JEE Mains 16 March 2021 Question Paper Shift 2 | JEE Main & Advanced Previous Year Papers PDF Download

 Page 1


1
SECTION-A
1. The following logic gate is equivalent to :
A
B
Y
(1) NOR Gate (2) OR Gate
(3) AND Gate (4) NAND Gate
Official Ans. by NTA (1)
Sol. Truth table for the given logic gate :
A BY
0 01
0 11
1 01
1 10
The truth table is similar to that of a NOR gate.
2. A large block of wood of mass M = 5.99 kg is
hanging from two long massless cords. A
bullet of mass m = 10g is fired into the block
and gets embedded in it. The (block + bullet)
then swing upwards, their centre of mass rising
a vertical distance h = 9.8 cm before the
(block + bullet) pendulum comes momentarily
to rest at the end of its arc. The speed of the
bullet just before collision is : (Take g = 9.8 ms
–2
)
h
M
m
v
(1) 841.4 m/s (2) 811.4 m/s
(3) 831.4 m/s (4) 821.4 m/s
Official Ans. by NTA (3)
FINAL JEE–MAIN EXAMINATION – MARCH, 2021
(Held On Tuesday 16
th
 March, 2021)    TIME : 3 : 00 PM   to  6 : 00 PM
PHYSICS TEST PAPER WITH ANSWER & SOLUTION
Sol. From energy conservation,
éù
êú
ëû
after bullet gets embedded till the
system comes momentarily at rest
(M + m)g h = 
+
2
1
1
(M m)v
2
[v
1
 is velocity after collision]
\ v
1
 = 2gh
Applying momentum conservation, (just
before and just after collision)
mv = (M + m)v
1
v = 
-
-
+ æö
= ´´´ ´
ç÷
èø ´
2
1 3
Mm6
v 2 9.8 9.8 10
m 10 10
»831.55 m / s
3. A charge Q is moving 
uur
dI
 distance in the
magnetic field 
r
B . Find the value of work done
by 
r
B
.
(1) 1 (2) Infinite
(3) Zero (4) –1
Official Ans. by NTA (3)
Sol. Since force on a point charge by magnetic field
is always perpendicular to vF qVB éù =´
ëû
r rr
r
\ Work by magnetic force on the point charge
is zero.
4. What will be the nature of flow of water from
a circular tap, when its flow rate increased from
0.18 L/min to 0.48 L/min ? The radius of the
tap and viscosity of water are 0.5 cm and
10
–3
 Pa s, respectively.
(Density of water : 10
3
 kg/m
3
)
(1) Unsteady to steady flow
(2) Remains steady flow
(3) Remains turbulent flow
(4) Steady flow to unsteady flow
Official Ans. by NTA (4)
Page 2


1
SECTION-A
1. The following logic gate is equivalent to :
A
B
Y
(1) NOR Gate (2) OR Gate
(3) AND Gate (4) NAND Gate
Official Ans. by NTA (1)
Sol. Truth table for the given logic gate :
A BY
0 01
0 11
1 01
1 10
The truth table is similar to that of a NOR gate.
2. A large block of wood of mass M = 5.99 kg is
hanging from two long massless cords. A
bullet of mass m = 10g is fired into the block
and gets embedded in it. The (block + bullet)
then swing upwards, their centre of mass rising
a vertical distance h = 9.8 cm before the
(block + bullet) pendulum comes momentarily
to rest at the end of its arc. The speed of the
bullet just before collision is : (Take g = 9.8 ms
–2
)
h
M
m
v
(1) 841.4 m/s (2) 811.4 m/s
(3) 831.4 m/s (4) 821.4 m/s
Official Ans. by NTA (3)
FINAL JEE–MAIN EXAMINATION – MARCH, 2021
(Held On Tuesday 16
th
 March, 2021)    TIME : 3 : 00 PM   to  6 : 00 PM
PHYSICS TEST PAPER WITH ANSWER & SOLUTION
Sol. From energy conservation,
éù
êú
ëû
after bullet gets embedded till the
system comes momentarily at rest
(M + m)g h = 
+
2
1
1
(M m)v
2
[v
1
 is velocity after collision]
\ v
1
 = 2gh
Applying momentum conservation, (just
before and just after collision)
mv = (M + m)v
1
v = 
-
-
+ æö
= ´´´ ´
ç÷
èø ´
2
1 3
Mm6
v 2 9.8 9.8 10
m 10 10
»831.55 m / s
3. A charge Q is moving 
uur
dI
 distance in the
magnetic field 
r
B . Find the value of work done
by 
r
B
.
(1) 1 (2) Infinite
(3) Zero (4) –1
Official Ans. by NTA (3)
Sol. Since force on a point charge by magnetic field
is always perpendicular to vF qVB éù =´
ëû
r rr
r
\ Work by magnetic force on the point charge
is zero.
4. What will be the nature of flow of water from
a circular tap, when its flow rate increased from
0.18 L/min to 0.48 L/min ? The radius of the
tap and viscosity of water are 0.5 cm and
10
–3
 Pa s, respectively.
(Density of water : 10
3
 kg/m
3
)
(1) Unsteady to steady flow
(2) Remains steady flow
(3) Remains turbulent flow
(4) Steady flow to unsteady flow
Official Ans. by NTA (4)
2
Sol. The nature of flow is determined by Reynolds
Number.
