jee-adv grand test solutions (p 1)
TRANSCRIPT
IIT Section
Subject Topic Grand Test – Paper I Date
C + M + P Grand Test – 08 IIT – GT – 08
12th May 2014 I20140508
1
Key Answers:
1. a 2. d 3. c 4. b 5. a 6. b 7. c 8. c 9. ab 10. ac
11. ac 12. a 13. d 14. a 15. b 16. b 17. b 18. a 19. 3 20. 8
21. 3 22. 8 23. 4 24. 9 25. 8 26. 6 27. 3 28. 5 29. c 30. a
31. b 32. a 33. b 34. a 35. a 36. d 37. abc 38. ad 39. abc 40. bc
41. ad 42. a 43. c 44. b 45. b 46. c 47. 2 48. 6 49. 7 50. 1
51. 4 52. 2 53. 2 54. 1 55. 4 56. 3 57. b 58. a 59. b 60. d
61. c 62. a 63. b 64. a 65. acd 66. a 67. ad 68. acd 69. ac 70. b
71. a 72. c 73. a 74. d 75. 6 76. 2 77. 7 78. 2 79. 2 80. 5
81. 6 82. 3 83. 6 84. 1
Solutions:
Chemistry
1.
2. Tertiary carbocation is formed first, which undergoes D 1,6- migration to give, more resonance
stablised carbocation, and it undergoes deprotonation in the last step.
3. 2 2 3 2 26 4 2 3NaOH S Na S O Na S H O
4. 2
X Na Ag CN Y = Ag
IIT Section
Subject Topic Grand Test – Paper I Date
C + M + P Grand Test – 08 IIT – GT – 08
12th May 2014 I20140508
2
5. log.
aE
3
1 15
303 3332 303 2 10
31log5 303 333 2.303 2 10
10.82 333 303
aE k cal mol
6.
2
wh
b
HKK
K C
214
4 3
10
4 10 3.3 10h
HK
6.5pH
7. thE H of graph (c) > thE H of graph (d)
8. .
. l
30 6 273
22 4373
9. In the first reaction, tertiary carbocation intermediate in formed and the product in a tertiary
bromide. In the second reaction 1NS hydrolysis takes place to give same carbocation
intermediate and the product is a monohydric alcohol.
10. Neighboring group participation of -bond pair of electrons of benzene ring.
11. 3
Bi OH ppt is formed, but 2
3 4Zn NH
soluble complex is formed.
Ferrous hydroxide green ppt, Ferric hydroxide brown ppt are formed.
12. O.N of Nickel in the complex is zero. Two electrons of 4s orbital are shifted to 3d - orbital so
that 3sp hybridistion occurs.
13. Hint No.of disintegrations per sec
Total no. of atoms in one gram of Ra
0.693
1600 365 24 60 60
or No. of disintegrations per sec 230.693 6.023 10
1600 365 24 60 60 226
14. Double inversion is taking place to give retention in the configuration, because OH group
present in C atom
IIT Section
Subject Topic Grand Test – Paper I Date
C + M + P Grand Test – 08 IIT – GT – 08
12th May 2014 I20140508
3
15.
16.
17.
nnC H COOH C H COOH
n
6 5 6 5
1
ni
11 1
1
n
n
1
11
18. In (a) .m 0 03
bT m
19. 2 3
0 0 0
/ /cell
Fe Fe Ag AgE E E
0.771 0.799 0.028volt
At equilibrium, 0cellE
3
0
2
0.05910 log
1cell
FeE
Fe Ag
H5C2 S
CH2
CH CH3
Cl
H5C2 S CH CH3
CH2 OH-
CH2 CH SC2H5
CH3
OH
(H5C2)2 N CH C2H5
CH2 OH-
CH2 CH N(C2H5)2
C2H5
OH
IIT Section
Subject Topic Grand Test – Paper I Date
C + M + P Grand Test – 08 IIT – GT – 08
12th May 2014 I20140508
4
0 10.0591logcellE
Ag
0.34Ag
0
log0.0591
3.0
nEK
K
20. Let the molecular formula be 22Ba OH xH O
Mol. Mass of 22. 137.4 2 16 2 1 18Ba OH xH O x
171.4 18x
Equilibrium mass of 22
171.4 18.
