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

 Page 1


 
    1 
FINAL JEE –MAIN EXAMINATION – JULY, 2022 
(Held On Tuesday 26
th
 July, 2022)              TIME : 3 : 00 PM  to  6 : 00 PM 
PHYSICS TEST PAPER WITH SOLUTION 
 
 
 
 
 
 
 
SECTION-A 
1. Two projectiles are thrown with same initial 
velocity making an angle of 45° and 30° with the 
horizontal respectively. The ratio of their 
respective ranges will be  
 (A) 1: 2   (B) 2 :1
 
 
 (C) 2: 3  (D) 3 :2 
 Official Ans. by NTA (C) 
  
Sol. Let projection speed is u  
 
22
12
u S i n( 9 0 º ) u s i n ( 6 0 º )
R ;R
gg
?? 
 
1
2
R 2
R 3
?  
2. In a Vernier Calipers. 10 divisions of Vernier scale 
is equal to the 9 divisions of main scale. When 
both jaws of Vernier calipers touch each other, the 
zero of the Vernier scale is shifted to the left of 
zero of the main scale and 4
th
 Vernier scale 
division exactly coincides with the main scale 
reading. One main scale division is equal to 1 mm. 
While measuring diameter of a spherical body, the 
body is held between two jaws. It is now observed 
that zero of the Vernier scale lies between 30 and 
31 divisions of main scale reading and 6
th
 Vernier 
scale division exactly. coincides with the main 
scale reading. The diameter of the spherical body 
will be : 
 (A) 3.02 cm   (B) 3.06 cm
 
 
 (C) 3.10 cm  (D) 3.20 cm 
 Official Ans. by NTA (C) 
  
Sol. 1 M.S.D = 1mm  
 9 M.S.D = 10 V.S.D 
 1 V.S.D = 0.9 M.S.D = 0.9 mm 
 L.C of vernier caliper = 1–0.9 = 0.1 mm = 0.01 cm 
 zero error =–(10–4)×0.1mm = – 0.6 mm 
 Reading = M.S.R + V.S.R – Zero error 
 = 3cm + 6 × 0.01 – [–0.06] 
 = 3 + 0.06 + 0.06 
 = 3.12 cm 
 Nearest given answer in the options is 3.10 
3. A ball of mass 0.15 kg hits the wall with its initial 
speed of 12 ms
-1
 and bounces back without 
changing its initial speed. If the force applied by 
the wall on the ball during the contact is 100 N. 
calculate the time duration of the contact of ball 
with the wall. 
 (A) 0.018 s   (B) 0.036 s
 
 
 (C) 0.009 s  (D) 0.072 s 
 Official Ans. by NTA (B) 
  
Sol. 
? ? i
ˆ
P 0.15 12 i ?? 
 
? ? f
ˆ
P 0.15 12 i ? ? ? 
 P 3.6 kg m / s ? ? ? 
 3.6 = F ? t 
 3.6 = 100 ? t 
 ? t = 0.036 sec  
4. A body of mass 8 kg and another of mass 2 kg are 
moving with equal kinetic energy. The ratio of 
their respective momenta will be : 
 (A)  1:1 (B) 2:1
 
(C) 1:4  (D) 4:1 
 Official Ans. by NTA (B) 
  
Sol. K.E = 
2
P
2m
 
 
? ?
2
1
1
P
K
28
? ; 
? ?
2
2
2
P
K
22
? 
 K
1 
= K
2
 
 So, 
 4P
2
2
 = P
1
2 
 
1
2
P
2
P
? 
5. Two uniformly charged spherical conductors A 
and B of radii 5 mm and 10 mm are separated by a 
distance of 2 cm. If the spheres are connected by a 
conducting wire, then in equilibrium condition, the 
ratio of the magnitudes of the electric fields at 
the surface of the sphere A and B will be : 
 (A)  1 : 2 (B) 2 : 1
 
(C) 1 : 1  (D) 1 : 4 
 Official Ans. by NTA (B) 
  
Page 2


 
    1 
FINAL JEE –MAIN EXAMINATION – JULY, 2022 
(Held On Tuesday 26
th
 July, 2022)              TIME : 3 : 00 PM  to  6 : 00 PM 
PHYSICS TEST PAPER WITH SOLUTION 
 
 
 
 
 
 
 
SECTION-A 
1. Two projectiles are thrown with same initial 
velocity making an angle of 45° and 30° with the 
horizontal respectively. The ratio of their 
respective ranges will be  
 (A) 1: 2   (B) 2 :1
 
 
 (C) 2: 3  (D) 3 :2 
 Official Ans. by NTA (C) 
  
Sol. Let projection speed is u  
 
22
12
u S i n( 9 0 º ) u s i n ( 6 0 º )
R ;R
gg
?? 
 
1
2
R 2
R 3
?  
2. In a Vernier Calipers. 10 divisions of Vernier scale 
is equal to the 9 divisions of main scale. When 
both jaws of Vernier calipers touch each other, the 
zero of the Vernier scale is shifted to the left of 
zero of the main scale and 4
th
 Vernier scale 
division exactly coincides with the main scale 
reading. One main scale division is equal to 1 mm. 
While measuring diameter of a spherical body, the 
body is held between two jaws. It is now observed 
that zero of the Vernier scale lies between 30 and 
31 divisions of main scale reading and 6
th
 Vernier 
scale division exactly. coincides with the main 
scale reading. The diameter of the spherical body 
will be : 
 (A) 3.02 cm   (B) 3.06 cm
 
 
 (C) 3.10 cm  (D) 3.20 cm 
 Official Ans. by NTA (C) 
  
Sol. 1 M.S.D = 1mm  
 9 M.S.D = 10 V.S.D 
 1 V.S.D = 0.9 M.S.D = 0.9 mm 
 L.C of vernier caliper = 1–0.9 = 0.1 mm = 0.01 cm 
 zero error =–(10–4)×0.1mm = – 0.6 mm 
 Reading = M.S.R + V.S.R – Zero error 
 = 3cm + 6 × 0.01 – [–0.06] 
 = 3 + 0.06 + 0.06 
 = 3.12 cm 
 Nearest given answer in the options is 3.10 
3. A ball of mass 0.15 kg hits the wall with its initial 
speed of 12 ms
-1
 and bounces back without 
changing its initial speed. If the force applied by 
the wall on the ball during the contact is 100 N. 
calculate the time duration of the contact of ball 
with the wall. 
 (A) 0.018 s   (B) 0.036 s
 
 
 (C) 0.009 s  (D) 0.072 s 
 Official Ans. by NTA (B) 
  
Sol. 
? ? i
ˆ
P 0.15 12 i ?? 
 
