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

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» 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 22

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.

Please comment the key only if you are sure about the answer, and also give the explanations regarding your answer.

Join all the groups Click Here

Comment the answers, solutions, and explanations....

^{nd}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**

We can complete the detailed answer key with your help.We can complete the detailed answer key with your help.

Please comment the key only if you are sure about the answer, and also give the explanations regarding your answer.

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Comment the answers, solutions, and explanations....

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UGC Qualified Group

ReplyDeletehai frnds please update key for csir net physics 22nd dec 2013

ReplyDeletehai frnds please update key for csir net chemical science 22nd dec -2013

ReplyDeletePlease upload question paper of csir net december 2013 physical science

ReplyDeletesend a scanned copy to physics43@gmail.com

Deletey dear

DeleteCSIR 2013 Dec 22 Physical Science

ReplyDeleteBooklet-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

i feel almost correct ...

Deleteexcept acce is zero

bcoz V*T increases initially and it decreases then after

SETA:17: 365

Delete19: S30W

I have doubt on 55.. the next number in the sequence

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

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

DeleteThird-First= 46-39=7

Fourth-Second= 50-42=8

So,

Fifth-Third=9

Fifth= third+9= 46+9=55

yes answer is 55 but my lgic was different..

Deletelook..

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

yes correct

Deleteadd first digit to that number...so 50+5=55

Delete18 is 320

DeleteI think x+y=2,

DeleteI think (14) cannot be determined option will be correct one.

DeletePARTA :

ReplyDeleteB- 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

Hi

ReplyDeletei think pi lies between 3 and 4

ReplyDeletewhat is the answer of series question in general science.

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

Deletesee pi * r^2 < (2r)^2

Delete=> pi<4

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

DeleteAny reply for part b or part c ?

ReplyDeleteSet A Part A

ReplyDelete1. 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

13: Max 32 (2,2,2,2,2)

Delete2. 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.

DeleteAnswer: 3

Set A part A

ReplyDelete15. (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

your explanation is not correct for ans 4.

Deletethe 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.

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

Deletemr.sant kumar just aad the initial digit in next number

Deleteseries ans,

ReplyDelete39, 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

yes

Delete39+3=42 42+4=46 46+4= 50 50+5=55 add first integer of the number to itself

Deletewhat about part B & C

ReplyDeleteFor a.b.c.d.e 32 is max possible 2 2 2 2 2 =10

ReplyDeletePART B SET A

ReplyDeleteQ.NO ANS

25. 1

28 2

29 3

31 1

32 1

35 4

36 4

37 4

44 4

Honey

ReplyDeletefull 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

PART C BOOKLET CODE A

ReplyDeleteQ.NO ANS

47 4

48 4

49 1

53 4

55 1

56 4

59 3

67 3

69 1

71 4

75 1

52. 2

Delete58. 2

66. 4

75. 1

book code c

ReplyDeleteplease....

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

ReplyDeleteprobabability between -a/2 and a/2 will be 1/2+1/pi

ReplyDeleteyes correct

Delete1/2 - 1/pi

Deletehow plz explain ...both of u

Deletepradeep ur answer is correct

Deleteyes its answer is 1/2 - 1/pi

Deleteanswer is 1/2 - 1/pi

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

the correct answer is 1/2+1/pi

Deletenet december 2013 exam answer key 99% correct.....

ReplyDeletebooklet 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

can you give like answers

Deletelittle bit correct,otherwise wrong

Deletewhich is wrong?????

DeletePLEASE LET ME KNOW.......WHICH ONE IS WRONG ANSWERS.....I POSTED.....HERE

DeleteI MY KNOWLEDGE ALL ARE RIGHT.....

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

DeleteElectric dipole moment create if there charge separation form,but in nucleus there only positive charge,,

Deletebut if we consider atom then electric dipole moment creates,,so ans is 0.

hi Mahesh pls send me the scanned copy of booklet A @ physics.hemant18@gmail.com

Deletei hv the booklet B so i m unable to comment on ur answer.

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

Deleteno of splitting is 6,,,b/c there is strong magnetic field ans spin is not zero.....so it splitting by

Delete2l+1 *2s+1

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

Delete.

