# ►CSIR UGC NET, 2013 June 23, Detailed Answer Key, Key, Answers, solutions, Explanations, Comment Here

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UGC CSIR NET 2013 June 23 Physics Answer Key Comment here with explanation

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section-c

ReplyDeletedegrees of freedom of rigid body in 6-Dspace

A,B,C. are spin half particles....possible spin of d when A=B+C+D

expectation value of Lx^2+Ly^2 for a wave function

digital electronics ans A

i think expectation value will be 2h^2

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DeleteThis comment has been removed by the author.

Deletefor rigid body, whatever be the dimensions it always have always 6 degrees of freedom

Deleteliquid's height is is 3h/2

DeleteBooklet B

DeleteQ-75

correct ans is A

Velocity and pressure is More than std and Humidity and temp is lesser.

density of state D(E) is proportional to Energy E.

Deletedg/g=dl/l+2dT/T

Delete70 rectangles

DeleteThis comment has been removed by the author.

Delete1-1/(e^(-1))

Deletehow 70, please help me, ii got 56

Deleteya its 70

Deleteharris how to get 70 rectangles pls explain..

Deletecount the number of rectangles using different combinations. eg.,1X4,4X1,2X4,4X2,3X4,4X3,etc. the total number will come to 70.

Delete# of degrees of freedom of a rigid body in d space dimension = d(d+1)/2

DeletePart-C digital electronics question option 1 is correct

in a given problem gates are NOT , NOR , OR and output is VP+TH so 0x0 + 1x1 = 0+1 = 1

section A

ReplyDeletesquare pyamid to b costructed using threadsin single strand...min cuts==3

largest number=3^100

square pyramid ans is 1

Deleteyes it will be 1

Deletelargest no 3^400

Deleteyes largest number is 3^400

Deletehow one for pyramid?

Deletehow to u got one in pyramid question pls give to detailed answer...

DeleteOption 2 is correct

Deletea^mn = (a^m)n 2^5=32 3^4=81 4^3=64 5^2=25 3^400 is the largest #

i also mark 1 for pyramid bt i think it must be 7.....

Deletebcoz 4 for lower base parts of rectangle nd 3 for above ...so it must be 7....

10^18 was answer of one ques of section a...no of nanoplates

ReplyDeleteMs white was wearig black dress

childcare facility...Answer A...only ratio can be calculated

please give me reason for 10^18

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Delete2nd and 3rd one are correct

Deleteyes it is 10^18

Delete1nm is 10^-9m not cm

Delete10^18 is right.. 1000nm3 is not equal to 10^-7cm3.it is cm3 not cm.(1000nm=1000*10^-7*10^-7*10^-7cm3)

DeleteFor child care facilities, none can be answered, as if we don't take survey on weekend, how can one get ratio for men and women in general??

Delete10^18 option is correct 1 cm^3= 10^-6 m^3 and 10X20X5 nm^3=1000X10^-27 m^3=10^-24 m^3

Delete# of unit cells=10^-6 / 10^-24=10^18

Ms Brown dress was White Ms Black dress was Brown Ms White dress was Black Option 3 is correct

series problem in section A: correct ans is 1

ReplyDeleteOption 2 is correct answer is 1

Delete(2-1)/1X2 + (3-2)/2X3 +(4-3)/3X4 + .......=1-1/2+1/2-1/3+1/3-.........-1/infinity=1+0=1

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DeleteThis comment has been removed by the author.

ReplyDeleteCSIR UGC NET, 2013 June 23, Detailed Answer Key, Key, Answers, solutions, Explanations, Comment Here

DeleteAre you sure your ans is correct as u given it ? have u cross checked.

DeleteDear check your answers:

DeleteSec C: 46. B, 48. B, 52. A, 54. B, 59. D, 60. C, 64. B, 70. C, 75. A

Sec B: 26. A, 31. A, 33. C, 35. B, 39. C, 41. A

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ReplyDeletesome are matched with this but some r not but m not 100% sure may be u r right how much u r sure about your answer

DeleteThree questions from section C,1 each from A and B are incorrect i think.

