JEE Main_29-07-2022_Shift-02
JEE-Main-29-07-2022-Shift-2 (Memory Based)
Physics
Question: Two plate have charge q1, q2 (q1 > q2) they are used to make capacitor. Find
potential difference?
Options:
(a) q1 + q2 / C
(b) ( q1 − q2 ) / 2C
IA
(c) q1 − q2 / C
q θ1 − θ 2
=
c
2c
ET
O
O
v=
SI
Answer: (b)
Solution:
θ −θ
q= 1 2
2
N
D
(d) q1 + q2 / 2C
Question: Linear momentum is increased by 20% then increase in kinetic energy?
Options:
(a) 40%
(b) 44%
(c) 50%
(d) 60%
Answer: (b)
Solution:
∆k k f − ki
=
ki
ki
Pf 2
m 1
= 2=
Pi 2
2n
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JEE Main_29-07-2022_Shift-02
2
 Pf 
2
=   − 1(1.2 ) =
− 1 1.44
P
 i 
Question: What is ratio of time t1 and t2 if t1 is time travelled from highest point to half of
distance and t2 the remaining half distance.
Options:
(a) t1 = 2t2
t1
(c)=
t2
(d)=
(
(
(
)
2 + 1) t
2 − 1) t
2 − 1 t2
2
IA
t1
(b)=
1
2
2⇒
t2
=
t1
(
)
2 −1
ET
O
O
t
1  t1 + t2 
= 
⇒ 1+ 2 =
2=

1  t1 
t1
2
SI
h 1 2
= gt1 ... (1)
2 2
1
2
=
h
g ( t1 + t2 ) ... ( 2 )
2
N
D
Answer: (d)
Solution:
Question: A current carrying wire x of 50 cm carring current 2A is parallel to another wire y
of length 5m and 3A current, has separation of 2m find force on wire y due to x.
Options:
(a) 1.4 x 10-5 N towards x
(b) 1.3 x 10-5 N towards y
(c) 1.4 x 10-5 N towards y
(d) 1.2 x 10-5 N towards x
Answer: (d)
Solution:
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JEE Main_29-07-2022_Shift-02
µ ii 
F =  0 1 2 l
 2π d 
N
D
IA
Question: Gravitation ka tha ki 1g ki body ko 3R from surface leke gye toh gain in potential
energy?
Options:
(a) 48 mJ
(b) 24 mJ
(c) 30 mJ
(d) 26 mJ
Answer: (a)
Solution:
∆U = U B − U A
GMm GMm
+
R
4R
GMm 3  Gm 
3
=  2  mR ×
R 4  R 
4
3
4
ET
O
O
= 10 ×1× 6400 ×10 ×
SI
−
Question: Time period of pendulum 10s. Its relative density is 5 it is immense in water. If
new time period is 5 x s. Find x.
Options:
(a) 5
(b) 3
(c) 2
(d) 4
Answer: (a)
Solution:
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JEE Main_29-07-2022_Shift-02
l
g eff
T ′ = 2π
= 2π
l
g
= 10 ×
5
2
=5
10 s= T= 2π
l
g
IA
mg=
mρ − B
eff
= 4 ρ vg
4
g
5
N
D
g eff =
Question: If α particle and proton are accelerated from same potential difference then the
ratio of their linear momenta.
Options:
(a) 2 2 :1
(c)
SI
(b) 2 2 : 3
2 :1
ET
O
O
(d) 2 : 2
Answer: (a)
Solution:
P
=
Pα
=
Pf
=
2mK
2π ( qv )
4 m 2e
× = 2 2 :1
m e
Question: Light ray from air enters a medium with 45o angle and it deviates 15o from its
original path. Find the refractive index of the medium.
Options:
(a) 2.314
(b) 1.414
(c) 1.314
(d) 1.333
Answer: (b)
Solution:
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JEE Main_29-07-2022_Shift-02
=
sin π µ sin 30°
µ= 2
1= x + y... (1)
ET
O
O
4x = 2 y
SI
N
D
IA
Question: Wire length of 1 m divided in x and y wire x stretched to twice, then stretched
wire is twice the resistance of y.
Options:
(a) 2: 1
(b) 1:2
(c) 4:1
(d) 1:4
Answer: (b)
Solution:
Length of x
Then
Length of y
 ρx 
 ρy
4
 = 2

