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## CSIR UGC NET, 2014 June 22, Question, Detailed Answer Key, Key, Answers, solutions, Explanations, Comment Here

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# ► CSIR UGC NET, 2014 June 22, Question, Detailed Answer Key, Key, Answers, solutions, Explanations, Comment Here

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UGC CSIR NET 2014 June 22 Physics Answer Key Comment here with explanation
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#### 184 comments :

1. Anonymous22 June, 2014

Pls comment answers along with questions

1. booklet A
21) 2
23) 4
25) 4
26) 3
27) 2
34) 1
35) 1
36) 3
39) 4
40) 3
50) 4
51) 4
54) 1
59) 3
61) 1
65) 4-

2. booklet A

PART A
1) 3
4) 2
5) 2
6) 2
7) 2
9) 4
11) 4
12) 3
13) 3
14) 4
15) 3
17) 2
18) 2
19) 2
20) 2

PART B
21) 2
23) 4
24) 2
25) 4
26) 3
27) 2
34) 1
35) 1
36) 3
39) 4
40) 3

PART C
50) 4
51) 4
54) 1
59) 3
61) 1
65) 4

3. many answers are wrongs

2. physical science set c
part( a)
1-3
3-1
4-2
5-3
7-4
8-2
9-2
11-4
12-1
13-3
14-2
16-2
17-2
18-2
19-2

1. 12 answer will be 40.5km

2. its 41.5 calculte it

3. 3-3 hoga 100% sure

4. no 12 answer is 40.5.... as in 30 minute the man covers 1.5km starting from rear end, after 500m he change his direction and moves to rear end again, after 500m he again moves towards front end of train, so after 30 min he is in front end of train, and in 30 minutes train covers a distance of 40km, so total distance between the observer and man is 40+.500=40.5km

3. This comment has been removed by the author.

4. please tell me cutoff marks

1. 70-75 for gen

2. Anonymous25 June, 2014

Last tym 82 gyi thi fir iss baar itni kam kyu

5. 1 b
8 d
9 b
10 b
11 c
13 d
14 b
15 b
16 b
17 d
20 b
25 c
26 c
28 c
29 b
31 b
33 b
34 d
35 c
36 b
37 c
38 d
40 c
43 a
44 d
46 c
50 d
51 d
52 d
56 a
63 d
65 b
72 d
74 d

1. which set no ? A B OR C

2. which set no

3. What is the SET of your answer key

6. shamshul haq23 June, 2014

part-c,
1-3
3-1
4-2
5-2
7-4
8-2
9-3
10-1
11-3
12-1
14-2
15-3
16-2
17-2
18-2
19-3
20-4

1. Fourier transform and will be option 2
In which both upper side in vertical form
You all can just go through web by just see the Fourier transform graph of cosine function

2. Matrices and is option four
[A B]=C, [B C]=0, [C A]=B
Just do some matrix multiplication and check the result

3. The relation best describes the dependence of n & M
n~M^4/9

4. X-component of velocity of electron is 4Eoa/h cut
Differentiate given equation with respect to Kx and put the given value(Ï€/a,0,0)
And result follows
Ve= dE/h cut dKx= 4Eoa/h cut

5. Anonymous03 July, 2014

V=-dE/h cut dkx = -4Eoa/h cut

7. Anonymous23 June, 2014

what will be the answer for the dependency of n(electron density) in Pressure of fermion gas?

8. Anonymous23 June, 2014

70-75 for LS and 80-85 for Jrf in general catagory

1. Keshav plz write booklet code

9. booklet B
part a

1) 2
2) 3
3) 2
4) 1
5) 2
6) 2
7) 4
9) 2
13) 4
14) 3
16) 2
17) 4
18) 3
19) 2
20) 2

10. Anonymous24 June, 2014

please comment answer keys of booklet code B

11. Please upload the answers of question booklet set "B" .

12. Anonymous24 June, 2014

What is the answer of vander wall equation ' s question???
I think a/27b2

1. it is 3b

2. This comment has been removed by the author.

13. Anonymous24 June, 2014

And answer of probability within the bohr radius is

1. Anonymous25 June, 2014

0.32.calculate groundstate probability with R10=2*a0^(-3/2)*exp(-r/a0).
u wil find.
the reverse of this question was asked on june 2011

14. DeltaC/C=?

1. It is 0.14
Since DelC/C=√{(0.01)^2+(0.01)^2}

2. Mr. Rajesh, can you explain this equation?
Did you notice that the error of mass is in milli gram.
The answer is 0.28% given in csir official website.

