JEE Mains 17 March 2021 Question Paper Shift 2 | JEE Main & Advanced Previous Year Papers PDF Download

 Page 1


1
SECTION-A
1. A rubber ball is released from a height of 5 m
above the floor. It bounces back repeatedly,
always rising to 
81
100
of the height through
which it falls. Find the average speed of the ball.
(Take g = 10 ms
–2
)
(1) 3.0 ms
–1
(2) 3.50 ms
–1
(3) 2.0 ms
–1
(4) 2.50 ms
–1
Official Ans. by NTA (4)
Sol. (4) v
0 
= 2gh
v = e 2gh = 2gh
Þ e = 0.9
S = h + 2e
2
h + 2e
4
 h +..........
t = 
2h
g
+ 2e
2h
g
+ 2e
2
2h
g
+..........
v
av
= 
s
t
= 2.5 m/s
2. If one mole of the polyatomic gas is having two
vibrational modes and b is the ratio of molar
specific heats for polyatomic gas 
P
V
C
C
æö
b=
ç÷
èø
then
the value of b is :
(1) 1.02 (2) 1.2
(3) 1.25 (4) 1.35
Official Ans. by NTA (2)
Sol. (2) f = 4 + 3 + 3 = 10
assuming non linear
p
v
C
2 12
1
C f 10
b= = +=
= 1.2
FINAL JEE–MAIN EXAMINATION – MARCH, 2021
(Held On Wednesday 17
th
 March, 2021)    TIME : 3 : 00 PM   to  6 : 00 PM
PHYSICS TEST PAPER WITH ANSWER & SOLUTION
3. A block of mass 1 kg attached to a spring is
made to oscillate with an initial amplitude of
12 cm. After 2 minutes the amplitude decreases
to 6 cm. Determine the value of the damping
constant for this motion. (take In 2 = 0.693)
(1) 0.69 × 10
2
 kg s
–1
(2) 3.3 × 10
2
 kg s
–1
(3) 1.16 × 10
2
 kg s
–1
(4) 5.7 × 10
–3 
kg s
–1
Official Ans. by NTA (NA)
Official Ans. by ALLEN (Bonus)
Sol. A = A
0
e
–gt
ln2 = 
b
120
2m
´
0.693 2 1
b
120
´´
=
1.16 × 10
–2 
kg/sec.
4. Which one of the following will be the output
of the given circuit ?
Y
B
A
(1) NOR Gate (2) NAND Gate
(3) AND Gate (4) XOR Gate
Official Ans. by NTA (4)
Sol. (4) Conceptual
5. An object is located at 2 km beneath the surface
of the water. If the fractional compression
V
V
D
is 1.36% , the ratio of hydraulic stress to
the corresponding hydraulic strain will be
___________.
[Given : density of water is 1000 kg m
–3 
and
g = 9.8 ms
–2
.]
(1) 1.96 × 10
7
 Nm
–2
(2) 1.44 × 10
7 
Nm
–2
(3) 2.26 × 10
9 
Nm
–2
(4) 1.44 × 10
9 
Nm
–2
Official Ans. by NTA (4)
Page 2


1
SECTION-A
1. A rubber ball is released from a height of 5 m
above the floor. It bounces back repeatedly,
always rising to 
81
100
of the height through
which it falls. Find the average speed of the ball.
(Take g = 10 ms
–2
)
(1) 3.0 ms
–1
(2) 3.50 ms
–1
(3) 2.0 ms
–1
(4) 2.50 ms
–1
Official Ans. by NTA (4)
Sol. (4) v
0 
= 2gh
v = e 2gh = 2gh
Þ e = 0.9
S = h + 2e
2
h + 2e
4
 h +..........
t = 
2h
g
+ 2e
2h
g
+ 2e
2
2h
g
+..........
v
av
= 
s
t
= 2.5 m/s
2. If one mole of the polyatomic gas is having two
vibrational modes and b is the ratio of molar
specific heats for polyatomic gas 
P
V
C
C
æö
b=
ç÷
èø
then
the value of b is :
(1) 1.02 (2) 1.2
(3) 1.25 (4) 1.35
Official Ans. by NTA (2)
Sol. (2) f = 4 + 3 + 3 = 10
assuming non linear
p
v
C
2 12
1
C f 10
b= = +=
= 1.2
FINAL JEE–MAIN EXAMINATION – MARCH, 2021
(Held On Wednesday 17
th
 March, 2021)    TIME : 3 : 00 PM   to  6 : 00 PM
PHYSICS TEST PAPER WITH ANSWER & SOLUTION
3. A block of mass 1 kg attached to a spring is
made to oscillate with an initial amplitude of
12 cm. After 2 minutes the amplitude decreases
to 6 cm. Determine the value of the damping
constant for this motion. (take In 2 = 0.693)
(1) 0.69 × 10
2
 kg s
–1
(2) 3.3 × 10
2
 kg s
–1
(3) 1.16 × 10
2
 kg s
–1
(4) 5.7 × 10
–3 
kg s
–1
Official Ans. by NTA (NA)
Official Ans. by ALLEN (Bonus)
Sol. A = A
0
e
–gt
ln2 = 
b
120
2m
´
0.693 2 1
b
120
´´
=
1.16 × 10
–2 
kg/sec.
