JEE Mains 17 March 2021 Question Paper Shift 1 | JEE Main & Advanced Previous Year Papers PDF Download

 Page 1


1
SECTION-A
1. A triangular plate is shown. A force =-
r
ˆˆ
F 4i 3j
is applied at point P. The torque at point P with
respect to point 'O' and 'Q' are :
10 cm
60°
60°
P
10 cm
10 cm
F
Y
X
O
Q
(1) – 15 – 20
3
, 15 – 20
3
(2) 15 + 20
3
, 15 – 20
3
(3) 15 – 20
3
, 15 + 20
3
(4) – 15 + 20
3
, 15 + 20
3
Official Ans. by NTA (1)
Sol. =-
r
ˆˆ
F 4i 3j
=+
r
1
ˆˆ
r 5i 5 3j & =-+
r
2
ˆˆ
r 5i 5 3j
Torque about 'O'
t =´
r
r r
O1
rF = ( )
--
ˆ
15 20 3 k
= ( )( )
+-
ˆ
15203k
Torque about 'Q'
t =´
r
r r
Q2
rF = ( )
-+
ˆ
15 20 3 k
=( )( )
--
ˆ
15203k
2. When two soap bubbles of radii a and
b (b > a) coalesce, the radius of curvature of
common surface is :
(1)
-
ab
ba
(2) 
+ ab
ab
(3) 
- ba
ab
(4)
+
ab
ab
Official Ans. by NTA (1)
(Held On Wednesday 17
th
 March, 2021)    TIME : 9 : 00 AM   to  12 : 00 NOON
PHYSICS TEST PAPER WITH ANSWER & SOLUTION
Sol. Excess pressure at common surface is given by
P
ex
 = 4T 
æö
-
ç÷
èø
11
ab
 = 
4T
r
\ 
=-
111
r ab
=
-
ab
r
ba
3. A polyatomic ideal gas has 24 vibrational
modes. What is the value of g ?
(1) 1.03 (2) 1.30
(3) 1.37 (4) 10.3
Official Ans. by NTA (1)
Sol. Since each vibrational mode has 2 degrees of
freedom hence total vibrational degrees of
freedom = 48
f = 3 + 3 + 48 = 54
g = 1 + 
2
f
 = 
28
27
 = 1.03
4. If an electron is moving in the n
th
 orbit of the
hydrogen atom, then its velocity (v
n
) for the n
th
orbit is given as :
(1) v
n
 µ n (2) v
n
 µ
1
n
(3) v
n
 µ n
2
(4) v
n
 µ
2
1
n
Official Ans. by NTA (2)
Sol. We know velocity of electron in n
th
 shell of
hydrogen atom is given by
p
=
2
2 kZe
v
nh
\ v µ 
1
n
FINAL JEE–MAIN EXAMINATION – MARCH, 2021
Page 2


1
SECTION-A
1. A triangular plate is shown. A force =-
r
ˆˆ
F 4i 3j
is applied at point P. The torque at point P with
respect to point 'O' and 'Q' are :
10 cm
60°
60°
P
10 cm
10 cm
F
Y
X
O
Q
(1) – 15 – 20
3
, 15 – 20
3
(2) 15 + 20
3
, 15 – 20
3
(3) 15 – 20
3
, 15 + 20
3
(4) – 15 + 20
3
, 15 + 20
3
Official Ans. by NTA (1)
Sol. =-
r
ˆˆ
F 4i 3j
=+
r
1
ˆˆ
r 5i 5 3j & =-+
r
2
ˆˆ
r 5i 5 3j
Torque about 'O'
t =´
r
r r
O1
rF = ( )
--
ˆ
15 20 3 k
= ( )( )
+-
ˆ
15203k
Torque about 'Q'
t =´
r
r r
Q2
rF = ( )
-+
ˆ
15 20 3 k
=( )( )
--
ˆ
15203k
2. When two soap bubbles of radii a and
b (b > a) coalesce, the radius of curvature of
common surface is :
(1)
-
ab
ba
(2) 
+ ab
ab
(3) 
- ba
ab
(4)
+
ab
ab
Official Ans. by NTA (1)
(Held On Wednesday 17
th
 March, 2021)    TIME : 9 : 00 AM   to  12 : 00 NOON
PHYSICS TEST PAPER WITH ANSWER & SOLUTION
Sol. Excess pressure at common surface is given by
P
ex
 = 4T 
æö
-
ç÷
èø
11
ab
 = 
4T
r
\ 
=-
111
r ab
=
-
ab
r
ba
3. A polyatomic ideal gas has 24 vibrational
modes. What is the value of g ?
(1) 1.03 (2) 1.30
(3) 1.37 (4) 10.3
Official Ans. by NTA (1)
Sol. Since each vibrational mode has 2 degrees of
freedom hence total vibrational degrees of
freedom = 48
f = 3 + 3 + 48 = 54
g = 1 + 
2
f
 = 
28
27
 = 1.03
4. If an electron is moving in the n
th
 orbit of the
hydrogen atom, then its velocity (v
n
) for the n
th
orbit is given as :
(1) v
n
 µ n (2) v
n
 µ
1
n
(3) v
n
 µ n
2
(4) v
n
 µ
2
1
n
Official Ans. by NTA (2)
Sol. We know velocity of electron in n
th
 shell of
hydrogen atom is given by
p
=
2
2 kZe
v
nh
\ v µ 
1
n
FINAL JEE–MAIN EXAMINATION – MARCH, 2021
2
5. An electron of mass m and a photon have same
energy E. The ratio of wavelength of electron
to that of photon is : (c being the velocity of
light)
(1) 
æö
ç÷
èø
1/2
1 2m
cE
(2) 
æö
ç÷
èø
1/2
1E
c 2m
(3) 
æö
ç÷
èø
1/2
E
2m
(4) c (2mE)
1/2
Official Ans. by NTA (2)
Sol.
l=
1
h
2mE
l=
2
hc
E
l æö
=
ç÷
l
èø
1/2
1
2
1E
c 2m
6. Two identical metal wires of thermal
conductivities K
1
 and K
2
 respectively are
connected in series. The effective thermal
conductivity of the combination is :
(1) 
+
12
12
2KK
KK
(2) 
+
12
12
KK
2KK
(3) 
+
12
12
KK
KK
(4) 
+
12
12
KK
KK
Official Ans. by NTA (1)
Sol.
K
1
K
2
l l
K
eq
2l
R
eff
 = 
1
KA
l
 + 
2
KA
l
 = 
eq
2
KA
l
=
+
12
eq
12
2KK
K
KK
7. The vernier scale used for measurement has a
positive zero error of 0.2 mm. If while taking
a measurement it was noted that '0' on the
vernier scale lies between 8.5 cm and 8.6 cm,
vernier coincidence is 6, then the correct value
of measurement is_____ cm.
(least count = 0.01 cm)
(1) 8.36 cm (2) 8.54 cm
(3) 8.58 cm (4) 8.56 cm
Official Ans. by NTA (2)
Sol. Positive zero error = 0.2 mm
Main scale reading = 8.5 cm
Vernier scale reading  = 6 × 0.01 = 0.06 cm
Final reading = 8.5 + 0.06 – 0.02 = 8.54 cm
8. An AC current is given by I = I
1
 sinwt + I
2
 coswt.
A hot wire ammeter will give a reading :
(1) 
-
22
12
II
2
(2) 
+
22
12
II
2
(3) 
+
12
II
2
(4) 
+
12
II
22
Official Ans. by NTA (2)
Sol. I = I
1
 sin wt + I
2
 cos wt
\ I
0
 =+
22
12
II
\ I
rms
 
