... 2 12 The value of the integral C z 3 dz , where C is a closed contour defined by the z 2 5z 6 equation 2 z 5 0, traversed in the anti-clockwise direction, is (b) 16 i (a) 16 i Ans: (c) 8 i (d) 2 i (a) Solution: z 2 5 z 6 0 z 2 2 z 3z 6 0 z z 2 3 z 2 0 z 3, 2 2 z 5 z 2.5 , only 2 will be inside. Residue z 2 z3 z 3 dz 8 2 i 8 16 i 8 2 c z 5z 6 z 3 z 2 z 2 2 3 NET/JRF (JUNE-2013) Q30. Given that H n x n 0 2 tn e t 2tx n! the value of H 4 0 is (a) 12 Ans: (b) 6 (c) 24 (d) 6 (a) Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 15 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Solution: H n x n 0 2 2 tn tn t4 t6 e t 2tx H n 0 e t 1 t 2 n! n! 2! 3! n 0 H 4 0 4 t 4 4! t H 4 0 12 . 4! 2! 2! Q31. on the xy -plane is at an angle of 120o with respect to i . The angle A unit vector n bn will be 60 o if and v an bi between the vectors u a i (a) b 3a / 2 Ans: (b) b 2a / 3 (c) b a / 2 (d) b a (c) bn bi , v an Solution: u ai n bn u v cos 60 a 2 i ab ba b 2 n .i an bi u v ai a 2 b 2 2ab cos120 cos60 a 2 2 cos120 2ab b 2 cos120 1 1 2 1 2 ab 1 2 2 2 2 a 2 b 2 2ab a b 2ab cos 60 a b 2ab a b 2 2 2 2 2 a2 b2 Q32. 5ab a b . 2 2 With z x iy, which of the following functions f x, y is NOT a (complex) analytic function of z ? (a) f x, y x iy 8 4 x 2 y 2 2ixy 3 7 (b) f x, y x iy 1 x iy 7 3 (c) f x, y x 2 y 2 2ixy 3 5 (d) f x, y 1 x iy 2 x iy 4 Ans: 6 (d) Solution: f x, y 1 x iy 2 x iy 4 6 1 x iy 2 x iy 4 6 Due to present of z x iy Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 16 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Q33. The solution of the partial differential equation 2 2 u x , t u x, t 0 t 2 x 2 satisfying the boundary u x,0 sin x / L and u 0, t 0 u L, t conditions and initial conditions u x, t t 0 sin 2 x / L is t (a) sin x / L cos t / L L sin 2 x / L cos 2 t / L 2 (b) 2 sin x / L cos t / L sin x / L cos 2 t / L (c) sin x / L cos 2 t / L (d) sin x / L cos t / L Ans: L sin 2 x / L sin t / L L sin 2 x / L sin 2 t / L 2 (d) Solution: x 2 x 2u 2u u 2 0 , u x,0 sin and sin 2 L t L t x This is a wave equation an t an t Bn sin So solution is given by u x, t An cos L L n 2 n x 2 n x f x sin dx, Bn g n sin dx L0 L an 0 L L with An Comparing a 2 L 2u 2u x 2 x 2 , We have a 1 and f x sin , g n sin , 2 t x L L 2 x L L 1 cos L n x 2 x 2 2 x L dx 2 L 1 (let n 1 ) An sin sin dx sin 2 dx L0 L L 0 2 L 2 L0 L L Putting n 2 Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 17 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES 4 x 1 cos 2 2 2 2 x n x 2 2 x L dx 2 L L sin dx Bn sin sin dx 2 0 L 2 0 2 2 2 2 an 0 L L L L Q34. L The solution of the differential equation dx x2 dt with the initial condition x 0 1 will blow up as t tends to (a) 1 Ans: (b) 2 (a) Solution: dx x 2 1 1 dx t C t C x 2 2 dt 2 1 x dt x x 0 1 Q35. (d) (c) 1 1 1 as t 1 x blows up 0 C C 1 t 1 x 1 x 1 t The inverse Laplace transforms of 1 is s s 1 2 1 2 t t e 2 (b) 1 2 t 1 e t 2 (c) t 1 e t (d) 1 2 t 1 e t 2 (a) Ans: (c) Solution: f s 1 t t 1 f t e t L 1 e dt e t s 1 s s 1 0 e t 0 t 1 1 t t t t t L 2 e 1 dt e t 0 e t 1. s s 1 0 1 Q36. The approximation cos 1 is valid up to 3 decimal places as long as is less than: (take 180o / 57.29o ) (a) 1.28 Ans: (b) 1.81 (c) 3.28 (d) 4.01 (b) Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 18 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Solution: cos 1 2 2! 