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CSIR UGC NET, 2014 Dec 21, Question Paper, Download, Answer Key, Answers, solutions, Explanations, Comment Here

► CSIR UGC NET, 2014 Dec 21, Question Paper, Download, Answer Key, Answers, solutions, Explanations, Comment Here

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UGC CSIR NET 2014 December 21 Answer Key Comment here with explanation
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Please comment the key only if you are sure about the answer, and also give the explanations regarding your answer.
Download Dec 2014 Physics NET question -C


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111 comments :

  1. This comment has been removed by the author.

    ReplyDelete
  2. Specific heat for diatomic 5/2 k

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

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  4. Replies
    1. how???? delB/delT=0

      Delete
    2. Shruti Gupta, your formula is absolute right according to this link see equation 2.14. so the calculation is found out to be Z=84.10. so the answer will be Z=84

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

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    4. https://www.google.co.in/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&cad=rja&uact=8&ved=0CBwQFjAA&url=http%3A%2F%2Fwww.springer.com%2Fcda%2Fcontent%2Fdocument%2Fcda_downloaddocument%2F9780387016726-c1.pdf%3FSGWID%3D0-0-45-166533-p34951025&ei=4jKZVKzbOomQuAS8mIDICQ&usg=AFQjCNG9YDMgv97TKCOUEqm3jwROOPNheQ&sig2=i3dYsK99wIkqAD2L2Nhjrw

      Delete
    5. delB/delZ=0
      => (-a/A)*2*(2Z-A)*2-(a/A^(1/3))*2Z=0
      => -0.44(2Z-A)=0.25Z
      =>-0.88Z+95.04=0.25Z
      =>0.88Z+0.25Z=95.04
      =>1.13Z=95.04
      =>Z=84.10

      Delete
    6. here first a= a(sys) and second a= a(c)

      Delete
  5. booklet c part A
    1 (2)
    2 (2)
    3(1)
    4(1)
    8(3)
    9(4)
    11(4)
    13(4)
    14(3)
    15(2)
    17(2)

    ReplyDelete
  6. BOOKLET C part B

    Q-24 CORRECT OPTION IS (1)equation of motion

    ReplyDelete
  7. BOOKLET -C Q-25 CORRECT OPTION IS (4) because 3 dimensional position vector is r=xi+yj+zk .so the divergence of position vector is 3 and curl is zero

    ReplyDelete
  8. BOOKLET -C Q -27
    CORRECT OPTION IS (3) because for free particle energy E=MC^2 and then solving it for velocity ...one finds option (3)as the right one.

    ReplyDelete
  9. 1 b
    2a
    3a
    4a
    5c
    6d
    7c
    9d
    10b
    13c
    15b
    16b
    17c
    19a
    20a
    21c
    25b
    26c
    27c
    29c
    30b
    31a
    32b
    33d
    34a
    35d
    36c
    37a
    38b
    40c
    41a
    42b
    43c
    44a
    45d
    46a
    47a
    50d
    51c
    52d
    55d
    58a
    59d
    61b
    63c
    64a
    65b
    67d
    68c
    69b
    71b
    72a
    73d
    74d
    75c

    ReplyDelete
    Replies
    1. this is booklet C

      Delete
    2. Can u pls explain anwer of 74is d how

      Delete
    3. Anwer of 67 is 1 because flux cannot pass thr superconductor in presence B meisner effect.

      Delete
  10. give the anwer of booklet-B

    ReplyDelete
    Replies
    1. http://physics43.files.wordpress.com/2014/12/net-physics-question-dec-2014-physicskerala.pdf

      Delete
  11. Booklet-C
    ***********************************************
    Part-A
    ---------------
    1(c)------> divisible by 3 and 7

    2(b)------->1Kg

    3(c)-------> both spheres will cool down at same rate (use Newton's law of cooling)

    4(a)-------> 3000

    5(a)-------> 2^n

    6(b)------->circle

    8(b)------> <1% gain (40/41 % gain)

    9(d)------> (O is missing)

    10(d)-----> proportional to surface area

    11(d)-----> 19

    15(b)------> 2001

    16(a)------> 1:1

    17(c)------> 25 students



    Part-B
    ---------------------

    24(a)-----> first equation of motion

    25(d)-----> del.r=3 and del(cross)r=0

    27(c)-----> total energy E= gamma*m0*c^2

    29(c)----> 32km

    31(a)----> i(1-p)p^-p

    33(c)----> -ihwx(d/dx)

    36(d)-----> r^-3 (quadrapole)

