JEE Main 2020 Physics September 2 Shift 2 Paper & Solutions | Mock Tests for JEE Main and Advanced 2025 PDF Download

 Page 1


JEE Main 2020 Paper
         2
nd
 September 2020 | (Shift-2), Physics     Page | 1
Date : 2nd September 2020
Time : 02 : 00 pm - 05 : 00 pm
Subject : Physics
1. If momentum (P), area (A) and time (T) are taken to be the fundamental quantities
then the dimensional formula for energy is:
(1) [P
1/2
 AT
–1
] (2) [PA
1/2
T
–1
] (3) [PA
1/2
T
–1
] (4) [P
2
AT
–2
]
Sol. (2)
[P] = MLT
–1
 ? momentum
[A] = M
0
L
2
T
0
 ? Area
[T] = M
0
L
0
T
1
 ? Time
Let [E] = P
x
A
y
T
z
ML
2
T
–2
 = [MLT
–1
]
x
 [L
2
]
y
 [T]
z
= M
x
 L
x+2y 
T
z-x
Comparing both sides :-
x = 1 .....(i)
x + 2y = 2 ? 1 + 2y= 2 or, y = 
1
2
.....(ii)
z – x = –2 ? z–1 = –2 or z=–1 .....(iii)
? [E] = [P
1
 A
1/2
 T
–1
]
2. Two uniform circular discs are rotating independently in the same direction around their
common axis passing through their centres. The moment of inertia and angular velocity
of the first disc are 0.1 kg –m
2
 and 10 rad s
–1
 respectively while those for the second
one are 0.2 kg–m
2
 and 5 rad s
–1
 respectively. At some instant they get stuck together
and start rotating as a single system about their common axis with some angular
speed. The Kinetic energy of the combined system is:
(1) 
2
J
3
(2) 
10
J
3
(3) 
5
J
3
(4) 
20
J
3
Sol. (4)
I
1
I
2
I
2
I
1
?
1
 + I
2
?
2
 = (I
1
+I
2
) ?
1 1 2 2
1 2
I I 0.1 10 0.2 5 1 1 2
I I 0.1 0.2 0.3 0.3
? ? ? ? ? ? ?
? ? ? ? ?
? ?
Page 2


JEE Main 2020 Paper
         2
nd
 September 2020 | (Shift-2), Physics     Page | 1
Date : 2nd September 2020
Time : 02 : 00 pm - 05 : 00 pm
Subject : Physics
1. If momentum (P), area (A) and time (T) are taken to be the fundamental quantities
then the dimensional formula for energy is:
(1) [P
1/2
 AT
–1
] (2) [PA
1/2
T
–1
] (3) [PA
1/2
T
–1
] (4) [P
2
AT
–2
]
Sol. (2)
[P] = MLT
–1
 ? momentum
[A] = M
0
L
2
T
0
 ? Area
[T] = M
0
L
0
T
1
 ? Time
Let [E] = P
x
A
y
T
z
ML
2
T
–2
 = [MLT
–1
]
x
 [L
2
]
y
 [T]
z
= M
x
 L
x+2y 
T
z-x
Comparing both sides :-
x = 1 .....(i)
x + 2y = 2 ? 1 + 2y= 2 or, y = 
1
2
.....(ii)
z – x = –2 ? z–1 = –2 or z=–1 .....(iii)
? [E] = [P
1
 A
1/2
 T
–1
]
2. Two uniform circular discs are rotating independently in the same direction around their
common axis passing through their centres. The moment of inertia and angular velocity
of the first disc are 0.1 kg –m
2
 and 10 rad s
–1
 respectively while those for the second
one are 0.2 kg–m
2
 and 5 rad s
–1
 respectively. At some instant they get stuck together
and start rotating as a single system about their common axis with some angular
speed. The Kinetic energy of the combined system is:
(1) 
2
J
3
(2) 
10
J
3
(3) 
5
J
3
(4) 
20
J
3
Sol. (4)
I
1
I
2
I
2
I
1
?
1
 + I
2
?
2
 = (I
1
+I
2
) ?
