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CSIR UGC NET, 2013 December 22, Question Paper, Download, Key, Answers, solutions, Explanations, Comment Here

►CSIR UGC NET, 2013 December 22, Question Paper, Download, Key, Answers, solutions, Explanations, Comment Here

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The single paper MCQ based test  held on Sunday, the 22nd December, 2013. The Multiple Choice Questions comprises the following six subjects, as here under: Life Sciences,Earth, Atmospheric, Ocean and Planetary Sciences,Mathematical Sciences,Chemical Science,Physical Sciences & Engineering Sciences.

The question paper is divided into three parts, (A, B & C) as per syllabus & Scheme of Exam.
Part ‘A’ shall be common to all subjects including Engineering Sciences. This part shall contain questions pertaining to General Aptitude with emphasis on Logical Reasoning, Graphical Analysis, Analytical and Numerical Ability, Quantitative Comparison, Series formation, Puzzles etc.
In Engineering Sciences, Part ‘B’ shall contain questions pertaining to Mathematics &  Engineering Aptitude & Part ‘C’ shall contain subject related Multiple Choice Questions (MCQs), generally covering the topics in the syllabus.
And for other disciplines:
# Part ‘B’ shall contain subject-related conventional Multiple Choice questions (MCQs), generally covering the topics given in the syllabus.
# Part ‘C’ shall contain higher value questions that may test the candidate’s knowledge of scientific concepts and/or application of the scientific concepts. The questions shall be of analytical nature where a candidate is expected to apply the scientific knowledge to arrive at the solution to the given scientific problem.
* Negative marking for wrong answers, wherever required, applicable as per scheme of Exam.

UGC CSIR NET 2013 December 22 PHYSICS Answer Key Comment here with explanation
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224 comments:

  1. hai frnds please update key for csir net physics 22nd dec 2013

    ReplyDelete
  2. hai frnds please update key for csir net chemical science 22nd dec -2013

    ReplyDelete
  3. Please upload question paper of csir net december 2013 physical science

    ReplyDelete
    Replies
    1. send a scanned copy to physics43@gmail.com

      Delete
  4. CSIR 2013 Dec 22 Physical Science

    Booklet-C
    Part-A

    (1)
    (2) x+y=4
    (3) 1 cm
    (4)
    (5) Acce is zero once on the path
    (6) 32 (a*b*c*....)
    (7) 8 (nine-digit positive.....)
    (8) (10,-7)
    (9) C is lying
    (10) 1/101
    (11) pi/2
    (12)
    (13) 55
    (14) 20 cm (perimeter)
    (15) 19 fishes
    (16) 61
    (17) 55 (last number in 10th set)
    (18) Rs.300
    (19)
    (20) pi<4

    ReplyDelete
    Replies
    1. i feel almost correct ...

      except acce is zero


      bcoz V*T increases initially and it decreases then after

      Delete
    2. SETA:17: 365
      19: S30W
      I have doubt on 55.. the next number in the sequence

      Delete
    3. Acc is zero once on the path. Because if you take the derivative (i.e., acc) at the peak (maximum), it is zero.

      Delete
    4. (13) 39,42,46,50,..... the last number is 55.

      Third-First= 46-39=7
      Fourth-Second= 50-42=8

      So,
      Fifth-Third=9

      Fifth= third+9= 46+9=55

      Delete
    5. yes answer is 55 but my lgic was different..
      look..
      diff b/w 39 and 42 is 3,42 and 46 is 4,46 and 50 is 4 so we can see that next term=(first digit of the number+number)
      so we get 50+5=55

      Delete
    6. add first digit to that number...so 50+5=55

      Delete
    7. I think (14) cannot be determined option will be correct one.

      Delete
  5. PARTA :

    B- SET
    Q.No ANS
    2---- 4
    3----3
    4-----3
    5----3
    6---3
    7----3
    8----2
    9---4
    10----3
    11----4
    12---1
    13----3
    14---4
    15---1
    16----2
    18----3
    19-----4

    ReplyDelete
  6. i think pi lies between 3 and 4

    ReplyDelete
    Replies
    1. what is the answer of series question in general science.

