ans full length test 2
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fiziks Institute for CSIR-UGC JRF/NET, GATE, IIT-JAM, GRE in PHYSICAL SCIENCES
Website: http://www.physicsbyfiziks.com Email: fiziks.physics@gmail.com 1
Head office: fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498
Branch office: Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16
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
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fiziks Institute for CSIR-UGC JRF/NET, GATE, IIT-JAM, GRE in PHYSICAL SCIENCES
Website: http://www.physicsbyfiziks.com Email: fiziks.physics@gmail.com 2
Head office: fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498
Branch office: Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16
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
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fiziks Institute for CSIR-UGC JRF/NET, GATE, IIT-JAM, GRE in PHYSICAL SCIENCES
Website: http://www.physicsbyfiziks.com Email: fiziks.physics@gmail.com 3
Head office: fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498
Branch office: Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16
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
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fiziks Institute for CSIR-UGC JRF/NET, GATE, IIT-JAM, GRE in PHYSICAL SCIENCES
Website: http://www.physicsbyfiziks.com Email: fiziks.physics@gmail.com 4
Head office: fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498
Branch office: Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16
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)
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fiziks Institute for CSIR-UGC JRF/NET, GATE, IIT-JAM, GRE in PHYSICAL SCIENCES
Website: http://www.physicsbyfiziks.com Email: fiziks.physics@gmail.com 5
Head office: fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498
Branch office: Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16
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.
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fiziks Institute for CSIR-UGC JRF/NET, GATE, IIT-JAM, GRE in PHYSICAL SCIENCES
Website: http://www.physicsbyfiziks.com Email: fiziks.physics@gmail.com 6
Head office: fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498
Branch office: Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16
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
-
fiziks Institute for CSIR-UGC JRF/NET, GATE, IIT-JAM, GRE in PHYSICAL SCIENCES
Website: http://www.physicsbyfiziks.com Email: fiziks.physics@gmail.com 7
Head office: fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498
Branch office: Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16
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
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fiziks Institute for CSIR-UGC JRF/NET, GATE, IIT-JAM, GRE in PHYSICAL SCIENCES
Website: http://www.physicsbyfiziks.com Email: fiziks.physics@gmail.com 8
Head office: fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498
Branch office: Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16
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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