JEE Main 2022 Question Paper 26 June Shift 2 | JEE Main & Advanced Previous Year Papers PDF Download

 Page 1


 1 
FINAL JEE –MAIN EXAMINATION – JUNE, 2022 
(Held On Sunday 26
th
 June, 2022)     TIME : 3:00 PM  to  06:00 PM 
PHYSICS TEST PAPER WITH SOLUTION 
 
 
SECTION-A 
1. The dimension of mutual inductance is :
(A) [ML
2
 T
–2
 A
–1
] (B) [ML
2
T
–3
A
–1
]
(C) [ML
2
T
–2
A
–2
] (D) [ML
2
T
–3
A
–2
]
Official Ans. by NTA (C) 
 
Sol. e
2
 : induced emf in secondary coil 
i
1
 : Current in primary coil 
M : Mutual inductance 
= [ML
2
T
–2
A
–2
] 
2. In the arrangement shown in figure a
1
,a
2
, a
3
 and a
4
are the accelerations of masses m
1
,m
2
,m
3
 and m
4
respectively. Which of the following relation is
true for this arrangement?
m
1
m
2
m
3
m
4
(A) 4a
1
 + 2a
2
 + a
3
 + a
4 
= 0
(B) a
1
 + 4a
2
 + 3a
3
 + a
4 
= 0
(C) a
1
 + 4a
2
 + 3a
3
 + 2a
4 
= 0
(D) 2a
1
 + 2a
2
 + 3a
3
 + a
4 
= 0
Official Ans. by NTA (A) 
 
Sol. 
Using costraint 
– 4Ta
1
 – 2Ta
2
 – Ta
3
 – Ta
4
 = 0
4a
1
 + 2a
2
 + a
3
 + a
4
 = 0
3. Arrange the four graphs in descending order of total
work done; where W
1
, W
2
, W
3
 and W
4
 are the work
done corresponding to figure a, b, c and d respectively.
Figure-c
F
F
x
–F
x
1
x
2
x
0
Figure-d
F
F
x
–F
x
1
x
2
x
0
x
3
(A) W
3
 > W
2
 > W
1 
> W
4
(B) W
3
 > W
2
 > W
4 
> W
1
(C) W
2
 > W
3
 > W
4 
> W
1 
(D) W
2
 > W
3
 > W
1 
> W
4
Official Ans. by NTA (A)
1
2
di
eM
dt
??
2
1
e
M
di
dt
??
? ?
? ?
2
11
W
e q
M
di di
dt dt
??
??
??
??
? ? ? ?
? ? ? ?
? ? ? ?
? ?
22
1
ML T
AT
AT
?
?
??
??
?
??
??
T.a 0 ?
?
Page 2


 1 
FINAL JEE –MAIN EXAMINATION – JUNE, 2022 
(Held On Sunday 26
th
 June, 2022)     TIME : 3:00 PM  to  06:00 PM 
PHYSICS TEST PAPER WITH SOLUTION 
 
 
SECTION-A 
1. The dimension of mutual inductance is :
(A) [ML
2
 T
–2
 A
–1
] (B) [ML
2
T
–3
A
–1
]
(C) [ML
2
T
–2
A
–2
] (D) [ML
2
T
–3
A
–2
]
Official Ans. by NTA (C) 
 
Sol. e
2
 : induced emf in secondary coil 
i
1
 : Current in primary coil 
M : Mutual inductance 
= [ML
2
T
–2
A
–2
] 
2. In the arrangement shown in figure a
1
,a
2
, a
3
 and a
4
are the accelerations of masses m
1
,m
2
,m
3
 and m
4
respectively. Which of the following relation is
true for this arrangement?
m
1
m
2
m
3
m
4
(A) 4a
1
 + 2a
2
 + a
3
 + a
4 
= 0
(B) a
1
 + 4a
2
 + 3a
3
 + a
4 
= 0
(C) a
1
 + 4a
2
 + 3a
3
 + 2a
4 
= 0
(D) 2a
1
 + 2a
2
 + 3a
3
 + a
4 
= 0
Official Ans. by NTA (A) 
 
Sol. 
Using costraint 
– 4Ta
1
 – 2Ta
2
 – Ta
3
 – Ta
4
 = 0
4a
1
 + 2a
2
 + a
3
 + a
4
 = 0
3. Arrange the four graphs in descending order of total
work done; where W
1
, W
2
, W
3
 and W
4
 are the work
done corresponding to figure a, b, c and d respectively.
Figure-c
F
F
x
–F
x
1
x
2
x
0
Figure-d
F
F
x
–F
x
1
x
2
x
0
x
3
(A) W
3
 > W
2
 > W
1 
> W
4
(B) W
3
 > W
2
 > W
4 
> W
1
(C) W
2
 > W
3
 > W
4 
> W
1 
(D) W
2
 > W
3
 > W
1 
> W
4
Official Ans. by NTA (A)
1
2
di
eM
dt
??
2
1
e
M
di
dt
??
? ?
? ?
2
11
W
e q
M
di di
dt dt
??
??
??
??
? ? ? ?
? ? ? ?
? ? ? ?
? ?
22
1
ML T
AT
AT
?
?
??
??
?
??
??
T.a 0 ?
?
 