? ? f
ˆ
P 0.15 12 i ? ? ? 
 P 3.6 kg m / s ? ? ? 
 3.6 = F ? t 
 3.6 = 100 ? t 
 ? t = 0.036 sec  
4. A body of mass 8 kg and another of mass 2 kg are 
moving with equal kinetic energy. The ratio of 
their respective momenta will be : 
 (A)  1:1 (B) 2:1
 
(C) 1:4  (D) 4:1 
 Official Ans. by NTA (B) 
  
Sol. K.E = 
2
P
2m
 
 
? ?
2
1
1
P
K
28
? ; 
? ?
2
2
2
P
K
22
? 
 K
1 
= K
2
 
 So, 
 4P
2
2
 = P
1
2 
 
1
2
P
2
P
? 
5. Two uniformly charged spherical conductors A 
and B of radii 5 mm and 10 mm are separated by a 
distance of 2 cm. If the spheres are connected by a 
conducting wire, then in equilibrium condition, the 
ratio of the magnitudes of the electric fields at 
the surface of the sphere A and B will be : 
 (A)  1 : 2 (B) 2 : 1
 
(C) 1 : 1  (D) 1 : 4 
 Official Ans. by NTA (B) 
  
 
2 
 
Sol. V
A
 = V
B 
 
AB
AB
KQ KQ
RR
? 
 
AA
BB
QR 1
Q R 2
?? 
 
AB
AB22
AB
KQ KQ
E ;E
RR
?? 
 
2
A A B B
2
B B A A
E Q R R 2
E Q R R 1
? ? ? ?  
6. The oscillating magnetic field in a plane 
electromagnetic wave is given by B
y
 = 5 × l0
-6
 sin 
l000 ? (5x - 4 × 10
8
 t)T. The amplitude of electric 
field will be : 
 (A) 15 × 10
2
 Vm
–1
  (B) 5 × 10
–6
 Vm
–1
 
 
 (C) 16 × 10
12
 Vm
–1
 (D) 4 × 10
2
 Vm
–1
 
 Official Ans. by NTA (D) 
  
Sol. B
0
=5×10
–6 
 v = Speed of wave = 
8
7
4 10
8 10
5
?
?? 
w
v
k
??
??
??
??
 
 E
0
 = vB
0
 = 40×10
1 
 
= 4 × 10
2 
V/m  
7. Light travels in two media M
1
 and M
2
 with speeds 
1.5 × 10
8
 ms
–1
 and 2.0 × 10
8
 ms
–1
 respectively. The 
critical angle between them is: 
 (A) 
1
3
tan
7
?
??
??
??
  (B) 
1
2
tan
3
?
??
??
??
  
 
 (C) 
1
3
cos
4
?
??
??
??
 (D) 
1
2
sin
3
?
??
??
??
 
 Official Ans. by NTA (A) 
  
Sol.
 
 
 
c
v
n
? 
 n
d
 sin i
c
 = n
r
 sin90º 
 sin i
c
 = 
d r
dr
v n
nv
? 
 sin i
c
 
8
8
1.5 10 1.5
2 10 2
?
??
?
 
 
c
3
sin i
4
? 
 tan i
c
 = 
22
33
7
43
?
?
 
 i
c
 = 
1
3
tan
7
?
??
??
??
  
8. A body is projected vertically upwards from the 
surface of earth with a velocity equal to one third 
of escape velocity. The maximum height attained 
by the body will be: 
(Take radius of earth = 6400 km and g=10 ms
–2
 ) 
 (A)  800 km  (B) 1600 km
 
 
 (C) 2133 km  (D) 4800 km  
 Official Ans. by NTA (A) 
  
Sol.  
 
e
2Gm
v
R
? 
 
2
e
V GMm 1 GMm
m
R 2 9 R h
?
? ? ?
?
 
 
2
e
V GM GM
R h R 18
??
?
 
 
GM GM GM
R h R 9R
??
?
 
 
GM 8GM
R h 9R
?
?
 
 
18
R h 9R
?
?
 
 9R = 8R + 8h 
 
R 6400
h 800km
88
? ? ? 
9. The maximum and minimum voltage of an 
amplitude modulated signal are 60 V and 20 V 
respectively. The percentage modulation index will 
be : 
 (A) 0.5%  (B) 50%
 
(C) 2%  (D) 30% 
 Official Ans. by NTA (B) 
  
Page 3


 
    1 
FINAL JEE –MAIN EXAMINATION – JULY, 2022 
(Held On Tuesday 26
th
 July, 2022)              TIME : 3 : 00 PM  to  6 : 00 PM 
PHYSICS TEST PAPER WITH SOLUTION 
 
 
 
 
 
 
 
SECTION-A 
1. Two projectiles are thrown with same initial 
velocity making an angle of 45° and 30° with the 
horizontal respectively. The ratio of their 
respective ranges will be  
 (A) 1: 2   (B) 2 :1
 
 
 (C) 2: 3  (D) 3 :2 
 Official Ans. by NTA (C) 
  
Sol. Let projection speed is u  
 
22
12
u S i n( 9 0 º ) u s i n ( 6 0 º )
R ;R
gg
?? 
 
