JEE Main 2022 Question Paper 25 June Shift 1 | JEE Main & Advanced Previous Year Papers PDF Download

 Page 1


1
Final JEE-Main Exam June, 2022/25-06-2022/Morning Session
SECTION-A
1. If
23
4
AB
Z
C
= , then the relative error in Z will
be :
(A)
ABC
A BC
D DD
++
(B) 
2 A 3B 4C
A BC
D DD
+-
(C) 
2A 3 B 4C
A BC
D DD
++
(D) 
A BC
A BC
D DD
+-
Official Ans. by NTA (C)
Sol.
23
4
AB
Z
C
=
In case of error
dZ 2dA 3dB 4dC
Z A BC
= ++
Z 2A 3B 4C
Z A BC
D D DD
= ++
2.
A
r
 is a vector quantity such that |A|
r
= non-
zero constant. Which of the following
expressions is true for 
A
r
?
(A)
A.A0 =
rr
(B)
A A0 ´<
rr
(C)
A A0 ´=
rr
(D)
A A0 ´>
rr
Official Ans. by NTA (C)
Sol. |A|0 ¹
r
ˆ A A |A||A|sin0 n 0 ´ = °=
r r rr
3. Which of the following relations is true for two
unit vectors  ˆ
A
 and  ˆ
B
 making an angle q  to
each other?
(A)
ˆˆ ˆˆ
|A B| |A B|tan
2
q
+ =-
(B)
ˆˆ ˆˆ
|A B| |A B|tan
2
q
- =+
(C)
ˆˆ ˆˆ
|A B| |A B|cos
2
q
+ =-
(D)
ˆˆ ˆˆ
|A B| |A B|cos
2
q
- =+
Official Ans. by NTA (B)
Sol.
22
ˆ ˆˆ ˆ ˆˆ
|A B| |A| |B| 2|A||B|cos += ++q
1 1 2cos = ++q
2(1 cos ) = +q
2
2 2cos
2
q
=´
2cos
2
q
=
22
ˆ ˆˆ ˆ ˆˆ
|A B| |A| |B| 2|A||B|cos -= +-q
2 2cos = -q
2sin
2
q
=
ˆˆ
|A B|
cot
ˆˆ
2 |A B|
+q
=
-
FINAL JEE–MAIN EXAMINATION – JUNE, 2022
(Held On Saturday 25
th
 June, 2022) TIME : 9 : 00 AM   to  12 : 00 PM
PHYSICS TEST PAPER WITH SOLUTION
Page 2


1
Final JEE-Main Exam June, 2022/25-06-2022/Morning Session
SECTION-A
1. If
23
4
AB
Z
C
= , then the relative error in Z will
be :
(A)
ABC
A BC
D DD
++
(B) 
2 A 3B 4C
A BC
D DD
+-
(C) 
2A 3 B 4C
A BC
D DD
++
(D) 
A BC
A BC
D DD
+-
Official Ans. by NTA (C)
Sol.
23
4
AB
Z
C
=
In case of error
dZ 2dA 3dB 4dC
Z A BC
= ++
Z 2A 3B 4C
Z A BC
D D DD
= ++
2.
A
r
 is a vector quantity such that |A|
r
= non-
zero constant. Which of the following
expressions is true for 
A
r
?
(A)
A.A0 =
rr
(B)
A A0 ´<
rr
(C)
A A0 ´=
rr
(D)
A A0 ´>
rr
Official Ans. by NTA (C)
Sol. |A|0 ¹
r
ˆ A A |A||A|sin0 n 0 ´ = °=
r r rr
3. Which of the following relations is true for two
unit vectors  ˆ
A
 and  ˆ
B
 making an angle q  to
each other?
(A)
ˆˆ ˆˆ
|A B| |A B|tan
2
q
+ =-
(B)
ˆˆ ˆˆ
|A B| |A B|tan
2
q
- =+
(C)
ˆˆ ˆˆ
|A B| |A B|cos
2
q
+ =-
(D)
ˆˆ ˆˆ
|A B| |A B|cos
2
q
- =+
Official Ans. by NTA (B)
Sol.