    39(b)---->(a+bp)pV

    44(d)----> 6.7 mA



    Part-C
    -----------------------

    47(a)----> 16I

    51(a)----> 4/9 (reflected/incident)

    53(d)----> (1/2m)Q^2 P^4 +(mw^2)P^-2

    58(b)----> -(2mb^2/pi^2h^2)

    64(c)----> x= root(h/2mw) and c=root(1/2)

    65(b)---> (**)(n+1/2)^2/3

    67(a)---> quantized flux is not hc/e

    69(b)---> 2pi/3 (r1^3+r2^3)/(r1+r2)^3









    ReplyDelete
    Replies
    1. 2(b)......plz calculate it...the ans should be 0 kg

      Delete
    2. for ques no 3 of part A, the ans is (b) ie smaller sphere because of stefans law

      Delete
    3. I have doubt on:
      Question (3), (33), (39)



      Correction:

      Question (16)-----> (b) ratio of volumes = 1:2



      Question (44)----> (c)
      **************************
      Diode resistance is 500 ohm. Total resistance is 1500 ohm.
      Now 0.7 volt is dropped across the diode. Remaining voltage= 10-0.7= 9.3 volt

      Current = (9.3/1500) = 6.2 mA

      Delete
  12. Where do I find the question?

    ReplyDelete
    Replies
    1. http://physics43.files.wordpress.com/2014/12/net-physics-question-dec-2014-physicskerala.pdf

      Delete
  13. In booklet c, 100% sure ans are
    5)4
    9) 4
    11) 4
    Part B
    24) 1
    25) 4
    27) 3
    31) 2
    36) 2
    38) 2
    40) 4
    Part C
    49) 2
    55) 2
    67) 3
    70) 1

    ReplyDelete
    Replies
    1. (1+x)^n=1+........ put x=1 we have 2^n

      Delete
    2. 70) 4th is also kinematically forbidden!

      Delete
  14. pl.dont publish incorrect answers..if you think that your answer is correct then also give explanation.

    ReplyDelete
  15. book let C
    ---------------------------------------------
    1) 2
    2) 2
    4) 1
    9) 4
    11) 4
    15) 2
    17) 3
    19) 1

    Explanation.
    QUESTION (1). let n=1
    So, n(n+1)(n+2)(n+3)(n+4)(n+5)(n+6)
    =1*2*3*4*5*6*7
    =5040 (It is divisible by both 3 and 7) 5040/3=1680 and 5040/7=720
    QUESTION (2). (70+72+74+76+78+80+82+84+86+88+90+92+94+100+79)/15=(1245/15)=83
    And (70+72+74+76+78+80+82+84+86+88+90+92+94)/13=(1066/13)=82
    So 83-82=1
    -------------------------------------------------------
    QUESTION (4).s=(a+b+c)/2, here a=50m, b=120m, c=130m
    So, s=(50+120+130)/2=(300/2)=150
    =>(Area)^2=s(s-a)(s-b)(s-c)
    => (Area)^2=150(150-50)(150-120)(150-130)
    => (Area)^2=150*100*20*30
    => (Area)^2=9000000
    => (Area)=squareroot(9000000)
    => (Area)=3000m^2
    -------------------------------------------------
    QUESTION (11). 2,3,4,7,6,11,8,15,10,?
    We need the 10th number i.e the even placed number
    See odd placed number are 2,4,6,8,10
    Even placed number are 3,7,11,15, (?). <----see these numbers are differ by 4
    Next number after 15, 15+4=19
    -------------------------------------------------------------
    QUESTION (15).see which one is smaller, (50*100)/150=33.33
    (75*100)/250=30
    (75*100)/200=37.5
    (50*100)/100=50
    So the least variability in (250+/-75) this is for the year 200
    -----------------------------------------------------------------
    QUESTION (17). let total students =x
    20%of x=x/5 <----- they got jobs in the first year
    Remaining students= x-(x/5)=4x/5
    20% of (4x/5)= (4x/25)
    jobless=x-((4x/25)+(x/5))=(16x/25)<-------- this is 64% of x
    So 64% are jobless
    In question jobless=16, so if 64%=16, then 100%=25, so x=25
    ------------------------------------------------------------------------------------------------
    QUESTION (19).
    2,5,10,17,28,41,
    The difference between consecutive numbers are respectively
    3,5,7,11,13 see these are composite numbers so,
    The next 3 numbers will be 58,77,100 they fulfill the sequence as the next composite numbers are 17,19,23

    ReplyDelete
  16. Book let C
    ----------------------
    (27) using m=M*sqrt(1-(v^2)/(c^2)) -----------------(1)
    as E=Mc^2
    M=E/(c^2)
    put this M in equation 1.
    m=(E/(c^2))*sqrt(1-(v^2)/(c^2))
    v=c*sqrt(1-(m*(c^2)/E)^2) <----------------option 3.