4. Which one of the following will be the output
of the given circuit ?
Y
B
A
(1) NOR Gate (2) NAND Gate
(3) AND Gate (4) XOR Gate
Official Ans. by NTA (4)
Sol. (4) Conceptual
5. An object is located at 2 km beneath the surface
of the water. If the fractional compression
V
V
D
is 1.36% , the ratio of hydraulic stress to
the corresponding hydraulic strain will be
___________.
[Given : density of water is 1000 kg m
–3 
and
g = 9.8 ms
–2
.]
(1) 1.96 × 10
7
 Nm
–2
(2) 1.44 × 10
7 
Nm
–2
(3) 2.26 × 10
9 
Nm
–2
(4) 1.44 × 10
9 
Nm
–2
Official Ans. by NTA (4)
2
Sol. (4) P = hrg
b = 
33
–2
p 2 10 10 9.8
V
1.36 10
V
´ ´´
=
D
´
= 1.44 × 10
9 
 N/m
2
6. A geostationary satellite is orbiting around an
arbitary planet 'P' at a height of 11R above the
surface of 'P' , R being the radius of 'P'. The time
period of another satellite in hours at a height
of 2R from the surface of 'P' is_________.'P'
has the time period of 24 hours.
(1) 6
2
(2) 
6
2
(3) 3 (4) 5
Official Ans. by NTA (3)
Sol. (3) T µ R
3/2
3/2
24 12R
T 3hr
T 3R
æö
= Þ=
ç÷
èø
7. A sound wave of frequency 245 Hz travels
with the speed of 300 ms
–1 
 along the positive
x-axis. Each point of the wave moves to and
fro through a total distance of 6 cm. What will
be the mathematical expression of this
travelling wave ?
(1) Y(x,t) = 0.03 [sin 5.1 x – (0.2 × 10
3
)t]
(2) Y(x,t) = 0.06 [sin 5.1 x – (1.5 × 10
3
)t]
(3) Y(x,t) = 0.06 [sin 0.8 x – (0.5 × 10
3
)t]
(4) Y(x,t) = 0.03 [sin 5.1 x – (1.5 × 10
3
)t]
Official Ans. by NTA (4)
Sol. (4) w = 2p f
= 1.5 × 10
3
A = 
6
2
= 3 cm = 0.03 m
8. Which one is the correct option for the two
different thermodynamic processes ?
(a) 
P
adiabatic
isothermal
V
(b) 
P
isothermal
adiabatic
T
(c) 
T
V
adiabatic
isothermal
(d) 
P
isothermal
adiabatic
T
(1) (c) and (a) (2) (c) and (d)
(3) (a) only (4) (b) and (c)
Official Ans. by NTA (2)
Sol. (2) Option (a) is wrong ; since in adiabatic
process V ¹ constant.
Option (b) is wrong, since in isothermal process
T = constant
Option (c) & (d) matches isothermes &
adiabatic formula :
TV
g–1
= constant & –1
T
p
g
g = constant
Page 3


1
SECTION-A
1. A rubber ball is released from a height of 5 m
above the floor. It bounces back repeatedly,
always rising to 
81
100
of the height through
which it falls. Find the average speed of the ball.
(Take g = 10 ms
–2
)
(1) 3.0 ms
–1
(2) 3.50 ms
–1
(3) 2.0 ms
–1
(4) 2.50 ms
–1
Official Ans. by NTA (4)
Sol. (4) v
0 
= 2gh
v = e 2gh = 2gh
Þ e = 0.9
S = h + 2e
2
h + 2e
4
 h +..........
t = 
2h
g
+ 2e
2h
g
+ 2e
2
2h
g
+..........
v
av
= 
s
t
= 2.5 m/s
2. If one mole of the polyatomic gas is having two
vibrational modes and b is the ratio of molar
specific heats for polyatomic gas 
P
V
C
C
æö
b=
ç÷
èø
then
the value of b is :
(1) 1.02 (2) 1.2
(3) 1.25 (4) 1.35
Official Ans. by NTA (2)
Sol. (2) f = 4 + 3 + 3 = 10
assuming non linear
p
v
C
2 12
1
C f 10
b= = +=
= 1.2
FINAL JEE–MAIN EXAMINATION – MARCH, 2021
(Held On Wednesday 17
th
 March, 2021)    TIME : 3 : 00 PM   to  6 : 00 PM
PHYSICS TEST PAPER WITH ANSWER & SOLUTION
3. A block of mass 1 kg attached to a spring is
made to oscillate with an initial amplitude of
12 cm. After 2 minutes the amplitude decreases
to 6 cm. Determine the value of the damping
constant for this motion. (take In 2 = 0.693)
(1) 0.69 × 10
2
 kg s
–1
(2) 3.3 × 10
2
 kg s
–1
(3) 1.16 × 10
2
 kg s
–1
(4) 5.7 × 10
–3 
kg s
–1
Official Ans. by NTA (NA)
Official Ans. by ALLEN (Bonus)
Sol. A = A
0
e
–gt
ln2 = 
b
120
2m
´
0.693 2 1
b
120
´´
=
1.16 × 10
–2 
kg/sec.