22
ˆ ˆˆ ˆ ˆˆ
|A B| |A| |B| 2|A||B|cos += ++q
1 1 2cos = ++q
2(1 cos ) = +q
2
2 2cos
2
q
=´
2cos
2
q
=
22
ˆ ˆˆ ˆ ˆˆ
|A B| |A| |B| 2|A||B|cos -= +-q
2 2cos = -q
2sin
2
q
=
ˆˆ
|A B|
cot
ˆˆ
2 |A B|
+q
=
-
FINAL JEE–MAIN EXAMINATION – JUNE, 2022
(Held On Saturday 25
th
 June, 2022) TIME : 9 : 00 AM   to  12 : 00 PM
PHYSICS TEST PAPER WITH SOLUTION
2
Final JEE-Main Exam June, 2022/25-06-2022/Morning Session
4. If force 
ˆ ˆˆ
F 3i 4j 2k =+-
r
 acts on a particle having
position vector 
ˆˆˆ
2i j 2k ++
 then, the torque
about the origin will be :-
(A)
ˆ ˆˆ
3i 4j 2k +-
(B) 
ˆ ˆˆ
10i 10j 5k - ++
(C) 
ˆ ˆˆ
10i 5j 10k +-
(D) 
ˆˆˆ
10i j 5k +-
Official Ans. by NTA (B)
Sol.
rF t=´
r
r r
ˆ ˆˆ
i jk
2 12
342
=
-
ˆ ˆˆ
i( 2 8) j( 4 6) k(8 3) = -- - -- + -
ˆ ˆˆ
10i 10j 5k =- ++
5. The height of any point P above the surface of
earth is equal to diameter of earth. The value
of acceleration due to gravity at point P will be
: (Given g = acceleration due to gravity at the
surface of earth)
(A) g/2
(B) g/4
(C) g/3
(D) g/9
Official Ans. by NTA (D)
Sol.
2
Gm
g
r
=
2
Gm
g
(3r)
¢ =
 
r
2r
2
Gm
g
9r
¢ =
g
g
9
¢ =
6. The terminal velocity (v
t
) of the spherical rain
drop depends on the radius (r) of the spherical
rain drop as:-
(A) r
1/2
(B) r
(C) r
2
(D) r
3
Official Ans. by NTA (C)
Sol.
2
pl
t
gr()
2
v
9
r -r
=
h
;    
2
t
vr µ
7. The relation between root mean square speed
(v
rms
) and most probable speed (v
p
) for the
molar mass M of oxygen gas molecule at the
temperature of 300 K will be :-
(A)
rmsp
2
vv
3
=
(B) 
rmsp
3
vv
2
=
(C) v
rms
 = v
p
(D) 
rmsp
1
vv
3
=
Official Ans. by NTA (B)
Sol.
rms
3RT
v
M
=
 and 
mp
2RT
v
M
=
Thus 
rms mp
3
vv
2
=
8. In the figure, a very large plane sheet of
positive charge is shown. P
1
 and P
2
 are two
points at distance l and 2l from the charge
distribution. If s is the surface charge density,
then the magnitude of electric fields E
1
 and E
2
at P
1
 and P
2
 respectively are :
(A) 
1 020
E / ,E /2 =s e =s e
(B) 
1 020
E2/,E/ = s e =se
(C) 
120
E E /2 = =se
(D) 
1 20
EE/ = =se
Official Ans. by NTA (C)
Page 3


1
Final JEE-Main Exam June, 2022/25-06-2022/Morning Session
SECTION-A
1. If
23
4
AB
Z
C
= , then the relative error in Z will
be :
(A)
ABC
A BC
D DD
++
(B) 
2 A 3B 4C
A BC
D DD
+-
(C) 
2A 3 B 4C
A BC
D DD
++
(D) 
A BC
A BC
D DD
+-
Official Ans. by NTA (C)
Sol.
23
4
AB
Z
C
=
In case of error
dZ 2dA 3dB 4dC
Z A BC
= ++
Z 2A 3B 4C
Z A BC
D D DD
= ++
2.
A
r
 is a vector quantity such that |A|
r
= non-
zero constant. Which of the following
expressions is true for 
A
r
?
(A)
A.A0 =
rr
(B)
A A0 ´<
rr
(C)
A A0 ´=
rr
(D)
A A0 ´>
rr
Official Ans. by NTA (C)
Sol. |A|0 ¹
r
ˆ A A |A||A|sin0 n 0 ´ = °=
r r rr
3. Which of the following relations is true for two
unit vectors  ˆ
A
 and  ˆ
B
 making an angle q  to
each other?
(A)
ˆˆ ˆˆ
|A B| |A B|tan
2
q
+ =-
(B)
ˆˆ ˆˆ
|A B| |A B|tan
2
q
- =+
(C)
ˆˆ ˆˆ
|A B| |A B|cos
2
q
+ =-
(D)
ˆˆ ˆˆ
|A B| |A B|cos
2
q
- =+
Official Ans. by NTA (B)
Sol.
