SrS60, ELITE, TARGET & LIIT - Jee-Main-GTM-22 - KEY & Sol'S - 220610 - 133348
SrS60, ELITE, TARGET & LIIT - Jee-Main-GTM-22 - KEY & Sol'S - 220610 - 133348
SrS60, ELITE, TARGET & LIIT - Jee-Main-GTM-22 - KEY & Sol'S - 220610 - 133348
CHEMISTRY
31) 2 32) 1 33) 1 34) 3 35) 2
36) 4 37) 3 38) 4 39) 4 40) 4
41) 1 42) 2 43) 4 44) 3 45) 4
46) 2 47) 3 48) 3 49) 3 50) 3
51) 6 52) 100 53) 50 54) 3 55) 4
56) 6 57) 4 58) 140 59) 4 60) 6
MATHEMATICS
61) 2 62) 4 63) 2 64) 2 65) 2
66) 4 67) 1 68) 2 69) 4 70) 2
71) 4 72) 3 73) 2 74) 4 75) 2
76) 3 77) 1 78) 3 79) 1 80) 1
81) 6 82) 3 83) 3 84) 2 85) 20
86) 1 87) 0 88) 0 89) 0 90) 10
SOLUTIONS
PHYSICS
1. Given velocity time relation is V 10 10t the relation between displacement and
Time taken by particle to hit the ground, t t1 t2 here t1is the time to leave the
block by the particle, t2 the time to hit the ground after leaving the block
L 2h 2L L 2L Mv
t1 ,t 2 t Vcm
v g g v g mM
The distance moved by the centre of mass in this time xcm = Vcm t
Mv L 2L
Vcm
m M v g
8. A
v1
v2
30o
Momentum will not change in the direction perpendicular to
line of impact( parallel to surfaces).
component of velocity along the incline surface ( = v1 sin ) will not change
v
v1 sin30 1 will be same as before collision.
2
9. dj
is perpendicular to A as well as J , rate of change of a vector is always
dt
perpendicular to the vector, then its magnitude shouldn’t change.
10.
B l
v
A
Components of velocity of A and velocity of B along the rod must be equal (since
v cos
length of rod doesn’t change) v A cos vB cos vB A s
cos
B
CHEMISTRY
1
31. M 3mol lit W 3 58.5gm
Wt of solution 1000 1.25 1250g Wt. of solvent 1250 175.5 1074.5g
1000
m 3 2.79m
1074.5
32. At constant volume, the density of the compound is directly proportional to molar
mass of the compound.
33. More the molar mass of compound,more will be boiling point.So boiling point
order will be Butane<1−chlorobutane<1−bromobutane<1−iodobutane.
As molecular mass is increasing so boiling point also increase.
34. ‘s’ is not derived from
35. Sodium rosinate enhances the lathering property of soap.
Lather is caused by increasing the surface tension of water (and then stirring it up).
The increased surface tension is also one of the mechanisms that soap uses to clean
things.
36. 2s – 2p mixing is observed in B2 , C2 and N 2
37. For H 2 , Z is always greater than 1 CO2 in more easily liquifiable than O2
38. Electron-donating groups (—OCH3, —CH3 etc.) tend to decrease and electron-
withdrawing groups (–NO2, —CN etc.) tend to increase the acidic character of
phenols. Since —OCH3 is a more powerful electron-donating group than —CH3
group, therefore, p-methylphenol is slightly more acidic that p-methoxyphenol while
p-nitrophenol is the strongest acid.Thus, option (d), i.e. p-methoxyphenol, p-
methylphenol, p-nitrophenol is correct.
39. Ionisation energy EO and electron affinity A are defined at at 0k. At other
temperatures heat capaities for reactants and products have to be taken into account
For Ionisation Cp O 5 / 2 R For electron gain Cp O 5 / 2 R
40. I2 and Na2CO3 react with acetophenone (C6H5COCH3) to give yellow ppt. of CHI3
but benzophenone (C6H5COC6H5) does not and hence can be used to distinguish
between them.
41. 2 H
MnO2 KOH
KNO3
MnO4 MnO4 MnO2
42. PbI 2 appears as golden spangles
KOH
NH COCl N C O
43. NH 2 KCl , H 2O
KOH
COCl2
KCl , H 2O
Phenylisocyanate
2 2
58. In addition results should be expressed to the least number of decimal places where
n
as in multiplication to least number of significant digits 33
20
59. P Br2
Ph CH 2 COOH CH3 CH 2 COOH Ph CH COOH CH 3 CH COOH
Br Br
d / d /
60. 6 XeF4 12 H 2O 4 Xe 2 XeO3 24 HF 3O2
MATHEMATICS
61. 2018 2019 1 2019
2
X' X
-1 1 Q
1 2
2
Y' Graph is as shown in figure. OPmax OQ 2 PQ 2 32 2
OPmax OPmax 9
63. sin x 0 x 0 x 0,1 is the solution set
6
64. C2 2! 1 5 C2 2! 2 53
65. Square the given equality to yield
1
3 4i
n 2
an bni an2 bn2 2anbni, so an bn
2
n
Im 3 4i and
n
an bn 1 3 4i 1 1 7
n 0 7
n
Im
2 n 0 7 n
Im
2 3 4i 16
.
