JEE Main 2022 Question Paper 26 July Shift 1 | 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 : 9 : 00 AM  to  12 : 00 NOON 
PHYSICS  TEST PAPER WITH SOLUTION 
 
 
 
 
 
 
 
SECTION-A 
1. Three masses M = 100 kg, m
1
 = 10 kg and 
m
2
=20 kg are arranged in a system as shown in 
figure. All the surfaces are frictionless and 
strings are inextensible and weightless. The 
pulleys are also weightless and frictionless. A 
force F is applied on the system so that the 
mass m
2
 moves upward with an acceleration of 
2 ms
-2
. The value of F is : 
(Take g = 10 ms
-2
) 
 
 
 (A) 3360 N (B) 3380 N 
 (C) 3120 N  (D) 3240 N 
 Official Ans. by NTA (C) 
  
Sol. Let acceleration of 100 kg block = a
1 
 FBD of 100 kg block w.r.t ground  
 
T
T
N
1
a
1
F
 
 F–T–N
1
 = 100 a
1
 ……..(i) 
 FBD of 20 block wrt 100kg 
 
T
20g
N
1
20a
1
2m/s
2
 
 T – 20g = 20(2) 
 T = 240      ….. (ii) 
 N
1
 = 20a
1
  ….. (iii) 
 FBD of 10 kg block wrt 100 kg  
 
T
10a
1
2m/s
2
  
 
1
10a 240 10(2) ?? 
 
2
1
a 26m / s ? 
 F 240 20(26) 100 26 ? ? ? ? 
 
? F = 3360 N 
2. A radio can tune to any station in 6 MHz to  
10 MHz band. The value of corresponding 
wavelength bandwidth will be : 
 (A) 4 m  (B) 20 m  
 (C) 30 m  (D) 50 m 
 Official Ans. by NTA (B ) 
 
Sol. Given: Frequency f
1 
= 6MHz   
 Frequency f
2 
= 10MHz 
 
1
1
c
f
?? 
 
2
2
c
f
?? 
 Wavelength bandwidth =  
 
- 
 
 = 20 m 
3. The disintegration rate of a certain radioactive 
sample at any instant is 4250 disintegrations 
per minute. 10 minutes later, the rate becomes 
2250 disintegrations per minute. The 
approximate decay constant is : 
(Take log
10
1.88 = 0.274) 
 (A) 0.02 min 
–1
 (B) 2.7 min 
–1
 
 (C) 0.063 min 
–1
 (D) 6.3 min 
–1
 
 Official Ans. by NTA (C ) 
  
Page 2


 
    1 
FINAL JEE –MAIN EXAMINATION – JULY, 2022 
(Held On Tuesday 26
th
July, 2022)           TIME : 9 : 00 AM  to  12 : 00 NOON 
PHYSICS  TEST PAPER WITH SOLUTION 
 
 
 
 
 
 
 
SECTION-A 
1. Three masses M = 100 kg, m
1
 = 10 kg and 
m
2
=20 kg are arranged in a system as shown in 
figure. All the surfaces are frictionless and 
strings are inextensible and weightless. The 
pulleys are also weightless and frictionless. A 
force F is applied on the system so that the 
mass m
2
 moves upward with an acceleration of 
2 ms
-2
. The value of F is : 
(Take g = 10 ms
-2
) 
 
 
 (A) 3360 N (B) 3380 N 
 (C) 3120 N  (D) 3240 N 
 Official Ans. by NTA (C) 
  
Sol. Let acceleration of 100 kg block = a
1 
 FBD of 100 kg block w.r.t ground  
 
T
T
N
1
a
1
F
 
 F–T–N
1
 = 100 a
1
 ……..(i) 
 FBD of 20 block wrt 100kg 
 
T
20g
N
1
20a
1
2m/s
2
 
 T – 20g = 20(2) 
 T = 240      ….. (ii) 
 N
1
 = 20a
1
  ….. (iii) 
 FBD of 10 kg block wrt 100 kg  
 
T
10a
1
2m/s
2
  
 
1
10a 240 10(2) ?? 
 
2
1
a 26m / s ? 
 F 240 20(26) 100 26 ? ? ? ? 
 
? F = 3360 N 
2. A radio can tune to any station in 6 MHz to  
10 MHz band. The value of corresponding 
wavelength bandwidth will be : 
 (A) 4 m  (B) 20 m  
 (C) 30 m  (D) 50 m 
 Official Ans. by NTA (B ) 
 
Sol. Given: Frequency f
1 
= 6MHz   
 Frequency f
2 
= 10MHz 
 
1
1
c
f
?? 
 
2
2
c
f
?? 
 Wavelength bandwidth =  
 
- 
 
 = 20 m 
3. The disintegration rate of a certain radioactive 
sample at any instant is 4250 disintegrations 
per minute. 10 minutes later, the rate becomes 
2250 disintegrations per minute. The 
approximate decay constant is : 
(Take log
10
1.88 = 0.274) 
 (A) 0.02 min 
–1
 (B) 2.7 min 
–1
 
 (C) 0.063 min 
–1
 (D) 6.3 min 
–1
 
 Official Ans. by NTA (C ) 
  
 
2 
 
?  
 