15. Anonymous24 June, 2014

SET C
Part A
1) 3
3) 1
4) 2
5) 3
6) 1
7) 4
9) 3
10) 3
12) 1
13) 3
14) 2
16) 1
17) 2
18) 2
Part B
23) 1
24) 4
27) 2
29) 4
31) 1
33) 2
34) 4
42) 3
43) 3
Part C
46) 4
50) 2
59) 4
61) 4
64) 1
72) 4

1. CODE C:

PART B:
21) 1
23) 1
26) 2
27) 2
28) 3
29) 4
30) 3
34) 4
36) 3
38) 1
42) 2 (not zero bcaz Z exist)
43) 3

PART C:

46) 4
47) 4
51) 4
55) 3
61) 4
62) 4
64) 2
65) 1

2. B-CODE c:

PART B:
21) 1
23) 1
26) 2
27) 2
28) 3
29) 4
30) 3
34) 4
36) 3
38) 1
42) 2 (not zero bcaz Z exist)
43) 3

PART C:

46) 4
47) 4
51) 4
55) 3
61) 4
62) 4
64) 2
65) 1

16. Anonymous24 June, 2014

What the answer of booklet- c . Question no. 61. V(x)= -§ (x) and ¥(x)=A e (a!x!). And H=b x2. Find ground state energy.

17. Set A
Section A
1. 3
2. 3
3. 2
4. 2
5. 2
6. 2
7. 3
9. 1
10. 1
12. 3
16. 2
17. 2
18. 3
19. 4
20. 2

Section B

25. 4
26. 3
27. 2
29. 4
31. 3
32. 4
34. 1
35. 2
36. 3
37. 1
38. 1
39. 2
40. 3
41. 1

Section C

52. 3
53. 2
59. 4
60. 1
61. 1
63. 4
65. 4
68. 2
69. 4
70. 4
71. 1
72. 2
73. 4

1. Anonymous26 June, 2014

are u sure b<-1for stable and unstable points

2. think carefully qn no 10.
the answer should be 4. not 1

3. Anonymous05 July, 2014

ans of 31-2,32-1

18. Anonymous24 June, 2014

Booklet B
Q22 Asystem can have 3 energylevels E=0,+E,-E. The level E=0 is double degenerate.........
ANS - Option 4

19. Anonymous24 June, 2014

n^2/3

20. Anonymous24 June, 2014

21. Anonymous24 June, 2014

plz give the answer keys of booklet series b for pfysical sciences june 2014

22. This comment has been removed by the author.

23. Anonymous24 June, 2014

booklet B
1.2
3.2
4.3,
5.2,
6.2,
9.2,
10.1,
11.3,
13.4,
14.2,
15.3,
16.2,
17.3,
18.1,
19.3,
22.4,
24.2,
26.3,
28.1,
29.3,
31.1,
32.1,
33.2,
34.4,
35.4,
37.2,
38.2,
43.1,
44.3,
46.3,
48.3,
49.4,
51.4,
52.1,
57.4,
63.4,
64.4,
65.4,
67.2,
69.1,
74.4
i'll be happy to see your suggestions...

1. i feel answer for 10 is option 4

2. book (b)
22.4
23.3
24.2
25.1
26.3
29.3
31.1
32.1
33.2
34.4
35.4
37.2
38.2
40.3
41.3
42.3
43.1
44.3
45.1
46.1
48.3
49.2
51.4
52.3
53.3
55.1
56.1
57.4
62.2
63.4
64.1
65.4
66.2
67.2
68.1
69.4
71.1
72.4
74.4

24. Fermi Pressure is number density times fermi energy i.e. Equivalent to n to the power five by three since fermi energy is itself n to the power two by three

