JEE Main 2021 Physics March 16 Shift 1 Paper & Solutions | Mock Tests for JEE Main and Advanced 2025 PDF Download

 Page 1


 
 
16
th
 March 2021|Shift 1
SECTION – A 
 
1. A 25 m long antenna is mounted on an antenna tower. The height of the antenna tower is 75 
m. The wavelength (in meter) of the signal transmitted by this antenna would be : 
 (1) 200 (2) 400 (3) 100 (4) 300 
Sol. (3) 
 Given that, height of peak of antenna : H = 25 m. 
 As, we know that 
 ? ? 4H 
 ? ? ? ? ?4×25 
 ? ? ? ? ?100 m 
 ? ?(3) 
 
2. A block of mass m slides along a floor while a force of magnitude F is applied to it at an angle 
? ?as shown in figure. The coefficient of kinetic friction is ?
K
. Then, the block's acceleration 'a' is 
given by : 
 (g is acceleration due to gravity) 
  
?
?
F
 
 (1) 
? ?
? ? ? ? ?
? ?
? ?
K
F F
cos g sin
m m
 (2) 
K
F F
cos g sin
m m
? ?
? ? ? ? ?
? ?
? ?
 
 (3) 
K
F F
cos g sin
m m
? ?
? ? ? ? ?
? ?
? ?
 (4) 
K
F F
cos g sin
m m
? ?
? ? ? ? ? ?
? ?
? ?
 
Sol. (1) 
 Drawing the FBD of the block. 
  
N
?
F
Fsin ?
Fcos ?
mg 
?
K
N
^
a
 
 ? ?N = mg – Fsin ?  …(1) 
 Also, Fcos ? – ?
K
N = m ?a …(2) 
 Substituting the value of N from eq. (1) in eq. (2) 
 ? Fcos ? – ?
K
(mg–Fsin ?) = m ?a 
 ? 
K
F F
a cos g sin
m m
? ?
? ? ? ? ? ?
? ?
? ?
 
 ? (1) 
Page 2


 
 
16
th
 March 2021|Shift 1
SECTION – A 
 
1. A 25 m long antenna is mounted on an antenna tower. The height of the antenna tower is 75 
m. The wavelength (in meter) of the signal transmitted by this antenna would be : 
 (1) 200 (2) 400 (3) 100 (4) 300 
Sol. (3) 
 Given that, height of peak of antenna : H = 25 m. 
 As, we know that 
 ? ? 4H 
 ? ? ? ? ?4×25 
 ? ? ? ? ?100 m 
 ? ?(3) 
 
2. A block of mass m slides along a floor while a force of magnitude F is applied to it at an angle 
? ?as shown in figure. The coefficient of kinetic friction is ?
K
. Then, the block's acceleration 'a' is 
given by : 
 (g is acceleration due to gravity) 
  
?
?
F
 
 (1) 
? ?
? ? ? ? ?
? ?
? ?
K
F F
cos g sin
m m
 (2) 
K
F F
cos g sin
m m
? ?
? ? ? ? ?
? ?
? ?
 
 (3) 
K
F F
cos g sin
m m
? ?
? ? ? ? ?
? ?
? ?
 (4) 
K
F F
cos g sin
m m
? ?
? ? ? ? ? ?
? ?
? ?
 
Sol. (1) 
 Drawing the FBD of the block. 
  
N
?
F
Fsin ?
Fcos ?
mg 
?
K
N
^
a
 
 ? ?N = mg – Fsin ?  …(1) 
 Also, Fcos ? – ?
K
N = m ?a …(2) 
 Substituting the value of N from eq. (1) in eq. (2) 
 ? Fcos ? – ?
K
(mg–Fsin ?) = m ?a 
 ? 
K
F F
a cos g sin
m m
? ?
? ? ? ? ? ?
? ?
? ?
 