R
e
 = 
vD r
h
density of fluid ; coefficient of
v velocity of flow               viscosity
D Diameter of pipe
r® h® éù
êú ®
êú
®
ëû
From NCERT
If R
e
 < 1000          ® flow is steady
1000 < R
e
 < 2000   ® flow becomes unsteady
R
e
 > 2000 ®   flow is turbulent
--
--
´´
=´´
p´ ´´
32
3
einitial 223
0.18 10 1 10
R 10
(0.5 10 ) 60 10
        = 382.16
--
--
´´
=´´
p´ ´´
32
3
e final 223
0.48 10 1 10
R 10
(0.5 10 ) 60 10
       = 1019.09
5. A mosquito is moving with a velocity
= ++
r
2
ˆˆ ˆ
v 0.5t i 3tj 9k m/s and accelerating in
uniform conditions. What will be the direction
of mosquito after 2s ?
(1) 
-
æö
ç÷
èø
1
2
tan
3
 from x-axis
(2) 
-
æö
ç÷
èø
1
2
tan
3
 from y-axis
(3) 
-
æö
ç÷
èø
1
5
tan
2
 from y-axis
(4) 
-
æö
ç÷
èø
1
5
tan
2
 from x-axis
Official Ans. by NTA (2)
Official Ans. by ALLEN (Bonus)
Sol. Given :
= ++
r
2
ˆˆ ˆ
v 0.5t i 3t j 9k
at t = 2
ˆˆ ˆ
v 2 i 6 j 9k = ++
r
\ Angle made by direction of motion of
mosquito will be,
-1
2
cos
11
 (from x-axis) = 
-1
117
tan
2
-1
6
cos
11
 (from y-axis) = 
1
85
tan
6
-
-1
9
cos
11
 (from z-axis) = 
1
40
tan
9
-
None of the option is matching.
Hence this question should be bonus.
6. Find out the surface charge density at the
intersection of point x = 3 m plane and x-axis,
in the region of uniform line charge of 8 nC/
m lying along the z-axis in free space.
(1) 0.424 nC m
–2
(2) 47.88 C/m
(3) 0.07 nC m
–2
(4) 4.0 nC m
–2
Official Ans. by NTA (1)
Sol.
ls
=
e
0
2K
r
(x = 3m)
-
s=´
9
2
C
0.424 10
m
7. The de-Broglie wavelength associated with an
electron and a proton were calculated by
accelerating them through same potential of
100 V. What should nearly be the ratio of their
wavelengths ? (m
P
 = 1.00727 u, m
e
 = 0.00055u)
(1) 1860 : 1 (2) (1860)
2
 : 1
(3) 41.4 : 1 (4) 43 : 1
Official Ans. by NTA (4)
Sol.
l= ==
h hh
mv 2mK 2mqV
l
=
l
12
21
m
m
l
= ==
l
e P
Pe
m
1831.4 42.79
m
Page 3


1
SECTION-A
1. The following logic gate is equivalent to :
A
B
Y
(1) NOR Gate (2) OR Gate
(3) AND Gate (4) NAND Gate
Official Ans. by NTA (1)
Sol. Truth table for the given logic gate :
A BY
0 01
0 11
1 01
1 10
The truth table is similar to that of a NOR gate.
2. A large block of wood of mass M = 5.99 kg is
hanging from two long massless cords. A
bullet of mass m = 10g is fired into the block
and gets embedded in it. The (block + bullet)
then swing upwards, their centre of mass rising
a vertical distance h = 9.8 cm before the
(block + bullet) pendulum comes momentarily
to rest at the end of its arc. The speed of the
bullet just before collision is : (Take g = 9.8 ms
–2
)
h
M
m
v
(1) 841.4 m/s (2) 811.4 m/s
(3) 831.4 m/s (4) 821.4 m/s
Official Ans. by NTA (3)
FINAL JEE–MAIN EXAMINATION – MARCH, 2021
(Held On Tuesday 16
th
 March, 2021)    TIME : 3 : 00 PM   to  6 : 00 PM
PHYSICS TEST PAPER WITH ANSWER & SOLUTION
Sol. From energy conservation,
éù
êú
ëû
after bullet gets embedded till the
system comes momentarily at rest
(M + m)g h = 
+
2
1
1
(M m)v
2
[v
1
 is velocity after collision]
\ v
1
 = 2gh
Applying momentum conservation, (just
before and just after collision)
mv = (M + m)v
1
v = 
-
-
+ æö
= ´´´ ´
ç÷
èø ´
2
1 3
Mm6
v 2 9.8 9.8 10
m 10 10
»831.55 m / s
3. A charge Q is moving 
uur
dI
 distance in the
magnetic field 
r
B . Find the value of work done
by 
r
B
.
(1) 1 (2) Infinite
(3) Zero (4) –1
Official Ans. by NTA (3)
Sol. Since force on a point charge by magnetic field
is always perpendicular to vF qVB éù =´
ëû
r rr
r
\ Work by magnetic force on the point charge
is zero.
4. What will be the nature of flow of water from
a circular tap, when its flow rate increased from
0.18 L/min to 0.48 L/min ? The radius of the
tap and viscosity of water are 0.5 cm and
10
–3
 Pa s, respectively.
