JEE Main 2022 Question Paper 27 July Shift 2 | JEE Main & Advanced Previous Year Papers PDF Download

 Page 1


 
    1 
FINAL JEE –MAIN EXAMINATION – JULY, 2022 
 (Held On Wednesday 27
th
 July, 2022)              TIME : 3 : 00 PM  to  6 : 00 PM 
PHYSICS TEST PAPER WITH SOLUTION 
 
 
 
 
 
 
 
SECTION-A 
1. An expression of energy density is given by 
?? ??
?
??
?? ?
x
u sin
kt
, where ?, ? are constants, x is 
displacement, k is Boltzmann constant and t is the 
temperature. The dimensions of ? will be : 
 (A) ?
2 –2 –1
[ML T ] (B) 
0 2 –2
[M L T ] 
 (C) 
0 0 0
[M L T ] (D) 
0 2 0
[M L T ] 
 Official Ans. by NTA (D) 
  
Sol. 
?
?
?
0 0 0
22
[L]
[M L T ]
[M L T ]
 
 
?
??
12
[ML T ] 
 
??
?
?
? ? ? ?
?
2 2 1 2 3
3 2 2
[M L T ] [M L T ][L ]
[L ] M L T
 
2. A body of mass 10 kg is projected at an angle of 
45° with the horizontal. The trajectory of the body 
is observed to pass through a point (20, 10). If T is 
the time of flight, then its momentum vector, at 
time ?
T
t
2
, is _________ 
 [Take g = 10 m/s
2
] 
 (A) ?
ˆˆ
100i (100 2 – 200)j 
 (B) ?
ˆˆ
100 2 i (100 – 200 2)j  
 (C) ?
ˆˆ
100 i (100 – 200 2)j  
 (D) ?
ˆˆ
100 2 i (100 2 – 200)j 
 Official Ans. by NTA (D) 
  
Sol. 
 
 u = 20 
 ??
(2)(20)
T 2 2
2 (10)
 
 
?
? ? ?
ˆˆ
v 10 2 i (10 2 10(2)] j 
 Momentum 
??
? ? ? ?
ˆˆ
p M v 100 2 i (100 2 200) j 
3. A block of mass M slides down on a rough 
inclined plane with constant velocity. The angle 
made by the incline plane with horizontal is ?. The 
magnitude of the contact force will be : 
 (A) Mg    
 (B) Mg cos ?    
 (C) ? ? ? Mgsin Mgcos   
 (D) ? ? ? Mgsin 1  
 Official Ans. by NTA (A) 
  
Sol.  
  
 N = Mgcos ? 
 f = Mgsin ? 
 
22
R N f ?? 
 R = Mg 
  
? ? ? ? ?
??
??
??
2
2
2
10x (10)(100)
y x 10 20
1
u
2u
2
Page 2


 
    1 
FINAL JEE –MAIN EXAMINATION – JULY, 2022 
 (Held On Wednesday 27
th
 July, 2022)              TIME : 3 : 00 PM  to  6 : 00 PM 
PHYSICS TEST PAPER WITH SOLUTION 
 
 
 
 
 
 
 
SECTION-A 
1. An expression of energy density is given by 
?? ??
?
??
?? ?
x
u sin
kt
, where ?, ? are constants, x is 
displacement, k is Boltzmann constant and t is the 
temperature. The dimensions of ? will be : 
 (A) ?
2 –2 –1
[ML T ] (B) 
0 2 –2
[M L T ] 
 (C) 
0 0 0
[M L T ] (D) 
0 2 0
[M L T ] 
 Official Ans. by NTA (D) 
  
Sol. 
?
?
?
0 0 0
22
[L]
[M L T ]
[M L T ]
 
 
?
??
12
[ML T ] 
 
??
?
?
? ? ? ?
?
2 2 1 2 3
3 2 2
[M L T ] [M L T ][L ]
[L ] M L T
 
2. A body of mass 10 kg is projected at an angle of 
45° with the horizontal. The trajectory of the body 
is observed to pass through a point (20, 10). If T is 
the time of flight, then its momentum vector, at 
time ?
T
t
2
, is _________ 
 [Take g = 10 m/s
2
] 
 (A) ?
ˆˆ
100i (100 2 – 200)j 
 (B) ?
ˆˆ
100 2 i (100 – 200 2)j  
 (C) ?
ˆˆ
100 i (100 – 200 2)j  
 (D) ?
ˆˆ
100 2 i (100 2 – 200)j 
 Official Ans. by NTA (D) 
  
Sol. 
 
 u = 20 
 ??
(2)(20)
T 2 2
2 (10)
 
 
?
? ? ?
ˆˆ
v 10 2 i (10 2 10(2)] j 
 Momentum 
??
? ? ? ?
ˆˆ
p M v 100 2 i (100 2 200) j 
3. A block of mass M slides down on a rough 
inclined plane with constant velocity. The angle 
made by the incline plane with horizontal is ?. The 
magnitude of the contact force will be : 
 (A) Mg    
 (B) Mg cos ?    
 (C) ? ? ? Mgsin Mgcos   
 (D) ? ? ? Mgsin 1  
 Official Ans. by NTA (A) 
  
Sol.  
  
