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Narayana IIT Academy INDIA
Sec: Sr. IIT_IZ JEE-MAIN Date: 03-12-17 Time: 07:30 AM to 10:30 AM RPTM-14 Max.Marks: 360
KEY SHEET
MATHS
1 3 2 1 3 2 4 2 5 1 6 3
7 1 8 4 9 2 10 2 11 2 12 2
13 2 14 3 15 3 16 3 17 2 18 2
19 1 20 2 21 3 22 1 23 3 24 4
25 2 26 1 27 3 28 2 29 1 30 2
CHEMISTRY
31 2 32 3 33 2 34 2 35 2 36 2
37 4 38 3 39 2 40 1 41 2 42 1
43 2 44 4 45 2 46 4 47 1 48 1
49 3 50 3 51 2 52 3 53 3 54 4
55 1 56 1 57 1 58 4 59 4 60 3
PHYSICS
61 3 62 4 63 1 64 1 65 1 66 2
67 4 68 3 69 2 70 4 71 2 72 3
73 1 74 3 75 4 76 2 77 3 78 2
79 1 80 3 81 3 82 3 83 4 84 3
85 3 86 3 87 2 88 2 89 2 90 2
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SOLUTIONS MATHS
1. Let BD n,CE n 1, AF n 2
BD BF n
CE CD n 1
AF AE n 2
a BD DC 2n 1
b 2n 3,c 2n 2
Largest side of the triangle = 2n + 3 = 15
02.
12 2
1 1 1 1Tr sin 1 1r r 1 rr 1
1 1r
1 1T sin sinr r 1
1 1Sn sin2 n 1
03. Take B as origin, BC as x-axis and take A as (h, k)
c 4,0
Area of 1ABC 4 k 2k2
D = (2, 0)
h 4 KE ,2 2
AD BE slope of AD slope of BE 1
ABC 11
n 2
n 1
r
A
EF
B CD x
A h, k
B 0,0 C 4,0
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04. 21 t 1 1 1 t 1
So we have a 0 a / 22
05. 10 10
1
n 1 m 1
MS tann
then 10 10
1
n 1 m 1
nS tanM
10 10
1 1
n 1 m 1
M n2S tan tann M
10 10
1
n 1 m 12S tan
2
2S 100 S 252
06. 1 1 2 23cos x sin 1 x 4x 1
Put 1 1 2x cos ,3cos cos sin sin 4cos 1
1 2sin 3sin 4sin
1sin sin 3
Equality holds when 0 and 32 2
30 cos 16 2
3x ,12
07. 2 2n 1 n n 1
1 12n 1 n n 1
n 1 1
2 2 2tan tan1 2 1 2 .2 4
08. I be the in centre, S be the circumcentre from given data Is = r
2 rR 2Rr r 2 1R
cos A cos B cos C 2
09. 0 0B C ABIC 180 902 2 2
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0BSC 360 2A
Hence, 0 0A90 360 2A2
0A 108
10. 2sin A sin B cos A cos B 2cos A cos B cos A cos B
cos A cos B NOT POSSIBLE
OR
tan A tan B
A B LC2 2
11. Conceptual
12. r 4S
1 2 3
1 2 3
r r r 4r r r
13. abcb c
sin B sin C sin C sin Asin B
sin 2C sin C sin C sin 3Csin 2C
0 0C 36 A 72
A
B C
S
P Q
R
1r 2r
3r
A
BC
D
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14. We have ck + bk = a � k = ab c
Also x y = b c k2
2
2
2 cos2Abc bc ax
b c b c
Hence x = 2
2sec2 ( )
Aab c
= DE ]
15. 2 2cos A 5/ 8,sin A 3/ 8
3 3sin A , bsin A 3 28 8
There is no triangle
16. BD : DC c b
acBDa b
AB BDsin ADB sin BAD
c acsin b c sin A / 2
b c B Csin sin A / 2 cosa 2
17. Graph of 1y sin sin x is
B
A
CD
A2
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3 4 5 2 2
18. xi sin 2
i J
xi xj cos 2
i J K
xi xj xk cos
1 2 3 4x x x x sin
4
1 1
i 1 1 2 3 4
xi xi xjxktan xi tan1 xi xj x x x x
1 sin 2 costan1 cos 2 sin 2
19 1 1 02 3 4 2 3sin sin 154 8
1 0 0 0sin cot 15 30 45 0
20. 23
1 2
n 1
cot n n 1
231
n 1
n 1 ncot tan1 n n 1
1 23 25cot tan25 23
21. Graph
No. of solutions = 3
22. In triangle ABD
BD = DC
y x
2 3 4
y x 2
y x 4
y 3 x
1y cos cos xxy 1
10
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0
AD BDsin B sin 30
1 ADBD2 sin B
(1)
In triangle ADC
0 0
CD ADsin 45 sin 105 B
0
ADCD2 sin 105 B
(2)
From 1 and 2
0
AD AD2sin B 2 sin 105 B
02 sin B sin 105 B
1sin B2 11 6 3
BC 2
23. a b c 2s
Then we have to find minimum value of
a b c s s s3s a s c s c s a s b s c
AM H.M
s s s3s a s b s c
s a s b s c3s s s
s s s 9s a s b s c
24. 2 R r
2 r 3 2 11 22
030 045
A
D C|| ||B
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25.