      Delete
    2. atleast one value is greater than 1, the question of cylender to be broken in 3 pieces, Q.1, Booklet -C , Part - A

      Delete
    3. see pi * r^2 < (2r)^2
      => pi<4

      Delete
    4. Even though pi lies b/w 3 nd 4,but it is early to 3 .

      Delete
  7. Any reply for part b or part c ?

    ReplyDelete
  8. Set A Part A
    1. Who lying C Because statement of C contradict others s
    2.
    3. Total area of shaded portion PI/2
    4. Series 39, 42,46,50 ...... 54
    12X3 3 , 13X3 3, 14X3 4, 15X3 5, 16X3 6
    5. Full price of book 300
    6. Perimeter 20 top short n bottom short n both vertical line can be shifted lower n outer position to get the square shape
    7. Fishes 19
    8. After tenth bounce 1024/ 2x2.....tentimes i.e. 1 Cm
    9. Centre of circle 10,-7
    10. At least one piece has volume greater than 1 cm2
    11. X Y =4
    13. Max 24(1,1,2,3,3) but no such options so 12 is also correct set
    14. Nine digit no. 8
    16. 55
    18. Area of square is a square is greater than the area of circle pi a square by 4 so
    pi is less than four
    ans. 3
    19. S 60 degree W ( B is at N 00 A N30E so from B west with 60 S)
    20. 61

    ReplyDelete
    Replies
    1. 2. one can have 3 full, 1 half n 3 empty, and other two can have 2 full, 3 half n 2 empty each. and these combination can have 3 ways.
      Answer: 3

      Delete
  9. Set A part A
    15. (2-1)/1X2 (3-2)/2X3 (4-3)/3X4 (5-4)/4X5 ....... (101 -100)/100X101
    1-1/2 1/2-1/3 1/3-1/4 1/4-1/5 ...... 1/100- 1/101
    (1-1/101)/100
    ans 1/101

    ReplyDelete
    Replies
    1. your explanation is not correct for ans 4.
      the series is 39 42 46 50......55
      difference between ist and third term=7
      2nd and fourth=8
      3rd and 5th=9
      and ans is 46 +9=55.

      Delete
    2. mr.sant kumar just aad the initial digit in next number

      Delete
    3. mr.sant kumar just aad the initial digit in next number

      Delete
  10. series ans,
    39, 42, 46, 50, 53,
    as 3+9=12,
    4+2=6,
    4+6=10,
    5+0=5
    see the first no is 12, second is 6,
    see the third is 10, fourth is 5,
    i mean, odd places are decreasing by 2, even are by 1,
    so the next no. in the series will be 53, as it sum to 5+3=8 [which is the fifth number (odd place)] and also less by 2 from the 6+4=10
    ANS: 53

    ReplyDelete
    Replies
    1. 39+3=42 42+4=46 46+4= 50 50+5=55 add first integer of the number to itself

      Delete
  11. For a.b.c.d.e 32 is max possible 2 2 2 2 2 =10

    ReplyDelete
  12. PART B SET A
    Q.NO ANS
    25. 1
    28 2
    29 3
    31 1
    32 1
    35 4
    36 4
    37 4
    44 4

    ReplyDelete
  13. Honey
    full 1, 2,3 can be taken half 5,3,2 can be taken so minimum 2 set
    A B C A B C
    H 2 2 3 1 3 3
    HF 3 3 1 5 1 1
    E 2 2 3 1 3 3
    E

    ReplyDelete
  14. PART C BOOKLET CODE A
    Q.NO ANS
    47 4
    48 4
    49 1
    53 4
    55 1
    56 4
    59 3
    67 3
    69 1
    71 4
    75 1

    ReplyDelete
  15. book code c
    please....

    ReplyDelete
  16. hi someone please upload full solutions of 22 december 2013 net exam of physics..