4. Which one of the following will be the output
of the given circuit ?
Y
B
A
(1) NOR Gate (2) NAND Gate
(3) AND Gate (4) XOR Gate
Official Ans. by NTA (4)
Sol. (4) Conceptual
5. An object is located at 2 km beneath the surface
of the water. If the fractional compression
V
V
D
is 1.36% , the ratio of hydraulic stress to
the corresponding hydraulic strain will be
___________.
[Given : density of water is 1000 kg m
–3 
and
g = 9.8 ms
–2
.]
(1) 1.96 × 10
7
 Nm
–2
(2) 1.44 × 10
7 
Nm
–2
(3) 2.26 × 10
9 
Nm
–2
(4) 1.44 × 10
9 
Nm
–2
Official Ans. by NTA (4)
2
Sol. (4) P = hrg
b = 
33
–2
p 2 10 10 9.8
V
1.36 10
V
´ ´´
=
D
´
= 1.44 × 10
9 
 N/m
2
6. A geostationary satellite is orbiting around an
arbitary planet 'P' at a height of 11R above the
surface of 'P' , R being the radius of 'P'. The time
period of another satellite in hours at a height
of 2R from the surface of 'P' is_________.'P'
has the time period of 24 hours.
(1) 6
2
(2) 
6
2
(3) 3 (4) 5
Official Ans. by NTA (3)
Sol. (3) T µ R
3/2
3/2
24 12R
T 3hr
T 3R
æö
= Þ=
ç÷
èø
7. A sound wave of frequency 245 Hz travels
with the speed of 300 ms
–1 
 along the positive
x-axis. Each point of the wave moves to and
fro through a total distance of 6 cm. What will
be the mathematical expression of this
travelling wave ?
(1) Y(x,t) = 0.03 [sin 5.1 x – (0.2 × 10
3
)t]
(2) Y(x,t) = 0.06 [sin 5.1 x – (1.5 × 10
3
)t]
(3) Y(x,t) = 0.06 [sin 0.8 x – (0.5 × 10
3
)t]
(4) Y(x,t) = 0.03 [sin 5.1 x – (1.5 × 10
3
)t]
Official Ans. by NTA (4)
Sol. (4) w = 2p f
= 1.5 × 10
3
A = 
6
2
= 3 cm = 0.03 m
8. Which one is the correct option for the two
different thermodynamic processes ?
(a) 
P
adiabatic
isothermal
V
(b) 
P
isothermal
adiabatic
T
(c) 
T
V
adiabatic
isothermal
(d) 
P
isothermal
adiabatic
T
(1) (c) and (a) (2) (c) and (d)
(3) (a) only (4) (b) and (c)
Official Ans. by NTA (2)
Sol. (2) Option (a) is wrong ; since in adiabatic
process V ¹ constant.
Option (b) is wrong, since in isothermal process
T = constant
Option (c) & (d) matches isothermes &
adiabatic formula :
TV
g–1
= constant & –1
T
p
g
g = constant
3
9. The velocity of a particle is v = v
0
 + gt + Ft
2
.
Its position is x = 0 at t = 0 ; then its
displacement after time (t = 1) is :
(1) v
0
 + g + F (2) v
0
 + 
g
2
 + 
F
3
(3) v
0
 + 
g
2
 + F (4) v
0
 + 2g + 3F
Official Ans. by NTA (2)
Sol. (2) v = v
0 
+ gt + Ft
2
ds
dt
= v
0 
+ gt + Ft
2
ds
ò
 = 
1
0
ò
(v
0 
+ gt + Ft
2
)dt
s = 
1
23
0
0
gt Ft
vt
23
éù
++
êú
ëû
s = v
0 
 + 
g
2
 + 
F
3
10. A carrier signal C(t) = 25 sin (2.512 × 10
10
 t)
is amplitude modulated by a message signal
m(t) = 5 sin (1.57 × 10
8 
t) and transmitted
through an antenna.What will be the bandwidth
of the modulated signal ?
