JEE Main 2020 Physics September 5 Shift 1 Paper & Solutions | Mock Tests for JEE Main and Advanced 2025 PDF Download

 Page 1


JEE Main 2020 Paper
         5
th
 September 2020 | (Shift-1), Physics     Page | 1
Date : 5th September 2020
Time : 09 : 00 am - 12 : 00 pm
Subject : Physics
1. Three different processes that can occur in an ideal monoatomic gas are shown in the P
vs V diagram. The paths are labelled as A ? B, A ? C and A ? D. The change in internal
energies during these process are taken as E
AB
 , E
AC
 and E
AD
 and the workdone as W
AB
,
W
AC
 and W
AD
. The correct relation between these parameters are :
(1) E
AB
 = E
AC
 = E
AD
, W
AB 
> 0, W
AC
 = 0, W
AD
 < 0
(2) E
AB
 > E
AC
 > E
AD
, W
AB 
< W
AC
 < W
AD
(3) E
AB
 < E
AC
 < E
AD
, W
AB 
> 0, W
AC
 > W
AD
(4) E
AB
 = E
AC
 = E
AD
, W
AB 
> 0, W
AC
 = 0, W
AD
 > 0
Sol. 1 (Bonus)
E
AB
 = E
AC
 = E
AD
?U = 
nfR
2
 (T
f
 – T
i
)
w
AB
 > 0 (+) V ?
w
AC
 = 0 V const.
w
AD
 < 0 (–) V ?
2. With increasing biasing voltage of a photodiode, the photocurrent magnitude :
(1) increases initially and saturates finally
(2) remains constant
(3) increases linearly
(4) increases initially and after attaining certain value, it decreases
Sol. 1
By theory
Page 2


JEE Main 2020 Paper
         5
th
 September 2020 | (Shift-1), Physics     Page | 1
Date : 5th September 2020
Time : 09 : 00 am - 12 : 00 pm
Subject : Physics
1. Three different processes that can occur in an ideal monoatomic gas are shown in the P
vs V diagram. The paths are labelled as A ? B, A ? C and A ? D. The change in internal
energies during these process are taken as E
AB
 , E
AC
 and E
AD
 and the workdone as W
AB
,
W
AC
 and W
AD
. The correct relation between these parameters are :
(1) E
AB
 = E
AC
 = E
AD
, W
AB 
> 0, W
AC
 = 0, W
AD
 < 0
(2) E
AB
 > E
AC
 > E
AD
, W
AB 
< W
AC
 < W
AD
(3) E
AB
 < E
AC
 < E
AD
, W
AB 
> 0, W
AC
 > W
AD
(4) E
AB
 = E
AC
 = E
AD
, W
AB 
> 0, W
AC
 = 0, W
AD
 > 0
Sol. 1 (Bonus)
E
AB
 = E
AC
 = E
AD
?U = 
nfR
2
 (T
f
 – T
i
)
w
AB
 > 0 (+) V ?
w
AC
 = 0 V const.
w
AD
 < 0 (–) V ?
2. With increasing biasing voltage of a photodiode, the photocurrent magnitude :
(1) increases initially and saturates finally
(2) remains constant
(3) increases linearly
(4) increases initially and after attaining certain value, it decreases
Sol. 1
By theory
JEE Main 2020 Paper
         5
th
 September 2020 | (Shift-1), Physics     Page | 2
3. A square loop of side 2a, and carrying current I, is kept in XZ plane with its centre at
origin. A long wire carrying the same current I is placed parallel to the z-axis and
passing through the point (0, b, 0), (b > > a). The magnitude of the torque on the loop
about z-axis is given by :
(1) 
2 3
0
2
2 I a
b
?
?
(2) 
2 3
0
2
I a
2 b
?
?
(3)
2 2
0
2 I a
b
?
?
(4)
2 2
0
I a
2 b
?
?
Sol. 3
M = I
2
 (2a)
2
 = 4a
2
I
2
(magnetic moment)
B = 
0 2
I
2 b
?
?
? = MB sin ?
? angle between B and M [ ? = 90°]
? ?= 4 (a
2
I
2
) 
0 1
I
2 b
?
?
? = 
2
0 1 2
2 I I a
b
?
?
 = 
2 2
0
2 I a
b
?
?
Page 3