26. F 1 2
F 2 F 1 F 1 4
F 3 F 2 F 1 8
a F 3 8
b F 1 F 3 2 8 10
c F 2 F 3 4 8 12
2 2 2b c acos A 3/ 4,cos 2A 1/8 cosC
2bc
27. 1 1ABC APB APC2 2
1 1 A 1 Abcsin A cpsin pbsin2 2 2 2 2
1 A 1 1 1cosP 2 2 b c
1 1 1a b c
28.
cos A cos B9991sin A sin B bcsin Acos C 2
sin c
O
B
C
x
x
21OAB x sin2
A
B C
R
P
Q
A2
A2
a
bc
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bccosA accos B9992 2abcosC
2
2
2
1998c 11998c
29. A Bcot cot 1C 22 2tan A B2 5cot cot2 2
s b s c s a s bA Ctan tan2 2 s s a s s c
5 2 s b6 5 s
2b a c
30. A BC cot cot2 2
s s a s s bC
s 1 cCs c s c s
CHEMISTRY 31. 0 t , NH3 is Stronger ligand than F-
splitting energy order II III I
32. Conceptual
33. Conceptual
34. Cl is weak ligand so no pairing of electrons take palce. so 42NiCl is tetrahedral.
3pph group is bulkier one so it favours tetrahedral geometry, through 3pph is strong field
ligand.
36. no. of moles of AgCl formula 28.7 0.2143.5
0.1 mole complex then 0.2 moles of tree Cl
37. (I) Fe+2 3d6 4s0 0 unpaired e-
II) Fe+2 3d5 4s0 1 unpaired e-
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III) 3Cv 3d3 4s0 3 unpaired e-
IV) 2Ni 3d8 2 unpaired e-
CN Strength of ligands
38. A) 3-
2 4 3Co C O EAN=36
B) 4( ) EAN=36Ni CO
C) 4
2 6EAN=32Ti H O
D) 4
6EAN=36Fe CN
45. More the electron density in central atom more will be the tendency for back bonding
then less will be the C – O bond energy
48. In the presence of Cl ligand cobalt forms tetrahedral 2 _4CoCl complex. Tetrahedral
complexes have dark colours compared to octahedral.
59. Order of strength of ligands is en>ox>NH3>F
PHYSICS 61. 3
Let be angle of projection, then
2
2 2gxh x tan
2u cos
For maximum ' x ' ,
dx 0d
2 2
2u gxh x 2 k k h u 2gk2g 2u
62. 4
PG PGx PGya g u ucos , u usin v PB PBx PBya g u ucos , u usin
So, wrt both frames projectile will follow parabolic path but parabolas would be
different
63. 1 2usint
gcos
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A C
B
and 2
2
2u sin cosAB
gcos
= ut3
64. 1
Putting BC = v = nv
1rmv 2rmv
1mv 2mv
D
C A
v (n+)v
B
we obtain
1
( 1) cot rmv n (b)
Using (a) & (b) we find
2 2(( 1) cot )rv v n v
2 2rv v 1 η-1 cot θ
65. 1
0
3 03 2
0 02 1
v t
v
vdv dvkv k dt vdH v v kt
66. 2
From the graph 2
25 / 6 5 / 6da d d sdt dt dt
Distance travelled from 0 to 6s = 31
5 6 3036
s m
Velocity at 256 15 /12
t s v t m s
From 6 to 12 s, 215 / 5 /u m s a m s
22
1 1802
s ut at m
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Total distance travelled = 30 180 210 m
67. 4
Acceleration can be written as 2 2 4 2a t or a t for t s
2 2t s or a t for t s
dv adt
12 /m s at the end of 4 's .