(1) 8 GHz
(2) 2.01 GHz
(3) 1987.5 MHz
(4) 50 MHz
Official Ans. by NTA (4)
Sol. (4) Band width = 2 f
m
w
m 
= 1.57 × 10
8 
 = 2pf
m
BW = 2f
m
 = 
8
10
Hz
2
= 50 MHz
11. Two cells of emf 2E and E with internal
resistance r
1
 and r
2
 respectively are connected
in series to an external resistor R (see figure).
The value of R, at which the potential difference
across the terminals of the first cell becomes
zero is
R
2E E
(1) r
1
 + r
2
(2) 
1
2
r
–r
2
(3) 
1
2
r
r
2
+
(4) r
1
 – r
2
Official Ans. by NTA (2)
Sol. (2) 
R
i
2E
r
1
r
2
E
i = 
12
3E
R rr ++
TPD = 2E – ir
1
 = 0
2E = ir
1
2E = 
1
12
3Er
R rr
´
++
2R + 2r
1
 + 2r
2
 = 3r
1
R = 
1
2
r
–r
2
12. A hairpin like shape as shown in figure is made
by bending a long current carrying wire. What
is the magnitude of a magnetic field at point P
which lies on the centre of the semicircle ?
I
I
I P
r
• (1) ( )
0
I
2–
4r
m
p
p
(2) ( )
0
I
2
4r
m
+p
p
(3) ( )
0
I
2
2r
m
+p
p
(4) ( )
0
I
2–
2r
m
p
p
Official Ans. by NTA (2)
Page 4


1
SECTION-A
1. A rubber ball is released from a height of 5 m
above the floor. It bounces back repeatedly,
always rising to 
81
100
of the height through
which it falls. Find the average speed of the ball.
(Take g = 10 ms
–2
)
(1) 3.0 ms
–1
(2) 3.50 ms
–1
(3) 2.0 ms
–1
(4) 2.50 ms
–1
Official Ans. by NTA (4)
Sol. (4) v
0 
= 2gh
v = e 2gh = 2gh
Þ e = 0.9
S = h + 2e
2
h + 2e
4
 h +..........
t = 
2h
g
+ 2e
2h
g
+ 2e
2
2h
g
+..........
v
av
= 
s
t
= 2.5 m/s
2. If one mole of the polyatomic gas is having two
vibrational modes and b is the ratio of molar
specific heats for polyatomic gas 
P
V
C
C
æö
b=
ç÷
èø
then
the value of b is :
(1) 1.02 (2) 1.2
(3) 1.25 (4) 1.35
Official Ans. by NTA (2)
Sol. (2) f = 4 + 3 + 3 = 10
assuming non linear
p
v
C
2 12
1
C f 10
b= = +=
= 1.2
FINAL JEE–MAIN EXAMINATION – MARCH, 2021
(Held On Wednesday 17
th
 March, 2021)    TIME : 3 : 00 PM   to  6 : 00 PM
PHYSICS TEST PAPER WITH ANSWER & SOLUTION
3. A block of mass 1 kg attached to a spring is
made to oscillate with an initial amplitude of
12 cm. After 2 minutes the amplitude decreases
to 6 cm. Determine the value of the damping
constant for this motion. (take In 2 = 0.693)
(1) 0.69 × 10
2
 kg s
–1
(2) 3.3 × 10
2
 kg s
–1
(3) 1.16 × 10
2
 kg s
–1
(4) 5.7 × 10
–3 
kg s
–1
Official Ans. by NTA (NA)
Official Ans. by ALLEN (Bonus)
Sol. A = A
0
e
–gt
ln2 = 
b
120
2m
´
0.693 2 1
b
120
´´
=
1.16 × 10
–2 
kg/sec.
4. Which one of the following will be the output
of the given circuit ?
Y
B
A
(1) NOR Gate (2) NAND Gate
(3) AND Gate (4) XOR Gate
Official Ans. by NTA (4)
Sol. (4) Conceptual
5. An object is located at 2 km beneath the surface
of the water. If the fractional compression
V
V
D
is 1.36% , the ratio of hydraulic stress to
the corresponding hydraulic strain will be
___________.
[Given : density of water is 1000 kg m
–3 
and
g = 9.8 ms
–2
.]
(1) 1.96 × 10
7
 Nm
–2
(2) 1.44 × 10
7 
Nm
–2
(3) 2.26 × 10
9 
Nm
–2
(4) 1.44 × 10
9 
Nm
–2
Official Ans. by NTA (4)
2
Sol. (4) P = hrg
b = 
33
–2
p 2 10 10 9.8
V
1.36 10
V
´ ´´
=
D
´
= 1.44 × 10
9 
 N/m
2
6. A geostationary satellite is orbiting around an
arbitary planet 'P' at a height of 11R above the
surface of 'P' , R being the radius of 'P'. The time
period of another satellite in hours at a height
of 2R from the surface of 'P' is_________.'P'
has the time period of 24 hours.