=
0
I
2
  
+
=
22
12
II
2
9. A modern grand-prix racing car of mass m is
travelling on a flat track in a circular arc of
radius R with a speed v. If the coefficient of
static friction between the tyres and the track
is m
s
, then the magnitude of negative lift F
L
acting downwards on the car is :
(Assume forces on the four tyres are identical
and g = acceleration due to gravity)
       
v
R
(1) 
æö
+
ç÷
m
èø
2
s
v
mg
R
(2) 
æö
-
ç÷
m
èø
2
s
v
mg
R
(3) 
æö
-
ç÷
m
èø
2
s
v
mg
R
(4) 
æö
-+
ç÷
m
èø
2
s
v
mg
R
Official Ans. by NTA (2)
Page 3


1
SECTION-A
1. A triangular plate is shown. A force =-
r
ˆˆ
F 4i 3j
is applied at point P. The torque at point P with
respect to point 'O' and 'Q' are :
10 cm
60°
60°
P
10 cm
10 cm
F
Y
X
O
Q
(1) – 15 – 20
3
, 15 – 20
3
(2) 15 + 20
3
, 15 – 20
3
(3) 15 – 20
3
, 15 + 20
3
(4) – 15 + 20
3
, 15 + 20
3
Official Ans. by NTA (1)
Sol. =-
r
ˆˆ
F 4i 3j
=+
r
1
ˆˆ
r 5i 5 3j & =-+
r
2
ˆˆ
r 5i 5 3j
Torque about 'O'
t =´
r
r r
O1
rF = ( )
--
ˆ
15 20 3 k
= ( )( )
+-
ˆ
15203k
Torque about 'Q'
t =´
r
r r
Q2
rF = ( )
-+
ˆ
15 20 3 k
=( )( )
--
ˆ
15203k
2. When two soap bubbles of radii a and
b (b > a) coalesce, the radius of curvature of
common surface is :
(1)
-
ab
ba
(2) 
+ ab
ab
(3) 
- ba
ab
(4)
+
ab
ab
Official Ans. by NTA (1)
(Held On Wednesday 17
th
 March, 2021)    TIME : 9 : 00 AM   to  12 : 00 NOON
PHYSICS TEST PAPER WITH ANSWER & SOLUTION
Sol. Excess pressure at common surface is given by
P
ex
 = 4T 
æö
-
ç÷
èø
11
ab
 = 
4T
r
\ 
=-
111
r ab
=
-
ab
r
ba
3. A polyatomic ideal gas has 24 vibrational
modes. What is the value of g ?
(1) 1.03 (2) 1.30
(3) 1.37 (4) 10.3
Official Ans. by NTA (1)
Sol. Since each vibrational mode has 2 degrees of
freedom hence total vibrational degrees of
freedom = 48
f = 3 + 3 + 48 = 54
g = 1 + 
2
f
 = 
28
27
 = 1.03
4. If an electron is moving in the n
th
 orbit of the
hydrogen atom, then its velocity (v
n
) for the n
th
orbit is given as :
(1) v
n
 µ n (2) v
n
 µ
1
n
(3) v
n
 µ n
2
(4) v
n
 µ
2
1
n
Official Ans. by NTA (2)
Sol. We know velocity of electron in n
th
 shell of
hydrogen atom is given by
p
=
2
2 kZe
v
nh
\ v µ 
1
n
FINAL JEE–MAIN EXAMINATION – MARCH, 2021
2
5. An electron of mass m and a photon have same
energy E. The ratio of wavelength of electron
to that of photon is : (c being the velocity of
light)
(1) 
æö
ç÷
èø
1/2
1 2m
cE
(2) 
æö
ç÷
èø
1/2
1E
c 2m
(3) 
æö
ç÷
èø
1/2
E
2m
(4) c (2mE)
1/2
Official Ans. by NTA (2)
Sol.
l=
1
h
2mE
l=
2
hc
E
l æö
=
ç÷
l
èø
1/2
1
2
1E
c 2m
6. Two identical metal wires of thermal
conductivities K
1
 and K
2
 respectively are
connected in series. The effective thermal
conductivity of the combination is :
(1) 
+
12
12
2KK
KK
(2) 
+
12
12
KK
2KK
(3) 
+
12
12
KK
KK
(4) 
+
12
12
KK
KK
Official Ans. by NTA (1)
Sol.
K
1
K
2
l l
K
eq
2l
R
eff
 = 
1
KA
l
 + 
2
KA
l
 = 
eq
2
KA
l
=
+
12
eq
12
2KK
K
KK
7. The vernier scale used for measurement has a
positive zero error of 0.2 mm. If while taking
a measurement it was noted that '0' on the
vernier scale lies between 8.5 cm and 8.6 cm,
vernier coincidence is 6, then the correct value
of measurement is_____ cm.
(least count = 0.01 cm)
(1) 8.36 cm (2) 8.54 cm