(Density of water : 10
3
 kg/m
3
)
(1) Unsteady to steady flow
(2) Remains steady flow
(3) Remains turbulent flow
(4) Steady flow to unsteady flow
Official Ans. by NTA (4)
2
Sol. The nature of flow is determined by Reynolds
Number.
R
e
 = 
vD r
h
density of fluid ; coefficient of
v velocity of flow               viscosity
D Diameter of pipe
r® h® éù
êú ®
êú
®
ëû
From NCERT
If R
e
 < 1000          ® flow is steady
1000 < R
e
 < 2000   ® flow becomes unsteady
R
e
 > 2000 ®   flow is turbulent
--
--
´´
=´´
p´ ´´
32
3
einitial 223
0.18 10 1 10
R 10
(0.5 10 ) 60 10
        = 382.16
--
--
´´
=´´
p´ ´´
32
3
e final 223
0.48 10 1 10
R 10
(0.5 10 ) 60 10
       = 1019.09
5. A mosquito is moving with a velocity
= ++
r
2
ˆˆ ˆ
v 0.5t i 3tj 9k m/s and accelerating in
uniform conditions. What will be the direction
of mosquito after 2s ?
(1) 
-
æö
ç÷
èø
1
2
tan
3
 from x-axis
(2) 
-
æö
ç÷
èø
1
2
tan
3
 from y-axis
(3) 
-
æö
ç÷
èø
1
5
tan
2
 from y-axis
(4) 
-
æö
ç÷
èø
1
5
tan
2
 from x-axis
Official Ans. by NTA (2)
Official Ans. by ALLEN (Bonus)
Sol. Given :
= ++
r
2
ˆˆ ˆ
v 0.5t i 3t j 9k
at t = 2
ˆˆ ˆ
v 2 i 6 j 9k = ++
r
\ Angle made by direction of motion of
mosquito will be,
-1
2
cos
11
 (from x-axis) = 
-1
117
tan
2
-1
6
cos
11
 (from y-axis) = 
1
85
tan
6
-
-1
9
cos
11
 (from z-axis) = 
1
40
tan
9
-
None of the option is matching.
Hence this question should be bonus.
6. Find out the surface charge density at the
intersection of point x = 3 m plane and x-axis,
in the region of uniform line charge of 8 nC/
m lying along the z-axis in free space.
(1) 0.424 nC m
–2
(2) 47.88 C/m
(3) 0.07 nC m
–2
(4) 4.0 nC m
–2
Official Ans. by NTA (1)
Sol.
ls
=
e
0
2K
r
(x = 3m)
-
s=´
9
2
C
0.424 10
m
7. The de-Broglie wavelength associated with an
electron and a proton were calculated by
accelerating them through same potential of
100 V. What should nearly be the ratio of their
wavelengths ? (m
P
 = 1.00727 u, m
e
 = 0.00055u)
(1) 1860 : 1 (2) (1860)
2
 : 1
(3) 41.4 : 1 (4) 43 : 1
Official Ans. by NTA (4)
Sol.
l= ==
h hh
mv 2mK 2mqV
l
=
l
12
21
m
m
l
= ==
l
e P
Pe
m
1831.4 42.79
m
3
8. For the given circuit, comment on the type of
transformer used :
P
~
220V
i
1
S
L
O
A
D
i
L 0.11A
60W
V
2
(1) Auxilliary transformer
(2) Auto transformer
(3) Step-up transformer
(4) Step down transformer
Official Ans. by NTA (3)
Sol. V
S
 = 
==
P 60
545.45
i 0.11
V
P
 = 220
V
S
 > V
P
Þ Step up transformer
9. The half-life of Au
198
 is 2.7 days. The activity
of 1.50 mg of Au
198
 if its atomic weight is
198 g mol
–1
 is, (N
A
 = 6 × 10
23
/mol)
(1) 240 Ci (2) 357 Ci
(3) 535 Ci (4) 252 Ci
Official Ans. by NTA (2)
Sol. A = lN  
æö
=
ç÷
èø l
1/2
ln2
t
N = nN
A
N = 
-
æö ´
ç÷
èø
3
A
1.5 10
N
198
A = 
æö
ç÷
èø
1/2
ln2
N
t
1 Curie = 3.7 × 10
10
 Bq
A = 365 Bq
10. Calculate the value of mean free path (l) for
oxygen molecules at temperature 27°C
and pressure 1.01 × 10
5
 Pa. Assume the
molecular diameter 0.3 nm and the gas is ideal.
(k = 1.38 × 10
–23
 JK
–1
)
(1) 58 nm (2) 32 nm
(3) 86 nm (4) 102 nm
Official Ans. by NTA (4)
Sol.
l=
p
2
A
RT
2 dNP
l =102 nm
11. The refractive index of a converging lens is 1.4.
What will be the focal length of this lens if it
is placed in a medium of same refractive
index ? (Assume the radii of curvature of the
faces of lens are R
1
 and R
2
 respectively)
(1) 1 (2) Infinite
(3) 
-
12
12
RR
RR
(4) Zero
Official Ans. by NTA (2)
Sol.