2 
 
Sol. Work done = area under F – x curve. Area below x 
–axis is negative & area above x-axis is positive. 
 so 
 W
3
 > W
2
 > W
1 
> W
4
 
4. Solid spherical ball is rolling on a frictionless 
horizontal plane surface about its axis of 
symmetry. The ratio of rotational kinetic energy of 
the ball to its total kinetic energy is :- 
 (A)  (B)  (C)  (D)   
 Official Ans. by NTA (B) 
  
Sol. K
total
 = K
rotational
 + K
Translational 
  
 
v
cm
 = R ? for pure rolling 
  
 K
Rot
 
 
 K
Total
  
  
5. Given below are two statements : One is labelled 
as Assertion A and the other is labelled as Reason 
R. 
 Assertion A : If we move from poles to equator, 
the direction of acceleration due to gravity of earth 
always points towards the center of earth without 
any variation in its magnitude. 
 Reason R : At equator, the direction of acceleration 
due to the gravity is towards the center of earth. 
 In the light of above statements, choose the correct 
answer from the options given below : 
 (A) Both A and R are true and R is the correct 
explanation of A. 
  (B) Both A and R are true but R is NOT the correct 
explanation of A. 
 (C) A is true but R is false 
 (D) A is false but R is true  
 Official Ans. by NTA (D) 
  
Sol. 
r
rw
2
?
R
w
g
 
 Effective acceleration due to gravity is the 
resultant of g & rw
2
 whose direction & magnitude 
depends upon ?. Hence assertion is false. 
 When ? = 0° (at equator), effective acceleration is 
radially inward. 
6. If ? is the density and ? is coefficient of viscosity 
of fluid which flows with a speed v in the pipe of 
diameter d, the correct formula for Reynolds 
number R
e
 is :  
 (A)   (B)   
 (C)   (D)   
 Official Ans. by NTA (C) 
  
Sol. Reynold’s number is given by 
   
 
  
7. A flask contains argon and oxygen in the ratio of 
3:2 in mass and the mixture is kept at 27°C. The 
ratio of their average kinetic energy per molecule 
respectively will be : 
 (A) 3 : 2  (B) 9 : 4  
 (C) 2 : 3  (D) 1 : 1  
 Official Ans. by NTA (D) 
  
Sol. Average K.E./molecule = 
 
 
   
 So, 
 
  
 
 
 
 
 
 
  
 
 
  
 
 
 
 
2
5
2
7
1
5
7
10
22
total cm cm
11
K I mV
22
? ? ?
2
cm
2
I mR
5
?
2 2 2
total cm cm cm
1 1 7
R mv mv mv
5 2 10
? ? ?
2
cm
Rot
2
Total
cm
1
mv
K 2
5
7
K7
mv
10
?
e
d
R
v
?
?
?
e
v
R
d
?
?
?
e
vd
R
?
?
?
e
R
vd
?
?
?
2
2 2 2 cm
rat cm cm 2
v 1 1 2 1
R I mR mv
2 2 5 2 R
? ? ? ? ? ?
2
cm
1
mv
5
Page 3


 1 
FINAL JEE –MAIN EXAMINATION – JUNE, 2022 
(Held On Sunday 26
th
 June, 2022)     TIME : 3:00 PM  to  06:00 PM 
PHYSICS TEST PAPER WITH SOLUTION 
 
 
SECTION-A 
1. The dimension of mutual inductance is :
(A) [ML
2
 T
–2
 A
–1
] (B) [ML
2
T
–3
A
–1
]
(C) [ML
2
T
–2
A
–2
] (D) [ML
2
T
–3
A
–2
]
Official Ans. by NTA (C) 
 
Sol. e
2
 : induced emf in secondary coil 
i
1
 : Current in primary coil 
M : Mutual inductance 
= [ML
2
T
–2
A
–2
] 
2. In the arrangement shown in figure a
1
,a
2
, a
3
 and a
4
are the accelerations of masses m
1
,m
2
,m
3
 and m
4
respectively. Which of the following relation is
true for this arrangement?
m
1
m
2
m
3
m
4
(A) 4a
1
 + 2a
2
 + a
3
 + a
4 
= 0
(B) a
1
 + 4a
2
 + 3a
3
 + a
4 
= 0
(C) a
1
 + 4a
2
 + 3a
3
 + 2a
4 
= 0
(D) 2a
1
 + 2a
2
 + 3a
3
 + a
4 
= 0
Official Ans. by NTA (A) 
 
Sol. 
Using costraint 
– 4Ta
1
 – 2Ta
2
 – Ta
3
 – Ta
4
 = 0
4a
1
 + 2a
2
 + a
3
 + a
4
 = 0
3. Arrange the four graphs in descending order of total
work done; where W
1
, W
2
, W
3
 and W
4
 are the work
done corresponding to figure a, b, c and d respectively.
Figure-c
F
F
x
–F
x
1
x
2
x
0
Figure-d
F
F
x
–F
x
1
x
2
x
0
x
3
(A) W
3
 > W
2
 > W
1 
> W
4
(B) W
3
 > W
2
 > W
4 
> W
1
(C) W
2
 > W
3
 > W
4 
> W
1 
(D) W
2
 > W
3
 > W
1 
> W
4
Official Ans. by NTA (A)
1
2
di
eM
dt
??
2
1
e
M
di
dt
??
? ?
? ?
2
11
W
e q
M
di di
dt dt
??
??
??
??
? ? ? ?
? ? ? ?
? ? ? ?
? ?
22
1
ML T
AT
AT
?
?
??
??
?
??
??
T.a 0 ?
?
 