1
7
66.
3/2 1
1 1 1 2 2 x 5
Area of R 2 2 xdx 2.2 .1.1 2
2 2 2 3 2
2
2 16 5 13
2
3 3 2 3
67. n n n
1/3 1 1/3 n n 1 1
Last term of 2
2
n
is tn 1 Cn 2
2
n /2
2
Also, we have
log 3 8
1 1 5/3 log 3 23
5/3 3 25 Thus,
5/3 3log 3 2
3
3
Sec: Sr.S60, ELITE, TARGET & LIIT Page 9
SRI CHAITANYA IIT ACADEMY, INDIA 10-06-22_ Sr.S60, ELITE, TARGET & LIIT_ Jee-Main_GTM-22_KEY &SOL’S
I .F e
1 sin 2 x
dx
e
ln 1 sin 2 x
f ' x
dx ln f x c
f x
1 sin 2 x
General solution is y. I .F . I .F . Qdx y 1 sin 2 x 1 sin 2 x 1 cossinx2 x dx
y 1 sin 2 x cos x dx sin x c When x 0, y 0 c 0
1 1 5 1 2
When x then y 1 y y
6 4 2 4 2 5
2
i.e. y
6 5
70.
Let P 5 cos ,
5 sin be a point lies on x 2 y 2 5 since RPQ is right angled at
point ‘P’ .So circumcenter of triangle PQR is midpoint of QR. So QR is chord of
x 5 cos y 5 sin
contact w.r.t point ‘P’ is S1 0 i.e,. 1 1
4 1
And equation of the chord QR, which is bisected at point M(h, k)
S1 S11
xh yk h 2 k 2 xh 4ky
2 2
2 1 2
4 1 4 1 h 4k h 4k 2
1 2 6 x y2
2 2 2 2
2
Variance 5.2 mean 5.2 41 x 2 y 2 80 26
5
2 2
x y 65 ………… (2)Solve (1) and (2) , we get
x 4,7
y 7,4
77. h
In BCD, tan 300 …………..(1)
d
H
In ABD, tan 600 …………….(2)
d
h
0
tan 30
(1) divided by (2) 0
d
tan 60 H
1 h H
h
3 3 H 3
A
600
H
Tower C
h
Pole
300 600
B D
d
78. a 2 c 2 c 2 b 2 a 2 b 2 2c 2
79. 1 2
1 0
2 1
1 For 1, we get x y z 1 and x y z 1.
Hence, system has no solution.
For 1, all three equations become x y z 1 , which represents coincident
planes.
1 2 1
80. 4 x 5 y 20 .......... i 3x 2 y 6 0 .......... ii
Line to 3x 2 y 6 0 and passes 1,1 is 2 x 3 y 5 ........... iii
35 33
On soling (i) & (iv) pt , 10 On solving (ii) & (iii) pt 13,
2 2
33
10 35 13 35
2
Side BC is y 10 x y 10 x
35 2 61 2
13
2
26 x 122 y 1675 0
Sec: Sr.S60, ELITE, TARGET & LIIT Page 12
SRI CHAITANYA IIT ACADEMY, INDIA 10-06-22_ Sr.S60, ELITE, TARGET & LIIT_ Jee-Main_GTM-22_KEY &SOL’S
81. 25 1
Probability of winning prize in single game 2
in this case 1st 4
25 25
games, must result in exactly 2 prizes and 5th game must result in prize
2 2 2
1 24 1
R. probability 4c2 . 6.
24
5
25 25 25 25
82. det A b2 3
2 3
det A b 3 b
b b b
Min value is 2 3
83. 1
lim x
4
x 1 x 2
lim x 1
1 4 1
x
1/2 1/2
2
x
4
x x 1 2x
x
1
1
x 4
1
2
1 1 1
lim x 1 1 . 4 ... 1
2 2 x 2 x
So, for above limit to exist and has value non-zero, we must have
1 4 3
84. Let g x x 1 x 2 3 x 2 at x 2 And h x x 1 x 2 4 x 3 at x 3
g x is non-differentiable but h x is differentiable and at x 3,
g x is differentiable but h x is non-differentiable at x 2.
f x is not differentiable at two points.
85. The image of focus with respect to tangents lie on directrix
equation of the directrix is 4 x y 17 0
4a 20 17
86. x 4 .e x .dx
I 2
e x
x 4
4 x3 12 x 2 24 x 24 72
Let
e x x 4 4 x 3 12 x 2 24 x 24 t x 4 .e x .dx dt
dt 1
I 2
c
t 72 t 72
ex
4 c
x 4 x 3 12 x 2 24 x 24 72.e x
g 0 96 f 0 1
87. 4a 1 6
x2 x 0 4a 1 5
a 1 a 1 4 I a 0, 2,6, 4
a 1 a 1