(3) 8.58 cm (4) 8.56 cm
Official Ans. by NTA (2)
Sol. Positive zero error = 0.2 mm
Main scale reading = 8.5 cm
Vernier scale reading  = 6 × 0.01 = 0.06 cm
Final reading = 8.5 + 0.06 – 0.02 = 8.54 cm
8. An AC current is given by I = I
1
 sinwt + I
2
 coswt.
A hot wire ammeter will give a reading :
(1) 
-
22
12
II
2
(2) 
+
22
12
II
2
(3) 
+
12
II
2
(4) 
+
12
II
22
Official Ans. by NTA (2)
Sol. I = I
1
 sin wt + I
2
 cos wt
\ I
0
 =+
22
12
II
\ I
rms
 
=
0
I
2
  
+
=
22
12
II
2
9. A modern grand-prix racing car of mass m is
travelling on a flat track in a circular arc of
radius R with a speed v. If the coefficient of
static friction between the tyres and the track
is m
s
, then the magnitude of negative lift F
L
acting downwards on the car is :
(Assume forces on the four tyres are identical
and g = acceleration due to gravity)
       
v
R
(1) 
æö
+
ç÷
m
èø
2
s
v
mg
R
(2) 
æö
-
ç÷
m
èø
2
s
v
mg
R
(3) 
æö
-
ç÷
m
èø
2
s
v
mg
R
(4) 
æö
-+
ç÷
m
èø
2
s
v
mg
R
Official Ans. by NTA (2)
3
Sol. m
s
N = 
2
mv
R
=
m
2
s
mv
N
R
 = mg + F
L
F
L
 = 
-
m
2
s
mv
mg
R
10. A car accelerates from rest at a constant rate a
for some time after which it decelerates at a
constant rate b to come to rest. If the total time
elapsed is t seconds, the total distance travelled
is :
(1) 
ab
a+b
2
4
t
()
(2) 
ab
a+b
2
2
t
()
(3) 
ab
a+b
2
t
2()
(4) 
ab
a+b
2
t
4()
Official Ans. by NTA (3)
Sol. v
0
 = at
1
 and 0 = v
0
 – bt
2
  Þ v
0
 = bt
2
t
1
 + t
2
 = t
æö
+=
ç÷
ab
èø
0
11
vt
      
v
v
0
time
t
Þ 
ab
=
a+b
0
t
v
Distance = area of v-t graph
= 
1
2
× t × v
0
 = 
1
2
 × t × 
ab
a+b
t
 = 
( )
ab
a+b
2
t
2
11. A solenoid of 1000 turns per metre has a core
with relative permeability 500. Insulated
windings of the solenoid carry an electric
current of 5A. The magnetic flux density
produced by the solenoid is :
(permeability of free space = 4p × 10
–7
 H/m)
(1) pT (2) 2 × 10
–3
 pT
(3) 
p
5
T (4) 10
–4
pT
Official Ans. by NTA (1)
Sol. B = mnI = m
0
m
r
nI
B = 4p × 10
–7
 × 500 × 1000 × 5
B = p Tesla
12. A mass M hangs on a massless rod of
length l which rotates at a constant angular
frequency. The mass M moves with steady
speed in a circular path of constant radius.
Assume that the system is in steady circular
motion with constant angular velocity w. The
angular momentum of M about point A is L
A
which lies in the positive z direction and the
angular momentum of M about B is L
B
. The
correct statement for this system is :
z
w
l
B
a
A
r
M
(1) L
A
 and L
B
 are both constant in magnitude
and direction
(2) L
B
 is constant in direction with varying
magnitude
(3) L
B
 is constant, both in magnitude and
direction
(4) L
A
 is constant, both in magnitude and
direction
Official Ans. by NTA (4)
Sol. We know, =´
r
r r
L m (r v)
Now with respect to A, we always get direction
of 
r
L along +ve z-axis and also constant
magnitude as mvr. But with respect to B, we
get constant magnitude but continuously
changing direction.
13. For what value of displacement the kinetic
energy and potential energy of a simple
harmonic oscillation become equal ?
(1) x = 0 (2) x = ± A
(3) x = ± 
A
2
(4) x = 
A
2
Official Ans. by NTA (3)
Page 4