2 
 
Sol. Work done = area under F – x curve. Area below x 
–axis is negative & area above x-axis is positive. 
 so 
 W
3
 > W
2
 > W
1 
> W
4
 
4. Solid spherical ball is rolling on a frictionless 
horizontal plane surface about its axis of 
symmetry. The ratio of rotational kinetic energy of 
the ball to its total kinetic energy is :- 
 (A)  (B)  (C)  (D)   
 Official Ans. by NTA (B) 
  
Sol. K
total
 = K
rotational
 + K
Translational 
  
 
v
cm
 = R ? for pure rolling 
  
 K
Rot
 
 
 K
Total
  
  
5. Given below are two statements : One is labelled 
as Assertion A and the other is labelled as Reason 
R. 
 Assertion A : If we move from poles to equator, 
the direction of acceleration due to gravity of earth 
always points towards the center of earth without 
any variation in its magnitude. 
 Reason R : At equator, the direction of acceleration 
due to the gravity is towards the center of earth. 
 In the light of above statements, choose the correct 
answer from the options given below : 
 (A) Both A and R are true and R is the correct 
explanation of A. 
  (B) Both A and R are true but R is NOT the correct 
explanation of A. 
 (C) A is true but R is false 
 (D) A is false but R is true  
 Official Ans. by NTA (D) 
  
Sol. 
r
rw
2
?
R
w
g
 
 Effective acceleration due to gravity is the 
resultant of g & rw
2
 whose direction & magnitude 
depends upon ?. Hence assertion is false. 
 When ? = 0° (at equator), effective acceleration is 
radially inward. 
6. If ? is the density and ? is coefficient of viscosity 
of fluid which flows with a speed v in the pipe of 
diameter d, the correct formula for Reynolds 
number R
e
 is :  
 (A)   (B)   
 (C)   (D)   
 Official Ans. by NTA (C) 
  
Sol. Reynold’s number is given by 
   
 
  
7. A flask contains argon and oxygen in the ratio of 
3:2 in mass and the mixture is kept at 27°C. The 
ratio of their average kinetic energy per molecule 
respectively will be : 
 (A) 3 : 2  (B) 9 : 4  
 (C) 2 : 3  (D) 1 : 1  
 Official Ans. by NTA (D) 
  
Sol. Average K.E./molecule = 
 
 
   
 So, 
 
  
 
 
 
 
 
 
  
 
 
  
 
 
 
 
2
5
2
7
1
5
7
10
22
total cm cm
11
K I mV
22
? ? ?
2
cm
2
I mR
5
?
2 2 2
total cm cm cm
1 1 7
R mv mv mv
5 2 10
? ? ?
2
cm
Rot
2
Total
cm
1
mv
K 2
5
7
K7
mv
10
?
e
d
R
v
?
?
?
e
v
R
d
?
?
?
e
vd
R
?
?
?
e
R
vd
?
?
?
2
2 2 2 cm
rat cm cm 2
v 1 1 2 1
R I mR mv
2 2 5 2 R
? ? ? ? ? ?
2
cm
1
mv
5
3 
8. The charge on capacitor of capacitance 15 ?F in the
figure given below is : 
V
– +
13V
10 F ? 15 F ? ???F
(A) 60 ?c (B) 130?c  (C) 260 ?c  (D) 585 ?c
Official Ans. by NTA (A) 
 
Sol. 
V
– +
13V
10 F ? 15 F ? ???F
Charge on each capacitor is same 
? they are in series. 
9. A parallel plate capacitor with plate area A and
plate separation d=2 m has a capacitance of 4 ?F. 
The new capacitance of the system if half of the 
space between them is filled with a dielectric 
material of dielectric constant K=3 (as shown in 
figure) will be : 
K = 3
d
(A) 2?F (B) 32 ?F (C) 6?F (D) 8 ?F
Official Ans. by NTA (C) 
 
Sol. 
C
1
 & C
2
 are in series 
10. Sixty four conducting drops each of radius 0.02 m
and each carrying a charge of 5 ?C are combined
to form a bigger drop. The ratio of surface density
of bigger drop to the smaller drop will be :
(A) 1 : 4 (B) 4 : 1 (C) 1 : 8 (D) 8 : 1
Official Ans. by NTA (B) 
 
Sol. Let R = radius of combined drop 
r = radius of smaller drop 
Volume will remain same 
R = 4r 
Q = 64q ; 
q : charge of smaller drop  
Q : Charge of combined drop 
eq
1 1 1 1 12 8 6 26
C 10 15 20 120 120
??
? ? ? ? ?
eq
60
CF
13
??
13 60
Q 60 C
13
?
? ? ?
0
original
A
C
d
?
?
K = 3 K
d/2 d/2
=
C
2
C
1
d/2 d/2
A A
00
1
A 2A
CC
d / 2 d
??
? ? ?
0 0 0
2
KA 2KA 6A
C 3C
d / 2 d d
? ? ?
? ? ? ?
12
new
12
CC C 3C 3C
C
C C C 3C 4
?
? ? ?
??
0
2A 3
4d
?
??
0
A 3
2d
?
??
new original
3
CC
2
?
3
4 6 F
2
? ? ? ?
33
44
R 64 r
33
? ? ? ?
2
2
bigger
2
smaller
2
Q
Qr
4R
·
q
q R
4r
?
?
??
?
?
2
2
r
64 4
16r
??
bigger
smaller
4
1
?
?
?
Page 4


