JEE Main 2022 Question Paper 25 July Shift 2 | JEE Main & Advanced Previous Year Papers PDF Download

 Page 1


1
Final JEE-Main Exam July, 2022/25-07-2022/Evening Session
SECTION-A
1. In AM modulation, a signal is modulated on a
carrier wave such that maximum and
minimum amplitude are found to be 6V and
2V respectively. The modulation index is
(A) 100% (B) 80%
(C) 60% (D) 50%
Official Ans. by NTA (D)
Sol. modulation index = 
max min.
max min .
VV
100%
VV
-
´
+
= 
62
100%
62
-
´
+
 = 50%
2. The electric current in a circular coil of 2 turns
produces a magnetic induction B
1
 at its centre.
The coil is unwound and is rewound into a
circular coil of 5 turns and the same current
produces a magnetic induction B
2
 at its centre.
The ratio of 
2
1
B
B
 is :
(A) 
5
2
(B) 
25
4
(C) 
5
4
(D) 
25
2
Official Ans. by NTA (B)
Sol. B = 
0
Ni
2R
m
B
1
 = 
10
1
Ni
2R
m
For N
2
 = 5
Radius of coil = R
2
 = 
11
2
NR
N
´
B
2
 = 
20
2
Ni
R
m
22 12 2
11 21 1
B NR NN
·
B NR NN
= =´
;     
2
1
B 25
B4
=
3. A drop of liquid of density r is floating half
immersed in a liquid of density s and surface
tension 7.5 × 10
–4
 Ncm
–1
. The radius of drop
in cm will be : (Take : g = 10 m/s
2
)
(A) 
15
2r-s
(B) 
15
r-s
(C) 
3
2 r-s
(D) 
3
202r-s
Official Ans. by NTA (A)
Sol.
mg
F +F
b T
Boyant force + surace tension = mg
V
g 2 RT Vg
2
s + p =r
33
(2)44
2RT · Rg;VR
233
r-s éù
p = p =p
êú
ëû
21
3
3T 3 7.5 10 N m
RR
(2)g (2)10
--
´´-
= Þ=
r-s r-s ´
R = 
3
20(2) r-s
 m = 
15
2r-s
 cm
FINAL JEE–MAIN EXAMINATION – JULY , 2022
(Held On Monday 25
th
 July, 2022)    TIME : 3 : 00 PM   to  6 : 00 PM
PHYSICS TEST PAPER WITH SOLUTION
Page 2


1
Final JEE-Main Exam July, 2022/25-07-2022/Evening Session
SECTION-A
1. In AM modulation, a signal is modulated on a
carrier wave such that maximum and
minimum amplitude are found to be 6V and
2V respectively. The modulation index is
(A) 100% (B) 80%
(C) 60% (D) 50%
Official Ans. by NTA (D)
Sol. modulation index = 
max min.
max min .
VV
100%
VV
-
´
+
= 
62
100%
62
-
´
+
 = 50%
2. The electric current in a circular coil of 2 turns
produces a magnetic induction B
1
 at its centre.
The coil is unwound and is rewound into a
circular coil of 5 turns and the same current
produces a magnetic induction B
2
 at its centre.
The ratio of 
2
1
B
B
 is :
(A) 
5
2
(B) 
25
4
(C) 
5
4
(D) 
25
2
Official Ans. by NTA (B)
Sol. B = 
0
Ni
2R
m
B
1
 = 
10
1
Ni
2R
m
For N
2
 = 5
Radius of coil = R
2
 = 
11
2
NR
N
´
B
2
 = 
20
2
Ni
R
m
22 12 2
11 21 1
B NR NN
·
B NR NN
= =´
;     
2
1
B 25
B4
=
3. A drop of liquid of density r is floating half
immersed in a liquid of density s and surface
tension 7.5 × 10
–4
 Ncm
–1
. The radius of drop
in cm will be : (Take : g = 10 m/s
2
)
(A) 
15
2r-s
(B) 
15
r-s
(C) 
3
2 r-s
(D) 
3
202r-s
Official Ans. by NTA (A)
Sol.
mg
F +F
b T
Boyant force + surace tension = mg
V
g 2 RT Vg
2
s + p =r
33
(2)44
2RT · Rg;VR
233
r-s éù
p = p =p
êú
ëû
21
3
3T 3 7.5 10 N m
RR
(2)g (2)10
--
´´-
= Þ=
r-s r-s ´
R = 
3
20(2) r-s
 m = 
15
2r-s
 cm
FINAL JEE–MAIN EXAMINATION – JULY , 2022
(Held On Monday 25
th
 July, 2022)    TIME : 3 : 00 PM   to  6 : 00 PM
PHYSICS TEST PAPER WITH SOLUTION
2
Final JEE-Main Exam July, 2022/25-07-2022/Evening Session
4. Two billiard balls of mass 0.05 kg each moving
in opposite directions with 10 ms
–1
 collide and
rebound with the same speed. If the time
duration of contact is t = 0.005 s, then what is
the force exerted on the ball due to each other?