 ? (1) 
 
 
3. Four equal masses, m each are placed at the corners of a square of length(l) as shown in the 
figure. The moment of inertia of the system about an axis passing through A and parallel to DB 
would be : 
  
D
C
m
m
m
m
l 
l 
l 
l 
A B
 
 (1) ml
2
 (2) 3ml
2
 (3) 3 ml
2
 (4) 2 ml
2
 
Sol. (2) 
  
D
C m
m
m
m
l 
l
l
l
A
B
d
 
 
2 2
AC l l ? ? 
 AC l 2 ? 
 
l 2
d
2
? 
 
l
d
2
? ? 
 Moment of inertia about the axis passing through A : 
 I = m(O)
2
 + m(d)
2
 + m(d)
2
 + M(AC)
2
 
 ? I = O + m
2
l
2
? ?
? ?
? ?
 + m
2
l
2
? ?
? ?
? ?
+ m
? ?
2
l 2 
 
2 2
2
ml ml
I 2ml
2 2
? ? ? ? 
 ? I = 3ml
2
 
 ? (2) 
  
Page 3


 
 
16
th
 March 2021|Shift 1
SECTION – A 
 
1. A 25 m long antenna is mounted on an antenna tower. The height of the antenna tower is 75 
m. The wavelength (in meter) of the signal transmitted by this antenna would be : 
 (1) 200 (2) 400 (3) 100 (4) 300 
Sol. (3) 
 Given that, height of peak of antenna : H = 25 m. 
 As, we know that 
 ? ? 4H 
 ? ? ? ? ?4×25 
 ? ? ? ? ?100 m 
 ? ?(3) 
 
2. A block of mass m slides along a floor while a force of magnitude F is applied to it at an angle 
? ?as shown in figure. The coefficient of kinetic friction is ?
K
. Then, the block's acceleration 'a' is 
given by : 
 (g is acceleration due to gravity) 
  
?
?
F
 
 (1) 
? ?
? ? ? ? ?
? ?
? ?
K
F F
cos g sin
m m
 (2) 
K
F F
cos g sin
m m
? ?
? ? ? ? ?
? ?
? ?
 
 (3) 
K
F F
cos g sin
m m
? ?
? ? ? ? ?
? ?
? ?
 (4) 
K
F F
cos g sin
m m
? ?
? ? ? ? ? ?
? ?
? ?
 
Sol. (1) 
 Drawing the FBD of the block. 
  
N
?
F
Fsin ?
Fcos ?
mg 
?
K
N
^
a
 
 ? ?N = mg – Fsin ?  …(1) 
 Also, Fcos ? – ?
K
N = m ?a …(2) 
 Substituting the value of N from eq. (1) in eq. (2) 
 ? Fcos ? – ?
K
(mg–Fsin ?) = m ?a 
 ? 
K
F F
a cos g sin
m m
? ?
? ? ? ? ? ?
? ?
? ?
 
 ? (1) 
 
 
3. Four equal masses, m each are placed at the corners of a square of length(l) as shown in the 
figure. The moment of inertia of the system about an axis passing through A and parallel to DB 
would be : 
  
D
C
m
m
m
m
l 
l 
l 
l 
A B
 
 (1) ml
2
 (2) 3ml
2
 (3) 3 ml
2
 (4) 2 ml
2
 
Sol. (2) 
  
D
C m
m
m
m
l 
l
l
l
A
B
d
 
 
2 2
AC l l ? ? 
 AC l 2 ? 
 
l 2
d
2
? 
 
l
d
2
? ? 
 Moment of inertia about the axis passing through A : 
 I = m(O)
2
 + m(d)
2
 + m(d)
2
 + M(AC)
2
 
 ? I = O + m
2
l
2
? ?
? ?
? ?
 + m
2
l
2
? ?
? ?
? ?
+ m
? ?
2
l 2 
 
2 2
2
ml ml
I 2ml
2 2
? ? ? ? 
 ? I = 3ml
2
 
 ? (2) 
  
 
 
16
th
 March 2021|Shift 1
 
4. The stopping potential in the context of photoelectric effect depends on the following property 
of incident electromagnetic radiation : 
 (1) Amplitude (2) Phase (3) Frequency (4) Intensity 
Sol. (3) 
 Stopping potential depends on frequency, according to Einstein's photoelectric equation. 
 h ? – h ?
0
 = eV 
 
0
h h
V
e e
? ? ? ? ? 
 ? (3) 
 
5. One main scale division of a vernier callipers is 'a' cm and n
th
 division of the vernier scale 
coincide with (n–1)
th
 division of the main scale. The least count of the callipers in mm is : 
 (1) 
n 1
a
10n
? ? ?
? ?
? ?
 (2) 
10a
n
 (3) 
? ?
10na
n 1 ?
 (4) 
? ?
10a
n 1 ?
 