2
xBa OH xH O
.
. x
30 789 2 20 10
171 4 18 4
x 8
Thus, 8g moles of water molecules are present in one g mole of the base
21. .
.x
x
2 5 3600 22 2
2 97 396500 177
22. pH of salt after hydrolysis may be calculated as,
1
log2
w apH pK pK C
4114 log10 log0.01
2
114 4 2
2
8
23. 3 2
x x
RT RT
m m
3 400 2 60
40
4 3030
4
R R
m
m
m
24. log P
400
10 9373
xP x 910 10 9
IIT Section
Subject Topic Grand Test – Paper I Date
C + M + P Grand Test – 08 IIT – GT – 08
12th May 2014 I20140508
5
25. SO Cl
PSO Cl
p pK
p
2 2
2 2
SOp
2
10 15
2
Total pressure SO Cl SO Clp p p 2 2 2 2
5 2 1 8
26. Hint: 6 5 2 6 5 2 26 2
Nitrobenzene Aniline
C H NO H C H NH H O
So, 6n
27. Hint: 1000
;Af
A B
wT K A solute B Solvent
m w
1200 10006 1.86
62 Bw
6000Bw g
weight of iceI total weight of 2H O - weight of 2H O at 6 C
9000 6000 3000g
3 kg ice
28. Hint: Original solution 1500
283..............( )760
wV S i
m
After dilution 2105.3
298......................( )760
wV S ii
m
or 2
1
105.3 298
500 283
V
V
or 2
1
298 5005
283 105.3
V
V
i.e., the solution is diluted to the extent that 2V is five times the original volume 1V
IIT Section
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12th May 2014 I20140508
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Mathematics
29. Let ,z x iy x y
,z x y
,iz y x
,z iz x y y x
21 2 1 2 2 2
1 3 1 3
1 1 33
2 2 2
x x y yx y z
x x y y
30. Solving two equations
2 21 2 3 1 1 0x x x x
Only one root 0
2 4 1 1 0
2
2 0 2
31. Total number of triangles 3
10C
number of triangles with one side common 1
10 6C
number of triangles with two sides common 10
3 1required number of traingles 10 10 6 10C C
32. Continuous at 0
10 lim sin 0 0n
xx n d
x
not differentiable at 10 0x f does not exist
1
0 0 0
1sin
0 1lim lim lim sin
0
n
n
x x x
xf x f x
xx x x
does not exist 1 0 1 0, 1n n n
33.
2 2 4, 1, 3x y P
TN TA AN
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12th May 2014 I20140508
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1
3m
1
1
yy m
m
21 1
3 32 3
1
2PTN TN AP
2 3 11 1
1
2
yy y m
m
21 1
2
ym
m
34.
1P be image of P in
1 , ,P A Q are collinear 1 1
13
5AP QPm m a
35. 2 1 ________ 1x t t by eliminating t
2 1 ________ 2y t t
2
x yx t x t td
2
21 1 2 2
4 2
x y x yx x y x y
. . 2L L R
36. cos2 sin 2 7x a x a
21 2sin sin 2 7x a x a
2 4 2 2 3sin
2 2
a a ax
4sin or 2
2
ax
1 sin 1x
41 1
2
a
2 6a
axisx
IIT Section
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12th May 2014 I20140508
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37. 1 1 1 1 2 1 1
b a c b b a c
2
2 1 1 2 1. ________ 1G E a
c b b bc b
3 2 1.G E
b a b
2
3 2___________ 2 c
abb
1 1 1 1 1 1 1 1 1 1.
2 2G E
a c c a c a c a
2 2
1 3 2 1________ 3
4b
acc a
38. 2 1
3 3 5 , 0 1n
Let 2 1
3 3 5 , 0 1n
Consider
22 112 3 3 5 .........
nn C
10 K
also 1 1 and an integer is an interger
0
10K is even
2 12 2 12 26 15 3nn
not divisible by 2 12 n
39. 1 2 3
1
2P E P E P E
1 2 2 3 3 1
1
4P E E P E E P E E
1 2 3
1
4P E E E
40.