(A) 100 N (B) 200 N
(C) 300 N (D) 400 N
Official Ans. by NTA (B)
Sol.
m = 0.05 kg m = 0.05 kg
10 m/s
10 m/s
10 m/s
10 m/s
Change in momentum of any one ball
P 2 0.05 10 D=´´
r
| P| D
r
 = 1
av
| P|
F
t
D
=
D
r
r
F
av.
 = 200 N
5. For a free body diagram shown in the figure,
the four forces are applied in the 'x' and 'y'
directions. What additional force must be
applied and at what angle with positive x-axis
so that the net acceleration of body is zero?
5N 6N
7N
8N
x
y
(A)
2
N, 45° (B) 
2
N, 135°
(C)
2
N
3
, 30° (D) 2 N, 45°
Official Ans. by NTA (A)
Sol. Let addition force required is = 
F
r
ˆ ˆ ˆˆ
F 5i 6i 7 j 8j 0 +- + -=
r
ˆˆ
F i j,|F|2 =+=
rr
Angle with x-axis:  tan q = 
y component 1
x component 1
=
q = 45°
6. Capacitance of an isolated conducting sphere
of radius R
1
 becomes n times when it is
enclosed by a concentric conducting sphere
of radius R
2
 connected to earth. The ratio of
their radii 
2
1
R
R
æö
ç÷
èø
 is:
(A)
n
n1 -
(B) 
2n
2n1 +
(C)
n1
n
+
(D) 
2n1
n
+
Official Ans. by NTA (A)
Sol. Capacitance of isolated Conducting sphere
= 4pe
0
R
1
By enclosing inside another sphere of radius
R
2
, new capacitance = 
0 12
21
4 RR
(R R)
pe
-
        Given:
0 12
01
21
4 RR
n4R
(R R)
pe
= ´ pe
-
Þ 
2
21
R
n
(R R)
=
-
 Þ 
2
1
2
1
R
R
n
R
1
R
=
æö
-
ç÷
èø
Þ 
22
11
RR
nn
RR
=-
 Þ 
2
1
Rn
R (n 1)
=
-
Page 3


1
Final JEE-Main Exam July, 2022/25-07-2022/Evening Session
SECTION-A
1. In AM modulation, a signal is modulated on a
carrier wave such that maximum and
minimum amplitude are found to be 6V and
2V respectively. The modulation index is
(A) 100% (B) 80%
(C) 60% (D) 50%
Official Ans. by NTA (D)
Sol. modulation index = 
max min.
max min .
VV
100%
VV
-
´
+
= 
62
100%
62
-
´
+
 = 50%
2. The electric current in a circular coil of 2 turns
produces a magnetic induction B
1
 at its centre.
The coil is unwound and is rewound into a
circular coil of 5 turns and the same current
produces a magnetic induction B
2
 at its centre.
The ratio of 
2
1
B
B
 is :
(A) 
5
2
(B) 
25
4
(C) 
5
4
(D) 
25
2
Official Ans. by NTA (B)
Sol. B = 
0
Ni
2R
m
B
1
 = 
10
1
Ni
2R
m
For N
2
 = 5
Radius of coil = R
2
 = 
11
2
NR
N
´
B
2
 = 
20
2
Ni
R
m
22 12 2
11 21 1
B NR NN
·
B NR NN
= =´
;     
2
1
B 25
B4
=
3. A drop of liquid of density r is floating half
immersed in a liquid of density s and surface
tension 7.5 × 10
–4
 Ncm
–1
. The radius of drop
in cm will be : (Take : g = 10 m/s
2
)
(A) 
15
2r-s
(B) 
15
r-s
(C) 
3
2 r-s
(D) 
3
202r-s
Official Ans. by NTA (A)
Sol.
mg
F +F
b T
Boyant force + surace tension = mg
V
g 2 RT Vg
2
s + p =r
33
(2)44
2RT · Rg;VR
233
r-s éù
p = p =p
êú
ëû
21
3
3T 3 7.5 10 N m
RR
(2)g (2)10
--
´´-
= Þ=
r-s r-s ´
R = 
3
20(2) r-s
 m = 
15
2r-s
 cm
FINAL JEE–MAIN EXAMINATION – JULY , 2022
(Held On Monday 25
th
 July, 2022)    TIME : 3 : 00 PM   to  6 : 00 PM
PHYSICS TEST PAPER WITH SOLUTION
2
Final JEE-Main Exam July, 2022/25-07-2022/Evening Session
4. Two billiard balls of mass 0.05 kg each moving
in opposite directions with 10 ms
–1
 collide and
rebound with the same speed. If the time
duration of contact is t = 0.005 s, then what is
the force exerted on the ball due to each other?