25. This comment has been removed by the author.

26. and 17 is 2

27. Plz give explanation of the aswer

28. Series ABCD answer is K

29. Derivate twice with respect to V and set it equal to zero and ans will be 3b

30. This comment has been removed by the author.

1. This s nuclear physics question. Erot= (h cross)square/2I*J(J+1). Delta J =2. Next excited state s 4. Ans 310kev

2. Yes you are right I also now referred basic nuclear physics by BN srivastava
Where it is given the excitation energy could be predicted by E I'/E l=I'(I'+1)/l(l+1)
So that I=0;2;4;6 of even parity
Ratio E4/E2=10/3
E6/E2=7
E8/E2=12
I will have even parity only if the body has a plane of symmetry and its intrinsic angular momentum is zero

31. Energy of photon will be share mass of higgs minus square mass of Z boson dived by twice the mass of higgs i.e. Equal to 30.1 MeV so its ans will be 30 MeV

32. Please amend to read share as square

33. Energy ground state will be 3 pi square h cut squareby 2 m a square i e ground level all quantum level is equal to one so share of each add will give 3

1. But the given Potential is not of a paricle in 3-D box.....

34. Anonymous24 June, 2014

booklet 2
22...2
28...1
29...4
31...1
33...3
34...4
34...4
37...2
39...4
52...2
53...3
55...3
58...1
63...4
65...4
71...1
72...2

35. what is cut off for obc

1. 85 to 90 for JRF

75 to 85 for LS

36. Area of triangle ABC/OST = 225

Since area of OST = 1/2(0.2)(0.2)=0.02( by similarity property check the side length ratio of triangle OPB and dotted triangle inside triangle TBR(let some point R after Pans meets T)

AREA OF ABC=9/2

37. Triangular series and is 80 ( multiply all and subtract it after adding all)

38. Minimum no of tiles and 12 because only one tile of size 5 unit can fit into it in any directing so it will be1+6+5=12

39. Weight of paper packet ans is 2.4 kg

40. Length of longer side of trapazium is 10 cm since put two vertically and one inverted between both and you will get it

41. Height of tree will be (10+17=27 cm) since angle between height and horizontal distance is tan 45°

1. Anonymous26 June, 2014

10+7 = 17

42. Series power of 17 and is 75 since after each interval of four you will get the sane 3 digut

43. Triangular series and is 80 since multiply each one and subtract after adding each

44. Time gap between min and hour hand is 12min since hour hand makes1/2 degree and min hand makes 6 degree I.e 5and 1/2 degree is varying every min so

45. No of shirts bought is 30

46. Ratio of weight to the square if the height is constant

1. Juhi Srivastava28 June, 2014

Proportionality can have an additive constant.therefore the ratio is constant only when the hight and weight is zero at the age zero

47. Rounded off chicory percentage is 14

48. 197o315 ....
Ans is 4

49. Relation between RD
RD of group > RD of group B

50. Prismatic crystal and 72 (4×16+8=72(

51. ABBCCD in it AD is a perfect square

52. 1/sin^2(z) contour integration |z|=1/2

53. Minimum no. Of tikes is 12 because only one tile if size 5 will fit into given size and sixsize2unit and five size one unit so total if 12

54. Ops amp when working as voltage follower then it matches high impedance source to low impedance source
This manely working as buffer for logic circuit
It is of two type
1. Current buffer
2. Voltage buffer
Its gain is unity I.e maximum transfer of source to load.

55. This comment has been removed by the author.

1. Anonymous04 July, 2014

according to the given potential this is 3D spherical box.
the GS energy pi^2 h(cut)^2/2ma^2 ------------griffith(2nd edition) page 153 example 4.1

2. according to the given potential this is 3D spherical box.
the GS energy pi^2 h(cut)^2/2ma^2 ------------griffith(2nd edition) page 153 example 4.1

56. Graph energy density versus wavelength at temp is and third option in which wavelength shifts towards the left side as temp increases
Refer quantum mechanics zeitilli where graph given is on energy density and frequency where it shifts towards right side on increase of temp

57. To get the oscillation RC network should give total of zero degree phase shifts with input I.e 360° so it requires 12 circuits as each one giving 30° shift

1. Sir,

We have used CE configuration here which add 180 phase shift. Do you think it reduces the requirement of RC circuits to 6 ??