JEE Main 2020 Paper
         5
th
 September 2020 | (Shift-1), Physics     Page | 1
Date : 5th September 2020
Time : 09 : 00 am - 12 : 00 pm
Subject : Physics
1. Three different processes that can occur in an ideal monoatomic gas are shown in the P
vs V diagram. The paths are labelled as A ? B, A ? C and A ? D. The change in internal
energies during these process are taken as E
AB
 , E
AC
 and E
AD
 and the workdone as W
AB
,
W
AC
 and W
AD
. The correct relation between these parameters are :
(1) E
AB
 = E
AC
 = E
AD
, W
AB 
> 0, W
AC
 = 0, W
AD
 < 0
(2) E
AB
 > E
AC
 > E
AD
, W
AB 
< W
AC
 < W
AD
(3) E
AB
 < E
AC
 < E
AD
, W
AB 
> 0, W
AC
 > W
AD
(4) E
AB
 = E
AC
 = E
AD
, W
AB 
> 0, W
AC
 = 0, W
AD
 > 0
Sol. 1 (Bonus)
E
AB
 = E
AC
 = E
AD
?U = 
nfR
2
 (T
f
 – T
i
)
w
AB
 > 0 (+) V ?
w
AC
 = 0 V const.
w
AD
 < 0 (–) V ?
2. With increasing biasing voltage of a photodiode, the photocurrent magnitude :
(1) increases initially and saturates finally
(2) remains constant
(3) increases linearly
(4) increases initially and after attaining certain value, it decreases
Sol. 1
By theory
JEE Main 2020 Paper
         5
th
 September 2020 | (Shift-1), Physics     Page | 2
3. A square loop of side 2a, and carrying current I, is kept in XZ plane with its centre at
origin. A long wire carrying the same current I is placed parallel to the z-axis and
passing through the point (0, b, 0), (b > > a). The magnitude of the torque on the loop
about z-axis is given by :
(1) 
2 3
0
2
2 I a
b
?
?
(2) 
2 3
0
2
I a
2 b
?
?
(3)
2 2
0
2 I a
b
?
?
(4)
2 2
0
I a
2 b
?
?
Sol. 3
M = I
2
 (2a)
2
 = 4a
2
I
2
(magnetic moment)
B = 
0 2
I
2 b
?
?
? = MB sin ?
? angle between B and M [ ? = 90°]
? ?= 4 (a
2
I
2
) 
0 1
I
2 b
?
?
? = 
2
0 1 2
2 I I a
b
?
?
 = 
2 2
0
2 I a
b
?
?
JEE Main 2020 Paper
         5
th
 September 2020 | (Shift-1), Physics     Page | 3
4. Assume that the displacement (s) of air is proportional to the pressure difference ( ?p)
created by a sound wave. Displacement(s) further depends on the speed of sound ( ?),
density of air ( ?) and the frequency (f). If ?p ? 10Pa, ? ? 300 m/s , ? ? 1 kg/m
3
 and
f ? 1000 Hz, then s will be of the order of (take the multiplicative constant to be 1)
(1) 1 mm (2) 10 mm (3) 
1
10
mm (4) 
3
100
 mm
Ans. 4
S
0
 = 
f 2 v
P
v
P
v
v
P
k
P
2 ? ?
?
?
? ?
?
?
?
?
?
?
?
?
? Proportionally constant = 1
S
0
= 
vf
P
?
?
= 
10
1 300 1000 ? ?
m
= 
300
10
mm
= 
90
3
~ 
100
3
mm
5. Two capacitors of capacitances C and 2C are charged to potential differences V and 2V,
respectively. These are then connected in parallel in such a manner that the positive
terminal of one is connected to the negative terminal of the other. The final energy of
this configuration is :
(1) zero (2) 
9
2
 CV
2
(3) 
25
6
 CV
2
(4) 
3
2
CV
2
Sol. 4
U
i
 = 
1
2
 Cv
2
 + 
1
2
 (2C) (2v)
2
= 
9
2
 Cv
2
q
1
 + q
2
 = q
1
'
 
+ q
2
'
–CV + (2C(2V) = (C + 2C)V'
V' = 
3CV
3C
 = V
U
f
 = 
1
2
 CV
2
 + 
1
2
 (2C)V
2
U
f
 = 
3
2
 CV
2
Page 4