1
2
R 2
R 3
?  
2. In a Vernier Calipers. 10 divisions of Vernier scale 
is equal to the 9 divisions of main scale. When 
both jaws of Vernier calipers touch each other, the 
zero of the Vernier scale is shifted to the left of 
zero of the main scale and 4
th
 Vernier scale 
division exactly coincides with the main scale 
reading. One main scale division is equal to 1 mm. 
While measuring diameter of a spherical body, the 
body is held between two jaws. It is now observed 
that zero of the Vernier scale lies between 30 and 
31 divisions of main scale reading and 6
th
 Vernier 
scale division exactly. coincides with the main 
scale reading. The diameter of the spherical body 
will be : 
 (A) 3.02 cm   (B) 3.06 cm
 
 
 (C) 3.10 cm  (D) 3.20 cm 
 Official Ans. by NTA (C) 
  
Sol. 1 M.S.D = 1mm  
 9 M.S.D = 10 V.S.D 
 1 V.S.D = 0.9 M.S.D = 0.9 mm 
 L.C of vernier caliper = 1–0.9 = 0.1 mm = 0.01 cm 
 zero error =–(10–4)×0.1mm = – 0.6 mm 
 Reading = M.S.R + V.S.R – Zero error 
 = 3cm + 6 × 0.01 – [–0.06] 
 = 3 + 0.06 + 0.06 
 = 3.12 cm 
 Nearest given answer in the options is 3.10 
3. A ball of mass 0.15 kg hits the wall with its initial 
speed of 12 ms
-1
 and bounces back without 
changing its initial speed. If the force applied by 
the wall on the ball during the contact is 100 N. 
calculate the time duration of the contact of ball 
with the wall. 
 (A) 0.018 s   (B) 0.036 s
 
 
 (C) 0.009 s  (D) 0.072 s 
 Official Ans. by NTA (B) 
  
Sol. 
? ? i
ˆ
P 0.15 12 i ?? 
 
? ? f
ˆ
P 0.15 12 i ? ? ? 
 P 3.6 kg m / s ? ? ? 
 3.6 = F ? t 
 3.6 = 100 ? t 
 ? t = 0.036 sec  
4. A body of mass 8 kg and another of mass 2 kg are 
moving with equal kinetic energy. The ratio of 
their respective momenta will be : 
 (A)  1:1 (B) 2:1
 
(C) 1:4  (D) 4:1 
 Official Ans. by NTA (B) 
  
Sol. K.E = 
2
P
2m
 
 
? ?
2
1
1
P
K
28
? ; 
? ?
2
2
2
P
K
22
? 
 K
1 
= K
2
 
 So, 
 4P
2
2
 = P
1
2 
 
1
2
P
2
P
? 
5. Two uniformly charged spherical conductors A 
and B of radii 5 mm and 10 mm are separated by a 
distance of 2 cm. If the spheres are connected by a 
conducting wire, then in equilibrium condition, the 
ratio of the magnitudes of the electric fields at 
the surface of the sphere A and B will be : 
 (A)  1 : 2 (B) 2 : 1
 
(C) 1 : 1  (D) 1 : 4 
 Official Ans. by NTA (B) 
  
 
2 
 
Sol. V
A
 = V
B 
 
AB
AB
KQ KQ
RR
? 
 
AA
BB
QR 1
Q R 2
?? 
 
AB
AB22
AB
KQ KQ
E ;E
RR
?? 
 
2
A A B B
2
B B A A
E Q R R 2
E Q R R 1
? ? ? ?  
6. The oscillating magnetic field in a plane 
electromagnetic wave is given by B
y
 = 5 × l0
-6
 sin 
l000 ? (5x - 4 × 10
8
 t)T. The amplitude of electric 
field will be : 
 (A) 15 × 10
2
 Vm
–1
  (B) 5 × 10
–6
 Vm
–1
 
 
 (C) 16 × 10
12
 Vm
–1
 (D) 4 × 10
2
 Vm
–1
 
 Official Ans. by NTA (D) 
  
Sol. B
0
=5×10
–6 
 v = Speed of wave = 
8
7
4 10
8 10
5
?
?? 
w
v
k
??
??
??
??
 
 E
0
 = vB
0
 = 40×10
1 
 
= 4 × 10
2 
V/m  
7. Light travels in two media M
1
 and M
2
 with speeds 
1.5 × 10
8
 ms
–1
 and 2.0 × 10
8
 ms
–1
 respectively. The 
critical angle between them is: 
 (A) 
1
3
tan
7
?
??
??
??
  (B) 
1
2
tan
3
?
??
??
??
  
 
 (C) 
1
3
cos
4
?
??
??
??
 (D) 
1
2
sin
3
?
??
??
??
 
 Official Ans. by NTA (A) 
  
Sol.
 
 
 
c
v
n
? 
 n
d
 sin i
c
 = n
r
 sin90º 
 sin i
c
 = 
d r
dr
v n
nv
? 
 sin i
c
 
8
8
1.5 10 1.5
2 10 2
?
??
?
 
 
c
3
sin i
4
? 
 tan i
c
 = 
22
33
7
43
?
?
 
 i
c
 = 
1
3
tan
7
?
??
??
??
  
8. A body is projected vertically upwards from the 
surface of earth with a velocity equal to one third 
of escape velocity. The maximum height attained 
by the body will be: 
(Take radius of earth = 6400 km and g=10 ms
–2
 ) 
 (A)  800 km  (B) 1600 km
 
 
 (C) 2133 km  (D) 4800 km  
 Official Ans. by NTA (A) 
  
Sol.  
 
e
2Gm
v
R
? 
 
2
e
V GMm 1 GMm
m
R 2 9 R h
?
? ? ?
?
 
 
2
e
V GM GM
R h R 18
??
?
 
 
GM GM GM
R h R 9R
??
?
 
 
GM 8GM
R h 9R
?
?
 
 
18
R h 9R
?
?
 
 9R = 8R + 8h 
 
R 6400
h 800km
88
? ? ? 
9. The maximum and minimum voltage of an 
amplitude modulated signal are 60 V and 20 V 
respectively. The percentage modulation index will 
be : 
 (A) 0.5%  (B) 50%
 
(C) 2%  (D) 30% 
 Official Ans. by NTA (B) 
  
 
 
3 
 
Sol. V
max
  = 60 
 V
min
 = 20 
 % modulation =  
 
max min
max min
VV 60 20 40
100 100 100
V V 60 20 80
?? ? ? ? ? ? ?
??
?? ? ? ? ?
??
? ? ? ?
??
 