22
ˆ ˆˆ ˆ ˆˆ
|A B| |A| |B| 2|A||B|cos += ++q
1 1 2cos = ++q
2(1 cos ) = +q
2
2 2cos
2
q
=´
2cos
2
q
=
22
ˆ ˆˆ ˆ ˆˆ
|A B| |A| |B| 2|A||B|cos -= +-q
2 2cos = -q
2sin
2
q
=
ˆˆ
|A B|
cot
ˆˆ
2 |A B|
+q
=
-
FINAL JEE–MAIN EXAMINATION – JUNE, 2022
(Held On Saturday 25
th
 June, 2022) TIME : 9 : 00 AM   to  12 : 00 PM
PHYSICS TEST PAPER WITH SOLUTION
2
Final JEE-Main Exam June, 2022/25-06-2022/Morning Session
4. If force 
ˆ ˆˆ
F 3i 4j 2k =+-
r
 acts on a particle having
position vector 
ˆˆˆ
2i j 2k ++
 then, the torque
about the origin will be :-
(A)
ˆ ˆˆ
3i 4j 2k +-
(B) 
ˆ ˆˆ
10i 10j 5k - ++
(C) 
ˆ ˆˆ
10i 5j 10k +-
(D) 
ˆˆˆ
10i j 5k +-
Official Ans. by NTA (B)
Sol.
rF t=´
r
r r
ˆ ˆˆ
i jk
2 12
342
=
-
ˆ ˆˆ
i( 2 8) j( 4 6) k(8 3) = -- - -- + -
ˆ ˆˆ
10i 10j 5k =- ++
5. The height of any point P above the surface of
earth is equal to diameter of earth. The value
of acceleration due to gravity at point P will be
: (Given g = acceleration due to gravity at the
surface of earth)
(A) g/2
(B) g/4
(C) g/3
(D) g/9
Official Ans. by NTA (D)
Sol.
2
Gm
g
r
=
2
Gm
g
(3r)
¢ =
 
r
2r
2
Gm
g
9r
¢ =
g
g
9
¢ =
6. The terminal velocity (v
t
) of the spherical rain
drop depends on the radius (r) of the spherical
rain drop as:-
(A) r
1/2
(B) r
(C) r
2
(D) r
3
Official Ans. by NTA (C)
Sol.
2
pl
t
gr()
2
v
9
r -r
=
h
;    
2
t
vr µ
7. The relation between root mean square speed
(v
rms
) and most probable speed (v
p
) for the
molar mass M of oxygen gas molecule at the
temperature of 300 K will be :-
(A)
rmsp
2
vv
3
=
(B) 
rmsp
3
vv
2
=
(C) v
rms
 = v
p
(D) 
rmsp
1
vv
3
=
Official Ans. by NTA (B)
Sol.
rms
3RT
v
M
=
 and 
mp
2RT
v
M
=
Thus 
rms mp
3
vv
2
=
8. In the figure, a very large plane sheet of
positive charge is shown. P
1
 and P
2
 are two
points at distance l and 2l from the charge
distribution. If s is the surface charge density,
then the magnitude of electric fields E
1
 and E
2
at P
1
 and P
2
 respectively are :
(A) 
1 020
E / ,E /2 =s e =s e
(B) 
1 020
E2/,E/ = s e =se
(C) 
120
E E /2 = =se
(D) 
1 20
EE/ = =se
Official Ans. by NTA (C)
3
Final JEE-Main Exam June, 2022/25-06-2022/Morning Session
Sol. As the sheet is very large 
E
r
 is independent of
distance from it.
Thus 12
0
EE
2
s
==
e
9. Match List-I with List-II
List-I List-II
(A) AC generator (I) Detects the presence
of current in the
circuit
(B) Galvanometer (II) Converts mechanical
energy into electrical
energy
(C) Transformer (III) Works on the principle
of resonance in AC
circuit
(D) Metal detector (IV) Changes an alternating
voltage for smaller or
greater value
Choose the correct answer from the options
given below :-
(A) (A)–(II), B–(I), (C)–(IV), (D)–(III)
(B) (A)–(II), B–(I), (C)–(III), (D)–(IV)
(C) (A)–(III), B–(IV), (C)–(II), (D)–(I)
(D) (A)–(III), B–(I), (C)–(II), (D)–(IV)
Official Ans. by NTA (A)
Sol. AC generator converts mechanical energy into
electrical energy. Galvanometer shows
deflection when current passes through it so it
is used to show presence of current in any wire.
Transformer is used to step up or step down
the voltage. Metals detectors contain inductor
coils and use principle of induction and
resonance in AC circuit.