(1) 6
2
(2) 
6
2
(3) 3 (4) 5
Official Ans. by NTA (3)
Sol. (3) T µ R
3/2
3/2
24 12R
T 3hr
T 3R
æö
= Þ=
ç÷
èø
7. A sound wave of frequency 245 Hz travels
with the speed of 300 ms
–1 
 along the positive
x-axis. Each point of the wave moves to and
fro through a total distance of 6 cm. What will
be the mathematical expression of this
travelling wave ?
(1) Y(x,t) = 0.03 [sin 5.1 x – (0.2 × 10
3
)t]
(2) Y(x,t) = 0.06 [sin 5.1 x – (1.5 × 10
3
)t]
(3) Y(x,t) = 0.06 [sin 0.8 x – (0.5 × 10
3
)t]
(4) Y(x,t) = 0.03 [sin 5.1 x – (1.5 × 10
3
)t]
Official Ans. by NTA (4)
Sol. (4) w = 2p f
= 1.5 × 10
3
A = 
6
2
= 3 cm = 0.03 m
8. Which one is the correct option for the two
different thermodynamic processes ?
(a) 
P
adiabatic
isothermal
V
(b) 
P
isothermal
adiabatic
T
(c) 
T
V
adiabatic
isothermal
(d) 
P
isothermal
adiabatic
T
(1) (c) and (a) (2) (c) and (d)
(3) (a) only (4) (b) and (c)
Official Ans. by NTA (2)
Sol. (2) Option (a) is wrong ; since in adiabatic
process V ¹ constant.
Option (b) is wrong, since in isothermal process
T = constant
Option (c) & (d) matches isothermes &
adiabatic formula :
TV
g–1
= constant & –1
T
p
g
g = constant
3
9. The velocity of a particle is v = v
0
 + gt + Ft
2
.
Its position is x = 0 at t = 0 ; then its
displacement after time (t = 1) is :
(1) v
0
 + g + F (2) v
0
 + 
g
2
 + 
F
3
(3) v
0
 + 
g
2
 + F (4) v
0
 + 2g + 3F
Official Ans. by NTA (2)
Sol. (2) v = v
0 
+ gt + Ft
2
ds
dt
= v
0 
+ gt + Ft
2
ds
ò
 = 
1
0
ò
(v
0 
+ gt + Ft
2
)dt
s = 
1
23
0
0
gt Ft
vt
23
éù
++
êú
ëû
s = v
0 
 + 
g
2
 + 
F
3
10. A carrier signal C(t) = 25 sin (2.512 × 10
10
 t)
is amplitude modulated by a message signal
m(t) = 5 sin (1.57 × 10
8 
t) and transmitted
through an antenna.What will be the bandwidth
of the modulated signal ?
(1) 8 GHz
(2) 2.01 GHz
(3) 1987.5 MHz
(4) 50 MHz
Official Ans. by NTA (4)
Sol. (4) Band width = 2 f
m
w
m 
= 1.57 × 10
8 
 = 2pf
m
BW = 2f
m
 = 
8
10
Hz
2
= 50 MHz
11. Two cells of emf 2E and E with internal
resistance r
1
 and r
2
 respectively are connected
in series to an external resistor R (see figure).
The value of R, at which the potential difference
across the terminals of the first cell becomes
zero is
R
2E E
(1) r
1
 + r
2
(2) 
1
2
r
–r
2
(3) 
1
2
r
r
2
+
(4) r
1
 – r
2
Official Ans. by NTA (2)
Sol. (2) 
R
i
2E
r
1
r
2
E
i = 
12
3E
R rr ++
TPD = 2E – ir
1
 = 0
2E = ir
1
2E = 
1
12
3Er
R rr
´
++
2R + 2r
1
 + 2r
2
 = 3r
1
R = 
1
2
r
–r
2
12. A hairpin like shape as shown in figure is made
by bending a long current carrying wire. What
is the magnitude of a magnetic field at point P
which lies on the centre of the semicircle ?
I
I
I P
r
• (1) ( )
0
I
2–
4r
m
p
p
(2) ( )
0
I
2
4r
m
+p
p
(3) ( )
0
I
2
2r
m
+p
p
(4) ( )
0
I
2–
2r
m
p
p
Official Ans. by NTA (2)
4
Sol. (2) B = 2 × B
st.wire 
+ B
loop
B = 2 × 
00
ii
4r 2r2
mm p æö
+
ç÷
pp
èø
B = ( )
0
i
2
4r
m
+p
p
13. The four arms of a Wheatstone bridge have
resistances as shown in the figure. A
galvanometer of 15 W resistance is connected
across BD. Calculate the current through the
galvanometer when a potential difference of
10V is maintained across AC.
