JEE Mains 18 March 2021 Question Paper Shift 2 | JEE Main & Advanced Previous Year Papers PDF Download

 Page 1


1
SECTION-A
1. Which of the following statements are correct?
(A) Electric monopoles do not exist whereas
magnetic monopoles exist.
(B) Magnetic field lines due to a solenoid at its
ends and outside cannot be completely
straight and confined.
(C) Magnetic field lines are completely confined
within a toroid.
(D) Magnetic field lines inside a bar magnet are
not parallel.
(E) c = – 1 is the condition for a perfect
diamagnetic material, where c is its
magnetic susceptibility.
Choose the correct answer from the options
given below :
(1) (C) and (E) only
(2) (B) and (D) only
(3) (A) and (B) only
(4) (B) and (C) only
Official Ans. by NTA (1)
Sol. Statement (C) is correct because, the magnetic
field outside the toroid is zero and they form
closed loops inside the toroid itself.
Statement (E) is correct because we know that
super conductors are materials inside which the
net magnetic field is always zero and they are
perfect diamagnetic.
µ
r
 = 1 + c
c = –1
µ
r
 = 0
For superconductors.
2. An object of mass m
1
 collides with another
object of mass m
2
, which is at rest. After the
collision the objects move with equal speeds
in opposite direction. The ratio of the masses
m
2
 : m
1
 is :
(1) 3 : 1 (2) 2 : 1
(3) 1 : 2 (4) 1 : 1
Official Ans. by NTA (1)
FINAL JEE–MAIN EXAMINATION – MARCH, 2021
(Held On Thursday 18
th
 March, 2021)    TIME : 3 : 00 PM   to  6 : 00 PM
PHYSICS TEST PAPER WITH ANSWER & SOLUTION
Sol.
m
1
m
2
v
1
v v
m
1
m
2
m
1
v
1
 = –m
1
v + m
2
v
=-+
2
1
1
m
v vv
m
( ) +
=
1
2
1
vv
m
vm
==
1
2v
e1
v
=
1
v
v
2
+
=
112
11
v v/2m
v/2m
=
2
1
m
3
m
3. For an adiabatic expansion of an ideal gas, the
fractional change in its pressure is equal to
(where g is the ratio of specific heats):
(1)
-g
dV
V
(2)
-g
V
dV
(3)
-
g
1 dV
V
(4) 
dV
V
Official Ans. by NTA (1)
Sol. PV
g
 = constant
Differentiating
g
=-
dPP
dVV
g
=-
dP dV
PV
Page 2


1
SECTION-A
1. Which of the following statements are correct?
(A) Electric monopoles do not exist whereas
magnetic monopoles exist.
(B) Magnetic field lines due to a solenoid at its
ends and outside cannot be completely
straight and confined.
(C) Magnetic field lines are completely confined
within a toroid.
(D) Magnetic field lines inside a bar magnet are
not parallel.
(E) c = – 1 is the condition for a perfect
diamagnetic material, where c is its
magnetic susceptibility.
Choose the correct answer from the options
given below :
(1) (C) and (E) only
(2) (B) and (D) only
(3) (A) and (B) only
(4) (B) and (C) only
Official Ans. by NTA (1)
Sol. Statement (C) is correct because, the magnetic
field outside the toroid is zero and they form
closed loops inside the toroid itself.
Statement (E) is correct because we know that
super conductors are materials inside which the
net magnetic field is always zero and they are
perfect diamagnetic.
µ
r
 = 1 + c
c = –1
µ
r
 = 0
For superconductors.
2. An object of mass m
1
 collides with another
object of mass m
2
, which is at rest. After the
collision the objects move with equal speeds
in opposite direction. The ratio of the masses
m
2
 : m
1
 is :
(1) 3 : 1 (2) 2 : 1
(3) 1 : 2 (4) 1 : 1
Official Ans. by NTA (1)
FINAL JEE–MAIN EXAMINATION – MARCH, 2021
(Held On Thursday 18
th
 March, 2021)    TIME : 3 : 00 PM   to  6 : 00 PM
PHYSICS TEST PAPER WITH ANSWER & SOLUTION
Sol.
m
1
m
2
v
1
v v
m
1
m
2
m
1
v
1
 = –m
1
v + m
2
v
=-+
2
1
1
m
v vv
m
( ) +
=
1
2
1
vv
m
vm
==
1
2v
e1
v
=
1
v
v
2
+
=
112
11
v v/2m
v/2m
=
2
1
m
3
m
3. For an adiabatic expansion of an ideal gas, the
fractional change in its pressure is equal to
(where g is the ratio of specific heats):
(1)
-g
dV
V
(2)
-g
V
dV
(3)
-
g
1 dV
V
(4) 
dV
V
Official Ans. by NTA (1)
Sol. PV
g
 = constant
Differentiating
g
=-
dPP
dVV
g
=-
dP dV
PV
2
4. A proton and an a-particle, having kinetic
energies K
p 
and K
a
, respectively, enter into a
magnetic field at right angles.
The ratio of the radii of trajectory of proton to
that of a-particle is 2 : 1. The ratio of K
p
 : K
a
is :
(1) 1 : 8 (2) 8 : 1
(3) 1 : 4 (4) 4 : 1
Official Ans. by NTA (4)
Sol.
==
mvp
r
qB qB
a
=
p
m
4
m
a
aa
==
pp
p
rp
q 2
r qp1
aa
æö
==
ç÷
èø
pp
p 2q
1
2
p q2
a
=
p
p
1
p
a
a a
=
2
pp
p
p
Kp
m
Km p
 = (1) (4)
5. A plane electromagnetic wave propagating
along y-direction can have the following pair
of electric field ( )
E
r
 and magnetic field ( )
B
r
components.
(1) E
y
, B
y
 or E
z
, B
z
(2) E
y
, B
x
 or E
x
, B
y
(3) E
x
, B
z
 or E
z
, B
x
(4) E
x
, B
y
 or E
y
, B
x
Official Ans. by NTA (3)
Sol.
E
z
y
x
B
K
B
z
y
x
E
K
6. Consider a uniform wire of mass M and length
L. It is bent into a semicircle. Its moment of
inertia about a line perpendicular to the plane
of the wire passing through the centre is :
(1) 
p
2
2
1 ML
4
(2) 
p
2
2
2 ML
5
(3) 
p
2
2
ML
(4) 
p
2
2
1 ML
2
Official Ans. by NTA (3)
Sol.
p= Þ=
p
L
r Lr
==
p
2
2
2
ML
I Mr
7. The velocity-displacement graph of a particle
is shown in the figure.
x
0
x
v
v
0
O
The acceleration-displacement graph of the
same particle is represented by :
(1) 
x
a
O
(2) 
x
a
O
(3) 
x
a
O
(4) 
x
a
O
Official Ans. by NTA (3)
Sol.
æö
=-+
ç÷
èø
0
0
0
v
v xv
x
Page 3