    ReplyDelete
  17. probabability between -a/2 and a/2 will be 1/2+1/pi

    ReplyDelete
    Replies
    1. 1/2 - 1/pi

      Delete
    2. how plz explain ...both of u

      Delete
    3. pradeep ur answer is correct

      Delete
    4. yes its answer is 1/2 - 1/pi

      Delete
    5. answer is 1/2 - 1/pi

      integration of cosx is +sinx (not -sinx). plz check your calculation who got 1/2 + 1/pi

      Delete
    6. the correct answer is 1/2+1/pi

      Delete
  18. net december 2013 exam answer key 99% correct.....
    booklet code c
    1. 1
    2. 4
    3. 3
    5. 2
    6. 2
    8. 3
    9. 3
    10. 4
    11. 2
    12. 4
    22. 2
    23. 1
    24. 4
    25. 4
    26. 2
    27. 4
    29. 2
    30. 2
    31. 1
    32. 2
    34. 2
    36. 3
    38. 3
    39. 1
    40. 3
    41. 2
    42. 2
    43. 3/4
    44. 4
    50. 4
    51. 2
    52. 2
    53. 1
    55. 1
    57. 2
    60. 4
    61. 2
    62. 1
    63. 4
    65. 4
    66. 1
    68. 3
    69. 4
    74. 4
    75. 3

    by Mahesh Saini

    ReplyDelete
    Replies
    1. little bit correct,otherwise wrong

      Delete
    2. PLEASE LET ME KNOW.......WHICH ONE IS WRONG ANSWERS.....I POSTED.....HERE
      I MY KNOWLEDGE ALL ARE RIGHT.....

      Delete
    3. At the ground state always normal zee man effect seen...so no of split is 3.

      Delete
    4. Electric dipole moment create if there charge separation form,but in nucleus there only positive charge,,
      but if we consider atom then electric dipole moment creates,,so ans is 0.

      Delete
    5. hi Mahesh pls send me the scanned copy of booklet A @ physics.hemant18@gmail.com
      i hv the booklet B so i m unable to comment on ur answer.

      Delete
    6. sir may you send the scanned copy of booklet B @ santkumar1210@gmail.com?

      Delete
    7. no of splitting is 6,,,b/c there is strong magnetic field ans spin is not zero.....so it splitting by


      2l+1 *2s+1

      Delete
    8. this formula is for whole 2P term... but not for 2P1/2.. so possibly 3 splits are correct
      .

      Delete
  19. PART C:

    1/root X, x and x2
    spectroscopic for the ground state --- 6
    w(k) =2w0Sin(ka/2)
    h2/2Ba2
    vector A.dl ===0
    16x6-12x2
    1/2+1/pi
    hSqrt(2)
    10-4E0Cos(pi
    9/8d2(2sqrt2-1) towards the centre
    4 charges ---- 2

    ReplyDelete
    Replies
    1. spectroscopic for the
      ground state --- 6

      Delete
  20. plz. provide the complete key of sec b&c of book let A

    ReplyDelete
  21. section A booklet A
    1-3
    3-4
    5-2
    6-1
    7-2
    8-3
    9-3
    10-4
    11-4
    12-3
    13-2
    16-3
    18-4
    19-1
    20-1
    14-1

    ReplyDelete
    Replies
    1. 18. is 4 not 3. As the circle is inside the cube, (pi*r^2)<(2r)^2 gives pi<4

      Delete
  22. please send any one key for set -B

    ReplyDelete
    Replies
    1. section B booklet B
      21-1
      23-2
      24-4
      25-4
      26-4
      38-2
      45-1

      Delete
    2. part C
      54-2
      55-1
      58-1
      66-4
      67-2
      68-1
      74-1

      Delete
    3. Answer of 54-2 is right....... I thinking ans is 4

      Delete
    4. yes answer is 4 of que 54

      Delete
    5. yes answer 4 is correct of que 52

      Delete
  23. booklet A
    part B
    21-1
    22-4
    24-2
    27-4
    30-3
    31-1
    32-1
    33-1
    34-2
    35-4
    36-4
    37-2
    38-3
    41-3
    part C
    49-1
    53-4
    54-3
    55-1
    58-2
    59-3
    71-4
    75-1

    ReplyDelete
    Replies
    1. Please clarify 30, 33, 41. All others i think are correct. Dont you think 41 is 1. Means tc>tb>ta... it is a graph of material changing phase from a to c where b phase is intermediate mixed phase. Anyhow temp is directly prop to energy as well as entropy. So increasing graph of energy and entropy mean increase in temperature.