    ReplyDelete
  17. Book ket C
    --------------------
    (16) the answer will be 1:2, option 2
    solution:
    if a rectangle of length d and breadth d/2 is revolved once completely around the length it will be a cylinder of radious d/2 and height d, volume=(pi)*(r^2)*h=pi(d/2)^2*d
    same rectangle of length d and breadth d/2 when revolved once completely around the breadth it will be a solid cicular sheet (like a birthday cake) of radious d and height d/2, volume=(pi)*(r^2)*h=pi(d)^2*d/2
    their ration= (pi(d/2)^2*(d))/(pi(d)^2*(d/2))=1:2

    ReplyDelete
  18. Book let C
    --------------------------------
    14. the answer will be 20, option 3
    the pattern will be like this LUNCH=V V V V V V V V V V V V V V V N N N N N
    DINNER=N N N N N N N N N V V V V V V V V V V V
    this combination matches with the question condition. 1) if he takes a non-veg (N) lunch, he will have only veg (V) for dinner
    2)he taken non-veg dinner for exactly 9 days
    3)he takes veg lunch for exactly 15 days
    4) he takes a total of 14 non-veg meals

    ReplyDelete
  19. what is the answer of question no 41 in booklet code C (phase space problem) ????

    ReplyDelete
  20. when csir will publish key ??

    ReplyDelete
  21. wa this dec paper tougher than june one??....what will be the cut off for gen this time?

    ReplyDelete
    Replies
    1. no this time paper is easier than previous one....so cut off will be around 80 for LS and 90 for JRF

      Delete
    2. Ya... It appeared tough this time... I expect the cut off for LS to be in the 60s,,,,

      Delete
  22. What is the intensity of radiation emitted by moving charges?

    ReplyDelete
  23. Question : A sphere is made up of very thin concentric shells of increasing radii. The mass of an arbitrarily chosen shell is

    My answer : (Mass)/(Volume) = Density = constant.....

    Therefore, Mass = constant * Volume....

    Hence, Mass is proportional to Volume (Option b)

    Please correct if I am wrong...

    ReplyDelete
  24. In Booklet C the answer of question (70) is option (4). Because free proton can not be converted into neutron, proton can be converted into neutron only inside the nucleus.

    ReplyDelete
    Replies
    1. Yes, Pranab Dhar. You are right. If any one want the reference they can check at following link.
      https://books.google.co.in/books?id=Il63XT5qbQEC&pg=PA483&lpg=PA483&dq=particle+physics+which+processes+are+forbidden+for+free+particles&source=bl&ots=mjNCtX1jrJ&sig=QgF8MV69_jHKqyC2JFRvNFt6mZ0&hl=en&sa=X&ei=vMmaVPLBEYfluQT8h4LgCQ&ved=0CC0Q6AEwAw#v=onepage&q=particle%20physics%20which%20processes%20are%20forbidden%20for%20free%20particles&f=false

      See equation 16-18. so option 4 is only correct answer.

      Delete
    2. Modern Physics By Bernstein

      Delete
    3. both the pion decay processes are allowed according to wikipedia. see http://en.wikipedia.org/wiki/Pion.

      Delete
  25. What is answer of 64 in booklet c

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

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  27. Booklet C Q. no 2 ans option (2) 1 kg. It is easy calculation

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

    ReplyDelete
    Replies
    1. Average 13 number =(70+72+74+76+78+80+82+84+86+88+90+92+94)/13=1066/13=82
      Now Average 15 number =(70+72+74+76+78+80+82+84+86+88+90+92+94+100+79)/15=1245/15=83

      So the average increases by 1

      Delete
  29. Anyone please upload question booklet B

    ReplyDelete
  30. Yes, u r right Shraddha. Q.no (67) correct option (1). Because the flux passing through the super conductor is quanised in units of hc/2e, not hc/e. This is due to formation of cooper pair by two electron of charge 2e.

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

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  32. What is answer of Q42 booklate c

    ReplyDelete
    Replies
    1. Answer of Q42 is 4.because given circuit is full wave rectifier using two op amp.function of first op amp is lilke buffer.so it help to keep positive half cycle of input signal.given 4options are not given this answer.so option 4 is right i think so pls iam wrong then give answer.