1 1 2 2
1 2
I I 0.1 10 0.2 5 1 1 2
I I 0.1 0.2 0.3 0.3
? ? ? ? ? ? ?
? ? ? ? ?
? ?
JEE Main 2020 Paper
         2
nd
 September 2020 | (Shift-2), Physics     Page | 2
20
3
? ?
Now find KE = 
2 2
1 2
1 1
I I
2 2
? ? ?
= ? ?
2
2
1 2
1 1 20
I I 0.3
2 2 3
? ?
? ? ? ? ?
? ?
? ?
= 
1 3 20 20
2 10 3 3
? ? ?
? ?
f
20
K.E.
3
?
3. A particle is moving 5 times as fast as an electron. The ratio of the de-Broglie wavelength
of the particle to that of the electron is 1.878 × 10
–4
. The mass of the particle is close to:
(1) 4.8 × 10
–27 
kg (2) 9.1 × 10
–31
 kg (3) 9.7 × 10
–28 
 kg (4) 1.2 × 10
–28
kg
Sol. (3)
h
P ?
?
h
P
? ?
4 Particle
e
1.878 10
?
?
? ?
?
4 4
particle particle
h Pe Pe
13.878 10 1.878 10
P h P
? ?
? ? ? ? ? ? ?
4 e e
p p
M .V
1.878 10
M .V
?
? ? ?
 
e e
p 4
p
M V
M
V 1.878 10
?
? ?
? ? ? ? ?
? ?
?
? ?
31
4
9.11 10 1
5 1.878 10
?
?
?
? ?
?
4
7.11 10 1
5 1.878 10
?
?
? ?
?
= 0.97 × 10
–27
 kg
= 9.7 ×10
–28
 kg
Page 3


JEE Main 2020 Paper
         2
nd
 September 2020 | (Shift-2), Physics     Page | 1
Date : 2nd September 2020
Time : 02 : 00 pm - 05 : 00 pm
Subject : Physics
1. If momentum (P), area (A) and time (T) are taken to be the fundamental quantities
then the dimensional formula for energy is:
(1) [P
1/2
 AT
–1
] (2) [PA
1/2
T
–1
] (3) [PA
1/2
T
–1
] (4) [P
2
AT
–2
]
Sol. (2)
[P] = MLT
–1
 ? momentum
[A] = M
0
L
2
T
0
 ? Area
[T] = M
0
L
0
T
1
 ? Time
Let [E] = P
x
A
y
T
z
ML
2
T
–2
 = [MLT
–1
]
x
 [L
2
]
y
 [T]
z
= M
x
 L
x+2y 
T
z-x
Comparing both sides :-
x = 1 .....(i)
x + 2y = 2 ? 1 + 2y= 2 or, y = 
1
2
.....(ii)
z – x = –2 ? z–1 = –2 or z=–1 .....(iii)
? [E] = [P
1
 A
1/2
 T
–1
]
2. Two uniform circular discs are rotating independently in the same direction around their
common axis passing through their centres. The moment of inertia and angular velocity
of the first disc are 0.1 kg –m
2
 and 10 rad s
–1
 respectively while those for the second
one are 0.2 kg–m
2
 and 5 rad s
–1
 respectively. At some instant they get stuck together
and start rotating as a single system about their common axis with some angular
speed. The Kinetic energy of the combined system is:
(1) 
2
J
3
(2) 
10
J
3
(3) 
5
J
3
(4) 
20
J
3
Sol. (4)
I
1
I
2
I
2
I
1
?
1
 + I
2
?
2
 = (I
1
+I
2
) ?
1 1 2 2
1 2
I I 0.1 10 0.2 5 1 1 2
I I 0.1 0.2 0.3 0.3
? ? ? ? ? ? ?
? ? ? ? ?
? ?
JEE Main 2020 Paper
         2
nd
 September 2020 | (Shift-2), Physics     Page | 2
20
3
? ?
Now find KE = 
2 2
1 2
1 1
I I
2 2
? ? ?
= ? ?