Deleteplease send answer key of booklet A

ReplyDeletefor booklet A

ReplyDeletePART B

26-2, 27- 2, 31-1, 33-3, 34-3, 40-1, 41-1, 42- 4, 44-1,

PART C

46-1, 47-2, 48-4, 50-1, 53-3, 58-1, 59-1, 61-3, 64-1, 68-3, 73-3, 75-3

for booklet A

DeletePART B

21-3, 22-2, 24-1, 25-1, 26-2, 27-2, 31-3, 34-3, 39-4, 41-4, 44-1

PART C

46-1, 48-4, 51-2, 54-1, 56-1, 57-4, 58-1, 59-1, 62-3, 65-2, 71-3, 73-4

PART A

20-4, 19-4, 16-3, 15-2, 14-3, 7-1, 6-3, 3-4, 2-2, 1-1

for booklet A

Delete7)2 14)4 16)3 18)2 19)4 20)4

23)3 24)4 26)3 27)2 34)1 36)4 39)4 42)2 44)1

46)1 52)2 54)4 56)2 59)4 61)3 62)4 64)1 65)2 73)4 75)3

this is exact key?

DeleteI feel 3-4 bits may be wrong that's it, except that everything is ok.

DeleteBooklet B

ReplyDeleteSection A

River flow will b downward i.e., South will b right answer.

no its east

Deleteyes south cos river's downsteam is from higher altitude to lower

Deletebut asked for direction of downstream water, if its south water must go into earth

DeleteArey baba...!!!

DeleteThe direction shown in the graph was for surface of earth...!!

And not for the height...!!!:-D

All path will be of same distance to reach from A to C

ReplyDeleteThe total distance travelled by a person is same for 2 paths

Delete10^18 will be right ans

ReplyDeletefor density of states it is constant for 2 dimensional fermi gas and for three dimensional

ReplyDeletefermi gas proportional to (E)^1/2

for 1-D it is proportional to 1/(E^1/2)

Deletefor 2-D it is ----------- E

for 3-D E^1/2

density of states of a electron gas in 2 dimension is independent of energy so option 4 is correct

DeleteSurvey

ReplyDeleteOption A can be answered only..

Weekend data can not be estimated from that information

yes its correct, only A can be estimated

Deletequestion for section c: expectation value of Lx^2+Ly^2 is 26hcut^2/25

ReplyDeleteyes tis s right

Delete3 power 400 and Density states constant , 100 rectangle, h/4 , independent of path...

ReplyDelete10 power 18; series =1;

ReplyDeletewhat is the equilibrium temperature ?

ReplyDeletein rest frame area of disk is : (1-v^2/c^2)^-1

P1V1+P2V2/p1v1/t1+p2v2/t2

Delete(1-v^2/c^2) i think

no of rectangles is 100

ice water temprature-0 degrees cos 800cal required by ice to melt while water gives 775 cal

Deleteyes u r right

Delete@ vineet pannu, can u explain how it is P1V1+P2V2/p1v1/t1+p2v2/t2

DeleteI also filled same choice but dont know the explanation.

10X80+10X1X(t-0)=30X1X(25-t) 8+t=75-3t 4t = -5 t = -1.25 0C

Deletefor expectation value, we have to write Lx^2+Ly^2= L^2-Lz^2

ReplyDeleteanswer is 26*hcut^2/25

temperature at eqlm: is -1.25 °C.

ReplyDeleteincrease in height of liquid in cylindrical vessel is : R/12

ReplyDeleteincrease in volume of water=volume of immersed part of sphere

pi*R^2*dH = (4/3*pi*(R/2)^3)/2 (since half of sphere is immersed )

so dH (increase in height of level of water)= R/12

Your procedure is exactly correct so correct option is 4 i.e. R/12

Deletemaximum area of triangle that can inscribed in semicircle is :

ReplyDelete=1/2*height*base

where, height = radius of semi circle

base = diameter of circle

finally,

1/2*R*2R

answer is R^2 ....option b

its true...!