D
C
10V
5W
10W
100W
60W
G
A
B
(1) 2.44 mA (2) 2.44 mA
(3) 4.87 mA (4) 4.87 mA
Official Ans. by NTA (3)
Sol. (3) 
D
C
10V
5W
10W
100W
60W
G
A
10 0
x
B
y
x–10 x–y x–0
100 15 10
++
= 0
53x – 20y = 30 .....(1)
y–10 y–x y–0
60 15 5
++
= 0
17 y – 4x = 10 ......(2)
on solving (1) & (2)
x = 0.865
y = 0.792
DV = 0.073 R = 15W
i = 4.87 mA
14. Two particles A and B of equal masses are
suspended from two massless springs of spring
constants K
1 
and K
2
 respectively.If the
maximum velocities during oscillations are
equal, the ratio of the amplitude of A and B is
(1) 
2
1
K
K
(2) 
1
2
K
K
(3) 
1
2
K
K
(4) 
2
1
K
K
Official Ans. by NTA (4)
Sol. (4) A
1
w
1
 = A
2
w
2
1
1
k
A
m
= 
2
2
k
A
m
1
2
A
A
= 
2
1
k
k
15. Match List-I with List-II
List-I List-II
(a) Phase difference (i) 
2
p
; current leads
between current and voltage
voltage in a purely
resistive AC circuit
(b) Phase difference (ii) zero
between current and
voltage in a pure
inductive AC circuit
(c) Phase difference (iii) 
2
p
; current lags
between current and                voltage
voltage in a pure
capacitive AC circuit
(d) Phase difference (iv) tan
–1
CL
X –X
R
æö
ç÷
èø
between current and
voltage in an LCR
series circuit
Choose the most appropriate answer from the
options given below :
(1) (a)–(i),(b)–(iii),(c)–(iv),(d)–(ii)
(2) (a)–(ii),(b)–(iv),(c)–(iii),(d)–(i)
(3) (a)–(ii),(b)–(iii),(c)–(iv),(d)–(i)
(4) (a)–(ii),(b)–(iii),(c)–(i),(d)–(iv)
Official Ans. by NTA (4)
Page 5


1
SECTION-A
1. A rubber ball is released from a height of 5 m
above the floor. It bounces back repeatedly,
always rising to 
81
100
of the height through
which it falls. Find the average speed of the ball.
(Take g = 10 ms
–2
)
(1) 3.0 ms
–1
(2) 3.50 ms
–1
(3) 2.0 ms
–1
(4) 2.50 ms
–1
Official Ans. by NTA (4)
Sol. (4) v
0 
= 2gh
v = e 2gh = 2gh
Þ e = 0.9
S = h + 2e
2
h + 2e
4
 h +..........
t = 
2h
g
+ 2e
2h
g
+ 2e
2
2h
g
+..........
v
av
= 
s
t
= 2.5 m/s
2. If one mole of the polyatomic gas is having two
vibrational modes and b is the ratio of molar
specific heats for polyatomic gas 
P
V
C
C
æö
b=
ç÷
èø
then
the value of b is :
(1) 1.02 (2) 1.2
(3) 1.25 (4) 1.35
Official Ans. by NTA (2)
Sol. (2) f = 4 + 3 + 3 = 10
assuming non linear
p
v
C
2 12
1
C f 10
b= = +=
= 1.2
FINAL JEE–MAIN EXAMINATION – MARCH, 2021
(Held On Wednesday 17
th
 March, 2021)    TIME : 3 : 00 PM   to  6 : 00 PM
PHYSICS TEST PAPER WITH ANSWER & SOLUTION
3. A block of mass 1 kg attached to a spring is
made to oscillate with an initial amplitude of
12 cm. After 2 minutes the amplitude decreases
to 6 cm. Determine the value of the damping
constant for this motion. (take In 2 = 0.693)
(1) 0.69 × 10
2
 kg s
–1
(2) 3.3 × 10
2
 kg s
–1
(3) 1.16 × 10
2
 kg s
–1
(4) 5.7 × 10
–3 
kg s
–1
Official Ans. by NTA (NA)
Official Ans. by ALLEN (Bonus)
Sol. A = A
0
e
–gt
ln2 = 
b
120
2m
´
0.693 2 1
b
120
´´
=
1.16 × 10
–2 
kg/sec.
4. Which one of the following will be the output
of the given circuit ?
Y
B
A
(1) NOR Gate (2) NAND Gate
(3) AND Gate (4) XOR Gate
Official Ans. by NTA (4)
Sol. (4) Conceptual
5. An object is located at 2 km beneath the surface
of the water. If the fractional compression
V
V
D
is 1.36% , the ratio of hydraulic stress to
the corresponding hydraulic strain will be
___________.
[Given : density of water is 1000 kg m
–3 
and
g = 9.8 ms
–2
.]