éùéù m
= --
êúêú
m
ëû ëû
L
S 12
1 11
1
F RR
If m
L
 = m
S
 Þ 
= Þ =¥
1
0F
F
12. In order to determine the Young's Modulus of
a wire of radius 0.2 cm (measured using a scale
of least count = 0.001 cm) and length 1m
(measured using a scale of least count = 1 mm),
a weight of mass 1kg (measured using a scale
of least count = 1g) was hanged to get the
elongation of 0.5 cm (measured using a scale
of least count 0.001 cm). What will be the
fractional error in the value of Young's
Modulus determined by this experiment ?
(1) 0.14%
(2) 0.9%
(3) 9%
(4) 1.4%
Official Ans. by NTA (4)
Sol. Y = 
==
p l
2
Stress FL mg.L
Strain Al R .
DDD D D
= + ++
l
l
Y m LR
2.
Y m LR
éù D æö
´= + ++
ç÷ êú
èø
ëû
Y 1 1 0.001 0.001
100 100 2
Y 1000 1000 0.2 0.5
            = 
+ ++ ==
1 1 1 14
1 1.4%
10 10 5 10
Page 4


1
SECTION-A
1. The following logic gate is equivalent to :
A
B
Y
(1) NOR Gate (2) OR Gate
(3) AND Gate (4) NAND Gate
Official Ans. by NTA (1)
Sol. Truth table for the given logic gate :
A BY
0 01
0 11
1 01
1 10
The truth table is similar to that of a NOR gate.
2. A large block of wood of mass M = 5.99 kg is
hanging from two long massless cords. A
bullet of mass m = 10g is fired into the block
and gets embedded in it. The (block + bullet)
then swing upwards, their centre of mass rising
a vertical distance h = 9.8 cm before the
(block + bullet) pendulum comes momentarily
to rest at the end of its arc. The speed of the
bullet just before collision is : (Take g = 9.8 ms
–2
)
h
M
m
v
(1) 841.4 m/s (2) 811.4 m/s
(3) 831.4 m/s (4) 821.4 m/s
Official Ans. by NTA (3)
FINAL JEE–MAIN EXAMINATION – MARCH, 2021
(Held On Tuesday 16
th
 March, 2021)    TIME : 3 : 00 PM   to  6 : 00 PM
PHYSICS TEST PAPER WITH ANSWER & SOLUTION
Sol. From energy conservation,
éù
êú
ëû
after bullet gets embedded till the
system comes momentarily at rest
(M + m)g h = 
+
2
1
1
(M m)v
2
[v
1
 is velocity after collision]
\ v
1
 = 2gh
Applying momentum conservation, (just
before and just after collision)
mv = (M + m)v
1
v = 
-
-
+ æö
= ´´´ ´
ç÷
èø ´
2
1 3
Mm6
v 2 9.8 9.8 10
m 10 10
»831.55 m / s
3. A charge Q is moving 
uur
dI
 distance in the
magnetic field 
r
B . Find the value of work done
by 
r
B
.
(1) 1 (2) Infinite
(3) Zero (4) –1
Official Ans. by NTA (3)
Sol. Since force on a point charge by magnetic field
is always perpendicular to vF qVB éù =´
ëû
r rr
r
\ Work by magnetic force on the point charge
is zero.
4. What will be the nature of flow of water from
a circular tap, when its flow rate increased from
0.18 L/min to 0.48 L/min ? The radius of the
tap and viscosity of water are 0.5 cm and
10
–3
 Pa s, respectively.
(Density of water : 10
3
 kg/m
3
)
(1) Unsteady to steady flow
(2) Remains steady flow
(3) Remains turbulent flow
(4) Steady flow to unsteady flow
Official Ans. by NTA (4)
2
Sol. The nature of flow is determined by Reynolds
Number.
R
e
 = 
vD r
h
density of fluid ; coefficient of
v velocity of flow               viscosity
D Diameter of pipe
r® h® éù
êú ®
êú
®
ëû
From NCERT
If R
e
 < 1000          ® flow is steady
1000 < R
e
 < 2000   ® flow becomes unsteady
R
e
 > 2000 ®   flow is turbulent
--
--
´´
=´´
p´ ´´
32
3
einitial 223
0.18 10 1 10
R 10
(0.5 10 ) 60 10
        = 382.16
--
--
´´
=´´
p´ ´´
32
3
e final 223
0.48 10 1 10
R 10
(0.5 10 ) 60 10
       = 1019.09
5. A mosquito is moving with a velocity
= ++
r
2
ˆˆ ˆ
v 0.5t i 3tj 9k m/s and accelerating in
uniform conditions. What will be the direction
of mosquito after 2s ?
(1) 
-
æö
ç÷
èø
1
2
tan
3
 from x-axis
(2) 
-
æö
ç÷
èø
1
2
tan
3
 from y-axis
(3) 
-
æö
ç÷
èø
1
5
tan
2
 from y-axis
(4) 
-
æö
ç÷
èø
1
5
tan
2
 from x-axis
Official Ans. by NTA (2)
Official Ans. by ALLEN (Bonus)
Sol. Given :
= ++
r
2
ˆˆ ˆ
v 0.5t i 3t j 9k
at t = 2
ˆˆ ˆ
v 2 i 6 j 9k = ++
r
\ Angle made by direction of motion of
mosquito will be,
-1
2
cos
11
 (from x-axis) = 
-1
117
tan
2
-1
6
cos
11
 (from y-axis) = 
1
85
tan
6
-
-1
9
cos
11
 (from z-axis) = 
1
40
tan
9
-
None of the option is matching.
