iitjee 2012 paper 2 solutions

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IIT-JEE 2012 PAPER- II CODE -7 (SOLUTION) Date: 08/ 04/ 2012

PHYSICS1.

l4

336 1000.3 4

4 512

= 15.2 cm

l

O

2.

rv 0

At t = 0

rv 0

AtT

t2

Hence (A)

3.

M = I (Area enclosed)

22a

I 4 a2 4

P Q



22a

I a k 2

4.

R/2

22 R

B 2 r j. r 4

0I

B 2 r

22 Rr4

2

2 RR

4

220

22

RI r4

BR

2 R 4

Rr2

B = 0 (D)

5.

Let the volume of the cylindrical shell be SV and the volume of the ‘hollow’ part inside it be 0V .

Therefore the total volume of the cylinder will be S 0V V . Now, let the volume of water inside

the hollow part be V ' . Since the cylinder is in equilibrium in a half submerged state,

s 0m s w w

V VW B V g V'g g

2

where w mand are the densities of water and the material

of the cylinder respectively. Rearranging the above expression we get,



0 mS

w

V 1V ' V

2 2

, which clearly shows that when the relative density is less than half, the

volume of water inside V’ must be greater than 0V

2

. Therefore ans is (A).

6.

Capacitors in series

P.d 16volt

charge on 3 F

48 C

7. (C)

Q U W

W 0

pQ nc T

5R

2 35 30 25R2

208J

8. (B)

P 0ˆˆV R1 V j

Q OˆV V j

Time of strike is T 8

Hence P lands in unshaded region & Q also in the unshaded region.

9. (D)

4 F

5 F

80 C

80 C



If mass of anti-neutrino taken into consideration the KE of electron will be less than max

KE

10. (C)

2 2 2

n p em c m c m c Te

6

Te 0.8 10 ev

2 2 2 6

n p em c m c m c 0.8 10 ev (1)

2 2 2

n p e anm c m c m c 3ev Te ' T (2)

Using (1) in (2)

6

an0.8 10 3ev Te ' T

6

anT 0.8 10

11. (D)

Angles rotated about instantaneous axes are same in both cases.

12. (A)

13. (B)

By definition;

c cn v

v n

14. (C)

1 21 sin sin

12

sinsin

As 20 0

so, only possible choice is (C)

15. (A), (B), (C)



Field due to F & C at O

1E 2K

towards OC

2E 2K

towards OE

Field due to A & D at OE

4K towards OD.

Resultant of 1 2E &E

will be

2K cos 3 2K cos 3

2K along OD

Hence, net field at O

4K 2K 6K along OD

Potential at O due to A, B, C, D, E, F is A B C D E FV ,V ,V ,V , V , V . Then,

A D A DV V V V 0 (1)

F C F CV V V V 0 (2)

EB B EV V V V 0 (3)

so, A B C D E FV V V V V V 0

All points on PR are equidistant from the pairs (A, D); (F, E); (B, C).

All these pairs have equal and opposite charges so their potentials add to zero on all points of PR.

16.

v gr

2

GMv R

R

2G 4v R

R 3

24v G R

3

v R

(B) and (D) are correct.



17.

Linear velocity of point o = 3R i

= 3 Ri (A) is correct

Linear velocity of P

p R Rv 3 R cos 60 i sin 60j2 2

11 R 3v i R j

4 2

Ans (B) is correct.

18.

Moment of inertia of P is more than that of Q

P Qa a

(D) is correct.

19.

If current is constant no change in magnetic flux through the loop. When current changes, change in magnetic flux

in left half side P and right half side Q of the loop are in opposite direction. Since net change in flux is zero therefor

no emf is induced.

P Q

(A) and (C) are correct.

20.

Current through 100 branch

2

12

C

20I

X R



12

26

20I

1100

100 100 10

20 2

10100 2

Voltage across 100 resistance

R

2V 100 10 2

10

(A,C) is correct.

CHEMISTRY

21. (D)

(s) 2(g) 2 2(g)

reaction reac tan t product

2C H C H

H (B.E.) (B.E.)

225 2(1414) (330) 2(350) (C C)

C C 815kJ

22. (A)

CH3 - CH2 - C - CH3

OCN-

CH3

- CH2

- C - CH3

O-

CN

CH3 - CH = CH - COOH

CN

95%

H2SO

4

23. (C)

For paramagnetic behavior no pairing is required in Ni+2 making it tetrahedral with sp3 hybridization and for

diamagnetic behavior d – e – s should be paired making it square planar with dsp2 hybridisation



24. (B)

Ag + NaCN + O2 + H2O

(Impure)( oxidising agent)

Na[ Ag(CN)2

] + NaOH

Zn + 2Na [ Ag(CN)2

]

+

2Ag

Na2

[ Zn (CN)4]

(Pure)

25. (C)

At room temperature white phosphorus to give PH3 and NaH2 PO2 ., but at higher temperature NaH2PO2 can

disproportionate to give PH3 and Na3PO4 . Therefore , the most appropriate option is (c) where the oxidation

state of phosphorus will be – 3 and +5.