? 50%  
10. A nucleus of mass M at rest splits into two parts 
having masses 
M'
3
and 
? ?
2M'
M' M
3
? . The ratio 
of de Broglie wavelength of two parts will be : 
 (A)  1 : 2  (B) 2 : 1
 
 
 (C) 1 : 1   (D) 2 : 3 
 Official Ans. by NTA (C) 
  
Sol.  
  
 
12
PP ? 
 Here P is momentum 
 So 
h
P
?? 
 Hence both will have same de broglie wavelength. 
11. An ice cube of dimensions 60 cm × 50 cm × 20 cm 
is placed in an insulation box of wall thickness  
1 cm. The box keeping the ice cube at 0°C of 
temperature is brought to a room of temperature 
40°C. The rate of melting of ice is approximately: 
(Latent heat of fusion of ice is 3.4 × 10
5
 J kg
–1
 and 
thermal conducting of insulation wall is  
0.05 Wm
–1
ºC
–1
) 
 (A)  61×10
–1
 kg s
–1
 (B) 61×10
–5
 kg s
–1
  
 (C) 208 kg s
–1
 (D) 30×10
–5
 kg s
–1
 
 Official Ans. by NTA (B) 
  
Sol.
 
 
 
dQ KA T
dt
?
? 
 A = 2 (0.6 × 0.5 + 0.5 × 0.2 + 0.2 × 0.6) 
 = 2(0.3 + 0.1 + 0.12) 
 = 2(0.4 + 0.12) 
 = 2(0.52) 
 = 1.04 m
2 
 
R
th
 = 
22
1 10 10
KA 0.05 1.04 0.052
??
?
??
?
 
 
2
2
th
dQ T 40 0.052
2.08 10 J / s
dt R 10
?
??
? ? ? ? 
 2.08 × 10
2
 = m × 3.4 × 10
5 
 
m = 
3
3
2.08
0.61 10
3.4 10
?
??
?
 kg/s 
 = 61 × 10
–5
 Kg/s 
12. A gas has n degrees of freedom. The ratio of 
specific heat of gas at constant volume to the 
specific heat of gas at constant pressure will be : 
 (A) 
n
n2 ?
  (B) 
n2
n
?
  
 (C) 
n
2n 2 ?
   (D) 
n
n2 ?
  
 Official Ans. by NTA (A) 
  
Sol. 
v
nR
C
2
? 
? ?
p
n 2 R
C
2
?
? 
 
v
p
C n
C n 2
?
?
  
13. A transverse wave is represented by y = 2sin 
? ? t kx ?? cm. The value of wavelength (in cm) for 
which the wave velocity becomes equal to the 
maximum particle velocity, will be ; 
 (A) 4 ? (B) 2 ?  
 (C) ?  (D) 2 
 Official Ans. by NTA (A) 
  
Sol. y = 2 sin ( ?t – kx) 
 Maximum particle velocity = A ? ?
? Wave velocity = 
k
?
 
 
k
?
= A ? ?
? k = 
1
A
 = 
2 ?
?
 
 ? = 2 ?A 
  = 4 ? cm ?
Page 4


 
    1 
FINAL JEE –MAIN EXAMINATION – JULY, 2022 
(Held On Tuesday 26
th
 July, 2022)              TIME : 3 : 00 PM  to  6 : 00 PM 
PHYSICS TEST PAPER WITH SOLUTION 
 
 
 
 
 
 
 
SECTION-A 
1. Two projectiles are thrown with same initial 
velocity making an angle of 45° and 30° with the 
horizontal respectively. The ratio of their 
respective ranges will be  
 (A) 1: 2   (B) 2 :1
 
 
 (C) 2: 3  (D) 3 :2 
 Official Ans. by NTA (C) 
  
Sol. Let projection speed is u  
 
22
12
u S i n( 9 0 º ) u s i n ( 6 0 º )
R ;R
gg
?? 
 
1
2
R 2
R 3
?  
2. In a Vernier Calipers. 10 divisions of Vernier scale 
is equal to the 9 divisions of main scale. When 
both jaws of Vernier calipers touch each other, the 
zero of the Vernier scale is shifted to the left of 
zero of the main scale and 4
th
 Vernier scale 
division exactly coincides with the main scale 
reading. One main scale division is equal to 1 mm. 
While measuring diameter of a spherical body, the 
body is held between two jaws. It is now observed 
that zero of the Vernier scale lies between 30 and 
31 divisions of main scale reading and 6
th
 Vernier 
scale division exactly. coincides with the main 
scale reading. The diameter of the spherical body 
will be : 
 (A) 3.02 cm   (B) 3.06 cm
 
 
 (C) 3.10 cm  (D) 3.20 cm 
 Official Ans. by NTA (C) 
  
Sol. 1 M.S.D = 1mm  
 9 M.S.D = 10 V.S.D 
 1 V.S.D = 0.9 M.S.D = 0.9 mm 
 L.C of vernier caliper = 1–0.9 = 0.1 mm = 0.01 cm 
 zero error =–(10–4)×0.1mm = – 0.6 mm 
 Reading = M.S.R + V.S.R – Zero error 
 = 3cm + 6 × 0.01 – [–0.06] 
 = 3 + 0.06 + 0.06 
 = 3.12 cm 
 Nearest given answer in the options is 3.10 
3. A ball of mass 0.15 kg hits the wall with its initial 
speed of 12 ms
-1
 and bounces back without 
changing its initial speed. If the force applied by 
the wall on the ball during the contact is 100 N. 
calculate the time duration of the contact of ball 
with the wall. 
 (A) 0.018 s   (B) 0.036 s
 
 
 (C) 0.009 s  (D) 0.072 s 
 Official Ans. by NTA (B) 
  
Sol. 
? ? i
ˆ
P 0.15 12 i ?? 
 