 N = Mgcos ? 
 f = Mgsin ? 
 
22
R N f ?? 
 R = Mg 
  
? ? ? ? ?
??
??
??
2
2
2
10x (10)(100)
y x 10 20
1
u
2u
2
 
2 
 
4. A block ‘A’ takes 2 s to slide down a frictionless 
incline of 30° and length ‘l’, kept inside a lift going 
up with uniform velocity ‘v’. If the incline is 
changed to 45°, the time taken by the block, to 
slide down the incline, will be approximately:  
 (A) 2.66 s  (B) 0.83 s  
 (C) 1.68 s  (D) 0.70 s  
 Official Ans. by NTA (C) 
 
Sol.  
  
 a = g sin ?  
 ??
2
1
gsin30 (2)
2
 ….(1) 
 ??
2
1
gsin 45 t
2
 ….(2) 
 
??
? ? ?
??
??
2
1 1
(4) t t 2 2 1.68
2 2
 
5. The velocity of the bullet becomes one third after it 
penetrates 4 cm in a wooden block. Assuming that 
bullet is facing a constant resistance during its 
motion in the block. The bullet stops completely 
after travelling at (4 + x) cm inside the block. The 
value of x is:  
 (A) 2.0  (B) 1.0  
 (C) 0. 5  (D) 1.5  
 Official Ans. by NTA (C) 
  
Sol.
 
 
 
??
? ? ? ? ?
??
??
2 22
2
8V V V
V 2a(4) a
9(8) 9 3
 
 ? ? ?
2
0 V 2a(4 x) 
 ? ?
??
??
??
??
2
2 V
V 2 (4 x)
9
 
  4.5 = 4 + x  
  x = 0.5 ?
6. A body of mass m is projected with velocity ?
e
v in 
vertically upward direction from the surface of the 
earth into space. It is given that 
e
v is escape 
velocity and ? < 1. If air resistance is considered to 
the negligible, then the maximum height from the 
centre of earth, to which the body can go, will be 
 (R : radius of earth) 
 (A) 
??
2
R
1
   (B) 
?
2
R
1–
 
 
 (C) 
?
R
1–
   (D) 
?
?
2
2
R
1–
  
 Official Ans. by NTA (B) 
  
Sol.
 
 
 
22
e
GMm 1 GMm
mV
R 2 h
? ? ? ? ? 
 
2
GMm 1 2GMm GMm
R 2 R h
? ? ? ? ? 
 
2
11
R R h
??
?? 
 
2
11
hR
??
? 
 
2
R
h
1
?
??
 
Page 3


 
    1 
FINAL JEE –MAIN EXAMINATION – JULY, 2022 
 (Held On Wednesday 27
th
 July, 2022)              TIME : 3 : 00 PM  to  6 : 00 PM 
PHYSICS TEST PAPER WITH SOLUTION 
 
 
 
 
 
 
 
SECTION-A 
1. An expression of energy density is given by 
?? ??
?
??
?? ?
x
u sin
kt
, where ?, ? are constants, x is 
displacement, k is Boltzmann constant and t is the 
temperature. The dimensions of ? will be : 
 (A) ?
2 –2 –1
[ML T ] (B) 
0 2 –2
[M L T ] 
 (C) 
0 0 0
[M L T ] (D) 
0 2 0
[M L T ] 
 Official Ans. by NTA (D) 
  
Sol. 
?
?
?
0 0 0
22
[L]
[M L T ]
[M L T ]
 
 
?
??
12
[ML T ] 
 
??
?
?
? ? ? ?
?
2 2 1 2 3
3 2 2
[M L T ] [M L T ][L ]
[L ] M L T
 
2. A body of mass 10 kg is projected at an angle of 
45° with the horizontal. The trajectory of the body 
is observed to pass through a point (20, 10). If T is 
the time of flight, then its momentum vector, at 
time ?
T
t
2
, is _________ 
 [Take g = 10 m/s
2
] 
 (A) ?
ˆˆ
100i (100 2 – 200)j 
 (B) ?
ˆˆ
100 2 i (100 – 200 2)j  
 (C) ?
ˆˆ
100 i (100 – 200 2)j  
 (D) ?
ˆˆ
100 2 i (100 2 – 200)j 
 Official Ans. by NTA (D) 
  
Sol. 
 
 u = 20 
 ??
(2)(20)
T 2 2
2 (10)
 
 
?
? ? ?
ˆˆ
v 10 2 i (10 2 10(2)] j 
 Momentum 
??
? ? ? ?
ˆˆ
p M v 100 2 i (100 2 200) j 
3. A block of mass M slides down on a rough 
inclined plane with constant velocity. The angle 
made by the incline plane with horizontal is ?. The 
magnitude of the contact force will be : 
 (A) Mg    
 (B) Mg cos ?    
 (C) ? ? ? Mgsin Mgcos   
 (D) ? ? ? Mgsin 1  
 Official Ans. by NTA (A) 
  
Sol.  
  
 N = Mgcos ? 
 f = Mgsin ? 
 