DeleteFor maximum area of triangle that canbe inscribed in semicircle is isosceles right angle triangle

Deleteso area = 1/2 X 2R X R = R^2 option 1 is correct

probability of some of 2 numbers selecting from 1-15

ReplyDelete(5,15)=(15,5)

(6,14)=(14,6)

(7,13)=(13,7)

(8.12)=(12,8)

(9,11)=(11,9)

total is 10

but number of ways we can arrange 2 numbers from 15 is 15P2 = 210

probability is = 10/210 = 1/21

ans = 1/21

yes 1/21

Deleteyes 1/21

Deleteits a correct one...!

Delete(5,15) (6,14) (7,13) (8,12) (9,11)

Delete# favorable events=5 and total # of events= 15 C 2 = 105

required probability=5/105=1/21 so option 2 is correct

differential equation is dx/dt=x^2 or 2dx/dt=x^2 in part B?

ReplyDeletepls let me know ?

it ws dx/dt=x^2

Deletebut i dnt knw ans...!:-D:-p

dx/x^2=dt -1/x=t+c x(t) = -1/(t+c) x(0)=-1/(0+c)=1 so c=-1

Deletex(t) = -1/(t-1) If t tends to 1 then x will be blow up so option 1 is correct

diff eqn for langrangian ques?

ReplyDelete-t^2/2m+c1t+c2

for lagrangian problem answer is -bt^2/2m + c1t + c2 by using generalized momentum and its derivative

DeleteWhat is the answer for the question, in which they asked about survey for health care service (23% men's and rest are women's???)???

ReplyDeletechoice A-only A can b answered

Deleteya but its not necessary that all the people participated in the survey, who came to the mall, then how we can tell the ratio of men's and women's in the mall??

DeleteOnly option A can be answered..

DeleteOption B cnt be cz of insufficient data..

What will b the ans of wavelength analog of layman series

ReplyDelete135 nm i suppose

Deleteyes its a 135 nm

Delete1/(1*2)+1/(2*3)+1/(3*4)+.....infinity what will b it's ans?

ReplyDelete1

DeleteI haven't done it in the exam but it will be 1.

DeleteIts a 1..!

DeleteCz 0.5+0.166+0.083+0.05+0.033+0.0239+.....=1

nth term Tn = 1/(n*(n+1)) = 1/n - 1/n+1.

DeleteSeries Sum= (1/1 + 1/2 + 1/3 +.....) - (1/2 + 1/3 + 1/4 +....) = 1 (only first term of first series)

can any one submit the key for booklet C

ReplyDeletefirst u provide then i cross check it ,in physical science

Deletepart A- of booklet C

Deletecan any one help me with the correctness of my answers, thanks in advnace

1.i did not answer, 2.1 3.1 5.3 6.3 7.4 8.2 10.4 12.3 13.1 14.3 `15.4 16.4 17.3

19.2 20.2

for booklet A

ReplyDeletePART B

21-3, 22-2, 24-1, 25-1, 26-2, 27-2, 31-3, 34-3, 39-4, 41-4, 44-1

PART C

46-1, 48-4, 51-2, 54-1, 56-1, 57-4, 58-1, 59-1, 62-3, 65-2, 71-3, 73-4

PART A

20-4, 19-4, 16-3, 15-2, 14-3, 7-1, 6-3, 3-4, 2-2, 1-1

ans of que 7 is 3*400

DeleteQ46...1 IS CORRECT SIR

DeleteB,qno: 1,ans is B

ReplyDeletedx/dt= x^2 as x(0)=1,ans is infinity

ReplyDeleteits answer is one. First integrate the function and then find the integration constant using initial condition

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Deletequestions are available in csir website

ReplyDeleteThis comment has been removed by the author.

Deleteyes question papers are now available on csir web but physics paper is not uploaded very properly.