Sol. (2) 
 n VSD = (n–1) MSD 
 
n 1
1 VSD MSD
n
? ? ?
?
? ?
? ?
 
 L ?C=1 MSD – 1 VSD 
 
n 1
1MSD MSD
n
? ? ?
? ?
? ?
? ?
 
 
MSD
1MSD 1MSD
n
? ? ? 
 
MSD
n
? 
 
a
cm
n
? 
 
10a
mm
n
? 
 ? (2) 
 
6. A plane electromagnetic wave of frequency 500 MHz is travelling in vacuum along y-direction. 
At a particular point in space and time, 
? 8
B 8.0 10 zT
?
? ?
?
. The value of electric field at this point 
is : 
 (speed of light = 3×10
8
 ms
–1
) 
 
? ? ?
x,y,z are unit vectors along x, y and z directions. 
 (1) 2.6
?
x V/m (2) –2.6
?
y V/m (3) 24
?
x V/m (4) –24
?
x V/m 
Sol. (4) 
 E
0
 = B ?C 
 E
0
 = (8×10
–8
)×(3×10
8
) 
 ? E
0
 = 24 
 Direction of wave travelling is in E B ?
? ? ?
 
 So 
?
? ?
? ?
x z y ? ? ? ? 
 ? 
?
E = –24
?
x V/m 
 ? (4) 
 
Page 4


 
 
16
th
 March 2021|Shift 1
SECTION – A 
 
1. A 25 m long antenna is mounted on an antenna tower. The height of the antenna tower is 75 
m. The wavelength (in meter) of the signal transmitted by this antenna would be : 
 (1) 200 (2) 400 (3) 100 (4) 300 
Sol. (3) 
 Given that, height of peak of antenna : H = 25 m. 
 As, we know that 
 ? ? 4H 
 ? ? ? ? ?4×25 
 ? ? ? ? ?100 m 
 ? ?(3) 
 
2. A block of mass m slides along a floor while a force of magnitude F is applied to it at an angle 
? ?as shown in figure. The coefficient of kinetic friction is ?
K
. Then, the block's acceleration 'a' is 
given by : 
 (g is acceleration due to gravity) 
  
?
?
F
 
 (1) 
? ?
? ? ? ? ?
? ?
? ?
K
F F
cos g sin
m m
 (2) 
K
F F
cos g sin
m m
? ?
? ? ? ? ?
? ?
? ?
 
 (3) 
K
F F
cos g sin
m m
? ?
? ? ? ? ?
? ?
? ?
 (4) 
K
F F
cos g sin
m m
? ?
? ? ? ? ? ?
? ?
? ?
 
Sol. (1) 
 Drawing the FBD of the block. 
  
N
?
F
Fsin ?
Fcos ?
mg 
?
K
N
^
a
 
 ? ?N = mg – Fsin ?  …(1) 
 Also, Fcos ? – ?
K
N = m ?a …(2) 
 Substituting the value of N from eq. (1) in eq. (2) 
 ? Fcos ? – ?
K
(mg–Fsin ?) = m ?a 
 ? 
K
F F
a cos g sin
m m
? ?
? ? ? ? ? ?
? ?
? ?
 