22
R N f ?? 
 R = Mg 
  
? ? ? ? ?
??
??
??
2
2
2
10x (10)(100)
y x 10 20
1
u
2u
2
 
2 
 
4. A block ‘A’ takes 2 s to slide down a frictionless 
incline of 30° and length ‘l’, kept inside a lift going 
up with uniform velocity ‘v’. If the incline is 
changed to 45°, the time taken by the block, to 
slide down the incline, will be approximately:  
 (A) 2.66 s  (B) 0.83 s  
 (C) 1.68 s  (D) 0.70 s  
 Official Ans. by NTA (C) 
 
Sol.  
  
 a = g sin ?  
 ??
2
1
gsin30 (2)
2
 ….(1) 
 ??
2
1
gsin 45 t
2
 ….(2) 
 
??
? ? ?
??
??
2
1 1
(4) t t 2 2 1.68
2 2
 
5. The velocity of the bullet becomes one third after it 
penetrates 4 cm in a wooden block. Assuming that 
bullet is facing a constant resistance during its 
motion in the block. The bullet stops completely 
after travelling at (4 + x) cm inside the block. The 
value of x is:  
 (A) 2.0  (B) 1.0  
 (C) 0. 5  (D) 1.5  
 Official Ans. by NTA (C) 
  
Sol.
 
 
 
??
? ? ? ? ?
??
??
2 22
2
8V V V
V 2a(4) a
9(8) 9 3
 
 ? ? ?
2
0 V 2a(4 x) 
 ? ?
??
??
??
??
2
2 V
V 2 (4 x)
9
 
  4.5 = 4 + x  
  x = 0.5 ?
6. A body of mass m is projected with velocity ?
e
v in 
vertically upward direction from the surface of the 
earth into space. It is given that 
e
v is escape 
velocity and ? < 1. If air resistance is considered to 
the negligible, then the maximum height from the 
centre of earth, to which the body can go, will be 
 (R : radius of earth) 
 (A) 
??
2
R
1
   (B) 
?
2
R
1–
 
 
 (C) 
?
R
1–
   (D) 
?
?
2
2
R
1–
  
 Official Ans. by NTA (B) 
  
Sol.
 
 
 
22
e
GMm 1 GMm
mV
R 2 h
? ? ? ? ? 
 
2
GMm 1 2GMm GMm
R 2 R h
? ? ? ? ? 
 
2
11
R R h
??
?? 
 
2
11
hR
??
? 
 
2
R
h
1
?
??
 
 
 
3 
 
?  
7. A steel wire of length 3.2 m (Y
S
 = 2.0 × 10
11
 Nm
–2
) 
and a copper wire of length 4.4 M  
(Y
C
 = 1.1 × 10
11
 Nm
–2
), both of radius 1.4 mm are 
connected end to end. When stretched by a load, 
the net elongation is found to be 1.4 mm. The load 
applied, in Newton, will be: (Given ??
22
7
)  
 (A) 360 (B) 180 (C) 1080  (D) 154  
 Official Ans. by NTA (D) 
  
Sol.
 
 
 ? ?
1
 + ? ?
2
 = ? ? 
  ? ? ?
12
1 1 2 2
FF
A y A y
 
 
2
12
1 1 2 2
F 1.54 10 154
A y A y
?
? ? ? ?
?
 
8. In 1
st
 case, Carnot engine operates between 
temperatures 300 K and 100 K. In 2
nd
 case, as 
shown in the figure, a combination of two engines 
is used. The efficiency of this combination (in 2
nd
 
case) will be : 
 
 (A) same as the 1
st
 case     
 (B) always greater than the 1
st
 case   
 (C) always less than the 1
st
 case   
 (D) may increase or decrease with respect to the 1
st
 
case  
 Official Ans. by NTA (C) 
  
Sol. First case : ? ? ? ?
100 2
1
300 3
 
 Second case : ?
net
 = ?
1
 + ?
2
 – ?
1
?
2 
 ? ? ? ?
1
200 1
1
300 3
 
 ? ? ? ?
2
100 1
1
200 2
 
 
net
1 1 1 2
3 2 6 3
? ? ? ? ? 
 ? (first case) = ? (second case) 
9. Which statements are correct about degrees of 
freedom?  
 A.  A molecule with n degrees of freedom has n
2
 
different ways of storing energy.    
 B. Each degree of freedom is associated with 
1
RT
2
 average energy per mole.  
 C. A monoatomic gas molecule has 1 rotational 
degree of freedom where as diatomic molecule 
has 2 rotational degrees of freedom   
 D. CH
4
 has a total to 6 degrees of freedom 
  Choose the correct answer from the option 
given below: 
 (A) B and C only (B) B and D only  
 (C) A and B only  (D) C and D only  
 Official Ans. by NTA (B) 
  
Sol. Methane molecule is tetrahedron 
 Degree of freedom due to rotation = 3 
 Degree of freedom due to translation = 3 
10. A charge of 4 ?C is to be divided into two. The 
distance between the two divided charges is 
constant. The magnitude of the divided charges so 
that the force between them is maximum, will be: 
 (A) 1 ?C and 3 ?C (B) 2 ?C and 2 ?C
 
 
 (C) 0 and 4 ?C (D) 1.5 ?C and 2.5 ?C  
 Official Ans. by NTA (B) 
  
Sol.  
 