Sol. At t=0  disintegration rate = 4250 dpm  
 At t=10   disintegration rate = 2250 dpm 
 
?
?
?t
o
A Ae 
 
? ? 10
2250 4250e
??
? 
 
? ? ?
4250
10 ln
2250
??
??
??
??
 
 
1
0.063min
?
? ? ? 
4. A parallel beam of light of wavelength 900 nm 
and intensity 100 Wm
-2
 is incident on a surface 
perpendicular to the beam. Tire number of 
photons crossing 1 cm
2
 area perpendicular to 
the beam in one second is : 
 (A) 3 × 10
16
 (B) 4.5 × 10
16
 
 (C) 4.5 × 10
17
 (D) 4.5 × 10
20
 
 Official Ans. by NTA (B ) 
  
Sol. Wavelength of incident beam 
9
900 10
?
? ? ? m 
Intensity of incident beam =I     W/m
2 
 
 No. of photons crossing per unit sec 
 
net
single photon
E IA
n
E hc
?
? ? ? 
 
? ? ? ? ? ?
49
16
34 8
100 1 10 900 10
4.5 10
6.62 10 3 10
??
?
??
? ? ?
? ? ?
 
5. In young’s double slit experiment, the fringe 
width is 12mm. If the entire arrangement is 
placed in water of refractive index 
4
3
,  then  
the fringe width becomes (in mm) 
 (A) 16  (B) 9 
 (C) 48  (D) 12 
 Official Ans. by NTA (B ) 
  
Sol. For a given light wavelength corresponding a 
medium of refractive index   
 
   
 = 
 
      
 
 
and we know that fringe width    
  
 
 
 Therefore,  
   
 = 
 
      
 
 = 
  
 
 
 = 9 mm  
6. The magnetic field of a plane electromagnetic 
wave is given by  
 
? ?
8 3 11
ˆ
B 2 10 sin 0.5 10 x 1.5 10 t jT
?
? ? ? ? ? 
 The amplitude of the electric field would be  
 (A) 
1
6Vm
?
along x-axis  
 (B)
 
1
3Vm
?
along z-axis 
 
 (C) 
1
6Vm
?
along z-axis  
 (D)
81
2 10 Vm
??
? along z-axis 
 Official Ans. by NTA (C) 
  
Sol. 
0
00
0
E
c E cB
B
? ? ?
 
 
? ? ? ?
88
0
E 3 10 2 10
?
? ? ? 
 
1
0
E 6Vm
?
? 
 As, B ? along y-axis 
 v ? along negative x-axis 
 hence 
0
E ? along z-axis  
7.  In a series LR circuit 
L
XR ? and power 
factor of the circuit is 
1
P . When capacitor with 
capacitance C such that 
LC
XX ? is put in 
series, the power factor becomes 
2
P . The ratio 
1
2
P
P
 is 
 (A)
1
2
  (B)
1
2
  
 (C) 
3
2
  (D)2 : 1  
 Official Ans. by NTA (B ) 
  
Page 3


 
    1 
FINAL JEE –MAIN EXAMINATION – JULY, 2022 
(Held On Tuesday 26
th
July, 2022)           TIME : 9 : 00 AM  to  12 : 00 NOON 
PHYSICS  TEST PAPER WITH SOLUTION 
 
 
 
 
 
 
 
SECTION-A 
1. Three masses M = 100 kg, m
1
 = 10 kg and 
m
2
=20 kg are arranged in a system as shown in 
figure. All the surfaces are frictionless and 
strings are inextensible and weightless. The 
pulleys are also weightless and frictionless. A 
force F is applied on the system so that the 
mass m
2
 moves upward with an acceleration of 
2 ms
-2
. The value of F is : 
(Take g = 10 ms
-2
) 
 
 
 (A) 3360 N (B) 3380 N 
 (C) 3120 N  (D) 3240 N 
 Official Ans. by NTA (C) 
  
Sol. Let acceleration of 100 kg block = a
1 
 FBD of 100 kg block w.r.t ground  
 
T
T
N
1
a
1
F
 
 F–T–N
1
 = 100 a
1
 ……..(i) 
 FBD of 20 block wrt 100kg 
 
T
20g
N
1
20a
1
2m/s
2
 
 T – 20g = 20(2) 
 T = 240      ….. (ii) 
 N
1
 = 20a
1
  ….. (iii) 
 FBD of 10 kg block wrt 100 kg  
 
T
10a
1
2m/s
2
  
 
1
10a 240 10(2) ?? 
 
2
1
a 26m / s ? 
 F 240 20(26) 100 26 ? ? ? ? 
 
? F = 3360 N 
2. A radio can tune to any station in 6 MHz to  
10 MHz band. The value of corresponding 
wavelength bandwidth will be : 
 (A) 4 m  (B) 20 m  
 (C) 30 m  (D) 50 m 
 Official Ans. by NTA (B ) 
 
Sol. Given: Frequency f
1 
= 6MHz   
 Frequency f
2 
= 10MHz 
 
1
1
c
f
?? 
 