      Delete
    2. i think ta>tc>tb at b temp is constant but remaing two ie a $ c is positive temps

      Delete
  24. can u plz tell answers for booklet code B

    ReplyDelete
  25. This comment has been removed by the author.

    ReplyDelete
  26. Is there any one who can give the solution of Computer Science SET-A & Expected cutoff of CSIR NET 22 DEC 2013?

    ReplyDelete
  27. kindly send the andwer key of maths net dec 2013

    ReplyDelete
  28. can any one tell the cut off mark
    for LS in physical science exam this time????

    ReplyDelete
  29. what was the cut off for all category for june 2013?
    pls share

    ReplyDelete
  30. This time the cut off marks for Physical Science will be between 75-80 for general candidates...cause the paper is easy compared to last year December paper...

    ReplyDelete
    Replies
    1. I think LS cut off 80-85 and JRF cut off 85-90 for Gen and LS 60-65 and JRF 65-70 for SC/ST

      Delete
    2. LS cutoff will be around 72-73, UGC JRF around 80 and CSIR around 90.

      Delete
    3. yes, may be close to 80 for LS and a ten more for jrf

      Delete
    4. around 80 for ls for jrf may be 90 or 95

      Delete
  31. BOOKLET C
    PART C
    50. 4
    51. 2
    56. 1
    61. 2
    62. 2
    64. 4
    65. 2

    ReplyDelete
    Replies
    1. Booklet C, Part-C
      Ques 62. = ?

      Ans:

      Lx^2 + Ly^2 = L^2 - Lz^2

      = = 1/2 ( - )
      = 1/2 [ l(l+1)*h^2 - m^2 * h^2 ]

      Now "m" can take value from -l to +l including 0.

      MIN = 1/2 [ l(l+1)*h^2 - l^2 * h^2] (taking m=l )
      = 1/2 [ l* h^2 ]

      MAX = 1/2 [ l(l+1)*h^2]


      So, 4 is correct.

      Delete
    2. Mistake.....

      5th line: I used expectation symbols. But webpage rendered it. Hope you can understand

      Delete
    3. 5th line: Lx^2 = Ly^2 = 1/2 [ L^2 - Lz^2]

      Delete
    4. what is the condition for being Lx=Ly is not equal to Lz
      as far as symmetry ans simultaneity is concern Lx=Ly=Lz

      Delete
  32. SET B PHYSICAL SCINCE ANSWERS
    47-2
    49-2
    50-1
    52-3
    53-3
    54-1
    55-1
    56-2
    57-3
    58-1
    59-1
    61-4
    62-4
    63-1
    64-4
    65-3
    66-3
    67-4
    71-1
    72-3
    73-4

    ReplyDelete
  33. setB PHY. SC. ANS
    21-1
    22-1
    23-2
    24-3
    25-4
    26-3
    27-4
    28-2
    29-4
    30-1
    31-1
    32-3
    33-2
    34-4
    35-3
    37-4
    38-3
    43--3
    44-1
    45-1

    ReplyDelete
    Replies
    1. hi Vivek . . how u got d ans. of 44 as 4.
      isn't is 1.i.e . 100:1
      pls explain. .

      Delete
  34. IF ANY ANSWER IS INCORRECT PLEASE SEND YOUR CLERIFICATION ABOUT ANS

    ReplyDelete
  35. BOOKLET C (PHYSICAL SCIENCES)

    PART A

    1- 4
    2- 4
    3- 3
    5- 2
    6- 2
    8- 3
    9- 3
    10-4
    12-4
    13-1
    16-1
    17-3
    18-2
    19-3
    20-3

    PART B

    21-1
    23-2
    24-4
    26-2
    30-2
    31-1
    33-2
    35-3
    38-1
    39-1
    40-4
    41-2
    43-1
    44-4
    45-3

    PART C

    46- 1
    47-2
    49-3
    51-2
    55-3
    57-3
    58-4
    60-2
    65-4
    67-2
    68-3
    69-4

    PLEASE SEND YOUR SUGGESTION FOR THE INCORRECT ANSWERS WITH EXPLANATION...