 1 
FINAL JEE –MAIN EXAMINATION – JUNE, 2022 
(Held On Sunday 26
th
 June, 2022)     TIME : 3:00 PM  to  06:00 PM 
PHYSICS TEST PAPER WITH SOLUTION 
 
 
SECTION-A 
1. The dimension of mutual inductance is :
(A) [ML
2
 T
–2
 A
–1
] (B) [ML
2
T
–3
A
–1
]
(C) [ML
2
T
–2
A
–2
] (D) [ML
2
T
–3
A
–2
]
Official Ans. by NTA (C) 
 
Sol. e
2
 : induced emf in secondary coil 
i
1
 : Current in primary coil 
M : Mutual inductance 
= [ML
2
T
–2
A
–2
] 
2. In the arrangement shown in figure a
1
,a
2
, a
3
 and a
4
are the accelerations of masses m
1
,m
2
,m
3
 and m
4
respectively. Which of the following relation is
true for this arrangement?
m
1
m
2
m
3
m
4
(A) 4a
1
 + 2a
2
 + a
3
 + a
4 
= 0
(B) a
1
 + 4a
2
 + 3a
3
 + a
4 
= 0
(C) a
1
 + 4a
2
 + 3a
3
 + 2a
4 
= 0
(D) 2a
1
 + 2a
2
 + 3a
3
 + a
4 
= 0
Official Ans. by NTA (A) 
 
Sol. 
Using costraint 
– 4Ta
1
 – 2Ta
2
 – Ta
3
 – Ta
4
 = 0
4a
1
 + 2a
2
 + a
3
 + a
4
 = 0
3. Arrange the four graphs in descending order of total
work done; where W
1
, W
2
, W
3
 and W
4
 are the work
done corresponding to figure a, b, c and d respectively.
Figure-c
F
F
x
–F
x
1
x
2
x
0
Figure-d
F
F
x
–F
x
1
x
2
x
0
x
3
(A) W
3
 > W
2
 > W
1 
> W
4
(B) W
3
 > W
2
 > W
4 
> W
1
(C) W
2
 > W
3
 > W
4 
> W
1 
(D) W
2
 > W
3
 > W
1 
> W
4
Official Ans. by NTA (A)
1
2
di
eM
dt
??
2
1
e
M
di
dt
??
? ?
? ?
2
11
W
e q
M
di di
dt dt
??
??
??
??
? ? ? ?
? ? ? ?
? ? ? ?
? ?
22
1
ML T
AT
AT
?
?
??
??
?
??
??
T.a 0 ?
?
 
2 
 
Sol. Work done = area under F – x curve. Area below x 
–axis is negative & area above x-axis is positive. 
 so 
 W
3
 > W
2
 > W
1 
> W
4
 
4. Solid spherical ball is rolling on a frictionless 
horizontal plane surface about its axis of 
symmetry. The ratio of rotational kinetic energy of 
the ball to its total kinetic energy is :- 
 (A)  (B)  (C)  (D)   
 Official Ans. by NTA (B) 
  
Sol. K
total
 = K
rotational
 + K
Translational 
  
 
v
cm
 = R ? for pure rolling 
  
 K
Rot
 
 
 K
Total
  
  
5. Given below are two statements : One is labelled 
as Assertion A and the other is labelled as Reason 
R. 
 Assertion A : If we move from poles to equator, 
the direction of acceleration due to gravity of earth 
always points towards the center of earth without 
any variation in its magnitude. 
 Reason R : At equator, the direction of acceleration 
due to the gravity is towards the center of earth. 
 In the light of above statements, choose the correct 
answer from the options given below : 
 (A) Both A and R are true and R is the correct 
explanation of A. 
  (B) Both A and R are true but R is NOT the correct 
explanation of A. 
 (C) A is true but R is false 
 (D) A is false but R is true  
 Official Ans. by NTA (D) 
  
Sol. 
r
rw
2
?
R
w
g
 
 Effective acceleration due to gravity is the 
resultant of g & rw
2
 whose direction & magnitude 
depends upon ?. Hence assertion is false. 
 When ? = 0° (at equator), effective acceleration is 
radially inward. 
6. If ? is the density and ? is coefficient of viscosity 
of fluid which flows with a speed v in the pipe of 
diameter d, the correct formula for Reynolds 
number R
e
 is :  
 (A)   (B)   
 (C)   (D)   
 Official Ans. by NTA (C) 
  
Sol. Reynold’s number is given by 
   
 
  
7. A flask contains argon and oxygen in the ratio of 
3:2 in mass and the mixture is kept at 27°C. The 
ratio of their average kinetic energy per molecule 
respectively will be : 
 (A) 3 : 2  (B) 9 : 4  
 (C) 2 : 3  (D) 1 : 1  
 Official Ans. by NTA (D) 
  