68. 3 8 2
100 25v v xx .
Acceleration 4625
a x
2100 0.64 /x m a m s
a x graph is a straight line with
maximum value = 0.64 2/m s
69. 2,1 2 1u u u v
1 2 2&P pv v u v v u
2 1 1 2 2 1v v u u v or v v v
70. Tension in the strings is zero.
Blocks fall freely
71. As the wedge moves towards right with an acceleration a, the block moves us the
wedge with an acceleration a, w.r.t the wedge
WGa
bWa
180
2 sin 2bg bw wga a a a
72. N=mg-ma
When a=g , N=o, at t= /g if a>g for /t g , N becomes –ve signifies downward
N.
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73. 1
Suppose M moves a distance x to the left w.r.t. wedge, then displacement of wedge
Mx8M
to the right.
Simultaneously, due to 2M, displacement of wedge
2M cos60º8M
to the left
Net displacement = 0.
74. (3)
40m/s
74º 37º E
N
|V – V |2 1
53º
40m/s
2 1V V 40sin 74 sin(90 37 )
2 140 2sin 37 cos37V V
cos 37
= 48 m/s
change in momentum = 13 48 = 16 kpm/s
avg16f
0.02 = 800 N.
75. (4)
a cosA
aA
aB90–
a sinB
52
21
21aA = 2aB.
76. 2
maxa g
2atgm
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2 mgta
77. 3
0.6 2 10 12f N
2 0.3 4 10 12f N
Net friction =24 N
,F f the blocks does is move
12 16 4T T N
2 4f T N
78. 2
2 80T F
80 70` 0.5T
45N
1 20T F
1 25 F N
1 0.5X m
79. (1)
Sol. Net retardation ext kf mga 4.5m
If body stop at time t, then
V u at
0 45 4.5t t 10sec
When block stops, Fext will try to bring the block back ward while frictional force will
oppose its motion, since block is stationary therefore at this moment, frictional force
will be static. Whose maximum value will be smg 30N . Since static frictional force is
self adjusting therefore it will be 25 N and block will not move after t = 10 sec.
S= 250 m
80. 3
2
2 2M
2
M g M g / 4a
M
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25g 1 5g 2.5m / sec4 2 4
1
kM
1
F faM
2
1 5g30 2 255 4 m / sec8 8
81. 3
1 22 sin cos cosmg mg
1 22 tan
1 2maxtan
2
82.
w t
2f m L
2ma m L
2 4t
t
83. 3
2 22 3 1 2.5 3
2 3T w r x N
84. 4
A A cmcma a , a
22 0
AVa r
2 20 02
r V V4r 4r
= 0
85. (3)
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Say at any instant, the velocity makes an angle with the x axis
ˆ ˆu u cos i sin j
dv d dˆ ˆa u sin i cos jdt dt dt
Now, dytan cos xdx
2 d dxsec sin xdt dt
2d dxcos sin xdt dt
Now, at x = /2, = 0o, dx udt
d udt
2a u
86. (3)
2mv mg
R
oo
RgRv Rg 1 Rg2R 2
87. 3
Component of a
along v
is the tangential acceleration
a.v .vv
i j . 4i 3j 4i 3j.4i 3j 4i 3j
27 4i 3j m / s
25
88. 2
1 1 18 3000 16 6002 60 2 2
N
200+80 =280
89. 2
Ang.velo about 0=08 rad/s
20.5 0.8acc
20.32 /m s
Speed = 0.5 0.8
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=0.4 m/s
90. Because one taxi leaves every 10min, at any instant there will be 11 taxies on the way
towards each station, one will be arriving and another leaving the other station. Figure
shows the location of taxies going from X to Y at the instant 2.00PM. The taxi which
leaves the station X at 0.00PM has just arrived at the station Y. Consider the taxi
leaving the station Y at 2.00PM.
It will meet all the 11 taxies marked 1 to 11 as well as 12 other taxies which would
leave the station X from 2.00PM to 3.50PM. When it arrives at the station X at 4.00PM,
there will be one more taxi leaving that station. However, it will not be counted among
the stations crossed by taxi under consideration. That is, it will cross 23 taxies leaving
the station X from 0.10PM to 3.50PM.