DeleteSome part of questions are missing in the scanned version of the paper

HERMITE FUNTION ANSWER H4(0)=12

ReplyDeletehermite function explain plz

Deletedx/dt=x^2

ReplyDelete-1/x=t+c

f(0)=1;

c=-1

x=-1/(t-1)

so...when t-----1.....x ----infinity..... blow up

I think in my question booklet (CODE B) was missing. Because in WKB method delta function potential (from PART C) is very famous problem to me. Also I have read the question paper very well especially PART-C. Is there any way too check my question paper again?

ReplyDeleteThis comment has been removed by the author.

ReplyDeleteThis comment has been removed by the author.

ReplyDeleteis it section C or section A parimi?

DeleteA

DeleteCsir kept all question papers on main website, but physics paper was not properly scanned.You will better go through the chemistry paper for aptitude (SECTION-A) part.

ReplyDeletethx srinu, how is IIIT

Deleteit is good,do u know about 3it?

Deletei know, 3it, i know ur campus, i even know ur director, Ibrahim khan

Deletemay i know who are u?

DeleteBooklet B of Physics (Or Booklet A of chemistry):

ReplyDeleteSec-A: 1.C, 2.A, 4.C, 5.A, 7.D, 8.D, 9.D, 10.A, 11.B, 12.C, 14.C, 15.D, 18.B, 19. D

Booklet B: For Sec-C, Please correct me if I am wrong:

Delete46. B, 48. B, 52. A, 54. B, 59. D, 60. C, 64. B, 70. C, 75. A

Any one can query me for any answer of Sec-A I mentioned here.

Deleteand also for SEc B and C if he remember the question(s)

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ReplyDeleteIN SECTION B THE CONVERSION TIME FOR THE ADC CONVERTER IN REMAINS UNCHANGED

ReplyDeleteits remain unchanged..

DeleteCz in SAR, input voltage has nothing to do with Conversation time...!

YES, Right choice. It will be unchanged because conversation time is characteristics of ADC/DAC. it does not depend upon input/output

Deleteohhh TY, even i answered it as unchanged, but i was in little confusion

Deleteany one have booklet c answer in physical science

ReplyDeleteCan any answer that question, with proper logic?

ReplyDeleteA=B+C+D

Booklet A, Q.31, relativistic motion of a disk,

ReplyDeleteSolution: key concept in the question is in relative motion, distance only in one dimension (say length or width) will be reduced by factor of [(1-u^2/v^2)]^1/2. so area will also be reduced with same factor. Ans is (1).

very right!!!!didnt think in tat way....

DeleteAny detailed ans for Q.64 Booklet-A ?

ReplyDeleteif A and B satisfy coomunication relation [A,B] = I, then

(i) [eA,B] = eA

(ii) [eA,B] = [eB,A]

(iii)[eA,B] = [e-B,A]

(iv) [eA,b] = 1

Please post here

option ii, as identity valids

Deleteoption ii, as identity valids

DeleteOption 1

Delete@ kissy, Can u please explain the validation of identity in option II

Deleteans is 1

DeleteOption 1 is correct

Delete[ f(A), B ] = [ A, B ] F'(A)

how option 1 is correct murty b

Deleteits e^A I did it in exams.

Deleteexpand e^A as 1+A+A^2/2!+....... and then solve we get answer a option

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DeleteThis comment has been removed by the author.

DeleteI have talked at CSIR, they said they will upload a new copy of scanned physics paper by tomorrow evening.

ReplyDeleteu did good job.

DeleteThanks dear

DeleteCapacitance in the parallel wires of 4cm diameter separated by distance 2m (as I remember) is 88.5 nF, I think.

ReplyDeleteCorrect me if it is wrong.

what about stable isobar ans??

ReplyDeletePlz friends check my Booklet A-part C ans: 48-4,50-1,53-3,54-3,61-3,64-4,65-2,72-2,73-2.

ReplyDeleteisobar answer 160 wala

ReplyDeletewhat was the answer of drude model question?

ReplyDeletewhat will be the cut off hey guys

ReplyDeleteThis comment has been removed by the author.

