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

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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

We can complete the 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.

Download Dec 2014 Physics NET question -C

Posted by our guest writer S . If there is any content violations Report the Administrator by clicking "Report Problem".

This comment has been removed by the author.

ReplyDeleteSpecific heat for diatomic 5/2 k

ReplyDelete%change in force 3

ReplyDeleteThis comment has been removed by the author.

DeleteL/r= 4pi

ReplyDeletestable isobar, Z= 92

ReplyDeletehow???? delB/delT=0

DeleteCorrect z=84

DeletedelB/delZ=0

DeleteShruti 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

DeleteThis comment has been removed by the author.

Deletehttps://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

DeletedelB/delZ=0

Delete=> (-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

here first a= a(sys) and second a= a(c)

Deletebooklet c part A

ReplyDelete1 (2)

2 (2)

3(1)

4(1)

8(3)

9(4)

11(4)

13(4)

14(3)

15(2)

17(2)

BOOKLET C part B

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

answer if Q 24 is 1

DeleteBOOKLET -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

ReplyDeleteBOOKLET -C Q -27

ReplyDeleteCORRECT 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.

1 b

ReplyDelete2a

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

this is booklet C

DeleteCan u pls explain anwer of 74is d how

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

Deletegive the anwer of booklet-B

ReplyDeletehttp://physics43.files.wordpress.com/2014/12/net-physics-question-dec-2014-physicskerala.pdf

DeleteBooklet-C

ReplyDelete***********************************************

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

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

Deletefor ques no 3 of part A, the ans is (b) ie smaller sphere because of stefans law

DeleteI have doubt on:

DeleteQuestion (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

Where do I find the question?

ReplyDeletehttp://physics43.files.wordpress.com/2014/12/net-physics-question-dec-2014-physicskerala.pdf

DeleteIn booklet c, 100% sure ans are

ReplyDelete5)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

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

Delete70) 4th is also kinematically forbidden!

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

ReplyDeletebook let C

ReplyDelete---------------------------------------------

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

Book let C

ReplyDelete----------------------

(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.

Book ket C

ReplyDelete--------------------

(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

Book let C

ReplyDelete--------------------------------

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

CORRECT ANSWER

Delete20

Deletewhat is the answer of question no 41 in booklet code C (phase space problem) ????

ReplyDeletewhen csir will publish key ??

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

ReplyDeleteno this time paper is easier than previous one....so cut off will be around 80 for LS and 90 for JRF

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

DeleteWhat is the intensity of radiation emitted by moving charges?

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

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

Therefore, Mass = constant * Volume....

Hence, Mass is proportional to Volume (Option b)

Please correct if I am wrong...

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.

ReplyDeleteYes, Pranab Dhar. You are right. If any one want the reference they can check at following link.

Deletehttps://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.

Modern Physics By Bernstein

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

DeleteWhat is answer of 64 in booklet c

ReplyDeleteThis comment has been removed by the author.

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

ReplyDeleteThis comment has been removed by the author.

ReplyDeleteAverage 13 number =(70+72+74+76+78+80+82+84+86+88+90+92+94)/13=1066/13=82

DeleteNow 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

Anyone please upload question booklet B

ReplyDeleteYes, 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.

ReplyDeleteThis comment has been removed by the author.

ReplyDeleteWhat is answer of Q42 booklate c

ReplyDeleteAnswer 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.

DeleteBooklet C-

ReplyDeleteQuestion 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).

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

ReplyDeleteI 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... :)

Deletegive answers of booklet A

ReplyDeleteanswer key of booklet A

ReplyDeletepart A

2 b

4 b

5 a

8 d

10 a

12 b

13 d

16 a

17 c

19 a

20 b

sorry it is not A it is booklet B

Deletepart B

ReplyDelete21 a

22 a

23 c

24 d

28 c

29 a

31 c

35 b

36 c

38 d

44 d

part C

ReplyDelete46 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

Hi Nivedita,

DeleteIf 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...

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.

ReplyDeletePhysical Sciences 39.5 (JRF) 34.5 (LS)

But why you people say it is 69?

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...

DeleteHi friends

ReplyDeletePls 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

Thanks Pranab... This helps..

DeleteMANY ANSWERS HERE ARE WRONG

DeletePlease post ur explanations. That will help...

DeleteCan u tell me where we get answer of booklate code c

DeleteShraddha, 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).

Deleteanswers keys given by career endeavour for booklet-B of physics 2014(II) 36,59,60,72 may be wrong, kindly confirm.

Deleteanswers keys should be 36-3, 59-1,60-2, 72-3, accrding to Griffith in introduction in quatum mechnics & others references.

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

DeleteIn 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?

ya in my opinion u r right

Deletebut 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

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

Deletehttp://www.algebra.com/algebra/homework/word/geometry/Geometry_Word_Problems.faq.question.101764.html

DeleteThis will give an insight...

Hi friends

DeleteI 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.

pranab dhar is absolutely right...

Deletewhat will be the expected cut off this time for general category?

ReplyDeleteIn 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....

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

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

ReplyDeletei 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

ReplyDeletefor nuclear reaction problem which one is the correct answer in my view only 2 nd eqn is correct

Deleteyes 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...

DeleteWhat about this argument then?

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

Thus Mass = Constant * Volume

Hence, Mass is proportional to Volume...

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.

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

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

i also think it

ReplyDeleteWhen csir key will publish

ReplyDeletesplitting 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 .

ReplyDeleteKartik same question asked in dec 2011.correct answer is ah^2.explaination

ReplyDeleteTotal 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.

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

Delete-ihwxd/dx option 3 should be the answer...

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 2ahË†2

DeleteQ 54 booklet c is wrong peasle cheke it

ReplyDeletebcz of given lorenz condition is wrong

ReplyDeleteWhat will be expected marks for general

ReplyDeletemay be 79

Deletegive the answer of poisson bracket of x,p,l

ReplyDeletegive answer of laurentz series question

ReplyDeleteWhat will be excepted cut off

ReplyDeleteIn the question mass of a shell is asked.

ReplyDeleteMass of a shell is = 4pi(r^2)*thickness*density

Now, thickness= constant, density = constant

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