 ? (1) 
 
 
3. Four equal masses, m each are placed at the corners of a square of length(l) as shown in the 
figure. The moment of inertia of the system about an axis passing through A and parallel to DB 
would be : 
  
D
C
m
m
m
m
l 
l 
l 
l 
A B
 
 (1) ml
2
 (2) 3ml
2
 (3) 3 ml
2
 (4) 2 ml
2
 
Sol. (2) 
  
D
C m
m
m
m
l 
l
l
l
A
B
d
 
 
2 2
AC l l ? ? 
 AC l 2 ? 
 
l 2
d
2
? 
 
l
d
2
? ? 
 Moment of inertia about the axis passing through A : 
 I = m(O)
2
 + m(d)
2
 + m(d)
2
 + M(AC)
2
 
 ? I = O + m
2
l
2
? ?
? ?
? ?
 + m
2
l
2
? ?
? ?
? ?
+ m
? ?
2
l 2 
 
2 2
2
ml ml
I 2ml
2 2
? ? ? ? 
 ? I = 3ml
2
 
 ? (2) 
  
 
 
16
th
 March 2021|Shift 1
 
4. The stopping potential in the context of photoelectric effect depends on the following property 
of incident electromagnetic radiation : 
 (1) Amplitude (2) Phase (3) Frequency (4) Intensity 
Sol. (3) 
 Stopping potential depends on frequency, according to Einstein's photoelectric equation. 
 h ? – h ?
0
 = eV 
 
0
h h
V
e e
? ? ? ? ? 
 ? (3) 
 
5. One main scale division of a vernier callipers is 'a' cm and n
th
 division of the vernier scale 
coincide with (n–1)
th
 division of the main scale. The least count of the callipers in mm is : 
 (1) 
n 1
a
10n
? ? ?
? ?
? ?
 (2) 
10a
n
 (3) 
? ?
10na
n 1 ?
 (4) 
? ?
10a
n 1 ?
 
Sol. (2) 
 n VSD = (n–1) MSD 
 
n 1
1 VSD MSD
n
? ? ?
?
? ?
? ?
 
 L ?C=1 MSD – 1 VSD 
 
n 1
1MSD MSD
n
? ? ?
? ?
? ?
? ?
 
 
MSD
1MSD 1MSD
n
? ? ? 
 
MSD
n
? 
 
a
cm
n
? 
 
10a
mm
n
? 
 ? (2) 
 
6. A plane electromagnetic wave of frequency 500 MHz is travelling in vacuum along y-direction. 
At a particular point in space and time, 
? 8
B 8.0 10 zT
?
? ?
?
. The value of electric field at this point 
is : 
 (speed of light = 3×10
8
 ms
–1
) 
 
? ? ?
x,y,z are unit vectors along x, y and z directions. 
 (1) 2.6
?
x V/m (2) –2.6
?
y V/m (3) 24
?
x V/m (4) –24
?
x V/m 
Sol. (4) 
 E
0
 = B ?C 
 E
0
 = (8×10
–8
)×(3×10
8
) 
 ? E
0
 = 24 
 Direction of wave travelling is in E B ?
? ? ?
 
 So 
?
? ?
? ?
x z y ? ? ? ? 
 ? 
?
E = –24
?
x V/m 
 ? (4) 
 
 
 
7. The maximum and minimum distances of a comet from the Sun are 1.6×10
12
 m and 8.0×10
10
 
m respectively. If the speed of the comet at the nearest point is 6×10
4
 ms
–1
, the speed at the 
farthest point is : 
 (1) 1.5×10
3
 m/s (2) 4.5×10
3
 m/s (3) 3.0×10
3
 m/s (4) 6.0×10
3
 m/s 
Sol. (3) 
  
r
1
r
2
v
1
=6×10
4
m/s
2
 
v
2
=?
1
 
 Let point 1 is nearest point, 
 and point 2 is farthest point. 
 Given, r
1
 = 8×10
10
m & r
2
 = 1.6×10
12
m 
 By angular momentum conservation 
 L
1
 = L
2
 
 mr
1
v
1
= mr
2
v
2
 
 
1 1
2
2
r v
v
r
? ? 
 
10 4
2 12
8 10 6 10
v
1.6 10
? ? ?
? ?
?
 
 ?v
2
 = 3.0×10
3
 m/s 
 ? ?(3) 
 
8. A block of 200 g mass moves with a uniform speed in a horizontal circular groove, with vertical 
side walls of radius 20 cm. If the block takes 40 s to complete one round, the normal force by 
the side walls of the groove is : 
 (1) 6.28×10
–3
 N (2) 0.0314 N (3) 9.859×10
–2
 N (4) 9.859×10
–4
 N 
Sol. (4) 
 Nsormal force will provide the necessary centripetal force. 
 ? N=m ?
2
R 
 Also; 
2
T
?
? ?
 