ReplyDeleteBooklet Code A (Find it on CSIR site)

ReplyDeletePART-A

Q1: Ans (4) [154]

Q2: Ans (4) [R/12]

Q4: Ans (2) [1]

Q5: Ans (2) [3h/2]

Q6: Ans (1) [R^2]

Q7: Ans (2) [3^400]

Q10: Ans (3) [436]

Q12: Ans (3) [R+Rtan]

Q14: Ans (3) [Ms. White's dress was black]

Q15: Ans (2) [1/21]

Q16: Ans (3)

Q19: Ans (4) [380]

PART-B

Q22: Ans (1) [0 C]

Q24: Ans (1)

Q31: Ans (1)

Q36: Ans (4) [1/3 & 2/3]

Q37: Ans (1) [b2=0 & b1=1]

Q40: Ans (1) [12]

Q41: Ans (4) [b=a]

Q44: Ans (1)

PART-C

Q46: Ans (3) [0,0,0] E=-del(-xyz)-d/dt(iA1+jA2+kA3)

=(iyz+jxz+kxy)-(iyz+jxz+kxy)

= i*0+j*0+k*0

Q48: Ans (4) [26h_cross^2/25]

Q49: Ans (2) [Check Dimension]

Q53: Ans (3) [See Griffith Q.Mech]

Q54: Ans (2) [If spin degeneracy is considered. Two fermions with (+-)1/2 spin can stay with E0 energy]

Ans (1) [If spin degeneracy is not considered. Only one fermion in each state]

Q56: Ans (4) [Put it in the equation]

Q59: Ans (1) [1] After integrating and using condition you will get x(t)=1/(1-t)

Q61: Ans (3) [t-1+e^(-1)] Find Laplace transform of Ans(3)

Q62: Ans (3) [d(d+1)/2] Find it on the internet.

Q65: Ans (4) [a=-2 & b=3] Use Hamilton's eqn of motions. Then,d/dt(Px-3y)=0 & d/dt(P_y+2x)=0

Q73: Ans (4) [Constant]

from where u got it

DeleteI got it from calculations & discussions of/with me and my friends.

DeleteI hope all answers are correct.

Hey For Q no 49 Ans should be option (3). Check the Dimension again. Option (2) will have s^-1 extra. It won't get cancelled. Check again...

DeleteIf you read the question, then you see : "The rate of total energy radiated per unit area"

DeleteThe rate of total energy= Energy per unit time (because of rate)=Power

So, Power per unit area is to be calculated.

See Larmor formula of "energy radiated by an accelerating charged particle (wikipedia)".

Larmor formula gives the total power radiated = Energy radiated per unit time.

Its dimension= (Charge^2)(Length^-1)(Time^-1). If you divide it by area, then dimension= (Charge^2)(Length^-3)(Time^-1) which gives you energy radiated per unit time per unit area (this is the question of NET) and it matches with option (2).

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DeleteQ 46...1

DeleteCorrection:

DeleteQ12: Ans(4) [R+Rcot]

booklet A

ReplyDeleteQ.23 Ans(3): Reason I dont know, please explain if someone knows

Q.24 Ans (1): As probability distribution and velocity both reduces with increment in temperature, (T1>T2, see question)

Q.28 Ans(1): conversation time is characteristics of device, does not depend upon input or output.

Q.29 Ans(2): Use V = V0*exp(-t/RC), put the values given, 15 = 150*exp(-1000/R*5*10^-6) => exp (2*10^8/R) = 10 => 2*10^8/R = 2.303 => R = 10^8 ohm

Q.31 Ans (1): As length reduction will only take place not width reduction.. Use relativity theory

Q.41 Ans (3): Use A.B = AB cos = dot product of A and B => [(a^2+b^2+2ab cos 120)]*cos 60 = 2ab+a^2 cos 120 + b^2 cos 120 => 1/2 (a^2+b^2-ab) = 2ab - a^2/2 - b^2/2 => 2a^2 + 2b^2 - 5ab = 0 => a =2b or b = 2a; from teh given options, option C is b = a/2