Sol. Average K.E./molecule = 
 
 
   
 So, 
 
  
 
 
 
 
 
 
  
 
 
  
 
 
 
 
2
5
2
7
1
5
7
10
22
total cm cm
11
K I mV
22
? ? ?
2
cm
2
I mR
5
?
2 2 2
total cm cm cm
1 1 7
R mv mv mv
5 2 10
? ? ?
2
cm
Rot
2
Total
cm
1
mv
K 2
5
7
K7
mv
10
?
e
d
R
v
?
?
?
e
v
R
d
?
?
?
e
vd
R
?
?
?
e
R
vd
?
?
?
2
2 2 2 cm
rat cm cm 2
v 1 1 2 1
R I mR mv
2 2 5 2 R
? ? ? ? ? ?
2
cm
1
mv
5
3 
8. The charge on capacitor of capacitance 15 ?F in the
figure given below is : 
V
– +
13V
10 F ? 15 F ? ???F
(A) 60 ?c (B) 130?c  (C) 260 ?c  (D) 585 ?c
Official Ans. by NTA (A) 
 
Sol. 
V
– +
13V
10 F ? 15 F ? ???F
Charge on each capacitor is same 
? they are in series. 
9. A parallel plate capacitor with plate area A and
plate separation d=2 m has a capacitance of 4 ?F. 
The new capacitance of the system if half of the 
space between them is filled with a dielectric 
material of dielectric constant K=3 (as shown in 
figure) will be : 
K = 3
d
(A) 2?F (B) 32 ?F (C) 6?F (D) 8 ?F
Official Ans. by NTA (C) 
 
Sol. 
C
1
 & C
2
 are in series 
10. Sixty four conducting drops each of radius 0.02 m
and each carrying a charge of 5 ?C are combined
to form a bigger drop. The ratio of surface density
of bigger drop to the smaller drop will be :
(A) 1 : 4 (B) 4 : 1 (C) 1 : 8 (D) 8 : 1
Official Ans. by NTA (B) 
 
Sol. Let R = radius of combined drop 
r = radius of smaller drop 
Volume will remain same 
R = 4r 
Q = 64q ; 
q : charge of smaller drop  
Q : Charge of combined drop 
eq
1 1 1 1 12 8 6 26
C 10 15 20 120 120
??
? ? ? ? ?
eq
60
CF
13
??
13 60
Q 60 C
13
?
? ? ?
0
original
A
C
d
?
?
K = 3 K
d/2 d/2
=
C
2
C
1
d/2 d/2
A A
00
1
A 2A
CC
d / 2 d
??
? ? ?
0 0 0
2
KA 2KA 6A
C 3C
d / 2 d d
? ? ?
? ? ? ?
12
new
12
CC C 3C 3C
C
C C C 3C 4
?
? ? ?
??
0
2A 3
4d
?
??
0
A 3
2d
?
??
new original
3
CC
2
?
3
4 6 F
2
? ? ? ?
33
44
R 64 r
33
? ? ? ?
2
2
bigger
2
smaller
2
Q
Qr
4R
·
q
q R
4r
?
?
??
?
?
2
2
r
64 4
16r
??
bigger
smaller
4
1
?
?
?
 
4 
 
11. The equivalent resistance between points A and B 
in the given network is : 
5?
10 ?
5 ?
5 ?
5?
5 ?
10 ?
10 ?
10 ?
A B
 
 (A) 65 ? ? (B) 20 ?  
 (C) 5 ?  (D) 2 ?  
 Official Ans. by NTA (C) 
  
Sol.  
 
10 ?
5 ?
5 ?
5 ?
5 ?
10 ?
10 ?
A B
10 ?
5 ?
5?
10 ?
10 ?
10 ?
B
=
A
?
5?
10 ?
10 ?
A
5 ?
5 ?
?
B
B
A
10 ?
10 ? 10 ?
5 ?
?
5 ?
10 ?
A
5 ?
B
?
10 ?
10 ?
A
B 
 R
AB
 = 5 ? 
12. A bar magnet having a magnetic moment of 2.0 × 
10
5
 JT
–1
, is placed along the direction of uniform 
magnetic field of magnitude B= 14 × 10
–5
 T. The 
work done in rotating the magnet slowly through 
60° from the direction of field is : 
 (A) 14 J (B) 8.4 J (C) 4 J (D) 1.4 J  
 Official Ans. by NTA (A) 
  
Sol. Work done = MB (cos??
 1
 – cos ??
 2
) 
  ?
 1
  = 0°, ?
2
 = 60° 
  = 2 × 10
5
 × 14 × 10
–5
 (1 – 1/2) 
  = 14 J 
13. Two coils of self inductance L
1
 and L
2
 are 
connected in series combination having mutual 
inductance of the coils as M. The equivalent self 
inductance of the combination will be :  
 
 (A)  (B) L
1
 + L
2
 + M   
 (C) L
1
 + L
2
 + 2M (D) L
1
 + L
2
 – 2M  
 Official Ans. by NTA (D) 
  