JEE Main 2020 Paper
         5
th
 September 2020 | (Shift-1), Physics     Page | 1
Date : 5th September 2020
Time : 09 : 00 am - 12 : 00 pm
Subject : Physics
1. Three different processes that can occur in an ideal monoatomic gas are shown in the P
vs V diagram. The paths are labelled as A ? B, A ? C and A ? D. The change in internal
energies during these process are taken as E
AB
 , E
AC
 and E
AD
 and the workdone as W
AB
,
W
AC
 and W
AD
. The correct relation between these parameters are :
(1) E
AB
 = E
AC
 = E
AD
, W
AB 
> 0, W
AC
 = 0, W
AD
 < 0
(2) E
AB
 > E
AC
 > E
AD
, W
AB 
< W
AC
 < W
AD
(3) E
AB
 < E
AC
 < E
AD
, W
AB 
> 0, W
AC
 > W
AD
(4) E
AB
 = E
AC
 = E
AD
, W
AB 
> 0, W
AC
 = 0, W
AD
 > 0
Sol. 1 (Bonus)
E
AB
 = E
AC
 = E
AD
?U = 
nfR
2
 (T
f
 – T
i
)
w
AB
 > 0 (+) V ?
w
AC
 = 0 V const.
w
AD
 < 0 (–) V ?
2. With increasing biasing voltage of a photodiode, the photocurrent magnitude :
(1) increases initially and saturates finally
(2) remains constant
(3) increases linearly
(4) increases initially and after attaining certain value, it decreases
Sol. 1
By theory
JEE Main 2020 Paper
         5
th
 September 2020 | (Shift-1), Physics     Page | 2
3. A square loop of side 2a, and carrying current I, is kept in XZ plane with its centre at
origin. A long wire carrying the same current I is placed parallel to the z-axis and
passing through the point (0, b, 0), (b > > a). The magnitude of the torque on the loop
about z-axis is given by :
(1) 
2 3
0
2
2 I a
b
?
?
(2) 
2 3
0
2
I a
2 b
?
?
(3)
2 2
0
2 I a
b
?
?
(4)
2 2
0
I a
2 b
?
?
Sol. 3
M = I
2
 (2a)
2
 = 4a
2
I
2
(magnetic moment)
B = 
0 2
I
2 b
?
?
? = MB sin ?
? angle between B and M [ ? = 90°]
? ?= 4 (a
2
I
2
) 
0 1
I
2 b
?
?
? = 
2
0 1 2
2 I I a
b
?
?
 = 
2 2
0
2 I a
b
?
?
JEE Main 2020 Paper
         5
th
 September 2020 | (Shift-1), Physics     Page | 3
4. Assume that the displacement (s) of air is proportional to the pressure difference ( ?p)
created by a sound wave. Displacement(s) further depends on the speed of sound ( ?),
density of air ( ?) and the frequency (f). If ?p ? 10Pa, ? ? 300 m/s , ? ? 1 kg/m
3
 and
f ? 1000 Hz, then s will be of the order of (take the multiplicative constant to be 1)
(1) 1 mm (2) 10 mm (3) 
1
10
mm (4) 
3
100
 mm
Ans. 4
S
0
 = 
f 2 v
P
v
P
v
v
P
k
P
2 ? ?
?
?
? ?
?
?
?
?
?
?
?
?
? Proportionally constant = 1
S
0
= 
vf
P
?
?
= 
10
1 300 1000 ? ?
m
= 
300
10
mm
= 
90
3
~ 
100
3
mm
5. Two capacitors of capacitances C and 2C are charged to potential differences V and 2V,
respectively. These are then connected in parallel in such a manner that the positive
terminal of one is connected to the negative terminal of the other. The final energy of
this configuration is :
(1) zero (2) 
9
2
 CV
2
(3) 
25
6
 CV
2
(4) 
3
2
CV
2
Sol. 4
U
i
 = 
1
2
 Cv
2
 + 
1
2
 (2C) (2v)
2
= 
9
2
 Cv
2
q
1
 + q
2
 = q
1
'
 