? ? f
ˆ
P 0.15 12 i ? ? ? 
 P 3.6 kg m / s ? ? ? 
 3.6 = F ? t 
 3.6 = 100 ? t 
 ? t = 0.036 sec  
4. A body of mass 8 kg and another of mass 2 kg are 
moving with equal kinetic energy. The ratio of 
their respective momenta will be : 
 (A)  1:1 (B) 2:1
 
(C) 1:4  (D) 4:1 
 Official Ans. by NTA (B) 
  
Sol. K.E = 
2
P
2m
 
 
? ?
2
1
1
P
K
28
? ; 
? ?
2
2
2
P
K
22
? 
 K
1 
= K
2
 
 So, 
 4P
2
2
 = P
1
2 
 
1
2
P
2
P
? 
5. Two uniformly charged spherical conductors A 
and B of radii 5 mm and 10 mm are separated by a 
distance of 2 cm. If the spheres are connected by a 
conducting wire, then in equilibrium condition, the 
ratio of the magnitudes of the electric fields at 
the surface of the sphere A and B will be : 
 (A)  1 : 2 (B) 2 : 1
 
(C) 1 : 1  (D) 1 : 4 
 Official Ans. by NTA (B) 
  
 
2 
 
Sol. V
A
 = V
B 
 
AB
AB
KQ KQ
RR
? 
 
AA
BB
QR 1
Q R 2
?? 
 
AB
AB22
AB
KQ KQ
E ;E
RR
?? 
 
2
A A B B
2
B B A A
E Q R R 2
E Q R R 1
? ? ? ?  
6. The oscillating magnetic field in a plane 
electromagnetic wave is given by B
y
 = 5 × l0
-6
 sin 
l000 ? (5x - 4 × 10
8
 t)T. The amplitude of electric 
field will be : 
 (A) 15 × 10
2
 Vm
–1
  (B) 5 × 10
–6
 Vm
–1
 
 
 (C) 16 × 10
12
 Vm
–1
 (D) 4 × 10
2
 Vm
–1
 
 Official Ans. by NTA (D) 
  
Sol. B
0
=5×10
–6 
 v = Speed of wave = 
8
7
4 10
8 10
5
?
?? 
w
v
k
??
??
??
??
 
 E
0
 = vB
0
 = 40×10
1 
 
= 4 × 10
2 
V/m  
7. Light travels in two media M
1
 and M
2
 with speeds 
1.5 × 10
8
 ms
–1
 and 2.0 × 10
8
 ms
–1
 respectively. The 
critical angle between them is: 
 (A) 
1
3
tan
7
?
??
??
??
  (B) 
1
2
tan
3
?
??
??
??
  
 
 (C) 
1
3
cos
4
?
??
??
??
 (D) 
1
2
sin
3
?
??
??
??
 
 Official Ans. by NTA (A) 
  
Sol.
 
 
 
c
v
n
? 
 n
d
 sin i
c
 = n
r
 sin90º 
 sin i
c
 = 
d r
dr
v n
nv
? 
 sin i
c
 
8
8
1.5 10 1.5
2 10 2
?
??
?
 
 
c
3
sin i
4
? 
 tan i
c
 = 
22
33
7
43
?
?
 
 i
c
 = 
1
3
tan
7
?
??
??
??
  
8. A body is projected vertically upwards from the 
surface of earth with a velocity equal to one third 
of escape velocity. The maximum height attained 
by the body will be: 
(Take radius of earth = 6400 km and g=10 ms
–2
 ) 
 (A)  800 km  (B) 1600 km
 
 
 (C) 2133 km  (D) 4800 km  
 Official Ans. by NTA (A) 
  
Sol.  
 
e
2Gm
v
R
? 
 
2
e
V GMm 1 GMm
m
R 2 9 R h
?
? ? ?
?
 
 
2
e
V GM GM
R h R 18
??
?
 
 
GM GM GM
R h R 9R
??
?
 
 
GM 8GM
R h 9R
?
?
 
 
18
R h 9R
?
?
 
 9R = 8R + 8h 
 
R 6400
h 800km
88
? ? ? 
9. The maximum and minimum voltage of an 
amplitude modulated signal are 60 V and 20 V 
respectively. The percentage modulation index will 
be : 
 (A) 0.5%  (B) 50%
 
(C) 2%  (D) 30% 
 Official Ans. by NTA (B) 
  
 
 
3 
 
Sol. V
max
  = 60 
 V
min
 = 20 
 % modulation =  
 
max min
max min
VV 60 20 40
100 100 100
V V 60 20 80
?? ? ? ? ? ? ?
??
?? ? ? ? ?
??
? ? ? ?
??
 
? 50%  
10. A nucleus of mass M at rest splits into two parts 
having masses 
M'
3
and 
? ?
2M'
M' M
3
? . The ratio 
of de Broglie wavelength of two parts will be : 
 (A)  1 : 2  (B) 2 : 1
 
 
 (C) 1 : 1   (D) 2 : 3 
 Official Ans. by NTA (C) 
  
Sol.  
  
 
12
PP ? 
 Here P is momentum 
 So 
h
P
?? 
 Hence both will have same de broglie wavelength. 
11. An ice cube of dimensions 60 cm × 50 cm × 20 cm 
is placed in an insulation box of wall thickness  
1 cm. The box keeping the ice cube at 0°C of 
temperature is brought to a room of temperature 
40°C. The rate of melting of ice is approximately: 
(Latent heat of fusion of ice is 3.4 × 10
5
 J kg
–1
 and 
thermal conducting of insulation wall is  
0.05 Wm
–1
ºC
–1
) 
 (A)  61×10
–1
 kg s
–1
 (B) 61×10
–5
 kg s
–1
  
 (C) 208 kg s
–1
 (D) 30×10
–5
 kg s
–1
 
 Official Ans. by NTA (B) 
  
Sol.
 
 
 
dQ KA T
dt
?
? 
 A = 2 (0.6 × 0.5 + 0.5 × 0.2 + 0.2 × 0.6) 
 = 2(0.3 + 0.1 + 0.12) 
 = 2(0.4 + 0.12) 
 = 2(0.52) 
 = 1.04 m
2 
 
R
th
 = 
22
1 10 10
KA 0.05 1.04 0.052
??
?
??
?
 