Sol. Current on both the inductor is in opposite 
direction. 
 Hence : 
 L
eq
 = L
1
 + L
2
 – 2M 
14. A metallic conductor of length 1m rotates in a 
vertical plane parallel to east-west direction about 
one of its end with angular velocity 5 rad/s. If the 
horizontal component of earth's magnetic field is 
0.2 × 10
–4
 T, then emf induced between the two 
ends of the conductor is :  
 (A) 5 ?V (B) 50 ?V (C) 5mV (D) 50mV  
 Official Ans. by NTA (B) 
  
Sol. emf induced between the two ends = 
 
 
  
 
 
 
 
15. Which is the correct ascending order of 
wavelengths? 
 (A) ?
visible
 < ?
X –ray
 < ?
gamma-ray
 < ?
microwave 
  (B) ?
gamma-ray
 < ?
X –ray
 < ?
visible
 < ?
microwave
  
 (C) ?
X-ray
 < ?
gamma-ray
 < ?
visible
 < ?
microwave 
 (D) ?
microwave
 < ?
visible
 < ?
gamma-ray
 < ?
X-ray 
 Official Ans. by NTA (B) 
  
 
12
1 1 1
L L M
??
4
46
0.2 10 5 1
0.5 10 50 10 V 50 V
2
?
??
? ? ?
? ? ? ? ? ?
Page 5


 1 
FINAL JEE –MAIN EXAMINATION – JUNE, 2022 
(Held On Sunday 26
th
 June, 2022)     TIME : 3:00 PM  to  06:00 PM 
PHYSICS TEST PAPER WITH SOLUTION 
 
 
SECTION-A 
1. The dimension of mutual inductance is :
(A) [ML
2
 T
–2
 A
–1
] (B) [ML
2
T
–3
A
–1
]
(C) [ML
2
T
–2
A
–2
] (D) [ML
2
T
–3
A
–2
]
Official Ans. by NTA (C) 
 
Sol. e
2
 : induced emf in secondary coil 
i
1
 : Current in primary coil 
M : Mutual inductance 
= [ML
2
T
–2
A
–2
] 
2. In the arrangement shown in figure a
1
,a
2
, a
3
 and a
4
are the accelerations of masses m
1
,m
2
,m
3
 and m
4
respectively. Which of the following relation is
true for this arrangement?
m
1
m
2
m
3
m
4
(A) 4a
1
 + 2a
2
 + a
3
 + a
4 
= 0
(B) a
1
 + 4a
2
 + 3a
3
 + a
4 
= 0
(C) a
1
 + 4a
2
 + 3a
3
 + 2a
4 
= 0
(D) 2a
1
 + 2a
2
 + 3a
3
 + a
4 
= 0
Official Ans. by NTA (A) 
 
Sol. 
Using costraint 
– 4Ta
1
 – 2Ta
2
 – Ta
3
 – Ta
4
 = 0
4a
1
 + 2a
2
 + a
3
 + a
4
 = 0
3. Arrange the four graphs in descending order of total
work done; where W
1
, W
2
, W
3
 and W
4
 are the work
done corresponding to figure a, b, c and d respectively.
Figure-c
F
F
x
–F
x
1
x
2
x
0
Figure-d
F
F
x
–F
x
1
x
2
x
0
x
3
(A) W
3
 > W
2
 > W
1 
> W
4
(B) W
3
 > W
2
 > W
4 
> W
1
(C) W
2
 > W
3
 > W
4 
> W
1 
(D) W
2
 > W
3
 > W
1 
> W
4
Official Ans. by NTA (A)
1
2
di
eM
dt
??
2
1
e
M
di
dt
??
? ?
? ?
2
11
W
e q
M
di di
dt dt
??
??
??
??
? ? ? ?
? ? ? ?
? ? ? ?
? ?
22
1
ML T
AT
AT
?
?
??
??
?
??
??
T.a 0 ?
?
 
2 
 
Sol. Work done = area under F – x curve. Area below x 
–axis is negative & area above x-axis is positive. 
 so 
 W
3
 > W
2
 > W
1 
> W
4
 
4. Solid spherical ball is rolling on a frictionless 
horizontal plane surface about its axis of 
symmetry. The ratio of rotational kinetic energy of 
the ball to its total kinetic energy is :- 
 (A)  (B)  (C)  (D)   
 Official Ans. by NTA (B) 
  
Sol. K
total
 = K
rotational
 + K
Translational 
  
 
v
cm
 = R ? for pure rolling 
  
 K
Rot
 
 
 K
Total
  
  
5. Given below are two statements : One is labelled 
as Assertion A and the other is labelled as Reason 
R. 
 Assertion A : If we move from poles to equator, 
the direction of acceleration due to gravity of earth 
always points towards the center of earth without 
any variation in its magnitude. 
 Reason R : At equator, the direction of acceleration 
due to the gravity is towards the center of earth. 
 In the light of above statements, choose the correct 
answer from the options given below : 
 (A) Both A and R are true and R is the correct 
explanation of A. 
  (B) Both A and R are true but R is NOT the correct 
explanation of A. 
 (C) A is true but R is false 
 (D) A is false but R is true  
 Official Ans. by NTA (D) 
  
Sol. 
r
rw
2
?
R
w
g
 
 Effective acceleration due to gravity is the 
resultant of g & rw
2
 whose direction & magnitude 
depends upon ?. Hence assertion is false. 
 When ? = 0° (at equator), effective acceleration is 
radially inward. 
6. If ? is the density and ? is coefficient of viscosity 
of fluid which flows with a speed v in the pipe of 
diameter d, the correct formula for Reynolds 
number R
e
 is :  
 (A)   (B)   
 (C)   (D)   
 Official Ans. by NTA (C) 
  