?
?
2
Kq(4 q)
F
d
 
 ? ? ?
2
dF K
[4 2q] 0
dq
d
 
 q = 2 
Page 4


 
    1 
FINAL JEE –MAIN EXAMINATION – JULY, 2022 
 (Held On Wednesday 27
th
 July, 2022)              TIME : 3 : 00 PM  to  6 : 00 PM 
PHYSICS TEST PAPER WITH SOLUTION 
 
 
 
 
 
 
 
SECTION-A 
1. An expression of energy density is given by 
?? ??
?
??
?? ?
x
u sin
kt
, where ?, ? are constants, x is 
displacement, k is Boltzmann constant and t is the 
temperature. The dimensions of ? will be : 
 (A) ?
2 –2 –1
[ML T ] (B) 
0 2 –2
[M L T ] 
 (C) 
0 0 0
[M L T ] (D) 
0 2 0
[M L T ] 
 Official Ans. by NTA (D) 
  
Sol. 
?
?
?
0 0 0
22
[L]
[M L T ]
[M L T ]
 
 
?
??
12
[ML T ] 
 
??
?
?
? ? ? ?
?
2 2 1 2 3
3 2 2
[M L T ] [M L T ][L ]
[L ] M L T
 
2. A body of mass 10 kg is projected at an angle of 
45° with the horizontal. The trajectory of the body 
is observed to pass through a point (20, 10). If T is 
the time of flight, then its momentum vector, at 
time ?
T
t
2
, is _________ 
 [Take g = 10 m/s
2
] 
 (A) ?
ˆˆ
100i (100 2 – 200)j 
 (B) ?
ˆˆ
100 2 i (100 – 200 2)j  
 (C) ?
ˆˆ
100 i (100 – 200 2)j  
 (D) ?
ˆˆ
100 2 i (100 2 – 200)j 
 Official Ans. by NTA (D) 
  
Sol. 
 
 u = 20 
 ??
(2)(20)
T 2 2
2 (10)
 
 
?
? ? ?
ˆˆ
v 10 2 i (10 2 10(2)] j 
 Momentum 
??
? ? ? ?
ˆˆ
p M v 100 2 i (100 2 200) j 
3. A block of mass M slides down on a rough 
inclined plane with constant velocity. The angle 
made by the incline plane with horizontal is ?. The 
magnitude of the contact force will be : 
 (A) Mg    
 (B) Mg cos ?    
 (C) ? ? ? Mgsin Mgcos   
 (D) ? ? ? Mgsin 1  
 Official Ans. by NTA (A) 
  
Sol.  
  
 N = Mgcos ? 
 f = Mgsin ? 
 
22
R N f ?? 
 R = Mg 
  
? ? ? ? ?
??
??
??
2
2
2
10x (10)(100)
y x 10 20
1
u
2u
2
 
2 
 
4. A block ‘A’ takes 2 s to slide down a frictionless 
incline of 30° and length ‘l’, kept inside a lift going 
up with uniform velocity ‘v’. If the incline is 
changed to 45°, the time taken by the block, to 
slide down the incline, will be approximately:  
 (A) 2.66 s  (B) 0.83 s  
 (C) 1.68 s  (D) 0.70 s  
 Official Ans. by NTA (C) 
 
Sol.  
  
 a = g sin ?  
 ??
2
1
gsin30 (2)
2
 ….(1) 
 ??
2
1
gsin 45 t
2
 ….(2) 
 
??
? ? ?
??
??
2
1 1
(4) t t 2 2 1.68
2 2
 
5. The velocity of the bullet becomes one third after it 
penetrates 4 cm in a wooden block. Assuming that 
bullet is facing a constant resistance during its 
motion in the block. The bullet stops completely 
after travelling at (4 + x) cm inside the block. The 
value of x is:  
 (A) 2.0  (B) 1.0  
 (C) 0. 5  (D) 1.5  
 Official Ans. by NTA (C) 
  
Sol.
 
 
 
??
? ? ? ? ?
??
??
2 22
2
8V V V
V 2a(4) a
9(8) 9 3
 
 ? ? ?
2
0 V 2a(4 x) 
 ? ?
??
??
??
??
2
2 V
V 2 (4 x)
9
 
  4.5 = 4 + x  
  x = 0.5 ?
6. A body of mass m is projected with velocity ?
e
v in 
vertically upward direction from the surface of the 
earth into space. It is given that 
e
v is escape 
velocity and ? < 1. If air resistance is considered to 
the negligible, then the maximum height from the 
centre of earth, to which the body can go, will be 
 (R : radius of earth) 
 (A) 
??
2
R
1
   (B) 
?
2
R
1–
 
 
 (C) 
?
R
1–
   (D) 
?
?
2
2
R
1–
  
 Official Ans. by NTA (B) 
  
Sol.
 
 
 
22
e
GMm 1 GMm
mV
R 2 h
? ? ? ? ? 
 