2
2
c
f
?? 
 Wavelength bandwidth =  
 
- 
 
 = 20 m 
3. The disintegration rate of a certain radioactive 
sample at any instant is 4250 disintegrations 
per minute. 10 minutes later, the rate becomes 
2250 disintegrations per minute. The 
approximate decay constant is : 
(Take log
10
1.88 = 0.274) 
 (A) 0.02 min 
–1
 (B) 2.7 min 
–1
 
 (C) 0.063 min 
–1
 (D) 6.3 min 
–1
 
 Official Ans. by NTA (C ) 
  
 
2 
 
?  
 
Sol. At t=0  disintegration rate = 4250 dpm  
 At t=10   disintegration rate = 2250 dpm 
 
?
?
?t
o
A Ae 
 
? ? 10
2250 4250e
??
? 
 
? ? ?
4250
10 ln
2250
??
??
??
??
 
 
1
0.063min
?
? ? ? 
4. A parallel beam of light of wavelength 900 nm 
and intensity 100 Wm
-2
 is incident on a surface 
perpendicular to the beam. Tire number of 
photons crossing 1 cm
2
 area perpendicular to 
the beam in one second is : 
 (A) 3 × 10
16
 (B) 4.5 × 10
16
 
 (C) 4.5 × 10
17
 (D) 4.5 × 10
20
 
 Official Ans. by NTA (B ) 
  
Sol. Wavelength of incident beam 
9
900 10
?
? ? ? m 
Intensity of incident beam =I     W/m
2 
 
 No. of photons crossing per unit sec 
 
net
single photon
E IA
n
E hc
?
? ? ? 
 
? ? ? ? ? ?
49
16
34 8
100 1 10 900 10
4.5 10
6.62 10 3 10
??
?
??
? ? ?
? ? ?
 
5. In young’s double slit experiment, the fringe 
width is 12mm. If the entire arrangement is 
placed in water of refractive index 
4
3
,  then  
the fringe width becomes (in mm) 
 (A) 16  (B) 9 
 (C) 48  (D) 12 
 Official Ans. by NTA (B ) 
  
Sol. For a given light wavelength corresponding a 
medium of refractive index   
 
   
 = 
 
      
 
 
and we know that fringe width    
  
 
 
 Therefore,  
   
 = 
 
      
 
 = 
  
 
 
 = 9 mm  
6. The magnetic field of a plane electromagnetic 
wave is given by  
 
? ?
8 3 11
ˆ
B 2 10 sin 0.5 10 x 1.5 10 t jT
?
? ? ? ? ? 
 The amplitude of the electric field would be  
 (A) 
1
6Vm
?
along x-axis  
 (B)
 
1
3Vm
?
along z-axis 
 
 (C) 
1
6Vm
?
along z-axis  
 (D)
81
2 10 Vm
??
? along z-axis 
 Official Ans. by NTA (C) 
  
Sol. 
0
00
0
E
c E cB
B
? ? ?
 
 
? ? ? ?
88
0
E 3 10 2 10
?
? ? ? 
 
1
0
E 6Vm
?
? 
 As, B ? along y-axis 
 v ? along negative x-axis 
 hence 
0
E ? along z-axis  
7.  In a series LR circuit 
L
XR ? and power 
factor of the circuit is 
1
P . When capacitor with 
capacitance C such that 
LC
XX ? is put in 
series, the power factor becomes 
2
P . The ratio 
1
2
P
P
 is 
 (A)
1
2
  (B)
1
2
  
 (C) 
3
2
  (D)2 : 1  
 Official Ans. by NTA (B ) 
  
 
 
3 
 
?  
Sol. In case of L-R circuit  
 
22
L
Z X R ?? & power factor  
 
1
R
P cos
Z
? ? ? 
 As  X
L
 = R 
 
?
 
Z 2R ? 
 
?
 11
R1
PP
2R 2
? ? ?  
 In case of L-C-R circuit  
 
? ?
2
2
LC
Z R X X ? ? ? 
 As 
LC
XX ?  
 
?Z = R 
 
?
 2
R
P cos 1
R
? ? ? ? 
 
?
 
1
2
P 1
P 2
? 
8. A charge particle is moving in a uniform 
magnetic field 
? ?
ˆˆ
2i 3j T ? . If it has an 
acceleration of 
? ?
2
ˆˆ
i 4j m / s ?? , then the value 
of ? will be  
 (A)  3 (B) 6 (C) 12 (D) 2 
 Official Ans. by NTA (B ) 
  
Sol. As 
? ?
F q v B ?? 
 
? ?
q
a v B
m
?? 
 So, a &B are ? to each other  
 Hence, a.B 0 ? 
 
? ? ? ?
ˆ ˆ ˆ ˆ
i 4j . 2i 3j 0 ? ? ? ? 
 ? ? ? ? ? ? 2 4 3 0 ? ? ? ? 
 