 A 
 A 
x 1
=
y 2
Question: At equilibrium Reaction force by inclined place.
Options:
(a) 30
(b) 40
(c) 50
(d) 10
Answer: (b)
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JEE Main_29-07-2022_Shift-02
N
D
SI
Question: Match the following
A – Torque, 1 – Nms-1
B – Stress, 2 – Jkg-1
C – Latent, 3 – Nm
D- Power, 4 – Nm-2
Options:
(a) A→1, B→4, C→3, D→2
(b) A→3, B→4, C→2, D→1
(c) A→1, B→3, C→2, D→4
(d) A→2, B→1, C→4, D→3
Answer: (b)
Solution:
A→3, B→4, C→2, D→1
IA
Solution:
ET
O
O
Question: Assertion: Constantan and magainin are used in resistance coil.
Reason: their temperature coefficient of resistance is low
Options:
(a) If both assertion and reason are true and the reason is the correct explanation of the
assertion.
(b) If both assertion and reason are true, but the reason is not the correct explanation of the
assertion.
(c) If assertion is true, but reason is false.
(d) If both the assertion and reason are false.
Answer: (a)
Solution:
α 0
=
R R0 (1 + α∆T )
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JEE Main_29-07-2022_Shift-02
Chemistry
N
D
ET
O
O
(b)
SI
Question: Hinsberg’s reagent isOptions:
(a)
IA
Question: Which of the following is not a natural polymer?
Options:
(a) Protein
(b) Rayon
(c) Starch
(d) Rubber
Answer: (b)
Solution: Rayon is a synthetic polymer.
(c)
(d)
Answer: (b)
Solution: Hinsberg’s reagent is benzenesulphonyl chloride
Question: In portland cement what enhances the settling time?
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JEE Main_29-07-2022_Shift-02
Options:
(a) CaSO4, ½H2O
(b) CaSO4, 2H2O
(c) CaCO3
(d) CaSO4
Answer: (b)
Solution: Gypsum (CaSO4, 2H2O) is usually added to prevent early hardening and increase
the settling time
ET
O
O
SI
N
D
IA
Question: Ethanol on reaction with conc. H2SO4 gives A, which on further reaction with
Baeyer's reagent will give:
Options:
(a) Ethane-1,2-diol
(b) Formaldehyde
(c) Formic acid
(d) Ethanoic acid
Answer: (a)
Solution:
Question: The sum of oxidation state (magnitude only) and coordination number of cobalt in
Na[Co(bpy)Cl4]
Options:
(a) 3
(b) 6
(c) 9
(d) 5
Answer: (c)
Solution: Oxidation number = x – 4 + 1 = 0
x =3
Coordination number = 6
Sum = 3 + 6 = 9
Question: Which of the following compound has O–O linkage
Options:
(a) H2SO4
(b) H2S2O8
(c) H2S2O7
(d) H2SO3
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JEE Main_29-07-2022_Shift-02
Answer: (b)
Solution:
SI
N
D
IA
Question: 200 ml of 0.01 M of HCl and 400 ml of 0.01 M of H2SO4 are mixed. What is the
final pH?
Options:
(a) 2
(b) 1
(c) 3
(d) 4
Answer: (a)
Solution:
2mmol + 8mmol 1
=
H+ =
600
60
+
pH = –log[H ]
1
= –log
60
pH = 1.77 ≈ 2
ET
O
O
Question: Which of the following ions has lowest value of hydration enthalpy in magnitude?
Options:
(a) Cr2+
(b) Mn2+
(c) Fe2+
(d) Co2+
Answer: (b)
Solution: Hydration enthalpy order Co2+ > Fe2+ > Cr2+ > Mn2+
Therefore, Mn2+ has largest hydration enthalpy
Question: HNO3 + KCl → KNO3 + Cl2 + NOCl + H2O. Find amount of HNO3 required to
make 110 g KNO3
Options:
(a) 91.5g
(b) 56.4g
(c) 14.7g
(d) 67.2g
Answer: (a)
Solution: 4HNO3 + 3KCl → 3KNO3 + Cl2 + NOCl + 2H2O
3 × 101 g of KNO3 – 4 × 63 g of HNO3
4 × 63 ×110
110 g of KNO3 –
= 91.5 g
3 ×101
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JEE Main_29-07-2022_Shift-02
Question: Number of chlorine atoms in Bithionol is
Answer: 4.00
Solution:
SI
N
D
IA
Question: How many among the following are sp3d2 hybridised?
BrF5, [ICl4]–, ICl3, ICl5, SF6, PCl5
Answer: 4.00
Solution:
1
BrF5 = (7 + 5) = 6 = sp2d2
2
1
[ICl4]– = (7 + 4 + 1) = 6 = sp3d2
2
1
ICl5 = (7 + 5) = 6 = sp3d2
2
1
SF6 = (6 + 6) = 6 = sp3d2
2
ET
O
O
Question: Weight of O2 is x gram and for Ne is 200 g. Total pressure is 25 bar and Partial
pressure of Ne 20 bar Find x =?
Answer: 80.00
Solution:
PNe = xNe Ptotal
20 4
=
25 5
200
4
20
=
xNe =
200 x 5
+
20 32
10
4
=
x 5
10 +
32
x
50 = 40 +
8
x = 80 g
x=
Ne
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JEE Main_29-07-2022_Shift-02
Mathematics
Question: The value of
20
∑(r
r =1
Options:
(a) 22!− 2 ⋅ ( 20 ) !
2
+ 1) ⋅ r ! is:
IA
(b) ( 22 ) !− 2 ( 21) !
(c) ( 22 ) !
(d) 2 ( 21) !
20
∑(r
2
N
D
Answer: (b)
Solution:
20
+ 1)=
r!
∑ ( ( r + 1)( r + 2 ) − 3 ( r + 1) + 2 ) r !
=
r 1=
r 1
20
r =1
=
20
SI
∑ ( ( r + 2 )!− 3 ( r + 1)!+ 2r !)
=
20
∑ ( ( r + 2 )!− ( r + 1)!) − 2∑ ( ( r + 1)!− r !)
r 1=
r 1
( 22!− 2!) − 2 ( 21!− 1!)
ET
O
O
=
= 22!− 2 × 2!− 2 + 2
=
( 22 )!− 2 ( 21)!
   
