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# ► 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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1. section-c
degrees 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

1. i think expectation value will be 2h^2

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4. for rigid body, whatever be the dimensions it always have always 6 degrees of freedom

5. liquid's height is is 3h/2

6. Booklet B
Q-75
correct ans is A
Velocity and pressure is More than std and Humidity and temp is lesser.

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

8. dg/g=dl/l+2dT/T

9. 70 rectangles

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11. 1-1/(e^(-1))

13. ya its 70

14. harris how to get 70 rectangles pls explain..

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

16. # of degrees of freedom of a rigid body in d space dimension = d(d+1)/2
Part-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

2. section A
square pyamid to b costructed using threadsin single strand...min cuts==3
largest number=3^100

1. square pyramid ans is 1

2. yes it will be 1

3. largest no 3^400

4. yes largest number is 3^400

5. how one for pyramid?

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

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

8. i also mark 1 for pyramid bt i think it must be 7.....
bcoz 4 for lower base parts of rectangle nd 3 for above ...so it must be 7....

3. 10^18 was answer of one ques of section a...no of nanoplates
Ms white was wearig black dress
childcare facility...Answer A...only ratio can be calculated

1. please give me reason for 10^18

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3. 2nd and 3rd one are correct

4. yes it is 10^18

5. 1nm is 10^-9m not cm

6. 10^18 is right.. 1000nm3 is not equal to 10^-7cm3.it is cm3 not cm.(1000nm=1000*10^-7*10^-7*10^-7cm3)

7. For 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??

8. 10^18 option is correct 1 cm^3= 10^-6 m^3 and 10X20X5 nm^3=1000X10^-27 m^3=10^-24 m^3
# 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

4. series problem in section A: correct ans is 1

1. Option 2 is correct answer is 1
(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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1. CSIR UGC NET, 2013 June 23, Detailed Answer Key, Key, Answers, solutions, Explanations, Comment Here

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

Sec 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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1. some 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

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

8. for booklet A
PART 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

1. for booklet A
PART 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

2. for booklet A
7)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

3. this is exact key?

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

9. Booklet B
Section A
River flow will b downward i.e., South will b right answer.

1. no its east

2. yes south cos river's downsteam is from higher altitude to lower

3. but asked for direction of downstream water, if its south water must go into earth

4. Arey baba...!!!

The direction shown in the graph was for surface of earth...!!
And not for the height...!!!:-D

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

1. The total distance travelled by a person is same for 2 paths

11. 10^18 will be right ans

12. for density of states it is constant for 2 dimensional fermi gas and for three dimensional
fermi gas proportional to (E)^1/2

1. for 1-D it is proportional to 1/(E^1/2)
for 2-D it is ----------- E
for 3-D E^1/2

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

13. Survey
Option A can be answered only..
Weekend data can not be estimated from that information

1. yes its correct, only A can be estimated

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

1. yes tis s right

15. 3 power 400 and Density states constant , 100 rectangle, h/4 , independent of path...

16. 10 power 18; series =1;

17. what is the equilibrium temperature ?

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

1. P1V1+P2V2/p1v1/t1+p2v2/t2
(1-v^2/c^2) i think
no of rectangles is 100

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

3. yes u r right

4. @ vineet pannu, can u explain how it is P1V1+P2V2/p1v1/t1+p2v2/t2
I also filled same choice but dont know the explanation.

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

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

19. temperature at eqlm: is -1.25 °C.

20. increase in height of liquid in cylindrical vessel is : R/12

increase 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

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

21. maximum area of triangle that can inscribed in semicircle is :

=1/2*height*base
where, height = radius of semi circle
base = diameter of circle

finally,
1/2*R*2R

1. its true...!

2. For maximum area of triangle that canbe inscribed in semicircle is isosceles right angle triangle
so area = 1/2 X 2R X R = R^2 option 1 is correct

22. probability of some of 2 numbers selecting from 1-15

(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

1. yes 1/21

2. yes 1/21

3. its a correct one...!

4. (5,15) (6,14) (7,13) (8,12) (9,11)
# favorable events=5 and total # of events= 15 C 2 = 105
required probability=5/105=1/21 so option 2 is correct

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

pls let me know ?