10. A long straight wire with a circular cross-
section having radius R, is carrying a steady
current I. The current I is uniformly distributed
across this cross-section. Then the variation of
magnetic field due to current I with distance r
(r < R) from its centre will be :-
(A)
2
Br µ
(B) 
Br µ
(C) 
2
1
B
r
µ
(D) 
1
B
r
µ
Official Ans. by NTA (B)
Sol. Use Ampere’s law
r
R
2
0 2
I
B.2r . .r
R
p=mp
p
Thus 
Br µ
11. If wattless current flows in the AC circuit, then
the circuit is
(A) Purely Resistive circuit
(B) Purely Inductive circuit
(C) LCR series circuit
(D) RC series circuit only
Official Ans. by NTA (B)
Sol. Purely Inductive circuit
2
p
q=
cos0
2
p
=
Average power = 0
12. The electric field in an electromagnetic wave
is given by 
1
E 56.5sin (t x/c)NC
-
= w- . Find
the intensity of the wave if it is propagating
along x-axis in the free space. (Given
122 12
0
8.85 10 C N m )
- --
e=´
(A) 5.65 Wm
–2
(B) 4.24 Wm
–2
(C) 1.9 × 10
–7
 Wm
–2
(D) 56.5 Wm
–2
Official Ans. by NTA (B)
Sol.
2
00
1
I Ec
2
=e
12 28
1
I (8.85 10 )(56.5) (3 10 )
2
-
= ´ ´ ´´
= 4.24 Wm
–2
.
Page 4


1
Final JEE-Main Exam June, 2022/25-06-2022/Morning Session
SECTION-A
1. If
23
4
AB
Z
C
= , then the relative error in Z will
be :
(A)
ABC
A BC
D DD
++
(B) 
2 A 3B 4C
A BC
D DD
+-
(C) 
2A 3 B 4C
A BC
D DD
++
(D) 
A BC
A BC
D DD
+-
Official Ans. by NTA (C)
Sol.
23
4
AB
Z
C
=
In case of error
dZ 2dA 3dB 4dC
Z A BC
= ++
Z 2A 3B 4C
Z A BC
D D DD
= ++
2.
A
r
 is a vector quantity such that |A|
r
= non-
zero constant. Which of the following
expressions is true for 
A
r
?
(A)
A.A0 =
rr
(B)
A A0 ´<
rr
(C)
A A0 ´=
rr
(D)
A A0 ´>
rr
Official Ans. by NTA (C)
Sol. |A|0 ¹
r
ˆ A A |A||A|sin0 n 0 ´ = °=
r r rr
3. Which of the following relations is true for two
unit vectors  ˆ
A
 and  ˆ
B
 making an angle q  to
each other?
(A)
ˆˆ ˆˆ
|A B| |A B|tan
2
q
+ =-
(B)
ˆˆ ˆˆ
|A B| |A B|tan
2
q
- =+
(C)
ˆˆ ˆˆ
|A B| |A B|cos
2
q
+ =-
(D)
ˆˆ ˆˆ
|A B| |A B|cos
2
q
- =+
Official Ans. by NTA (B)
Sol.
22
ˆ ˆˆ ˆ ˆˆ
|A B| |A| |B| 2|A||B|cos += ++q
1 1 2cos = ++q
2(1 cos ) = +q
2
2 2cos
2
q
=´
2cos
2
q
=
22
ˆ ˆˆ ˆ ˆˆ
|A B| |A| |B| 2|A||B|cos -= +-q
2 2cos = -q
2sin
2
q
=
ˆˆ
|A B|
cot
ˆˆ
2 |A B|
+q
=
-
FINAL JEE–MAIN EXAMINATION – JUNE, 2022
(Held On Saturday 25
th
 June, 2022) TIME : 9 : 00 AM   to  12 : 00 PM
PHYSICS TEST PAPER WITH SOLUTION
2
Final JEE-Main Exam June, 2022/25-06-2022/Morning Session
4. If force 
ˆ ˆˆ
F 3i 4j 2k =+-
r
 acts on a particle having
position vector 
ˆˆˆ
2i j 2k ++
 then, the torque
about the origin will be :-
(A)
ˆ ˆˆ
3i 4j 2k +-
(B) 
ˆ ˆˆ
10i 10j 5k - ++
(C) 
ˆ ˆˆ
10i 5j 10k +-
(D) 
ˆˆˆ
10i j 5k +-
Official Ans. by NTA (B)
Sol.
rF t=´
r
r r
ˆ ˆˆ
i jk
2 12
342
=
-
ˆ ˆˆ
i( 2 8) j( 4 6) k(8 3) = -- - -- + -
ˆ ˆˆ
10i 10j 5k =- ++
5. The height of any point P above the surface of
earth is equal to diameter of earth. The value
of acceleration due to gravity at point P will be
: (Given g = acceleration due to gravity at the
surface of earth)
(A) g/2
(B) g/4
(C) g/3
(D) g/9
Official Ans. by NTA (D)
Sol.
2
Gm
g
r
=
2
Gm
g
(3r)
¢ =
 
r
2r
2
Gm
g
9r
¢ =
g
g
9
¢ =
6. The terminal velocity (v
t
) of the spherical rain
drop depends on the radius (r) of the spherical
rain drop as:-
(A) r
1/2
(B) r
(C) r
2
(D) r
3
Official Ans. by NTA (C)
Sol.
