Sec: Sr. IIT-IZ(L25) JEE MAIN Dt: 24-12-2018 Time : 3 ... · NARAYANA IIT ACADEMY 24-12-18_Sr.IIT-IZ(L25)_Jee-Mains_key&sol Don’t wait to reach your goal to be proud of yourself

NARAYANA IIT ACADEMY 24-12-18_Sr.IIT-IZ(L25)_Jee-Mains_key&sol

Don’t wait to reach your goal to be proud of yourself. Be proud of every step you take towardsreaching the goal. Page 1

Sec: Sr. IIT-IZ(L25) JEE MAIN Dt: 24-12-2018

Time : 3 Hours CTM Max Marks : 360

Key & Solutions

MATHS

1 2 3 4 5 6 7 8 9 10 B B D D C D D A B D 11 12 13 14 15 16 17 18 19 20 A B C B B B C C A A 21 22 23 24 25 26 27 28 29 30 B C C D A C C C C D

PHYSICS

31 32 33 34 35 36 37 38 39 40 A A C C C A A D D B 41 42 43 44 45 46 47 48 49 50 B B C C B A A B A B 51 52 53 54 55 56 57 58 59 60 A B A B A A C A D B

CHEMISTRY

61 62 63 64 65 66 67 68 69 70 C B B D D C C C D B 71 72 73 74 75 76 77 78 79 80 A C C B B A B B C D 81 82 83 84 85 86 87 88 89 90 D B D C B C A A A B



MATHS

1. Let x1 be the number of station before the first halting station, x2 between first and second, x3 between second and third, x4 between third and fourth and x5 on the right of the fourth station.

Then , 1 5 2 3 40, 0, , , 1x x x x x such that 1 2 3 4 5 8x x x x x …..(i)

The total number of ways is the number of solutions of the above equation.

Let 2 2 3 3 4 41, 1, 1.y x y x y x Then (i) becomes

1 2 3 4 5 5x y y y x ….(ii)

Where 2 3 4, , 0y y y

The number of solutions of (ii) are

5 5 1 9

5 1 4C C

2.

3. Given that AC=CB=h

,CPA BPC

2 4AP AB h

1, tan4

In APC



30o

10m

P

A

M

1, tan( )2

In ABP

tan tan 11 tan tan 2

1 2tan9

4.



5. We have, 22cos sin 2

22(1 sin ) sin 2

22sin sin 0

sin (2sin 1) 0

Now, following two cases arises:

Case I: When sin 0 (2sin 1) 0and

1 5sin 0 sin2 6 6

and

5 3:2 6 2 2

A B B

Case II: when sin 0 (2sin 1) 0and

1sin 0 sin2

and

sin 0 2

3 3:2 2 2

A B B

Thus 5 3: :2 6 2

A B

6.



7. We have, ( )3 2 4f x x

( )33 4 2 ( ) log (4 2 )f x x xf x

Clearly, f(x) will be defined. If 4 2 0x

22 4 2 2 2x x x

: 2A x R x

Now, 3( ) log (4 2 )xf x

| 2 log 2( ) 0 2(4 2 ) log3

x

xf x x

|( ) sin ( 2, )f x is increa g on

( ) : ( 2) ( )Rangeof f x y f y f

3 3: log 4 : log 4B y y x x

8. 2 8 .......(1)y x

30o30o

l

l

X

Y

A

C

B

AB BC CA l



The coordinates of B are ( cos30 , sin 30 )o ol l

3. , ,2 2

l li e

Since the point B lies on (1)

2 38 16 3

2 2l l or l

9.

Clearly the radius of the circumcircle of the square ABCD

So, the equation of the circumcircle is

10.



11. Let 2( ) 1f y y my . Since 1 lies between the roots of the equation

2 1 0y my

2 4 0 1 1 0m and my [ 0 (1) 0]Disc and f

( 2 2) 2 2m or m and m m

Now, using A.M. –G.M. inequality, we get

2

2

16 4 14 02 216

x xx

x

2 2

4 40 1 0

16 16

mx x

x x

12.



