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

► 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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183 comments :

  1. Anonymous22 June, 2014

    Pls comment answers along with questions

    ReplyDelete
    Replies
    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-

      Delete
    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

      Delete
    3. many answers are wrongs

      Delete
  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

    ReplyDelete
    Replies
    1. 12 answer will be 40.5km

      Delete
    2. its 41.5 calculte it

      Delete
    3. 3-3 hoga 100% sure

      Delete
    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

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

    ReplyDelete
  4. please tell me cutoff marks

    ReplyDelete
    Replies
    1. Anonymous25 June, 2014

      Last tym 82 gyi thi fir iss baar itni kam kyu

      Delete
  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

    ReplyDelete
  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

    ReplyDelete
    Replies
    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

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

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

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

      Delete
    5. Anonymous03 July, 2014

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

      Delete
  7. Anonymous23 June, 2014

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

    ReplyDelete
  8. Anonymous23 June, 2014

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

    ReplyDelete
    Replies
    1. Keshav plz write booklet code

      Delete
  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

    ReplyDelete
  10. Anonymous24 June, 2014

    please comment answer keys of booklet code B

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

    ReplyDelete
  12. Anonymous24 June, 2014

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

    ReplyDelete
  13. Anonymous24 June, 2014

    And answer of probability within the bohr radius is

    ReplyDelete
    Replies
    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

      Delete
  14. DeltaC/C=?

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

      Delete
    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.

      Delete
  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

    ReplyDelete
    Replies
    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

      Delete
    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

      Delete
  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.

    ReplyDelete
  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

    ReplyDelete
    Replies
    1. Anonymous26 June, 2014

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

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

      Delete
    3. Anonymous05 July, 2014

      ans of 31-2,32-1

      Delete
  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

    ReplyDelete
  19. Anonymous24 June, 2014

    please apload answer key of set -c

    ReplyDelete
  20. Anonymous24 June, 2014

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

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

    ReplyDelete
  22. 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...

    ReplyDelete
    Replies
    1. i feel answer for 10 is option 4

      Delete
    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

      Delete
  23. 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

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

    ReplyDelete
  25. Plz give explanation of the aswer

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

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

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

      Delete
    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

      Delete
  28. 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

    ReplyDelete
  29. Please amend to read share as square

    ReplyDelete
  30. 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

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

      Delete
  31. 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

    ReplyDelete
  32. 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

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

    ReplyDelete
  34. 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

    ReplyDelete
  35. Weight of paper packet ans is 2.4 kg

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

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

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

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

    ReplyDelete
  40. 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

    ReplyDelete
  41. Ratio of weight to the square if the height is constant

    ReplyDelete
    Replies
    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

      Delete
  42. Rounded off chicory percentage is 14

    ReplyDelete
  43. Relation between RD
    RD of group > RD of group B

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

    ReplyDelete
  45. ABBCCD in it AD is a perfect square

    ReplyDelete
  46. 1/sin^2(z) contour integration |z|=1/2

    ReplyDelete
  47. 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

    ReplyDelete
  48. 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.

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

    ReplyDelete
    Replies
    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

      Delete
    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

      Delete
  50. 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

    ReplyDelete
  51. 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

    ReplyDelete
    Replies
    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 ??

      Plz reply

      Delete
    2. Anonymous26 June, 2014

      ya u r right

      Delete
  52. 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

    ReplyDelete
  53. Effective capacitance is unaffected

    ReplyDelete
  54. 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(πx1/a)sin(2πx2/a) - sin(2πx1/a)sin(πx2/a))

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

    ReplyDelete
    Replies
    1. Anonymous26 June, 2014

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

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

      Delete
  56. 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

    ReplyDelete
  57. 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

    ReplyDelete
  58. 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

    ReplyDelete
    Replies
    1. Anonymous26 June, 2014

      solving we get d/5

      Delete
  59. 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

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

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

    ReplyDelete
    Replies
    1. Anonymous26 June, 2014

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

      Delete
    2. Anonymous26 June, 2014

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

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

      Delete
    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

      Delete
    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

      Delete
    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

      Delete
    7. I.e option four MU zero K/ 4π along Z

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

      Delete
  62. 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

    ReplyDelete
    Replies
    1. Anonymous01 July, 2014

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

      Delete
  63. 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

    ReplyDelete
  64. The period of oscillation of simple pendulum under gravity is given by T=2π√(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

    ReplyDelete
    Replies
    1. Anonymous26 June, 2014

      plz give all answers from Part C,

      Delete
  65. Juhi Srivastava26 June, 2014

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

    ReplyDelete
  66. 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

    ReplyDelete
  67. 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
    Answer with reason
    J

    ReplyDelete
  68. most of the answers are sure to be correct

    ReplyDelete
  69. 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

    ReplyDelete
  70. 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

    ReplyDelete
    Replies
    1. Anonymous28 June, 2014

      plz explain Q.52of code A

      Delete
    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)

      Delete
  71. 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

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

    ReplyDelete
  73. 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.

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

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

    ReplyDelete
  76. energy eigenvalue of wave function Asin^3(πx/a) is 9/10(π^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

    ReplyDelete
  77. Anonymous01 July, 2014

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

    ReplyDelete
  78. Anonymous01 July, 2014

    first order correction to ground state us option 4

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

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

    ReplyDelete
    Replies
    1. Anonymous03 July, 2014

      no it is 3/8

      Delete
    2. that is not correct answerasir

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

      Delete
    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.

      Delete
  81. 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

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

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

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

    ReplyDelete
  85. 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 Schrödinger equation
    so option first will be the correct choice

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

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

      Delete
  86. Anonymous02 July, 2014

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

    ReplyDelete
  87. Anonymous03 July, 2014

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

    ReplyDelete
  88. 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)

    ReplyDelete
  89. #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

    ReplyDelete
  90. 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.

    ReplyDelete
    Replies
    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.

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

    ReplyDelete
  92. Anonymous05 July, 2014

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

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

    ReplyDelete
    Replies
    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

      Delete
  94. 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.

    ReplyDelete
  95. 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

    ReplyDelete
  96. 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

    ReplyDelete
  97. 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)

    ReplyDelete
    Replies
    1. Anonymous14 July, 2014

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

      Delete
    2. Anonymous15 July, 2014

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

      Delete
  98. 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

    ReplyDelete
  99. 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

    ReplyDelete
  100. 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 ?)

    ReplyDelete

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