12
6
2
? ? ? ? ? 
9. 
XY
B and B are the magnetic field at the centre 
of two coils of two coils X and Y respectively, 
each carrying equal current. If coil X has 200 
turns and 20 cm radius and coil Y has 400 
turns and 20 cm radius, the ratio of 
XY
B and B 
is  
 (A) 1 : 1   (B) 1 : 2   
 (C)2 : 1  (D) 4 : 1  
 Official Ans. by NTA (B ) 
  
Sol. At centre 
0
i
BN
2R
? ??
?
??
??
 
 
0
x
i
B 200
2 20cm
? ??
?
??
?
??
 
 
0
y
i
B 400
2 20cm
? ??
?
??
?
??
 
 
x
y
B 1
B2
? 
10. The current I in the given circuit will be : 
 
 (A)  10A  (B) 20 A 
 
 
 (C)  4A   (D) 40A 
 Official Ans. by NTA (A ) 
  
Sol.  
 Given circuit is balanced wheat stone bridge 
 Hence 2 ? can be neglected  
 
net
R4 ?? 
 
40
I
4
? 
 I = 10A  
Page 4


 
    1 
FINAL JEE –MAIN EXAMINATION – JULY, 2022 
(Held On Tuesday 26
th
July, 2022)           TIME : 9 : 00 AM  to  12 : 00 NOON 
PHYSICS  TEST PAPER WITH SOLUTION 
 
 
 
 
 
 
 
SECTION-A 
1. Three masses M = 100 kg, m
1
 = 10 kg and 
m
2
=20 kg are arranged in a system as shown in 
figure. All the surfaces are frictionless and 
strings are inextensible and weightless. The 
pulleys are also weightless and frictionless. A 
force F is applied on the system so that the 
mass m
2
 moves upward with an acceleration of 
2 ms
-2
. The value of F is : 
(Take g = 10 ms
-2
) 
 
 
 (A) 3360 N (B) 3380 N 
 (C) 3120 N  (D) 3240 N 
 Official Ans. by NTA (C) 
  
Sol. Let acceleration of 100 kg block = a
1 
 FBD of 100 kg block w.r.t ground  
 
T
T
N
1
a
1
F
 
 F–T–N
1
 = 100 a
1
 ……..(i) 
 FBD of 20 block wrt 100kg 
 
T
20g
N
1
20a
1
2m/s
2
 
 T – 20g = 20(2) 
 T = 240      ….. (ii) 
 N
1
 = 20a
1
  ….. (iii) 
 FBD of 10 kg block wrt 100 kg  
 
T
10a
1
2m/s
2
  
 
1
10a 240 10(2) ?? 
 
2
1
a 26m / s ? 
 F 240 20(26) 100 26 ? ? ? ? 
 
? F = 3360 N 
2. A radio can tune to any station in 6 MHz to  
10 MHz band. The value of corresponding 
wavelength bandwidth will be : 
 (A) 4 m  (B) 20 m  
 (C) 30 m  (D) 50 m 
 Official Ans. by NTA (B ) 
 
Sol. Given: Frequency f
1 
= 6MHz   
 Frequency f
2 
= 10MHz 
 
1
1
c
f
?? 
 
2
2
c
f
?? 
 Wavelength bandwidth =  
 
- 
 
 = 20 m 
3. The disintegration rate of a certain radioactive 
sample at any instant is 4250 disintegrations 
per minute. 10 minutes later, the rate becomes 
2250 disintegrations per minute. The 
approximate decay constant is : 
(Take log
10
1.88 = 0.274) 
 (A) 0.02 min 
–1
 (B) 2.7 min 
–1
 
 (C) 0.063 min 
–1
 (D) 6.3 min 
–1
 
 Official Ans. by NTA (C ) 
  
 
2 
 
?  
 
Sol. At t=0  disintegration rate = 4250 dpm  
 At t=10   disintegration rate = 2250 dpm 
 
?
?
?t
o
A Ae 
 
? ? 10
2250 4250e
??
? 
 
? ? ?
4250
10 ln
2250
??
??
??
??
 
 
1
0.063min
?
? ? ? 
4. A parallel beam of light of wavelength 900 nm 
and intensity 100 Wm
-2
 is incident on a surface 
perpendicular to the beam. Tire number of 
photons crossing 1 cm
2
 area perpendicular to 
the beam in one second is : 
 (A) 3 × 10
16
 (B) 4.5 × 10
16
 
 (C) 4.5 × 10
17
 (D) 4.5 × 10
20
 
 Official Ans. by NTA (B ) 
  
Sol. Wavelength of incident beam 
9
900 10
?
? ? ? m 
Intensity of incident beam =I     W/m
2 
 
 No. of photons crossing per unit sec 
 
net
single photon
E IA
n
E hc
?
? ? ? 
 
? ? ? ? ? ?
49
16
34 8
100 1 10 900 10
4.5 10
6.62 10 3 10
??
?
??
? ? ?
? ? ?
 