13. We find that 0cos 40 cos80 cos160 2cos 60 cos 20 cos 20 0......( )o o o o o i

Now, 0cos 40 cos80 cos80 cos160 cos160 cos 40o o o o o

01 (2cos 40 cos80 2cos80 cos160 2cos160 cos 40 )2

o o o o o

01 (cos120 cos 40 cos 240 cos80 cos 200 cos120 )2

o o o o o

01 1 1 1cos 40 cos80 cos 2002 2 2 2

o o

01 3 32cos 60 cos 20 cos 202 2 4

o o

[using (i)]

Also, cos 40 cos80 cos160o o o

3

3

sin(2 40 ) sin 320 12 sin 40 8(sin 40 ) 8

o o

o o

1 sin 2cos cos 2 cos 4 .....cos(2 )2 sin

nn

n for all n N

So the equation having 0cos 40 ,cos80 cos160o o and as its roots is given by

3 2 33 10 0 8 6 1 04 8

x x x or x x

The equation having

3 2sec 40 ,sec80 sec160 6 8 0sec 40 sec80 sec160 6

o o o

o o o

and as its roots is x x

14.



15.

16.

17. We have, (2 1) 0....(1)x y x y

Clearly, (i) represents a family of lines passing through the point of intersection of the lines.



1 10 2 1 0 . ., ,3 3

x y and x y i e

The required line passes through 1 1,3 3

and is perpendicular to the line joining (1,4)

And 1 1,3 3

So, its equation is 1 4 13 11 3

y x

12 33 7x y

18.

19. Let 99sin(101 )sinI x xdx

99sin(100 )sinI x x xdx



20.

Integrating we get, 3y x C

It passes through (1,1), 1C

Putting C=1 in (i), we get 3 1y x as the equation of the curve. Clearly, it passes through (1/8,2) Also, equation of normal is given by



( )dxY y X xdy

Equation of normal at (1,1) is 3 2 0x y

O

Y'

Y

B

XA

X'

P(x,y)

y=f(x)

21. 2 2 |{[ ( )] [ ( )] } 2[ ( ) ( ). ( )]d f x f x f x x xdx

2[ ( ). ( ) ( ). ( )]f x x x f x | |[ ( ) ( ) ( ) ( )] 0f x x and x f x

2 2[ ( )] [ ( )] tanf x x cons t

2 2 2 2 2 | 2[ (10) ] [ (10] [ (3) [ (3) [ (3)] [ (3)]f f f f

25 16 9

22.

23. We have, 1s t

1 122 1

ds dsdt dt st

2

2 2 2 3

1 1 1 1.2 2 2 4

d s dsdt s dt s s s

3 32

2

12 22

d s dsdt s dt

24. We know that the lines



1 1 1

1 1 1

x x y y z zl m n

And 2 2 2

2 2 2

x x y y z zl m n

are coplanar, iff

2 1 2 1 2 1

1 1 1

2 2 2

0x x y y z zl m nl m n

So, the given lines will be coplanar iff

2

1 2 4 3 5 41 1 0 3 0 0, 3

2 1k k k k

k

25.

26.



27.

12 ( 1)2i

nx x dn n n d

2 2 2 2 2 22 [1 2 3 ...... ]ix x d n

2 ( 1)(2 1)26

n n nd

28.



and max. 3

2

1

( ) (3) ( 3 2)f x f t t t dt

34

3 2

1

24t t t

Since f(x) is continuous on [1,3], Therefore, the range of

1( ) [ (2), (3)] , 24

f x f f

29.

30.



PHYSICS

31. The body gains velocity due to friction. The acceleration due to friction.

forceof friction mga gmass m

Further, 0v at

Therefore, 0 0v vta g

From work energy theorem,

Work done by force of friction=change in kinetic energy

Or 20

12

W mv

Mean power Wt

From Eq.s (i) and (ii)

012meanP mgv

Substituting the values, we have 1 0.27 1.0 9.8 1.52meanP

2.0W

32. 212AllW mv

212F mg NW W W mv

21 1 1(5 5) 10 5 0

2 2 2v

14.14 /v m s

33. 2/ 2

m mdm d d

2( ) (1 cos )imgRdU dm gh d



/2

0

2 12i i

mgRU dU

21mgR

Now, i i f fU K U K

22 11 02

mgR mv

Or 22 1v gR

34. At C, potential energy is minimum So it is stable equilibrium position.

Further,

( )dUF Slopeof U rgraphdr

Negative force mean attraction and positive force means repulsion.