26. (D)

Hybridization of Xe in XeO2F2 is sp3 d . due to one lone pair on Xe the shape of molecule is “ sew – saw”

27. (A)

Tb = Kb × m

2 = 0.76 x m

m = 2 / 0.76 = 200/76

m = 2 /0.76

B

2

760 p 0.76X2 1000760

0.76 18

760 p 36

760 – 36 = p

724 = p



28. (B)

O

COOH

O

+ CO2

- keto acid gives CO2 on heating .

29. (D)

cellG 2FE

2 96500 0.059

11.387

11.4kJ

30. (B)

Emf for concentration cell

0.059 sE 0 log

2 0.001

0.059 =0.0591 s

log2 0.001

s2 log

0.001

10 -2 =5

3

ss 10

10

Ksp = 4s3 = 4 ( 10-5 ) 3 = 4 × 10 – 15

31. (A)

I

CHO



CHO

(CH3CO)

2

CH3COONa

CH= CH - C - OH

O

H2 , Pd - C

CH2CH

2- CH

2OHSOCl2

CH2

Cl

AlCl3

(K)

32. (A)

Since J give NaHCO3 so it should be a carboxylic acid. Since Bayer’s test is positive so C – C bond should be

present so J is unsaturated carboxylic acid & with given set of reagents Benzaldehyde (I) form cinnamic acid

(J).

33. (C)

CaOCl2 + CH3COOH (CH3COO)2 Ca + Cl2

Cl2 + 2KI I2 + 2KCl

I2 + 2Na2 S2O3 2NaI + Na2 S4O6

No. of equation of Na2S2O3 = number of equation of Cl2

2

48 0.25no.of equi.of Cl

1000

= 12 x 10 – 3

n-factor of Cl2 = 2

Number of moles Cl2 = 6 × 10 – 3

Number of moles CaOCl2 = 6 × 10 – 3

325 M6 10

1000

6M 0.24M

25



34. (A)

2 22HOCl Cl O H O

35. (ACD)

3 6 4 6 3KI K [Fe(CN) ] K Fe(CN) KI (Brownish yellow solution)

K4 [Fe(CN)6 ] + ZnSO4 Zn2 [ Fe(CN)6 ] + K2SO4

( White ppt )

KI3 + Na2 S2O3 KI + Na2S4O6 ( colour less)

36. (ACD)

O

CH3

O

LiAlH4

CH2OH

CH2OH

CH3

excess

(CH3CO)

2O

CH2- O - COCH3

CH2- O - COCH3

CH3

CrO3

/ H+

COOH

COOH

CH3

(V) optically inactive

(W)(U)

37. A, B, C

M and N are diastereomers

M and O are identical

M and P are enantiomers

M and Q are diastereomers

38. B, D



Graphite has higher electrical conductivity but diamond has higher thermal conductivity. C–C bond order in

graphite is more than one due to delocalized π electrons.

39. A, C

40. A, D

Mathematics

41. Let

2

2

2dxxcos

x

xlnxI

Also

2

2

2dxxcos

x

xlnxI

Adding

2

2

2 dxxcosx2I2

2

0

2 dyxcosx2I

Integrating using By parts

42

2

I

(B)

42. Consider the2021 a

1...,

a

1,

a

1.AP

In this A.P.

d)120(51

251

1925

4d

Now, if an > 0, 0a

1

n



01925

4)1n(

5

1

a

1

n

Solving 14

95n

Since In

25n

least n = 25

(D)

43. Consider k ˆ3 j2in1

k ˆ jin2

111

321

ˆˆˆ

21

k ji

nn

= k ˆ j11i5

Thus, the plane is 5x – 11y + z = d

3

2

1115

d)1(1)1(11)3(5

222

Thus, we get d = 17

5x – 11y + z = 17

(A)

44. Simplifying

P

tanP2sinPsin2

P2sinPsin2 2

By cosine role,

)2 / 5()2 / 7(22)2 / 5()2 / 7(PCos

222

=)5)(7(2

162549

35

29

2 / Ptan1

2 / Ptan12

2



32

32 / Ptan2

Also )()()( csbsass

= 22

1

2

34

= 6

Comparing points we get it as

2

4

3

(C)

45. Let k ˆ3 j2i7r

The vector can be as

a

r

a

Now 29|ba|

Also 29|r|

rba

4)k 3 j2i7(.r

(C)

46. P = –I = (PT = 2P + I P = 2PT + I = 2(2P + I) + I 3P + 3I = 0 P = –I)

100

010

001

P is a solution

(D)

47. 0)1a1(x)1a12(x)1a13( 2 ….(1)

Under limit 0a



Using series expansion

...!2

a)1n(nna1)a1(

2n

na1)a1( n

=0)1a16(x)1a12(x)1a13( 2

Applying this result in eq (1)

06

a

2

axx

3

a 2

2x2 + 3x + 1 = 0

2x2 + 2x + x + 1 = 0

2x(x + 1) + 1 (x + 1) = 0

(2x + 1) (x + 1) = 0

2

1x or x = –1

(B)

48.

