JEE Main 2020 Physics September 5 Shift 2 Paper & Solutions | Mock Tests for JEE Main and Advanced 2025 PDF Download

 Page 1


JEE Main 2020 Paper
         5
th
 September 2020 | (Shift-2), Physics     Page | 1
Date : 5th September 2020
Time : 02 : 00 pm - 05 : 00 pm
Subject : Physics
1. A ring is hung on a nail. It can oscillate, without slipping or sliding (i) in its plane with
a time period T
1
 and, (ii) back and forth in a direction perpendicular to its plane, with a
period T
2
. The ratio 
1
2
T
T
 will be:
(1) 
3
2
(2) 
2
3
(3) 
2
3
(4) 
2
3
SOl. 3
T
1
 = 
?
?
2 2
(mR mR )
2
mgR
T
1
 = 2 ? 
2R
g
T
2
 = 2 ? 
cm
I
mgL
T
2 
= 2 ? 
2
3mR /2
mgR
 = 
3R
2
2g
?
COM
COM
Back and forth
1
2
T
T
 =  
4 2
3
3
?
2. The correct match between the entries in column I and column II are:
I         II
     Radiation Wavelength
(a) Microwave (i)   100 m
(b) Gamma rays (ii)  10
–15
 m
(c)  A.M. radio waves (iii) 10
–10
 m
(d) X-rays (iv)  10
–3
 m
(1) (a) - (ii), (b)-(i), (c)-(iv), (d)-(iii) (2) (a)-(iii), (b)-(ii), (c)-(i), (d)-(iv)
(3) (a)-(iv), (b)-(ii), (c)-(i), (d)-(iii) (4) (a)-(i),(b)-(iii), (c)-(iv), (d)-(ii)
Sol. 3
By theory
Page 2


JEE Main 2020 Paper
         5
th
 September 2020 | (Shift-2), Physics     Page | 1
Date : 5th September 2020
Time : 02 : 00 pm - 05 : 00 pm
Subject : Physics
1. A ring is hung on a nail. It can oscillate, without slipping or sliding (i) in its plane with
a time period T
1
 and, (ii) back and forth in a direction perpendicular to its plane, with a
period T
2
. The ratio 
1
2
T
T
 will be:
(1) 
3
2
(2) 
2
3
(3) 
2
3
(4) 
2
3
SOl. 3
T
1
 = 
?
?
2 2
(mR mR )
2
mgR
T
1
 = 2 ? 
2R
g
T
2
 = 2 ? 
cm
I
mgL
T
2 
= 2 ? 
2
3mR /2
mgR
 = 
3R
2
2g
?
COM
COM
Back and forth
1
2
T
T
 =  
4 2
3
3
?
2. The correct match between the entries in column I and column II are:
I         II
     Radiation Wavelength
(a) Microwave (i)   100 m
(b) Gamma rays (ii)  10
–15
 m
(c)  A.M. radio waves (iii) 10
–10
 m
(d) X-rays (iv)  10
–3
 m
(1) (a) - (ii), (b)-(i), (c)-(iv), (d)-(iii) (2) (a)-(iii), (b)-(ii), (c)-(i), (d)-(iv)
(3) (a)-(iv), (b)-(ii), (c)-(i), (d)-(iii) (4) (a)-(i),(b)-(iii), (c)-(iv), (d)-(ii)
Sol. 3
By theory
JEE Main 2020 Paper
         5
th
 September 2020 | (Shift-2), Physics     Page | 2
3. In an experiment to verify Stokes law, a small spherical ball of radius r and density ? falls
under gravity through a distance h in air before entering a tank of water. If the terminal
velocity of the ball inside water is same as its velocity just before entering the water
surface, then the value of h is proportional to : (ignore viscosity of air)
(1) r
4
(2) r (3) r
3
(4) r
2
Sol. 1
V
T
 = 2gh
2
9
 r
2
 
? ? ? ? ?
?
g
 = 2gh
r
2
 ? 
h
 ? r
4
 ? ?h
h ? r
4
4. Ten charges are placed on the circumference of a circle of radius R with constant angular
separation between successive charges. Alternate charges 1, 3, 5, 7, 9 have charge (+q)
each, while 2, 4, 6, 8, 10 have charge (–q) each. The potential V and the electric field E
at the centre of the circle are respectively: (Take V= 0 at infinity)
(1) V = 0; E = 0 (2) 2
0 0
10q 10q
V ;E
4 R 4 R
? ?
? ? ? ?
(3) 2
0
10q
V 0;E
4 R
? ?
? ?
(4) 
0
10q
V ;E 0
4 R
? ?
? ?
Sol. 1
v
net
 = 5 
kq
R
? ?
? ?
? ?
 + 
5k( q)
R
? ? ?
? ?
? ?
v
net
 = 0 [Q
net
 = 0]
E
net
 = 0 by symmetry
Page 3