1. it ws dx/dt=x^2

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

2. dx/x^2=dt -1/x=t+c x(t) = -1/(t+c) x(0)=-1/(0+c)=1 so c=-1
x(t) = -1/(t-1) If t tends to 1 then x will be blow up so option 1 is correct

24. diff eqn for langrangian ques?
-t^2/2m+c1t+c2

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

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

1. choice A-only A can b answered

2. ya 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??

3. Only option A can be answered..
Option B cnt be cz of insufficient data..

26. What will b the ans of wavelength analog of layman series

1. 135 nm i suppose

2. yes its a 135 nm

27. 1/(1*2)+1/(2*3)+1/(3*4)+.....infinity what will b it's ans?

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

2. Its a 1..!
Cz 0.5+0.166+0.083+0.05+0.033+0.0239+.....=1

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

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

28. can any one submit the key for booklet C

1. first u provide then i cross check it ,in physical science

2. part A- of booklet C
can 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

29. for booklet A
PART 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

1. ans of que 7 is 3*400

2. Q46...1 IS CORRECT SIR

30. B,qno: 1,ans is B

31. dx/dt= x^2 as x(0)=1,ans is infinity

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

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32. questions are available in csir website

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2. yes question papers are now available on csir web but physics paper is not uploaded very properly.
Some part of questions are missing in the scanned version of the paper

1. hermite function explain plz

34. dx/dt=x^2
-1/x=t+c
f(0)=1;
c=-1
x=-1/(t-1)
so...when t-----1.....x ----infinity..... blow up

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1. is it section C or section A parimi?

38. Csir 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.

1. thx srinu, how is IIIT

2. it is good,do u know about 3it?

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

4. may i know who are u?

39. Booklet B of Physics (Or Booklet A of chemistry):
Sec-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

1. Booklet B: For Sec-C, Please correct me if I am wrong:
46. B, 48. B, 52. A, 54. B, 59. D, 60. C, 64. B, 70. C, 75. A

2. Any one can query me for any answer of Sec-A I mentioned here.
and also for SEc B and C if he remember the question(s)

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41. IN SECTION B THE CONVERSION TIME FOR THE ADC CONVERTER IN REMAINS UNCHANGED

1. its remain unchanged..
Cz in SAR, input voltage has nothing to do with Conversation time...!

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

3. ohhh TY, even i answered it as unchanged, but i was in little confusion

42. any one have booklet c answer in physical science

43. Can any answer that question, with proper logic?
A=B+C+D

44. Booklet A, Q.31, relativistic motion of a disk,
Solution: 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).

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

45. Any detailed ans for Q.64 Booklet-A ?
if 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

1. option ii, as identity valids

2. option ii, as identity valids

3. Option 1

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

5. ans is 1

6. Option 1 is correct
[ f(A), B ] = [ A, B ] F'(A)

7. how option 1 is correct murty b

8. its e^A I did it in exams.
expand e^A as 1+A+A^2/2!+....... and then solve we get answer a option

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

1. u did good job.

2. Thanks dear

47. Capacitance in the parallel wires of 4cm diameter separated by distance 2m (as I remember) is 88.5 nF, I think.
Correct me if it is wrong.

48. what about stable isobar ans??

49. Plz 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.

51. what was the answer of drude model question?

52. what will be the cut off hey guys

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54. Booklet Code A (Find it on CSIR site)

PART-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]

1. from where u got it

2. I got it from calculations & discussions of/with me and my friends.
I hope all answers are correct.

3. 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...

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

The 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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8. Correction:

Q12: Ans(4) [R+Rcot]

55. booklet A

Q.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

56. Booklet A

Q.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

1. that is Q.75, not Q.74

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

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

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57. no...its 4/9 for Sa+Sb+Sc invarient
because....
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

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

59. degrees of freedom is D*(D+1)/2

60. COde A
Sec. 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

1. please explain sec c- 71-3

part 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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2. not 44th but for 54th

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

63. In CSIR NET is there section-wise cutoff or not? Please reply me?

1. cumulative total not section wise

64. booklet A:
section 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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2. 9-4 since 60/2=30 60/3=20 60/4=15 60/5=12 60/6=10

65. what is the amswer of 25 question in booklet A?

1. what will be the cutoff

2. may be 70 marks

3. around 70 marks

66. around 70 marks

67. can some one pot the key for booklet C please

68. pronab choubey 30 june 2013 what is the answer of 67 question in booklet A