narayana iit academy · 12/11/2017 · narayana iit academy 11-12-17_sr.iit_iz_jee-adv...

Narayana IIT Academy INDIA

Sec: Sr. IIT_IZ Jee-Advanced Date: 11-12-17 Time: 09:00 AM to 12:00 Noon 2014_P1 MODEL Max.Marks:180

KEY SHEET

PHYSICS 1 BC 2 AB 3 C 4 ABCD 5 BC 6 AC 7 BD 8 ABCD 9 ABD 10 A 11 2 12 5 13 2 14 9 15 5 16 6 17 4 18 3 19 4 20 4

CHEMISTRY 21 AD 22 CD 23 ABC 24 BCD 25 AD

26 ABC 27 A 28 ABD 29 BC 30 BC

31 7 32 8 33 9 34 4 35 8

36 4 37 1 38 6 39 8 40 2

MATHS 41 BD 42 BCD 43 ABCD 44 ABC 45 ABC

46 BCD 47 AC 48 BD 49 ABCD 50 ABD

51 5 52 7 53 8 54 1 55 3

56 8 57 5 58 5 59 9 60 7

Narayana IIT Academy 11-12-17_Sr.IIT_IZ_JEE-ADV (2014_P1)_RPTA-15_Key & Sol’s

Sec: Sr. IIT_IZ

SOLUTIONS PHYSICS

1. Conceptual

2. Conceptual

3. From the principle of conservation of momentum, we have total final momentum =

total initial momentum.

Momentum conservation is possible in cases (C) and (D). IN case (C), the two masses

should move mutually perpendicular to particle A. In case (D), particle B must move

with velocity v in the original direction of motion of A. Hence, the correct choices are

(C) and (D).

4. A, B, D sin sin , cos cosv eu v u

u

v

2 2 2cos sinv u e

2 2 21 sin sinu e

2 21 1 sinu e

sin sin sin 1I m v u mu e

Ratio of 2

2 2 2

2

12 cos sin12

mvKE e

mu

5.

This is basically the energy of oscillation of the particle. K, U and e at mean position

and extreme position are shown in figure.

K= 100J = maximum

U = 60J = maximum K= 160J = maximum

E = 160J constant E = 160 J = constant


Sec: Sr. IIT_IZ

6.

7.

8. Let the deformation in each spring be respectively as shown

Let the block be displaced by x.

Then

Also

From equation (i), (ii) and (iii)


Sec: Sr. IIT_IZ

9. Conceptual

10. Conceptual

11. Conceptual

12. Conceptual

13. Conceptual

14. colliding with the wall its vertical component yv of velocity will remain unchanged

(component along common tangent direction remains unchanged) while horizontal

component xv is reversed remaining same in magnitude. Thus, path of the particle

will be as shown in figure.

(b)Total number of collisions with the walls before the ball comes back to the ground

are nine.

15. 020 2 cos 45 20 /xu m s

020 2 sin 45 20 /yu m s

After 1 s, horizontal component remains unchanged while vertical component

becomes

20 10 10 /y yv u gt m s

Due to explosion, one part comes to rest. Hence, from conservation of linear

momentum, vertical component of second part will become 1 20 /yv m s . Therefore,

maximum height attained by the second part will be

1 2H h h

Here,

1h height attained in 1 s

2120 1 10 1 152

m

And 2h height attained in 1 s

221 20

202 2 10

yvm

g


Sec: Sr. IIT_IZ

15 20 35H m

16. sin sin , cos cosv u v eu

u

3

4

5 v

x

y

line 4 3 12 0x y

2 2 2cos sinv u e

2 2

2 4 310 0.75 6 2 /5 5

m s

17. angle between initial velocities be and the situation is as shown in Fig. Conserve

the momentum along X-and Y-directions.

m

m v v

/ 2v

2m

X

Y

cos 2 cos ,2vmv mv m For X-axis

sin 3 sin ,2Vmv m For Y-axis

Solving above equations, 060

Required angle 0 060 120

18. [ constrained relation for pulley 1]