JEE Main 2020 Paper
         5
th
 September 2020 | (Shift-2), Physics     Page | 1
Date : 5th September 2020
Time : 02 : 00 pm - 05 : 00 pm
Subject : Physics
1. A ring is hung on a nail. It can oscillate, without slipping or sliding (i) in its plane with
a time period T
1
 and, (ii) back and forth in a direction perpendicular to its plane, with a
period T
2
. The ratio 
1
2
T
T
 will be:
(1) 
3
2
(2) 
2
3
(3) 
2
3
(4) 
2
3
SOl. 3
T
1
 = 
?
?
2 2
(mR mR )
2
mgR
T
1
 = 2 ? 
2R
g
T
2
 = 2 ? 
cm
I
mgL
T
2 
= 2 ? 
2
3mR /2
mgR
 = 
3R
2
2g
?
COM
COM
Back and forth
1
2
T
T
 =  
4 2
3
3
?
2. The correct match between the entries in column I and column II are:
I         II
     Radiation Wavelength
(a) Microwave (i)   100 m
(b) Gamma rays (ii)  10
–15
 m
(c)  A.M. radio waves (iii) 10
–10
 m
(d) X-rays (iv)  10
–3
 m
(1) (a) - (ii), (b)-(i), (c)-(iv), (d)-(iii) (2) (a)-(iii), (b)-(ii), (c)-(i), (d)-(iv)
(3) (a)-(iv), (b)-(ii), (c)-(i), (d)-(iii) (4) (a)-(i),(b)-(iii), (c)-(iv), (d)-(ii)
Sol. 3
By theory
JEE Main 2020 Paper
         5
th
 September 2020 | (Shift-2), Physics     Page | 2
3. In an experiment to verify Stokes law, a small spherical ball of radius r and density ? falls
under gravity through a distance h in air before entering a tank of water. If the terminal
velocity of the ball inside water is same as its velocity just before entering the water
surface, then the value of h is proportional to : (ignore viscosity of air)
(1) r
4
(2) r (3) r
3
(4) r
2
Sol. 1
V
T
 = 2gh
2
9
 r
2
 
? ? ? ? ?
?
g
 = 2gh
r
2
 ? 
h
 ? r
4
 ? ?h
h ? r
4
4. Ten charges are placed on the circumference of a circle of radius R with constant angular
separation between successive charges. Alternate charges 1, 3, 5, 7, 9 have charge (+q)
each, while 2, 4, 6, 8, 10 have charge (–q) each. The potential V and the electric field E
at the centre of the circle are respectively: (Take V= 0 at infinity)
(1) V = 0; E = 0 (2) 2
0 0
10q 10q
V ;E
4 R 4 R
? ?
? ? ? ?
(3) 2
0
10q
V 0;E
4 R
? ?
? ?
(4) 
0
10q
V ;E 0
4 R
? ?
? ?
Sol. 1
v
net
 = 5 
kq
R
? ?
? ?
? ?
 + 
5k( q)
R
? ? ?
? ?
? ?
v
net
 = 0 [Q
net
 = 0]
E
net
 = 0 by symmetry
JEE Main 2020 Paper
         5
th
 September 2020 | (Shift-2), Physics     Page | 3
5. A spaceship in space sweeps stationary interplanetary dust. As a result, its mass increases
at a rate 
? ? dM t
dt
= bv
2
 (t), where v(t) is its instantaneous velocity. The instantaneous
acceleration of the satellite is :
(1) –bv
3
(t) (2) 
3
bv
–
M(t)
(3) 
3
2bv
–
M(t)
(4) 
3
bv
–
2M(t)
Sol. 2
dM(t)
dt
 = -bv
2
in free space
no external force
so there in only thrust force on rocket
f
in
 = 
dM
dt
 (V
rel
)
Ma = 
v
) t (
bv
2
?
?
?
?
?
?
?
? ?
a = 
3
bv
M(t)
?
6. Two different wires having lengths L
1
 and L
2
, and respective temperature coefficient of
linear expansion ?
1
 and ?
2
, are joined end-to-end. Then the effective temperature
coefficient of linear expansion is:
(1) 
2
1 1 2 2
1
L L
L L
? ? ?
?
(2) 
1 2
2 ? ?
(3) 
? ?
1
2
1 2
2
1 2
2 1
L L
4
L L
? ?
? ? ?
?
(4) 
1 2
2
? ? ?
Sol. 1
L'
1
 = L
1
 (1 + ?
1
?T)
L'
2
 = L
2
 (1 + ?
2
?T)
L'+L
2
'
 
