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1. (C)  2. (A)  3. B)   4. (D)  5. (B)  6. (C)  7. (A)
8. (B)  9. (D)  10. (B) 11. (D) 12. (C) 13. (B) 14. (C)
15. (D) 16. (A) 17. (A) 18. (D) 19. (C) 20. (D) 21. (A)
22. (B) 23. (C) 24. (C) 25. (D) 26. (D) 27. (B) 28. (A)
29. (C) 30. (A) 31. (D) 32. (C) 33. (A) 34. (C) 35. (D)
36. (B) 37. (A) 38. (D) 39. (A) 40. (B) 41. (C) 42. (D)
43. (C) 44. (B) 45. (D) 46. (C) 47. (A) 48. (D) 49. (B)
50. (A) 51. (B) 52. (C) 53. (B) 54. (A) 55. (D) 56. (B)
57. (C) 58. (B) 59. (A) 60. (D) 61. (C) 62. (A) 63. (D)
64. (C) 65. (D)

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1. i need answer key of gate 2011 physics

what is the answer of the question from countour integration?

3. answer is zero because poles are at Z=3.14 WHICH DOES NOT LIE WITHIN UNIT CIRCLE.

4. y rnt you puting solutions on this site...the link is not serving the purpose

5. when will we get the solutions key for 2011 physics?also,what is the answer for the linked answer questions on harmonic oscillator,and the one on electromagnetism?

6. Answers to Genral Aptitude(GA) Questions (56 - 65)
56. (A) Hyperbolic
57. (A) To visit
58. (B) q^2 = PR
59. (A) Incomprehensible
60. (D) Separate
61. (C) 6
62. (A) 5
63. (A) P
64. (A) How to write a letter of...
65. (D) 7.29 litres

If you have any doubts mail me at futurehelp at gmail.com

All the best

48. (C) f(z) has a pole of order 2...
49. (B) zero
50. (C)
51. (B)
52. (C) -1
53. (B) 3hbar/2w
54. (A)
55. (D)

9. Hi Chetan, it seems that you have gotten many wrong... the answer to the corrected ones should be

(56) restrained
(57) visiting
(58) (B) is fine
(60) Split
(61) (C) is fine
(62) (A) is fine
(63) There is an inverse proportionality involved here. The microbe quantity was mg required to destroy some half the mass of body. Clearly, more of it means less potency.
(65) (D) is fine

10. arunkumar33015@gmail.com14 February, 2011

answer of hormonic oscillator is α = -1 and 9Ћw/2

11. I owe this to physicskerala since I downloaded a lot of stuff from here.

1 (C)
2 (A)
3 (B)
4 (D)
5 (B)
6 (C)
7 (A)
8 (B)
9
10 (B)
11 (D)
12 (C)
13 (B)
14 (A)
15 (D)
16 (A)
17 (A)
18 (D)
19 (C)
20 (D)
21 (A)
22 (B)
23
24 (C)
25 (D)

12. And the last set of answers

26 (D)
27 (B)
28 (A)
29 (A)
30 (A)
31 (D)
32 (C)
33 (A)
34 (C)
35 (D)
36 (B)
37 (C)
38 (D)
39 (A)
40
41 (C)
42 (C)
43 (C)
44 (B)
45
46 (C)
47 (A)
48 (C)
49 (B)
50 (A)
51 (B)
52 (C)
53 (D)
54 (A)
55 (D)

13. @arun kumar: thank you for that solution!:) and may i know when the solutions would be out for physics?also,i would request someone to provide the answer for the relativity question,i somehow could not arrive at any of the options given there!:( thanks again!

14. @Basudev
Great work Mr. Basudev,
Q 53) Its B
Becoz, you have to normalize the wave function while calculating energy eigen value,

Q 48) I think its D
Becoz the given fn sin(z)/(z-pi)^2 is not diverging function. It has to be differentiated,

Q 14) Its C
Just check it

Q 10) Its C,
there won't be any phase shift between input and output voltages in common emitter amplifier,

Q 60) You can find that both separate and split are the antonyms to amalgamate,

Q 45) I think Its D but not sure

Q 40) Its B

Q 9) D

cheers....

15. i think
9. D

16. 4.A bcz entropy of the universe increases n work done in a cyclic process is zero.
9.D bcoz in superconductors energy gap is very less.
14.A fine
22.A right ans n B IS ONLY FOR SINGLE e- system.
23.C IS RIGHT ans for optical branches, B FOR ACOUSTICAL branches.
40. B
42.i think its B.
43.its D BCZ IN rotational Raman specrum first stoke line is at 6B.
45.ITS D ONLY.