1
SECTION-A
1. A triangular plate is shown. A force =-
r
ˆˆ
F 4i 3j
is applied at point P. The torque at point P with
respect to point 'O' and 'Q' are :
10 cm
60°
60°
P
10 cm
10 cm
F
Y
X
O
Q
(1) – 15 – 20
3
, 15 – 20
3
(2) 15 + 20
3
, 15 – 20
3
(3) 15 – 20
3
, 15 + 20
3
(4) – 15 + 20
3
, 15 + 20
3
Official Ans. by NTA (1)
Sol. =-
r
ˆˆ
F 4i 3j
=+
r
1
ˆˆ
r 5i 5 3j & =-+
r
2
ˆˆ
r 5i 5 3j
Torque about 'O'
t =´
r
r r
O1
rF = ( )
--
ˆ
15 20 3 k
= ( )( )
+-
ˆ
15203k
Torque about 'Q'
t =´
r
r r
Q2
rF = ( )
-+
ˆ
15 20 3 k
=( )( )
--
ˆ
15203k
2. When two soap bubbles of radii a and
b (b > a) coalesce, the radius of curvature of
common surface is :
(1)
-
ab
ba
(2) 
+ ab
ab
(3) 
- ba
ab
(4)
+
ab
ab
Official Ans. by NTA (1)
(Held On Wednesday 17
th
 March, 2021)    TIME : 9 : 00 AM   to  12 : 00 NOON
PHYSICS TEST PAPER WITH ANSWER & SOLUTION
Sol. Excess pressure at common surface is given by
P
ex
 = 4T 
æö
-
ç÷
èø
11
ab
 = 
4T
r
\ 
=-
111
r ab
=
-
ab
r
ba
3. A polyatomic ideal gas has 24 vibrational
modes. What is the value of g ?
(1) 1.03 (2) 1.30
(3) 1.37 (4) 10.3
Official Ans. by NTA (1)
Sol. Since each vibrational mode has 2 degrees of
freedom hence total vibrational degrees of
freedom = 48
f = 3 + 3 + 48 = 54
g = 1 + 
2
f
 = 
28
27
 = 1.03
4. If an electron is moving in the n
th
 orbit of the
hydrogen atom, then its velocity (v
n
) for the n
th
orbit is given as :
(1) v
n
 µ n (2) v
n
 µ
1
n
(3) v
n
 µ n
2
(4) v
n
 µ
2
1
n
Official Ans. by NTA (2)
Sol. We know velocity of electron in n
th
 shell of
hydrogen atom is given by
p
=
2
2 kZe
v
nh
\ v µ 
1
n
FINAL JEE–MAIN EXAMINATION – MARCH, 2021
2
5. An electron of mass m and a photon have same
energy E. The ratio of wavelength of electron
to that of photon is : (c being the velocity of
light)
(1) 
æö
ç÷
èø
1/2
1 2m
cE
(2) 
æö
ç÷
èø
1/2
1E
c 2m
(3) 
æö
ç÷
èø
1/2
E
2m
(4) c (2mE)
1/2
Official Ans. by NTA (2)
Sol.
l=
1
h
2mE
l=
2
hc
E
l æö
=
ç÷
l
èø
1/2
1
2
1E
c 2m
6. Two identical metal wires of thermal
conductivities K
1
 and K
2
 respectively are
connected in series. The effective thermal
conductivity of the combination is :
(1) 
+
12
12
2KK
KK
(2) 
+
12
12
KK
2KK
(3) 
+
12
12
KK
KK
(4) 
+
12
12
KK
KK
Official Ans. by NTA (1)
Sol.
K
1
K
2
l l
K
eq
2l
R
eff
 = 
1
KA
l
 + 
2
KA
l
 = 
eq
2
KA
l
=
+
12
eq
12
2KK
K
KK
7. The vernier scale used for measurement has a
positive zero error of 0.2 mm. If while taking
a measurement it was noted that '0' on the
vernier scale lies between 8.5 cm and 8.6 cm,
vernier coincidence is 6, then the correct value
of measurement is_____ cm.
(least count = 0.01 cm)
(1) 8.36 cm (2) 8.54 cm
(3) 8.58 cm (4) 8.56 cm
Official Ans. by NTA (2)
Sol. Positive zero error = 0.2 mm
Main scale reading = 8.5 cm
Vernier scale reading  = 6 × 0.01 = 0.06 cm
Final reading = 8.5 + 0.06 – 0.02 = 8.54 cm
8. An AC current is given by I = I
1
 sinwt + I
2
 coswt.
A hot wire ammeter will give a reading :
(1) 
-
22
12
II
2
(2) 
+
22
12
II
2
(3) 
+
12
II
2
(4) 
+
12
II
22
Official Ans. by NTA (2)
Sol. I = I
1
 sin wt + I
2
 cos wt
\ I
0
 =+
22
12
II
\ I
rms
 