2
GMm 1 2GMm GMm
R 2 R h
? ? ? ? ? 
 
2
11
R R h
??
?? 
 
2
11
hR
??
? 
 
2
R
h
1
?
??
 
 
 
3 
 
?  
7. A steel wire of length 3.2 m (Y
S
 = 2.0 × 10
11
 Nm
–2
) 
and a copper wire of length 4.4 M  
(Y
C
 = 1.1 × 10
11
 Nm
–2
), both of radius 1.4 mm are 
connected end to end. When stretched by a load, 
the net elongation is found to be 1.4 mm. The load 
applied, in Newton, will be: (Given ??
22
7
)  
 (A) 360 (B) 180 (C) 1080  (D) 154  
 Official Ans. by NTA (D) 
  
Sol.
 
 
 ? ?
1
 + ? ?
2
 = ? ? 
  ? ? ?
12
1 1 2 2
FF
A y A y
 
 
2
12
1 1 2 2
F 1.54 10 154
A y A y
?
? ? ? ?
?
 
8. In 1
st
 case, Carnot engine operates between 
temperatures 300 K and 100 K. In 2
nd
 case, as 
shown in the figure, a combination of two engines 
is used. The efficiency of this combination (in 2
nd
 
case) will be : 
 
 (A) same as the 1
st
 case     
 (B) always greater than the 1
st
 case   
 (C) always less than the 1
st
 case   
 (D) may increase or decrease with respect to the 1
st
 
case  
 Official Ans. by NTA (C) 
  
Sol. First case : ? ? ? ?
100 2
1
300 3
 
 Second case : ?
net
 = ?
1
 + ?
2
 – ?
1
?
2 
 ? ? ? ?
1
200 1
1
300 3
 
 ? ? ? ?
2
100 1
1
200 2
 
 
net
1 1 1 2
3 2 6 3
? ? ? ? ? 
 ? (first case) = ? (second case) 
9. Which statements are correct about degrees of 
freedom?  
 A.  A molecule with n degrees of freedom has n
2
 
different ways of storing energy.    
 B. Each degree of freedom is associated with 
1
RT
2
 average energy per mole.  
 C. A monoatomic gas molecule has 1 rotational 
degree of freedom where as diatomic molecule 
has 2 rotational degrees of freedom   
 D. CH
4
 has a total to 6 degrees of freedom 
  Choose the correct answer from the option 
given below: 
 (A) B and C only (B) B and D only  
 (C) A and B only  (D) C and D only  
 Official Ans. by NTA (B) 
  
Sol. Methane molecule is tetrahedron 
 Degree of freedom due to rotation = 3 
 Degree of freedom due to translation = 3 
10. A charge of 4 ?C is to be divided into two. The 
distance between the two divided charges is 
constant. The magnitude of the divided charges so 
that the force between them is maximum, will be: 
 (A) 1 ?C and 3 ?C (B) 2 ?C and 2 ?C
 
 
 (C) 0 and 4 ?C (D) 1.5 ?C and 2.5 ?C  
 Official Ans. by NTA (B) 
  
Sol.  
 
?
?
2
Kq(4 q)
F
d
 
 ? ? ?
2
dF K
[4 2q] 0
dq
d
 
 q = 2 
 
4 
 
11. A. The drift velocity of electrons decreases with 
the increase in the temperature of conductor. 
 B. The drift velocity is inversely proportional to 
the area of cross-section of given conductor. 
 C. The drift velocity does not depend on the 
applied potential difference to the conductor. 
 D. The drift velocity of electron is inversely 
proportional to the length of the conductor. 
 E. The drift velocity increases with the increase in 
the temperature of conductor.  
 Choose the correct answer from the options given 
below: 
 (A) A and B only  (B) A and D only  
 (C) B and E only (D) B and C only  
 Official Ans. by NTA (B) 
  
Sol. Drift velocity = 
? ??
??
??
e
E
m
 
 
d
eV
v
m
?? ? ? ? ?
?
? ? ? ?
? ? ? ?
 
 ?V = Potential difference applied across the wire 
 As temperature increases, relaxation time 
decreases, hence V
d
 decreases. 
 As per formula, ?
d
1
V 
 
d
I
v
neA
? , as it is not mentioned that current is at 
steady state neither it is mentioned that n is 
constant for given conductor. So it can't be said 
that v
d
 is inversely proportional to A. 
 I = neAv
d
 = 
VV
A
R
?
?
 
 
d
V
v
ne
?
?
      
V
E
??
?
??
??
 
 
d
eE
v
m
?
?  
 ? decrease with temperature increase. 
 First and fourth statements are correct. 
12. A compass needle of oscillation magnetometer 
oscillates 20 times per minute at a place P of dip 
30°. The number of oscillations per minute become 
10 at another place Q of 60° dip. The ratio of the 
total magnetic field at the two places (B
Q
 : B
P
) is: 
 (A) 3 : 4  (B) 4: 3  
 (C) 3 :2  (D) 2: 3 
 Official Ans. by NTA (A) 
  
Sol. ??
H
I
T2
BM
 
 
1
P
I
T 3sec 2
(B cos30 )M
? ? ?
?
 
 
2
Q
I
T 6sec 2
(B cos60 )M
? ? ?
?
 
 
? ?
Q
P
31
B / 2
6
3
B
2
??
??
??
??
??
 