PART C:

ReplyDelete1/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

spectroscopic for the

Deleteground state --- 6

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

ReplyDeletesection A booklet A

ReplyDelete1-3

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

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

Deleteplease send any one key for set -B

ReplyDeletesection B booklet B

Delete21-1

23-2

24-4

25-4

26-4

38-2

45-1

part C

Delete54-2

55-1

58-1

66-4

67-2

68-1

74-1

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

Deleteyes answer is 4 of que 54

Deleteyes answer 4 is correct of que 52

Deletebooklet A

ReplyDeletepart 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

SET A

Delete52- 2

66- 4

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.

Deletei think ta>tc>tb at b temp is constant but remaing two ie a $ c is positive temps

Deletecan u plz tell answers for booklet code B

ReplyDeleteThis comment has been removed by the author.

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

ReplyDeletekindly send the andwer key of maths net dec 2013

ReplyDeletecan any one tell the cut off mark

ReplyDeletefor LS in physical science exam this time????

what was the cut off for all category for june 2013?

ReplyDeletepls share

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...

ReplyDeleteI 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

DeleteLS cutoff will be around 72-73, UGC JRF around 80 and CSIR around 90.

Deleteyes, may be close to 80 for LS and a ten more for jrf

Deletearound 80 for ls for jrf may be 90 or 95

DeleteBOOKLET C

ReplyDeletePART C

50. 4

51. 2

56. 1

61. 2

62. 2

64. 4

65. 2

Booklet C, Part-C

DeleteQues 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.

Mistake.....

Delete5th line: I used expectation symbols. But webpage rendered it. Hope you can understand

& = ?

Delete5th line: Lx^2 = Ly^2 = 1/2 [ L^2 - Lz^2]

Deletewhat is the condition for being Lx=Ly is not equal to Lz

Deleteas far as symmetry ans simultaneity is concern Lx=Ly=Lz

SET B PHYSICAL SCINCE ANSWERS

ReplyDelete47-2

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

setB PHY. SC. ANS

ReplyDelete21-1

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

hi Vivek . . how u got d ans. of 44 as 4.

Deleteisn't is 1.i.e . 100:1

pls explain. .

IF ANY ANSWER IS INCORRECT PLEASE SEND YOUR CLERIFICATION ABOUT ANS

ReplyDeleteBOOKLET C (PHYSICAL SCIENCES)

ReplyDeletePART 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...

Pradeep

DeleteQ.46- Ans is 3 not 1.

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)

ReplyDeleteso its ans is option 4 for set A

Ya the ans is 4...

DeleteThis comment has been removed by the author.

ReplyDelete27. poission bracket value of {A,{B,C}}-{{A,B},C} = {{C,A},B}

ReplyDeletesince {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

Yes it right

Delete28. LAGRANGIAN IS OPTION 2. L= - m square root of (1- derivative of X square) - V(x) refer Page 1096 mathematical physics by BS Rajput

ReplyDeleteya its right... will see same ans from h goldstein ... classical mechanics p. No. 344

DeleteThis comment has been removed by the author.

ReplyDeleteSET B

ReplyDeletePart 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

33 have Vo= 5 volt

DeleteAnd 69 have ans 4

33 have Vo= 5 volt

DeleteAnd 69 have ans 4

Can you explain your answer?

Delete31. force between two parallel conductor I1*I2/ D

ReplyDelete32. 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

above 45 th answer is correct

DeleteSET -B

ReplyDeletePart- 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

hai thank

ReplyDeleteset c

ReplyDeleteplease 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

A peturbation v=aL^2 is .........applied on 2p state(spin should not be considered).....

ReplyDeleteWHAT IS THE ANS OF THIS QUESTION?(EXPLANATION REQUIRED)

v= aL^2 nd 2p state is given , so l=1 nd for 1st order correction , eigen value of l^2 is

Deletel(l+1)h^2 .... answr is 2h^2

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

ReplyDeleteset a

ReplyDelete1.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

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

ReplyDeleteThis comment has been removed by the author.

ReplyDeletedr. Mahesh pls allow me to access ur google drive.....

ReplyDeleteThis comment has been removed by the author.

