JEE Mains 18 March 2021 Question Paper Shift 1 | JEE Main & Advanced Previous Year Papers PDF Download

 Page 1


1
SECTION-A
1. An oil drop of radius 2 mm with a density 3g
cm
–3
 is held stationary under a constant electric
field 3.55 × 10
5
 V m
–1
 in the Millikan's oil drop
experiment. What is the number of excess
electrons that the oil drop will possess ?
(consider g = 9.81 m/s
2
)
(1) 48.8 × 10
11
(2) 1.73 × 10
10
(3) 17.3 × 10
10
(4) 1.73 × 10
12
Official Ans. by NTA (2)
Sol. qE = Mg
neE = 
æö
r p´
ç÷
èø
3
4
rg
3
n × 1.6 × 10
–19
 × 3.55 × 10
5
= 3 × 10
3
 × 
4
3
 × p × (2 × 10
–3
)
3 
× 9.81
n = 173 × 10
(3 – 9 – 5 + 19)
n = 1.73 × 10
10
2. Match List–I with List–II.
List–I
(a) 10 km height over earth's surface
(b) 70 km height over earth's surface
(c) 180 km height over earth's surface
(d) 270 km height over earth's surface
List–II
(i) Thermosphere
(ii) Mesosphere
(iii) Stratosphere
(iv) Troposphere
(1) (a)–(iv), (b)–(iii), (c)–(ii), (d)–(i)
(2) (a)–(i), (b)–(iv), (c)–(iii), (d)–(ii)
(3) (a)–(iii), (b)–(ii), (c)–(i), (d)–(iv)
(4) (a)–(ii), (b)–(i), (c)–(iv), (d)–(iii)
Official Ans. by NTA (1)
FINAL JEE–MAIN EXAMINATION – MARCH, 2021
(Held On Thursday 18
th
 March, 2021)    TIME : 9 : 00 AM   to  12 : 00 NOON
PHYSICS TEST PAPER WITH ANSWER & SOLUTION
Sol. Order of atmosphere stratification from bottom
Troposphere, stralospherre, Mesosphre,
Thermosphere
(a) ® (iv)
(b) ® (iii)
(c) ® (ii)
(d) ® (i)
3. Imagine that the electron in a hydrogen atom
is replaced by a muon (µ). The mass of muon
particle is 207 times that of an electron and
charge is equal to the charge of an electron. The
ionization potential of this hydrogen atom will
be :-
(1) 13.6 eV (2) 2815.2 eV
(3) 331.2 eV (4) 27.2 eV
Official Ans. by NTA (2)
Sol.
µ
1
E
r
µ
1
r
m
E µ m
Ionization potential = 13.6 × 
( )
( )
m
e
Mass eV
Mass
= 13.6 × 207 eV = 2815.2 eV
4. A plane electromagnetic wave of frequency
100 MHz is travelling in vacuum along the x-
direction. At a particular point in space and
time, 
–8
ˆ
B 2.0 10 kT =´
r
. (where, ˆ
k
 is unit
vector along z-direction) What is 
E
r
 at this
point ?
(1) 0.6
ˆ
j
 V/m (2) 6.0
ˆ
k V/m
(3) 6.0
ˆ
j V/m (4) 0.6
ˆ
k V/m
Official Ans. by NTA (3)
Page 2