2
2
1 2
1 1 20
I I 0.3
2 2 3
? ?
? ? ? ? ?
? ?
? ?
= 
1 3 20 20
2 10 3 3
? ? ?
? ?
f
20
K.E.
3
?
3. A particle is moving 5 times as fast as an electron. The ratio of the de-Broglie wavelength
of the particle to that of the electron is 1.878 × 10
–4
. The mass of the particle is close to:
(1) 4.8 × 10
–27 
kg (2) 9.1 × 10
–31
 kg (3) 9.7 × 10
–28 
 kg (4) 1.2 × 10
–28
kg
Sol. (3)
h
P ?
?
h
P
? ?
4 Particle
e
1.878 10
?
?
? ?
?
4 4
particle particle
h Pe Pe
13.878 10 1.878 10
P h P
? ?
? ? ? ? ? ? ?
4 e e
p p
M .V
1.878 10
M .V
?
? ? ?
 
e e
p 4
p
M V
M
V 1.878 10
?
? ?
? ? ? ? ?
? ?
?
? ?
31
4
9.11 10 1
5 1.878 10
?
?
?
? ?
?
4
7.11 10 1
5 1.878 10
?
?
? ?
?
= 0.97 × 10
–27
 kg
= 9.7 ×10
–28
 kg
JEE Main 2020 Paper
         2
nd
 September 2020 | (Shift-2), Physics     Page | 3
4. A potentiometer wire PQ of 1 m length is connected to a standard cell E
1
. Another cell E
2
of emf 1.02 V is connected with a resistance ‘r’ and switch S (as shown in figure). With
switch S open, the null positon is obtained at a distance of 49 cm from Q. The potential
gradient in the potentiometer wire is:
(1) 0.03V/cm (2) 0.02 V/cm (3) 0.04 V/cm (4) 0.01 V/cm
Sol. (2)
G
E
1
J
r
P Q
E
2
1.02 V
S
PQ = 1m
QJ = 49 cm
?
 PJ = 51 cm
v 1.02
51
? ?
?
 0.02 v/cm
5. In the following digital circuit, what will be the output at ‘Z’, when the input (A,B) are
(1,0), (0,0), (1,1), (0,1):
Z
A
B
(1) 0,1,0,0 (2) 1,1,0,1 (3) 0,0,1,0 (4) 1,0,1,1
Page 4


JEE Main 2020 Paper
         2
nd
 September 2020 | (Shift-2), Physics     Page | 1
Date : 2nd September 2020
Time : 02 : 00 pm - 05 : 00 pm
Subject : Physics
1. If momentum (P), area (A) and time (T) are taken to be the fundamental quantities
then the dimensional formula for energy is:
(1) [P
1/2
 AT
–1
] (2) [PA
1/2
T
–1
] (3) [PA
1/2
T
–1
] (4) [P
2
AT
–2
]
Sol. (2)
[P] = MLT
–1
 ? momentum
[A] = M
0
L
2
T
0
 ? Area
[T] = M
0
L
0
T
1
 ? Time
Let [E] = P
x
A
y
T
z
ML
2
T
–2
 = [MLT
–1
]
x
 [L
2
]
y
 [T]
z
= M
x
 L
x+2y 
T
z-x
Comparing both sides :-
x = 1 .....(i)
x + 2y = 2 ? 1 + 2y= 2 or, y = 
1
2
.....(ii)
z – x = –2 ? z–1 = –2 or z=–1 .....(iii)
? [E] = [P
1
 A
1/2
 T
–1
]
2. Two uniform circular discs are rotating independently in the same direction around their
common axis passing through their centres. The moment of inertia and angular velocity
of the first disc are 0.1 kg –m
2
 and 10 rad s
–1
 respectively while those for the second
one are 0.2 kg–m
2
 and 5 rad s
–1
 respectively. At some instant they get stuck together
and start rotating as a single system about their common axis with some angular
speed. The Kinetic energy of the combined system is:
(1) 
2
J
3
(2) 
10
J
3
(3) 
5
J
3
(4) 
20
J
3
Sol. (4)
I
1
I
2
I
2
I
1
?