= L
1
 + L
2
 + L
1
?
1
?T + L
2
?
2
?T
= (L
1
 + L
2
) 
1 1 2 2
1 2
L L
1 T
L L
? ? ? ? ? ? ?
? ?
? ? ? ?
?
? ? ? ? ? ?
= (L
1
 + L
2
) [1 + ?
eq
?T)
So, ?
eq
 = 
1 1 2 2
1 2
L L
L L
? ? ?
?
Page 4


JEE Main 2020 Paper
         5
th
 September 2020 | (Shift-2), Physics     Page | 1
Date : 5th September 2020
Time : 02 : 00 pm - 05 : 00 pm
Subject : Physics
1. A ring is hung on a nail. It can oscillate, without slipping or sliding (i) in its plane with
a time period T
1
 and, (ii) back and forth in a direction perpendicular to its plane, with a
period T
2
. The ratio 
1
2
T
T
 will be:
(1) 
3
2
(2) 
2
3
(3) 
2
3
(4) 
2
3
SOl. 3
T
1
 = 
?
?
2 2
(mR mR )
2
mgR
T
1
 = 2 ? 
2R
g
T
2
 = 2 ? 
cm
I
mgL
T
2 
= 2 ? 
2
3mR /2
mgR
 = 
3R
2
2g
?
COM
COM
Back and forth
1
2
T
T
 =  
4 2
3
3
?
2. The correct match between the entries in column I and column II are:
I         II
     Radiation Wavelength
(a) Microwave (i)   100 m
(b) Gamma rays (ii)  10
–15
 m
(c)  A.M. radio waves (iii) 10
–10
 m
(d) X-rays (iv)  10
–3
 m
(1) (a) - (ii), (b)-(i), (c)-(iv), (d)-(iii) (2) (a)-(iii), (b)-(ii), (c)-(i), (d)-(iv)
(3) (a)-(iv), (b)-(ii), (c)-(i), (d)-(iii) (4) (a)-(i),(b)-(iii), (c)-(iv), (d)-(ii)
Sol. 3
By theory
JEE Main 2020 Paper
         5
th
 September 2020 | (Shift-2), Physics     Page | 2
3. In an experiment to verify Stokes law, a small spherical ball of radius r and density ? falls
under gravity through a distance h in air before entering a tank of water. If the terminal
velocity of the ball inside water is same as its velocity just before entering the water
surface, then the value of h is proportional to : (ignore viscosity of air)
(1) r
4
(2) r (3) r
3
(4) r
2
Sol. 1
V
T
 = 2gh
2
9
 r
2
 
? ? ? ? ?
?
g
 = 2gh
r
2
 ? 
h
 ? r
4
 ? ?h
h ? r
4
4. Ten charges are placed on the circumference of a circle of radius R with constant angular
separation between successive charges. Alternate charges 1, 3, 5, 7, 9 have charge (+q)
each, while 2, 4, 6, 8, 10 have charge (–q) each. The potential V and the electric field E
at the centre of the circle are respectively: (Take V= 0 at infinity)
(1) V = 0; E = 0 (2) 2
0 0
10q 10q
V ;E
4 R 4 R
? ?
? ? ? ?
(3) 2
0
10q
V 0;E
4 R
? ?
? ?
(4) 
0
10q
V ;E 0
4 R
? ?
? ?
Sol. 1
v
net
 = 5 
kq
R
? ?
? ?
? ?
 + 
5k( q)
R
? ? ?
? ?
? ?
v
net
 = 0 [Q
net
 = 0]
E
net
 = 0 by symmetry
JEE Main 2020 Paper
         5
th
 September 2020 | (Shift-2), Physics     Page | 3
5. A spaceship in space sweeps stationary interplanetary dust. As a result, its mass increases
at a rate 
? ? dM t
dt
= bv
2
 (t), where v(t) is its instantaneous velocity. The instantaneous
acceleration of the satellite is :
(1) –bv
3
(t) (2) 
3
bv
–
M(t)
(3) 
3
2bv
–
M(t)
(4) 
3
bv
–
2M(t)
Sol. 2
dM(t)
dt
 = -bv
2
in free space
no external force
so there in only thrust force on rocket
f
in
 = 
dM
dt
 (V
rel
)
Ma = 
v
) t (
bv
2
?
?
?
?
?
?
?
? ?
a = 
3
bv
M(t)
?
6. Two different wires having lengths L
1
 and L
2
, and respective temperature coefficient of
linear expansion ?
1
 and ?
2
, are joined end-to-end. Then the effective temperature
coefficient of linear expansion is:
(1) 
2
1 1 2 2
1
L L
L L
? ? ?
?
(2) 
1 2
2 ? ?
(3) 
? ?
1
2
1 2
2
1 2
2 1
L L
4
L L
? ?
? ? ?
?
(4) 
1 2
2
? ? ?
Sol. 1
L'
1
 = L
1
 (1 + ?
1
?T)
L'
2
 = L
2
 (1 + ?
2
?T)
L'+L
2
'
 
