key & solutions€¦ · narayana iit academy 27-10-18_sr.iit-iz(l25)_jee-mains_key &sol’s...
Post on 16-Oct-2020
0 Views
Preview:
TRANSCRIPT
NARAYANA IIT ACADEMY 27-10-18_Sr.IIT-IZ(L25)_Jee-Mains_key &sol’s
1 | P a g e
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
NARAYANA IIT ACADEMY 27-10-18_Sr.IIT-IZ(L25)_Jee-Mains_key &sol’s
2 | P a g e
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
NARAYANA IIT ACADEMY 27-10-18_Sr.IIT-IZ(L25)_Jee-Mains_key &sol’s
3 | P a g e
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
NARAYANA IIT ACADEMY 27-10-18_Sr.IIT-IZ(L25)_Jee-Mains_key &sol’s
4 | P a g e
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.
NARAYANA IIT ACADEMY 27-10-18_Sr.IIT-IZ(L25)_Jee-Mains_key &sol’s
5 | P a g e
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
NARAYANA IIT ACADEMY 27-10-18_Sr.IIT-IZ(L25)_Jee-Mains_key &sol’s
6 | P a g e
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
NARAYANA IIT ACADEMY 27-10-18_Sr.IIT-IZ(L25)_Jee-Mains_key &sol’s
7 | P a g e
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
NARAYANA IIT ACADEMY 27-10-18_Sr.IIT-IZ(L25)_Jee-Mains_key &sol’s
8 | P a g e
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
NARAYANA IIT ACADEMY 27-10-18_Sr.IIT-IZ(L25)_Jee-Mains_key &sol’s
9 | P a g e
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
NARAYANA IIT ACADEMY 27-10-18_Sr.IIT-IZ(L25)_Jee-Mains_key &sol’s
10 | P a g e
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
NARAYANA IIT ACADEMY 27-10-18_Sr.IIT-IZ(L25)_Jee-Mains_key &sol’s
11 | P a g e
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
NARAYANA IIT ACADEMY 27-10-18_Sr.IIT-IZ(L25)_Jee-Mains_key &sol’s
12 | P a g e
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
NARAYANA IIT ACADEMY 27-10-18_Sr.IIT-IZ(L25)_Jee-Mains_key &sol’s
13 | P a g e
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
NARAYANA IIT ACADEMY 27-10-18_Sr.IIT-IZ(L25)_Jee-Mains_key &sol’s
14 | P a g e
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
NARAYANA IIT ACADEMY 27-10-18_Sr.IIT-IZ(L25)_Jee-Mains_key &sol’s
15 | P a g e
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
NARAYANA IIT ACADEMY 27-10-18_Sr.IIT-IZ(L25)_Jee-Mains_key &sol’s
16 | P a g e
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
NARAYANA IIT ACADEMY 27-10-18_Sr.IIT-IZ(L25)_Jee-Mains_key &sol’s
17 | P a g e
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
NARAYANA IIT ACADEMY 27-10-18_Sr.IIT-IZ(L25)_Jee-Mains_key &sol’s
18 | P a g e
84. Conceptual
85. Conceptual
86. Conceptual
87. Conceptual
88. Conceptual
89. Conceptual
90. Conceptual
top related