=
0
I
2
  
+
=
22
12
II
2
9. A modern grand-prix racing car of mass m is
travelling on a flat track in a circular arc of
radius R with a speed v. If the coefficient of
static friction between the tyres and the track
is m
s
, then the magnitude of negative lift F
L
acting downwards on the car is :
(Assume forces on the four tyres are identical
and g = acceleration due to gravity)
       
v
R
(1) 
æö
+
ç÷
m
èø
2
s
v
mg
R
(2) 
æö
-
ç÷
m
èø
2
s
v
mg
R
(3) 
æö
-
ç÷
m
èø
2
s
v
mg
R
(4) 
æö
-+
ç÷
m
èø
2
s
v
mg
R
Official Ans. by NTA (2)
3
Sol. m
s
N = 
2
mv
R
=
m
2
s
mv
N
R
 = mg + F
L
F
L
 = 
-
m
2
s
mv
mg
R
10. A car accelerates from rest at a constant rate a
for some time after which it decelerates at a
constant rate b to come to rest. If the total time
elapsed is t seconds, the total distance travelled
is :
(1) 
ab
a+b
2
4
t
()
(2) 
ab
a+b
2
2
t
()
(3) 
ab
a+b
2
t
2()
(4) 
ab
a+b
2
t
4()
Official Ans. by NTA (3)
Sol. v
0
 = at
1
 and 0 = v
0
 – bt
2
  Þ v
0
 = bt
2
t
1
 + t
2
 = t
æö
+=
ç÷
ab
èø
0
11
vt
      
v
v
0
time
t
Þ 
ab
=
a+b
0
t
v
Distance = area of v-t graph
= 
1
2
× t × v
0
 = 
1
2
 × t × 
ab
a+b
t
 = 
( )
ab
a+b
2
t
2
11. A solenoid of 1000 turns per metre has a core
with relative permeability 500. Insulated
windings of the solenoid carry an electric
current of 5A. The magnetic flux density
produced by the solenoid is :
(permeability of free space = 4p × 10
–7
 H/m)
(1) pT (2) 2 × 10
–3
 pT
(3) 
p
5
T (4) 10
–4
pT
Official Ans. by NTA (1)
Sol. B = mnI = m
0
m
r
nI
B = 4p × 10
–7
 × 500 × 1000 × 5
B = p Tesla
12. A mass M hangs on a massless rod of
length l which rotates at a constant angular
frequency. The mass M moves with steady
speed in a circular path of constant radius.
Assume that the system is in steady circular
motion with constant angular velocity w. The
angular momentum of M about point A is L
A
which lies in the positive z direction and the
angular momentum of M about B is L
B
. The
correct statement for this system is :
z
w
l
B
a
A
r
M
(1) L
A
 and L
B
 are both constant in magnitude
and direction
(2) L
B
 is constant in direction with varying
magnitude
(3) L
B
 is constant, both in magnitude and
direction
(4) L
A
 is constant, both in magnitude and
direction
Official Ans. by NTA (4)
Sol. We know, =´
r
r r
L m (r v)
Now with respect to A, we always get direction
of 
r
L along +ve z-axis and also constant
magnitude as mvr. But with respect to B, we
get constant magnitude but continuously
changing direction.
13. For what value of displacement the kinetic
energy and potential energy of a simple
harmonic oscillation become equal ?
(1) x = 0 (2) x = ± A
(3) x = ± 
A
2
(4) x = 
A
2
Official Ans. by NTA (3)
4
Sol. KE = PE
1
2
mw
2
 (A
2
 – x
2
) = 
1
2
 mw
2
 x
2
A
2
 – x
2
 = x
2
2x
2
 = A
2
x = ± 
A
2
14. A Carnot's engine working between 400 K and
800 K has a work output of 1200 J per cycle.
The amount of heat energy supplied to the
engine from the source in each cycle is :
(1) 3200 J (2) 1800 J
(3) 1600 J (4) 2400 J
Official Ans. by NTA (4)
Sol. h = 
=
22
11
TQ
TQ
 = 
-
1
1
QW
Q
     (Q W = Q
1
 – Q
2
)
=-
1
400 W
1
800 Q
= -=
1
W 11
1
Q 22
Q
1
 = 2W = 2400 J
15. The thickness at the centre of a plano convex
lens is 3 mm and the diameter is 6 cm. If the
speed of light in the material of the lens is
2 × 10
8
 ms
–1
. The focal length of the lens is
_____.
(1) 0.30 cm (2) 15 cm
(3) 1.5 cm (4) 30 cm
Official Ans. by NTA (4)
Sol. R
2
 = r
2
 + (R – t)
2
R
2
 = r
2
 + R
2
 + t
2
 – 2Rt      
R
r
t
Neglecting t
2
, we get
=
2
r
R
2t
\ 
1
f
= (m – 1) 
æö
-
ç÷
¥
èø
11
R
 = 
m-1
R
f = 
m-
R
1
 = 
m-
2
r
2t ( 1)
 = 
-
-
´
æö
´´ ´-
ç÷
èø
22
3
(3 10)
3
23101
2
-
-
´
=
´´
4
3
9 10
6101
 × 2
f = 0.3 m = 30 cm
16. The output of the given combination gates
represents :
A
B
Y
(1) XOR Gate (2) NAND Gate
(3) AND Gate (4) NOR Gate
Official Ans. by NTA (2)
Sol. By De Morgan's theorem, we have
A
B
Y
A
B
A·B = NAND
A·B
17. A boy is rolling a 0.5 kg ball on the frictionless
floor with the speed of 20 ms
–1
. The ball gets
deflected by an obstacle on the way. After
deflection it moves with 5% of its initial kinetic
energy. What is the speed of the ball now ?
(1) 19.0 ms
–1
(2) 4.47 ms
–1
(3) 14.41 ms
–1
(4) 1.00 ms
–1
Official Ans. by NTA (2)
Sol. Given, m = 0.5 kg  and  u = 20 m/s
Initial kinetic energy (k
i
) = 
2
1
mu
2
= 
1
2
× 0.5 × 20 × 20 = 100 J
After deflection it moves with 5% of k
i
\ k
f
 = 
5
100
 × k
i
  Þ 
5
100
 × 100
Þ k
f
 = 5 J
Now, let the final speed be 'v' m/s, then :
k
f
 = 5 = 
1
2
mv
2
Þ v
2
 = 20
Þ v = 
20
 = 4.47 m/s
18. Which level of the single ionized carbon has
the same energy as the ground state energy of
hydrogen atom?
(1) 1 (2) 6
(3) 4 (4) 8
Official Ans. by NTA (2)
Page 5