6

16sameareD,D,DAll3

3

1

6

29056C

differentisothersame

areD,D,Dof twoAny2

2

1

6

3120456differentareD,D,D1

)P(outcomefavourableof obabilityPr)N(casesof .NoCasei

321

1

3321

321

ii

Probability =

i

ii

N

NP

=690120

6

16

3

190

2

1120

= 216

13060

=216

91

Hence, answer is (A)

49. P : f(x) + 2x = 2(1 + x2)

(1 – x)2sin2x + x2 + 2x = 2(1 + x2)



2

22

)x1(

)x1(1xsin

not possible

P is false

Q : 2f(x) + 1 = 2x(1 + x)

]1,0[

)x1(2

1x2xsin

2

2

Has a solution

Q is true

(C)

50. )x(f .)x(h)x(f xln1x

1x2)x(g

Where xln1x1x2)x(h

2

2

2)x1(x

)1x(

x

1

)x1(

4)x(h

1x0

h(x) is s.d.

0)x(hlim1x

)1(h)x(h

)0)x(f As(0)x(g 0)x(h

g(x) is S.d.

(B)

51. 12 M A MxY

options fromcase M and samebewill M M x M y

M Mx y0

1)3–(

12

2

2

Comparing the equation (1) and (2)

1MM31M222

M31M2 substitutions in (1)

22 M91M

22

1M



8

92

22

1

x y

622 x y

(D)

52. x2 + y2 = 4 at )1,3(P

3m4y3x

3

1m1

0c3yx

1

)3(1

303

2

c

3 + c = 2

c = –1

Possible Equation of L 13yx

(A)

53. The fi rst digit has to be 1 and the next digit can be either 0 or 1

Case (i) 2nd digit is ‘0’

digits2n

.............101

an – 2

Case (ii) 2nd digit is ‘1’

1

1

.............11

n

n

a

digits

an = an–1 + an–2

(A)

54. For b6 the number has the last digit is 1 and first digit 1.

1……1

Case (i) no ‘O’s used

1.

Case (ii) One ‘O’ used

4C1 = 4



Case (iii) two ‘O’s used

1 x 1 x 1 x 1

We have to choose two gaps to put zeros = 3C2

3

b6 = 1 + 4 + 3 = 8

(B)

55.

(–1, –1, 0)

(1, –1, 0)

Ca

b

k ˆ2 jk i2a

k ˆk j2i5b

i2c

0]cba[

a , b , c coplanar

0

002

k 25

2k 2

k2 = 4

k = ±2

If k = 2

ban1

= )k ˆ2 j2i5()k ˆ2 j2i2(

= k ˆ6 j6

y – z = –1

If k = –2

)ˆ2ˆ2ˆ5()ˆ2ˆ2ˆ2(2 k jik jin



= k ˆ14 j14

y + z = –1

(B, C)

56. det(adjA) = (det A)2 ( it is a 3 × 3 matrix)

4

311

712

441

)adjA(det

det(A) = ±2

(A, D)

57. 3 / 14cos

3 / 112cos22

3 / 22cos 2

3

22cos

3

21cos2

2

2

3

21

cos2

3

21

2sec

2

Taking posit ive sign, f (1/3) =

3

21

22

2

=

3

22

3

212

=2

31



f(1/3) =

3

21

22

2

=

322

3

212

= 12

3

(A, B)

58.)(

)()|(

Y P

Y X PY X P

=)Y(P

61

2

1

3

12

6

1)Y(P

Similarly P (Y| X) =)X(P

)YX(P

2

13

6

1)X(P

)YX(P)Y(P)X(P)YX(P

=6

1

3

1

2

1

=3

2

Also

6

1

3

1.

2

1)Y(P.)X(P

)Y(P.)X(P)YX(P

X and Y are independent events.

Thus, XCand Y will also be independent

2

1)X(P1)X(P C



6

1

3

1.

2

1)Y(P.)X(P)YX(P CC

(A) And (B)

59. 0)3x()2x(e)x(f 2x

for, 3x,2x0)x(f

Also )3x()2x(x2e)5x2(e)x(f 22 xx

= ))5x2()3x()2x(x2(e2x

)2(f = 0e))1(0(e 44

So, maxima of x = 2 (A)

0e)10(e)3(f 99

So, minima at x = 3 (D)

0)x(f in )3,2(x

So, f(x) is decreasing in (2, 3) (B)

0)2(f , 0)3(f

Hence by Rolle’s theorem

0)c(f for at least one )3,2(c

hence, (C)

(A, B, C, D)

60. As sin x and xcos are continuous functions and an, bn are real number

f(x) is continuous for )n2,1n2(]1n2,n2[x

Now, nn a)n2(sina)n2(f

1bxcoslimb)x(f lim nn2x

nn2x

an – bn = 1

Now, for )2n2,1n2(x

xcosb)x(f 1n

)1n2sin(a)1n2(f n

= an



xcoslimb)x(f lim)1n2(x

1n)1n2(x

= bn + 1 – 1

an – bn + 1 = –1

By taking m = n + 1

am – 1 –bm = –1 or an – 1 – bn = –1

(B, D)

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iitjee 2012 paper 2 solutions