17. answer of 4 is D
Carnot engine is an ideal engine. Look T-S diagram

18. Kalyani,
For ques.43, Ans is C.As you said the first stoke/anti stoke line is at 6B from origninal frequency.In question, 2B=20/cm. B=10/cm.

19. For 43, ans is c.

20. for ques.44, ans is D. because vol. of primitive cell of b.c.c lattice is (a^3)/2. please check it & reply

21. 40. B

23. 48.) D, simple pole at z=pi

24. @sree..
48.) D, simple pole at z=pi. function sin(z) has a zero at z= pi, so in the lorentz series expansion should terminate at the first term of the principal part.

check it by expanding the function about z=pi.

49.) B, singular point is outside the contour of integration

25. thank you all,but may i know what the cut off is expected to be?i mean,was this paper tough or easy?i personally felt it was a difficult one,so im expecting cut off to be lower,say,25 or so!

26. I think the cutoff may be 25(+or-)3. The questions are somewhat easy compare last two years. Anyway, its not necessay think of it becaues whan the questions are easy, propotionaly there is a propability for increase in negative marks(due to careless mistakes).

27. 14)c since [Lx,Lz}= -ih(bar)Ly

4)C. Since, entropy never decreases in any process. Also, entropy of universe always increases.

16)B. Since, there will be no potential from grounded plane. so, V due to just 2q charge.

22)B

30)C

44)C. Since, volume=a^3 & d=lambda

53)B

57)visiting

60)separate

63)S. As toxicity(mg needed) becomes inversely proportional to danger.

"entropy never decreases in any process" is wrong idea.

entropy can increase or decrease in process.

entropy of the universe is increasing.

4. D

31. @sreejith P: yeah..you are correct..i typed it mistakenly..but have a look at my other answers.

32. @basudev what would u tell about Q33

Look back ques.44, The Answer is D since the B.C.C primitive cell contains 2 atoms. The answer is a^3 /2.

34. 16) C. Grounded plane contributes a potential (see image problems). So final potential is due to plane and 2q.

35. can anyone explain 29.

36. can anyone expalin me ques no. 36?

37. Hi, for problem number 44, I am pretty sure that the answer is (B), since for a BCC lattice, there is only (110) reflection. The (100) does not exist. Do the structure factor calculation to convince yourself. But the volume is indeed a^3/2.

For (42) I think the answer should be (c) since there are 3 possible ways in which the 2 p electrons can arrange themselves. Thereafter, the d electron can of course arrange itself in 5 ways. The multiplicity due to spin up and down should also be taken in consideration but the answer 30 does not exist. I think 15 should be the answer

For (39) the distribution of electrons with energy is indeed considered to find out the missing mass due to antineutrinos. The fellows at CERN (most likely) happened to do the extrapolation and initially found a negative value for mass^2 of the antineutrino. This was corrected later of course.

38. 55) A. what do u think?

39. You are right about (14), the answer is indeed (C). It was a typo.

The answer to (10) should be (B). Remember the load line and how the input signal is reflected across the load line to get the output.

For (37), I have a correction. the answer is actually (A). Since the charge would be residing on the surface of the sphere, and the field inside the sphere would be 0, you have to use E-0 = rho/e0 = del(V)/del(r)

40. For (48) the answer is indeed (D). I got that wrong.

I also missed the normalisation in problem (53). The answer is (B)

For (36), the N1 = k (degeneracy) exp(-E/kT), so that N1 = k*4*exp(-E/kT)
and N2 = k*2*exp(-2E/kT).
Do N2/N1 and plug that into N1+N2=N. You can get N2.

For (29), it is just a partial derivative F=-del(V)/del(q)

41. @Aleena the answer to (55) is indeed (D). Do the ExB/mu. It works out.

For (50), the number of microstates n = 2 Integral(dx dp)/h = (1/pi) L*k. Plug the value of k in terms of energy E and then take the derivative dn/dE and represent back in terms of k. The answer is (B)

42. @basudev:
kindly explain how the answer to Q5) is B.
since for specific heat delta T is in the denominator & for entropy just T so I think answer should be D.But want to know more about it.

43. The answer to (33) is (A). You can perform the integral

44. Can anyone give me an idea about the rank estimattor for the total marks of 29

45. for Q24 i think its A because fermi level depends on electron only

46. @Chandan The entropy is the first derivative of gibbs free energy. The first derivatives for first order phase transitions are discontinuous. Further, the specific heat being the derivative of entropy (remember Cv = dQ/dt = T dS/dt), simply blows out at Tc. The answer is indeed (B)

47. @Abhi The fermi energy level goes as Ef = (Ec+Ev)/2 + KT ln(mh/me)... so you can see that it depends both upon mass of holes and mass of electrons.