1
SECTION-A
1. An oil drop of radius 2 mm with a density 3g
cm
–3
 is held stationary under a constant electric
field 3.55 × 10
5
 V m
–1
 in the Millikan's oil drop
experiment. What is the number of excess
electrons that the oil drop will possess ?
(consider g = 9.81 m/s
2
)
(1) 48.8 × 10
11
(2) 1.73 × 10
10
(3) 17.3 × 10
10
(4) 1.73 × 10
12
Official Ans. by NTA (2)
Sol. qE = Mg
neE = 
æö
r p´
ç÷
èø
3
4
rg
3
n × 1.6 × 10
–19
 × 3.55 × 10
5
= 3 × 10
3
 × 
4
3
 × p × (2 × 10
–3
)
3 
× 9.81
n = 173 × 10
(3 – 9 – 5 + 19)
n = 1.73 × 10
10
2. Match List–I with List–II.
List–I
(a) 10 km height over earth's surface
(b) 70 km height over earth's surface
(c) 180 km height over earth's surface
(d) 270 km height over earth's surface
List–II
(i) Thermosphere
(ii) Mesosphere
(iii) Stratosphere
(iv) Troposphere
(1) (a)–(iv), (b)–(iii), (c)–(ii), (d)–(i)
(2) (a)–(i), (b)–(iv), (c)–(iii), (d)–(ii)
(3) (a)–(iii), (b)–(ii), (c)–(i), (d)–(iv)
(4) (a)–(ii), (b)–(i), (c)–(iv), (d)–(iii)
Official Ans. by NTA (1)
FINAL JEE–MAIN EXAMINATION – MARCH, 2021
(Held On Thursday 18
th
 March, 2021)    TIME : 9 : 00 AM   to  12 : 00 NOON
PHYSICS TEST PAPER WITH ANSWER & SOLUTION
Sol. Order of atmosphere stratification from bottom
Troposphere, stralospherre, Mesosphre,
Thermosphere
(a) ® (iv)
(b) ® (iii)
(c) ® (ii)
(d) ® (i)
3. Imagine that the electron in a hydrogen atom
is replaced by a muon (µ). The mass of muon
particle is 207 times that of an electron and
charge is equal to the charge of an electron. The
ionization potential of this hydrogen atom will
be :-
(1) 13.6 eV (2) 2815.2 eV
(3) 331.2 eV (4) 27.2 eV
Official Ans. by NTA (2)
Sol.
µ
1
E
r
µ
1
r
m
E µ m
Ionization potential = 13.6 × 
( )
( )
m
e
Mass eV
Mass
= 13.6 × 207 eV = 2815.2 eV
4. A plane electromagnetic wave of frequency
100 MHz is travelling in vacuum along the x-
direction. At a particular point in space and
time, 
–8
ˆ
B 2.0 10 kT =´
r
. (where, ˆ
k
 is unit
vector along z-direction) What is 
E
r
 at this
point ?
(1) 0.6
ˆ
j
 V/m (2) 6.0
ˆ
k V/m
(3) 6.0
ˆ
j V/m (4) 0.6
ˆ
k V/m
Official Ans. by NTA (3)
2
Sol. E = BC = 6
(Dir. of wave) || ( )
´ EB
rr
=´
ˆˆ ˆ
i jk
=
ˆ
E 6j V/m
r
5. A thin circular ring of mass M and radius r is
rotating about its axis with an angular speed w.
Two particles having mass m each are now
attached at diametrically opposite points. The
angular speed of the ring will become :
(1) 
w
+
M
Mm
(2) 
+
w
M 2m
M
(3) 
w
+
M
M 2m
(4) 
-
w
+
M 2m
M 2m
Official Ans. by NTA (3)
Sol. Using conservation of angular momentum
(Mr
2
)w = (Mr
2
 + 2mr
2
)w'
w
w=
+
M
'
M 2m
6. Four identical long solenoids A, B, C and D are
connected to each other as shown in the figure.
If the magnetic field at the center of A is 3T,
the field at the center of C would be : (Assume
that the magnetic field is confined with in the
volume of respective solenoid).
B
C
D
A
i
(1) 12T (2) 6T
(3) 9T (4) 1T
Official Ans. by NTA (4)
Sol.
B
C
D
A
i
i/3
i/3
i/3
f µ i
Þ B µ i
so, field at centre of C = 
=
3
1T
3
7. The time period of a simple pendulum is given
by 
=p T2
g
l
. The measured value of the
length of pendulum is 10 cm known to a 1mm
accuracy. The time for 200 oscillations of the
pendulum is found to be 100 second using a
clock of 1s resolution. The percentage accuracy
in the determination of 'g' using this pendulum
is 'x'. The value of 'x' to the nearest integer is:-
(1) 2% (2) 3%
(3) 5% (4) 4%
Official Ans. by NTA (2)
Sol. g = 
p
2
2
4
T
l
D DD
=+
gT
2
gT
l
l
 = 
æö
ç÷
+
ç÷
ç÷
èø
1
0.1
200
2
10 0.5
D
=+
g 11
g 100 50
D
´=
g
100 3%
g
8. A constant power delivering machine has
towed a box, which was initially at rest, along
a horizontal straight line. The distance moved
by the box in time 't' is proportional to :-
(1) t
2/3
(2) t
3/2
(3) t (4) t
1/2
Official Ans. by NTA (2)
Page 3