    ReplyDelete
  36. 22. the expression del square by del x1 square + del square by del x2square + del square by del x3 square +del square by del x4 square ) * 1 /(X1+X2+X3+X4) IS EQUAL TO -16 (X1^2+X2^2=x3^2+X4^2)
    so its ans is option 4 for set A

    ReplyDelete
  37. This comment has been removed by the author.

    ReplyDelete
  38. 27. poission bracket value of {A,{B,C}}-{{A,B},C} = {{C,A},B}

    since {A,B}= - {B,A}
    and {{A,B},C} + {{C,A},B} + {B,C},A} = 0

    {{A,B},C} - {B,{C,A}} - {A,{B,C},}=0

    {A,{B,C}}-{{A,B},C} = - {B,{C,A}} ={{C,A},B}

    ANS 4



    ReplyDelete
  39. 28. LAGRANGIAN IS OPTION 2. L= - m square root of (1- derivative of X square) - V(x) refer Page 1096 mathematical physics by BS Rajput

    ReplyDelete
    Replies
    1. ya its right... will see same ans from h goldstein ... classical mechanics p. No. 344

      Delete
  40. This comment has been removed by the author.

    ReplyDelete
  41. SET B
    Part C
    52 4
    54 4
    61 4
    63 3
    65 3
    66 4
    67 1
    69 3
    71 2
    72 2
    73 4

    Part B
    21 1
    23 2
    24 4
    25 4
    29 2
    30 2
    33 3
    41 1
    42 1
    45 1


    Part A
    04 3
    11 4
    12 4
    14 4
    15 1

    ReplyDelete
    Replies
    1. 33 have Vo= 5 volt
      And 69 have ans 4

      Delete
    2. 33 have Vo= 5 volt
      And 69 have ans 4

      Delete
    3. Can you explain your answer?

      Delete
  42. 31. force between two parallel conductor I1*I2/ D

    32. The following transformation leave the electric and magnetic field unchanged
    option 1. A' = A + Del Psi
    V' = V - Del psi by Del t

    refered page 493 UGC CSIR NET SERIES TUTOR GAUGE TRANSFORMATION

    34. Probability of getting +h/2 and - h/2 for Sz ans . option 2. 2/11 and 9/11
    36. The expectation value of Lz = 5/18 h cut

    45. JK flip flop can act as :
    1.D or T type flip flop when both its inputs are J=1,K=1 and its output compliments
    2. RS fip-flop when J=1 and K=1 condition is avoided
    refered digital electronics by RK Gaur

    so ans is option 4 follows the inputs and the cct acts like in RS-flip flop


    ReplyDelete
  43. SET -B

    Part- A
    01-2
    03-3
    04-3
    06-3
    07-3
    08-2
    10-2
    12-1
    15-1
    19-4
    20-4

    Part -B

    21-1
    22-2
    23 -2
    25- 4
    26-3
    27-4
    31-4
    32-1
    35-1
    36-2
    45-4

    Part -C

    46- 2
    53-2
    54-2
    55-1
    57-1
    58-1
    59-2
    60-1
    61-4
    67- 1
    69-4

    ReplyDelete
  44. set c

    please upload the correct key for the following questions of SET

    1. 21
    2. 40
    3. 41
    4. 42
    5.45
    6.49
    7.55
    8.67
    9. 69
    10.75

    ReplyDelete
  45. A peturbation v=aL^2 is .........applied on 2p state(spin should not be considered).....
    WHAT IS THE ANS OF THIS QUESTION?(EXPLANATION REQUIRED)

    ReplyDelete
    Replies
    1. v= aL^2 nd 2p state is given , so l=1 nd for 1st order correction , eigen value of l^2 is
      l(l+1)h^2 .... answr is 2h^2

      Delete
  46. ans is2ah^2 becoz 1st order pertubation is its expectation value and 2nd order pertubation is zero

