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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

Narayana IIT Academy 03-12-17_Sr.IIT_IZ_JEE-MAIN_RPTM-14_Key&Sol’s

Sec: Sr.IIT_IZ

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


Sec: Sr.IIT_IZ

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


Sec: Sr.IIT_IZ

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


Sec: Sr.IIT_IZ

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


Sec: Sr.IIT_IZ

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


Sec: Sr.IIT_IZ

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


Sec: Sr.IIT_IZ

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


Sec: Sr.IIT_IZ

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-


Sec: Sr.IIT_IZ

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


Sec: Sr.IIT_IZ

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


Sec: Sr.IIT_IZ

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.


Sec: Sr.IIT_IZ

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


Sec: Sr.IIT_IZ

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


Sec: Sr.IIT_IZ

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)


Sec: Sr.IIT_IZ

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


Sec: Sr.IIT_IZ

=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.

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