 
2
2
th
dQ T 40 0.052
2.08 10 J / s
dt R 10
?
??
? ? ? ? 
 2.08 × 10
2
 = m × 3.4 × 10
5 
 
m = 
3
3
2.08
0.61 10
3.4 10
?
??
?
 kg/s 
 = 61 × 10
–5
 Kg/s 
12. A gas has n degrees of freedom. The ratio of 
specific heat of gas at constant volume to the 
specific heat of gas at constant pressure will be : 
 (A) 
n
n2 ?
  (B) 
n2
n
?
  
 (C) 
n
2n 2 ?
   (D) 
n
n2 ?
  
 Official Ans. by NTA (A) 
  
Sol. 
v
nR
C
2
? 
? ?
p
n 2 R
C
2
?
? 
 
v
p
C n
C n 2
?
?
  
13. A transverse wave is represented by y = 2sin 
? ? t kx ?? cm. The value of wavelength (in cm) for 
which the wave velocity becomes equal to the 
maximum particle velocity, will be ; 
 (A) 4 ? (B) 2 ?  
 (C) ?  (D) 2 
 Official Ans. by NTA (A) 
  
Sol. y = 2 sin ( ?t – kx) 
 Maximum particle velocity = A ? ?
? Wave velocity = 
k
?
 
 
k
?
= A ? ?
? k = 
1
A
 = 
2 ?
?
 
 ? = 2 ?A 
  = 4 ? cm ?
 
4 
 
14.  A battery of 6 V is connected to the circuit as 
shown below. The current I drawn from the battery 
is : 
 
 (A)  1A (B) 2A  
 (C) 
6
A
11
   (D) 
4
A
3
 
 Official Ans. by NTA (A) 
  
Sol. Balanced wheat stone bridge in circuit so there is 
no current in 5 ? ?resistor so it can be removed 
from the circuit. 
 
 
eq
6 12
R2
6 12
?
??
?
 
 
6 12
2
18
?
?? 
 
eq
R6 ?? 
 
eq
V6
I 1 Amp.
R6
? ? ? 
15. A source of potential difference V is connected to 
the combination of two identical capacitors as 
shown in the figure. When key 'K’ is closed, the 
total energy stored across the combination is E
1
. 
Now key ‘K’ is opened and dielectric of dielectric 
constant 5 is introduced between the plates of the 
capacitors. The total energy stored across the 
combination is now E
2
. The ratio E
1
/E
2
 will be : 
 
 (A) 
1
10
  (B) 
2
5
  
 (C) 
5
13
  (D) 
5
26
 
 Official Ans. by NTA (C) 
  
Sol.  
 (1) Switch is closed 
  C
eq
 = 2C 
 Energy 
2
1 eq
1
E C V
2
? 
 = 
2
1
2C V
2
? 
 E
1
 = CV
2
 
 (ii) When switch is opened charge on right 
capacitor remain CV while potential on left 
capacitor remain same 
  Dielectric K = 5  
 C
’
 = KC 
 C
’
 = 5C 
 ? ?
? ?
? ?
2
2
2
CV
1
E 5C V
2 2 5C
?? 
 
22
2
5CV CV
E
2 10
?? 
Page 5


 
    1 
FINAL JEE –MAIN EXAMINATION – JULY, 2022 
(Held On Tuesday 26
th
 July, 2022)              TIME : 3 : 00 PM  to  6 : 00 PM 
PHYSICS TEST PAPER WITH SOLUTION 
 
 
 
 
 
 
 
SECTION-A 
1. Two projectiles are thrown with same initial 
velocity making an angle of 45° and 30° with the 
horizontal respectively. The ratio of their 
respective ranges will be  
 (A) 1: 2   (B) 2 :1
 
 
 (C) 2: 3  (D) 3 :2 
 Official Ans. by NTA (C) 
  
Sol. Let projection speed is u  
 
22
12
u S i n( 9 0 º ) u s i n ( 6 0 º )
R ;R
gg
?? 
 
1
2
R 2
R 3
?  
2. In a Vernier Calipers. 10 divisions of Vernier scale 
is equal to the 9 divisions of main scale. When 
both jaws of Vernier calipers touch each other, the 
zero of the Vernier scale is shifted to the left of 
zero of the main scale and 4
th
 Vernier scale 
division exactly coincides with the main scale 
reading. One main scale division is equal to 1 mm. 
While measuring diameter of a spherical body, the 
body is held between two jaws. It is now observed 
that zero of the Vernier scale lies between 30 and 
31 divisions of main scale reading and 6
th
 Vernier 
scale division exactly. coincides with the main 
scale reading. The diameter of the spherical body 
will be : 
 (A) 3.02 cm   (B) 3.06 cm
 
 
 (C) 3.10 cm  (D) 3.20 cm 
 Official Ans. by NTA (C) 
  
Sol. 1 M.S.D = 1mm  
 9 M.S.D = 10 V.S.D 
 1 V.S.D = 0.9 M.S.D = 0.9 mm 
 L.C of vernier caliper = 1–0.9 = 0.1 mm = 0.01 cm 
 zero error =–(10–4)×0.1mm = – 0.6 mm 
 Reading = M.S.R + V.S.R – Zero error 
 = 3cm + 6 × 0.01 – [–0.06] 
 = 3 + 0.06 + 0.06 
 = 3.12 cm 
 Nearest given answer in the options is 3.10 
3. A ball of mass 0.15 kg hits the wall with its initial 
speed of 12 ms
-1
 and bounces back without 
changing its initial speed. If the force applied by 
the wall on the ball during the contact is 100 N. 
calculate the time duration of the contact of ball 
with the wall. 
 (A) 0.018 s   (B) 0.036 s
 
 
 (C) 0.009 s  (D) 0.072 s 
 Official Ans. by NTA (B) 
  
Sol. 
? ? i
ˆ
P 0.15 12 i ?? 
 