1
 + I
2
?
2
 = (I
1
+I
2
) ?
1 1 2 2
1 2
I I 0.1 10 0.2 5 1 1 2
I I 0.1 0.2 0.3 0.3
? ? ? ? ? ? ?
? ? ? ? ?
? ?
JEE Main 2020 Paper
         2
nd
 September 2020 | (Shift-2), Physics     Page | 2
20
3
? ?
Now find KE = 
2 2
1 2
1 1
I I
2 2
? ? ?
= ? ?
2
2
1 2
1 1 20
I I 0.3
2 2 3
? ?
? ? ? ? ?
? ?
? ?
= 
1 3 20 20
2 10 3 3
? ? ?
? ?
f
20
K.E.
3
?
3. A particle is moving 5 times as fast as an electron. The ratio of the de-Broglie wavelength
of the particle to that of the electron is 1.878 × 10
–4
. The mass of the particle is close to:
(1) 4.8 × 10
–27 
kg (2) 9.1 × 10
–31
 kg (3) 9.7 × 10
–28 
 kg (4) 1.2 × 10
–28
kg
Sol. (3)
h
P ?
?
h
P
? ?
4 Particle
e
1.878 10
?
?
? ?
?
4 4
particle particle
h Pe Pe
13.878 10 1.878 10
P h P
? ?
? ? ? ? ? ? ?
4 e e
p p
M .V
1.878 10
M .V
?
? ? ?
 
e e
p 4
p
M V
M
V 1.878 10
?
? ?
? ? ? ? ?
? ?
?
? ?
31
4
9.11 10 1
5 1.878 10
?
?
?
? ?
?
4
7.11 10 1
5 1.878 10
?
?
? ?
?
= 0.97 × 10
–27
 kg
= 9.7 ×10
–28
 kg
JEE Main 2020 Paper
         2
nd
 September 2020 | (Shift-2), Physics     Page | 3
4. A potentiometer wire PQ of 1 m length is connected to a standard cell E
1
. Another cell E
2
of emf 1.02 V is connected with a resistance ‘r’ and switch S (as shown in figure). With
switch S open, the null positon is obtained at a distance of 49 cm from Q. The potential
gradient in the potentiometer wire is:
(1) 0.03V/cm (2) 0.02 V/cm (3) 0.04 V/cm (4) 0.01 V/cm
Sol. (2)
G
E
1
J
r
P Q
E
2
1.02 V
S
PQ = 1m
QJ = 49 cm
?
 PJ = 51 cm
v 1.02
51
? ?
?
 0.02 v/cm
5. In the following digital circuit, what will be the output at ‘Z’, when the input (A,B) are
(1,0), (0,0), (1,1), (0,1):
Z
A
B
(1) 0,1,0,0 (2) 1,1,0,1 (3) 0,0,1,0 (4) 1,0,1,1
JEE Main 2020 Paper
         2
nd
 September 2020 | (Shift-2), Physics     Page | 4
Sol. (3)
Z
A
C
D
E
B
A B C A.B D A B E C.D Z C E
1 0 1 1 1 0
0 0 1 0 0 0
0 0 1 0 0 0
1 1 0 1 0 1
0 1 1 1 1 0
? ? ? ? ? ?
6. A wire carrying current I is bent in the shape ABCDEFA as shown, where rectangle
ABCDA and ADEFA are perpendicular to each other. If the sides of the rectangles are of
lengths a and b, then the magnitude and direction of magnetic moment of the loop
ABCDEFA  is:
(1) 
ˆ
j 2k
2 abl,along
5 5
? ?
? ? ?
? ?
? ?
?
(2) 
ˆ
j 2k
abl,along
5 5
? ?
? ? ?
? ?
? ?
?
(3) 
?
j k
2 abl,along
2 2
? ?
? ? ?
? ?
? ?
?
(4) 
?
j k
abl,along
2 2
? ?
? ? ?
? ?
? ?
?
Sol. (3)
LOOP = ABCD
? ?
1
ˆ
M abI k ?
?
Loop DEFA
? ?