1
SECTION-A
1. An oil drop of radius 2 mm with a density 3g
cm
–3
 is held stationary under a constant electric
field 3.55 × 10
5
 V m
–1
 in the Millikan's oil drop
experiment. What is the number of excess
electrons that the oil drop will possess ?
(consider g = 9.81 m/s
2
)
(1) 48.8 × 10
11
(2) 1.73 × 10
10
(3) 17.3 × 10
10
(4) 1.73 × 10
12
Official Ans. by NTA (2)
Sol. qE = Mg
neE = 
æö
r p´
ç÷
èø
3
4
rg
3
n × 1.6 × 10
–19
 × 3.55 × 10
5
= 3 × 10
3
 × 
4
3
 × p × (2 × 10
–3
)
3 
× 9.81
n = 173 × 10
(3 – 9 – 5 + 19)
n = 1.73 × 10
10
2. Match List–I with List–II.
List–I
(a) 10 km height over earth's surface
(b) 70 km height over earth's surface
(c) 180 km height over earth's surface
(d) 270 km height over earth's surface
List–II
(i) Thermosphere
(ii) Mesosphere
(iii) Stratosphere
(iv) Troposphere
(1) (a)–(iv), (b)–(iii), (c)–(ii), (d)–(i)
(2) (a)–(i), (b)–(iv), (c)–(iii), (d)–(ii)
(3) (a)–(iii), (b)–(ii), (c)–(i), (d)–(iv)
(4) (a)–(ii), (b)–(i), (c)–(iv), (d)–(iii)
Official Ans. by NTA (1)
FINAL JEE–MAIN EXAMINATION – MARCH, 2021
(Held On Thursday 18
th
 March, 2021)    TIME : 9 : 00 AM   to  12 : 00 NOON
PHYSICS TEST PAPER WITH ANSWER & SOLUTION
Sol. Order of atmosphere stratification from bottom
Troposphere, stralospherre, Mesosphre,
Thermosphere
(a) ® (iv)
(b) ® (iii)
(c) ® (ii)
(d) ® (i)
3. Imagine that the electron in a hydrogen atom
is replaced by a muon (µ). The mass of muon
particle is 207 times that of an electron and
charge is equal to the charge of an electron. The
ionization potential of this hydrogen atom will
be :-
(1) 13.6 eV (2) 2815.2 eV
(3) 331.2 eV (4) 27.2 eV
Official Ans. by NTA (2)
Sol.
µ
1
E
r
µ
1
r
m
E µ m
Ionization potential = 13.6 × 
( )
( )
m
e
Mass eV
Mass
= 13.6 × 207 eV = 2815.2 eV
4. A plane electromagnetic wave of frequency
100 MHz is travelling in vacuum along the x-
direction. At a particular point in space and
time, 
–8
ˆ
B 2.0 10 kT =´
r
. (where, ˆ
k
 is unit
vector along z-direction) What is 
E
r
 at this
point ?
(1) 0.6
ˆ
j
 V/m (2) 6.0
ˆ
k V/m
(3) 6.0
ˆ
j V/m (4) 0.6
ˆ
k V/m
Official Ans. by NTA (3)
2
Sol. E = BC = 6
(Dir. of wave) || ( )
´ EB
rr
=´
ˆˆ ˆ
i jk
=
ˆ
E 6j V/m
r
5. A thin circular ring of mass M and radius r is
rotating about its axis with an angular speed w.
Two particles having mass m each are now
attached at diametrically opposite points. The
angular speed of the ring will become :
(1) 
w
+
M
Mm
(2) 
+
w
M 2m
M
(3) 
w
+
M
M 2m
(4) 
-
w
+
M 2m
M 2m
Official Ans. by NTA (3)
Sol. Using conservation of angular momentum
(Mr
2
)w = (Mr
2
 + 2mr
2
)w'
w
w=
+
M
'
M 2m
6. Four identical long solenoids A, B, C and D are
connected to each other as shown in the figure.
If the magnetic field at the center of A is 3T,
the field at the center of C would be : (Assume
that the magnetic field is confined with in the
volume of respective solenoid).
B
C
D
A
i
(1) 12T (2) 6T
(3) 9T (4) 1T
Official Ans. by NTA (4)
Sol.
B
C
D
A
i
i/3
i/3
i/3
f µ i
Þ B µ i
so, field at centre of C = 
=
3
1T
3
7. The time period of a simple pendulum is given
by 
=p T2
g
l
. The measured value of the
length of pendulum is 10 cm known to a 1mm
accuracy. The time for 200 oscillations of the
pendulum is found to be 100 second using a
clock of 1s resolution. The percentage accuracy
in the determination of 'g' using this pendulum
is 'x'. The value of 'x' to the nearest integer is:-
(1) 2% (2) 3%
(3) 5% (4) 4%
Official Ans. by NTA (2)
Sol. g = 
p
2
2
4
T
l
D DD
=+
gT
2
gT
l
l
 = 
æö
ç÷
+
ç÷
ç÷
èø
1
0.1
200
2
10 0.5
D
=+
g 11
g 100 50
D
´=
g
100 3%
g
8. A constant power delivering machine has
towed a box, which was initially at rest, along
a horizontal straight line. The distance moved
by the box in time 't' is proportional to :-
(1) t
2/3
(2) t
3/2
(3) t (4) t
1/2
Official Ans. by NTA (2)
3
Sol. P = C
FV = C
M
=
dV
VC
dt
µ
2
V
t
2
V µ t
1/2
µ
1/2
dx
t
dt
x µ t
3/2
9. What will be the average value of energy along
one degree of freedom for an ideal gas in
thermal equilibrium at a temperature T ? (k
B 
is
Boltzmann constant)
(1) 
B
1
kT
2
(2) 
B
2
kT
3
(3) 
B
3
kT
2
(4) k
B
T
Official Ans. by NTA (1)
Sol. Energy associated with each degree of
freedom per molecule = 
B
1
kT
2
.
10. A radioactive sample disintegrates via two
independent decay processes having half lives
() 1
1/2
T and 
() 2
1/2
T respectively. The effective half-
life 
1/2
T of the nuclei is :
(1) None of the above (2) 
() ()
=+
12
1/2 1/2 1/2
T TT
(3) 
( ) ()
() ()
=
+
12
1/2 1/2
1/2 12
1/2 1/2
TT
T
TT
(4) 
() ()
() ()
+
=
-
12
1/2 1/2
1/2 12
1/2 1/2
TT
T
TT
Official Ans. by NTA (3)
Sol. l
eq
 = l
1
 + l
2
() ()
=+
12
1/2 1/2 1/2
1 11
T
TT
( ) ()
() ()
=
+
12
1/2 1/2
1/2 12
1/2 1/2
TT
T
TT
11. The P-V diagram of a diatomic ideal gas system
going under cyclic process as shown in figure.
The work done during an adiabatic process CD
is (use g = 1.4) :
A
B
D
C
1 3 4
200N/m
2
100N/m
2
P
V(m)
3
(1) –500 J (2) –400 J
(3) 400 J (4) 200 J
Official Ans. by NTA (1)
Sol. Adiabatic process is from C to D
-
=
-g
2 2 11
P V PV
WD
1
= 
-
-g
D D CC
P V PV
1
= 
-
-
200(3) (100)(4)
1 1.4
= –500 J Ans. (1)
12. In Young's double slit arrangement, slits are
separated by a gap of 0.5 mm, and the screen
is placed at a distance of 0.5 m from them. The
distance between the first and the third bright
fringe formed when the slits are illuminated by
a monochromatic light of 5890 Å is :-
(1) 1178 × 10
–9
 m (2) 1178 × 10
–6
 m
(3) 1178 × 10
–12
 m (4) 5890 × 10
–7
 m
Official Ans. by NTA (2)
Sol.
-
-
l ´´
b==
´
10
3
D 5890 10 0.5
d 0.5 10
= 589 × 10
–6
 m
Distance between first and third bright fringe
is 2b = 2×589 × 10
–6 
m
= 1178 × 10
–6
 m Ans. (2)
Page 4