1
SECTION-A
1. A triangular plate is shown. A force =-
r
ˆˆ
F 4i 3j
is applied at point P. The torque at point P with
respect to point 'O' and 'Q' are :
10 cm
60°
60°
P
10 cm
10 cm
F
Y
X
O
Q
(1) – 15 – 20
3
, 15 – 20
3
(2) 15 + 20
3
, 15 – 20
3
(3) 15 – 20
3
, 15 + 20
3
(4) – 15 + 20
3
, 15 + 20
3
Official Ans. by NTA (1)
Sol. =-
r
ˆˆ
F 4i 3j
=+
r
1
ˆˆ
r 5i 5 3j & =-+
r
2
ˆˆ
r 5i 5 3j
Torque about 'O'
t =´
r
r r
O1
rF = ( )
--
ˆ
15 20 3 k
= ( )( )
+-
ˆ
15203k
Torque about 'Q'
t =´
r
r r
Q2
rF = ( )
-+
ˆ
15 20 3 k
=( )( )
--
ˆ
15203k
2. When two soap bubbles of radii a and
b (b > a) coalesce, the radius of curvature of
common surface is :
(1)
-
ab
ba
(2) 
+ ab
ab
(3) 
- ba
ab
(4)
+
ab
ab
Official Ans. by NTA (1)
(Held On Wednesday 17
th
 March, 2021)    TIME : 9 : 00 AM   to  12 : 00 NOON
PHYSICS TEST PAPER WITH ANSWER & SOLUTION
Sol. Excess pressure at common surface is given by
P
ex
 = 4T 
æö
-
ç÷
èø
11
ab
 = 
4T
r
\ 
=-
111
r ab
=
-
ab
r
ba
3. A polyatomic ideal gas has 24 vibrational
modes. What is the value of g ?
(1) 1.03 (2) 1.30
(3) 1.37 (4) 10.3
Official Ans. by NTA (1)
Sol. Since each vibrational mode has 2 degrees of
freedom hence total vibrational degrees of
freedom = 48
f = 3 + 3 + 48 = 54
g = 1 + 
2
f
 = 
28
27
 = 1.03
4. If an electron is moving in the n
th
 orbit of the
hydrogen atom, then its velocity (v
n
) for the n
th
orbit is given as :
(1) v
n
 µ n (2) v
n
 µ
1
n
(3) v
n
 µ n
2
(4) v
n
 µ
2
1
n
Official Ans. by NTA (2)
Sol. We know velocity of electron in n
th
 shell of
hydrogen atom is given by
p
=
2
2 kZe
v
nh
\ v µ 
1
n
FINAL JEE–MAIN EXAMINATION – MARCH, 2021
2
5. An electron of mass m and a photon have same
energy E. The ratio of wavelength of electron
to that of photon is : (c being the velocity of
light)
(1) 
æö
ç÷
èø
1/2
1 2m
cE
(2) 
æö
ç÷
èø
1/2
1E
c 2m
(3) 
æö
ç÷
èø
1/2
E
2m
(4) c (2mE)
1/2
Official Ans. by NTA (2)
Sol.
l=
1
h
2mE
l=
2
hc
E
l æö
=
ç÷
l
èø
1/2
1
2
1E
c 2m
6. Two identical metal wires of thermal
conductivities K
1
 and K
2
 respectively are
connected in series. The effective thermal
conductivity of the combination is :
(1) 
+
12
12
2KK
KK
(2) 
+
12
12
KK
2KK
(3) 
+
12
12
KK
KK
(4) 
+
12
12
KK
KK
Official Ans. by NTA (1)
Sol.
K
1
K
2
l l
K
eq
2l
R
eff
 = 
1
KA
l
 + 
2
KA
l
 = 
eq
2
KA
l
=
+
12
eq
12
2KK
K
KK
7. The vernier scale used for measurement has a
positive zero error of 0.2 mm. If while taking
a measurement it was noted that '0' on the
vernier scale lies between 8.5 cm and 8.6 cm,
vernier coincidence is 6, then the correct value
of measurement is_____ cm.
(least count = 0.01 cm)
(1) 8.36 cm (2) 8.54 cm
(3) 8.58 cm (4) 8.56 cm
Official Ans. by NTA (2)
Sol. Positive zero error = 0.2 mm
Main scale reading = 8.5 cm
Vernier scale reading  = 6 × 0.01 = 0.06 cm
Final reading = 8.5 + 0.06 – 0.02 = 8.54 cm
8. An AC current is given by I = I
1
 sinwt + I
2
 coswt.
A hot wire ammeter will give a reading :
(1) 
-
22
12
II
2
(2) 
+
22
12
II
2
(3) 
+
12
II
2
(4) 
+
12
II
22
Official Ans. by NTA (2)
Sol. I = I
1
 sin wt + I
2
 cos wt
\ I
0
 =+
22
12
II
\ I
rms
 