AC a b
A B
CD
a
b
IIT Section
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12th May 2014 I20140508
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2 2 2AC a b a b
2 216 4 16 112
4 7AC
also BC a b
4 3BC
41. 0 1, 1 2x x
2 3 2 3x x x x
2 3
3 42 2x x I I
1 2I I
42.
50 2
100 !100
50 !C
7 2
100 100100 ! 16
7 7E
7 2
50 5050 ! 8
7 7E
Exponent of 50
7 in100 0C
43. 2 108 ! 104E
5 108 ! 25E
member of zero’s = exponent of 10 = min 104, 25 25
44. 18 8 4 2 1 1 1 120 ! 2 3 5 7 11 13 17 19
4 14 8 2 1 1 1 110 2 3 7 11 13 17 19
It ends with 4 as non-zero digit
Exponent of 12 = 48
The D.E. for 31 21 2 3
m xm x m xy c e c e c e is
3 2
1 2 3 1 2 2 3 3 1 1 2 33 2
0 7 6
0
b c d
d y d y dym m m m m m m m m m m m y
dxdx dx
1a
45. order=3, degree=1
46. 1 0 6 7a b d
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47. 2 2 2 22 21 2 1 2 1 2az bz bz az a b z z
2 2 2 215 13a b
225 169
394
1 1 1
2 4 8
= 394 197k , 2k
48. 2 1000
1000. 1 1 .......
1 1 1
x x xG E x
x x x
1000
10001
11
11
x
xx
x
x
100110011001
1001
11
1
x xx
x
1001 10011 x x
coefficient of
50
50 1001 !1001
50! 951 !Cx
1001, 50, 951V
R.V 1001 100 28543 3954 659 6k k
49. 2 5 7A A I
3 2 180 35A A A A I
5 3 2 18 35 5 7A A A A I A I
149 385A I aA bI
149, 385a b
2 3 1453, 207 7a b k k
50. 1nf x x
10 1001f
31 10 1nx
3n
IIT Section
Subject Topic Grand Test – Paper I Date
C + M + P Grand Test – 08 IIT – GT – 08
12th May 2014 I20140508
11
320 20 1 8000 1f
8000 1k k
51. L.H.L = . .2
f R H L
3
2 2
2 2
1 sin1 sinPut 2
3cos 2x y
b xxLt a Lt x y
x x
2
20
2
1 cos1 sin 1 sin sin 2
3 1 sin 1 sin yx
ybx x x
Lt a Ltx x y
1
3 4
3 2 2a b
1 1
2 8 24
ba a
b
5/35/3 81 324
8 8 8
b
a
52. cos sinx a t t t
sin cosy a t t t
cos sin cos
sin cos sin
a t t t tdy
dx a t t t t
sintan
cos
dyat tt
dtat tdx
dt
2
2 2
2
1sec sec
cos
d y dtt tdx at tdx
3sec t
at
2 2
2 2
/3
8 24120 24 120
3t
d y d ya
adx dxa
2880 1440 , 2k k
53. 1 23 12 9 0f x x x
2 4 3 0x x
1, 3 but 3 0, 2x
0 1f
1 1 6 9 1 5f
2 8 24 18 1f 3
5, 1
4 8 625 1 626 313 2k k
IIT Section
Subject Topic Grand Test – Paper I Date
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12th May 2014 I20140508
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54. Let 2 2 2x y a be the circle
2 2 2 0x y a
Put 2 2
2
1 1, 0mi mi ami mi
1 2 3 4 1m m m m
55. Let 3 4, 5 6, 2 1P r r r be a point in the first line. This lies on both planes of second line
3 3 4 2 5 6 2 1 5 0r r r
2r 2,4, 3P in 2 3 4 0x y z k 4 12 12 0k
4k
56. 1 1 1sin sin sin 2 12
x y x y z
P.V 216
2500 33
7500 72 7428 2476 , 3k k
4 2 2 1 0mi a mi
1m
2m
3m
4m
IIT Section
Subject Topic Grand Test – Paper I Date
C + M + P Grand Test – 08 IIT – GT – 08
12th May 2014 I20140508
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Physics:
57.