 
? ? ? ?
2
2
4
N 0.2 0.2
T
? ? ?
?
? ?
? ?
 
 
? ?
? ?
2
2
4 3.14
N 0.2 0.2
40
?
? ? ? ? 
 ? N=9.859×10
–4 
N 
 ? (4) 
  
Page 5


 
 
16
th
 March 2021|Shift 1
SECTION – A 
 
1. A 25 m long antenna is mounted on an antenna tower. The height of the antenna tower is 75 
m. The wavelength (in meter) of the signal transmitted by this antenna would be : 
 (1) 200 (2) 400 (3) 100 (4) 300 
Sol. (3) 
 Given that, height of peak of antenna : H = 25 m. 
 As, we know that 
 ? ? 4H 
 ? ? ? ? ?4×25 
 ? ? ? ? ?100 m 
 ? ?(3) 
 
2. A block of mass m slides along a floor while a force of magnitude F is applied to it at an angle 
? ?as shown in figure. The coefficient of kinetic friction is ?
K
. Then, the block's acceleration 'a' is 
given by : 
 (g is acceleration due to gravity) 
  
?
?
F
 
 (1) 
? ?
? ? ? ? ?
? ?
? ?
K
F F
cos g sin
m m
 (2) 
K
F F
cos g sin
m m
? ?
? ? ? ? ?
? ?
? ?
 
 (3) 
K
F F
cos g sin
m m
? ?
? ? ? ? ?
? ?
? ?
 (4) 
K
F F
cos g sin
m m
? ?
? ? ? ? ? ?
? ?
? ?
 
Sol. (1) 
 Drawing the FBD of the block. 
  
N
?
F
Fsin ?
Fcos ?
mg 
?
K
N
^
a
 
 ? ?N = mg – Fsin ?  …(1) 
 Also, Fcos ? – ?
K
N = m ?a …(2) 
 Substituting the value of N from eq. (1) in eq. (2) 
 ? Fcos ? – ?
K
(mg–Fsin ?) = m ?a 
 ? 
K
F F
a cos g sin
m m
? ?
? ? ? ? ? ?
? ?
? ?
 
 ? (1) 
 
 
3. Four equal masses, m each are placed at the corners of a square of length(l) as shown in the 
figure. The moment of inertia of the system about an axis passing through A and parallel to DB 
would be : 
  
D
C
m
m
m
m
l 
l 
l 
l 
A B
 
 (1) ml
2
 (2) 3ml
2
 (3) 3 ml
2
 (4) 2 ml
2
 
Sol. (2) 
  
D
C m
m
m
m
l 
l
l
l
A
B
d
 
 
2 2
AC l l ? ? 
 AC l 2 ? 
 
l 2
d
2
? 
 
l
d
2
? ? 
 Moment of inertia about the axis passing through A : 
 I = m(O)
2
 + m(d)
2
 + m(d)
2
 + M(AC)
2
 
 ? I = O + m
2
l
2
? ?
? ?
? ?
 + m
2
l
2
? ?
? ?
? ?
+ m
? ?
2
l 2 
 
2 2
2
ml ml
I 2ml
2 2
? ? ? ? 
 ? I = 3ml
2
 
 ? (2) 
  
 
 
16
th
 March 2021|Shift 1
 
4. The stopping potential in the context of photoelectric effect depends on the following property 
of incident electromagnetic radiation : 
 (1) Amplitude (2) Phase (3) Frequency (4) Intensity 
Sol. (3) 
 Stopping potential depends on frequency, according to Einstein's photoelectric equation. 
 h ? – h ?
0
 = eV 
 
0
h h
V
e e
? ? ? ? ? 
 ? (3) 
 
5. One main scale division of a vernier callipers is 'a' cm and n
th
 division of the vernier scale 
coincide with (n–1)
th
 division of the main scale. The least count of the callipers in mm is : 
 (1) 
n 1
a
10n
? ? ?
? ?
? ?
 (2) 
10a
n
 (3) 
? ?
10na
n 1 ?
 (4) 
? ?
10a
n 1 ?
 