Hence this question should be bonus.
6. Find out the surface charge density at the
intersection of point x = 3 m plane and x-axis,
in the region of uniform line charge of 8 nC/
m lying along the z-axis in free space.
(1) 0.424 nC m
–2
(2) 47.88 C/m
(3) 0.07 nC m
–2
(4) 4.0 nC m
–2
Official Ans. by NTA (1)
Sol.
ls
=
e
0
2K
r
(x = 3m)
-
s=´
9
2
C
0.424 10
m
7. The de-Broglie wavelength associated with an
electron and a proton were calculated by
accelerating them through same potential of
100 V. What should nearly be the ratio of their
wavelengths ? (m
P
 = 1.00727 u, m
e
 = 0.00055u)
(1) 1860 : 1 (2) (1860)
2
 : 1
(3) 41.4 : 1 (4) 43 : 1
Official Ans. by NTA (4)
Sol.
l= ==
h hh
mv 2mK 2mqV
l
=
l
12
21
m
m
l
= ==
l
e P
Pe
m
1831.4 42.79
m
3
8. For the given circuit, comment on the type of
transformer used :
P
~
220V
i
1
S
L
O
A
D
i
L 0.11A
60W
V
2
(1) Auxilliary transformer
(2) Auto transformer
(3) Step-up transformer
(4) Step down transformer
Official Ans. by NTA (3)
Sol. V
S
 = 
==
P 60
545.45
i 0.11
V
P
 = 220
V
S
 > V
P
Þ Step up transformer
9. The half-life of Au
198
 is 2.7 days. The activity
of 1.50 mg of Au
198
 if its atomic weight is
198 g mol
–1
 is, (N
A
 = 6 × 10
23
/mol)
(1) 240 Ci (2) 357 Ci
(3) 535 Ci (4) 252 Ci
Official Ans. by NTA (2)
Sol. A = lN  
æö
=
ç÷
èø l
1/2
ln2
t
N = nN
A
N = 
-
æö ´
ç÷
èø
3
A
1.5 10
N
198
A = 
æö
ç÷
èø
1/2
ln2
N
t
1 Curie = 3.7 × 10
10
 Bq
A = 365 Bq
10. Calculate the value of mean free path (l) for
oxygen molecules at temperature 27°C
and pressure 1.01 × 10
5
 Pa. Assume the
molecular diameter 0.3 nm and the gas is ideal.
(k = 1.38 × 10
–23
 JK
–1
)
(1) 58 nm (2) 32 nm
(3) 86 nm (4) 102 nm
Official Ans. by NTA (4)
Sol.
l=
p
2
A
RT
2 dNP
l =102 nm
11. The refractive index of a converging lens is 1.4.
What will be the focal length of this lens if it
is placed in a medium of same refractive
index ? (Assume the radii of curvature of the
faces of lens are R
1
 and R
2
 respectively)
(1) 1 (2) Infinite
(3) 
-
12
12
RR
RR
(4) Zero
Official Ans. by NTA (2)
Sol.
éùéù m
= --
êúêú
m
ëû ëû
L
S 12
1 11
1
F RR
If m
L
 = m
S
 Þ 
= Þ =¥
1
0F
F
12. In order to determine the Young's Modulus of
a wire of radius 0.2 cm (measured using a scale
of least count = 0.001 cm) and length 1m
(measured using a scale of least count = 1 mm),
a weight of mass 1kg (measured using a scale
of least count = 1g) was hanged to get the
elongation of 0.5 cm (measured using a scale
of least count 0.001 cm). What will be the
fractional error in the value of Young's
Modulus determined by this experiment ?
(1) 0.14%
(2) 0.9%
(3) 9%
(4) 1.4%
Official Ans. by NTA (4)
Sol. Y = 
==
p l
2
Stress FL mg.L
Strain Al R .
DDD D D
= + ++
l
l
Y m LR
2.
Y m LR
éù D æö
´= + ++
ç÷ êú
èø
ëû
Y 1 1 0.001 0.001
100 100 2
Y 1000 1000 0.2 0.5
            = 
+ ++ ==
1 1 1 14
1 1.4%
10 10 5 10
4
13. A bimetallic strip consists of metals A and B.
It is mounted rigidly as shown. The metal A has
higher coefficient of expansion compared to
that of metal B. When the bimetallic strip is
placed in a cold both, it will :
A B
(1) Bend towards the right
(2) Not bend but shrink
(3) Neither bend nor shrink
(4) Bend towards the left
Official Ans. by NTA (4)
Sol. a
A
 > a
B
Length of both strips will decrease
DL
A
 > DL
B
14. A resistor develops 500 J of thermal energy in
20s when a current of 1.5 A is passed through
it. If the current is increased from 1.5 A to 3A,
what will be the energy developed in 20 s.