2. Anonymous26 June, 2014

ya u r right

58. Thermodynamic equilibrium then value of psi is equal to derivative of F with respect to psi and make it equal to zero so and is +/-(a/3b)^1/4

59. Effective capacitance is unaffected

60. Composite wave function for fermion is even after ignoring the spin there is at least two quantum no is required to give the sum of square is equal to 5.so option first
2/a(sin(Ï€x1/a)sin(2Ï€x2/a) - sin(2Ï€x1/a)sin(Ï€x2/a))

61. Hamiltonian is p ^2/2(m-lambda q)
Since p=del H/del q dot
H=sum of p into q dot minus L

1. Anonymous26 June, 2014

After calculation option 4 is matching ...is it wrong ?

2. Hamiltonian is function of (p, q , t)
In or option it is function of p & derivative of q so please check

62. The line integral is equal to surface integral of curl of given vector into ds
Since here curl is zero so and will be zero

63. P5(-1)= -1
First put the value of x=-1 and n=5 and expand both sides of power series and equate the co efficient so and will come as -1

64. Image problem speed if charged particle first evaluate potential then calculate velocity by using 1mv^2 formula
Velocity is proportional to square root of distance from image charge

1. Anonymous26 June, 2014

solving we get d/5

65. So and is d/4

66. Please do not write options give correct explanations so that hardel 75questions will be covered in few posts then in reply give your justification if you found some mistakes

67. Eigen value of matrix is -7;0;+7
Since determinant is zero
Sum of principal diagonal element is zero

68. This comment has been removed by the author.

1. Anonymous26 June, 2014

if u solve its double derivative and put value then ans is 1

2. Anonymous26 June, 2014

are you sure the answer is 1/2 check it is 1

3. This comment has been removed by the author.

4. Integral of square root of X between zero and one will be 0.657
Here h=1-0/4=0.25
Yo=√Xo=0
Y1=√X1=√0.25
Y2=√X2=√0.5
Y3=√X3=√0.75
Y4=√X4=√1
Now
I=h/3[Yo+Y4+2Y2+4(Y1+Y3)]
=0.25[o+1+√2+2+2√3]/3
=0.25/3[3+1.414+3.64]
=0.25/3[0.7.878]
=2.626/4
=0.656

5. Charge density will be + epsilon zero phi zero by e into r square
By using Poisson distribution for potential in spherical polar coordinate
1/r^2 del by del r( r^2 del phi by del r)= - rho by epsilon zero

Result is positive value as option one

6. Magnetic vector potential answer is option four
-√c square t square minus a square. To. +√c square t square minus a square
T dz/(a square+ z square)^ 1/2
Kindly refer Griffith electrodynamics fourth edition page no. 447

7. I.e option four MU zero K/ 4Ï€ along Z

8. could u please tell the page number for 3ed griffith....or at least the chapter?

69. BKLET B KEY
1B
2C
4C
5B
6B
7D
8A
9B
10D
11D
12B
13D
14B
15C
16B
17D
19A
PART B
22B
24D
25A
26B
28A
29C
30D
31A
32A
33B
34D
35D
36C
37B
38B
39D
40C
42C
43C
44B
45B
PART C
46C
50B
52C
53D
54D
55C
56A
57C
58A
63D
67D
69D
72A
74C

1. Anonymous01 July, 2014

is the ans for q.54 is option D. please let us know the reference.

70. V=Vo/1-c.v/c^2 which is non relativistic formula
When source of light is moving radially away from the observer. V=Vo√(c-v/c+v) I.e Vo>V and non relativistic freq = Voc/c+v
If source of light is moving radially outward V=Vo√c+v/c-v I.e V>Vo and non relativistic freq=Voc/c-v
Invade light source moving transversely with respect to the observer c.v=0
I.e V=Vo√(1-v^2/c^2)
Non relativistic freq=Vo
Doppler effect does not exist in non relativistic .
In relativity V=Vo√(1-v^2/c^2)/1-c.v/c^2
So and will be f(1-v^2/c^2)^1/2

71. The period of oscillation of simple pendulum under gravity is given by T=2Ï€√(L/g)
Here additional acceleration is given horizontal so it will add vectorially with vertical acceleion and its resultant magnitude will be equal to 2g
So time period will be T/√2 and will make 60° with vertical

1. Anonymous26 June, 2014

plz give all answers from Part C,

72. Juhi Srivastava26 June, 2014

Please someone provide answers for question no. 47 and 67 of booklet A.

