key & solutions€¦ · narayana iit academy 27-10-18_sr.iit-iz(l25)_jee-mains_key &sol’s...

NARAYANA IIT ACADEMY 27-10-18_Sr.IIT-IZ(L25)_Jee-Mains_key &sol’s

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Sec: Sr.IIT-IZ(L25) JEE MAIN Dt: 27-10-2018

Time : 3 Hours RPTM Max Marks : 360

Key & Solutions

Maths

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

Physics

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

Chemistry

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


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MATHS SOLUTIONS

1. Using square root formula

3 4 2 3 4 2i i and i i

2.

3 2 3

( 2 ) 3 2 3 2 2 4

z i z i i

z i i z i i

3. 1 2 1 2arg( ) arg( ) arg( )3 4 12

z z z z

4.

0

A(z)

( )iB ze

Area of the triangle OAB

1 ( )( )sin2

OA OB

21 sin sin2 2

i zz ze


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5. (1,0), ( 1,0),C(0, 1)A B form a triangle, hence z is circumcentre of ABC

6. 1 2 3 1 2 3

1 1 1......... ......4 4 4

x x x cis

7. 10 23 2,w w w w

8.

3

33 3 3

3 3 3 3

1 36 6 3 3 32 2

3 3 3

1

i

iiz i

iz e i

iz i

ee e

e e

9. 2

1

2525

iz ez

(by coni’ s theorem)

2 25 15(3 4 )( 1) z 102 2

z i i

10. Let 2 2 2 2z x iy z x y i xy

2

1z

z is real, 2 22 ( 1) ( ) 0xy x y x y

11. cos cos cos sinA sinB sinC 0A B C

(cos sinA) (cosB sinB) (cosC isinC) 0 i0A i

12.

10

1

2 [ 1]11

(0 1)k

ki cis i sumof roots

i i


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13. 3 3z i hence 1 3

2i

is a complex cube roots of unity

1 1 2 3c c c c

2 2

2

3 1 10 ( 1) 3 (1 2 )0

k k z

14. sec sec , ,arg ( )2

z as z

Also cos( ) cos( ) and sin( ) sin( )

15. Since the equation has real solution

Z x iy x

2 ( ) 0x p iq x r is

2 0x px r o and qx s

From 2nd, x=-s/q and putting in first, we get 2 2pqs s q r which is the required condition.

16.

Let the bird flight at B, the top of the tree BD, and ‘O’ be the observer. Then

45oBOD and BD=20 mts. Now the bird flying horizontally reaches M in 1 second.


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30oMON where MN perpendicular to ON Now BD=MN=20 mts. From ,BOD

2045 20o BDTan mts

OD OD from

20, 30

20o MNMON Tan

ON DN

20 3 1DN

20(0.732) 14.64mts BM

Speed of bird = distance/Time =14.64 m/s

17.

Let BD=h, In right angled triangle ACD

sin , cos / 22

r AD r ecAD

In right angled triangle ABD

sin ; cos sincos / 2 2

h h r ecr ec


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18.

Formula: ( )cot cotA ncotBm n m

( )cot cot 30 cot 60o ox x x x

19.

1 13 3 2cot cot , cot5 5 5

2cot5

160cot cot

dtake h

20.

Let ABC be the triangular park, AP be the pole at A,D be the midpoint of BC, Let each side of the equalilateral triangle ABC be ‘a’ then

2 2

2 2 2 2 3 34 4 2a aAD AB BD a AD a


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And since AP=h and ADP we have cotAD h

3 2cot a cot

2 3a h h

21.

2 2 21 2 1 2 1 2( ) ( ) ( ) 2( )(os )cos60os s os os os

2 2(72) (96) (72)(96)

2 2

1 2(72) (96) (72)(96), find s s

22.

Let BC be the declivity and BA be the tower. Using sine rule in ,ABC

We have sin 75 sin 30o o

BC AB


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80sin 30 40 2 2 40( 6 2)

sin 75 3 1

o

oAB

23.

Height of tower DE=h mtrs.

Let CD xmtrs

tan cot .....(1)

tan 45 .....(2)o

h a b x ha b x

h b x hb x

tan(90 ) cot ....(3) tanh x hx

From (1) and (2) cota h h

From (2) and (3) tanb h h

2 2h ah h hb h h ba h a h

abab bh ah h

a b


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24.

OAB is equilateral.

OA OB AB a

tan 303

o h aha

25.

p q p p q ( )P p q T T F T T T F F F T F T T T T F F T T T

26. p: it is raining

Q: I will not come Given statement p q

The contra positive is q p

If I will come then it is not raining

27. Apply the formula ( ) p qp q

As ( ( )) p (q r)p q r

p ( q r)

28. p: x+2=6; q : x=4

Converse of p q is q p ; If x=4 then x+2=6


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29. ( ) ( p q)p q

( ) ( )p q p q

( )p q q p

30.

p q q p q ( )p q p q

T T F F T T T F T T F F F T F T F F F F T F T T

PHYSICS

31. 2 2

48

(10) 1500 10 15 102 2 5 10

L dg mY

32. Energy stored per unit volume 12

Stress Strain

2 21 1' mod ( )2 2

Young s ulus Strain Y x

33. max. /maxstress stress mg AY strain

strain Y Y

6 23 10A m

11 3 62 10 10 3 10

10Y strain Am

g

60kg

34. Increase in tension of wire YA

6 11 2 48 10 2.2 10 10 10 5 8.8N

35. Initial length (circumference) of the ring 2 r

Final length (circumference) of the ring 2 R


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Change in length 2 2R r

Strain 2 ( )