2
ˆ
M abI j ?
?
? ? 1 2
ˆ ˆ
M M M abI j k ? ? ? ?
? ? ?
M 2abI ?
?
direction = along 
ˆ ˆ
j k
2 2
? ?
? ? ?
? ?
? ?
2
abl, along 
ˆ ˆ
j k
2 2
? ?
? ? ?
? ?
? ?
Page 5


JEE Main 2020 Paper
         2
nd
 September 2020 | (Shift-2), Physics     Page | 1
Date : 2nd September 2020
Time : 02 : 00 pm - 05 : 00 pm
Subject : Physics
1. If momentum (P), area (A) and time (T) are taken to be the fundamental quantities
then the dimensional formula for energy is:
(1) [P
1/2
 AT
–1
] (2) [PA
1/2
T
–1
] (3) [PA
1/2
T
–1
] (4) [P
2
AT
–2
]
Sol. (2)
[P] = MLT
–1
 ? momentum
[A] = M
0
L
2
T
0
 ? Area
[T] = M
0
L
0
T
1
 ? Time
Let [E] = P
x
A
y
T
z
ML
2
T
–2
 = [MLT
–1
]
x
 [L
2
]
y
 [T]
z
= M
x
 L
x+2y 
T
z-x
Comparing both sides :-
x = 1 .....(i)
x + 2y = 2 ? 1 + 2y= 2 or, y = 
1
2
.....(ii)
z – x = –2 ? z–1 = –2 or z=–1 .....(iii)
? [E] = [P
1
 A
1/2
 T
–1
]
2. Two uniform circular discs are rotating independently in the same direction around their
common axis passing through their centres. The moment of inertia and angular velocity
of the first disc are 0.1 kg –m
2
 and 10 rad s
–1
 respectively while those for the second
one are 0.2 kg–m
2
 and 5 rad s
–1
 respectively. At some instant they get stuck together
and start rotating as a single system about their common axis with some angular
speed. The Kinetic energy of the combined system is:
(1) 
2
J
3
(2) 
10
J
3
(3) 
5
J
3
(4) 
20
J
3
Sol. (4)
I
1
I
2
I
2
I
1
?
1
 + I
2
?
2
 = (I
1
+I
2
) ?
1 1 2 2
1 2
I I 0.1 10 0.2 5 1 1 2
I I 0.1 0.2 0.3 0.3
? ? ? ? ? ? ?
? ? ? ? ?
? ?
JEE Main 2020 Paper
         2
nd
 September 2020 | (Shift-2), Physics     Page | 2
20
3
? ?
Now find KE = 
2 2
1 2
1 1
I I
2 2
? ? ?
= ? ?
2
2
1 2
1 1 20
I I 0.3
2 2 3
? ?
? ? ? ? ?
? ?
? ?
= 
1 3 20 20
2 10 3 3
? ? ?
? ?
f
20
K.E.
3
?
3. A particle is moving 5 times as fast as an electron. The ratio of the de-Broglie wavelength
of the particle to that of the electron is 1.878 × 10
–4
. The mass of the particle is close to:
(1) 4.8 × 10
–27 
kg (2) 9.1 × 10
–31
 kg (3) 9.7 × 10
–28 
 kg (4) 1.2 × 10
–28
kg
Sol. (3)
h
P ?
?
h
P
? ?
4 Particle
e
1.878 10
?
?
? ?
?
4 4
particle particle
h Pe Pe
13.878 10 1.878 10
P h P
? ?
? ? ? ? ? ? ?
4 e e
p p
M .V
1.878 10
M .V
?
? ? ?
 
e e
p 4
p
M V
M
V 1.878 10
?
? ?
? ? ? ? ?
? ?
?
? ?
31
4
9.11 10 1
5 1.878 10
?
?
?
? ?
?
4
7.11 10 1
5 1.878 10
?
?
? ?
?
