KEY SHEET (KEY & SOLUTIONS)

Paper – II JEE-Advanced FST - 7

KEY SHEET (KEY & SOLUTIONS)

CHEMISTRY PHYSICS MATHEMATICS QNO KEY QNO KEY QNO KEY

1 B 21 D 41 D

2 A 22 B 42 D

3 A 23 D 43 A

4 C 24 A 44 B

5 A 25 C 45 C

6 B 26 A 46 A

7 C 27 A 47 A

8 C 28 C 48 A

9 ACD 29 BC 49 A

10 AB 30 CD 50 ABC

11 AC 31 CD 51 ACD

12 ABC 32 ACD 52 BC

13 1 33 5 53 8

14 5 34 8 54 3

15 1 35 2 55 5

16 4 36 4 56 1

17 8 37 3 57 9

18 7 38 3 58 7

19 A – PS; B –P; C – QR; D - RST

39 A – P ; B – RS ; C – QT ; D - PR

59 A – R ; B – S ; C – P ; D - Q

20 A – Q;

B – T; C – R; D – T

40 A –Q;B–Q;

C – R ; D -S

60 A – P,Q,R ; B – P ; C – Q ; D – Q,S

Page 2

HINTS & SOLUTIONS

CHEMISTRY

1. Arylamines form stable diazonium salts which couple with activated benzene rings to give azo dyes.

2. Given is the structure of -D-fructopyranose, thus it is a ketohexose.

3. Let x be the row where both the gases meet from the side of X.

Then 176 200

300 44x x

x

4. 2A t+ On differentiating this equation

1dA dA2Adt dt 2

A

5. Abstraction of most acidic alpha hydrogen followed by nucleophilic substitution.

6. 3 2 22FeCl 6KI 2FeI I + 6KCl

11. (A) It reacts with 2 moles of MeMgBr to give 1 mol of t-butyl alcohol and another mol of ethanol

(gives iodoform). (B) It give two moles of the same product propan-2-ol(both give idoform). (C) It gives one mole of proan-2-ol (gives iodoform) and one mole of proan-1-ol. (D) It gives one mole of 2-methylbutan-2-ol and 1 mol of methanol (both don’t give iodofom) 12. 2

22Cu 4I I 2CuI

2CuCl Cu 2CuCl

In fact, in the above reaction Cu[CuCl2] is formed fist which on dilution gives precipitate of

CuCl.

222Cu RCHO RCOO Cu O (not balanced)

2 2Cu F CuF

Page 3

13.

p = 20, q = 12, r = 30

14. 2 2

3 5 2 2 3 5Pb [Co(NH ) I]I PbI +[Co(NH ) I]0.1 2 0.025 20 2 5V V

15. AgCl + 1e– Ag + Cl– E° = 0.2 V

Ag Ag+ + 1e– E° = – 0.79 V –––––––––––––––––––––––––––––––––––––––– AgCl Ag+ + Cl– E° = – 0.59 V

E° = 0.059

n log K – 0.59 =

0.0591

log KSP

Ksp = 10–10 Now solubility of AgCl in 0.1 M AgNO3

S (S + 0.1) = 10–10 S = 10–9 mol/L Hence 1 mole dissolves in 109 L solution hence in 106 L amount that dissolves in 1 m mol.

16. Cu2+, Cd2+, Hg2+, Bi3+

17. Analysis of Ksp values of both the compounds reveal that TlBr is more is about 105 times more soluble than Hg2Br2 making [Br-] in the solution 10-3 M.

22 2 2Hg Br Hg 2Br �

x 310 2 2 24 2 3 2 24 2 2 18

2 2 2[Hg ][Br ] 4 10 [Hg ][10 ] 4 10 [Hg ] 2 10

Page 4

PHYSICS

21. 2 2 221/ 2extreme

y c hE U T lx l

ANS: D

22. S is very sensitive to T. It becomes zero at critical point.

ANS: B

23. 2 2He airm R O m R , air Hm m

ANS: D

24. B finally comes to rest only when 0,1e when 1,e B will come back to original position but when 1,e it

will stop at a place right of original place.