   
   
  
Question: If a b c = 14 and a × b ⋅ b × c + b × c ⋅ ( c × a ) + ( c × a ) ⋅ a × b =
168 and

 
  
a , b , c are coplanar, concurrent and make equal angles with each other, then a + b + c is
(
)(
) (
)
(
)
equal to:
Options:
(a) 14
(b) 16
(c) 10
(d) 12
Answer: (b)
Solution:
  
 a , b , c are coplanar and make equal angle with each other (say θ )
So, θ= 60°
   
   
a ×b ⋅ b ×c = a ×b b ×c
(
)(
)
 
 
(a a × b and b × c will be parallel)
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JEE Main_29-07-2022_Shift-02

 2 
b c sin 2 θ 14sin 2 θ b
= a=
3   
a+b +c =
168
4
  
⇒a+b +c =
16
So, 14 ×
(
)
ET
O
O
SI
N
D
IA
14 and intersect plane
Question: A perpendicular drawn from (1, 2, 3) to the plane x + 2 y + z =
at Q . R be a point on plane such that PR makes an angle 60° with the plane, then area of
∆PQR is:
Options:
(a) 3 sq. units
(b) 3 sq. units
3
sq. units
(c)
2
(d) 4 sq. units
Answer: (a)
Solution:
 QR
= PQ ⋅ cot 60=
°
Also, l
=
1 + 4 + 3 − 14
=
1+ 4 +1
l
3
6
1 l
6
=
= 3
Area of ∆PQR = l ⋅
2
3 2 3
 x2 + x 
x
−x
Question: The number of solution of the equation 2 cos 
=
 4 + 4 is/are:
6


Options:
(a) 1
(b) 0
(c) 3
(d) Infinite
Answer: (a)
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JEE Main_29-07-2022_Shift-02
Solution:
 x2 + x 
x
−x
=
2 cos 
 4 +4
6


 x2 + x 
Equality holds when 4 x + 4− x =
2 and cos 
 =1
 6 
 x2 + x 
4 x + 4− x =
2 gives x = 0 for which cos 
 =1
 6 
So, there exist only one solution x = 0 .
)
Also,
 2  2 2 2
⇒ a + b + a ⋅b =
a b
2
⇒ 75 + 9 =
6a
ET
O
O
 2 84
⇒ a = =14
6
N
D
(
SI
Answer: 14.00
Solution:
2
 2 2
  2 2
a + b = a + 2a ⋅ b + b = a + 2 b
2
 
⇒ b = 2 a ⋅b = 6
IA
2
 
 2 2
 
 2
Question: Let a , b are two vectors and a ⋅ b= 3, a × b = 75, and a + b = a + 2 b , then
2
a is equal to ____.
Question: If sum and product of mean and variance in a binomial distribution are 82.5 and
1350 respectively, then n is equal to ___.
(where n is number of trial in binomial distribution).
Answer: 96.00
Solution:
 Mean and variance are the roots of
x 2 − 82.5 x + 1350 =
0
= np
= 60
So, mean
= npq
= 22.5
and variance
22.5 3
=
60 8
5
60
n = 96
So, p = and =
5
8
8
Question: The number of numbers lying between 1024 and 23146 which are divisible by 55
and made from 2, 3, 4, 5, 6 without repetition, is ____.
Answer: 6.00
⇒=
q
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JEE Main_29-07-2022_Shift-02
Solution:
We will solve this in two cases:
Case I:
When number has 4 digits (say abcd )
Here d is fixed as 5.
So, a, b, c can be
( 6, 4,3) , ( 3, 4, 6 ) , ( 2,3, 6 ) , ( 6,3, 2 ) , ( 3, 2, 4 )
or ( 4, 2,3) only
ET
O
O
SI
N
D
IA
Number of numbers possible = 6
Case II:
When number has 5 digits.
No such number is possible because even last number formed is greater than 23146.
Total number of such number = 6
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