      Delete
  33. Booklet C-
    Question Number 65 (The WKB approximation problem)
    answer will be option (2) .
    (65) 2.
    For reference check zettili quantum mechanics solution manual. question number 9-38, chapter 9.
    This question was taken from Zettli's book. its available in NET for free download(The book but not the solution manual).

    ReplyDelete
  34. What is the cut off for JRF in GEN category this time???

    ReplyDelete
    Replies
    1. I strongly feel that it may not be more than that of June 2014. It was 69 in June for GEN category. I expect it to be less than that. Fingers crossed... :)

      Delete
  35. give answers of booklet A

    ReplyDelete
  36. answer key of booklet A
    part A
    2 b
    4 b
    5 a
    8 d
    10 a
    12 b
    13 d
    16 a
    17 c
    19 a
    20 b

    ReplyDelete
  37. part B
    21 a
    22 a
    23 c
    24 d
    28 c
    29 a
    31 c
    35 b
    36 c
    38 d
    44 d

    ReplyDelete
  38. part C
    46 a
    48 a
    51 d
    59 c
    60 c
    65 b
    66 c
    71 a
    75 d

    i have doubt in 60.if there is any mistake then please tell me

    ReplyDelete
    Replies
    1. Hi Nivedita,

      If answer key for a particular booklet code is given, it will never help people with other booklet codes. Moreover, It will be very helpful if you can explain your answers for the questions that were not discussed above. Please try to explain which ever you can. Thanks in advance...

      Delete
  39. As published by website: Minimum cut–off percentage for the award of fellowship/lectureship in different disciplines in the Joint CSIR –UGC test for Junior Research Fellowship and Eligibility for Lectureship held on 22nd June, 2014.

    Physical Sciences 39.5 (JRF) 34.5 (LS)

    But why you people say it is 69?

    ReplyDelete
    Replies
    1. You are talking in terms of %. I was mentioning in terms of marks out of 200. That's it. Hence, for June 2014 exam, the cut off for JRF was 79 and the cut off for LS was 69. It's as simple as that...

      Delete
  40. Hi friends
    Pls verify your answer with answer key given by Career Endeavour at folloeing link

    http://www.careerendeavour.com/fck/file/DECEMBER/CSIR-NET-JRF%20ANSWER%20KEY%20BOOKLET%20B%20_PHYSICAL%20SCIENCES_(1).pdf

    http://www.careerendeavour.com/fck/file/DECEMBER/01(1).pdf

    ReplyDelete
    Replies
    1. Thanks Pranab... This helps..

      Delete
    2. MANY ANSWERS HERE ARE WRONG

      Delete
    3. Please post ur explanations. That will help...

      Delete
    4. Can u tell me where we get answer of booklate code c

      Delete
    5. Shraddha, Please find two web links shared by Pranab.. It has questions as well as answers for booklet code B.. You can go through that and make ur own key for booklet code C (though some people say that it has many wrong answers).

      Delete
    6. answers keys given by career endeavour for booklet-B of physics 2014(II) 36,59,60,72 may be wrong, kindly confirm.
      answers keys should be 36-3, 59-1,60-2, 72-3, accrding to Griffith in introduction in quatum mechnics & others references.

      Delete
    7. I too have doubts in many questions. Will check...

      In one of the questions in part A, How can mass depend upon surface area? Can anyone please explain?
      In my opinion, mass depends upon volume. The following is my explanation.

      Mass/Volume = density = constant
      Hence, Mass = Constant * Volume
      Thus Mass is proportional to Volume..

      What do u people say?

      Delete
    8. ya in my opinion u r right

      Delete
    9. but i think it should be the radius.....coz in both the cases whether its volume or surface are only radius is variable but all the other terms are constant....i think option c is correct

      Delete
    10. question itself says increasing radii, mass proportional to radius ..all the other things are constants...hance option-c should be correct.

      Delete
    11. http://www.algebra.com/algebra/homework/word/geometry/Geometry_Word_Problems.faq.question.101764.html

      This will give an insight...

      Delete
    12. Hi friends
      I think the answer of question related to mass of the shell (q.no 20 in booklet B) will be option (4) i.e the mass of any shell is proportional to surface area of each shell.

      Explanation:
      Let us consider a shell of radius r and negligibly small thickness dr , then its volume = 4*pi*r^2*dr
      therefore its mass = 4*pi*r^2*dr*d (d= density)
      From this it seems the mass is directly proportional to r^2 (as dr is arbitrary, d is constant)
      But proportional to r^2 is not given in any option. So it must be proportional to surface area S (=4*pi*r^2).
      Here I also want to mention that all the answer key given by Career Endeavour may not be correct.Last time also this happened.
      But more than 90% answer are correct.