    ReplyDelete
  47. set a
    1.c
    2.
    3.4
    4.4
    5.2
    6.3
    7
    8..3
    9.3
    10/4
    11.4
    12.4
    13.
    14.
    15.2
    16
    17.
    18.3
    19.3
    20.1
    part B
    21.
    22.
    23.
    24.
    25
    26.
    27
    28
    29. 3
    30
    31. 1
    32. 1
    33 2
    34
    35
    36
    37 2
    38 4
    39 2
    40 4
    41
    42
    43 3
    44
    45
    Part C
    46. 2
    47 1
    48 4
    49
    50 3
    51 4
    52 1
    53
    54
    55
    56
    57
    58
    59 3
    60
    61
    62
    63 3
    64 4
    65
    66
    67
    689
    70
    71
    72
    73
    74
    75



    a5

    ReplyDelete
  48. Plz dnt update ur all ans..only update dose wch r true 100%

    ReplyDelete
  49. This comment has been removed by the author.

    ReplyDelete
  50. dr. Mahesh pls allow me to access ur google drive.....

    ReplyDelete
    Replies
    1. This comment has been removed by the author.

      Delete
  51. This comment has been removed by the author.

    ReplyDelete
  52. Hi friends, I am sharing with you a link that contains net DEC 2013 question paper( booklet code A) with answers. The blue circled answers are correct according to me . If any mistake found don't forget to comment below.

    by Maahiya


    https://drive.google.com/file/d/0B6uGxbVkmL9leElmR3g4dGlpN00/edit?usp=sharing

    ReplyDelete
    Replies
    1. Answer for 6th question is 20cm, simple logic (since it is right angles, perimeter is same as the 5cm sided square)

      Delete
    2. hi Mahesh
      ans. for Q. 69 should be 3 as it is strong field , so no. of splitting would be N = 2*L+1

      Delete
    3. yes 3 is correct because ...question is about 2P1/2 only ....but not about whole term 2P...in 2P term there are two fine structure levels 2P3/2 and 2P1/2...both have split in 3 and 3 ..so total 6 as according to formula (2S+1)*(2*L+1). But here it is asked about 2p1/2 level not term ...so it is 2L+1 = 3.

      Delete
  53. for terminal velocity answer is 8...because time to reach half of its terminal velocity is directly prop to mass of the body

    ReplyDelete
    Replies
    1. terminal velocity directly proportional to mass of the body.... but inversely proportional to its radius... So correct answer is 4.. option 3.

      Delete
  54. set A
    PART A
    1-3
    4-4
    5-2
    7-2
    8-3
    9-3
    12-2
    14-1
    16-3
    20-1
    part B
    22-4
    25-1
    27-4
    30-3
    31-1
    32-1
    34-2
    35-4
    36-4
    38-3
    44-2
    45-4

    ReplyDelete
    Replies
    1. how did u get option 2 for question 44

      Delete
  55. hi frnds plz explain possible states for three fermions in 2 states

    ReplyDelete
    Replies
    1. two electrons in ground state with up and down spin and other electron in either up or down in excited..........same for two in excited and one in ground state with either spin up or down..............

      Delete
    2. Hi Mahesh
      ans. for Q.60 is 1/4 or 1/8 ?????
      is it electric dipole or A Quadrupole ????

      Delete
  56. electrostatics potential ratio V(2r)/V(r) is

    ReplyDelete
  57. This comment has been removed by the author.

    ReplyDelete
  58. plz explain question 75 in set C

    ReplyDelete
  59. 29. Angular velocity of ring when reaches the base
    height of inclined plane(h-R)
    the loss of P.E = gain of K.E
    mg(h-R) = 1/2 m V ^ 2 1/2 I w^2
    since Iw =m R^2 about centre of ring
    and also V= RW
    so we get square root (g(h-R)/R^2)
    this is option 3

    referred properties of matter by brijpal and N.subrahmanyam page 463 s.chand

    ReplyDelete
  60. 53. Del square psi plus 2m by h cut square(E-V) psi = 0 is time independent schrodinger equation

    solve it for given psi = A exp( - X^4)
    give V(X) = 16 X^3- 12 X^2
    option 4

    ReplyDelete
  61. Use given value of h cut=2m=1

    ReplyDelete
  62. CAN ANY ONE TELL THE ANSWER FOR THE FOLLOWING QUESTIONS...