5. In young’s double slit experiment, the fringe 
width is 12mm. If the entire arrangement is 
placed in water of refractive index 
4
3
,  then  
the fringe width becomes (in mm) 
 (A) 16  (B) 9 
 (C) 48  (D) 12 
 Official Ans. by NTA (B ) 
  
Sol. For a given light wavelength corresponding a 
medium of refractive index   
 
   
 = 
 
      
 
 
and we know that fringe width    
  
 
 
 Therefore,  
   
 = 
 
      
 
 = 
  
 
 
 = 9 mm  
6. The magnetic field of a plane electromagnetic 
wave is given by  
 
? ?
8 3 11
ˆ
B 2 10 sin 0.5 10 x 1.5 10 t jT
?
? ? ? ? ? 
 The amplitude of the electric field would be  
 (A) 
1
6Vm
?
along x-axis  
 (B)
 
1
3Vm
?
along z-axis 
 
 (C) 
1
6Vm
?
along z-axis  
 (D)
81
2 10 Vm
??
? along z-axis 
 Official Ans. by NTA (C) 
  
Sol. 
0
00
0
E
c E cB
B
? ? ?
 
 
? ? ? ?
88
0
E 3 10 2 10
?
? ? ? 
 
1
0
E 6Vm
?
? 
 As, B ? along y-axis 
 v ? along negative x-axis 
 hence 
0
E ? along z-axis  
7.  In a series LR circuit 
L
XR ? and power 
factor of the circuit is 
1
P . When capacitor with 
capacitance C such that 
LC
XX ? is put in 
series, the power factor becomes 
2
P . The ratio 
1
2
P
P
 is 
 (A)
1
2
  (B)
1
2
  
 (C) 
3
2
  (D)2 : 1  
 Official Ans. by NTA (B ) 
  
 
 
3 
 
?  
Sol. In case of L-R circuit  
 
22
L
Z X R ?? & power factor  
 
1
R
P cos
Z
? ? ? 
 As  X
L
 = R 
 
?
 
Z 2R ? 
 
?
 11
R1
PP
2R 2
? ? ?  
 In case of L-C-R circuit  
 
? ?
2
2
LC
Z R X X ? ? ? 
 As 
LC
XX ?  
 
?Z = R 
 
?
 2
R
P cos 1
R
? ? ? ? 
 
?
 
1
2
P 1
P 2
? 
8. A charge particle is moving in a uniform 
magnetic field 
? ?
ˆˆ
2i 3j T ? . If it has an 
acceleration of 
? ?
2
ˆˆ
i 4j m / s ?? , then the value 
of ? will be  
 (A)  3 (B) 6 (C) 12 (D) 2 
 Official Ans. by NTA (B ) 
  
Sol. As 
? ?
F q v B ?? 
 
? ?
q
a v B
m
?? 
 So, a &B are ? to each other  
 Hence, a.B 0 ? 
 
? ? ? ?
ˆ ˆ ˆ ˆ
i 4j . 2i 3j 0 ? ? ? ? 
 ? ? ? ? ? ? 2 4 3 0 ? ? ? ? 
 
12
6
2
? ? ? ? ? 
9. 
XY
B and B are the magnetic field at the centre 
of two coils of two coils X and Y respectively, 
each carrying equal current. If coil X has 200 
turns and 20 cm radius and coil Y has 400 
turns and 20 cm radius, the ratio of 
XY
B and B 
is  
 (A) 1 : 1   (B) 1 : 2   
 (C)2 : 1  (D) 4 : 1  
 Official Ans. by NTA (B ) 
  
Sol. At centre 
0
i
BN
2R
? ??
?
??
??
 
 
0
x
i
B 200
2 20cm
? ??
?
??
?
??
 
 
0
y
i
B 400
2 20cm
? ??
?
??
?
??
 
 
x
y
B 1
B2
? 
10. The current I in the given circuit will be : 
 
 (A)  10A  (B) 20 A 
 
 
 (C)  4A   (D) 40A 
 Official Ans. by NTA (A ) 
  
Sol.  
 Given circuit is balanced wheat stone bridge 
 Hence 2 ? can be neglected  
 
net
R4 ?? 
 
40
I
4
? 
 I = 10A  
 
4 
 
?  
 
11. The total charge on the system of capacitance 
1 2 3
C 1 F, C 2 F,C 4 F ? ? ? ? ? ? and 
4
C 3 F ?? 
connected in parallel is  
 (Assume a battery of 20V is connected to the 
combination) 
 (A) 200 C ? (B) 200C 
 (C) 10 C ?  (D) 10C 
 Official Ans. by NTA (A) 
  
Sol.  
 
20V
q C = 2 F 
2  2 
?
q C = 4 F 
3  3 
?
q C = 3 F 
4  4 
?
q C = 1 F 
1   1 
?
 
 Total charge = 
1 2 2 4
q q q q ??? 
 1 20 2 20 4 20 3 20 200 C ? ? ? ? ? ? ? ? ? ? 
12. When a particle executes simple Harmonic 
motion, the nature of graph of velocity as 
function of displacement will be : 
 (A) Circular  (B)Ellipitical  
 (C) Sinusoidal (D) Straight line 
 Official Ans. by NTA (B ) 
  
Sol. For a particle in SHM, its speed depends on 
position as  
 
22
v A x ? ? ? 
 Where ? is angular frequency and A is amplitude 
 Now 
2 2 2 2 2
v A x ? ? ? ? 
 So, 
? ? ? ?
22
22
vx
1
AA
??
?
 