Sol. Reynold’s number is given by 
   
 
  
7. A flask contains argon and oxygen in the ratio of 
3:2 in mass and the mixture is kept at 27°C. The 
ratio of their average kinetic energy per molecule 
respectively will be : 
 (A) 3 : 2  (B) 9 : 4  
 (C) 2 : 3  (D) 1 : 1  
 Official Ans. by NTA (D) 
  
Sol. Average K.E./molecule = 
 
 
   
 So, 
 
  
 
 
 
 
 
 
  
 
 
  
 
 
 
 
2
5
2
7
1
5
7
10
22
total cm cm
11
K I mV
22
? ? ?
2
cm
2
I mR
5
?
2 2 2
total cm cm cm
1 1 7
R mv mv mv
5 2 10
? ? ?
2
cm
Rot
2
Total
cm
1
mv
K 2
5
7
K7
mv
10
?
e
d
R
v
?
?
?
e
v
R
d
?
?
?
e
vd
R
?
?
?
e
R
vd
?
?
?
2
2 2 2 cm
rat cm cm 2
v 1 1 2 1
R I mR mv
2 2 5 2 R
? ? ? ? ? ?
2
cm
1
mv
5
3 
8. The charge on capacitor of capacitance 15 ?F in the
figure given below is : 
V
– +
13V
10 F ? 15 F ? ???F
(A) 60 ?c (B) 130?c  (C) 260 ?c  (D) 585 ?c
Official Ans. by NTA (A) 
 
Sol. 
V
– +
13V
10 F ? 15 F ? ???F
Charge on each capacitor is same 
? they are in series. 
9. A parallel plate capacitor with plate area A and
plate separation d=2 m has a capacitance of 4 ?F. 
The new capacitance of the system if half of the 
space between them is filled with a dielectric 
material of dielectric constant K=3 (as shown in 
figure) will be : 
K = 3
d
(A) 2?F (B) 32 ?F (C) 6?F (D) 8 ?F
Official Ans. by NTA (C) 
 
Sol. 
C
1
 & C
2
 are in series 
10. Sixty four conducting drops each of radius 0.02 m
and each carrying a charge of 5 ?C are combined
to form a bigger drop. The ratio of surface density
of bigger drop to the smaller drop will be :
(A) 1 : 4 (B) 4 : 1 (C) 1 : 8 (D) 8 : 1
Official Ans. by NTA (B) 
 
Sol. Let R = radius of combined drop 
r = radius of smaller drop 
Volume will remain same 
R = 4r 
Q = 64q ; 
q : charge of smaller drop  
Q : Charge of combined drop 
eq
1 1 1 1 12 8 6 26
C 10 15 20 120 120
??
? ? ? ? ?
eq
60
CF
13
??
13 60
Q 60 C
13
?
? ? ?
0
original
A
C
d
?
?
K = 3 K
d/2 d/2
=
C
2
C
1
d/2 d/2
A A
00
1
A 2A
CC
d / 2 d
??
? ? ?
0 0 0
2
KA 2KA 6A
C 3C
d / 2 d d
? ? ?
? ? ? ?
12
new
12
CC C 3C 3C
C
C C C 3C 4
?
? ? ?
??
0
2A 3
4d
?
??
0
A 3
2d
?
??
new original
3
CC
2
?
3
4 6 F
2
? ? ? ?
33
44
R 64 r
33
? ? ? ?
2
2
bigger
2
smaller
2
Q
Qr
4R
·
q
q R
4r
?
?
??
?
?
2
2
r
64 4
16r
??
bigger
smaller
4
1
?
?
?
 
4 
 
11. The equivalent resistance between points A and B 
in the given network is : 
5?
10 ?
5 ?
5 ?
5?
5 ?
10 ?
10 ?
10 ?
A B
 
 (A) 65 ? ? (B) 20 ?  
 (C) 5 ?  (D) 2 ?  
 Official Ans. by NTA (C) 
  
Sol.  
 
10 ?
5 ?
5 ?
5 ?
5 ?
10 ?
10 ?
A B
10 ?
5 ?
5?
10 ?
10 ?
10 ?
B
=
A
?
5?
10 ?
10 ?
A
5 ?
5 ?
?
B
B
A
10 ?
10 ? 10 ?
5 ?
?
5 ?
10 ?
A
5 ?
B
?
10 ?
10 ?
A
B 
 R
AB
 = 5 ? 
12. A bar magnet having a magnetic moment of 2.0 × 
10
5
 JT
–1
, is placed along the direction of uniform 
magnetic field of magnitude B= 14 × 10
–5
 T. The 
work done in rotating the magnet slowly through 
60° from the direction of field is : 
 (A) 14 J (B) 8.4 J (C) 4 J (D) 1.4 J  
 Official Ans. by NTA (A) 
  
Sol. Work done = MB (cos??
 1
 – cos ??
 2
) 
  ?
 1
  = 0°, ?
2
 = 60° 
  = 2 × 10
5
 × 14 × 10
–5
 (1 – 1/2) 
  = 14 J 
13. Two coils of self inductance L
1
 and L
2
 are 
connected in series combination having mutual 
inductance of the coils as M. The equivalent self 
inductance of the combination will be :  
 