      Delete
    13. pranab dhar is absolutely right...

      Delete
  41. what will be the expected cut off this time for general category?

    ReplyDelete
    Replies
    1. In a particular website, it is given that 29% is the expected cut off for LS for general this time... I am not able to find it now....

      Delete
    2. may be 75 for JRF (Gen) ie, 37.5% & 65 for NET(Gen) ie, 32.5%.

      Delete
  42. many answers are wrong in career endeavour key....i wonder what they teach to their students

    ReplyDelete
  43. i think answer of que no 10 of part A in booklet C is c..............mass is proportional to radius.......coz its varable rest terms are constant whether its volume or surface are

    ReplyDelete
    Replies
    1. for nuclear reaction problem which one is the correct answer in my view only 2 nd eqn is correct

      Delete
    2. yes correct, b'coz itself in the question it is written increasing radii...all the other things ie, are constants, answer should be mass be proportional to radii...

      Delete
    3. What about this argument then?

      Mass/Volume = Density = constant (for any material)

      Thus Mass = Constant * Volume

      Hence, Mass is proportional to Volume...

      Delete
    4. when we expand the volume term ie.in case sphere it is 4/3 pie R^3 in this R is variable everything else(4/3 pie) come out from integral & radius term will integrate to make it to the volume.

      Delete
    5. the problem in this website (http://www.algebra.com/algebra/homework/word/geometry/Geometry_Word_Problems.faq.question.101764.html) says

      "Since volume is proportional to the cube of the radii, then mass is also proportional to the cube of the radii"

      Delete
  44. i also think it

    ReplyDelete
  45. When csir key will publish

    ReplyDelete
  46. splitting of energy levels due spin-spin interaction b/w elctron & proton should be 2ah^2 & not ah^2 acc. to stndard book griffith "introduction to quantum mechanics "hence ans should Q-72-option2 & not option3 of booklet-B 2014(II) as given by career endeavour.kindly cross-check .

    ReplyDelete
  47. Kartik same question asked in dec 2011.correct answer is ah^2.explaination
    Total spin s=s[electron]+s(proton)
    squaring above eqn
    s^2=se^2+sp^2+2s(e)s(p)
    s(e)s(p)=1/2[s^2-s(e)^2-s(p)^2]
    H=as(e)s(p)=a/2[s^2-s(e)^2-s(p)^2]
    S(e)^2=s(p)^2=s(s+1)h^2=3/4h^2
    H=a/2(s^2-3/4h^2-3/4h^2)
    =a/2[s^2-3/4h^2]
    3s1gives s=1;s^2=s(s+1)h^2=2h^2
    1so gives 0h^2
    H1=a/2[2-3/2]h^2=a/4h^2 for 3s1
    H2=a/2[0-3/2]h^2=-3/4ah^2 for 1so
    splitting between 3s1and1so is
    H=H1-H2=(1/4+3/4)ah^2=ah^2
    So answer is ah^2
    Here pls consider h as h cross.on mobile there is no symbol for h cross. So i used h only.


    ReplyDelete
    Replies
    1. thanks for the explanation shraddha plz, also discuss Q59 in booklet B, expectation value of =(h/mw)^1/2 or =(h/2mw)^1/2 with Co=1/root2 in 1-D harmonic oscillator.& Q36 in booklet B, represenation of hamiltonian H=wxp quantum-mechanically isn't
      -ihwxd/dx option 3 should be the answer...

      Delete
    2. But Shraddha...In december 2011 they specifically mentioned those two levels 3S1 and 1S0 . But here in this problem they told that ground state is splitting Hence F=0,1 .There is a theory that Difference in consecutive splitted levels is directly proportional to F+1 ,where F----> highest number so here =1 so Difference in energy level is directly proportional to 2 . so ans is 2ahˆ2

      Delete
  48. Q 54 booklet c is wrong peasle cheke it

    ReplyDelete
  49. bcz of given lorenz condition is wrong

    ReplyDelete
  50. What will be expected marks for general

    ReplyDelete
  51. give the answer of poisson bracket of x,p,l

    ReplyDelete
  52. give answer of laurentz series question

    ReplyDelete
  53. What will be excepted cut off

    ReplyDelete
  54. In the question mass of a shell is asked.
    Mass of a shell is = 4pi(r^2)*thickness*density

    Now, thickness= constant, density = constant

    So, mass is proportional to surface (or r^2)

    ReplyDelete

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