 So graph between v and x is elliptical  
13. 7 mole of certain monoatomic ideal gas 
undergoes a temperature increase of 40K at 
constant pressure. The increase in the internal 
energy of the gas in this process is  
 (Given R = 8.3
11
JK mol
??
) 
 (A)  5810 J   (B) 3486 J  
 (C) 11620J  (D) 6972 J 
 Official Ans. by NTA (B ) 
  
Sol. For a quasi-static process the change in internal 
energy of an ideal gas is  
 
V
U nC T ? ? ? 
 
3R
nT
2
? ? ? ? 
 [molar heat capacity at constant volume for mono 
atomic gas = 
3R
2
] 
 
3
U 7 8.3 40 3486J
2
? ? ? ? ? ?
 
14. A monoatomic gas at pressure P and volume V 
is suddenly compressed to one eighth of its 
original volume. The final pressure at constant 
entropy will be: 
 (A) P  (B) 8P (C) 32P (D) 64 P 
 Official Ans. by NTA (C ) 
  
Sol. Constant entropy means process is adiabatic  
 PV
? 
= constant
 
 
1
2
V
V
8
? 
 
1 1 2 2
P V P V
??
? 
 
5/3
1
1 1 2
V
P V P
8
?
??
?
??
??
 
 
5/3
5/3 21
11
PV
PV
32
? 
 P
2 
= 32P
1 
Page 5


 
    1 
FINAL JEE –MAIN EXAMINATION – JULY, 2022 
(Held On Tuesday 26
th
July, 2022)           TIME : 9 : 00 AM  to  12 : 00 NOON 
PHYSICS  TEST PAPER WITH SOLUTION 
 
 
 
 
 
 
 
SECTION-A 
1. Three masses M = 100 kg, m
1
 = 10 kg and 
m
2
=20 kg are arranged in a system as shown in 
figure. All the surfaces are frictionless and 
strings are inextensible and weightless. The 
pulleys are also weightless and frictionless. A 
force F is applied on the system so that the 
mass m
2
 moves upward with an acceleration of 
2 ms
-2
. The value of F is : 
(Take g = 10 ms
-2
) 
 
 
 (A) 3360 N (B) 3380 N 
 (C) 3120 N  (D) 3240 N 
 Official Ans. by NTA (C) 
  
Sol. Let acceleration of 100 kg block = a
1 
 FBD of 100 kg block w.r.t ground  
 
T
T
N
1
a
1
F
 
 F–T–N
1
 = 100 a
1
 ……..(i) 
 FBD of 20 block wrt 100kg 
 
T
20g
N
1
20a
1
2m/s
2
 
 T – 20g = 20(2) 
 T = 240      ….. (ii) 
 N
1
 = 20a
1
  ….. (iii) 
 FBD of 10 kg block wrt 100 kg  
 
T
10a
1
2m/s
2
  
 
1
10a 240 10(2) ?? 
 
2
1
a 26m / s ? 
 F 240 20(26) 100 26 ? ? ? ? 
 
? F = 3360 N 
2. A radio can tune to any station in 6 MHz to  
10 MHz band. The value of corresponding 
wavelength bandwidth will be : 
 (A) 4 m  (B) 20 m  
 (C) 30 m  (D) 50 m 
 Official Ans. by NTA (B ) 
 
Sol. Given: Frequency f
1 
= 6MHz   
 Frequency f
2 
= 10MHz 
 
1
1
c
f
?? 
 
2
2
c
f
?? 
 Wavelength bandwidth =  
 
- 
 
 = 20 m 
3. The disintegration rate of a certain radioactive 
sample at any instant is 4250 disintegrations 
per minute. 10 minutes later, the rate becomes 
2250 disintegrations per minute. The 
approximate decay constant is : 
(Take log
10
1.88 = 0.274) 
 (A) 0.02 min 
–1
 (B) 2.7 min 
–1
 
 (C) 0.063 min 
–1
 (D) 6.3 min 
–1
 
 Official Ans. by NTA (C ) 
  
 
2 
 
?  
 
Sol. At t=0  disintegration rate = 4250 dpm  
 At t=10   disintegration rate = 2250 dpm 
 
?
?
?t
o
A Ae 
 
? ? 10
2250 4250e
??
? 
 
? ? ?
4250
10 ln
2250
??
??
??
??
 
 
1
0.063min
?
? ? ? 
4. A parallel beam of light of wavelength 900 nm 
and intensity 100 Wm
-2
 is incident on a surface 
perpendicular to the beam. Tire number of 
photons crossing 1 cm
2
 area perpendicular to 
the beam in one second is : 
 (A) 3 × 10
16
 (B) 4.5 × 10
16
 
 (C) 4.5 × 10
17
 (D) 4.5 × 10
20
 
 Official Ans. by NTA (B ) 
  
Sol. Wavelength of incident beam 
9
900 10
?
? ? ? m 
Intensity of incident beam =I     W/m
2 
 
 No. of photons crossing per unit sec 
 
net
single photon
E IA
n
E hc
?
? ? ? 
 