=
0
I
2
  
+
=
22
12
II
2
9. A modern grand-prix racing car of mass m is
travelling on a flat track in a circular arc of
radius R with a speed v. If the coefficient of
static friction between the tyres and the track
is m
s
, then the magnitude of negative lift F
L
acting downwards on the car is :
(Assume forces on the four tyres are identical
and g = acceleration due to gravity)
       
v
R
(1) 
æö
+
ç÷
m
èø
2
s
v
mg
R
(2) 
æö
-
ç÷
m
èø
2
s
v
mg
R
(3) 
æö
-
ç÷
m
èø
2
s
v
mg
R
(4) 
æö
-+
ç÷
m
èø
2
s
v
mg
R
Official Ans. by NTA (2)
3
Sol. m
s
N = 
2
mv
R
=
m
2
s
mv
N
R
 = mg + F
L
F
L
 = 
-
m
2
s
mv
mg
R
10. A car accelerates from rest at a constant rate a
for some time after which it decelerates at a
constant rate b to come to rest. If the total time
elapsed is t seconds, the total distance travelled
is :
(1) 
ab
a+b
2
4
t
()
(2) 
ab
a+b
2
2
t
()
(3) 
ab
a+b
2
t
2()
(4) 
ab
a+b
2
t
4()
Official Ans. by NTA (3)
Sol. v
0
 = at
1
 and 0 = v
0
 – bt
2
  Þ v
0
 = bt
2
t
1
 + t
2
 = t
æö
+=
ç÷
ab
èø
0
11
vt
      