= L
1
 + L
2
 + L
1
?
1
?T + L
2
?
2
?T
= (L
1
 + L
2
) 
1 1 2 2
1 2
L L
1 T
L L
? ? ? ? ? ? ?
? ?
? ? ? ?
?
? ? ? ? ? ?
= (L
1
 + L
2
) [1 + ?
eq
?T)
So, ?
eq
 = 
1 1 2 2
1 2
L L
L L
? ? ?
?
JEE Main 2020 Paper
         5
th
 September 2020 | (Shift-2), Physics     Page | 4
7. In the circuit, given in the figure currents in different branches and value of one resistor
are shown. Then potential at point B with respect to the point A is:
2A
F
B
2V
C
1V
A
1A
E
D
2
(1) +2 V (2) –2 V (3) +1 V (4) –1 V
Sol. 3
Let V
A
 = 0
applying KVL from A to B via ACDB
1 2 2
A B
V V ? ? ? ?
V
B
 =  1V
8. The velocity (v) and time (t) graph of a body in a straight line motion is shown in the
figure. The point S is at 4.333 seconds. The total distance covered by the body in 6 s is:
–2
0
2
4
1 2 3 4 5 6
A B
S D
C
t (in s)
v (m/s)
(1) 
37
m
3
(2) 
49
m
4
(3) 12 m (4) 11 m
Sol. 1
Page 5


JEE Main 2020 Paper
         5
th
 September 2020 | (Shift-2), Physics     Page | 1
Date : 5th September 2020
Time : 02 : 00 pm - 05 : 00 pm
Subject : Physics
1. A ring is hung on a nail. It can oscillate, without slipping or sliding (i) in its plane with
a time period T
1
 and, (ii) back and forth in a direction perpendicular to its plane, with a
period T
2
. The ratio 
1
2
T
T
 will be:
(1) 
3
2
(2) 
2
3
(3) 
2
3
(4) 
2
3
SOl. 3
T
1
 = 
?
?
2 2
(mR mR )
2
mgR
T
1
 = 2 ? 
2R
g
T
2
 = 2 ? 
cm
I
mgL
T
2 
= 2 ? 
2
3mR /2
mgR
 = 
3R
2
2g
?
COM
COM
Back and forth
1
2
T
T
 =  
4 2
3
3
?
2. The correct match between the entries in column I and column II are:
I         II
     Radiation Wavelength
(a) Microwave (i)   100 m
(b) Gamma rays (ii)  10
–15
 m
(c)  A.M. radio waves (iii) 10
–10
 m
(d) X-rays (iv)  10
–3
 m
(1) (a) - (ii), (b)-(i), (c)-(iv), (d)-(iii) (2) (a)-(iii), (b)-(ii), (c)-(i), (d)-(iv)
(3) (a)-(iv), (b)-(ii), (c)-(i), (d)-(iii) (4) (a)-(i),(b)-(iii), (c)-(iv), (d)-(ii)
Sol. 3
By theory
JEE Main 2020 Paper
         5
th
 September 2020 | (Shift-2), Physics     Page | 2
3. In an experiment to verify Stokes law, a small spherical ball of radius r and density ? falls
under gravity through a distance h in air before entering a tank of water. If the terminal
velocity of the ball inside water is same as its velocity just before entering the water
surface, then the value of h is proportional to : (ignore viscosity of air)
(1) r
4
(2) r (3) r
3
(4) r
2
Sol. 1
V
T
 = 2gh
2
9
 r
2
 