 (A)  (B) L
1
 + L
2
 + M   
 (C) L
1
 + L
2
 + 2M (D) L
1
 + L
2
 – 2M  
 Official Ans. by NTA (D) 
  
Sol. Current on both the inductor is in opposite 
direction. 
 Hence : 
 L
eq
 = L
1
 + L
2
 – 2M 
14. A metallic conductor of length 1m rotates in a 
vertical plane parallel to east-west direction about 
one of its end with angular velocity 5 rad/s. If the 
horizontal component of earth's magnetic field is 
0.2 × 10
–4
 T, then emf induced between the two 
ends of the conductor is :  
 (A) 5 ?V (B) 50 ?V (C) 5mV (D) 50mV  
 Official Ans. by NTA (B) 
  
Sol. emf induced between the two ends = 
 
 
  
 
 
 
 
15. Which is the correct ascending order of 
wavelengths? 
 (A) ?
visible
 < ?
X –ray
 < ?
gamma-ray
 < ?
microwave 
  (B) ?
gamma-ray
 < ?
X –ray
 < ?
visible
 < ?
microwave
  
 (C) ?
X-ray
 < ?
gamma-ray
 < ?
visible
 < ?
microwave 
 (D) ?
microwave
 < ?
visible
 < ?
gamma-ray
 < ?
X-ray 
 Official Ans. by NTA (B) 
  
 
12
1 1 1
L L M
??
4
46
0.2 10 5 1
0.5 10 50 10 V 50 V
2
?
??
? ? ?
? ? ? ? ? ?
 
 
5 
 
Sol. From electromagnetic wave spectrum. 
 ? ?increases
 
??? ? ?
?-ray x-rays ultra 
violet 
visible infrared microwave Radio 
wave 
 ?
gamma-ray
 < ?
X –ray
 < ?
visible
 < ?
microwave 
16. For a specific wavelength 670 nm of light coming 
from a galaxy moving with velocity v, the 
observed wavelength is 670.7 nm.  
 The value of v is : 
 (A) 3 × 10
8
 ms
–1
 (B) 3 × 10
10
 ms
–1
  
 (C) 3.13 × 10
5
 ms
–1
(D) 4.48 × 10
5
 ms
–1
  
 Official Ans. by NTA (C) 
  
Sol. ?
emitted
 = 670 nm 
 ?
obs
 = 670.7 nm 
 v = ? 
 c = 3 × 10
8
 m/s 
 If v << c 
  
  
 V = 3.13 × 10
5
 m/s 
17. A metal surface is illuminated by a radiation of 
wavelength 4500 Å. The ejected photo-electron 
enters a constant magnetic field of 2 mT making an 
angle of 90° with the magnetic field. If it starts 
revolving in a circular path of radius 2 mm, the 
work function of the metal is approximately :  
 (A) 1.36 eV (B) 1.69 eV (C) 2.78 eV (D) 2.23 eV  
 Official Ans. by NTA (A) 
  
Sol. ? = 4500 Å 
 B = 2mT, R = 2mm 
  
  
  
 K = 2.25 × 10
–19
 J  
   
  
 ? = E – K = (2.76 – 1.40) eV = 1.36 eV 
18. A radioactive nucleus can decay by two different 
processes. Half-life for the first process is 3.0 
hours while it is 4.5 hours for the second process. 
The effective half- life of the nucleus will be : 
 (A) 3.75 hours (B) 0.56 hours  
 (C) 0.26 hours (D) 1.80 hours  
 Official Ans. by NTA (D) 
  
Sol. ?
eq
 = ?
1
 + ?
2
 
  
  
  
19. The positive feedback is required by an amplifier 
to act an oscillator. The feedback here means : 
 (A) External input is necessary to sustain ac signal 
in output. 
 (B) A portion of the output power is returned back 
to the input.  
 (C) Feedback can be achieved by LR network. 
 (D) The base-collector junction must be forward 
biased.  
 Official Ans. by NTA (B) 
  
Sol. When the amplifier connects with positive 
feedback, it acts as the oscillator the feedback here 
is positive feedback which means some amount of 
voltage is given to the input.  
obs emitted
emitted
v
c
? ? ?
?
?
670.7 670 v
670 c
?
?
? ?
2
2Km
R
qB
qBR
K
2m
?
?
? ?
2
19 3 3
31
1.6 10 2 10 2 10
K
2 9.1 10
? ? ?
?
? ? ? ? ?
?
??
? ?
2
50
31
6.4
10
K
2 9.1 10
?
?
??
?
19
19
2.25 10
eV 1.40eV
1.6 10
?
?
?
??
?
12400
E 2.76eV
4500
??
? ? ? ? ? ?
1/2 1/2 1/2
eq 1 2
ln 2 ln 2 ln 2
t t t
??
? ?
? ? ? ?
? ? ? ?
1/2 1/2
12
1/2
eq
1/2 1/2
12
tt
t
tt
?
?
?
3 4.5 3 4.5 3 3
1.8 hr
3 4.5 7.5 5
? ? ?
? ? ? ?
?