2
pl
t
gr()
2
v
9
r -r
=
h
;    
2
t
vr µ
7. The relation between root mean square speed
(v
rms
) and most probable speed (v
p
) for the
molar mass M of oxygen gas molecule at the
temperature of 300 K will be :-
(A)
rmsp
2
vv
3
=
(B) 
rmsp
3
vv
2
=
(C) v
rms
 = v
p
(D) 
rmsp
1
vv
3
=
Official Ans. by NTA (B)
Sol.
rms
3RT
v
M
=
 and 
mp
2RT
v
M
=
Thus 
rms mp
3
vv
2
=
8. In the figure, a very large plane sheet of
positive charge is shown. P
1
 and P
2
 are two
points at distance l and 2l from the charge
distribution. If s is the surface charge density,
then the magnitude of electric fields E
1
 and E
2
at P
1
 and P
2
 respectively are :
(A) 
1 020
E / ,E /2 =s e =s e
(B) 
1 020
E2/,E/ = s e =se
(C) 
120
E E /2 = =se
(D) 
1 20
EE/ = =se
Official Ans. by NTA (C)
3
Final JEE-Main Exam June, 2022/25-06-2022/Morning Session
Sol. As the sheet is very large 
E
r
 is independent of
distance from it.
Thus 12
0
EE
2
s
==
e
9. Match List-I with List-II
List-I List-II
(A) AC generator (I) Detects the presence
of current in the
circuit
(B) Galvanometer (II) Converts mechanical
energy into electrical
energy
(C) Transformer (III) Works on the principle
of resonance in AC
circuit
(D) Metal detector (IV) Changes an alternating
voltage for smaller or
greater value
Choose the correct answer from the options
given below :-
(A) (A)–(II), B–(I), (C)–(IV), (D)–(III)
(B) (A)–(II), B–(I), (C)–(III), (D)–(IV)
(C) (A)–(III), B–(IV), (C)–(II), (D)–(I)
(D) (A)–(III), B–(I), (C)–(II), (D)–(IV)
Official Ans. by NTA (A)
Sol. AC generator converts mechanical energy into
electrical energy. Galvanometer shows
deflection when current passes through it so it
is used to show presence of current in any wire.
Transformer is used to step up or step down
the voltage. Metals detectors contain inductor
coils and use principle of induction and
resonance in AC circuit.
10. A long straight wire with a circular cross-
section having radius R, is carrying a steady
current I. The current I is uniformly distributed
across this cross-section. Then the variation of
magnetic field due to current I with distance r
(r < R) from its centre will be :-
(A)
2
Br µ
(B) 
Br µ
(C) 
2
1
B
r
µ
(D) 
1
B
r
µ
Official Ans. by NTA (B)
Sol. Use Ampere’s law
r
R
2
0 2
I
B.2r . .r
R
p=mp
p
Thus 
Br µ
11. If wattless current flows in the AC circuit, then
the circuit is
(A) Purely Resistive circuit
(B) Purely Inductive circuit
(C) LCR series circuit
(D) RC series circuit only
Official Ans. by NTA (B)
Sol. Purely Inductive circuit
2
p
q=
cos0
2
p
=
Average power = 0
12. The electric field in an electromagnetic wave
is given by 
1
E 56.5sin (t x/c)NC
-
= w- . Find
the intensity of the wave if it is propagating
along x-axis in the free space. (Given
122 12
0
8.85 10 C N m )
- --
e=´
(A) 5.65 Wm
–2
(B) 4.24 Wm
–2
(C) 1.9 × 10
–7
 Wm
–2
(D) 56.5 Wm
–2
Official Ans. by NTA (B)
Sol.
2
00
1
I Ec
2
=e
12 28
1
I (8.85 10 )(56.5) (3 10 )
2
-
= ´ ´ ´´
= 4.24 Wm
–2
.
4
Final JEE-Main Exam June, 2022/25-06-2022/Morning Session
13. The two light beams having intensities I and
9I interfere to produce a fringe pattern on a
screen. The phase difference between the
beams is 
2
p
 at point P and p at point Q. Then
the difference between the resultant intensities
at P and Q will be :
(A) 2 I (B) 6 I
(C) 5 I (D) 7 I
Official Ans. by NTA (B)
Sol.
P
I I 9I 2 I 9Icos
2
p
=+ +´
I
P
 = 10 I
Q
I I 9I 2 I 9Icos =++ ´p
= 10 I – 6I = 4I
PQ
I I 10I 4I 6I \ - = -=
14. A light wave travelling linearly in a medium
of dielectric constant 4, incident on the
horizontal interface separating medium with air.