2 4! ....... 1 cos 1 when 1.81o 100 2 2! .0314 JRF/NET-(DEC-2013) Q37. xy , then the integral yz If A i jxz k A dl (where C is along the perimeter of a C rectangular area bounded by x 0, x a and y 0, y b ) is (a) Ans: 1 3 a b3 2 (b) ab 2 a 2 b (c) a 3 b3 (d) 0 (d) A d l A .d a 0 since A 0 . C Q38. S If A, B and C are non-zero Hermitian operators, which of the following relations must be false? (a) A, B C Ans: (b) AB BA C (c) ABA C (d) A B C (a) Solution: A, B C AB BA C ( AB BA) C (( AB) ( BA) ) C ( B A ) ( A B ) C Hence A,B and C are hermitian then BA AB C A, B C Q39. Which of the following functions cannot be the real part of a complex analytic function of z x iy ? (a) x 2 y Ans: (b) x 2 y 2 (c) x 3 3xy 2 (d) 3 x 2 y y y 3 (a) Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 19 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Solution: Let x 2 y be real part of a complex function. Use Milne Thomson s method to write analytic complex function. The real part of that function should be (1) but that is not the case. So this cannot be real part of an analytic function. Also, z 2 x iy x 2 y 2 2ixy , Real part option (2) 2 z 3 x iy x 3 iy 3 3ixy x iy 3 x 3 iy 3 3ix 2 y 3xy 2 , Real part option (3) Q40. The expression 2 2 2 2 1 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 2 3 4 1 2 3 4 1 is proportional to (a) x1 x2 x3 x4 2 2 2 (c) x12 x2 x3 x4 Ans: (b) x1 x 2 x3 x 4 3 / 2 2 2 2 (d) x12 x2 x3 x4 2 (b) 1 2 Solution: 2 2 2 x1 x1 x 2 x3 x 4 2 x1 2 2 2 2 x1 x2 x3 x4 2 x 2 x 2 x 2 x 2 2 1 2 2x x x 2 x 2 x 2 x 2 2 2 3 4 1 1 1 2 3 4 2 1 2 2 2 2 4 x12 x1 x2 x3 x4 x 2 x 2 x 2 x 2 2 4 x 2 8x 2 2 x 2 x 2 x 2 x 2 2 3 4 1 1 2 3 4 2 1 1 2 2 2 2 3 2 2 2 2 3 x1 x2 x3 x4 x1 x2 x3 x4 Now similarly solving all and add up then we get 2 2 2 2 1 1 1 1 x 2 x 2 x 2 x 2 x2 x2 x2 x2 2 3 4 1 2 3 4 1 2 2 2 8 x12 x2 x3 x4 8 x12 x22 x32 x42 x 2 1 2 2 2 x2 x3 x4 3 0 also if all x1 , x2 , x3 , x4 becomes zero it should be infinity. Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 20 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES 2 2 2 2 1 So x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x1 x 2 x3 x4 2 3 4 1 2 3 4 1 dx Q41. Given that the integral 2 , the value of 2 2y 0 y x y3 (a) Ans: (b) 4y (c) 3 y dx 2 0 x2 2 is 8y3 (d) 2 y3 (b) Solution: y 0 dx 2 x2 2 1 dx , pole is of 2nd order at x iy , residue 1/ 4iy 3 2 2 2 2 y x Integral 1/ 2* 2 i *1/ (4iy 3 ) / (4 y 3 ) Q42. The Fourier transform of the derivative of the Dirac - function, namely x , is proportional to (a) 0 Ans: (b) 1 (c) sin k (d) ik (d) Solution: Fourier transform of x H K x e ikx dx ike k 0 ik Q43. Consider an n n n 1 matrix A , in which Aij is the product of the indices i and j (namely Aij ij ). The matrix A (a) has one degenerate eigevalue with degeneracy n 1 (b) has two degenerate eigenvalues with degeneracies 2 and n 2 (c) has one degenerate eigenvalue with degeneracy n (d) does not have any degenerate eigenvalue Ans: (a) 1 2 Solution: If matrix is 2 2 let then eigen value is given by 2 4 Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 21 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES 