JEE Main 2020 September 2 Question Paper Shift 2 | JEE Main & Advanced Previous Year Papers PDF Download

 Page 1


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
? ? ? ? ? ? ?
? ? ? ? ?
? ?
20
3
? ?
PHYSICS _ 2 Sep. _ SHIFT - 2
Page 2


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
? ? ? ? ? ? ?
? ? ? ? ?
? ?
20
3
? ?
PHYSICS _ 2 Sep. _ SHIFT - 2
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
? ? ?
9
400
2
3 . 0
?
? ?
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


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
? ? ? ? ? ? ?
? ? ? ? ?
? ?
20
3
? ?
PHYSICS _ 2 Sep. _ SHIFT - 2
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
? ? ?
9
400
2
3 . 0
?
? ?
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
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


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
? ? ? ? ? ? ?
? ? ? ? ?
? ?
20
3
? ?
PHYSICS _ 2 Sep. _ SHIFT - 2
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
? ? ?
9
400
2
3 . 0
?
? ?
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
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
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


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
? ? ? ? ? ? ?
? ? ? ? ?
? ?
20
3
? ?
PHYSICS _ 2 Sep. _ SHIFT - 2
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
? ? ?
9
400
2
3 . 0
?
? ?
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
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
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
? ?
? ? ?
? ?
? ?
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 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 2j ? ? ? ?
? ?
 = 
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
?