18 10 15 Sr.iit Iz Co Spark Jee Adv2011 p2 Rpta 9 Key Sols

Narayana IIT Academy 18-10-15_Sr.IIT-IZ-CO-SPARK_JEE-ADV_(2011_P2)_RPTA-9_Key &Sol’s

Sr.IIT-IZ-CO-SPARK_P2_Solutions

NARAYANA IIT ACADEMY INDIA

Sec: Sr.IIT-IZ-CO-SPARK JEE–ADVANCE Date: 18-10-15 Time: 3 Hours 2011-P2-Model Max Marks: 240

KEY & SOLUTIONS CHEMISTRY

1 D 2 A 3 D 4 B 5 B 6 D 7 C 8 B 9 A 10 BCD 11 CD 12 A 13 2 14 3 15 0 16 5 17 0 18 3 19 A-PT,

B-PS, C-PQ, D-PR

20 A-PQST, B-PQR, C-PS, D-PS

PHYSICS

21 D 22 C 23 A 24 C 25 B 26 B 27 C 28 C 29 AD 30 AD 31 AC 32 BC 33 1 34 9 35 7 36 5 37 9 38 5 39 A-S;

B-QT; C-R; D-PT

40 A-PRT; B-QS; C-Q; D-S;

MATHEMATICS

41 B 42 C 43 A 44 B 45 C 46 B 47 C 48 D 49 BC 50 ABCD 51 BD 52 AD 53 2 54 2 55 1 56 4 57 1 58 3 59 A-R,

B-P, C-PQR, D-S

60 A-R, B-RS, C-RS, D-PR



CHEMISTRY

1. Due to hydrogen bond it can’t liberate 2H

2.

62O S S

O

O

3.

N N

O. .:

. .

O: . .:

O. .

:

Asymm. form N O N

OO. .

::

. .

. .. .. .. . Symm.form

Lone pair is eight

5. X is 24 6S O :

O S S S S O

O

O

O

O 6. 4SiF can’t under go complete hydrolysis : 4 2 4

SiF H O Si OH HF 4 2 6SiF 2HF H SiF 7. X is 3 2C O : O C C C O .

No carbon is 2sp 12. 3O can act as bleaching in dry state .Moist 2SO and 2Cl moist acts as bleaching agent

14. 5 3 10X Na P O

NaO P O P O P ONa

O O O

ONa ONa ONa 16. Cu S C KBr KI, , , , are oxidized by 2 4conc H SO.

17. 3PH forms phosphonium salt only with strong anhydrous acid.



PHYSICS 21. conceptual 22. conceptual 23. conceptual 24. conceptual 25. conceptual 26. conceptual 27. conceptual 28. conceptual 29.

A

B

C

D

Corresponding displacementwave

Aliter: p = p0 cos xw t

v

s - 0xs sin w tv

0ds xv S W cos w tdt v

V mirrors p At A & C max imumslope

max speed

A – Slope – Ve V Ve

C – Slope +Ve Vis Ve

B,D slope zero V = 0

30. conceptual 31. conceptual 32. conceptual

33. pPv

20

2 Independent of wavelength.

34. xx x=9

12



35.

I AI AI A

AA

2

21 1

25 5

15

25

5

36. A sin . .cm

A . .

2 7 5 3 560

3 5 2 4 95 5

37. Let the Time when the race car emitted a pulse corresponding to 1.5 kHz be t, Then the time it reaches the observer is equal to

s s

s

tt

v v

ttv

2

2

11000 10 10002 10

5 10

s

s

v.v t

1 510

Solving for velocity of sound we get vs=360 m/s 38. B,C,D,E,F are all either at rest of moving down. 39. Conceptual 40.

P

displacement wave at t = t3

C corresponding displacementwave

Adisplacement is towards right (+ve) displacement is towards

left (-ve) At t = t1 particle is at extreme hence speed is zero At t = t2 particle is at mean position hence speed is maximum

A is moving towards –Ve (left) C is moving towards +Ve (right) t = t3 displacement of C is +Ve & it is increasing V is +Ve towards right.

Displacement of A is – ve



MATHEMATICS

41. 2

2

22 2y 4y 9a c . b d 3

4x 16x 25 3

2 22y 4y 9 4x 16x 25 63 . This is possible only when y 1, x 2

42. Let the new plane is (3x 8y 15z 91) (5x 17y 29z 2) 0 . It should be same as

x y z k . Hence 3 5 8 17 15 29 91 2 1 ; k 1841 1 1 k 2

.

43. 1cos3

44. 3 1NP AB AF4 2

45. Direction ratios are proportional to 2, 7,5

49. We have f ' g; g ' h; h ' f hence a f i g j h k;

b g i h j f k;

c h i f j g k

Consider 3 3 3A f g h 3fgh we can see that dA 0dx

. We can see that A=1.

50. a 4 1 x 00 2 3 y 13 0 b z 2

1

a 4 1 0 4 10 2 3 30 2ab, 1 2 3 4b 203 0 b 2 0 6

2 6a ab 3, 3 4a 12

If ab 15 ¸system has unique solution

If ab = 15, 1 2 360 20, 6a 18, 4a 12a

For a = 3, 1 2 3 0

For a 3 or b 5, at least one of 1 2 3, , is non zero

51. Take i, jas sides of right triangle.

52. AB AC AD a

. AB AC b;AB AD c

AB.a 3 AC.a AD.a 3

; 1 1AB AC a b ; AB AD a c3 3



53. We have 1AD.BC AD BC 4d.(c b) d c b d b4

as the triangle ABC is

equilateral.

This gives 14 3 b 6 b cos 6 b cos4

54. 323 59m 32 44

9

. (2mi j k).(i 2mj 3k) | 4m 3 | 2

56. Take r x y z( )

57. Direction ratios of the line of intersection of 2x y z 0 and x y 2z 0 are 1, 3,1

58. P(3, 5, 2)

59. We have 2 2 2ab bc ca 1;(a b) (b c) (c a) 1 . This gives 2 3a2

. Hence

2 3 7(a b c) 22 2

. We know, if 2 2 2x y z 0;x y z 1 then 2 39xyz2

2 1 7OP OQ OR a a ab2 2

60. A) 3x y x y a b x y a b2

and x y a

then 1 1x a (a b)2 3

then

1 4p 5q

B) 16 ( 2)2a 3b r 4c 3b 2c r 4a 2c 4a r 3b 48[a b c] r [a b c] r5

implies 5, 3

C) Given a c, b c c p(a b)

. We have c.c 36(a b) (b c) b bp p

Hence 6m14p

We have 2 224 3c p(a b) 36 p 9.25. p9.25 2 14

this gives m 4

D) pq=6, p+q=r+1=5 then the value of 3p 2q is 0 or 5

18 10 15 Sr.iit Iz Co Spark Jee Adv2011 p2 Rpta 9 Key Sols

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