? ? ? ? ?
?
g
 = 2gh
r
2
 ? 
h
 ? r
4
 ? ?h
h ? r
4
4. Ten charges are placed on the circumference of a circle of radius R with constant angular
separation between successive charges. Alternate charges 1, 3, 5, 7, 9 have charge (+q)
each, while 2, 4, 6, 8, 10 have charge (–q) each. The potential V and the electric field E
at the centre of the circle are respectively: (Take V= 0 at infinity)
(1) V = 0; E = 0 (2) 2
0 0
10q 10q
V ;E
4 R 4 R
? ?
? ? ? ?
(3) 2
0
10q
V 0;E
4 R
? ?
? ?
(4) 
0
10q
V ;E 0
4 R
? ?
? ?
Sol. 1
v
net
 = 5 
kq
R
? ?
? ?
? ?
 + 
5k( q)
R
? ? ?
? ?
? ?
v
net
 = 0 [Q
net
 = 0]
E
net
 = 0 by symmetry
JEE Main 2020 Paper
         5
th
 September 2020 | (Shift-2), Physics     Page | 3
5. A spaceship in space sweeps stationary interplanetary dust. As a result, its mass increases
at a rate 
? ? dM t
dt
= bv
2
 (t), where v(t) is its instantaneous velocity. The instantaneous
acceleration of the satellite is :
(1) –bv
3
(t) (2) 
3
bv
–
M(t)
(3) 
3
2bv
–
M(t)
(4) 
3
bv
–
2M(t)
Sol. 2
dM(t)
dt
 = -bv
2
in free space
no external force
so there in only thrust force on rocket
f
in
 = 
dM
dt
 (V
rel
)
Ma = 
v
) t (
bv
2
?
?
?
?
?
?
?
? ?
a = 
3
bv
M(t)
?
6. Two different wires having lengths L
1
 and L
2
, and respective temperature coefficient of
linear expansion ?
1
 and ?
2
, are joined end-to-end. Then the effective temperature
coefficient of linear expansion is:
(1) 
2
1 1 2 2
1
L L
L L
? ? ?
?
(2) 
1 2
2 ? ?
(3) 
? ?
1
2
1 2
2
1 2
2 1
L L
4
L L
? ?
? ? ?
?
(4) 
1 2
2
? ? ?
Sol. 1
L'
1
 = L
1
 (1 + ?
1
?T)
L'
2
 = L
2
 (1 + ?
2
?T)
L'+L
2
'
 
= L
1
 + L
2
 + L
1
?
1
?T + L
2
?
2
?T
= (L
1
 + L
2
) 
1 1 2 2
1 2
L L
1 T
L L
? ? ? ? ? ? ?
? ?
? ? ? ?
?
? ? ? ? ? ?
= (L
1
 + L
2
) [1 + ?
eq
?T)
So, ?
eq
 = 
1 1 2 2
1 2
L L
L L
? ? ?
?
JEE Main 2020 Paper
         5
th
 September 2020 | (Shift-2), Physics     Page | 4
7. In the circuit, given in the figure currents in different branches and value of one resistor
are shown. Then potential at point B with respect to the point A is:
2A
F
B
2V
C
1V
A
1A
E
D
2
(1) +2 V (2) –2 V (3) +1 V (4) –1 V
Sol. 3
Let V
A
 = 0
applying KVL from A to B via ACDB
1 2 2
A B
V V ? ? ? ?
V
B
 =  1V
8. The velocity (v) and time (t) graph of a body in a straight line motion is shown in the
figure. The point S is at 4.333 seconds. The total distance covered by the body in 6 s is:
–2
0
2
4
1 2 3 4 5 6
A B
S D
C
t (in s)
v (m/s)
(1) 
37
m
3
(2) 
49
m
4
(3) 12 m (4) 11 m
Sol. 1
JEE Main 2020 Paper
         5
th
 September 2020 | (Shift-2), Physics     Page | 5
distance = area under graph
= 
1
2
 (4) 
13
1
3
? ?
?
? ?
? ?
 + 
1 13
6 2
2 3
? ? ? ?
? ?
? ? ? ?
? ? ? ?
= 2 × 
16
3
 + 
5
3
= 
32
3
 + 
5
3
 = 
37
3
 m
9. An infinitely long straight wire carrying current I, one side opened rectangular loop and
a conductor C with a sliding connector are located in the same plane, as shown in the
figure. The connector has length l and resistance R. It slides to the right with a velocity
v. The resistance of the conductor and the self inductance of the loop are negligible. The
induced current in the loop, as a function of separation r, between the connector and the
straight wire is:
v
l R
C
r
I
one side opened long
conducting wire loop
(1) 
0
Ivl
2 Rr
?
?
(2) 
0
Ivl
Rr
?
?
(3) 
0
2 Ivl
Rr
?
?
(4) 
0
Ivl
4 Rr
?
?
Sol. 1
B = 
0
I
2 r
? ? ?
? ?
?
? ?
induced emf
e = Bvl
l
v
r
I = 
0
I
2 r
?
?
 V.l
= 
?
?
0
Ivl
2 r
induced current i = 
e
R
 = 
?
?
0
Ivl
2 rR