1. u can verify easily

73. Anonymous26 June, 2014

x(t)= (A+Bt) exp(-t)
first initial condition gives A=0 ,
second initial condition gives B=1
so, x(t) = texp(-t)
now for maximum differentiate the solution
-t*expt(-t)+exp(-t)=0
exp(-t) * (1-t)=0
so t=1

74. Anonymous26 June, 2014

A spectral line due to a
transition from an electronic
state p to an s state splits
into three zeeman lines in
presence of a strong
magnetic field. At
intermediate field strengths
the number of spertral line is
1. 10
2. 3
3. 6
4. 9
J

1. ans will be 6

75. most of the answers are sure to be correct

76. Re(sin(X+iY))= sinXcos(iy)+cosXsin(iy)
Cos(iY)= Cos h y ,
Sin(iY)= i Sin(iY)
So its real part is equal to SinXCosh Y~phi
Del phi by del X = CosXCosh Y
Del square by del X square= - Sin X Cosh Y
Del phi by del Y = SinX Sin h Y
Del square by del Y=SinX Cos h Y
Del square phi by del Z square= - Cos(z-vt)
Del square phi by del t square/v^2= - Cos(z-vt)
So and is option no. One in which
1/ v^2 del square phi by del t square is equal to ( del square by del X square plus del square by del Y square plus del z by del square) phi

77. The mass of meson will be 283MeV/C^2
First calculate the separation between quark and anti quark
Derivate the given equation with respect to r and put it equal to zero, it will give r equal to (b/a)^1/2
& corresponding potential at that separation will be equal to 2√ab
Put the values and it will come as 282.8MeV which the energy by which meson is bounded so its mass will be 283 MeV / C^2 option 2

1. Anonymous28 June, 2014

plz explain Q.52of code A

2. Equate dx/dt =0 you get the value of b u can calculate x= -1 it is classical mechanics gupta kumar
central force problem stable orbit for circle unstable hyperbola may be option (1)

78. Born approximation ans will be option one
-4m (bits)(MU) by h cut square ( b square plus mu square) ^2
F(thita)=- 2m/ h cut square integral ( zero to infinity) (bits) V(r) Sin qr/qr . r^2 .Dr
Her V(r)= (bita) e to the power minus mu r
Use e to the power (iqr)
& integral of r to the power n into e to the power minus at is equal to factorial n by alpha to the power (n+1)
Then equate each side for imaginary part which is and as option one

79. Dear Rajesh may I know answer of d question to find d scattering amplitude??? Just give d right option

80. Booklet C q.45 Ans.1.6 since CE amp will provide phase shift 180 degrees and 6 RC network provide 180 degrees phase shift so total phase shift would be 0 or 360 degrees which is required for oscillation.

81. Friends please give correct explanation of question no. 25 and 56 of booklet b..

82. Plz answer of Bcc lattice wavefunction ques 56 booklet b and ground state energy in 3D booklet b quest no 25

83. energy eigenvalue of wave function Asin^3(Ï€x/a) is 9/10(Ï€^2hcut^2/ma^2)
use sin3x=3sinx-4sin^3x
normalize the function it will give A=16/5a
integrate between zero to a for obtained function.
refer Griffith 2.36 problem

84. Anonymous01 July, 2014

Plz what is the answer for part C first order correction to the ground state?

85. Anonymous01 July, 2014

first order correction to ground state us option 4

86. scattering amplitude is
-4mbu/h^2(b^2 u^2)^2
option one

87. random walker ans is( 1/2)^4=1/16

1. Anonymous03 July, 2014

no it is 3/8

2. that is not correct answerasir

6/16 or 3/8 is the correct solution in random walk problem refer solutions of rief book

3. Anonymous05 July, 2014

I think there are only 4 cases (LLRR,LRLR,RLRL,RRLL).so fans shld b 4/16=1/4.plz explain other two cases if there r.