35. F i jx y

( 3 4 )i j

Since particle was initially at rest. So it will move in the direction of force. We can see that initial velocity is in the direction of PO. So the particle will cross the X-axis at origin.

i i f fK U K U

0 (3 6 4 8) (3 0 4 0)

50f

f

Kor K J

36. 13 7

12 6dU a bFdx x x

At equilibrium, 1/620 aF or x

b

At this value of x, we can see that 2

2

d Udx

is positive. So, potential energy is minimum of

equilibrium is stable.



37. (a) Velocity is decreasing. Therefore , acceleration (or net force) is opposite to the direction of motion. (b) and (c) some other forces ( other than friction) may also act which retard the motion.

38.

v

T

mg 2 2v gh

2mvT mg

l

or (2 )m ghT mgl

21 hmgl

39. cos( )d components of T are cancelled and sin( )d components towards centre provide the necessary centripetal force to small portion PQ.

d

d d

d

T P QT

22 sin( ) ( )( )PQT d m R

For small angle sin d d

22 (2 )( )(2 )2mT d R n

22 mn R

Substituting the values we get,



2(2 )(2 )(300 / 60) (0.25)

250T

N

40. Particle breaks off the sphere at 2cos3

g

The tangential acceleration at this instant is

2 4sin 1 cos 19

g g g

53

g

41. 3 0.54

y x

37o

r

0.553o

Slope 3tan4

37o

0.5sin 53 0.4or m

2(3)(5)(0.4) 6 /L mvr kg m s

42. 1 2 3ABI I I I

A B

34a

1 2

3



2

7 / 2

Fm mgM

(or) 2 1

Net work done by friction in pure rolling is zero.

43. In case of pure rolling can be obtained about bottommost point, about which torque of friction is zero

F

a

f

2

(2 ) 43 / 2 3F R F

mR mR

43

FRm

f

44. 2 2( ) mdT dm dx xt

O

x

T+dT T

A

2

0

T x

t

mdT xdxt

2

22

1 12

xT m tt

45. 2

2 54 2lr t l

2 2

21 22[ ] 2

3 12ml mlI I I mr

2l

r

l1

2



2103

ml

46. (2 3 ) /F mE i j N kg

We can check that path given (there fore displacement) is perpendicular to force

0W

47. Weight =Upthrust

334

Mg a g

1/3

43Ma

48.

mg

N 2a

Relative acceleration along the inclined plane

cos cosr

ma mgam

cos sina g

3 110 3 102 2

210 /m s

2 2 1 110 5r

st sa

49. 12 6

12 6f i

av

v v v vat

210 205 /

6m s



50. y components of moments are cancelled and x-components are added.

v

dm

v

d

dm

y

x

0 0

( )( sin )xP dP dm v

0( sin )M d v

2Mv

51.

R

2R

2R

4 4(2 )2

R rr

Now 2 ,2 and R are in parallel.

52. 1

2

1lr Rl

1y y xR Rx x

53. Applying Kirchhoff’s loop law in outermost loop, we have

15 2 2.5 3 1 18 03 2q q

Solving this equation we get

30q C

54. They have a common potential in the beginning. This implies that only B has the charge in the beginning.



A

B

0B BKq KqVa b

or BA

KqKq a aVb

Now A BB

Kq KqVb b

1a aV V Vb b

55.

045 0452a

C

0 004 sin 45 sin 454 / 2c

IBa

02 2 Ia

56.

vR

2/ 2v v

l l

2

2B le



22

2

vB ll Bvl

57. 35 10 2000 10LX L

6

1 1 102000 50 10LX

C

,L cSince X X circuit is in resonance.

Z=R=(6+4)=10

20 / 2

1.41410

rmsrms

VI AZ

This is also the reading of ammeter.