(A) 100 N (B) 200 N
(C) 300 N (D) 400 N
Official Ans. by NTA (B)
Sol.
m = 0.05 kg m = 0.05 kg
10 m/s
10 m/s
10 m/s
10 m/s
Change in momentum of any one ball
P 2 0.05 10 D=´´
r
| P| D
r
 = 1
av
| P|
F
t
D
=
D
r
r
F
av.
 = 200 N
5. For a free body diagram shown in the figure,
the four forces are applied in the 'x' and 'y'
directions. What additional force must be
applied and at what angle with positive x-axis
so that the net acceleration of body is zero?
5N 6N
7N
8N
x
y
(A)
2
N, 45° (B) 
2
N, 135°
(C)
2
N
3
, 30° (D) 2 N, 45°
Official Ans. by NTA (A)
Sol. Let addition force required is = 
F
r
ˆ ˆ ˆˆ
F 5i 6i 7 j 8j 0 +- + -=
r
ˆˆ
F i j,|F|2 =+=
rr
Angle with x-axis:  tan q = 
y component 1
x component 1
=
q = 45°
6. Capacitance of an isolated conducting sphere
of radius R
1
 becomes n times when it is
enclosed by a concentric conducting sphere
of radius R
2
 connected to earth. The ratio of
their radii 
2
1
R
R
æö
ç÷
èø
 is:
(A)
n
n1 -
(B) 
2n
2n1 +
(C)
n1
n
+
(D) 
2n1
n
+
Official Ans. by NTA (A)
Sol. Capacitance of isolated Conducting sphere
= 4pe
0
R
1
By enclosing inside another sphere of radius
R
2
, new capacitance = 
0 12
21
4 RR
(R R)
pe
-
        Given:
0 12
01
21
4 RR
n4R
(R R)
pe
= ´ pe
-
Þ 
2
21
R
n
(R R)
=
-
 Þ 
2
1
2
1
R
R
n
R
1
R
=
æö
-
ç÷
èø
Þ 
22
11
RR
nn
RR
=-
 Þ 
2
1
Rn
R (n 1)
=
-
3
Final JEE-Main Exam July, 2022/25-07-2022/Evening Session
7. The ratio of wavelengths of proton and
deuteron accelerated by potential V
p
 and V
d
 is
1:2
. Then, the ratio of V
p
 to V
d
 will be
(A) 1 : 1 (B) 2:1
(C) 2 : 1 (D) 4 : 1
Official Ans. by NTA (D)
Sol. Kinetic energy gained by a charged particle
accelerated by a potential V is qV
KE = qV
Þ  
2
p
qV
2m
=  Þ  p 2mqV =
p = 
h
l
,  thus 
h
2mqV
l=
now 
p
dd
d pp
mV
mV
l
=
l
Þ  
d
p
1 2V
V 2
=
Þ  
p
d
V
4
V
=
8. For an object placed at a distance 2.4 m from a
lens, a sharp focused image is observed on a
screen placed at a distance 12 cm from the lens.
A glass plate of refractive index 1.5 and
thickness 1 cm is introduced between lens and
screen such that the glass plate plane faces
parallel to the screen. By what distance should
the object be shifted so that a sharp focused
image is observed again on the screen?
(A) 0.8 m (B) 3.2 m
(C) 1.2 m (D) 5.6 m
Official Ans. by NTA (B)
Sol.
//////////////////
Screen
O
1
2.4m 12 cm
Applying lens formula
1 11
0.12 2.4 f
+=
  Þ  
1 210
f 24
=
Upon putting the glass slab, shift of image is
11
x t 1 cm
3
æö
D= -=
ç÷
m
èø
Now v = 
1 35
12 cm
33
-=
Again apply lens formula
1 1 1 210
0.12 u f 24
+ ==
Solving u = – 5.6 m
Thus shift of object is
5.6 – 2.4 = 3.2 m
9. Light wave traveling in air along x-direction
is given by E
y
 = 540 sin p × 10
4
(x – ct) Vm
–1
.
Then, the peak value of magnetic field of wave
will be (Given c = 3 × 10
8
 ms
–1
)
(A) 18 × 10
–7
 T (B) 54 × 10
–7
 T
(C) 54 × 10
–8
 T (D) 18 × 10
–8
 T
Official Ans. by NTA (A)
Sol. E
y
 = 540 sin p × 10
4
(x – ct) Vm
–1
E
0
 = 540 Vm
–1
B
0
 = 
0
E
C
 = 
7
8
540
18 10 T
3 10
-
=´
´
10. When you walk through a metal detector
carrying a metal object in your pocket, it raises
an alarm. This phenomenon works on
(A) Electromagnetic induction
(B) Resonance in ac circuits
(C) Mutual induction in ac circuits
(D) interference of electromagnetic waves
Official Ans. by NTA (B)
)
Sol. Metal detector works on the principle of
transmitting an electromagnetic signal and
analyses a return signal from the target. So it
works on the principle of resonance in AC
circuit.