48. For Q.48 ans is C I think.. Remember f(z) has a pole of order 2 at z=pi.... Any controversial??????

49. what is the answer of Que 11.
how to calculate largest eigen value.

3.B, 6.C, 11.D, 12.C, 15.D,19.C, 26.D, 29.A ,30.A, 36.B, 40.B, 46.C, 52.C , 54.A, 58.B, 61.C, 62.A, 63.D,64.C

51. Some of the answers I’m quite sure about is given.CHECK IT out
3.B
6.C
11.D
12.C
15.D
19.C
26.D
29.A
30.A
36.B
40.B
46.C
52.C
54.A
58.B
61.C
62.A
63.D
64.C
65.D
56.B
60.B

52. ANONYMOUS :

WHAT'S THE RANK NEEDED TO GET THRO IISc???

53. For 32, B is correct since both the bodies have same mass and velocity they will attain no net velocity after sticking together.So their mass will be rest mass only.

54. I went through the discussion.It was very helpful and explainatory.But I have few doubts:
4(a),as (d) is always true entropy of the universe certainly increases
10 Is it(c)?
22.Is it (a) or (b)?
25. I dont think (d) is correct.The answer can be either(a) or (b)
27.The answer should be (d) as the cylinder is centred at origin.
35.Is the answer (d),as ln3 is not convincing.
36.Ithink the answer (c) as it is doubly dengenerate for upper level.
39.I think the answer is (b) as n,p,e are all fermions.But according to the equation n becomes boson in absence of antineutriono.
43.(d).I think the answer is (d).
47(c)
48. Answer is (c).As for residue we need to take derivative of denominator and only then we will get ans (d).But for finding pole we dont have to do this.
50.(b)
63.(a).Ratio between toxicity:potency:growth is highest for P.

55. What is expected cut-off for this year?

56. basudev...
can u tell me how u get answer of 18 and 21.....i think answer of 18-B and 21-B....EXPLAIN UR ANSWERS....

57. 16.) D
total configuration according to method of images is
k(-q^2 + (2q)^2 + -((2q)^2)/4) = k (q^2)/2

58. basudev..........how u got the answers of 18-D and 21-A ...explain ur answers ...i think 18-B and 21-B should be the right answer....

59. To Chintoo...
For Eigenvalue problem
Trace= sum of diagonal elements( say a+b+c=11)
also
The absolute value of the determinant is the product of the eigenvalues of the matrix
ie, a*b*c= 36
Three unknowns and two equations. So Lets try trail and error method
Now possible values for a,b and c is
9,4 and 1 for this 9+4+1=14. Our trace is 11 this is not the right combination
another possibility is
9+2+2= 14 it wont
12+3+1= 16 it wont
6+3+2= 11 it works
highest eigenvalue is 6

For Q. 48 and is C only
balagangadhar said its not diverging and has to be differentiated...
Of course you are wright.
But we don't care about the Residue here.. we have to find POLE and OREDER first. then according to the ORDER only we should differentiate.. What U say...! I don't know why basudev changed his mind eventually.

61. @shishir: (18) should have the sum of two torques, one m x B due to the local field and the second, the torque about the origin due to the non uniform field, r x F.

For (21) the total angular momentum of the highest state is l+(1/2). You can check that it will have 2l+2 states.

62. @Amlan: For 16, you have to form a couple of image charges, of charge -q at -.5 m and -2q at -1.5m. All the forces point in the same direction. Answer is indeed (A)

For 20, this is actually a tad confusing since all the methods mentioned can be used to detect neutrons. However, the 2 most likely answers are (B) and (D) since ionisation happens through secondary particles. However, according to wikipedia, direct scintillation is the most preferred method.

For 25, the voltage across the 10 V zener has to be 10 V. Obviously, the current is 5 mA.

For 27, you can transform the integral into integral((del.r)dV). del.r = 3 and integral(dV) = pi*r^2*h. Answer is (B).

For 35, the number of available states is 3^N. Entropy is indeed k ln(3^N) = N k ln(3).

For 39, am absolutely positive that it is (A).

For 43, the answer is (c). Look at the discussion above.

47 is (A). It is just a comparator.

For 63, I am not sure why you have to take the ratio between the toxicity and potency and growth. The facet mentioned is proportionality. How exactly does the ratio follow?