1
SECTION-A
1. An oil drop of radius 2 mm with a density 3g
cm
–3
 is held stationary under a constant electric
field 3.55 × 10
5
 V m
–1
 in the Millikan's oil drop
experiment. What is the number of excess
electrons that the oil drop will possess ?
(consider g = 9.81 m/s
2
)
(1) 48.8 × 10
11
(2) 1.73 × 10
10
(3) 17.3 × 10
10
(4) 1.73 × 10
12
Official Ans. by NTA (2)
Sol. qE = Mg
neE = 
æö
r p´
ç÷
èø
3
4
rg
3
n × 1.6 × 10
–19
 × 3.55 × 10
5
= 3 × 10
3
 × 
4
3
 × p × (2 × 10
–3
)
3 
× 9.81
n = 173 × 10
(3 – 9 – 5 + 19)
n = 1.73 × 10
10
2. Match List–I with List–II.
List–I
(a) 10 km height over earth's surface
(b) 70 km height over earth's surface
(c) 180 km height over earth's surface
(d) 270 km height over earth's surface
List–II
(i) Thermosphere
(ii) Mesosphere
(iii) Stratosphere
(iv) Troposphere
(1) (a)–(iv), (b)–(iii), (c)–(ii), (d)–(i)
(2) (a)–(i), (b)–(iv), (c)–(iii), (d)–(ii)
(3) (a)–(iii), (b)–(ii), (c)–(i), (d)–(iv)
(4) (a)–(ii), (b)–(i), (c)–(iv), (d)–(iii)
Official Ans. by NTA (1)
FINAL JEE–MAIN EXAMINATION – MARCH, 2021
(Held On Thursday 18
th
 March, 2021)    TIME : 9 : 00 AM   to  12 : 00 NOON
PHYSICS TEST PAPER WITH ANSWER & SOLUTION
Sol. Order of atmosphere stratification from bottom
Troposphere, stralospherre, Mesosphre,
Thermosphere
(a) ® (iv)
(b) ® (iii)
(c) ® (ii)
(d) ® (i)
3. Imagine that the electron in a hydrogen atom
is replaced by a muon (µ). The mass of muon
particle is 207 times that of an electron and
charge is equal to the charge of an electron. The
ionization potential of this hydrogen atom will
be :-
(1) 13.6 eV (2) 2815.2 eV
(3) 331.2 eV (4) 27.2 eV
Official Ans. by NTA (2)
Sol.
µ
1
E
r
µ
1
r
m
E µ m
Ionization potential = 13.6 × 
( )
( )
m
e
Mass eV
Mass
= 13.6 × 207 eV = 2815.2 eV
4. A plane electromagnetic wave of frequency
100 MHz is travelling in vacuum along the x-
direction. At a particular point in space and
time, 
–8
ˆ
B 2.0 10 kT =´
r
. (where, ˆ
k
 is unit
vector along z-direction) What is 
E
r
 at this
point ?
(1) 0.6
ˆ
j
 V/m (2) 6.0
ˆ
k V/m
(3) 6.0
ˆ
j V/m (4) 0.6
ˆ
k V/m
Official Ans. by NTA (3)
2
Sol. E = BC = 6
(Dir. of wave) || ( )
´ EB
rr
=´
ˆˆ ˆ
i jk
=
ˆ
E 6j V/m
r
5. A thin circular ring of mass M and radius r is
rotating about its axis with an angular speed w.
Two particles having mass m each are now
attached at diametrically opposite points. The
angular speed of the ring will become :
(1) 
w
+
M
Mm
(2) 
+
w
M 2m
M
(3) 
w
+
M
M 2m
(4) 
-
w
+
M 2m
M 2m
Official Ans. by NTA (3)
Sol. Using conservation of angular momentum
(Mr
2
)w = (Mr
2
 + 2mr
2
)w'
w
w=
+
M
'
M 2m
6. Four identical long solenoids A, B, C and D are
connected to each other as shown in the figure.
If the magnetic field at the center of A is 3T,
the field at the center of C would be : (Assume
that the magnetic field is confined with in the
volume of respective solenoid).
B
C
D
A
i
(1) 12T (2) 6T
(3) 9T (4) 1T
Official Ans. by NTA (4)
Sol.
B
C
D
A
i
i/3
i/3
i/3
f µ i
Þ B µ i
so, field at centre of C = 
=
3
1T
3
7. The time period of a simple pendulum is given
by 
=p T2
g
l
. The measured value of the
length of pendulum is 10 cm known to a 1mm
accuracy. The time for 200 oscillations of the
pendulum is found to be 100 second using a
clock of 1s resolution. The percentage accuracy
in the determination of 'g' using this pendulum
is 'x'. The value of 'x' to the nearest integer is:-
(1) 2% (2) 3%
(3) 5% (4) 4%
Official Ans. by NTA (2)
Sol. g = 
p
2
2
4
T
l
D DD
=+
gT
2
gT
l
l
 = 
æö
ç÷
+
ç÷
ç÷
èø
1
0.1
200
2
10 0.5
D
=+
g 11
g 100 50
D
´=
g
100 3%
g
8. A constant power delivering machine has
towed a box, which was initially at rest, along
a horizontal straight line. The distance moved
by the box in time 't' is proportional to :-
(1) t
2/3
(2) t
3/2
(3) t (4) t
1/2
Official Ans. by NTA (2)
3
Sol. P = C
FV = C
M
=
dV
VC
dt
µ
2
V
t
2
V µ t
1/2
µ
1/2
dx
t
dt
x µ t
3/2
9. What will be the average value of energy along
one degree of freedom for an ideal gas in
thermal equilibrium at a temperature T ? (k
B 
is
Boltzmann constant)
(1) 
B
1
kT
2
(2) 
B
2
kT
3
(3) 
B
3
kT
2
(4) k
B
T
Official Ans. by NTA (1)
Sol. Energy associated with each degree of
freedom per molecule = 
B
1
kT
2
.
10. A radioactive sample disintegrates via two
independent decay processes having half lives
() 1
1/2
T and 
() 2
1/2
T respectively. The effective half-
life 
1/2
T of the nuclei is :
(1) None of the above (2) 
() ()
=+
12
1/2 1/2 1/2
T TT
(3) 
( ) ()
() ()
=
+
12
1/2 1/2
1/2 12
1/2 1/2
TT
T
TT
(4) 
() ()
() ()
+
=
-
12
1/2 1/2
1/2 12
1/2 1/2
TT
T
TT
Official Ans. by NTA (3)
Sol. l
eq
 = l
1
 + l
2
() ()
=+
12
1/2 1/2 1/2
1 11
T
TT
( ) ()
() ()
=
+
12
1/2 1/2
1/2 12
1/2 1/2
TT
T
TT
11. The P-V diagram of a diatomic ideal gas system
going under cyclic process as shown in figure.
The work done during an adiabatic process CD
is (use g = 1.4) :
A
B
D
C
1 3 4
200N/m
2
100N/m
2
P
V(m)
3
(1) –500 J (2) –400 J
(3) 400 J (4) 200 J
Official Ans. by NTA (1)
Sol. Adiabatic process is from C to D
-
=
-g
2 2 11
P V PV
WD
1
= 
-
-g
D D CC
P V PV
1
= 
-
-
200(3) (100)(4)
1 1.4
= –500 J Ans. (1)
12. In Young's double slit arrangement, slits are
separated by a gap of 0.5 mm, and the screen
is placed at a distance of 0.5 m from them. The
distance between the first and the third bright
fringe formed when the slits are illuminated by
a monochromatic light of 5890 Å is :-
(1) 1178 × 10
–9
 m (2) 1178 × 10
–6
 m
(3) 1178 × 10
–12
 m (4) 5890 × 10
–7
 m
Official Ans. by NTA (2)
Sol.
-
-
l ´´
b==
´
10
3
D 5890 10 0.5
d 0.5 10
= 589 × 10
–6
 m
Distance between first and third bright fringe
is 2b = 2×589 × 10
–6 
m
= 1178 × 10
–6
 m Ans. (2)
4
13. A particle is travelling 4 times as fast as an
electron. Assuming the ratio of de-Broglie
wavelength of a particle to that of electron is
2 : 1, the mass of the particle is :-
(1) 
1
16
 times the mass of e
–
(2) 8 times the mass of e
–
(3) 16 times the mass of e
–
(4) 
1
8
 times the mass of e
–
Official Ans. by NTA (4)
Sol.
l=
h
p
l
==
l
p
e ee
e p pp
p mv
p mv
æö
=
ç÷
èø
ee
pe
mv
2
m 4v
\=
e
p
m
m
8
Ans. (4)
14. The position, velocity and acceleration of a
particle moving with a constant acceleration
can be represented by :
(1)
position
t
x(t)
 