2changein length R r R roriginal length r r

Now Young’s modulus/ A // L (R ) / r

F F AEr

R rF AE

r

36. 2 2

2

1 ( ) 12 2

stress F A LU volumeY A Y

37. 2 10 6 3 2

22

2 10 10 (10 ) 2 102 2 50 10

YAlW JL

38. 1 /V VC V C P VK P

54 10 100 100 0.4cc

39. 3

9200 10 10 2 10/ / 0.1/100P h gK

V V V V

40. The pressure exerted by a 3000m column of water on the bottom layer is

P h g

3 2 7 23000 1000 10 3 10m kgm ms Nm

Fractional compression V

V

is

7 2

29 2

3 10 1.36 10 1.36%2.2 10

V P NmV B Nm

41. (1 2 )dV dLV L


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3 3 12 2 10 4 10 0.52

dVV

Percentage change in volume 14 10 0.4%

42 2 (1 ) 3 2 (1 )Y

3 112 2

Now substituting the value of in the following expression

3 (1 2 )3(1 2 )

YY K K

43. Angle of shear 14 10 30 0.12

100o or

L

44.

4

. tan2rC Cons tL

4 4 '

0 0( ) ( / 2) ( )2 2( / 2)

r rl l

0 00

( ) 82 16 9


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45. Speed of sound in stretched string ......( )Tv i

Where T is the tension in the string and is mass per unit length.

According to Hooke’s law, ........( )F x T x ii

From (i) and (ii), v x

1 1.5 1.22v v v

46. Total force at height 3L/4 from its lower end

= Weight suspended + Weight of 3/4 of the wire

1 (3 / 4)W W

Hence stress= 1 (3 / 4)W WS

47. Temperature of interface 1 1 2 2

1 2

K KK K

11 2

2

1[ 4 ]4

K If K K then K KK

0 4 100 80

5oK K C

K

48. Rate of heat flow 2 2

1 2( )Q k r rt L L

2 21 1 2

2 12 2 1

1 2 1 22 1 2

Q r l Q QQ r l


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49.

lRKA

B C C DA B T T T TT TR R R

60 ...(i)60 T 240...(ii)

B B C

B C

T T TT

Solving (i) and (ii)

T 120oB C

50. Here, 34 4 10x mm m

32oT C

Transmit heat per hours

200 1000 4.2200 / / 233.33 /

60 60Q kcal h J s J sT

2 4 25 5 10A cm m

We know that, Q TKAT x

Thermal conductivity of material, /

( / )Q TK

A T x

or 3

4

233.33 4 10 58.33 /5 10 32

oK W m C


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51. 2 1

1

2

1500 5000 30002500

om m

T AT

52. 1

1 11

3000 32000 2

mm m

m

TT TT

53. 4Q P A Tt

54. Power radiated by sun at 4 2( 273) 4ot C t r

Power received by a unit surface 4 2

2

( 273) 44

t rR

2 4

2

( 273)r tR

55. Loss of heat 4 40(T )Q A T t

Rate of loss of heat 4 40

Q A T Tt

4 8 4 410 10 1 5.67 10 (273 127) (273 27)

=0.99 W.

56. 42

42

4 22 1

sphere

Disc

E r TE ATE r T

57. According to Newton’s law of cooling

Rate of cooling Mean temperature difference

1 202

Fall in temperatureTime

1 2 1 2 1 2

1 2 3

1 2 3

2 2 2T T T


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58. Rate of cooling (here it is rate of loss of heat)

1 1( ) ( )c cdQ d dmc W m c m cdt dt dt

60 55(0.5 2400 0.2 900) 11560

dQ Jdt s

59. Rate of cooling 4 4

0( )A T Tt mc

4 40, , (T ) tanmt t T arecons t

A

3

1 12

2 2

m Volume a t at t aA Area a t a

22

100 1 200 .2

t st

60. According to Newton’s law of cooling

1 2 1 202

Kt

For first process: 0

(80 64) 80 64 ....( )5 2

K i

For second process: 0(80 52) 80 52 ....( )

10 2K ii

For third process: 0(80 ) 80 ....( )

15 2K iii

On solving equation (i) and (ii), we get 1

15K and 0 24oC . Putting these

values in equation (iii) we get 42.7o C


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CHEMISTRY

61. Conceptual

62. Conceptual

63. Conceptual

64. Conceptual

65. Conceptual

66. Conceptual

67. No. of soap ions per micelle 17

19

4 10 2501.6 10

0.004 250 1CMC M

68. Conceptual

69. Conceptual

70. Conceptual

71 Conceptual

72. 32 8

73. Conceptual

74. Conceptual

75. Conceptual

76. Because of Hydrogen bonding

77. Conceptual

78. Conceptual

79. Conceptual

80. Conceptual

81. Conceptual

82. Conceptual

83. Conceptual


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84. Conceptual

85. Conceptual

86. Conceptual

87. Conceptual

88. Conceptual

89. Conceptual

90. Conceptual

key & solutions€¦ · narayana iit academy 27-10-18_sr.iit-iz(l25)_jee-mains_key &sol’s...

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