The angle of incidence for which the total
intensity of incident wave will be reflected back
into the same medium will be (Given : relative
permeability of medium µ
r
 = 1)
(A) 10° (B) 20°
(C) 30° (D) 60°
Official Ans. by NTA (D)
Sol. For total internal reflection, 
C
i >q
C
sini sin Þ >q
R
D
sini
m
Þ>
m
.............(1)
Also 
rr
m= mÎ
R
D
111
2 41
m´
==
m ´
From (1), 
1
sini
2
>
i 30 Þ >°
, i = 60°
15. Given below are two statements :-
Statement I : Davisson-Germer experiment
establishes the wave nature of electrons.
Statement II : If electrons have wave nature,
they can interfere and show diffraction.
In the light of the above statements choose the
correct answer from the options given below:-
(A) Both Statement I and Statement II are true
(B) Both Statement I and Statement II are false
(C) Statement I is true but Statement II is false
(D) Statement I is false but Statement II is
true
Official Ans. by NTA (A)
Sol. In Davisson-Germer experiment the electrons
exhibit diffraction there by proving that
electrons have wave nature. Hence both
statement are correct.
Sol. Both the options are correct by concept.
16. The ratio for the speed of the electron in the 3
rd
orbit of He
+
 to the speed of the electron in the
3
rd
 orbit of hydrogen atom will be :-
(A) 1 : 1 (B) 1 : 2
(C) 4 : 1 (D) 2 : 1
Official Ans. by NTA (D)
Sol.
Z
vZ
n
µµ
 (n = constant)
He He
HH
vZ
2
v Z1
++
Þ ==
17. The photodiode is used to detect the optiocal
signals. These diodes are preferably operated
in reverse biased mode because.
(A) fractional change in majority carriers
produce higher forward bias current
(B) fractional change in majority carriers
produce higher reverse bias current
(C) fractional change in minority carriers
produce higher forward bias current
(D) fractional change in minority carriers
produce higher reverse bias current
Official Ans. by NTA (D)
Page 5


1
Final JEE-Main Exam June, 2022/25-06-2022/Morning Session
SECTION-A
1. If
23
4
AB
Z
C
= , then the relative error in Z will
be :
(A)
ABC
A BC
D DD
++
(B) 
2 A 3B 4C
A BC
D DD
+-
(C) 
2A 3 B 4C
A BC
D DD
++
(D) 
A BC
A BC
D DD
+-
Official Ans. by NTA (C)
Sol.
23
4
AB
Z
C
=
In case of error
dZ 2dA 3dB 4dC
Z A BC
= ++
Z 2A 3B 4C
Z A BC
D D DD
= ++
2.
A
r
 is a vector quantity such that |A|
r
= non-
zero constant. Which of the following
expressions is true for 
A
r
?
(A)
A.A0 =
rr
(B)
A A0 ´<
rr
(C)
A A0 ´=
rr
(D)
A A0 ´>
rr
Official Ans. by NTA (C)
Sol. |A|0 ¹
r
ˆ A A |A||A|sin0 n 0 ´ = °=
r r rr
3. Which of the following relations is true for two
unit vectors  ˆ
A
 and  ˆ
B
 making an angle q  to
each other?
(A)
ˆˆ ˆˆ
|A B| |A B|tan
2
q
+ =-
(B)
ˆˆ ˆˆ
|A B| |A B|tan
2
q
- =+
(C)
ˆˆ ˆˆ
|A B| |A B|cos
2
q
+ =-
(D)
ˆˆ ˆˆ
|A B| |A B|cos
2
q
- =+
Official Ans. by NTA (B)
Sol.
22
ˆ ˆˆ ˆ ˆˆ
|A B| |A| |B| 2|A||B|cos += ++q
1 1 2cos = ++q
2(1 cos ) = +q
2
2 2cos
2
q
=´
2cos
2
q
=
22
ˆ ˆˆ ˆ ˆˆ
|A B| |A| |B| 2|A||B|cos -= +-q
2 2cos = -q
2sin
2
q
=
ˆˆ
|A B|
cot
ˆˆ
2 |A B|
+q
=
-
FINAL JEE–MAIN EXAMINATION – JUNE, 2022
(Held On Saturday 25
th
 June, 2022) TIME : 9 : 00 AM   to  12 : 00 PM
PHYSICS TEST PAPER WITH SOLUTION
2
Final JEE-Main Exam June, 2022/25-06-2022/Morning Session
4. If force 
ˆ ˆˆ
F 3i 4j 2k =+-
r
 acts on a particle having
position vector 
ˆˆˆ
2i j 2k ++
 then, the torque
about the origin will be :-
(A)
ˆ ˆˆ
3i 4j 2k +-
(B) 
ˆ ˆˆ
10i 10j 5k - ++
(C) 
ˆ ˆˆ
10i 5j 10k +-
(D) 
ˆˆˆ
10i j 5k +-
Official Ans. by NTA (B)
Sol.
rF t=´
r
r r
ˆ ˆˆ
i jk
2 12
342
=
-
ˆ ˆˆ
i( 2 8) j( 4 6) k(8 3) = -- - -- + -
ˆ ˆˆ
10i 10j 5k =- ++
5. The height of any point P above the surface of
earth is equal to diameter of earth. The value
of acceleration due to gravity at point P will be
: (Given g = acceleration due to gravity at the
surface of earth)
(A) g/2
(B) g/4
(C) g/3
(D) g/9
Official Ans. by NTA (D)
Sol.