DeleteThis comment has been removed by the author.

ReplyDeleteHi 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.

ReplyDeleteby Maahiya

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

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

Deletehi Mahesh

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

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.

Deleteshare

ReplyDeleteResistance Voltage Currentmeaningfor terminal velocity answer is 8...because time to reach half of its terminal velocity is directly prop to mass of the body

ReplyDeletei think this is right .

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

Deleteset A

ReplyDeletePART 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

how did u get option 2 for question 44

Deletehi frnds plz explain possible states for three fermions in 2 states

ReplyDeletetwo 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..............

DeleteHi Mahesh

Deleteans. for Q.60 is 1/4 or 1/8 ?????

is it electric dipole or A Quadrupole ????

electrostatics potential ratio V(2r)/V(r) is

ReplyDeleteThis comment has been removed by the author.

ReplyDeleteplz explain question 75 in set C

ReplyDelete29. Angular velocity of ring when reaches the base

ReplyDeleteheight 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

are u sure??

Delete53. Del square psi plus 2m by h cut square(E-V) psi = 0 is time independent schrodinger equation

ReplyDeletesolve it for given psi = A exp( - X^4)

give V(X) = 16 X^3- 12 X^2

option 4

Use given value of h cut=2m=1

ReplyDeleteCAN ANY ONE TELL THE ANSWER FOR THE FOLLOWING QUESTIONS...

ReplyDelete1- 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

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

ReplyDeleteNo... 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....

Deletepotential is given ...so force F=-Del(V).

Deleteso 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.

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

Deleteposition vector is in "z" direction?

Deletefor Carnot engine problem ...answer is sqrt (T1*T2)

ReplyDeleteThe 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...

Deletefor three identical spin 1/2 fermions ...answer is 2 possibilities

ReplyDeletehi Vijay

Deletewhat is the ans. of Q.60 .

i.e, it is electric dipole or Quadrupole???????

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

Deletebooklet code c

ReplyDeletesection 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

if any answer is incorrect please clarify.........

Deletebook let code B.

DeletePART 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

ans . series 39.42,46,50,53

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

ReplyDeleteAns is55 ...... 39+3=42. 42+4=46. 46+4=50. 50+5=55. add no with its first digit

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

ReplyDeletebooklet c

ReplyDeletepart 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

This comment has been removed by the author.

Deleteplz 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

Deletebooklet code c

ReplyDelete46.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

Q.46 ans 3 is not 1

Deleteans 3 is correct

Delete49.option 1 is correct. the net force is along x axis. f(net)= 2k(sqrt(L^2-X^2)-L^2)sin(theta).

ReplyDeletesin(theta)=theta(small).

theta= X/L= sin(theta).

substituting and using binomial theorem we have f(net)= kx^3/l^2

48. option 1 is correct.

ReplyDeletefor 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.

52. option 4 is correct.

ReplyDeletenear 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

57. option 4 is correct

ReplyDeletemonopole= 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

yes it is correct

Delete75. option 3 is correct

ReplyDeletethis 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.

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

Deletei think ans is 4

Deleteyou can chk griffiths book. chapter 3 image problem aptly expalins what i have said.

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

Deleteans is option 2

Deleteyes i am agree with the shiv kumar

Deleteall d ans are for code C

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

ReplyDelete@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

ReplyDeletein 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

Deletewhat about question 23 about analytic functions? i got 3 answers as correct..

Deleteans is option a. any analytic function should satisfy the laplace eqn. first option does not satisfy

Deletewhat is the ans for Fourier transform of first derivative of Dirac delta function??

ReplyDeleteik

ReplyDeleteis it correct ans?

Deleteyes..because fourier transform of dirac delta function is 1.so use the property of derivative of fourier transform.answer is ik

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

ReplyDeleteAns: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.

DeleteQ.45 ans is 3 ....... 100% correct.

Deletewhat is the ans for q no 70 in book let c( digital electronics sop method)?

ReplyDeletewhat is the answer for perimeter problem in part a

DeleteThis comment has been removed by the author.

Deleteq no 70 ans

DeleteBbarDbar+BD

PERIMETER PROB ANS 20

is it correct ans for qno 70?

Delete