(1) 1500 J (2) 1000 J
(3) 500 J (4) 2000 J
Official Ans. by NTA (4)
Sol. 500 = (1.5)
2
 × R × 20
E = (3)
2
 × R × 20
E = 2000 J
15. Statement I : A cyclist is moving on an
unbanked road with a speed of
7 kmh
–1
 and takes a sharp
circular turn along a path of
radius of 2m without reducing
the speed. The static friction
coefficient is 0.2. The cyclist
will not slip and pass the curve
(g = 9.8 m/s
2
)
Statement II : If the road is banked at an angle
of 45°, cyclist can cross the
curve of 2m radius with the
speed of 18.5 kmh
–1
 without
slipping.
In the light of the above statements, choose the
correct answer from the options given below.
(1) Statement I is incorrect and statement II is
correct
(2) Statement I is correct and statement II is
incorrect
(3) Both statement I and statement II are false
(4) Both statement I and statement II are true
Official Ans. by NTA (4)
Sol. Statement I :
= m = ´´
max
v Rg (0.2) 2 9.8
v
max
 = 1.97 m/s
7 km/h = 1.944 m/s
Speed is lower than v
max
, hence it can take safe
turn.
Statement II
v
max
  = 
tan
Rg
1 tan
éù q+m
êú
-mq
ëû
= 
+ éù
´=
êú
-
ëû
1 0.2
2 9.8 5.42 m / s
1 0.2
18.5 km/h = 5.14 m/s
Speed is lower than v
max
, hence it can take safe
turn.
16. Two identical antennas mounted on identical
towers are separated from each other by a
distance of 45 km. What should nearly be
the minimum height of receiving antenna
to receive the signals in line of sight ?
(Assume radius of earth is 6400 km)
(1) 19.77 m (2) 39.55 m
(3) 79.1 m (4) 158.2 m
Official Ans. by NTA (2)
Sol. D = 2 2Rh
h = =@
´
22
D 45
km 39.55m
8R 8 6400
Page 5


1
SECTION-A
1. The following logic gate is equivalent to :
A
B
Y
(1) NOR Gate (2) OR Gate
(3) AND Gate (4) NAND Gate
Official Ans. by NTA (1)
Sol. Truth table for the given logic gate :
A BY
0 01
0 11
1 01
1 10
The truth table is similar to that of a NOR gate.
2. A large block of wood of mass M = 5.99 kg is
hanging from two long massless cords. A
bullet of mass m = 10g is fired into the block
and gets embedded in it. The (block + bullet)
then swing upwards, their centre of mass rising
a vertical distance h = 9.8 cm before the
(block + bullet) pendulum comes momentarily
to rest at the end of its arc. The speed of the
bullet just before collision is : (Take g = 9.8 ms
–2
)
h
M
m
v
(1) 841.4 m/s (2) 811.4 m/s
(3) 831.4 m/s (4) 821.4 m/s
Official Ans. by NTA (3)
FINAL JEE–MAIN EXAMINATION – MARCH, 2021
(Held On Tuesday 16
th
 March, 2021)    TIME : 3 : 00 PM   to  6 : 00 PM
PHYSICS TEST PAPER WITH ANSWER & SOLUTION
Sol. From energy conservation,
éù
êú
ëû
after bullet gets embedded till the
system comes momentarily at rest
(M + m)g h = 
+
2
1
1
(M m)v
2
[v
1
 is velocity after collision]
\ v
1
 = 2gh
Applying momentum conservation, (just
before and just after collision)
mv = (M + m)v
1
v = 
-
-
+ æö
= ´´´ ´
ç÷
èø ´
2
1 3
Mm6
v 2 9.8 9.8 10
m 10 10
»831.55 m / s
3. A charge Q is moving 
uur
dI
 distance in the
magnetic field 
r
B . Find the value of work done
by 
r
B
.
(1) 1 (2) Infinite
(3) Zero (4) –1
Official Ans. by NTA (3)
Sol. Since force on a point charge by magnetic field
is always perpendicular to vF qVB éù =´
ëû
r rr
r
\ Work by magnetic force on the point charge
is zero.
4. What will be the nature of flow of water from
a circular tap, when its flow rate increased from
0.18 L/min to 0.48 L/min ? The radius of the
tap and viscosity of water are 0.5 cm and
10
–3
 Pa s, respectively.
(Density of water : 10
3
 kg/m
3
)
(1) Unsteady to steady flow
(2) Remains steady flow
(3) Remains turbulent flow
(4) Steady flow to unsteady flow
Official Ans. by NTA (4)
2
Sol. The nature of flow is determined by Reynolds
Number.
R
e
 = 
vD r
h
density of fluid ; coefficient of
v velocity of flow               viscosity
D Diameter of pipe
r® h® éù
êú ®
êú
®
ëû
From NCERT
If R
e
 < 1000          ® flow is steady
1000 < R
e
 < 2000   ® flow becomes unsteady
R
e
 > 2000 ®   flow is turbulent
--
--
´´
=´´
p´ ´´
32
3
einitial 223
0.18 10 1 10
R 10
(0.5 10 ) 60 10
        = 382.16
--
--
´´
=´´
p´ ´´
32
3
e final 223
0.48 10 1 10
R 10
(0.5 10 ) 60 10
       = 1019.09
5. A mosquito is moving with a velocity
= ++
r
2
ˆˆ ˆ
v 0.5t i 3tj 9k m/s and accelerating in
uniform conditions. What will be the direction
of mosquito after 2s ?
