► CSIR UGC NET, 2014 Dec 21, Question Paper, Download, Answer Key, Answers, solutions, Explanations, Comment Here
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Download Dec 2014 Physics NET question -C
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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 2ahˆ2
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)