For 48, I would so dearly love the answer to be (C) but the thing is that you wont have to go as far as differentiating the function. You can expand the sine function about z=pi, in which case, the only term that blows out is 1/(z-pi). The rest of them go to zero. The answer has a very good chance of being (D)

63. basudev....i think u r very good student of physics and u r going to be in top 10 ranks..........
thanks for ur comment on ...
Q NO.18 & 21..u r correct....... explain Q no. 28-A and 48-C.....
i think u r correct and i also got the same answers .... but lots of comment shows different answers...

64. for Q no. 20 the correct answer must be D... refer KAPLAN nuclear physics pg no.-563-564 ,for Q no. 32 the answer is C...refer modern physics schaum series (pg no-53 & solved problem no.- 8.26).......
for Q no. 23 answer is C...refer kittel(7 th ed.- solid state physics)pg no.-104.........
Q no. 38-D is the correct answer..this Question has been repeated..see(gate-2007 & Q no.60)..

65. @copra
i dont understand the meaning of -
'then according to the ORDER only we should differentiate.. '
and 'its not diverging and has to be differentiated... '
this is hilarious.

whether we care about residue or not, if you want to classify the singular points, we should stick to Lorentz series. there are many ways of getting the ans here i wil list 2.

simplest way is look at the numerator of the function it has a simple zero at z=pi. so overall the function has singularity at z=pi of order one.
simplest example is ' (sin(z))/z ', it doesnt have singular point. the num and denominator has zero at z=0.
please expand it and see, no principal part for lorentz series.

when we consider lorentz series
expand
you can see that it terminates at the first principal part.

so ans for 48.) D

@basudev . u did a great job of collecting the ans and active in discussion

as u said we have to place the image charges in those positions.
so by couloub's law, the total force acting on q is,(it obeys superposition)

k(-q^2 + (2q)^2 + -((2q)^2)/4) = k(q^2)/2
so the ans is 16.) D

similarly, last yrs ans-key has lot of mistakes

66. For Q.37, i got A..anybody else got same?

67. basudev...
jest 2011... only 3 marks questions....
for relativity problem 0.74c is the speed ....for carbon dating problem 16800 is the answer....for displacement current problem it is inserted to make total current continous.....
what will be the solution for ...
poison bracket problem..(2x+2 or 0)..... for non commutator problem..i think eigen ket will never be equal...what will be the answer for...parity conservation problem and for complex analysis problem..
what will be the ratio of Ne to He in the larger partition.....
what will be the answer for ESR spectroscopy problem....what will be the answer for maximum value of angle ...red, green ,yellow or violet........for n two leVEL SYSTEM,THE SYSTEM IN THE EXCITED STATE WILL BE n/2 ,n, 0, 3n/4....
for forced oscillation problem what will be answer...and for two body system what will be the no of independent coordinates 4,5,6.... for entropy problem what will be the temperature...3,1/3,1,ln2

68. 47,D
what is the answer of qns no.36??

69. @copra
i dont understand the meaning of -
'then according to the ORDER only we should differentiate.. '
and 'its not diverging and has to be differentiated... '
this is hilarious.

whether we care about residue or not, if you want to classify the singular points, we should stick to Lorentz series. there are many ways of getting the ans here i wil list 2.

simplest way is look at the numerator of the function it has a simple zero at z=pi. so overall the function has singularity at z=pi of order one.
simplest example is ' (sin(z))/z ', it doesnt have singular point. the num and denominator has zero at z=0.
please expand it and see, no principal part for lorentz series.

when we consider lorentz series
expand
you can see that it terminates at the first principal part.

so ans for 48.) D

@basudev . u did a great job of collecting the ans and active in discussion

as u said we have to place the image charges in those positions.
so by couloub's law, the total force acting on q is,(it obeys superposition)

k(-q^2 + (2q)^2 + -((2q)^2)/4) = k(q^2)/2
so the ans is 16.) D

similarly, last yrs ans-key has lot of mistakes

70. @dreamz: My friend, you are putting the repulsive interaction between the +q and 2q charges as +ve. Thing is, this force is going to be directed in the same direction as the ones for the negative charges since this positive charge is located in the opposite side of the negative charges.

@shishir: Thanks for your wishes. About JEST,unfortunately I could not appear for it due to a freak registration related incident. I can't be of much help on that. Sorry about it.

71. By the way, about 28, (y+1)dy=-xdx, so that y^2/2 + y = -x^2 +c.... rearrange terms, and a circle with center at x=0 and y=-1/2 shows up with a variable radius.

72. To basudev.....
Yes man U r ryt... I caught on the wright trap.. Q. 48 the ans id D. and I had a Nice discussion with U and some other Guys... Thanks to the blogspot owner and Friends whoever posted their view.

73. @creepy: (37) is indeed (A). I corrected my answer in the discussion above.