1
SECTION-A
1. Which of the following statements are correct?
(A) Electric monopoles do not exist whereas
magnetic monopoles exist.
(B) Magnetic field lines due to a solenoid at its
ends and outside cannot be completely
straight and confined.
(C) Magnetic field lines are completely confined
within a toroid.
(D) Magnetic field lines inside a bar magnet are
not parallel.
(E) c = – 1 is the condition for a perfect
diamagnetic material, where c is its
magnetic susceptibility.
Choose the correct answer from the options
given below :
(1) (C) and (E) only
(2) (B) and (D) only
(3) (A) and (B) only
(4) (B) and (C) only
Official Ans. by NTA (1)
Sol. Statement (C) is correct because, the magnetic
field outside the toroid is zero and they form
closed loops inside the toroid itself.
Statement (E) is correct because we know that
super conductors are materials inside which the
net magnetic field is always zero and they are
perfect diamagnetic.
µ
r
 = 1 + c
c = –1
µ
r
 = 0
For superconductors.
2. An object of mass m
1
 collides with another
object of mass m
2
, which is at rest. After the
collision the objects move with equal speeds
in opposite direction. The ratio of the masses
m
2
 : m
1
 is :
(1) 3 : 1 (2) 2 : 1
(3) 1 : 2 (4) 1 : 1
Official Ans. by NTA (1)
FINAL JEE–MAIN EXAMINATION – MARCH, 2021
(Held On Thursday 18
th
 March, 2021)    TIME : 3 : 00 PM   to  6 : 00 PM
PHYSICS TEST PAPER WITH ANSWER & SOLUTION
Sol.
m
1
m
2
v
1
v v
m
1
m
2
m
1
v
1
 = –m
1
v + m
2
v
=-+
2
1
1
m
v vv
m
( ) +
=
1
2
1
vv
m
vm
==
1
2v
e1
v
=
1
v
v
2
+
=
112
11
v v/2m
v/2m
=
2
1
m
3
m
3. For an adiabatic expansion of an ideal gas, the
fractional change in its pressure is equal to
(where g is the ratio of specific heats):
(1)
-g
dV
V
(2)
-g
V
dV
(3)
-
g
1 dV
V
(4) 
dV
V
Official Ans. by NTA (1)
Sol. PV
g
 = constant
Differentiating
g
=-
dPP
dVV
g
=-
dP dV
PV
2
4. A proton and an a-particle, having kinetic
energies K
p 
and K
a
, respectively, enter into a
magnetic field at right angles.
The ratio of the radii of trajectory of proton to
that of a-particle is 2 : 1. The ratio of K
p
 : K
a
is :
(1) 1 : 8 (2) 8 : 1
(3) 1 : 4 (4) 4 : 1
Official Ans. by NTA (4)
Sol.
==
mvp
r
qB qB
a
=
p
m
4
m
a
aa
==
pp
p
rp
q 2
r qp1
aa
æö
==
ç÷
èø
pp
p 2q
1
2
p q2
a
=
p
p
1
p
a
a a
=
2
pp
p
p
Kp
m
Km p
 = (1) (4)
5. A plane electromagnetic wave propagating
along y-direction can have the following pair
of electric field ( )
E
r
 and magnetic field ( )
B
r
components.
(1) E
y
, B
y
 or E
z
, B
z
(2) E
y
, B
x
 or E
x
, B
y
(3) E
x
, B
z
 or E
z
, B
x
(4) E
x
, B
y
 or E
y
, B
x
Official Ans. by NTA (3)
Sol.
E
z
y
x
B
K
B
z
y
x
E
K
6. Consider a uniform wire of mass M and length
L. It is bent into a semicircle. Its moment of
inertia about a line perpendicular to the plane
of the wire passing through the centre is :
(1) 
p
2
2
1 ML
4
(2) 
p
2
2
2 ML
5
(3) 
p
2
2
ML
(4) 
p
2
2
1 ML
2
Official Ans. by NTA (3)
Sol.
p= Þ=
p
L
r Lr
==
p
2
2
2
ML
I Mr
7. The velocity-displacement graph of a particle
is shown in the figure.
x
0
x
v
v
0
O
The acceleration-displacement graph of the
same particle is represented by :
(1) 
x
a
O
(2) 
x
a
O
(3) 
x
a
O
(4) 
x
a
O
Official Ans. by NTA (3)
Sol.
æö
=-+
ç÷
èø
0
0
0
v
v xv
x
3
=
vdv
a
dx
éù æ ö éù
=- +-
êú ç÷ êú
êú è ø ëû ëû
00
0
00
vv
a xv
xx
æö
=-
ç÷
èø
2
2
00
00
vv
ax
xx
8. The correct relation between a (ratio of
collector current to emitter current) and b (ratio
of collector current to base current) of a
transistor is :
(1) 
a
b=
+a 1
(2) 
b
a=
-a 1
(3) 
b=
-a
1
1
(4) 
b
a=
+b 1
Official Ans. by NTA (4)
Sol.
a= b=
CC
EB
II
,
II
I
E
 = I
B
 + I
C
a==
+
+
C
B
BC
C
I 1
I
II
1
I
a=
+
b
1
1
1
b
a=
+b 1
9. Three rays of light, namely red (R), green (G)
and blue (B) are incident on the face PQ of a
right angled prism PQR as shown in figure.
P
R
Q
B
G
R
The refractive indices of the material of the
prism for red, green and blue wavelength are
1.27, 1.42 and 1.49 respectively. The colour
of the ray(s) emerging out of the face PR is :
(1) green (2) red
(3) blue and green (4) blue
Official Ans. by NTA (2)
Sol.
P
R
Q
45°
Assuming that the right angled prism is an
isoceles prism, so the other angles will be 45°
each.
Þ Each incident ray will make an angle of 45°
with the normal at face PR.
Þ The wavelength corresponding to which the
incidence angle is less than the critical angle,
will pass through PR.
Þ q
C
 = critical angle
Þ 
-
æö
q=
ç÷
m
èø
1
C
1
sin
Þ If q
C
 ³ 45°
the light ray will pass
Þ 
()
-
æö
q = =°
ç÷
èø
1
C
Red
1
sin 51.94
1.27
Page 4