ANS: A

25. 22 ; / min

24 60V d where w rad

3 / , 10.8cm s d m

ANS: C

26. GCD of 2 kHz and 1.2 kHz 400 Hz .

2000 Hz Corresponds to second over tone or 122

loops

5 8 202

l cm cm

ANS: A

Page 5

27.

Let 1 2 3,x x and x be the deformations of the springs.

1 32 2

x xx i

0 1 3 1 1 3 3 1 30 3F F or k x k x or x x ii

ANS: A

28. 2 , , 0 2BV r I a BI r mgR

ANS: C

29. If 1 20,T a a a say

1 2 11 2

2

202

T k x mg ma k k mgT mg x or k kmg k x T ma x

.

If the string becomes slack, 1 2a a & 1 22mgk k

x

ANS: B,C

30. 1 2120 , 5 & 0.3 150 22.52

P N s P N s I N s

1 11 22.5 , 17.5P N s P N s . Both reverse the direction of motion

1/ 2 0.3 150

1/ 2 0.21/ 2 150

reforms

a deform

I tVe t sV I t

ANS: C,D

Page 6

31. 0 0 0 002 3 4

1

10 10 0 0 4BB

V V V V V or V Vc c c c

.

One more condition is required to solve for 0 & BV V

ANS: C,D

32.

2

2

2

cm

cm

mg T ma i

a rd ii

mrTr iii

ANS: A,C,D

33.

1v 5v

Ans: 5

34. 1 25 5/ /4 1

k N cm k N cm

1 25 0.8

5 / 4 5 /NF k k x x cm

N cm

ANS: 8

35. 2

0 , 22 e

mV mg V gRR

ANS: 2

Page 7

36. 2, tan2 2

vp dV p dV R dV V TC C or Cono tn dT n dT dT T V

ANS: 4

37. 20 1&

2QE

c LC

2 4 & 3f i f i iE E E E E E

3WE

ANS: 3

38. 2 2dV dD d lV D lv D l

1 , 0.5 , 10 , 50dD mn dl mm D mm l mm

ANS: 3

39. Let sinS a t kx

cossp B Bak t kxx

cossV a t kxt

&P V are in phase.

ANS: A-P, B-R,S C-Q,T D-P,R

40. 0, 0b a a b a bP P V V V V , pressure in the tube must be non-zero every where

0 1 2 1, 2 , 0bP h h g V gh h

ANS: A-Q, B-Q, C-R, D-S

Page 8

MATHS 41. 2a b c , . . . 2a b b c c a

2

a b c =

a. a a. b a.c

b.a b.b b.c

c.a c.b c.c

= 4 2 22 4 2 322 2 4

volume = 1 2 2a b c6 3 Ans (D)

42. 2 2 1 2 2 1dyy x x y xdx

23 2 1AI BI CI x y

3 22

Ax Bx Cx dx x y

y

3 22 2Ax Bx Cx dyx xy

y dx

3 2 2 22 12 1

2

x

Ax Bx Cx x x x x

2

3 53 22xx x

53, , 22

A B C

52

A B C

Ans (D)

Page 9

43 22 2

2 0 0

( ) (1 1 ) 2 62

x

tf x t dt tLt

2

3

x

f xLt

Hence discontinuous at 2x

Ans (A)

44. Let r and x be the radius of K and M

Let P be the centre of K

, PC r x 22

rxPD , CD x

2 2 2 4 PC PD CD r x

Ratio of areas = 16

Ans (B)

45. I = 1

3 23

0

2 3 3 1 x x x dx

= 1

3 33

0

(1 ) x x dx

= 1

3 33

0

(1 ) x x dx

= - I

0I

Ans (c)

Page 10

46.