74. Good to see that the answers have finally been summarized. I find the answer to 29 as (C) particularly interesting since the generalised potential has dissipation and is not a conservative potential. I missed this point. Having said this, I am not sure if the person posting the answers understood why the answers were going to be such.

For example, the answer to 16, being a sitter problem for high school students, has not been provided. Clearly, the person putting the answer up does not understand it.

Coming to 27, which as I mentioned in my solution, has to be transformed into a volume integral to become a 3 second problem, has been wrongly evaluated. The answer should be (B).

For (39), the existence of antineutrinos certainly weren't inferred from parity violation experiments. Look up any text book. Nowhere does it say that. It cannot be forward backward asymmetry of electrons!

Coming to 42, could anybody explain why there should be 150 combinations? I can't get that figure by any stretch of imagination! Dont tell me you put it up simply because someone suggested the answer.

By the way, for 56, it is actually straight out of barrons GRE book. Type the sentence asked on google and the answer would show up straightaway. It is indeed (B).

75. Check this out:

http://en.wikipedia.org/wiki/Beta_decay

Neutrinos in beta decay
Historically, the study of beta decay provided the first physical evidence of the neutrino. In 1911 Lise Meitner and Otto Hahn performed an experiment that showed that the energies of electrons emitted by beta decay had a continuous rather than discrete spectrum. This was in apparent contradiction to the law of conservation of energy, as it appeared that energy was lost in the beta decay process. A second problem was that the spin of the Nitrogen-14 atom was 1, in contradiction to the Rutherford prediction of ½.

76. The answer key has some errors. I wud like to correct few which i am confident. 16. C ; 17. C (Refer D.J.Griffith page no. 226, example 5.8); 28. B(I am not sure).

77. One error in the answer key is ques.4. I strongly believe that the ans is C.

78. @Remya: I am afraid the problem you mentioned in Griffiths is not quite equivalent to the problem in the paper. There is an issue of handedness involved. The system of coordinates has to behave like a right handed system by convention. The answer given in the key is correct.

So goes for both 16 and 28 (which I explicitly mentioned how to do in the discussion above).

79. @remya
i believe
answer of question 4 is D
Carnot engine is an ideal theoretical engine.
its cycle is ideal and reversible,
look its TS diagram

80. basudev,
You are right. I just missed them. The answers for 16 and 17 is A. 28 ofcourse I was not sure. Thanks anyway.

81. About the Qus. no. 54 & 55
We are give ans (A) of Qus.(54)
Then direction of properation
is -(i+j)/2^1/2.
OR Poynting Vector Ans (D).
Poynting Vector & Direction of properation is same OR Hold left hand rule.
It's combnation is correct.

But, When
we are take Ans (B) of Qus.54. Then direction of properation
is (i-j)/2^1/2.
OR Poynting Vector Ans (A).
Poynting Vector & Direction of properation is same OR Hold left hand rule.

it's combnation is also correct.
so plz tell me.

What is this?

i thing ans(D)
Because in oscillator system have orthonormalized states so
0th level energy is 1h/2.
1st level energy is 3h/2.
2ed level energy is 5h/2.

every state have unity probability.
So simply add every state energy.
Get total energy.
Plz tell me.
How can get ans.(B)
thanks

83. @Anuj: Instead of sitting and guessing, please kindly plug the magnetic field into the Maxwell's equations. There is only one unique solution.

84. @basudev you can say right.
but when we take electric field (B).
and use Farday law (Mexwell 2ed eq.)
or solved same pattern then we get also this magnetic field.
again problem.

85. @Anuj: I dont follow what you said. By the way, Faraday's law merely states del.E = rho/epsilon. Where is the connection between magnetic and electric fields?

86. @basudev farday law are

Curl E = dB/dt.

solved by this eq.
get it problem.
because Maxwall have linear Independence solution.
So maxwall eq.s have one or more sol.
Thanks.

87. To Anuj Kumar

Try to find the E( Electric field) first using the above relation. Don't try to fix the answer arbitrarily.

The second combination You mentioned above holds good the left hand rule. But Our aim is to find correct answer. not the combination of answers

88. @ CoPRA
I am also find a right ans.
But when i am use maxwell forth eq.(Ampeer law) for free space then we get sol. (A).
or we are going with mexwell third eq.(farday law) for free space then get ans. (B).
Our both sol.'s are not satisfies Mexwell first or second eq.(Gauss or free magnetic Pole)
So How can use maxwell eq. solved this qus.

89. Anonymous02 July, 2011

90. When $$a \ne 0$$, there are two solutions to $$ax^2 + bx + c = 0$$ and they are

$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$

91. 42. D
50. A
How?