2 1 0 (1 )(4 ) 4 0 0,5 4 2 1 2 3 If If matrix is 3 3 let 2 4 6 then eigen value is given by 3 6 9 1 2 3 3 4 6 0 6 9 2 1 4 9 36 2 18 2 9 3 12 3 4 1 2 13 36 36 2 18 18 2 3 12 12 3 0 2 13 3 13 2 13 0 3 14 2 0 0, 0, 14 i.e. has one degenerate eigenvalue with degeneracy 2. Thus one can generalized that for n dimensional matrix has one degenerate eigevalue with degeneracy n 1 . Q44. Three sets of data A, B and C from an experiment, represented by , and , are plotted on a log-log scale. Each of these are fitted with straight lines as shown in the figure. 1000 100 C B 10 1 0.1 0.1 1 10 A 100 1000 The functional dependence y x for the sets A, B and C are respectively (a) Ans: x, x and x 2 x (b) , x and 2 x 2 (c) 1 , x and x 2 x2 (d) 1 x , x and x 2 (d) Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 22 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES JRF/NET-(JUNE-2014) Q45. Consider the differential equation d 2x dx 2 x 0 2 dt dt 0 1 . The solution x t attains its maximum with the initial conditions x 0 0 and x value when t is (a) 1/2 Ans: (b) 1 (c) 2 (d) (b) Solution: d 2x dx 2 2 x 0 m2 2m 1 0 m 1 0 m 1, 1 2 dt dt x c1 c2t e t since x 0 0 0 c1 x c2 te t 0 t t c2 x te e 0 1 1 c2 x te t Since x For maxima or minima x te t e t 0 x 0 x e t 1 t e t 0, 1 t 0 t , t 1 x e t 1 1 t e t 1 e t t 1 e t x 1 e 1 0e t 0 Q46. Consider the matrix 2i 3i 0 M 2i 0 6i 3i 6i 0 The eigenvalues of M are (a) 5, 2, 7 Ans: (b) 7, 0, 7 (c) 4i, 2i, 2i (d) 2, 3, 6 (b) 2i 3i 2i 3i 0 0 Solution: M 2i 0 6i , M 2i 0 6i 3i 6i 0 3i 6i 0 Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 23 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES M M Matrix is Hermitian so roots are real and trace = 0. 1 2 3 0, 1 2 3 0 7, 0, 7 Q47. If C is the contour defined by z 1 , the value of the integral 2 dz C sin 2 z is (b) 2 i (a) Ans: (c) 0 (d) i (c) 1 z 2 3 5 z z 1 1 sin z z .... 2 2 3 5 sin z z3 z5 z .... 3 5 Solution: f z 1 sin 2 z 1 1 z2 z4 1 .... 3 5 sin 2 z z 2 Q48. Given (a) 0.26 Ans: n 0 2 dz 0 C sin 2 z Pn x t n 1 2 xt t 2 1 / 2 , for t 1 , the value of P5 1 is (b) 1 (c) 0.5 (d) 1 (d) Pn 1 1 if n is odd P 5 1 1 Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 24 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Q49. The graph of a real periodic function f x for the range , is shown below f x x Which of the following graphs represents the real part of its Fourier transform? (a) Re f k (b) Re f k k (c) k Re f k (d) Re f k k Ans: k (b) Solution: This is cosine function f x A cos x F k A k k0 k k0 2 NET/JRF (DEC-2014) Q50. Let r denote the position vector of any point in three-dimensional space, and r r . Then (a) r 0 and r r / r (c) r 3 and 2 r r / r 2 (b) r 0 and 2 r 0 (d) r 3 and r 0 Ans: (d) yy zz Solution: r xx Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 25 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES x y z r 1 1 1 3 x y z x y z z y x z y x y z 0 r / x / y / z x y z z x x y x y z Q51. a The column vector b is a simultaneous eigenvector of a 0 0 1 A 0 1 0 and 1 0 0 0 1 1 B 1 0 1 if 1 1 0 (a) b 0 or a 0 (c) b 2a or b a Ans: (b) b a or b 2a (d) b a / 2 or b a / 2 (b) Solution: Let b a 0 0 1 a a 0 1 1 a a a 0 1 0 a a and 1 0 1 a a a 1 0 0 a a 1 