Page 4


1
Final JEE-Main Exam July, 2022/25-07-2022/Evening Session
SECTION-A
1. In AM modulation, a signal is modulated on a
carrier wave such that maximum and
minimum amplitude are found to be 6V and
2V respectively. The modulation index is
(A) 100% (B) 80%
(C) 60% (D) 50%
Official Ans. by NTA (D)
Sol. modulation index = 
max min.
max min .
VV
100%
VV
-
´
+
= 
62
100%
62
-
´
+
 = 50%
2. The electric current in a circular coil of 2 turns
produces a magnetic induction B
1
 at its centre.
The coil is unwound and is rewound into a
circular coil of 5 turns and the same current
produces a magnetic induction B
2
 at its centre.
The ratio of 
2
1
B
B
 is :
(A) 
5
2
(B) 
25
4
(C) 
5
4
(D) 
25
2
Official Ans. by NTA (B)
Sol. B = 
0
Ni
2R
m
B
1
 = 
10
1
Ni
2R
m
For N
2
 = 5
Radius of coil = R
2
 = 
11
2
NR
N
´
B
2
 = 
20
2
Ni
R
m
22 12 2
11 21 1
B NR NN
·
B NR NN
= =´
;     
2
1
B 25
B4
=
3. A drop of liquid of density r is floating half
immersed in a liquid of density s and surface
tension 7.5 × 10
–4
 Ncm
–1
. The radius of drop
in cm will be : (Take : g = 10 m/s
2
)
(A) 
15
2r-s
(B) 
15
r-s
(C) 
3
2 r-s
(D) 
3
202r-s
Official Ans. by NTA (A)
Sol.
mg
F +F
b T
Boyant force + surace tension = mg
V
g 2 RT Vg
2
s + p =r
33
(2)44
2RT · Rg;VR
233
r-s éù
p = p =p
êú
ëû
21
3
3T 3 7.5 10 N m
RR
(2)g (2)10
--
´´-
= Þ=
r-s r-s ´
R = 
3
20(2) r-s
 m = 
15
2r-s
 cm
FINAL JEE–MAIN EXAMINATION – JULY , 2022
(Held On Monday 25
th
 July, 2022)    TIME : 3 : 00 PM   to  6 : 00 PM
PHYSICS TEST PAPER WITH SOLUTION
2
Final JEE-Main Exam July, 2022/25-07-2022/Evening Session
4. Two billiard balls of mass 0.05 kg each moving
in opposite directions with 10 ms
–1
 collide and
rebound with the same speed. If the time
duration of contact is t = 0.005 s, then what is
the force exerted on the ball due to each other?
(A) 100 N (B) 200 N
(C) 300 N (D) 400 N
Official Ans. by NTA (B)
Sol.
m = 0.05 kg m = 0.05 kg
10 m/s
10 m/s
10 m/s
10 m/s
Change in momentum of any one ball
P 2 0.05 10 D=´´
r
| P| D
r
 = 1
av
| P|
F
t
D
=
D
r
r
F
av.
 = 200 N
5. For a free body diagram shown in the figure,
the four forces are applied in the 'x' and 'y'
directions. What additional force must be
applied and at what angle with positive x-axis
so that the net acceleration of body is zero?