(1) 1.96 × 10
7
 Nm
–2
(2) 1.44 × 10
7 
Nm
–2
(3) 2.26 × 10
9 
Nm
–2
(4) 1.44 × 10
9 
Nm
–2
Official Ans. by NTA (4)
2
Sol. (4) P = hrg
b = 
33
–2
p 2 10 10 9.8
V
1.36 10
V
´ ´´
=
D
´
= 1.44 × 10
9 
 N/m
2
6. A geostationary satellite is orbiting around an
arbitary planet 'P' at a height of 11R above the
surface of 'P' , R being the radius of 'P'. The time
period of another satellite in hours at a height
of 2R from the surface of 'P' is_________.'P'
has the time period of 24 hours.
(1) 6
2
(2) 
6
2
(3) 3 (4) 5
Official Ans. by NTA (3)
Sol. (3) T µ R
3/2
3/2
24 12R
T 3hr
T 3R
æö
= Þ=
ç÷
èø
7. A sound wave of frequency 245 Hz travels
with the speed of 300 ms
–1 
 along the positive
x-axis. Each point of the wave moves to and
fro through a total distance of 6 cm. What will
be the mathematical expression of this
travelling wave ?
(1) Y(x,t) = 0.03 [sin 5.1 x – (0.2 × 10
3
)t]
(2) Y(x,t) = 0.06 [sin 5.1 x – (1.5 × 10
3
)t]
(3) Y(x,t) = 0.06 [sin 0.8 x – (0.5 × 10
3
)t]
(4) Y(x,t) = 0.03 [sin 5.1 x – (1.5 × 10
3
)t]
Official Ans. by NTA (4)
Sol. (4) w = 2p f
= 1.5 × 10
3
A = 
6
2
= 3 cm = 0.03 m
8. Which one is the correct option for the two
different thermodynamic processes ?
(a) 
P
adiabatic
isothermal
V
(b) 
P
isothermal
adiabatic
T
(c) 
T
V
adiabatic
isothermal
(d) 
P
isothermal
adiabatic
T
(1) (c) and (a) (2) (c) and (d)
(3) (a) only (4) (b) and (c)
Official Ans. by NTA (2)
Sol. (2) Option (a) is wrong ; since in adiabatic
process V ¹ constant.
Option (b) is wrong, since in isothermal process
T = constant
Option (c) & (d) matches isothermes &
adiabatic formula :
TV
g–1
= constant & –1
T
p
g
g = constant
3
9. The velocity of a particle is v = v
0
 + gt + Ft
2
.
Its position is x = 0 at t = 0 ; then its
displacement after time (t = 1) is :
(1) v
0
 + g + F (2) v
0
 + 
g
2
 + 
F
3
(3) v
0
 + 
g
2
 + F (4) v
0
 + 2g + 3F
Official Ans. by NTA (2)
Sol. (2) v = v
0 
+ gt + Ft
2
ds
dt
= v
0 
+ gt + Ft
2
ds
ò
 = 
1
0
ò
(v
0 
+ gt + Ft
2
)dt
s = 
1
23
0
0
gt Ft
vt
23
éù
++
êú
ëû
s = v
0 
 + 
g
2
 + 
F
3
10. A carrier signal C(t) = 25 sin (2.512 × 10
10
 t)
is amplitude modulated by a message signal
m(t) = 5 sin (1.57 × 10
8 
t) and transmitted
through an antenna.What will be the bandwidth
of the modulated signal ?
(1) 8 GHz
(2) 2.01 GHz
(3) 1987.5 MHz
(4) 50 MHz
Official Ans. by NTA (4)
Sol. (4) Band width = 2 f
m
w
m 
= 1.57 × 10
8 
 = 2pf
m
BW = 2f
m
 = 
8
10
Hz
2
= 50 MHz
11. Two cells of emf 2E and E with internal
resistance r
1
 and r
2
 respectively are connected
in series to an external resistor R (see figure).
The value of R, at which the potential difference
across the terminals of the first cell becomes
zero is
R
2E E
(1) r
1
 + r
2
(2) 
1
2
r
–r
2
(3) 
1
2
r
r
2
+
(4) r
1
 – r
2
Official Ans. by NTA (2)
Sol. (2) 
R
i
2E
r
1
r
2
E
i = 
12
3E
R rr ++
TPD = 2E – ir
1
 = 0
2E = ir
1
2E = 
1
12
3Er
R rr
´
++
2R + 2r
1
 + 2r
2
 = 3r
1
R = 
1
2
r
–r
2
12. A hairpin like shape as shown in figure is made
by bending a long current carrying wire. What
is the magnitude of a magnetic field at point P
which lies on the centre of the semicircle ?