 
Q
P
B
3
6
3B
??
?
??
??
??
 
 
Q
P
B
3
4B
? 
 
QP
B : B 3 : 4 ? 
13. A cyclotron is used to accelerate protons. If the 
operating magnetic field is 1.0 T and the radius of 
the cyclotron 'dees' is 60 cm, the kinetic energy of 
the accelerated protons in MeV will be : 
 [use m
p
 = 1.6 × 10
–27
 kg, e = 1.6 × 10
–19
 C] 
 (A) 12 (B) 18 
 (C) 16  (D) 32 
 Official Ans. by NTA (B) 
  
Sol. Kinetic energy of electron in cyclotron  
 = 
??
??
??
??
2 2 2
0
q B r
2m
 
 = 18 MeV 
Page 5


 
    1 
FINAL JEE –MAIN EXAMINATION – JULY, 2022 
 (Held On Wednesday 27
th
 July, 2022)              TIME : 3 : 00 PM  to  6 : 00 PM 
PHYSICS TEST PAPER WITH SOLUTION 
 
 
 
 
 
 
 
SECTION-A 
1. An expression of energy density is given by 
?? ??
?
??
?? ?
x
u sin
kt
, where ?, ? are constants, x is 
displacement, k is Boltzmann constant and t is the 
temperature. The dimensions of ? will be : 
 (A) ?
2 –2 –1
[ML T ] (B) 
0 2 –2
[M L T ] 
 (C) 
0 0 0
[M L T ] (D) 
0 2 0
[M L T ] 
 Official Ans. by NTA (D) 
  
Sol. 
?
?
?
0 0 0
22
[L]
[M L T ]
[M L T ]
 
 
?
??
12
[ML T ] 
 
??
?
?
? ? ? ?
?
2 2 1 2 3
3 2 2
[M L T ] [M L T ][L ]
[L ] M L T
 
2. A body of mass 10 kg is projected at an angle of 
45° with the horizontal. The trajectory of the body 
is observed to pass through a point (20, 10). If T is 
the time of flight, then its momentum vector, at 
time ?
T
t
2
, is _________ 
 [Take g = 10 m/s
2
] 
 (A) ?
ˆˆ
100i (100 2 – 200)j 
 (B) ?
ˆˆ
100 2 i (100 – 200 2)j  
 (C) ?
ˆˆ
100 i (100 – 200 2)j  
 (D) ?
ˆˆ
100 2 i (100 2 – 200)j 
 Official Ans. by NTA (D) 
  
Sol. 
 
 u = 20 
 ??
(2)(20)
T 2 2
2 (10)
 
 
?
? ? ?
ˆˆ
v 10 2 i (10 2 10(2)] j 
 Momentum 
??
? ? ? ?
ˆˆ
p M v 100 2 i (100 2 200) j 
3. A block of mass M slides down on a rough 
inclined plane with constant velocity. The angle 
made by the incline plane with horizontal is ?. The 
magnitude of the contact force will be : 
 (A) Mg    
 (B) Mg cos ?    
 (C) ? ? ? Mgsin Mgcos   
 (D) ? ? ? Mgsin 1  
 Official Ans. by NTA (A) 
  
Sol.  
  
 N = Mgcos ? 
 f = Mgsin ? 
 
22
R N f ?? 
 R = Mg 
  
? ? ? ? ?
??
??
??
2
2
2
10x (10)(100)
y x 10 20
1
u
2u
2
 
2 
 
4. A block ‘A’ takes 2 s to slide down a frictionless 
incline of 30° and length ‘l’, kept inside a lift going 
up with uniform velocity ‘v’. If the incline is 
changed to 45°, the time taken by the block, to 
slide down the incline, will be approximately:  
 (A) 2.66 s  (B) 0.83 s  
 (C) 1.68 s  (D) 0.70 s  
 Official Ans. by NTA (C) 
 
Sol.  
  
 a = g sin ?  
 ??
2
1
gsin30 (2)
2
 ….(1) 
 ??
2
1
gsin 45 t
2
 ….(2) 
 
??
? ? ?
??
??
2
1 1
(4) t t 2 2 1.68
2 2
 
5. The velocity of the bullet becomes one third after it 
penetrates 4 cm in a wooden block. Assuming that 
bullet is facing a constant resistance during its 
motion in the block. The bullet stops completely 
after travelling at (4 + x) cm inside the block. The 
value of x is:  
 (A) 2.0  (B) 1.0  
 (C) 0. 5  (D) 1.5  
 Official Ans. by NTA (C) 
  
Sol.
 