5N 6N
7N
8N
x
y
(A)
2
N, 45° (B) 
2
N, 135°
(C)
2
N
3
, 30° (D) 2 N, 45°
Official Ans. by NTA (A)
Sol. Let addition force required is = 
F
r
ˆ ˆ ˆˆ
F 5i 6i 7 j 8j 0 +- + -=
r
ˆˆ
F i j,|F|2 =+=
rr
Angle with x-axis:  tan q = 
y component 1
x component 1
=
q = 45°
6. Capacitance of an isolated conducting sphere
of radius R
1
 becomes n times when it is
enclosed by a concentric conducting sphere
of radius R
2
 connected to earth. The ratio of
their radii 
2
1
R
R
æö
ç÷
èø
 is:
(A)
n
n1 -
(B) 
2n
2n1 +
(C)
n1
n
+
(D) 
2n1
n
+
Official Ans. by NTA (A)
Sol. Capacitance of isolated Conducting sphere
= 4pe
0
R
1
By enclosing inside another sphere of radius
R
2
, new capacitance = 
0 12
21
4 RR
(R R)
pe
-
        Given:
0 12
01
21
4 RR
n4R
(R R)
pe
= ´ pe
-
Þ 
2
21
R
n
(R R)
=
-
 Þ 
2
1
2
1
R
R
n
R
1
R
=
æö
-
ç÷
èø
Þ 
22
11
RR
nn
RR
=-
 Þ 
2
1
Rn
R (n 1)
=
-
3
Final JEE-Main Exam July, 2022/25-07-2022/Evening Session
7. The ratio of wavelengths of proton and
deuteron accelerated by potential V
p
 and V
d
 is
1:2
. Then, the ratio of V
p
 to V
d
 will be
(A) 1 : 1 (B) 2:1
(C) 2 : 1 (D) 4 : 1
Official Ans. by NTA (D)
Sol. Kinetic energy gained by a charged particle
accelerated by a potential V is qV
KE = qV
Þ  
2
p
qV
2m
=  Þ  p 2mqV =
p = 
h
l
,  thus 
h
2mqV
l=
now 
p
dd
d pp
mV
mV
l
=
l
Þ  
d
p
1 2V
V 2
=
Þ  
p
d
V
4
V
=
8. For an object placed at a distance 2.4 m from a
lens, a sharp focused image is observed on a
screen placed at a distance 12 cm from the lens.
A glass plate of refractive index 1.5 and
thickness 1 cm is introduced between lens and
screen such that the glass plate plane faces
parallel to the screen. By what distance should
the object be shifted so that a sharp focused
image is observed again on the screen?
(A) 0.8 m (B) 3.2 m
(C) 1.2 m (D) 5.6 m
Official Ans. by NTA (B)
Sol.
//////////////////
Screen
O
1
2.4m 12 cm
Applying lens formula
1 11
0.12 2.4 f
+=
  Þ  
1 210
f 24
=
Upon putting the glass slab, shift of image is
11
x t 1 cm
3
æö
D= -=
ç÷
m
èø
Now v = 
1 35
12 cm
33
-=
Again apply lens formula
1 1 1 210
0.12 u f 24
+ ==
Solving u = – 5.6 m
Thus shift of object is
5.6 – 2.4 = 3.2 m
9. Light wave traveling in air along x-direction
is given by E
y
 = 540 sin p × 10
4
(x – ct) Vm
–1
.
Then, the peak value of magnetic field of wave
will be (Given c = 3 × 10
8
 ms
–1
)
(A) 18 × 10
–7
 T (B) 54 × 10
–7
 T
(C) 54 × 10
–8
 T (D) 18 × 10
–8
 T
Official Ans. by NTA (A)
Sol. E
y
 = 540 sin p × 10
4
(x – ct) Vm
–1
E
0
 = 540 Vm
–1
B
0
 = 
0
E
C
 = 
7
8
540
18 10 T
3 10
-
=´
´
10. When you walk through a metal detector
carrying a metal object in your pocket, it raises
an alarm. This phenomenon works on
(A) Electromagnetic induction
(B) Resonance in ac circuits
(C) Mutual induction in ac circuits
(D) interference of electromagnetic waves
Official Ans. by NTA (B)
)
Sol. Metal detector works on the principle of
transmitting an electromagnetic signal and
analyses a return signal from the target. So it
works on the principle of resonance in AC
circuit.
4
Final JEE-Main Exam July, 2022/25-07-2022/Evening Session
11. An electron with energy 0.1 keV moves at
right angle to the earth's magnetic field of
1 × 10
–4
 
 
Wbm
–2
. The frequency of revolution
of the electron will be
(Take mass of electron = 9.0 × 10
–31
 kg)
(A) 1.6 × 10
5
 Hz (B) 5.6 × 10
5
 Hz
(C) 2.8 × 10
6
 Hz (D) 1.8 × 10
6
 Hz
Official Ans. by NTA (C)
Sol. f = 
1 eB
T 2m
=
p
  = 
194
31
1.6 10 10
2 9 10
--
-
´´
p´´
 = 
6
2.8 10 Hz ´
12. A current of 15 mA flows in the circuit as
shown in figure. The value of potential
difference between the points A and B will be
10 kW
5 kW
5 kW
10 kW
15mA
A B
(A) 50V (B) 75V
(C) 150V (D) 275V
Official Ans. by NTA (D)
Sol.
10 kW
5 kW
5 kW
10 kW
A B
i=15mA
i=15mA
i
1
1
5
i 15mA 5mA
105
= ´=
+
V
A
 – 5i –10i
1
 – 10i = V
B
V
A
 – V
B
 = 75 + 50 + 150 = 275 V
13. The length of a seconds pendulum at a height
h = 2R from earth surface will be:
(Given: R=Radius of earth and acceleration due
to gravity at the surface of earth g = p
2
m/s
–2
)
(A)
2
m
9
(B) 
4
m
9
(C)
8
m
9
(D) 
1
m
9
Official Ans. by NTA (D)
Sol.