(1) 
-
æö
ç÷
èø
1
2
tan
3
 from x-axis
(2) 
-
æö
ç÷
èø
1
2
tan
3
 from y-axis
(3) 
-
æö
ç÷
èø
1
5
tan
2
 from y-axis
(4) 
-
æö
ç÷
èø
1
5
tan
2
 from x-axis
Official Ans. by NTA (2)
Official Ans. by ALLEN (Bonus)
Sol. Given :
= ++
r
2
ˆˆ ˆ
v 0.5t i 3t j 9k
at t = 2
ˆˆ ˆ
v 2 i 6 j 9k = ++
r
\ Angle made by direction of motion of
mosquito will be,
-1
2
cos
11
 (from x-axis) = 
-1
117
tan
2
-1
6
cos
11
 (from y-axis) = 
1
85
tan
6
-
-1
9
cos
11
 (from z-axis) = 
1
40
tan
9
-
None of the option is matching.
Hence this question should be bonus.
6. Find out the surface charge density at the
intersection of point x = 3 m plane and x-axis,
in the region of uniform line charge of 8 nC/
m lying along the z-axis in free space.
(1) 0.424 nC m
–2
(2) 47.88 C/m
(3) 0.07 nC m
–2
(4) 4.0 nC m
–2
Official Ans. by NTA (1)
Sol.
ls
=
e
0
2K
r
(x = 3m)
-
s=´
9
2
C
0.424 10
m
7. The de-Broglie wavelength associated with an
electron and a proton were calculated by
accelerating them through same potential of
100 V. What should nearly be the ratio of their
wavelengths ? (m
P
 = 1.00727 u, m
e
 = 0.00055u)
(1) 1860 : 1 (2) (1860)
2
 : 1
(3) 41.4 : 1 (4) 43 : 1
Official Ans. by NTA (4)
Sol.
l= ==
h hh
mv 2mK 2mqV
l
=
l
12
21
m
m
l
= ==
l
e P
Pe
m
1831.4 42.79
m
3
8. For the given circuit, comment on the type of
transformer used :
P
~
220V
i
1
S
L
O
A
D
i
L 0.11A
60W
V
2
(1) Auxilliary transformer
(2) Auto transformer
(3) Step-up transformer
(4) Step down transformer
Official Ans. by NTA (3)
Sol. V
S
 = 
==
P 60
545.45
i 0.11
V
P
 = 220
V
S
 > V
P
Þ Step up transformer
9. The half-life of Au
198
 is 2.7 days. The activity
of 1.50 mg of Au
198
 if its atomic weight is
198 g mol
–1
 is, (N
A
 = 6 × 10
23
/mol)
(1) 240 Ci (2) 357 Ci
(3) 535 Ci (4) 252 Ci
Official Ans. by NTA (2)
Sol. A = lN  
æö
=
ç÷
èø l
1/2
ln2
t
N = nN
A
N = 
-
æö ´
ç÷
èø
3
A
1.5 10
N
198
A = 
æö
ç÷
èø
1/2
ln2
N
t
1 Curie = 3.7 × 10
10
 Bq
A = 365 Bq
10. Calculate the value of mean free path (l) for
oxygen molecules at temperature 27°C
and pressure 1.01 × 10
5
 Pa. Assume the
molecular diameter 0.3 nm and the gas is ideal.
(k = 1.38 × 10
–23
 JK
–1
)
(1) 58 nm (2) 32 nm
(3) 86 nm (4) 102 nm
Official Ans. by NTA (4)
Sol.
l=
p
2
A
RT
2 dNP
l =102 nm
11. The refractive index of a converging lens is 1.4.
What will be the focal length of this lens if it
is placed in a medium of same refractive
index ? (Assume the radii of curvature of the
faces of lens are R
1
 and R
2
 respectively)
(1) 1 (2) Infinite
(3) 
-
12
12
RR
RR
(4) Zero
Official Ans. by NTA (2)
Sol.
éùéù m
= --
êúêú
m
ëû ëû
L
S 12
1 11
1
F RR
If m
L
 = m
S
 Þ 
= Þ =¥
1
0F
F
12. In order to determine the Young's Modulus of
a wire of radius 0.2 cm (measured using a scale
of least count = 0.001 cm) and length 1m
(measured using a scale of least count = 1 mm),
a weight of mass 1kg (measured using a scale
of least count = 1g) was hanged to get the
elongation of 0.5 cm (measured using a scale
of least count 0.001 cm). What will be the
fractional error in the value of Young's
Modulus determined by this experiment ?
(1) 0.14%
(2) 0.9%
(3) 9%
(4) 1.4%
Official Ans. by NTA (4)
Sol. Y = 
==
p l
2
Stress FL mg.L
Strain Al R .
DDD D D
= + ++
l
l
Y m LR
2.
Y m LR
éù D æö
´= + ++
ç÷ êú
èø
ëû
Y 1 1 0.001 0.001
100 100 2
Y 1000 1000 0.2 0.5
            = 
+ ++ ==
1 1 1 14
1 1.4%
10 10 5 10
4
13. A bimetallic strip consists of metals A and B.
It is mounted rigidly as shown. The metal A has
higher coefficient of expansion compared to
that of metal B. When the bimetallic strip is
placed in a cold both, it will :
A B
(1) Bend towards the right
(2) Not bend but shrink
(3) Neither bend nor shrink
(4) Bend towards the left
Official Ans. by NTA (4)
Sol. a
A
 > a
B
Length of both strips will decrease
DL
A
 > DL
B
14. A resistor develops 500 J of thermal energy in
20s when a current of 1.5 A is passed through
it. If the current is increased from 1.5 A to 3A,
what will be the energy developed in 20 s.
