KEY SHEET (KEY & SOLUTIONS)
Transcript of 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
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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
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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
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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
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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
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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
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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
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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)
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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)
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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)
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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
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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
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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
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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
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1 sin 44
Range 1 1,4 4
(D) 1 1
1 1 01 1 1
1