    1- PENDULUM CONSISTING A RING OF MASS M (I THINK 2 pi sqrt (l
    +R)/g should be correct answer as the effective length will be l+R )

    2- RANDOM WALK BY CHILD N LARGE STEPS (sqrt2N x 10^-2 a in north east direction is right)

    3- FOR POINT CHARGES THE ANSWER SHOULD BE 1/8 AS FOR r>>a
    it will behave as a quadrupole and for that V prop 1/r^3

    ReplyDelete
  63. for particle is moving in potential V=x^2+y^2+z^2/2 ..answer is only Lz is conserved

    ReplyDelete
    Replies
    1. No... None is answer... because Lx,Ly and Lz don't commute with the hamiltonian... For constant of motion the angular momentum should commute with the hamiltonian....

      Delete
    2. potential is given ...so force F=-Del(V).

      so force F= -(2xi+2yj+zk)

      now find the Torque T= rxF= yzi-xzj+ (0)k
      now compare with T = Txi+Tyj+Tzk

      Tx (x component of torque) = yz
      Ty = - xz
      Tz = 0

      so Z component of torque is zero ..i.e z component of angular momentum is conserved.

      Delete
    3. r is a postion vector... we can't say it is in z direction only... but classical approach is correct...

      Delete
    4. position vector is in "z" direction?

      Delete
  64. for Carnot engine problem ...answer is sqrt (T1*T2)

    ReplyDelete
    Replies
    1. The answer is (t1+t2)/2 because both the bodies are of equal heat capacity... so the energy lost by first body to decrease one degree is exactly equal to increase temp of second bodyy by one degree...

      Delete
  65. for three identical spin 1/2 fermions ...answer is 2 possibilities

    ReplyDelete
    Replies
    1. hi Vijay
      what is the ans. of Q.60 .
      i.e, it is electric dipole or Quadrupole???????

      Delete
    2. Fermion answer is 2 possibilities...... refer p.no. 325 statistical mechanics by reif

      Delete
  66. booklet code c
    section A
    2. 4
    4. 3
    5. 1
    6. 2
    8. 3
    9. 3
    11. 4
    14. 4
    16. 1
    18. 2

    Section B
    23. 1
    24 3
    25. 4
    26. 3
    28. 1
    29. 4
    31. 1
    33. 4
    34. 2
    39. 1
    40. 3
    41. 2
    42. 2
    section c
    47. 1
    51. 2
    52. 2
    53. 1
    56. 1
    57. 4
    59.4
    64. 4
    65. 3
    67. 3
    68. 2
    73. 4
    74. 3
    75. 3

    ReplyDelete
    Replies
    1. if any answer is incorrect please clarify.........

      Delete
    2. book let code B.
      PART A
      2. 2
      3. 3
      4. 3
      5. 3
      6. 3
      7. 3
      8. 2
      9. 4
      10. 3
      11. 4
      12. 1
      14. 4
      15. 1
      16. 2
      17. 1
      18. 3
      19. 4
      20. 2

      Delete
  67. ans . series 39.42,46,50,53

    ReplyDelete
  68. ans. correct ans is 53 in series 39,42,46,50....

    ReplyDelete
    Replies
    1. Ans is55 ...... 39+3=42. 42+4=46. 46+4=50. 50+5=55. add no with its first digit

      Delete
  69. Please upload full answer,not 1,2,3,4 etc . because we've not all A,B,C question booklet. ..

    ReplyDelete
  70. booklet c
    part A
    3-4
    4-2
    6-2
    7-1
    8-3
    9-3
    11-4
    12-1
    13-4
    15-2
    16-1
    17-3
    18-2
    19-3
    20-3
    part-B
    22-2
    25-4
    26-3
    27-4
    31-1
    34-2
    37-4
    39-1
    45-2
    part -C
    46-1
    49-3
    50-4
    51-2
    56-1
    57-1
    59-4
    60-2
    61-3
    62-4
    64-4
    65-2
    67-3
    73-1
    74-2
    75-1

    ReplyDelete
    Replies
    1. This comment has been removed by the author.