L
T2
g
=p
,  
2
2
GMg
g'
9R 99
p
= ==
2 = 
2
L
29 p´
p
Þ  1 = 
3
L p´
p
  Þ  
1
Lm
9
=
14. Sound travels in a mixture of two moles of
helium and n moles of hydrogen. If rms speed
of gas molecules in the mixture is 
2
 times
the speed of sound, then the value of n will be
(A) 1 (B) 2
(C) 3 (D) 4
Official Ans. by NTA (B)
Sol. v
s
 = 
RT
M
g
v
rms
 = 
3RT
M
s
rms
v
v3
g
=
 = 
1
2
  Þ  
1
32
g
=
  Þ  g = 
3
2
g = 
mix.
2
1
f
+
f
mix.
 = 
2 3 n5
n2
´ +´
+
 = 
6 n5
(n 2)
+´
+
g = 
2(n 2)
1
6 n5
+
+
+´
 = 
6 5n 2n4
6 5n
+ ++
+
g = 
7n 10
6 5n
+
+
 = 
3
2
14n + 20 = 18 + 15n
n = 2
Page 5


1
Final JEE-Main Exam July, 2022/25-07-2022/Evening Session
SECTION-A
1. In AM modulation, a signal is modulated on a
carrier wave such that maximum and
minimum amplitude are found to be 6V and
2V respectively. The modulation index is
(A) 100% (B) 80%
(C) 60% (D) 50%
Official Ans. by NTA (D)
Sol. modulation index = 
max min.
max min .
VV
100%
VV
-
´
+
= 
62
100%
62
-
´
+
 = 50%
2. The electric current in a circular coil of 2 turns
produces a magnetic induction B
1
 at its centre.
The coil is unwound and is rewound into a
circular coil of 5 turns and the same current
produces a magnetic induction B
2
 at its centre.
The ratio of 
2
1
B
B
 is :
(A) 
5
2
(B) 
25
4
(C) 
5
4
(D) 
25
2
Official Ans. by NTA (B)
Sol. B = 
0
Ni
2R
m
B
1
 = 
10
1
Ni
2R
m
For N
2
 = 5
Radius of coil = R
2
 = 
11
2
NR
N
´
B
2
 = 
20
2
Ni
R
m
22 12 2
11 21 1
B NR NN
·
B NR NN
= =´
;     
2
1
B 25
B4
=
3. A drop of liquid of density r is floating half
immersed in a liquid of density s and surface
tension 7.5 × 10
–4
 Ncm
–1
. The radius of drop
in cm will be : (Take : g = 10 m/s
2
)
(A) 
15
2r-s
(B) 
15
r-s
(C) 
3
2 r-s
(D) 
3
202r-s
Official Ans. by NTA (A)
Sol.
mg
F +F
b T
Boyant force + surace tension = mg
V
g 2 RT Vg
2
s + p =r
33
(2)44
2RT · Rg;VR
233
r-s éù
p = p =p
êú
ëû
21
3
3T 3 7.5 10 N m
RR
(2)g (2)10
--
´´-
= Þ=
r-s r-s ´
R = 
3
20(2) r-s
 m = 
15
2r-s
 cm
FINAL JEE–MAIN EXAMINATION – JULY , 2022
(Held On Monday 25
th
 July, 2022)    TIME : 3 : 00 PM   to  6 : 00 PM
PHYSICS TEST PAPER WITH SOLUTION
2
Final JEE-Main Exam July, 2022/25-07-2022/Evening Session
4. Two billiard balls of mass 0.05 kg each moving
in opposite directions with 10 ms
–1
 collide and
rebound with the same speed. If the time
duration of contact is t = 0.005 s, then what is
the force exerted on the ball due to each other?
(A) 100 N (B) 200 N
(C) 300 N (D) 400 N
Official Ans. by NTA (B)
Sol.
m = 0.05 kg m = 0.05 kg
10 m/s
10 m/s
10 m/s
10 m/s
Change in momentum of any one ball
P 2 0.05 10 D=´´
r
| P| D
r
 = 1
av
| P|
F
t
D
=
D
r
r
F
av.
 = 200 N
5. For a free body diagram shown in the figure,
the four forces are applied in the 'x' and 'y'
directions. What additional force must be
applied and at what angle with positive x-axis
so that the net acceleration of body is zero?