velocity
t
v(t)
 
acceleration
t
a(t)
(2) 
position
t
x(t)
 
velocity
t
v(t)
acceleration
t
a(t)
(3) 
position
t
x(t)
 
velocity
t
v(t)
acceleration
t
a(t)
(4) 
position
t
x(t)
velocity
t
v(t)
 
acceleration
t
a(t)
Official Ans. by NTA (2)
Sol. Option (2) represent correct graph for particle
moving with constant acceleration, as for
constant acceleration velocity time graph is
straight line with positive slope and x-t graph
should be an opening upward parabola.
15. In the experiment of Ohm's law, a potential
difference of 5.0 V is applied across the end of
a conductor of length 10.0 cm and diameter of
5.00 mm. The measured current in the
conductor is 2.00 A. The maximum
permissible percentage error in the resistivity
of the conductor is :-
(1) 3.9 (2) 8.4
(3) 7.5 (4) 3.0
Official Ans. by NTA (1)
Sol.
r
==
V
R
AI
l
p
r==
2
AV dV
I 4I ll
æö p
=
ç÷
èø
2
d
A
4
\
Dr D D DD
= + ++
r
2d VI
d VI
l
l
Dr æö
= + ++
ç÷
r
èø
0.01 0.1 0.01 0.1
2
5.00 5.0 2.00 10.0
Dr
r
 = 0.004 + 0.02 + 0.005 + 0.01
Dr
r
 = 0.039
% error = 
Dr
r
×100 = 0.039 × 100 = 3.90%
Ans. (1)
16. In a scries LCR resonance circuit, if we
change the resistance only, from a lower to
higher value :
(1) The bandwidth of resonance circuit will
increase.
(2) The resonance frequency will increase.
(3) The quality factor will increase.
(4) The quality factor and the resonance
frequency will remain constant.
Official Ans. by NTA (1)
Sol. Bandwidth = R/L
Bandwidth µ R
So bandwidth will increase
Page 5