2
2
1
2 4
2 2
lw
TPo h pg
R
2
2
2
3
2 4
2 2
lw
TPo h g p
R
Solving we get 2 2
2 1
4
4
l w R Th h
prg
58. When spring extended by x
20m L x kx
202
m Lx
K m
The new length L will be 0L L x
Tension in the spring 20T m L x
22
0 21
mm L
K m
202
m KLT
K m
59. Emf induced Blv
Q Cv
Q CBlv
60. I max 2
R
at 0t
IR
at t so charge on the capacitor is C , when current is 50% of maximum current.
61. Let the length of edge be ' 'x
Net magnitude of displacement from A to 3D x
Total time 1 2 3
x x x
v v v
Average velocity
1 2 3 1 2 3
3 3
1 1 1x x x
v v v v v v
IIT Section
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12th May 2014 I20140508
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62. When frequency is less 1
wlwc
As frequency increases, capacitive reactance decreases but inductive reactance increases. So phase
difference decreases. Then L becomes greater than 1
c. Now as frequency increases, the
magnitude of phase difference again increases.
63. 2T V
2TV C
or 1PV C
1
2 1
1
nR T TW
450W R
64. Let at any time the speed of the block along the induce up wards be V . from 2N L
sin cosp mdv
mg mgv dt
The speed is maximum when 0dv
dt
maxsin cos
pV
mg mg
65. Charge of 1l is 1 1 1q l rd
Charge of 2l 2is 2 2 2q l r d
1 1
2 2
q r
q r
1 1
1 2 2
11 1
kq k r d k dE
rr r
2 2
2 2 2
22 2
kq k r d k dE
rr r
So net electric field can’t be zero
1E is more, so direction is towards 2l
Potential due to 1l is 1 1
1
1 1
kq k r dv k d
r r
Potential due to 2l is 2 2
2
2 2
kq k r dv
r r
2v k d 1 2v v
IIT Section
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12th May 2014 I20140508
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66. EMF across 2
2
BwlPR
EMF across 2
2
BwlPQ
So emf across 0QR
67. At
1, maxz x Upper end open lower end closed.
0.99 , minz m x
left2 2
0.80.8
m
0.99 0.01 1m
5, 5 five half loops Second overtone
4 4
ll
68. Velocity is more in rarer medium
v f , where f is constant.
69. sinX A t
12
sin2
AA t
T
112
Tt
1 22
sin2
AA t t
T
1 28
Tt t
70. FBD of small section of ring subtending angle d at center of ring is as shown
,22 2
N N mdTd g
2
mgT
71. T
v
22
mggR
m
R
72. Since there is no friction between ring and cone the tension in the ring will remain same
73. 2 1
3
4v v
We know that
IIT Section
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12th May 2014 I20140508
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1
2
i r i
vA A A
v …(1)
2 1
2 1
r i
v vA A
v v
and 2
2 1
2i i
vA A
v v
…(2)
From equation (1) and (2), 1: 7r
i
A
A
74. 6 : 7i
i
A
A
75. Let x be the temperature of block in steady state10 5 3
0 6x x x
x CR R R
76. 1 0
s
v vn n
v v
77. 1 2 3 0Q Q Q
1 2 02
Q QIR
c c
3 2 2 0
2
Q QIR
c c
1 2 3
5 7, and
4 2 4
IRCQ IRC Q Q IRC
78. When S is closed 1t RC
When S is opened 1
2
2
3 2
2 3
tRCt
t
79. Y bright n D
d
Y Dark 1
2
Dn
d
80. 2
2 2
1 2
1 113.6E z
n n
2
2 2
1 147.2 13.6 5
2 3z z
81. 5 12 12 3 8C BV V
6C BV V V
82. 17h ev
IIT Section
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12th May 2014 I20140508
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Now for metal maxh KE
17 2ev ev
15ev
3K
83. In steady state let T be the temperature of all the three blocks
10 2 5 2 5 0mc T mc T mc T
84. Considering reflection at the curved surface
1 2
3; 1
2
20 ; 10u cm v xcm
20R cm
from 2 1 2 1
v u R
1 3 0.5
10 40 20x
1 1 3
10 40 40x
10 10x
1x