5N 6N
7N
8N
x
y
(A)
2
N, 45° (B) 
2
N, 135°
(C)
2
N
3
, 30° (D) 2 N, 45°
Official Ans. by NTA (A)
Sol. Let addition force required is = 
F
r
ˆ ˆ ˆˆ
F 5i 6i 7 j 8j 0 +- + -=
r
ˆˆ
F i j,|F|2 =+=
rr
Angle with x-axis:  tan q = 
y component 1
x component 1
=
q = 45°
6. Capacitance of an isolated conducting sphere
of radius R
1
 becomes n times when it is
enclosed by a concentric conducting sphere
of radius R
2
 connected to earth. The ratio of
their radii 
2
1
R
R
æö
ç÷
èø
 is:
(A)
n
n1 -
(B) 
2n
2n1 +
(C)
n1
n
+
(D) 
2n1
n
+
Official Ans. by NTA (A)
Sol. Capacitance of isolated Conducting sphere
= 4pe
0
R
1
By enclosing inside another sphere of radius
R
2
, new capacitance = 
0 12
21
4 RR
(R R)
pe
-
        Given:
0 12
01
21
4 RR
n4R
(R R)
pe
= ´ pe
-
Þ 
2
21
R
n
(R R)
=
-
 Þ 
2
1
2
1
R
R
n
R
1
R
=
æö
-
ç÷
èø
Þ 
22
11
RR
nn
RR
=-
 Þ 
2
1
Rn
R (n 1)
=
-
3
Final JEE-Main Exam July, 2022/25-07-2022/Evening Session
7. The ratio of wavelengths of proton and
deuteron accelerated by potential V
p
 and V
d
 is
1:2
. Then, the ratio of V
p
 to V
d
 will be
(A) 1 : 1 (B) 2:1
(C) 2 : 1 (D) 4 : 1
Official Ans. by NTA (D)
Sol. Kinetic energy gained by a charged particle
accelerated by a potential V is qV
KE = qV
Þ  
2
p
qV
2m
=  Þ  p 2mqV =
p = 
h
l
,  thus 
h
2mqV
l=
now 
p
dd
d pp
mV
mV
l
=
l
Þ  
d
p
1 2V
V 2
=
Þ  
p
d
V
4
V
=
8. For an object placed at a distance 2.4 m from a
lens, a sharp focused image is observed on a
screen placed at a distance 12 cm from the lens.
A glass plate of refractive index 1.5 and
thickness 1 cm is introduced between lens and
screen such that the glass plate plane faces
parallel to the screen. By what distance should
the object be shifted so that a sharp focused
image is observed again on the screen?
(A) 0.8 m (B) 3.2 m
(C) 1.2 m (D) 5.6 m
Official Ans. by NTA (B)
Sol.
//////////////////
Screen
O
1
2.4m 12 cm
Applying lens formula
1 11
0.12 2.4 f
+=
  Þ  
1 210
f 24
=
Upon putting the glass slab, shift of image is
11
x t 1 cm
3
æö
D= -=
ç÷
m
èø
Now v = 
1 35
12 cm
33
-=
Again apply lens formula
1 1 1 210
0.12 u f 24
+ ==
Solving u = – 5.6 m
Thus shift of object is
5.6 – 2.4 = 3.2 m
9. Light wave traveling in air along x-direction
is given by E
y
 = 540 sin p × 10
4
(x – ct) Vm
–1
.
Then, the peak value of magnetic field of wave
will be (Given c = 3 × 10
8
 ms
–1
)
(A) 18 × 10
–7
 T (B) 54 × 10
–7
 T
(C) 54 × 10
–8
 T (D) 18 × 10
–8
 T
Official Ans. by NTA (A)
Sol. E
y
 = 540 sin p × 10
4
(x – ct) Vm
–1
E
0
 = 540 Vm
–1
B
0
 = 
0
E
C
 = 
7
8
540
18 10 T
3 10
-
=´
´
10. When you walk through a metal detector
carrying a metal object in your pocket, it raises
an alarm. This phenomenon works on
(A) Electromagnetic induction
(B) Resonance in ac circuits
(C) Mutual induction in ac circuits
(D) interference of electromagnetic waves
Official Ans. by NTA (B)
)
Sol. Metal detector works on the principle of
transmitting an electromagnetic signal and
analyses a return signal from the target. So it
works on the principle of resonance in AC
circuit.
4
Final JEE-Main Exam July, 2022/25-07-2022/Evening Session
11. An electron with energy 0.1 keV moves at
right angle to the earth's magnetic field of
1 × 10
–4
 
 
Wbm
–2
. The frequency of revolution
of the electron will be
(Take mass of electron = 9.0 × 10
–31
 kg)
(A) 1.6 × 10
5
 Hz (B) 5.6 × 10
5
 Hz
(C) 2.8 × 10
6
 Hz (D) 1.8 × 10
6
 Hz
Official Ans. by NTA (C)
Sol. f = 
1 eB
T 2m
=
p
  = 
194
31
1.6 10 10
2 9 10
--
-
´´
p´´
 = 
6
2.8 10 Hz ´
12. A current of 15 mA flows in the circuit as
shown in figure. The value of potential
difference between the points A and B will be
10 kW
5 kW
5 kW
10 kW
15mA
A B
(A) 50V (B) 75V
(C) 150V (D) 275V
Official Ans. by NTA (D)
Sol.