= 0.97 × 10
–27
 kg
= 9.7 ×10
–28
 kg
JEE Main 2020 Paper
         2
nd
 September 2020 | (Shift-2), Physics     Page | 3
4. A potentiometer wire PQ of 1 m length is connected to a standard cell E
1
. Another cell E
2
of emf 1.02 V is connected with a resistance ‘r’ and switch S (as shown in figure). With
switch S open, the null positon is obtained at a distance of 49 cm from Q. The potential
gradient in the potentiometer wire is:
(1) 0.03V/cm (2) 0.02 V/cm (3) 0.04 V/cm (4) 0.01 V/cm
Sol. (2)
G
E
1
J
r
P Q
E
2
1.02 V
S
PQ = 1m
QJ = 49 cm
?
 PJ = 51 cm
v 1.02
51
? ?
?
 0.02 v/cm
5. In the following digital circuit, what will be the output at ‘Z’, when the input (A,B) are
(1,0), (0,0), (1,1), (0,1):
Z
A
B
(1) 0,1,0,0 (2) 1,1,0,1 (3) 0,0,1,0 (4) 1,0,1,1
JEE Main 2020 Paper
         2
nd
 September 2020 | (Shift-2), Physics     Page | 4
Sol. (3)
Z
A
C
D
E
B
A B C A.B D A B E C.D Z C E
1 0 1 1 1 0
0 0 1 0 0 0
0 0 1 0 0 0
1 1 0 1 0 1
0 1 1 1 1 0
? ? ? ? ? ?
6. A wire carrying current I is bent in the shape ABCDEFA as shown, where rectangle
ABCDA and ADEFA are perpendicular to each other. If the sides of the rectangles are of
lengths a and b, then the magnitude and direction of magnetic moment of the loop
ABCDEFA  is:
(1) 
ˆ
j 2k
2 abl,along
5 5
? ?
? ? ?
? ?
? ?
?
(2) 
ˆ
j 2k
abl,along
5 5
? ?
? ? ?
? ?
? ?
?
(3) 
?
j k
2 abl,along
2 2
? ?
? ? ?
? ?
? ?
?
(4) 
?
j k
abl,along
2 2
? ?
? ? ?
? ?
? ?
?
Sol. (3)
LOOP = ABCD
? ?
1
ˆ
M abI k ?
?
Loop DEFA
? ?
2
ˆ
M abI j ?
?
? ? 1 2
ˆ ˆ
M M M abI j k ? ? ? ?
? ? ?
M 2abI ?
?
direction = along 
ˆ ˆ
j k
2 2
? ?
? ? ?
? ?
? ?
2
abl, along 
ˆ ˆ
j k
2 2
? ?
? ? ?
? ?
? ?
JEE Main 2020 Paper
         2
nd
 September 2020 | (Shift-2), Physics     Page | 5
7. A small point mass carrying some positive charge on it, is released from the edge of a
table. There is a uniform electric field in this region in the horizontal direction. Which of
the following options then correctly describe the trajectory of the mass? (Curves are
drawn schematically and are not to scale).
E
x
y
(1) 
y
x
(2) 
y
x
(3) 
y
x
(4) 
y
x
Sol. (3)
a
g
a
net
a
E
x x
y
Since it is released from rest.
And a
net
 is constant.
If will have straight line path along net ‘a’.
8. In a plane electromagnetic wave, the directions of electric field and magnetic field are
represented by k
ˆ
 and 2i–2 j
? ?
, respectively. What is the unit vector along direction of
propagation of the wave.
(1) ? ?
1
i j
2
?
? ?
(2) ? ?
1
2i j
5
?
? ?
(3) ? ?
1
i 2 j
5
?
? ?
(4) 
?
? ?
1
j k
2
?
?
Sol. (1)
? ?
ˆ ˆ ˆ
E B k 2i 2 j ? ? ? ?
? ?
 = 
ˆ ˆ ˆ ˆ
2k i 2k j ? ? ? = 
ˆ ˆ
2j 2i ?
unit vector along ? ?
1
ˆ ˆ
E B 2i 2j
2 2
? ? ?
? ?
 = ? ?
1
ˆ ˆ
i j
2
?
C = 
? ? j
ˆ
i
ˆ
2
1
?