88. first order correction is
b/2a^2
use integration of x^n.exp(-ax)= factorial n by a to the power (n 1)
on normalizing the value of A^2=2a

89. This comment has been removed by the author.

90. (del P) (Del r)= h cut
as we have r = a zero so
del p= h cut / a zero

91. RC network n CE combination will produce oscillation with 6 RC network .thanks for ur attention .

92. in normalized wave function is y expv[_ mW/2h( 2x^2 y^2)]
differentiate w.r.t x and y partially twice and use two dimensional SchrÃ¶dinger equation
so option first will be the correct choice

1. rajeshji , what is the answer of q. 28 70 73 of booklet series C

2. PLEASE GIVE THE ANSWER OF QN 72,69,65,62,51,45,36,30,28 OF SET A

93. Anonymous02 July, 2014

Did anyone try the sum on fractional error in measuring concentration?
Its 39 in booklet code C.

94. Anonymous03 July, 2014

Please ans which option is true for horses,donkeys, monkeys part A question.

95. generating function
ans option second q^2 P
Q= q^2
P=p/2q
time independent generating function will satisfy the condition of Q=del F /Del P
p = del F/del q
check it is satisfied by
q square P

just write p = 2qP ( which is give as P = p/2q)

96. #given time derivative is a function of x
#put it equal to zero
# equate the values of x
#evaluate derivative of function x at those fixed points
#if you find it is greater than one then it represent the unstable condition
# if it is less than one it shows stable condition
# in given option best match is b is less than minus one I.e option third

97. Anonymous05 July, 2014

What is the answer for transformer primary and secondary current problem, SET-B (the uploaded question) question Number 31?
Also explain the answer.

1. Anonymous05 July, 2014

ans. b.apply Faraday's law.current in secondary coil will b in such a direction that it opposes the change in current in primary coil.

98. Please give the answers of QUESTION NUMBERS 30,36,45,51,62,65,66,69,72 of Question Booklet ; A

99. Anonymous05 July, 2014

what is the answer key of question numbers 30,36,45,5162,65,66,69,72 of SET A

100. dependence of viscosity on M ....Is it 4/9 or 3?? Plzz reply...

1. I think its 3. First have a look at this link:
http://en.wikipedia.org/wiki/Mark%E2%80%93Houwink_equation

Now let a denote the dependence of eta on M then we can write:-
eta = k * M^a where k is a constant

Then we get
eta1/M1^a = eta2/M2^a

So, a = Log[eta1/eta2]/Log[M1/M2]

If you plugin any two pairs of data from the given list into the above eqn you always get nearly equal to 3

101. Anonymous07 July, 2014

Hi All,

CODE B:PART C: Q NO:46. here my doubt is, we need to consider average energy(total energy) or Ground sate energy, please clarify my doubt.

102. Poisson distribution result is option first {C2,C3}=C1
{C3,C1}=C2
use {A,B}= (del A/del x1)(del B/del p1) (del A/del x2)(del B/del p 2) (del A/del x3)(del B/ del p 3) -(del A/delp 1)(del B/del x1)-(del A/del p 2)(del B/ del x 3)-(del A / del p3)(delB/del x3)
which is satisfied by option first

103. Poisson distribution result is option first {C2,C3}=C1
{C3,C1}=C2
use {A,B}= (del A/del x1)(del B/del p1) (del A/del x2)(del B/del p 2) (del A/del x3)(del B/ del p 3) -(del A/delp 1)(del B/del x1)-(del A/del p 2)(del B/ del x 3)-(del A / del p3)(delB/del x3)
which is satisfied by option first

104. The Gibbs potential answer is zero
use 1/T= (del S / del U) at constant V
P/T = ( del S / del V) at constant U
put it in given equation of G
you will find result as
G = U- TS TS/4=U- 3TS/4
= 0(on putting the values of T and S from above)

1. Anonymous14 July, 2014

plz comment 49 and 65 Q.no's of bcode A

2. Anonymous15 July, 2014

Plz sir answer me the question of No.-66,73 of Set-C.

105. Anonymous14 July, 2014

How much does the total angular momentum quantum number J change in the transition of Cr (3d6) atom as it ionizes to Cr2+ (3d4) ? [Set-C, Q-73] plz answer it someone.
1. Increases by 2
2. Decreases by 2
3. Decreases by 4
4. Does not change

106. the total angular quantum number does not change as the value of L is same for both state and spin is also same in both state

107. Anonymous23 July, 2014

A particle of mass m in three dimensions
is in the potential--for this Q csir ans is option 1, please share me the scenario or reference book or URL for conversion from 3d to 1d at V(r)=0 and r<a (here r is radius if i am correct ?)

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