10 kW
5 kW
5 kW
10 kW
A B
i=15mA
i=15mA
i
1
1
5
i 15mA 5mA
105
= ´=
+
V
A
 – 5i –10i
1
 – 10i = V
B
V
A
 – V
B
 = 75 + 50 + 150 = 275 V
13. The length of a seconds pendulum at a height
h = 2R from earth surface will be:
(Given: R=Radius of earth and acceleration due
to gravity at the surface of earth g = p
2
m/s
–2
)
(A)
2
m
9
(B) 
4
m
9
(C)
8
m
9
(D) 
1
m
9
Official Ans. by NTA (D)
Sol.
L
T2
g
=p
,  
2
2
GMg
g'
9R 99
p
= ==
2 = 
2
L
29 p´
p
Þ  1 = 
3
L p´
p
  Þ  
1
Lm
9
=
14. Sound travels in a mixture of two moles of
helium and n moles of hydrogen. If rms speed
of gas molecules in the mixture is 
2
 times
the speed of sound, then the value of n will be
(A) 1 (B) 2
(C) 3 (D) 4
Official Ans. by NTA (B)
Sol. v
s
 = 
RT
M
g
v
rms
 = 
3RT
M
s
rms
v
v3
g
=
 = 
1
2
  Þ  
1
32
g
=
  Þ  g = 
3
2
g = 
mix.
2
1
f
+
f
mix.
 = 
2 3 n5
n2
´ +´
+
 = 
6 n5
(n 2)
+´
+
g = 
2(n 2)
1
6 n5
+
+
+´
 = 
6 5n 2n4
6 5n
+ ++
+
g = 
7n 10
6 5n
+
+
 = 
3
2
14n + 20 = 18 + 15n
n = 2
5
Final JEE-Main Exam July, 2022/25-07-2022/Evening Session
15. Let h
1
 is the efficiency of an engine at
T
1
 = 447°C and T
2
 = 147°C while h
2
 is the
efficiency at T
1
 = 947°C and T
2
 = 47°C. The
ratio 
1
2
h
h
 will be :
(A) 0.41 (B) 0.56
(C) 0.73 (D) 0.70
Official Ans. by NTA (B)
Sol. Efficiency h = 
L
H
T
1
T
-
1
147 273
1
447 273
+
h =-
+
 = 
420
1
720
-
1
300
720
h=
2
47 273 320
11
947 273 1220
+
h =- =-
+
2
900
1220
h=
1
2
300 1220 122
720 900 72 3
h
= ´=
h´
1
2
0.56
h
=
h
16. An object is taken to a height above the surface
of earth at a distance 
5
4
R from the centre of
the earth. Where radius of earth, R = 6400 km.
The percentage decrease in the weight of the
object will be
(A) 36% (B) 50%
(C) 64% (D) 25%
Official Ans. by NTA (A)
Sol.
R/4
5R/4
g
eff
 = 
2
g
h
1
R
æö
+
ç÷
èø
;  g
eff
 = 
2
g 16g
25
1
1
4
=
æö
+
ç÷
èø
change = 
eff
gg
100
g
-
´ = 
16
1
25
100
1
-
´
      = 
9
100
25
-
´
 = – 36%
Hence % decrease in the weight = 36%
17. A bag of sand of mass 9.8 kg is suspended by
a rope. A bullet of 200 g travelling with speed
10 ms
–1
 gets embedded in it, then loss of kinetic
energy will be
(A) 4.9 J (B) 9.8 J
(C) 14.7 (D) 19.6 J
Official Ans. by NTA (B)
Sol. P
i
 = P
f
 (no any external force)
0.2 × 10 = 10 × v
v = 0.2 m/sec
Loss in K.E. = ()
2
11
(0.2) 10 10 0.2
22
2
´ ´ -´
= [ ]
1
10 (0.2) 10 0.2
2
´´-
= 9.8 J
18. A ball is projected from the ground with a
speed 15 ms
–1
 at an angle q with horizontal so
that its range and maximum height are equal,
then'tan q' will be equal to
(A) 
1
4
(B) 
1
2
(C) 2 (D) 4
Official Ans. by NTA (D)
Sol. R = H
2
x yy
2v vv
g 2g
´
=
v
x
 = 
y
v
4
;  u cos q = 
u sin
4
q
tan q = 4