v
v
0
time
t
Þ 
ab
=
a+b
0
t
v
Distance = area of v-t graph
= 
1
2
× t × v
0
 = 
1
2
 × t × 
ab
a+b
t
 = 
( )
ab
a+b
2
t
2
11. A solenoid of 1000 turns per metre has a core
with relative permeability 500. Insulated
windings of the solenoid carry an electric
current of 5A. The magnetic flux density
produced by the solenoid is :
(permeability of free space = 4p × 10
–7
 H/m)
(1) pT (2) 2 × 10
–3
 pT
(3) 
p
5
T (4) 10
–4
pT
Official Ans. by NTA (1)
Sol. B = mnI = m
0
m
r
nI
B = 4p × 10
–7
 × 500 × 1000 × 5
B = p Tesla
12. A mass M hangs on a massless rod of
length l which rotates at a constant angular
frequency. The mass M moves with steady
speed in a circular path of constant radius.
Assume that the system is in steady circular
motion with constant angular velocity w. The
angular momentum of M about point A is L
A
which lies in the positive z direction and the
angular momentum of M about B is L
B
. The
correct statement for this system is :
z
w
l
B
a
A
r
M
(1) L
A
 and L
B
 are both constant in magnitude
and direction
(2) L
B
 is constant in direction with varying
magnitude
(3) L
B
 is constant, both in magnitude and
direction
(4) L
A
 is constant, both in magnitude and
direction
Official Ans. by NTA (4)
Sol. We know, =´
r
r r
L m (r v)
Now with respect to A, we always get direction
of 
r
L along +ve z-axis and also constant
magnitude as mvr. But with respect to B, we
get constant magnitude but continuously
changing direction.
13. For what value of displacement the kinetic
energy and potential energy of a simple
harmonic oscillation become equal ?
(1) x = 0 (2) x = ± A
(3) x = ± 
A
2
(4) x = 
A
2
Official Ans. by NTA (3)
4
Sol. KE = PE
1
2
mw
2
 (A
2
 – x
2
) = 
1
2
 mw
2
 x
2
A
2
 – x
2
 = x
2
2x
2
 = A
2
x = ± 
A
2
14. A Carnot's engine working between 400 K and
800 K has a work output of 1200 J per cycle.
The amount of heat energy supplied to the
engine from the source in each cycle is :
(1) 3200 J (2) 1800 J
(3) 1600 J (4) 2400 J
Official Ans. by NTA (4)
Sol. h = 
=
22
11
TQ
TQ
 = 
-
1
1
QW
Q
     (Q W = Q
1
 – Q
2
)
=-
1
400 W
1
800 Q
= -=
1
W 11
1
Q 22
Q
1
 = 2W = 2400 J
15. The thickness at the centre of a plano convex
lens is 3 mm and the diameter is 6 cm. If the
speed of light in the material of the lens is
2 × 10
8
 ms
–1
. The focal length of the lens is
_____.
(1) 0.30 cm (2) 15 cm
(3) 1.5 cm (4) 30 cm
Official Ans. by NTA (4)
Sol. R
2
 = r
2
 + (R – t)
2
R
2
 = r
2
 + R
2
 + t
2
 – 2Rt      
R
r
t
Neglecting t
2
, we get
=
2
r
R
2t
\ 
1
f
= (m – 1) 
æö
-
ç÷
¥
èø
11
R
 = 
m-1
R
f = 
m-
R
1
 = 
m-
2
r
2t ( 1)
 = 
-
-
´
æö
´´ ´-
ç÷
èø
22
3
(3 10)
3
23101
2
-
-
´
=
´´
4
3
9 10
6101
 × 2
f = 0.3 m = 30 cm
16. The output of the given combination gates
represents :
A
B
Y
(1) XOR Gate (2) NAND Gate
(3) AND Gate (4) NOR Gate
Official Ans. by NTA (2)
Sol. By De Morgan's theorem, we have
A
B
Y
A
B
A·B = NAND
A·B
17. A boy is rolling a 0.5 kg ball on the frictionless
floor with the speed of 20 ms
–1
. The ball gets
deflected by an obstacle on the way. After
deflection it moves with 5% of its initial kinetic
energy. What is the speed of the ball now ?
(1) 19.0 ms
–1
(2) 4.47 ms
–1
(3) 14.41 ms
–1
(4) 1.00 ms
–1
Official Ans. by NTA (2)
Sol. Given, m = 0.5 kg  and  u = 20 m/s
Initial kinetic energy (k
i
) = 
2
1
mu
2
= 
1
2
× 0.5 × 20 × 20 = 100 J
After deflection it moves with 5% of k
i
\ k
f
 = 
5
100
 × k
i
  Þ 
5
100
 × 100
Þ k
f
 = 5 J
Now, let the final speed be 'v' m/s, then :
k
f
 = 5 = 
1
2
mv
2
Þ v
2
 = 20
Þ v = 
20
 = 4.47 m/s
18. Which level of the single ionized carbon has
the same energy as the ground state energy of
hydrogen atom?
(1) 1 (2) 6
(3) 4 (4) 8
Official Ans. by NTA (2)
5
Sol. Energy of H-atom is E = – 13.6 Z
2
/n
2
for H-atom Z = 1 & for ground state, n = 1
Þ E = – 13.6 ×
2
2
1
1
 = – 13.6 eV
Now for carbon atom (single ionised), Z = 6
E = – 13.6 
2
2
Z
n
 = – 13.6        (given)
Þ n
2
 = 6
2
 Þ n = 6
19. Two ideal polyatomic gases at temperatures
T
1
 and T
2
 are mixed so that there is no loss of
energy. If F
1
 and F
2
, m
1
 and m
2
, n
1
 and n
2
 be
the degrees of freedom, masses, number of
molecules of the first and second gas
respectively, the temperature of mixture of
these two gases is :
(1) 
+
+
1 1 22
12
nT nT
nn
(2) 
+
+
1 1 1 2 22
1 1 22
n FT n FT
n F nF
(3) 
+
+
1 1 1 2 22
12
n FT n FT
FF
(4) 
+
+
1 1 1 2 22
12
n FT n FT
nn
Official Ans. by NTA (2)
Sol. Let the final temperature of the mixture be T.
Since, there is no loss in energy.
DU = 0
Þ 
1
F
2
n
1
R DT + 
2
F
2
 n
2
RDT = 0
Þ 
1
F
2
n
1
R (T
1
 – T) + 
2
F
2
n
2
R (T
2
 – T) = 0
Þ T = 
+
+
11 1 222
1 1 22
F n RT F n RT
Fn R FnR
Þ 
+
+
1 1 1 2 22
1 1 22
FnT FnT
Fn Fn
20. A current of 10A exists in a wire of
crosssectional area of 5 mm
2
 with a drift
velocity of 2 × 10
–3
 ms
–1
. The number of free
electrons in each cubic meter of the wire is ___.
(1) 2 × 10
6
(2) 625 × 10
25
(3) 2 × 10
25
(4) 1 × 10
23
Official Ans. by NTA (2)
Sol. i = 10A, A = 5 mm
2
 = 5 × 10
–6
 m
2
and v
d
 = 2 × 10
–3
 m/s
We know,  i = neAvd
\ 10 = n × 1.6 × 10
–19
 × 5 × 10
–6
 × 2 × 10
–3
Þ n = 0.625 × 10
28
 = 625 × 10
25
SECTION-B
1. For VHF signal broadcasting, ____ km
2
 of
maximum service area will be covered by an
antenna tower of height 30m, if the receiving
antenna is placed at ground. Let radius of the
earth be 6400 km. (Round off to the Nearest
Integer) (Take p as 3.14)
Official Ans. by NTA (1206)
Sol. d = 
2Rh
A = pd
2
A = p2Rh
= 3.14 × 2 × 6400 × 
30
1000
A = 1205.76 km
2
A 
;
 1206 km
2
2. The angular speed of truck wheel is increased
from 900 rpm to 2460 rpm in 26 seconds. The
number of revolutions by the truck engine
during this time is _______.
(Assuming the acceleration to be uniform).
Official Ans. by NTA (728)
Sol. We know, 
w +w æö
q=
ç÷
èø
12
t
2
Let number of revolutions be N
\ 2pN = 2p 
æö +
´
ç÷
´
èø
900 2460
26
602
N = 728
3. The equivalent resistance of series combination
of two resistors is 's'. When they are connected
in parallel, the equivalent resistance is 'p'.
If s = np, then the minimum value for n is ___.
(Round off to the Nearest Integer)
Official Ans. by NTA (4)