1 0 a a a Let b 2a 0 0 1 a a 0 1 1 a a a 0 1 0 2a 2a and 1 0 1 2a 2a 1 2a 1 0 0 a a 1 1 0 a a a For other combination above relation is not possible. Q52. The principal value of the integral (a) 2 Ans: (b) sin 2 x dx is x3 (c) (d) 2 (a) Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 26 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Solution: Let f z ei 2 z z3 ei 2 z 1 finite and 0 z 0 is pole of order 3. z3 iz d2 3 e z 0 2 dz 2 z3 lim z 0 z 0 f z lim z 0 z 0 3 Residue R Q53. 1 lim 2! z 0 3 f x dx i R i 2 2 i Im. Part 2 f x dx 2 The Laurent series expansion of the function f z e z e1 / z about z 0 is given by n 1 1 only if 0 z 1 z n z n! (a) (b) (c) Ans: zn n n! for all z n 1 1 z n for all 0 z n 0 z n! (d) zn n n! only if z 1 n 0 (c) n 1 1 1 1 z2 z .... n 0 n Solution: e z 1 z .... n 0 and e1/ z 1 2 z 2! z z n! 2! n! 1 1 for all 0 z f z ez e1/ z n 0 z n n z n ! Q54. Two independent random variables m and n , which can take the integer values 0, 1, 2, ..., , follow the Poisson distribution, with distinct mean values and respectively. Then (a) the probability distribution of the random variable l m n is a binomial distribution. (b) the probability distribution of the random variable r m n is also a Poisson distribution. (c) the variance of the random variable l m n is equal to (d) the mean value of the random variable r m n is equal to 0. Ans: (c) 2 2 n Solution: l2 m Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 27 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Q55. 1 of a complex ln 1 z z z re i r 0, . The singularities of f z are as follows: (a) branch points at z 1 and z ; and a pole at z 0 only for 0 2 Consider the function f z variable (b) branch points at z 1 and z ; and a pole at z 0 for all other than 0 2 (c) branch points at z 1 and z ; and a pole at z 0 for all (d) branch points at z 0, z 1 and z . Ans: None of the above is correct 1 1 z 2 z3 z z2 Solution: For f z ln 1 z z ..... 1 ..... z z 2 3 2 3 There is no principal part and when z 0 , f z 1. So there is removable singularity at z 0 . Also z 1 and z is Branch point. Q56. 1 n x 2n 1 n ! n 1 ! 2 satisfies the differential equation (b) x 2 d2 f df 2x x2 1 f 0 2 dx dx (d) x 2 d2 f df x x2 1 f 0 2 dx dx The function f x n 0 (a) x 2 (c) x 2 Ans: d2 f df x x2 1 f 0 2 dx dx d2 f df x x2 1 f 0 2 dx dx (c) Solution: 2 n 1 n 1 x is f x n 0 n ! n 1 ! 2 generating function (Bessel Function of first kind) d2 f df x x 2 n 2 f 0 , put n 1 . 2 dx dx Let and be complex numbers. Which of the following sets of matrices forms a which satisfies the differential equation x 2 Q57. group under matrix multiplication? (a) 0 0 * * (c) * , where is real Ans: 1 (b) , where 1 2 (d) * * , where 2 1 (d) Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 28 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Solution: 2 2 1 * * x , p, L are (where is the Levi-Civita symbol, x , p , L i jk i jk i j k 3 Q58. The expression i , j , k 1 the position, momentum and angular momentum respectively, and A, B represents the Poisson Bracket of A and B ) simplifies to (a) 0 Ans: (b) 6 (c) x , p L (d) x p (b) Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 29

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