 
 
??
? ? ? ? ?
??
??
2 22
2
8V V V
V 2a(4) a
9(8) 9 3
 
 ? ? ?
2
0 V 2a(4 x) 
 ? ?
??
??
??
??
2
2 V
V 2 (4 x)
9
 
  4.5 = 4 + x  
  x = 0.5 ?
6. A body of mass m is projected with velocity ?
e
v in 
vertically upward direction from the surface of the 
earth into space. It is given that 
e
v is escape 
velocity and ? < 1. If air resistance is considered to 
the negligible, then the maximum height from the 
centre of earth, to which the body can go, will be 
 (R : radius of earth) 
 (A) 
??
2
R
1
   (B) 
?
2
R
1–
 
 
 (C) 
?
R
1–
   (D) 
?
?
2
2
R
1–
  
 Official Ans. by NTA (B) 
  
Sol.
 
 
 
22
e
GMm 1 GMm
mV
R 2 h
? ? ? ? ? 
 
2
GMm 1 2GMm GMm
R 2 R h
? ? ? ? ? 
 
2
11
R R h
??
?? 
 
2
11
hR
??
? 
 
2
R
h
1
?
??
 
 
 
3 
 
?  
7. A steel wire of length 3.2 m (Y
S
 = 2.0 × 10
11
 Nm
–2
) 
and a copper wire of length 4.4 M  
(Y
C
 = 1.1 × 10
11
 Nm
–2
), both of radius 1.4 mm are 
connected end to end. When stretched by a load, 
the net elongation is found to be 1.4 mm. The load 
applied, in Newton, will be: (Given ??
22
7
)  
 (A) 360 (B) 180 (C) 1080  (D) 154  
 Official Ans. by NTA (D) 
  
Sol.
 
 
 ? ?
1
 + ? ?
2
 = ? ? 
  ? ? ?
12
1 1 2 2
FF
A y A y
 
 
2
12
1 1 2 2
F 1.54 10 154
A y A y
?
? ? ? ?
?
 
8. In 1
st
 case, Carnot engine operates between 
temperatures 300 K and 100 K. In 2
nd
 case, as 
shown in the figure, a combination of two engines 
is used. The efficiency of this combination (in 2
nd
 
case) will be : 
 
 (A) same as the 1
st
 case     
 (B) always greater than the 1
st
 case   
 (C) always less than the 1
st
 case   
 (D) may increase or decrease with respect to the 1
st
 
case  
 Official Ans. by NTA (C) 
  
Sol. First case : ? ? ? ?
100 2
1
300 3
 
 Second case : ?
net
 = ?
1
 + ?
2
 – ?
1
?
2 
 ? ? ? ?
1
200 1
1
300 3
 
 ? ? ? ?
2
100 1
1
200 2
 
 
net
1 1 1 2
3 2 6 3
? ? ? ? ? 
 ? (first case) = ? (second case) 
9. Which statements are correct about degrees of 
freedom?  
 A.  A molecule with n degrees of freedom has n
2
 
different ways of storing energy.    
 B. Each degree of freedom is associated with 
1
RT
2
 average energy per mole.  
 C. A monoatomic gas molecule has 1 rotational 
degree of freedom where as diatomic molecule 
has 2 rotational degrees of freedom   
 D. CH
4
 has a total to 6 degrees of freedom 
  Choose the correct answer from the option 
given below: 
 (A) B and C only (B) B and D only  
 (C) A and B only  (D) C and D only  
 Official Ans. by NTA (B) 
  
Sol. Methane molecule is tetrahedron 
 Degree of freedom due to rotation = 3 
 Degree of freedom due to translation = 3 
10. A charge of 4 ?C is to be divided into two. The 
distance between the two divided charges is 
constant. The magnitude of the divided charges so 
that the force between them is maximum, will be: 
 (A) 1 ?C and 3 ?C (B) 2 ?C and 2 ?C
 
 
 (C) 0 and 4 ?C (D) 1.5 ?C and 2.5 ?C  
 Official Ans. by NTA (B) 
  
Sol.  
 
?
?
2
Kq(4 q)
F
d
 
 ? ? ?
2
dF K
[4 2q] 0
dq
d
 
 q = 2 
 
4 
 
11. A. The drift velocity of electrons decreases with 
the increase in the temperature of conductor. 
 B. The drift velocity is inversely proportional to 
the area of cross-section of given conductor. 
 C. The drift velocity does not depend on the 
applied potential difference to the conductor. 
 D. The drift velocity of electron is inversely 
proportional to the length of the conductor. 
 E. The drift velocity increases with the increase in 
the temperature of conductor.  
 Choose the correct answer from the options given 
below: 
 (A) A and B only  (B) A and D only  
 (C) B and E only (D) B and C only  
 Official Ans. by NTA (B) 
  
Sol. Drift velocity = 
? ??
??
??
e
E
m
 
 
d
eV
v
m
?? ? ? ? ?
?
? ? ? ?
? ? ? ?
 
 ?V = Potential difference applied across the wire 
 As temperature increases, relaxation time 
decreases, hence V
d
 decreases. 
 As per formula, ?
d
1
V 
 
d
I
v
neA
? , as it is not mentioned that current is at 
steady state neither it is mentioned that n is 
constant for given conductor. So it can't be said 
that v
d
 is inversely proportional to A. 
 I = neAv
d
 = 
VV
A
R
?
?
 
 
d
V
v
ne
?
?
      