? ? f
ˆ
P 0.15 12 i ? ? ? 
 P 3.6 kg m / s ? ? ? 
 3.6 = F ? t 
 3.6 = 100 ? t 
 ? t = 0.036 sec  
4. A body of mass 8 kg and another of mass 2 kg are 
moving with equal kinetic energy. The ratio of 
their respective momenta will be : 
 (A)  1:1 (B) 2:1
 
(C) 1:4  (D) 4:1 
 Official Ans. by NTA (B) 
  
Sol. K.E = 
2
P
2m
 
 
? ?
2
1
1
P
K
28
? ; 
? ?
2
2
2
P
K
22
? 
 K
1 
= K
2
 
 So, 
 4P
2
2
 = P
1
2 
 
1
2
P
2
P
? 
5. Two uniformly charged spherical conductors A 
and B of radii 5 mm and 10 mm are separated by a 
distance of 2 cm. If the spheres are connected by a 
conducting wire, then in equilibrium condition, the 
ratio of the magnitudes of the electric fields at 
the surface of the sphere A and B will be : 
 (A)  1 : 2 (B) 2 : 1
 
(C) 1 : 1  (D) 1 : 4 
 Official Ans. by NTA (B) 
  
 
2 
 
Sol. V
A
 = V
B 
 
AB
AB
KQ KQ
RR
? 
 
AA
BB
QR 1
Q R 2
?? 
 
AB
AB22
AB
KQ KQ
E ;E
RR
?? 
 
2
A A B B
2
B B A A
E Q R R 2
E Q R R 1
? ? ? ?  
6. The oscillating magnetic field in a plane 
electromagnetic wave is given by B
y
 = 5 × l0
-6
 sin 
l000 ? (5x - 4 × 10
8
 t)T. The amplitude of electric 
field will be : 
 (A) 15 × 10
2
 Vm
–1
  (B) 5 × 10
–6
 Vm
–1
 
 
 (C) 16 × 10
12
 Vm
–1
 (D) 4 × 10
2
 Vm
–1
 
 Official Ans. by NTA (D) 
  
Sol. B
0
=5×10
–6 
 v = Speed of wave = 
8
7
4 10
8 10
5
?
?? 
w
v
k
??
??
??
??
 
 E
0
 = vB
0
 = 40×10
1 
 
= 4 × 10
2 
V/m  
7. Light travels in two media M
1
 and M
2
 with speeds 
1.5 × 10
8
 ms
–1
 and 2.0 × 10
8
 ms
–1
 respectively. The 
critical angle between them is: 
 (A) 
1
3
tan
7
?
??
??
??
  (B) 
1
2
tan
3
?
??
??
??
  
 
 (C) 
1
3
cos
4
?
??
??
??
 (D) 
1
2
sin
3
?
??
??
??
 
 Official Ans. by NTA (A) 
  
Sol.
 
 
 
c
v
n
? 
 n
d
 sin i
c
 = n
r
 sin90º 
 sin i
c
 = 
d r
dr
v n
nv
? 
 sin i
c
 
8
8
1.5 10 1.5
2 10 2
?
??
?
 
 
c
3
sin i
4
? 
 tan i
c
 = 
22
33
7
43
?
?
 
 i
c
 = 
1
3
tan
7
?
??
??
??
  
8. A body is projected vertically upwards from the 
surface of earth with a velocity equal to one third 
of escape velocity. The maximum height attained 
by the body will be: 
(Take radius of earth = 6400 km and g=10 ms
–2
 ) 
 (A)  800 km  (B) 1600 km
 
 
 (C) 2133 km  (D) 4800 km  
 Official Ans. by NTA (A) 
  
Sol.  
 
e
2Gm
v
R
? 
 
2
e
V GMm 1 GMm
m
R 2 9 R h
?
? ? ?
?
 
 
2
e
V GM GM
R h R 18
??
?
 
 
GM GM GM
R h R 9R
??
?
 
 
GM 8GM
R h 9R
?
?
 
 
18
R h 9R
?
?
 
 9R = 8R + 8h 
 
R 6400
h 800km
88
? ? ? 
9. The maximum and minimum voltage of an 
amplitude modulated signal are 60 V and 20 V 
respectively. The percentage modulation index will 
be : 
 (A) 0.5%  (B) 50%
 
(C) 2%  (D) 30% 
 Official Ans. by NTA (B) 
  
 
 
3 
 
Sol. V
max
  = 60 
 V
min
 = 20 
 % modulation =  
 
max min
max min
VV 60 20 40
100 100 100
V V 60 20 80
?? ? ? ? ? ? ?
??
?? ? ? ? ?
??
? ? ? ?
??
 
? 50%  
10. A nucleus of mass M at rest splits into two parts 
having masses 
M'
3
and 
? ?
2M'
M' M
3
? . The ratio 
of de Broglie wavelength of two parts will be : 
 (A)  1 : 2  (B) 2 : 1
 
 
 (C) 1 : 1   (D) 2 : 3 
 Official Ans. by NTA (C) 
  
Sol.  
  
 
12
PP ? 
 Here P is momentum 
 So 
h
P
?? 
 Hence both will have same de broglie wavelength. 
11. An ice cube of dimensions 60 cm × 50 cm × 20 cm 
is placed in an insulation box of wall thickness  
1 cm. The box keeping the ice cube at 0°C of 
temperature is brought to a room of temperature 
40°C. The rate of melting of ice is approximately: 
(Latent heat of fusion of ice is 3.4 × 10
5
 J kg
–1
 and 
thermal conducting of insulation wall is  
0.05 Wm
–1
ºC
–1
) 
 (A)  61×10
–1
 kg s
–1
 (B) 61×10
–5
 kg s
–1
  
 (C) 208 kg s
–1
 (D) 30×10
–5
 kg s
–1
 
 Official Ans. by NTA (B) 
  
Sol.
 
 
 
dQ KA T
dt
?
? 
 A = 2 (0.6 × 0.5 + 0.5 × 0.2 + 0.2 × 0.6) 
 = 2(0.3 + 0.1 + 0.12) 
 = 2(0.4 + 0.12) 
 = 2(0.52) 
 = 1.04 m
2 
 
R
th
 = 
22
1 10 10
KA 0.05 1.04 0.052
??
?
??
?
 