I
I
I P
r
• (1) ( )
0
I
2–
4r
m
p
p
(2) ( )
0
I
2
4r
m
+p
p
(3) ( )
0
I
2
2r
m
+p
p
(4) ( )
0
I
2–
2r
m
p
p
Official Ans. by NTA (2)
4
Sol. (2) B = 2 × B
st.wire 
+ B
loop
B = 2 × 
00
ii
4r 2r2
mm p æö
+
ç÷
pp
èø
B = ( )
0
i
2
4r
m
+p
p
13. The four arms of a Wheatstone bridge have
resistances as shown in the figure. A
galvanometer of 15 W resistance is connected
across BD. Calculate the current through the
galvanometer when a potential difference of
10V is maintained across AC.
D
C
10V
5W
10W
100W
60W
G
A
B
(1) 2.44 mA (2) 2.44 mA
(3) 4.87 mA (4) 4.87 mA
Official Ans. by NTA (3)
Sol. (3) 
D
C
10V
5W
10W
100W
60W
G
A
10 0
x
B
y
x–10 x–y x–0
100 15 10
++
= 0
53x – 20y = 30 .....(1)
y–10 y–x y–0
60 15 5
++
= 0
17 y – 4x = 10 ......(2)
on solving (1) & (2)
x = 0.865
y = 0.792
DV = 0.073 R = 15W
i = 4.87 mA
14. Two particles A and B of equal masses are
suspended from two massless springs of spring
constants K
1 
and K
2
 respectively.If the
maximum velocities during oscillations are
equal, the ratio of the amplitude of A and B is
(1) 
2
1
K
K
(2) 
1
2
K
K
(3) 
1
2
K
K
(4) 
2
1
K
K
Official Ans. by NTA (4)
Sol. (4) A
1
w
1
 = A
2
w
2
1
1
k
A
m
= 
2
2
k
A
m
1
2
A
A
= 
2
1
k
k
15. Match List-I with List-II
List-I List-II
(a) Phase difference (i) 
2
p
; current leads
between current and voltage
voltage in a purely
resistive AC circuit
(b) Phase difference (ii) zero
between current and
voltage in a pure
inductive AC circuit
(c) Phase difference (iii) 
2
p
; current lags
between current and                voltage
voltage in a pure
capacitive AC circuit
(d) Phase difference (iv) tan
–1
CL
X –X
R
æö
ç÷
èø
between current and
voltage in an LCR
series circuit
Choose the most appropriate answer from the
options given below :
(1) (a)–(i),(b)–(iii),(c)–(iv),(d)–(ii)
(2) (a)–(ii),(b)–(iv),(c)–(iii),(d)–(i)
(3) (a)–(ii),(b)–(iii),(c)–(iv),(d)–(i)
(4) (a)–(ii),(b)–(iii),(c)–(i),(d)–(iv)
Official Ans. by NTA (4)
5
Sol. (4) (a) 
I
V = V
R
(b) 
I
V
L
(c) 
I
V
C
(d) tan f = 
L C LC
R
V –V X –X
VR
=
16. Two identical blocks A and B each of mass m
resting on the smooth horizontal floor are
connected by a light spring of natural length L
and spring constant K. A third block C of mass
m moving with a speed v along the line joining
A and B collides with A.The maximum
compression in the spring is
m m m
C A B
(1) 
M
v
2K
(2) 
mv
2K
(3) 
mv
K
(4) 
m
2K
Official Ans. by NTA (1)
Sol. (1) C comes to rest
V
cm
 of A & B = 
v
2
Þ
1
2
is 
2
ret
1
v
2
=
kx
2
x = 
2
v
k
m´
= 
m
v
2k
17. The atomic hydrogen emits a line spectrum
consisting of various series.Which series of
hydrogen atomic spectra is lying in the visible
region ?
(1) Brackett series (2) Paschen series
(3) Lyman series (4) Balmer series
Official Ans. by NTA (4)
Sol. (4) Conceptual
18. Two identical photocathodes receive the light
of frequencies f
1
 and f
2
 respectively. If the
velocities of the photo-electrons coming out are
v
1
 and v
2
 respectively, then
(1) 
2
1
v – 
2
2
v = 
2h
m
[f
1 
– f
2
]
(2) 
2
1
v + 
2
2
v = 
2h
m
[f
1 
+ f
2
]
(3) v
1 
+ v
2 
=
 ( )
1
2
12
2h
ff
m
éù
+
êú
ëû
(4) v
1 
– v
2 
= ( )
1/2
12
2h
f –f
m
éù
êú
ëû
Official Ans. by NTA (1)
Sol. (1) 
2
1
1
mv
2
= hf
1
 – f
2
2
1
mv
2
= hf
2
–f
( )
22
1 2 12
2h
v –v f –f
m
=
19. What happens to the inductive reactance and
the current in a purely inductive circuit if the
frequency is halved ?
(1) Both, inductive reactance and current will
be halved.
(2) Inductive reactance will be halved and
current will be doubled.
(3) Inductive reactance will be doubled and
current will be halved.
(4) Both, inducting reactance and current will
be doubled.
Official Ans. by NTA (2)
Sol. (2) X
L 
= wL
i = 
0
v
L w