+ q
2
'
–CV + (2C(2V) = (C + 2C)V'
V' = 
3CV
3C
 = V
U
f
 = 
1
2
 CV
2
 + 
1
2
 (2C)V
2
U
f
 = 
3
2
 CV
2
JEE Main 2020 Paper
         5
th
 September 2020 | (Shift-1), Physics     Page | 4
6. A helicopter rises from rest on the ground vertically upwards with a constant acceleration
g. A food packet is dropped from the helicopter when it is at a height h. The time taken
by the packet to reach the ground is close to [g is the acceleration due to gravity] :
(1) t = 3.4 
h
g
? ?
? ?
? ?
(2) t =
2h
3g
(3) t =
2
3
h
g
? ?
? ?
? ?
(4) t = 1.8 
h
g
Sol. 1
V
B
2
 = 0
2
 + 2gh
V
B
 = 2gh
–h = (V
B
)t – 
1
2
 gt
2
–h = 2gh t  – 
1
2
 gt
2
gt
2
 – 2 2gh t – 2h = 0
t = 
? ? 2gh 8gh 8gh
2g
 = 
2 2gh 16gh
2g
?
 = 
2gh 2 gh
g
?
t = 
2h
g
 + 
h
2
g
 = 
h
g
 ? ?
2 2 ?
 = 3.4 
h
g
7. A bullet of mass 5 g, travelling with a speed of 210 m/s, strikes a fixed wooden target.
One half of its kinetic energy is converted into heat in the bullet while the other half is
converted into heat in the wood. The rise of temperature of the bullet if the specific heat
of its material is 0.030 cal/(g – °C) (1 cal = 4.2 × 10
7
 ergs) close to :
(1) 38.4°C (2) 87.5°C (3) 83.3°C (4)119.2°C
Sol. 2
2
1
m
2
? ?
?
? ?
? ?
 × 
1
2
 = ms ?T s = 0.03 cal/gm°C
2
4
?
 = 126 × ?T = ?
?
3
0.03 4.2J
10 kgC
?
2
 = 4 × 126 × ?T = 126 J/kg°C
(210)
2
 = 4 × 126 × ?T
210 × 210 = 4 × 126 × ?T
44100 = 504 × ?T
?T = 
504
44100
 = 87.5°C
Page 5


JEE Main 2020 Paper
         5
th
 September 2020 | (Shift-1), Physics     Page | 1
Date : 5th September 2020
Time : 09 : 00 am - 12 : 00 pm
Subject : Physics
1. Three different processes that can occur in an ideal monoatomic gas are shown in the P
vs V diagram. The paths are labelled as A ? B, A ? C and A ? D. The change in internal
energies during these process are taken as E
AB
 , E
AC
 and E
AD
 and the workdone as W
AB
,
W
AC
 and W
AD
. The correct relation between these parameters are :
(1) E
AB
 = E
AC
 = E
AD
, W
AB 
> 0, W
AC
 = 0, W
AD
 < 0
(2) E
AB
 > E
AC
 > E
AD
, W
AB 
< W
AC
 < W
AD
(3) E
AB
 < E
AC
 < E
AD
, W
AB 
> 0, W
AC
 > W
AD
(4) E
AB
 = E
AC
 = E
AD
, W
AB 
> 0, W
AC
 = 0, W
AD
 > 0
Sol. 1 (Bonus)
E
AB
 = E
AC
 = E
AD
?U = 
nfR
2
 (T
f
 – T
i
)
w
AB
 > 0 (+) V ?
w
AC
 = 0 V const.
w
AD
 < 0 (–) V ?
2. With increasing biasing voltage of a photodiode, the photocurrent magnitude :
(1) increases initially and saturates finally
(2) remains constant
(3) increases linearly
(4) increases initially and after attaining certain value, it decreases
Sol. 1
By theory
JEE Main 2020 Paper
         5
th
 September 2020 | (Shift-1), Physics     Page | 2
3. A square loop of side 2a, and carrying current I, is kept in XZ plane with its centre at
origin. A long wire carrying the same current I is placed parallel to the z-axis and
passing through the point (0, b, 0), (b > > a). The magnitude of the torque on the loop
about z-axis is given by :
(1) 
2 3
0
2
2 I a
b
?
?
(2) 
2 3
0
2
I a
2 b
?
?
(3)
2 2
0
2 I a
b
?
?
(4)
2 2
0
I a
2 b
?
?
Sol. 3
M = I
2
 (2a)
2
 = 4a
2
I
2
(magnetic moment)
B = 
0 2
I
2 b
?
?
? = MB sin ?
? angle between B and M [ ? = 90°]
? ?= 4 (a
2
I
2
) 
0 1
I
2 b
?
?
? = 
2
0 1 2
2 I I a
b
?
?
 = 
2 2
0
2 I a
b
?
?
JEE Main 2020 Paper
         5
th
 September 2020 | (Shift-1), Physics     Page | 3
4. Assume that the displacement (s) of air is proportional to the pressure difference ( ?p)
created by a sound wave. Displacement(s) further depends on the speed of sound ( ?),
density of air ( ?) and the frequency (f). If ?p ? 10Pa, ? ? 300 m/s , ? ? 1 kg/m
3
 and
f ? 1000 Hz, then s will be of the order of (take the multiplicative constant to be 1)
(1) 1 mm (2) 10 mm (3) 
1
10
mm (4) 
3
100
 mm
Ans. 4
S
0
 = 
f 2 v
P
v
P
v
v
P
k
P
2 ? ?
?
?
? ?
?
?
?
?
?
?
?
?
? Proportionally constant = 1
S
0
= 
vf
P
?
?
= 
10
1 300 1000 ? ?
m
= 
300
10
mm
= 
90
3
~ 
100
3
mm
5. Two capacitors of capacitances C and 2C are charged to potential differences V and 2V,
respectively. These are then connected in parallel in such a manner that the positive
terminal of one is connected to the negative terminal of the other. The final energy of
this configuration is :
(1) zero (2) 
9
2
 CV
2
(3) 
25
6
 CV
2
(4) 
3
2
CV
2
Sol. 4
U
i
 = 
1
2
 Cv
2
 + 
1
2
 (2C) (2v)
2
= 
9
2
 Cv
2
q
1
 + q
2
 = q
1
'
 