2
Gm
g
r
=
2
Gm
g
(3r)
¢ =
 
r
2r
2
Gm
g
9r
¢ =
g
g
9
¢ =
6. The terminal velocity (v
t
) of the spherical rain
drop depends on the radius (r) of the spherical
rain drop as:-
(A) r
1/2
(B) r
(C) r
2
(D) r
3
Official Ans. by NTA (C)
Sol.
2
pl
t
gr()
2
v
9
r -r
=
h
;    
2
t
vr µ
7. The relation between root mean square speed
(v
rms
) and most probable speed (v
p
) for the
molar mass M of oxygen gas molecule at the
temperature of 300 K will be :-
(A)
rmsp
2
vv
3
=
(B) 
rmsp
3
vv
2
=
(C) v
rms
 = v
p
(D) 
rmsp
1
vv
3
=
Official Ans. by NTA (B)
Sol.
rms
3RT
v
M
=
 and 
mp
2RT
v
M
=
Thus 
rms mp
3
vv
2
=
8. In the figure, a very large plane sheet of
positive charge is shown. P
1
 and P
2
 are two
points at distance l and 2l from the charge
distribution. If s is the surface charge density,
then the magnitude of electric fields E
1
 and E
2
at P
1
 and P
2
 respectively are :
(A) 
1 020
E / ,E /2 =s e =s e
(B) 
1 020
E2/,E/ = s e =se
(C) 
120
E E /2 = =se
(D) 
1 20
EE/ = =se
Official Ans. by NTA (C)
3
Final JEE-Main Exam June, 2022/25-06-2022/Morning Session
Sol. As the sheet is very large 
E
r
 is independent of
distance from it.
Thus 12
0
EE
2
s
==
e
9. Match List-I with List-II
List-I List-II
(A) AC generator (I) Detects the presence
of current in the
circuit
(B) Galvanometer (II) Converts mechanical
energy into electrical
energy
(C) Transformer (III) Works on the principle
of resonance in AC
circuit
(D) Metal detector (IV) Changes an alternating
voltage for smaller or
greater value
Choose the correct answer from the options
given below :-
(A) (A)–(II), B–(I), (C)–(IV), (D)–(III)
(B) (A)–(II), B–(I), (C)–(III), (D)–(IV)
(C) (A)–(III), B–(IV), (C)–(II), (D)–(I)
(D) (A)–(III), B–(I), (C)–(II), (D)–(IV)
Official Ans. by NTA (A)
Sol. AC generator converts mechanical energy into
electrical energy. Galvanometer shows
deflection when current passes through it so it
is used to show presence of current in any wire.
Transformer is used to step up or step down
the voltage. Metals detectors contain inductor
coils and use principle of induction and
resonance in AC circuit.
10. A long straight wire with a circular cross-
section having radius R, is carrying a steady
current I. The current I is uniformly distributed
across this cross-section. Then the variation of
magnetic field due to current I with distance r
(r < R) from its centre will be :-
(A)
2
Br µ
(B) 
Br µ
(C) 
2
1
B
r
µ
(D) 
1
B
r
µ
Official Ans. by NTA (B)
Sol. Use Ampere’s law
r
R
2
0 2
I
B.2r . .r
R
p=mp
p
Thus 
Br µ
11. If wattless current flows in the AC circuit, then
the circuit is
(A) Purely Resistive circuit
(B) Purely Inductive circuit
(C) LCR series circuit
(D) RC series circuit only
Official Ans. by NTA (B)
Sol. Purely Inductive circuit
2
p
q=
cos0
2
p
=
Average power = 0
12. The electric field in an electromagnetic wave
is given by 
1
E 56.5sin (t x/c)NC
-
= w- . Find
the intensity of the wave if it is propagating
along x-axis in the free space. (Given
122 12
0
8.85 10 C N m )
- --
e=´
(A) 5.65 Wm
–2
(B) 4.24 Wm
–2
(C) 1.9 × 10
–7
 Wm
–2
(D) 56.5 Wm
–2
Official Ans. by NTA (B)
Sol.
2
00
1
I Ec
2
=e
12 28
1
I (8.85 10 )(56.5) (3 10 )
2
-
= ´ ´ ´´
= 4.24 Wm
–2
.