[ constrained relation for pulley 2]

From above two equations

[ Netwons II law for block A ] II

[ Netwons II law for block B] III


Sec: Sr. IIT_IZ

From equation I, II and III

19. = /I = 2 3 gL MLMg

2 3 2 L

acm = L/2 = 3g/4 = Mg NM

N = Mg4

20. Conceptual

CHEMISTRY 21. AD

Hint: conceptual

22. D,C

No.of m.e of 2 4H SO = no. of m.e of KI

23. A,B,C

Hint: Conceptual

24. : B,C,D

Hint: conceptual

25. (A D)

Hint: conceptual

26. A,B,C

Hint: Conceptual

27. A

Hint: use h2mK

28. ABD

Hint: Conceptual

29. B, C

Hint: conceptual

30. BC

Hint: conceptual


Sec: Sr. IIT_IZ

31. 7

HINT: CONCEPTUAL.

32. 8

HINT: FOR 10 ML OF SOL” X = 32 mg

33. 9

CONCEPTUAL

34. 4

CONCEPTUAL

35. 8

Hint: use n n 12

36. 4

Hint: 2 r n

37. 1

Hint: 3 2 23CN 10H 7NO 3CO 10NO 5H O 38. 6

39. (8)

n 2n 2C H be the alkane.

n 1n

2 7n 4

n 2

40. 2

Hint: Equate the milliequivalents or use balancing of reactions.

MATHS

41. . . .m m m

r mn r

r n r n r n

r m m n mC Cm r n r m r m r r n n m n

.2m m nnC

42. 11.n n

r rnC Cr

.

43. Conceptual

44. Req. numbers = 5 4 4 5 4 5 24 4 3 1 2 2 14 2 2P P C C C C C .


Sec: Sr. IIT_IZ

45. 1

1nr

nr

C n rrC

and 1

1.n nr r

nC Cr

.

46. Conceptual.

47. Number of distinct terms = 28

48. 7 2 11 , 7 2 11 , 0 1n n

I f F F

220 22 7 7 2 11 ....n n n nI f F C C

f F is an integer and 0,2 1f F f F and I = even integer-1

49. When z = n + 1 we can choose x, y from {1, 2, …….n}

when z = n + 1, x, y can be chosen in n2-ways and if z = n, x, y can be chosen in

(n-1)2 ways and so on.

22 2 11 ..... 1 1 2 16

n n n n n

Ways of choosing triplets

x = y < z, x < y < z, y < z < x can be chosen in 1 1 12 3 3, ,n n nC C C ways.

There are 1 1 2 1 2 12 3 2 3 3 22 2n n n n n nC C C C C C .

50. a) 5 3 5 3 51 23 .2 .1 150C C

b) 53

d) 5! 5!3 1502!2!1! 3!1!1!


Sec: Sr. IIT_IZ

51. x y z

73 73 70, 71, 72,73 Õ 32 4 2C =10

23 22 20, 21, 22 Õ 6-1 = 5

Total number of numbers = 10 x 5 = 50.

52. 17256 = (290-1)128 =1000k + 2128 128126 127290 (290) 1C C

1 2 35, 6, 1a a a .

53. 1 1n naf n

a

.

54. Conceptual

55. 5

20 182 1

12r

rC

.

56. N = 1440 = 25.32.5 no. of divisors of N = 36.

Product of divisors of N is 18 90 36 181440 2 .3 .5 30P k .

57. 2 5 5, 4 5a b .

58. Conceptual

59. 15 14 131 2 ...... 15 1 1.2 .....x x x x x x

Put 1x req. sum = 16 1 120 .

60. 22007 2007 2007 2007a b ab a b .

narayana iit academy · 12/11/2017 · narayana iit academy 11-12-17_sr.iit_iz_jee-adv...

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