Sol. (2) 
 n VSD = (n–1) MSD 
 
n 1
1 VSD MSD
n
? ? ?
?
? ?
? ?
 
 L ?C=1 MSD – 1 VSD 
 
n 1
1MSD MSD
n
? ? ?
? ?
? ?
? ?
 
 
MSD
1MSD 1MSD
n
? ? ? 
 
MSD
n
? 
 
a
cm
n
? 
 
10a
mm
n
? 
 ? (2) 
 
6. A plane electromagnetic wave of frequency 500 MHz is travelling in vacuum along y-direction. 
At a particular point in space and time, 
? 8
B 8.0 10 zT
?
? ?
?
. The value of electric field at this point 
is : 
 (speed of light = 3×10
8
 ms
–1
) 
 
? ? ?
x,y,z are unit vectors along x, y and z directions. 
 (1) 2.6
?
x V/m (2) –2.6
?
y V/m (3) 24
?
x V/m (4) –24
?
x V/m 
Sol. (4) 
 E
0
 = B ?C 
 E
0
 = (8×10
–8
)×(3×10
8
) 
 ? E
0
 = 24 
 Direction of wave travelling is in E B ?
? ? ?
 
 So 
?
? ?
? ?
x z y ? ? ? ? 
 ? 
?
E = –24
?
x V/m 
 ? (4) 
 
 
 
7. The maximum and minimum distances of a comet from the Sun are 1.6×10
12
 m and 8.0×10
10
 
m respectively. If the speed of the comet at the nearest point is 6×10
4
 ms
–1
, the speed at the 
farthest point is : 
 (1) 1.5×10
3
 m/s (2) 4.5×10
3
 m/s (3) 3.0×10
3
 m/s (4) 6.0×10
3
 m/s 
Sol. (3) 
  
r
1
r
2
v
1
=6×10
4
m/s
2
 
v
2
=?
1
 
 Let point 1 is nearest point, 
 and point 2 is farthest point. 
 Given, r
1
 = 8×10
10
m & r
2
 = 1.6×10
12
m 
 By angular momentum conservation 
 L
1
 = L
2
 
 mr
1
v
1
= mr
2
v
2
 
 
1 1
2
2
r v
v
r
? ? 
 
10 4
2 12
8 10 6 10
v
1.6 10
? ? ?
? ?
?
 
 ?v
2
 = 3.0×10
3
 m/s 
 ? ?(3) 
 
8. A block of 200 g mass moves with a uniform speed in a horizontal circular groove, with vertical 
side walls of radius 20 cm. If the block takes 40 s to complete one round, the normal force by 
the side walls of the groove is : 
 (1) 6.28×10
–3
 N (2) 0.0314 N (3) 9.859×10
–2
 N (4) 9.859×10
–4
 N 
Sol. (4) 
 Nsormal force will provide the necessary centripetal force. 
 ? N=m ?
2
R 
 Also; 
2
T
?
? ?
 
 
? ? ? ?
2
2
4
N 0.2 0.2
T
? ? ?
?
? ?
? ?
 
 
? ?
? ?
2
2
4 3.14
N 0.2 0.2
40
?
? ? ? ? 
 ? N=9.859×10
–4 
N 
 ? (4) 
  
 
 
16
th
 March 2021|Shift 1
9. An RC circuit as shown in the figure is driven by a AC source generating a square wave. The 
output wave pattern monitored by CRO would look close to : 
  
R
C
 
~
 
CRO
 
 (1)
 
 
 (2) 
 
 
 (3) 
 
 
 (4) 
 
 
Sol. (2) 
 Assuming AC start with positive voltage, when +ve voltage is across input then the capacitor 
start charging, trying to reach saturation value, till there is +ve voltage across input, when –ve 
voltage of AC appears across input, the capacitor starts discharging till there is –ve voltage 
across input and this process of charging and discharging keeps on going alternatively. 
  
+ –
For charging 
+
–
+
– 
increasing current
 
+ –
For discharging
+
–
– 
+
decreasing current
 
  
charging
discharging
 
 ? (2)