 
2
2
th
dQ T 40 0.052
2.08 10 J / s
dt R 10
?
??
? ? ? ? 
 2.08 × 10
2
 = m × 3.4 × 10
5 
 
m = 
3
3
2.08
0.61 10
3.4 10
?
??
?
 kg/s 
 = 61 × 10
–5
 Kg/s 
12. A gas has n degrees of freedom. The ratio of 
specific heat of gas at constant volume to the 
specific heat of gas at constant pressure will be : 
 (A) 
n
n2 ?
  (B) 
n2
n
?
  
 (C) 
n
2n 2 ?
   (D) 
n
n2 ?
  
 Official Ans. by NTA (A) 
  
Sol. 
v
nR
C
2
? 
? ?
p
n 2 R
C
2
?
? 
 
v
p
C n
C n 2
?
?
  
13. A transverse wave is represented by y = 2sin 
? ? t kx ?? cm. The value of wavelength (in cm) for 
which the wave velocity becomes equal to the 
maximum particle velocity, will be ; 
 (A) 4 ? (B) 2 ?  
 (C) ?  (D) 2 
 Official Ans. by NTA (A) 
  
Sol. y = 2 sin ( ?t – kx) 
 Maximum particle velocity = A ? ?
? Wave velocity = 
k
?
 
 
k
?
= A ? ?
? k = 
1
A
 = 
2 ?
?
 
 ? = 2 ?A 
  = 4 ? cm ?
 
4 
 
14.  A battery of 6 V is connected to the circuit as 
shown below. The current I drawn from the battery 
is : 
 
 (A)  1A (B) 2A  
 (C) 
6
A
11
   (D) 
4
A
3
 
 Official Ans. by NTA (A) 
  
Sol. Balanced wheat stone bridge in circuit so there is 
no current in 5 ? ?resistor so it can be removed 
from the circuit. 
 
 
eq
6 12
R2
6 12
?
??
?
 
 
6 12
2
18
?
?? 
 
eq
R6 ?? 
 
eq
V6
I 1 Amp.
R6
? ? ? 
15. A source of potential difference V is connected to 
the combination of two identical capacitors as 
shown in the figure. When key 'K’ is closed, the 
total energy stored across the combination is E
1
. 
Now key ‘K’ is opened and dielectric of dielectric 
constant 5 is introduced between the plates of the 
capacitors. The total energy stored across the 
combination is now E
2
. The ratio E
1
/E
2
 will be : 
 
 (A) 
1
10
  (B) 
2
5
  
 (C) 
5
13
  (D) 
5
26
 
 Official Ans. by NTA (C) 
  
Sol.  
 (1) Switch is closed 
  C
eq
 = 2C 
 Energy 
2
1 eq
1
E C V
2
? 
 = 
2
1
2C V
2
? 
 E
1
 = CV
2
 
 (ii) When switch is opened charge on right 
capacitor remain CV while potential on left 
capacitor remain same 
  Dielectric K = 5  
 C
’
 = KC 
 C
’
 = 5C 
 ? ?
? ?
? ?
2
2
2
CV
1
E 5C V
2 2 5C
?? 
 
22
2
5CV CV
E
2 10
?? 
 
 
5 
 
 
2
2
13CV
E
5
?   
 
2
1
2
2
E CV 5
13CV E 13
5
?? 
 
1
2
E 5
E 13
?  
16. Two concentric circular loops of radii r
1
=30 cm 
and r
2
=50 cm are placed in X-Y plane as shown in 
the figure. A current I = 7A is flowing through 
them in the direction as shown in figure. The net 
magnetic moment of this system of two circular 
loops is approximately : 
 
 (A) 
2
7
ˆ
k Am
2
  (B) 
2
7
ˆ
k Am
2
?  
 (C) 
2
ˆ
7k Am  (D) –
2
ˆ
7 k Am  
 Official Ans. by NTA (B) 
  
Sol.  
 Magnetic moment  
 ? ? ? ?
22
ˆˆ
M i 0.5 k i 0.3 k ? ? ? ? ? 
 
22 25 9
ˆ
M 7 k
7 100 100
??
? ? ? ?
??
??
 
 = 
16
ˆ
22 k
100
??
?
??
??
 
 
ˆ
M 3.52k ??  Am
2
 
 
7
ˆ
k
2
?? Am
2
 
17. A velocity selector consists of electric field 
ˆ
E Ek ? and magnetic field 
ˆ
B Bj ? with B=12 mT. 
The value E required for an electron of energy  
728 eV moving along the positive x-axis to pass 
undeflected is : 
  (Given, mass of electron = 9.1×10
–31
kg) 
 (A) 192 kVm
–1
  (B) 192 m Vm
–1
  
 (C) 9600 kVm
–1
 (D) 16 kVm
–1
 
 Official Ans. by NTA (A) 
  
Sol. 
ˆ
E E k ? B = 12 mT 
 
ˆ
B B j ? Energy = 728 eV 
 Energy = 
2
1
mv
2
 
 
31 2
1
728eV 9.1 10 v
2
?
? ? ? ? 
 
19 31 2
1
728 1.6 10 9.1 10 v
2
??
? ? ? ? ? ? 
 v = 16 × 10
6
 m/s 
 E = vB 
 E = 16×10
6
 × 12 × 10
–3 
 
E = 192 × 10
3
 V/m 
18. Two masses M
1
 and .M
2
 are tied together at the 
two ends of a light inextensible string that passes 
over a frictionless pulley. When the mass M
2
 is 
twice that of M
1
. the acceleration of the system is 
a
1
. When the mass M
2 
is thrice that of M
1
. The 
acceleration of The system is a
2
. The ratio 
1
2
a
a
 will 
be: 
 
 (A) 
1
3
  (B) 
2
3
  
 (C) 
3
2
  (D) 
1
2
 
 Official Ans. by NTA (B) 
  
Sol. 
21
12
m g m g
a
mm
?
?
?
 
Case 1 M
2
 = 2m
1 
  
a
1
 = 
11
1
2m g m g
3m
?
 
  a
1
 = g/3