? ? ? ? ? ?
49
16
34 8
100 1 10 900 10
4.5 10
6.62 10 3 10
??
?
??
? ? ?
? ? ?
 
5. In young’s double slit experiment, the fringe 
width is 12mm. If the entire arrangement is 
placed in water of refractive index 
4
3
,  then  
the fringe width becomes (in mm) 
 (A) 16  (B) 9 
 (C) 48  (D) 12 
 Official Ans. by NTA (B ) 
  
Sol. For a given light wavelength corresponding a 
medium of refractive index   
 
   
 = 
 
      
 
 
and we know that fringe width    
  
 
 
 Therefore,  
   
 = 
 
      
 
 = 
  
 
 
 = 9 mm  
6. The magnetic field of a plane electromagnetic 
wave is given by  
 
? ?
8 3 11
ˆ
B 2 10 sin 0.5 10 x 1.5 10 t jT
?
? ? ? ? ? 
 The amplitude of the electric field would be  
 (A) 
1
6Vm
?
along x-axis  
 (B)
 
1
3Vm
?
along z-axis 
 
 (C) 
1
6Vm
?
along z-axis  
 (D)
81
2 10 Vm
??
? along z-axis 
 Official Ans. by NTA (C) 
  
Sol. 
0
00
0
E
c E cB
B
? ? ?
 
 
? ? ? ?
88
0
E 3 10 2 10
?
? ? ? 
 
1
0
E 6Vm
?
? 
 As, B ? along y-axis 
 v ? along negative x-axis 
 hence 
0
E ? along z-axis  
7.  In a series LR circuit 
L
XR ? and power 
factor of the circuit is 
1
P . When capacitor with 
capacitance C such that 
LC
XX ? is put in 
series, the power factor becomes 
2
P . The ratio 
1
2
P
P
 is 
 (A)
1
2
  (B)
1
2
  
 (C) 
3
2
  (D)2 : 1  
 Official Ans. by NTA (B ) 
  
 
 
3 
 
?  
Sol. In case of L-R circuit  
 
22
L
Z X R ?? & power factor  
 
1
R
P cos
Z
? ? ? 
 As  X
L
 = R 
 
?
 
Z 2R ? 
 
?
 11
R1
PP
2R 2
? ? ?  
 In case of L-C-R circuit  
 
? ?
2
2
LC
Z R X X ? ? ? 
 As 
LC
XX ?  
 
?Z = R 
 
?
 2
R
P cos 1
R
? ? ? ? 
 
?
 
1
2
P 1
P 2
? 
8. A charge particle is moving in a uniform 
magnetic field 
? ?
ˆˆ
2i 3j T ? . If it has an 
acceleration of 
? ?
2
ˆˆ
i 4j m / s ?? , then the value 
of ? will be  
 (A)  3 (B) 6 (C) 12 (D) 2 
 Official Ans. by NTA (B ) 
  
Sol. As 
? ?
F q v B ?? 
 
? ?
q
a v B
m
?? 
 So, a &B are ? to each other  
 Hence, a.B 0 ? 
 
? ? ? ?
ˆ ˆ ˆ ˆ
i 4j . 2i 3j 0 ? ? ? ? 
 ? ? ? ? ? ? 2 4 3 0 ? ? ? ? 
 
12
6
2
? ? ? ? ? 
9. 
XY
B and B are the magnetic field at the centre 
of two coils of two coils X and Y respectively, 
each carrying equal current. If coil X has 200 
turns and 20 cm radius and coil Y has 400 
turns and 20 cm radius, the ratio of 
XY
B and B 
is  
 (A) 1 : 1   (B) 1 : 2   
 (C)2 : 1  (D) 4 : 1  
 Official Ans. by NTA (B ) 
  
Sol. At centre 
0
i
BN
2R
? ??
?
??
??
 
 
0
x
i
B 200
2 20cm
? ??
?
??
?
??
 
 
0
y
i
B 400
2 20cm
? ??
?
??
?
??
 
 
x
y
B 1
B2
? 
10. The current I in the given circuit will be : 
 
 (A)  10A  (B) 20 A 
 
 
 (C)  4A   (D) 40A 
 Official Ans. by NTA (A ) 
  
Sol.  
 Given circuit is balanced wheat stone bridge 
 Hence 2 ? can be neglected  
 
net
R4 ?? 
 
40
I
4
? 
 I = 10A  
 
4 
 
?  
 
11. The total charge on the system of capacitance 
1 2 3
C 1 F, C 2 F,C 4 F ? ? ? ? ? ? and 
4
C 3 F ?? 
connected in parallel is  
 (Assume a battery of 20V is connected to the 
combination) 
 (A) 200 C ? (B) 200C 
 (C) 10 C ?  (D) 10C 
 Official Ans. by NTA (A) 
  
Sol.  
 
