ans full length test 2

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Answer of Full Length Test-02 PART A

Q1: Ans: (d) Q2: Ans: (d) Q3: Ans: (c)

Q4: Ans: (c) Q5: Ans: (c) Q6: Ans: (c)

Q7: Ans: (b) Q8: Ans: (a) Q9: Ans: (a)

Q10: Ans: (b) Q11: Ans: (d) Q12: Ans: (a)

Q13: Ans: (d) Q14: Ans: (b) Q15: Ans: (c)

Q16: Ans: (c) Q17: Ans: (b) Q18: Ans: (b)

Q19: Ans: (c) Q20: Ans: (c)

PART B

Q21. Ans: (a)

Q22. Ans: (b)

Q23. Ans: (d)

Q24. Ans: (a)

n 2i 2j k3

Q25. Solution: (d)

Q26. Ans (c):

2 21 21T m(x x )2

and 2 2 21 1 2 2 1 21 1 1V k x k (x x ) kx2 2 2

2 2 2 21 1 2 2 1 1 2 1 21 1 1k x k (x x 2x x ) k x2 2 2

2 21 2 1 1 2 2 1 21 1 1x (k k )x (k k )x k 2x x2 2 2

m 0

T0 m

and 1 2 22 1 2

k k kV

k k k





For Normal frequencies.

2(V T) 0

2

1 2 22

2 1 2

k k m k0

k k k m

2 2 21 2 2(k k m) k 0

2 21 2 2 1 2 2(k k m k )(k k m k ) 0

2 21 2 1(k 2k m)(k m) 0

11km

and 1 22k 2k

m

Q27. Solution: (c)

dF F[H, F]dt t

P P 0m m

dF 0dt

F = constant of motion.

Q28. Solution: (c)

Q29. Solution: (b)

1 2 31 2 9H E E E

14 7 14 0 00

E E1 2 9E14 7 4 14 9

003E1 1 1E

14 14 14 14

Q30. Solution: (b)

z[L cos ]

z zL cos cos L z z zL cos cos L cos L

zL cos i cos

i sin i sin





Q31. Ans: (c)

Solution: The possible energy for given system in .4,3,1 (For FD statistics).

Q32. Ans: (a)

E = P E = PE cos

PEcos /kTsin d0

z e

2kT PEz sin hPE kT

Q33. Solution: (b)

5/2p

G 5 apS R R lnT 2 (RT)

pS 5C T RT 2

Q34. Ans: (a)

Q35. Ans: (c)

0

2

2220 16

1923

113

4 QQQQQF

Q36. Ans: (d)

Q: 37. Ans: (c)

Solution: 2.4232

10102,108

715 vcn

kvk

Q38. Ans: (d) Q39. Ans: (c) Q40. Ans: (b)

Q41: Ans: (d) Q42: Ans: (c)

Q43: Ans: (d)

mgZzyxmL 22221 22

21 yxaZ

Which has cylindrical symmetries x r cos , y r sin , z=z 21Z a r2

x r cos r sin y rsin r cos Z a rr so 2 2 2 2 2 21L m 1 a r r r gar2





Q44: Ans: (a)

Solution:

22

121

FF

TTET

Q45: Ans : (c)

PART C Q46. Ans: (c)

Solution: Separation of variables with u(x,t) (x)G(t) gives dGdt

(k 1)G and d2dx2

.

The boundary conditions gives G(t)(0) 0

G(t) ddx

(L) G(t)(L) G(t) ddx

(L) (L)

0

So for non- trivial solutions

(0) and ddx

(L)(L) 0 .

Q47. Ans: (a)

Q48. Ans: (d) Degeneracy is 2S+1 =3

Q49. Ans: (a)

m nn m0 0m n n m

| w || |E E

qE n | n 1 n 1 | n 12m

Where W = qEx 1qE (a a)2m

m,m 1 m,m 1m | X | n n n 12m

Q50. Ans: (b)





Q51. Ans: (a)

Let the frequency of the photon be v and the momentum of the atom in the excited state be p.

The conservation laws of the energy and momentum give

2/12222 cpMhMc Conservation of Energy p

ch

Conservation of Momentum

And hence,

Mh

c2

12

Q52. Ans: (c)

, here n is the dimensionality

Q53. Ans: (b)

Q54. Ans: (d) Hints: E1 =3.27MeV, E2 =4.03Mev, E3 =17.59MeV

Q55. Ans: (c) Q56. Ans: (d)

Hints:

(a) ut (pux )xpuxx 0uxt pxux ut 0Since 02 p0 0, the equation is parabolic.

(b) utt c2uxx u

utt 0utx c2uxx u 0

Since 02 (1)(c 2) 0, the equation is hyperbolic.

(c) (qux)x (qut )t 0quxx 0uxt qutt qxux qtut 0Since 02 qq 0, the equation is elliptic.

uttx uxx 0 is wave equation which is hyperbolic.





Q57. Ans: (c)

2

2

PH2mR

J P2

2

2 2 2 2

J JH E4 2mR 8m R

2E J

4 mR

Q58. Ans: (c)

Q59. Ans: (c)

Solution: (a) Forbidden as it violates J = 0, 1 00

(b) Forbidden as it violates J = 0, 1 (Here J = 2)

(c) Forbidden as it violates L = 0, 1 (Here L = 2)

Q60. Ans: (b)

Q61. Ans: c

Q62. Ans: (b)

Q63. Ans: (d) Q64. Ans: (c)

Q65. Ans -: (b) Solution: If the electron gas confined in one dimension rod of length L, then number of

possible states in between k and k + dk

dkLdkkg

Since, 2

22 22 mEk

mkE

dEEmdk 2/12/1

2 212

dEEmLdEEg 2/12/1

2 212)(

Since, electrons have spin 1/2. So multiply above equation with 2.

dEEmLdEEg 2/12/1

2

2)(

At T = 00 K. The total number of electron between E = 0 and EF is





dEEgEFNFE

0

FE

dEEmLN0

2/12/1

2

2

(Since F(E) = 1 at T=0K)

22222

2/12/1

2

2/12/1

2

2222

222/1

2

LN

mLN

mE

EmLEmL

F

FF

Q: 66. Ans: (b) Q: 67. (A) Ans: (a) 00 4.

Senc adEQ

(B) Ans: (a)

Q68. (A) Ans: (c)

Hints: (i) Forbidden; muon and electron lepton numbers are not conserved.

(ii) Forbidden; baryon number not conserved. (iii) Allowed; by weak interaction as S 1 , but not by strong because

strangeness is not conserved.

(B) Ans: (c)

Hints:

p x1 K0

B : 0 1 x1 0

S : 0 0 x1 (1)

p p x2 K p

B :1 1 x1 0 1

S : 0 0 x1 (1) 0

K n x3 0 0

B : 0 1 x1 0 0

S :1 0 x1 0 0





Q69. Ans: (d)

Hints: Vc 1

40

Z Z e2

R R, where Z and Z are the atomic numbers of the two

nuclei and R and R are their effective radii. Classically, the distance of closest approach is R R .

Q70. (A) Ans: (c) (B) Ans: (b)

There are 12 members in the lepton family: e ,e, , ,

, , and their antiparticles. Out of these only 10 are discovered so far. , and its antiparticle are theoritically predicted but not discovered yet.

Q71: Ans: (d)

Q72: Ans: (c)

Q73: Ans: (b)

Q74: Ans: (a)

.mod.

21.7,15.6,8.6,4.14,

2,2

1010

20110110

22

22

usedbecaneTEHencecmcmcmcmSo

bn

amb

namcf mnmn

Q75: Ans: (a)

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ans full length test 2

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