(1) 1500 J (2) 1000 J
(3) 500 J (4) 2000 J
Official Ans. by NTA (4)
Sol. 500 = (1.5)
2
 × R × 20
E = (3)
2
 × R × 20
E = 2000 J
15. Statement I : A cyclist is moving on an
unbanked road with a speed of
7 kmh
–1
 and takes a sharp
circular turn along a path of
radius of 2m without reducing
the speed. The static friction
coefficient is 0.2. The cyclist
will not slip and pass the curve
(g = 9.8 m/s
2
)
Statement II : If the road is banked at an angle
of 45°, cyclist can cross the
curve of 2m radius with the
speed of 18.5 kmh
–1
 without
slipping.
In the light of the above statements, choose the
correct answer from the options given below.
(1) Statement I is incorrect and statement II is
correct
(2) Statement I is correct and statement II is
incorrect
(3) Both statement I and statement II are false
(4) Both statement I and statement II are true
Official Ans. by NTA (4)
Sol. Statement I :
= m = ´´
max
v Rg (0.2) 2 9.8
v
max
 = 1.97 m/s
7 km/h = 1.944 m/s
Speed is lower than v
max
, hence it can take safe
turn.
Statement II
v
max
  = 
tan
Rg
1 tan
éù q+m
êú
-mq
ëû
= 
+ éù
´=
êú
-
ëû
1 0.2
2 9.8 5.42 m / s
1 0.2
18.5 km/h = 5.14 m/s
Speed is lower than v
max
, hence it can take safe
turn.
16. Two identical antennas mounted on identical
towers are separated from each other by a
distance of 45 km. What should nearly be
the minimum height of receiving antenna
to receive the signals in line of sight ?
(Assume radius of earth is 6400 km)
(1) 19.77 m (2) 39.55 m
(3) 79.1 m (4) 158.2 m
Official Ans. by NTA (2)
Sol. D = 2 2Rh
h = =@
´
22
D 45
km 39.55m
8R 8 6400
5
17. The magnetic field in a region is given by
æö
=
ç÷
èø
r
0
x
ˆ
BBk
a
. A square loop of side d is placed
with its edges along the x and y axes. The loop
is moved with a constant velocity =
r
0
ˆ
v vi . The
emf induced in the loop is :
d
d
y
x
z
(1) 
2
00
B vd
2a
(2) 
00
B vd
2a
(3) 
2
00
B vd
a
(4) 
2
00
B vd
2a
Official Ans. by NTA (3)
Sol. E
1
 = 
0
0
B (x d)
vd
a
+
=
0
20
B (x)
E vd
a
   
E
2
E
1
x x+d
E
net
 = E
1
 – E
2
E
net
 = 
2
00
Bvd
a
18. Amplitude of a mass-spring system, which is
executing simple harmonic motion
decreases with time. If mass = 500g, Decay
constant = 20 g/s then how much time is
required for the amplitude of the system to drop
to half of its initial value ? (ln 2 = 0.693)
(1) 34.65 s (2) 17.32 s
(3) 0.034 s (4) 15.01 s
Official Ans. by NTA (1)
Sol. A = A
0
e
–gt
 = 
-
bt
2m
0
Ae
bt
0 2m
0
A
Ae
2
-
=
=
bt
ln2
2m
t = 
´´
=
2m 2 500 0.693
ln2
b 20
t = 34.65 second.
19. Calculate the time interval between 33% decay
and 67% decay if half-life of a substance is 20
minutes.
(1) 60 minutes (2) 20 minutes
(3) 40 minutes (4) 13 minutes
Official Ans. by NTA (2)
Sol. N
1
 = 
1
t
0
Ne
-l
-l
=
1
t 1
0
N
e
N
-l
=
1
t
0.67 e
ln(0.67) = –lt
1
-l
=
2
t
20
N Ne
-l
=
2
t 2
0
N
e
N
-l
=
2
t
0.33 e
ln(0.33) = –lt
2
ln(0.67) – ln(0.33) = lt
1 
– lt
2
l(t
1
 – t
2
) = 
æö
ç÷
èø
0.67
ln
0.33
l -@
12
(t t ) ln2
-=
l
;
1 2 1/2
ln2
ttt
Half life = t
1/2
 = 20 minutes.
20. Red light differs from blue light as they
have :
(1) Different frequencies and different wavelengths
(2) Different frequencies and same wavelengths
(3) Same frequencies and same wavelengths
(4) Same frequencies and different wavelengths
Official Ans. by NTA (1)
Sol. Red light and blue light have different
wavelength and different frequency.
SECTION-B
1. The energy dissipated by a resistor is 10 mJ  in
1s when an electric current of 2 mA flows
through it. The resistance is _______ W.
(Round off to the Nearest Integer)
Official Ans. by NTA (2500)
Sol. Ans. (2500)
Q = i
2
 RT
R = 
-
-
´
=
´´
3
26
Q 10 10
it 4101
 = 2500 W