+ q
2
'
–CV + (2C(2V) = (C + 2C)V'
V' = 
3CV
3C
 = V
U
f
 = 
1
2
 CV
2
 + 
1
2
 (2C)V
2
U
f
 = 
3
2
 CV
2
JEE Main 2020 Paper
         5
th
 September 2020 | (Shift-1), Physics     Page | 4
6. A helicopter rises from rest on the ground vertically upwards with a constant acceleration
g. A food packet is dropped from the helicopter when it is at a height h. The time taken
by the packet to reach the ground is close to [g is the acceleration due to gravity] :
(1) t = 3.4 
h
g
? ?
? ?
? ?
(2) t =
2h
3g
(3) t =
2
3
h
g
? ?
? ?
? ?
(4) t = 1.8 
h
g
Sol. 1
V
B
2
 = 0
2
 + 2gh
V
B
 = 2gh
–h = (V
B
)t – 
1
2
 gt
2
–h = 2gh t  – 
1
2
 gt
2
gt
2
 – 2 2gh t – 2h = 0
t = 
? ? 2gh 8gh 8gh
2g
 = 
2 2gh 16gh
2g
?
 = 
2gh 2 gh
g
?
t = 
2h
g
 + 
h
2
g
 = 
h
g
 ? ?
2 2 ?
 = 3.4 
h
g
7. A bullet of mass 5 g, travelling with a speed of 210 m/s, strikes a fixed wooden target.
One half of its kinetic energy is converted into heat in the bullet while the other half is
converted into heat in the wood. The rise of temperature of the bullet if the specific heat
of its material is 0.030 cal/(g – °C) (1 cal = 4.2 × 10
7
 ergs) close to :
(1) 38.4°C (2) 87.5°C (3) 83.3°C (4)119.2°C
Sol. 2
2
1
m
2
? ?
?
? ?
? ?
 × 
1
2
 = ms ?T s = 0.03 cal/gm°C
2
4
?
 = 126 × ?T = ?
?
3
0.03 4.2J
10 kgC
?
2
 = 4 × 126 × ?T = 126 J/kg°C
(210)
2
 = 4 × 126 × ?T
210 × 210 = 4 × 126 × ?T
44100 = 504 × ?T
?T = 
504
44100
 = 87.5°C
JEE Main 2020 Paper
         5
th
 September 2020 | (Shift-1), Physics     Page | 5
8. A wheel is rotating freely with an angular speed ? on a shaft. The moment of inertia of
the wheel is I and the moment of inertia of the shaft is negligible. Another wheel of
moment of inertia 3I initially at rest is suddenly coupled to the same shaft. The resultant
fractional loss in the kinetic energy of the system is :
(1) 
3
4
(2) 0 (3) 
5
6
(4)
1
4
Sol. 1
k
i
 = 
1
2
 I ?
2
k
f
 = 
1
2
 (4I) ( ?')
2
= 2I 
2
4
? ? ?
? ?
? ?
 = 
1
8
 I ?
2
A.M.C
I ? = (I+3I) ?'
?' = 
I
4I
?
 = 
4
?
= 
i f
i
K K
K
?
 ? 
2 2
2
1 1
I I
2 8
1
I
2
? ? ?
?
2
2
I
2
1
I
8
3
?
?
 = 
4
3
9. A balloon is moving up in air vertically above a point A on the ground. When it is at a
height h
1
 , a girl standing at a distance d(point B) from A (see figure) sees it at an angle
45° with respect to the vertical. When the balloon climbs up a further height h
2
, it is
seen at an angle 60° with respect to the vertical if the girl moves further by a distance
2.464 d (point C). Then the height h
2
 is (given tan 30° = 0.5774) :
(1) 0.464 d (2) d (3) 0.732 d (4) 1.464 d
Sol. 2