      Delete
    2. plz brother upload booklet C paper or give the answer hint because we dont have C paper booklet. So it creates problem to match exact questions and answers

      Delete
  71. booklet code c
    46.option 3 (time period of a physical pendulum is 2*pi*sqrt(I/MGL). I is moment of inertia about the axis of rotation.L is the distance of the centre of mass about the point of suspension). I HERE IS MR^2 + M(R+L)^2. LENGTH L is L+R

    ReplyDelete
  72. 49.option 1 is correct. the net force is along x axis. f(net)= 2k(sqrt(L^2-X^2)-L^2)sin(theta).
    sin(theta)=theta(small).
    theta= X/L= sin(theta).
    substituting and using binomial theorem we have f(net)= kx^3/l^2

    ReplyDelete
  73. 48. option 1 is correct.
    for a classical particle for each quadratic term we have in the energy expression there is a energy contribution of 1/2kT.
    kinetic energy contribution = 3/2 kT( as 3 quadratic contribution.v(x)^2 ,v(y)^2),v(z)^2
    potential energy contribution= 3/2 kT. 3 dim potential. 3 quadratic terms.

    ReplyDelete
  74. 52. option 4 is correct.
    near the bottom of band k is small.
    so coska can be approximated as 1- ka^2
    so expanding cosk(x)a+ cosk(y)a+cosk(z)a= 3- (k(x)a^2+k(y)a^2+k(z)a^2)/2
    =3- (ka)^2/2
    m*= hbar^2/ d^2(E)/dk^2=hbar^2/Ba^2

    ReplyDelete
  75. 57. option 4 is correct
    monopole= sum of charges= 0
    dipole= sum of ( magnitude of charge * its position vector)= 0
    so its a quadrapole. V varies as inverse of distance cube

    ReplyDelete
  76. 75. option 3 is correct
    this is an image problem
    here we can replace the two grounded infinite plates by two equal and opposite charges i.e -q at distances 2d from the charge q (to keep the potential boundary conditions same. ). now solving for the new problem at hand we find net resultant force due to the new configuration. this gives d ans.

    ReplyDelete
    Replies
    1. i think 2 negative and one positive charge is induced.because 2 negatve charge and original positive charge wont yield boundary condition

      Delete
    2. you can chk griffiths book. chapter 3 image problem aptly expalins what i have said.

      Delete
    3. sory nagesh i am wrong.ur concept is right. but i feel ans should be option b.

      Delete
    4. yes i am agree with the shiv kumar

      Delete
  77. @shivakumar : I have also written 52 ans as 4 option. But csir previous ans says it as 3Ba^2. How is that possible?

    ReplyDelete
  78. @shivakumar 52 ans is 4th option are sure? Coz its a previous year csir question and in earlier key they gave the ans 3Ba^2

    ReplyDelete
    Replies
    1. in the earlier question k was at the boundary of the boundary of first brillouin zones so k is not small. but dis yrs question its in the bottom of the band so k is small so energy expression can be approximated by taylor series expansion of cos function

      Delete
    2. what about question 23 about analytic functions? i got 3 answers as correct..

      Delete
    3. ans is option a. any analytic function should satisfy the laplace eqn. first option does not satisfy

      Delete
  79. what is the ans for Fourier transform of first derivative of Dirac delta function??

    ReplyDelete
  80. Replies
    1. yes..because fourier transform of dirac delta function is 1.so use the property of derivative of fourier transform.answer is ik

      Delete
  81. can some one tell the answer of q 45 of booklet c. it is the v-i characteristics of zener diode in series question

    ReplyDelete
    Replies
    1. Ans:B . Because in forward characteristics I is increasing. But in reverse first very low value than break down occur. So at first forward is dominant. But after that both ate high and opposite. So we get constant graph.

      Delete
    2. Q.45 ans is 3 ....... 100% correct.

      Delete
  82. what is the ans for q no 70 in book let c( digital electronics sop method)?

    ReplyDelete
    Replies
    1. what is the answer for perimeter problem in part a

      Delete
    2. This comment has been removed by the author.

      Delete
    3. q no 70 ans
      BbarDbar+BD

      PERIMETER PROB ANS 20

      Delete
    4. is it correct ans for qno 70?

      Delete

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