1
SECTION-A
1. An oil drop of radius 2 mm with a density 3g
cm
–3
 is held stationary under a constant electric
field 3.55 × 10
5
 V m
–1
 in the Millikan's oil drop
experiment. What is the number of excess
electrons that the oil drop will possess ?
(consider g = 9.81 m/s
2
)
(1) 48.8 × 10
11
(2) 1.73 × 10
10
(3) 17.3 × 10
10
(4) 1.73 × 10
12
Official Ans. by NTA (2)
Sol. qE = Mg
neE = 
æö
r p´
ç÷
èø
3
4
rg
3
n × 1.6 × 10
–19
 × 3.55 × 10
5
= 3 × 10
3
 × 
4
3
 × p × (2 × 10
–3
)
3 
× 9.81
n = 173 × 10
(3 – 9 – 5 + 19)
n = 1.73 × 10
10
2. Match List–I with List–II.
List–I
(a) 10 km height over earth's surface
(b) 70 km height over earth's surface
(c) 180 km height over earth's surface
(d) 270 km height over earth's surface
List–II
(i) Thermosphere
(ii) Mesosphere
(iii) Stratosphere
(iv) Troposphere
(1) (a)–(iv), (b)–(iii), (c)–(ii), (d)–(i)
(2) (a)–(i), (b)–(iv), (c)–(iii), (d)–(ii)
(3) (a)–(iii), (b)–(ii), (c)–(i), (d)–(iv)
(4) (a)–(ii), (b)–(i), (c)–(iv), (d)–(iii)
Official Ans. by NTA (1)
FINAL JEE–MAIN EXAMINATION – MARCH, 2021
(Held On Thursday 18
th
 March, 2021)    TIME : 9 : 00 AM   to  12 : 00 NOON
PHYSICS TEST PAPER WITH ANSWER & SOLUTION
Sol. Order of atmosphere stratification from bottom
Troposphere, stralospherre, Mesosphre,
Thermosphere
(a) ® (iv)
(b) ® (iii)
(c) ® (ii)
(d) ® (i)
3. Imagine that the electron in a hydrogen atom
is replaced by a muon (µ). The mass of muon
particle is 207 times that of an electron and
charge is equal to the charge of an electron. The
ionization potential of this hydrogen atom will
be :-
(1) 13.6 eV (2) 2815.2 eV
(3) 331.2 eV (4) 27.2 eV
Official Ans. by NTA (2)
Sol.
µ
1
E
r
µ
1
r
m
E µ m
Ionization potential = 13.6 × 
( )
( )
m
e
Mass eV
Mass
= 13.6 × 207 eV = 2815.2 eV
4. A plane electromagnetic wave of frequency
100 MHz is travelling in vacuum along the x-
direction. At a particular point in space and
time, 
–8
ˆ
B 2.0 10 kT =´
r
. (where, ˆ
k
 is unit
vector along z-direction) What is 
E
r
 at this
point ?
(1) 0.6
ˆ
j
 V/m (2) 6.0
ˆ
k V/m
(3) 6.0
ˆ
j V/m (4) 0.6
ˆ
k V/m
Official Ans. by NTA (3)
2
Sol. E = BC = 6
(Dir. of wave) || ( )
´ EB
rr
=´
ˆˆ ˆ
i jk
=
ˆ
E 6j V/m
r
5. A thin circular ring of mass M and radius r is
rotating about its axis with an angular speed w.
Two particles having mass m each are now
attached at diametrically opposite points. The
angular speed of the ring will become :
(1) 
w
+
M
Mm
(2) 
+
w
M 2m
M
(3) 
w
+
M
M 2m
(4) 
-
w
+
M 2m
M 2m
Official Ans. by NTA (3)
Sol. Using conservation of angular momentum
(Mr
2
)w = (Mr
2
 + 2mr
2
)w'
w
w=
+
M
'
M 2m
6. Four identical long solenoids A, B, C and D are
connected to each other as shown in the figure.
If the magnetic field at the center of A is 3T,
the field at the center of C would be : (Assume
that the magnetic field is confined with in the
volume of respective solenoid).
B
C
D
A
i
(1) 12T (2) 6T
(3) 9T (4) 1T
Official Ans. by NTA (4)
Sol.
B
C
D
A
i
i/3
i/3
i/3
f µ i
Þ B µ i
so, field at centre of C = 
=
3
1T
3
7. The time period of a simple pendulum is given
by 
=p T2
g
l
. The measured value of the