Booklet A

ReplyDeleteQ.52: Ans (3) 2/9 : at any stage (say SA = 1, SB = 1, SC = -1), there are only two other possibilities fpr making SA+SB_SC = invariant, i.e. (SA = -1, SB= 1, SC =1 and SA =1, SB =-1, SC =1) including one condition that it can remain same, total probability of finding above mentioned conditions is: 1/3+ 1/3, including that these events can occur even with probability of 1/3 of total possible conditions, complete probability is 1/3(1/3+1/3) = 2/9

Q. 58 , Ans (1), It is explained earlier in this forum

Q. 59, Ans (1), It is also explained earlier in this forum

Q. 62, Little confusion, as there are number of sources on internet report it 6 also. I went with 6, option B

Q. 70, Ans (2): as in normal case C = EA/D, use this equation considering area of the wire(s) is length *width = 0.75km * 4cm, you will get the answer (correct me if wrong)

Q. 73, Ans (3): as it is proportional to E in 2-D, to E^1/2 in 3-D. Though I filled there E^1/2

Q. 74, I am confused between (3) or (4), I went with (3)

I have also attempted Q. 64 with ans (2) and Q.56 with Ans (3) dont know these are right or wrong

that is Q.75, not Q.74

Deletefrom where i can get the exact key..when the official key of csir will be released?/?

Deleteas csir said (the official key),they will provide in the first week of July on csir main website.

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Deleteno...its 4/9 for Sa+Sb+Sc invarient

ReplyDeletebecause....

probability for every S to change to -S is 1/3

...........................not to change is 2/3

there are three outcomes for an event Sa+Sb+Sb=constant(1).

(1)if all S's will not change, then Sa+Sb+sc=constant

then probability is = 2/3*2/3*2/3=8/27

(2)now, if Sa change to -1 and Sc change to +1 then Sa+Sb+sc= constant=1,

so probability is = 1/3*2/3*1/3=2/27

(3) next if Sa remains same,Sb change, and Sc change , then probability= 2/3*1/3*1/3=2/27

so total probability is = (8/27)+(2/27)+(2/27)=12/27=4/9, because three are independent events, so we have to add.............ans is 4/9

[e^A,B]=e^A is correct

ReplyDeletedegrees of freedom is D*(D+1)/2

ReplyDeleteCOde A

ReplyDeleteSec. B

21-3, 24-1, 26-2, 32-2, 38-3, 39-4, 42-2, 43-3,

Sec c..

49-2, 50-3, 51-2, 54-2, 57-4, 59-1, 63-3, 66-3, 71-3

please explain sec c- 71-3

Deletebooklet A) answers

ReplyDeletepart c......

48) 4

52) 4

53) 3 (you can check last december paper)

54) 4

58) 1

59) 1

61) 3

62) 3

64) 1

68) 1

73) 4

part b.........

23) 4

24) 4

27) 1

28) 1

31) 4

36) 4

37) 1

40 ) 1

44) 3

part a..........

1) 4

2) 4

3) 4

4) 2

6) 1

7) 2

8) 3

9) 2

10) 3

12) 4

14) 3

15) 2

16) 3

18) 2

19) 4

20) 4

..........these answers are all 100 % correct......if have guys any queries regarding to my answers pls post here......

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Deletenot 44th but for 54th

DeleteKey for June 2013 JRF/NET exam question papers will be uploaded in the first week of July 2013.

ReplyDeleteIn CSIR NET is there section-wise cutoff or not? Please reply me?

ReplyDeletecumulative total not section wise

Deletebooklet A:

ReplyDeletesection A:

1-1

2-4

3-4

4-2

5-2

6-1

7-2

8-3

9-1

10-3

11-3

12----

13-1

14-1

15-2

16-3

17----

18---

19-4

20-4

expected key

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Delete5-4

Delete9-4 since 60/2=30 60/3=20 60/4=15 60/5=12 60/6=10

Deletewhat is the amswer of 25 question in booklet A?

ReplyDeletewhat will be the cutoff

Deletemay be 70 marks

Deletearound 70 marks

Deletearound 70 marks

ReplyDeletecan some one pot the key for booklet C please

ReplyDeletepronab choubey 30 june 2013 what is the answer of 67 question in booklet A

ReplyDelete