4
Final JEE-Main Exam June, 2022/25-06-2022/Morning Session
13. The two light beams having intensities I and
9I interfere to produce a fringe pattern on a
screen. The phase difference between the
beams is 
2
p
 at point P and p at point Q. Then
the difference between the resultant intensities
at P and Q will be :
(A) 2 I (B) 6 I
(C) 5 I (D) 7 I
Official Ans. by NTA (B)
Sol.
P
I I 9I 2 I 9Icos
2
p
=+ +´
I
P
 = 10 I
Q
I I 9I 2 I 9Icos =++ ´p
= 10 I – 6I = 4I
PQ
I I 10I 4I 6I \ - = -=
14. A light wave travelling linearly in a medium
of dielectric constant 4, incident on the
horizontal interface separating medium with air.
The angle of incidence for which the total
intensity of incident wave will be reflected back
into the same medium will be (Given : relative
permeability of medium µ
r
 = 1)
(A) 10° (B) 20°
(C) 30° (D) 60°
Official Ans. by NTA (D)
Sol. For total internal reflection, 
C
i >q
C
sini sin Þ >q
R
D
sini
m
Þ>
m
.............(1)
Also 
rr
m= mÎ
R
D
111
2 41
m´
==
m ´
From (1), 
1
sini
2
>
i 30 Þ >°
, i = 60°
15. Given below are two statements :-
Statement I : Davisson-Germer experiment
establishes the wave nature of electrons.
Statement II : If electrons have wave nature,
they can interfere and show diffraction.
In the light of the above statements choose the
correct answer from the options given below:-
(A) Both Statement I and Statement II are true
(B) Both Statement I and Statement II are false
(C) Statement I is true but Statement II is false
(D) Statement I is false but Statement II is
true
Official Ans. by NTA (A)
Sol. In Davisson-Germer experiment the electrons
exhibit diffraction there by proving that
electrons have wave nature. Hence both
statement are correct.
Sol. Both the options are correct by concept.
16. The ratio for the speed of the electron in the 3
rd
orbit of He
+
 to the speed of the electron in the
3
rd
 orbit of hydrogen atom will be :-
(A) 1 : 1 (B) 1 : 2
(C) 4 : 1 (D) 2 : 1
Official Ans. by NTA (D)
Sol.
Z
vZ
n
µµ
 (n = constant)
He He
HH
vZ
2
v Z1
++
Þ ==
17. The photodiode is used to detect the optiocal
signals. These diodes are preferably operated
in reverse biased mode because.
(A) fractional change in majority carriers
produce higher forward bias current
(B) fractional change in majority carriers
produce higher reverse bias current
(C) fractional change in minority carriers
produce higher forward bias current
(D) fractional change in minority carriers
produce higher reverse bias current
Official Ans. by NTA (D)
5
Final JEE-Main Exam June, 2022/25-06-2022/Morning Session
Sol. Very small change in minority charge carriers
produces high value of reverse bias current.
18. A signal of 100 THz frequency can be
transmitted with maximum efficiency by :
(A) Coaxial cable
(B) Optical fibre
(C) Twisted pair of copper wires
(D) Water
Official Ans. by NTA (B)
Sol. Optical fibre frequency range is 1 THz to 1000
THz.
19. The difference of speed of light in the two
media A and B (v
A
 – v
B
) is 2.6 × 10
7
 m/s. If the
refractive index of medium B is 1.47, then the
ratio of refractive index of medium B to
medium A is : (Given : speed of light in vacuum
c = 3 × 10
8
 ms
–1
)
(A) 1.303
(B) 1.318
(C) 1.13
(D) 0.12
Official Ans. by NTA (C)
Sol.
c
v =
m
8
87
B
3 10
v 2.04 10 20.4 10 m/s
1.47
´
Þ = = ´ =´
7
AB
v v 2.6 10 m/s - =´ Q
77
A
v (20.4 2.6) 10 23 10 m/s \ = + ´ =´
7
BA
7
AB
v 23 10
1.13
v 20.4 10
m´
\===
m ´
20. A teacher in his physics laboratory allotted an
experiment to determine the resistance (G) of
a galvanometer. Students took the observations
for 
1
3
 deflection in the galvanometer. Which
of the below is true for measuring value of G?
(A)
1
3
 deflection method cannot be used for
determining the resistance of the
galvanometer.
(B) 
1
3
 deflection method can be used and in
this case the G equals to twice the value of
shunt resistance(s).
(C)
1
3
 deflection method can be used and in
this case, the G equals to three times the
value of shunt resistance(s)
(D) 
1
3
 deflection method can be used and in
this case the G value equals to the shunt
resistance(s).
Official Ans. by NTA (B)
Sol. In galvanometer
gg
(I I)S IG Þ -=
G Ig
S
I
g
I
S
I SG
=
+
1S
S G 3S G 2S
3 SG
Þ = Þ + = Þ=
+