length of pendulum is 10 cm known to a 1mm
accuracy. The time for 200 oscillations of the
pendulum is found to be 100 second using a
clock of 1s resolution. The percentage accuracy
in the determination of 'g' using this pendulum
is 'x'. The value of 'x' to the nearest integer is:-
(1) 2% (2) 3%
(3) 5% (4) 4%
Official Ans. by NTA (2)
Sol. g = 
p
2
2
4
T
l
D DD
=+
gT
2
gT
l
l
 = 
æö
ç÷
+
ç÷
ç÷
èø
1
0.1
200
2
10 0.5
D
=+
g 11
g 100 50
D
´=
g
100 3%
g
8. A constant power delivering machine has
towed a box, which was initially at rest, along
a horizontal straight line. The distance moved
by the box in time 't' is proportional to :-
(1) t
2/3
(2) t
3/2
(3) t (4) t
1/2
Official Ans. by NTA (2)
3
Sol. P = C
FV = C
M
=
dV
VC
dt
µ
2
V
t
2
V µ t
1/2
µ
1/2
dx
t
dt
x µ t
3/2
9. What will be the average value of energy along
one degree of freedom for an ideal gas in
thermal equilibrium at a temperature T ? (k
B 
is
Boltzmann constant)
(1) 
B
1
kT
2
(2) 
B
2
kT
3
(3) 
B
3
kT
2
(4) k
B
T
Official Ans. by NTA (1)
Sol. Energy associated with each degree of
freedom per molecule = 
B
1
kT
2
.
10. A radioactive sample disintegrates via two
independent decay processes having half lives
() 1
1/2
T and 
() 2
1/2
T respectively. The effective half-
life 
1/2
T of the nuclei is :
(1) None of the above (2) 
() ()
=+
12
1/2 1/2 1/2
T TT
(3) 
( ) ()
() ()
=
+
12
1/2 1/2
1/2 12
1/2 1/2
TT
T
TT
(4) 
() ()
() ()
+
=
-
12
1/2 1/2
1/2 12
1/2 1/2
TT
T
TT
Official Ans. by NTA (3)
Sol. l
eq
 = l
1
 + l
2
() ()
=+
12
1/2 1/2 1/2
1 11
T
TT
( ) ()
() ()
=
+
12
1/2 1/2
1/2 12
1/2 1/2
TT
T
TT
11. The P-V diagram of a diatomic ideal gas system
going under cyclic process as shown in figure.
The work done during an adiabatic process CD
is (use g = 1.4) :
A
B
D
C
1 3 4
200N/m
2
100N/m
2
P
V(m)
3
(1) –500 J (2) –400 J
(3) 400 J (4) 200 J
Official Ans. by NTA (1)
Sol. Adiabatic process is from C to D
-
=
-g
2 2 11
P V PV
WD
1
= 
-
-g
D D CC
P V PV
1
= 
-
-
200(3) (100)(4)
1 1.4
= –500 J Ans. (1)
12. In Young's double slit arrangement, slits are
separated by a gap of 0.5 mm, and the screen
is placed at a distance of 0.5 m from them. The
distance between the first and the third bright
fringe formed when the slits are illuminated by
a monochromatic light of 5890 Å is :-
(1) 1178 × 10
–9
 m (2) 1178 × 10
–6
 m
(3) 1178 × 10
–12
 m (4) 5890 × 10
–7
 m
Official Ans. by NTA (2)
Sol.
-
-
l ´´
b==
´
10
3
D 5890 10 0.5
d 0.5 10
= 589 × 10
–6
 m
Distance between first and third bright fringe
is 2b = 2×589 × 10
–6 
m
= 1178 × 10
–6
 m Ans. (2)
4
13. A particle is travelling 4 times as fast as an
electron. Assuming the ratio of de-Broglie
wavelength of a particle to that of electron is
2 : 1, the mass of the particle is :-
(1) 
1
16
 times the mass of e
–
(2) 8 times the mass of e
–
(3) 16 times the mass of e
–
(4) 
1
8
 times the mass of e
–
Official Ans. by NTA (4)
Sol.
l=
h
p
l
==
l
p
e ee
e p pp
p mv
p mv
æö
=
ç÷
èø
ee
pe
mv
2
m 4v
\=
e
p
m
m
8
Ans. (4)
14. The position, velocity and acceleration of a
particle moving with a constant acceleration
can be represented by :
(1)
position
t
x(t)
 