C E

A C' B

d

C ' is the reflection of C about AE

Area ACE = Area 'AC E and 'AC AC

Sum of the area = ( )2d AB AC

752

Ans (A)

47.Let y mx c be the line through the points then the abscise of those point will be the roots of the equation 4 3 23 20 12 96 100 x x x x mx c

Sum of the roots = 203

AM = 53

Ans (A)

48. Angle subtended by 1z and 1z at 0 is 2n

, tan 2 1 , 8

, 8

n, 8 n

O

1z

1z

Ans (A)

Page 11

49. 2

0

cos cos

nnI x n x dx

2

11

0

cos cos( 1)

n

nI x n x dx

= 2

1

0

cos (cos cos sin sin )

n x nx x nx x dx

= 2

1

0

(cos sin )sin

nnI x x nx dx

= 22

00

cos cossin cos

n n

nx xI nx n nx dx

n n

= 2 nI

Ans(A,C)

50. Conceptual

51. (A,C,D)

52. Conceptual

53. Let ( )kp be the prob. That the bag has exactly k white badly then

2( )p 2 3

2 12 3 3 2 4 1

2 1 2 1 2 1

15

C CC C C C C C

3 42 2( ) ( )3 5

p p

Reqd prob = 1 2 1 2.(0) .15 5 2 5

= 35

Ans = 8

Page 12

54. 2 3 80 1 2 3 8......f x A A x A x A x A x

2 2 3 2 80 1 2 3 8......f wx A A wx A w x A x A w x

2 2 30 1 2 3 .....f w x A A w x A wx A x

2 3 60 3 63f x f wx f w x A A x A x

3n Ans = 3

55. Draw the diagram

56.

1 2

3 20

tan sin ln 10

1

x

x

x ae b x c xLt a

x x

=

1 12

20

1 cos 13

x

x b x c xLt

x

=

22 2 2

20

1 2 1 2 2 1 12

3x

xx a x x b c x xLt

x

1 2 0a b c 4 0a c 1b

1a b

Ans. 1

57. 1. 2. 1 3. 2 ...... .1nS n n n n

1. ncoeff of x is

2 1 2 11 2 3 ....... 1 2 3 .......n nx x nx x x nx

= 1ncoeff of x is 221 2 3 .......x x

Coeff of 41 1nx is x

=

31

1 24 2

6n C

n n nH n

25 25 26 27 9325 6 325S

Ans. 9

Page 13

58. 2 0f

21 3

1x ydy

dx x

let 1 , 3x X y Y

dY Y dY YX XdX X dX X

2 1

dYX YdXX

1d Y

dX X

Y X cX

23 1 1y x c x

Substituting (2, 0)

-3 = 9 + 3c

C = -4

Y = 21 3 3 2x x x x Area = 2

2

0

423

x x dx

Ans. 7

59. A – r ; B – s ; C – p ; D - q

(A) Adding ln 4 8 ln 2x x

Subtrating 6ln 6 ln 1y y ln 3xy

B)

From graph we get 4 points

C) 2 3 2

20

2 6 9 52 5

x x x dxx x

2

20

5 12 1

1 4x

x dxx

Putting x – 1 = t

Page 14

1

21

52 04

tt dtt

D)

22 2214 4 42

xxx x

2x x is min at 12

x

min1 32 2

f f

Ans A r, B s, C p, D q

60. A – p,q,r ; B – p ; C – q ; D – q,s

(A) 0

sinx

x tf x e t dt

= 0

sinx

x te e t dt

0

' sin sinx

x tf x e e t dt x 0

'' sin sin cosx

x tf x e e t dt x x

'' sin cosf x f x x x Range = 2, 2

B) 1tanx t 2

11

dxdt t

21dy dytdx dt

The given equation is

2 1 2 2d dx dy1 t y tan t 1 t 1 t 0dx dt dt

2 2 1 2 2d dx dy1 t 1 t y tan t 0dt dt dt

1 t 1 t

2

2 12

d y dy1 t 2t y tan t 1dt dt

(c) Put a cos ,b sin

3 3 2 2b ab sin cos cos sina

1 sin 2 cos 22

Page 15

1 sin 44

Range 1 1,4 4

(D) 1 1

1 1 01 1 1

1