V
E
??
?
??
??
 
 
d
eE
v
m
?
?  
 ? decrease with temperature increase. 
 First and fourth statements are correct. 
12. A compass needle of oscillation magnetometer 
oscillates 20 times per minute at a place P of dip 
30°. The number of oscillations per minute become 
10 at another place Q of 60° dip. The ratio of the 
total magnetic field at the two places (B
Q
 : B
P
) is: 
 (A) 3 : 4  (B) 4: 3  
 (C) 3 :2  (D) 2: 3 
 Official Ans. by NTA (A) 
  
Sol. ??
H
I
T2
BM
 
 
1
P
I
T 3sec 2
(B cos30 )M
? ? ?
?
 
 
2
Q
I
T 6sec 2
(B cos60 )M
? ? ?
?
 
 
? ?
Q
P
31
B / 2
6
3
B
2
??
??
??
??
??
 
 
Q
P
B
3
6
3B
??
?
??
??
??
 
 
Q
P
B
3
4B
? 
 
QP
B : B 3 : 4 ? 
13. A cyclotron is used to accelerate protons. If the 
operating magnetic field is 1.0 T and the radius of 
the cyclotron 'dees' is 60 cm, the kinetic energy of 
the accelerated protons in MeV will be : 
 [use m
p
 = 1.6 × 10
–27
 kg, e = 1.6 × 10
–19
 C] 
 (A) 12 (B) 18 
 (C) 16  (D) 32 
 Official Ans. by NTA (B) 
  
Sol. Kinetic energy of electron in cyclotron  
 = 
??
??
??
??
2 2 2
0
q B r
2m
 
 = 18 MeV 
 
 
5 
 
?  
14. A series LCR circuit has L = 0.01 H, R = 10 ? and 
C = 1 ?F and it is connected to ac voltage of 
amplitude (V
m
) 50 V. At frequency 60% lower 
than resonant frequency, the amplitude of current 
will be approximately : 
 (A) 466 mA   (B) 312 mA  
 (C) 238 mA   (D) 196 mA 
 Official Ans. by NTA (C) 
  
Sol. Resonant frequency, 
4
0
1
10 rad/sec
LC
? ? ? 
 
4
' .4 10 4000 rad/sec ? ? ? ? 
 ??
?
0
0
22
CL
V
i 238 mA
R (X' – X' )
 
15. Identify the correct statements from the following 
descriptions of various properties of 
electromagnetic waves. 
 A. In a plane electromagnetic wave electric field 
and magnetic field must be perpendicular to 
each other and direction of propagation of 
wave should be along electric field or magnetic 
field. 
 B. The energy in electromagnetic wave is divided 
equally between electric and magnetic fields. 
 C. Both electric field and magnetic field are 
parallel to each other and perpendicular to the 
direction of propagation of wave. 
 D. The electric field, magnetic field and direction 
of propagation of wave must be perpendicular 
to each other. 
 E. The ratio of amplitude of magnetic field to the 
amplitude of electric field is equal to speed of 
light. 
 Choose the most appropriate answer from the 
options given below:  
(A) D only  
(B) B and D only 
(C) B, C and E only 
(D) A, B and E only 
 Official Ans. by NTA (B) 
  
Sol. Second and fourth statements are correct. 
16. Two coherent sources of light interfere. The 
intensity ratio of two sources is 1 : 4. For this 
interference pattern if the value of 
?
?
max min
max min
II
II
 is 
equal to 
??
??
21
3
, then 
?
?
 will be : 
 (A) 1.5 (B) 2 (C) 0.5 (D) 1 
 Official Ans. by NTA (B) 
  
Sol. ?
1
2
I 1
I4
 
 ?
21
I 4I 
 ? ? ? ?
max 1 1 1 1 1
I I 4I 2 I 4I 9I 
 ? ? ?
min 1 1 1 1 1
I I 4I – 2 I 4I I 
 
? ??
? ? ? ?
??
11
11
9I I 10 5 2 1
9I – I 8 4 1
 
 ? ? ? ? 21 
 
?
? ? ?
?
2
2
1
 
17. With reference to the observations in photo-electric 
effect, identify the correct statements from below: 
 A. The square of maximum velocity of 
photoelectrons varies linearly with frequency 
of incident light. 
 B. The value of saturation current increases on 
moving the source of light away from the metal 
surface. 
 C. The maximum kinetic energy of photo-electrons 
decreases on decreasing the power of LED 
(light emitting diode) source of light. 
 D. The immediate emission of photo-electrons out 
of metal surface can not be explained by 
particle nature of light/electromagnetic waves. 
 E. Existence of threshold wavelength can not be 
explained by wave nature of 
light/electromagnetic waves. 
 Choose the correct answer from the options given 
below: 
 (A)  A and B only  (B) A and E only 
 (C) C and E only (D) D and E only 
 Official Ans. by NTA (B) 
  
Sol. ??
2
max
1
mV hf –
2
 
 Photoelectric effect can be explained by particle 
nature of light. Threshold ? is max wavelength at 
which emission takes place.