velocity
t
v(t)
 
acceleration
t
a(t)
(2) 
position
t
x(t)
 
velocity
t
v(t)
acceleration
t
a(t)
(3) 
position
t
x(t)
 
velocity
t
v(t)
acceleration
t
a(t)
(4) 
position
t
x(t)
velocity
t
v(t)
 
acceleration
t
a(t)
Official Ans. by NTA (2)
Sol. Option (2) represent correct graph for particle
moving with constant acceleration, as for
constant acceleration velocity time graph is
straight line with positive slope and x-t graph
should be an opening upward parabola.
15. In the experiment of Ohm's law, a potential
difference of 5.0 V is applied across the end of
a conductor of length 10.0 cm and diameter of
5.00 mm. The measured current in the
conductor is 2.00 A. The maximum
permissible percentage error in the resistivity
of the conductor is :-
(1) 3.9 (2) 8.4
(3) 7.5 (4) 3.0
Official Ans. by NTA (1)
Sol.
r
==
V
R
AI
l
p
r==
2
AV dV
I 4I ll
æö p
=
ç÷
èø
2
d
A
4
\
Dr D D DD
= + ++
r
2d VI
d VI
l
l
Dr æö
= + ++
ç÷
r
èø
0.01 0.1 0.01 0.1
2
5.00 5.0 2.00 10.0
Dr
r
 = 0.004 + 0.02 + 0.005 + 0.01
Dr
r
 = 0.039
% error = 
Dr
r
×100 = 0.039 × 100 = 3.90%
Ans. (1)
16. In a scries LCR resonance circuit, if we
change the resistance only, from a lower to
higher value :
(1) The bandwidth of resonance circuit will
increase.
(2) The resonance frequency will increase.
(3) The quality factor will increase.
(4) The quality factor and the resonance
frequency will remain constant.
Official Ans. by NTA (1)
Sol. Bandwidth = R/L
Bandwidth µ R
So bandwidth will increase
5
17. An AC source rated 220 V, 50 Hz is connected
to a resistor. The time taken by the current to
change from its maximum to the rms value
is :
(1) 2.5 ms (2) 25 ms
(3) 2.5 s (4) 0.25 ms
Official Ans. by NTA (1)
Sol. i = i
0
 cos(wt)
i = i
0
 at t = 0
=
0
i
i
2
 at 
p
w= t
4
pp
===
wp
1
t
4 4(2 f) 8f
==
1
t 2.5 ms
400
18. Your friend is having eye sight problem. She
is not able lo see clearly a distant uniform
window mesh and it appears to her as non-
uniform and distorted. The doctor diagnosed
the problem as :
(1) Astigmatism
(2) Myopia with Astigmatism
(3) Presbyopia with Astigmatism
(4) Myopia and hypermetropia
Official Ans. by NTA (2)
Sol. If distant objects are blurry then problem is
Myopia.
If objects are distorted then problem is
Astigmatism
19. A loop of flexible wire of irregular shape
carrying current is placed in an external
magnetic field. Identify the effect of the field
on the wire.
(1) Loop assumes circular shape with its plane
normal to the field.
(2) Loop assumes circular shape with its plane
parallel to the field.
(3) Wire gets stretched to become straight.
(4) Shape of the loop remains unchanged.
Official Ans. by NTA (1)
Sol. Every part (d l) of the wire is pulled by force
i(d l)B acting perpendicular to current &
magnetic field giving it a shape of circle.
20. The time period of a satellite in a circular orbit
of radius R is T. The period of another satellite
in a circular orbit of radius 9R is :
(1) 9 T (2) 27 T
(3) 12 T (4) 3 T
Official Ans. by NTA (2)
Sol.
µ
23
TR
æ ö æö
=
ç ÷ ç÷
è ø èø
23
T' 9R
TR
T'
2
 = T
2
 × 9
3
T' = T ×3
3
T' = 27 T
SECTION-B
1. A particle performs simple harmonic motion
with a period of 2 second. The time taken by
the particle to cover a displacement equal to
half of its amplitude from the mean position
is 
1
s
a
. The value of 'a' to the nearest integer
is _______ .
Official Ans. by NTA (6)
Sol.
O
A/2
t = T/12
T = 2sec.
t = 
=
21
126
\ Correct answer = 6.00
2. The circuit shown in the figure consists of a
charged capacitor of capacity 3 µF and a charge
of 30 µC. At time t = 0, when the key is closed,
the value of current flowing through the 5 MW
resistor is 'x' µ-A. The value of 'x to the nearest
integer is _______.
C = 3µF
5MW
q = 30 µC
Official Ans. by NTA (2)
Sol. i
0
 = 
-
= =´
´
6
6
V 30/3
2 10
R 5 10
\ Ans. = 2.00