58.

A

B

V

P

0V 05V

06P

03P

Specific heat

1v

Q WC U W CT T T

For the given process

00 0 0

94 182PW v PV W area of P V graph

Also, 2 1T T T

0 0 0 0 0 06 5 3 27P V P V PVR R R

And, 32vC R

0 0

0 0

183 3 2 13272 2 3 6v

PVW R R R RC CPVTR



59. As is increased, there will be a value of above which photoelectron will be cease to come out show photocurrent will becomes zero

60. Conceptual

CHEMISTRY

61.

3 2

100 56? 0.959

CaCO CaO COg g

g

100 0.959 1.7156

gm

22 3 3 2

111 100? 1.71

CaCl CO CaCO Clg g

g

111 1.71 1.89100

gm

21.89% 100 45%4.22

CaCl in the mixture

62. Conceptual

63. Conceptual

64. 5BrF and 4XeOF have 3 2sp d hybridization hence their shape is square pyramidal.

65. 2

0.7 0.0821 3003H unknown

nP P

60.7 0.73088.21

n

0.0308n

0.7 1033.750.0308 2

x x

66. As water freezes from liquid to solid, randomness decreases i.e., systemS decreases . Heat released during the process is absorbed by the surroundings hence surroundingsS increases.

67. 103 3 1.5 10spNaCl AgNO AgCl NaNO K



2 21[ ] 10 0.5 102

Ag M M

2 21[ ] 10 0.5 102

Cl M M

2 2 5[ ][ ] (0.5 10 ) (0.5 10 ) 2.5 10ip ip spK Ag Cl K K

When ip spK K it results in precipitation.

68. High concentration of heavy water retards the growth of the plants.

69. 2 2 22 2Sn H O SnO Hsteam

Tin decomposes steam to form tin dioxide and hydrogen gas.

70. Conceptual

71. Conceptual

72.

2NaNH

223 2 3( ) ( ) NaNHCH CH Br CH Br CH C CH

2 53 2 3 3

C H BrCH C CCH CH CH C CN a

Total no of moles of 2 2 1 3NaNH

73. Conceptual

74. 2 (43%)H O and (57%)HI make maximum boiling azeotropic mixture on boiling at 400 K.

75. Electrolyte X is strong electrolyte as on dilution the number of ions remain same, only interionic attraction decreases and hence not much increase in m is seen. While m for a weak electrolyte increases significantly.

76. 2.303 log akt a x

For 7/8 of the reaction to complete 7/8t t

7 / 8 / 8a x a a a



7/82.303 2.303log log8

/ 8at

k a k

For half of the reaction to complete, 1/2t t

/ 2 / 2x a a a

1/22.303 2.303log log 2

/ 2at

k a k

3

7/8

1/2

log8 log 2 3log 2 3log 2 log 2 log 2

tt

77. Conceptual

78. Conceptual

79. Conceptual

80. Conceptual

81. Conceptual

82. Conceptual

83. Conceptual

84. Conceptual

85.

3CH3CH

3CH3CH C OH

H

(X)

2 2 7K Cr OH C O

(Y)

2 2NH CONHNH HCl+

3CH COONa

3 2 2( )CH C NNHCONH

(Z)



86.

33 2 3 2

PClCH CH COOH CH CH COCl 6 6C H3AlCl(A)

2 3COCH CH

(B)

2 2,

NH NHbase heat

(C)

2 2 3CH CH CH

87.

2NH2Br

KOH(i)

2NH3CHCl

KOHNC

2 /H Pd

(ii)

(iii)

3NHCH

O

88. Conceptual

89. Conceptual

90. Detergents having straight hydrocarbon chains are easily degraded by microorganisms while detergents with branched hydrocarbon chains are generally non-biodegradable.

Documents

Sec: Sr. IIT-IZ(L25) JEE MAIN Dt: 24-12-2018 Time : 3 ... · NARAYANA IIT ACADEMY 24-12-18_Sr.IIT-IZ(L25)_Jee-Mains_key&sol Don’t wait to reach your goal to be proud of yourself