20V
q C = 2 F 
2  2 
?
q C = 4 F 
3  3 
?
q C = 3 F 
4  4 
?
q C = 1 F 
1   1 
?
 
 Total charge = 
1 2 2 4
q q q q ??? 
 1 20 2 20 4 20 3 20 200 C ? ? ? ? ? ? ? ? ? ? 
12. When a particle executes simple Harmonic 
motion, the nature of graph of velocity as 
function of displacement will be : 
 (A) Circular  (B)Ellipitical  
 (C) Sinusoidal (D) Straight line 
 Official Ans. by NTA (B ) 
  
Sol. For a particle in SHM, its speed depends on 
position as  
 
22
v A x ? ? ? 
 Where ? is angular frequency and A is amplitude 
 Now 
2 2 2 2 2
v A x ? ? ? ? 
 So, 
? ? ? ?
22
22
vx
1
AA
??
?
 
 So graph between v and x is elliptical  
13. 7 mole of certain monoatomic ideal gas 
undergoes a temperature increase of 40K at 
constant pressure. The increase in the internal 
energy of the gas in this process is  
 (Given R = 8.3
11
JK mol
??
) 
 (A)  5810 J   (B) 3486 J  
 (C) 11620J  (D) 6972 J 
 Official Ans. by NTA (B ) 
  
Sol. For a quasi-static process the change in internal 
energy of an ideal gas is  
 
V
U nC T ? ? ? 
 
3R
nT
2
? ? ? ? 
 [molar heat capacity at constant volume for mono 
atomic gas = 
3R
2
] 
 
3
U 7 8.3 40 3486J
2
? ? ? ? ? ?
 
14. A monoatomic gas at pressure P and volume V 
is suddenly compressed to one eighth of its 
original volume. The final pressure at constant 
entropy will be: 
 (A) P  (B) 8P (C) 32P (D) 64 P 
 Official Ans. by NTA (C ) 
  
Sol. Constant entropy means process is adiabatic  
 PV
? 
= constant
 
 
1
2
V
V
8
? 
 
1 1 2 2
P V P V
??
? 
 
5/3
1
1 1 2
V
P V P
8
?
??
?
??
??
 
 
5/3
5/3 21
11
PV
PV
32
? 
 P
2 
= 32P
1 
 
 
5 
 
?  
15. A water drop of radius 1cm is broken into  
729 equal droplets. If surface tension of water 
is 75 dyne/cm, then the gain in surface energy 
upto first decimal place will be :  
 [Given 3.1 ?? 4] 
 (A)
4
8.5 10 J
?
? (B)
4
8.2 10 J
?
?  
 (C) 
4
7.5 10 J
?
? (D)
4
5.3 10 J
?
? 
 Official Ans. by NTA (C ) 
  
Sol. Initial surface energy = TA 
 Where T is surface tension and A is surface area 
 
? ?
5
2
2
i 2
75 10 N
U 4 1 10
10 m
?
?
?
?? ?
??
? ? ? ?
??
??
??
??
 
 
34
75 10 4 10
??
? ? ? ? ?
7
942 10 J
?
?? 
 To get final radius of drops by volume 
conservation  
 
33
44
R 729 r
33
??
? ? ?
??
??
 
 R = Initial radius 
 r = final radius  
 
? ?
1/3
R R 1
r cm
99
729
? ? ? 
 Final surface energy 
 ? ?
f
U 729 T A ? 
 
2
5
2
2
75 10 N 1
729 4 10
10 m 9
?
?
?
??
?? ? ??
? ? ? ?
??
?? ??
??
?? ??
??
 
 
4
3
4 10
729 75 10
81
?
?
?? ??
? ? ?
??
??
 
 
7
9 942 10 J
?
?? ??
??
 
 Gain in surface energy  
 
77
U 9 942 10 942 10
??
? ? ? ? ? ? 
 
7
8 942 10 J
?
? ? ?
7
7536 10 J
?
?? 
 
4
7.5 10 J
?
?? 
16. The percentage decrease in the weight of a 
rocket, when taken to a height of 32 km above 
the surface of earth will, be : 
 (Radius of earth = 6400km) 
 (A) 1 %  (B) 3%  
 (C) (D) 0.5% 
 Official Ans. by NTA (A) 
  
Sol. Acceleration due to gravity at a height h<< R is  
 
2h
g ' g 1
R
??
??
??
??
 
? ? ?
g 2h
gR
?
? ?
? ? ??
g 2h
100 100
gR
?
? ? ? ?
?
32
2 100 1%
6400
? ? ? ? ? ?
17.  As per the given figure, two blocks each of 
mass 250g are connected to a spring of spring 
constant 
1
2Nm
?
. If both are given velocity v 
in opposite directions, then maximum 
elongation of the spring is : 
 
 (A)
v
22
  (B)
v
2
  
 (C) 
v
4
  (D)
v
2
  
 Official Ans. by NTA (B) 
  
Sol. 
l
0
x
2
x
2
v
v=0 v=0
 
 using energy conservation  
 
22
11
mv 2 kx
22
??  
 
?
 
22
11
v 2 x
42
? ? ? 
? 
v
x
2
? ?