1
SECTION-A
1. Which of the following statements are correct?
(A) Electric monopoles do not exist whereas
magnetic monopoles exist.
(B) Magnetic field lines due to a solenoid at its
ends and outside cannot be completely
straight and confined.
(C) Magnetic field lines are completely confined
within a toroid.
(D) Magnetic field lines inside a bar magnet are
not parallel.
(E) c = – 1 is the condition for a perfect
diamagnetic material, where c is its
magnetic susceptibility.
Choose the correct answer from the options
given below :
(1) (C) and (E) only
(2) (B) and (D) only
(3) (A) and (B) only
(4) (B) and (C) only
Official Ans. by NTA (1)
Sol. Statement (C) is correct because, the magnetic
field outside the toroid is zero and they form
closed loops inside the toroid itself.
Statement (E) is correct because we know that
super conductors are materials inside which the
net magnetic field is always zero and they are
perfect diamagnetic.
µ
r
 = 1 + c
c = –1
µ
r
 = 0
For superconductors.
2. An object of mass m
1
 collides with another
object of mass m
2
, which is at rest. After the
collision the objects move with equal speeds
in opposite direction. The ratio of the masses
m
2
 : m
1
 is :
(1) 3 : 1 (2) 2 : 1
(3) 1 : 2 (4) 1 : 1
Official Ans. by NTA (1)
FINAL JEE–MAIN EXAMINATION – MARCH, 2021
(Held On Thursday 18
th
 March, 2021)    TIME : 3 : 00 PM   to  6 : 00 PM
PHYSICS TEST PAPER WITH ANSWER & SOLUTION
Sol.
m
1
m
2
v
1
v v
m
1
m
2
m
1
v
1
 = –m
1
v + m
2
v
=-+
2
1
1
m
v vv
m
( ) +
=
1
2
1
vv
m
vm
==
1
2v
e1
v
=
1
v
v
2
+
=
112
11
v v/2m
v/2m
=
2
1
m
3
m
3. For an adiabatic expansion of an ideal gas, the
fractional change in its pressure is equal to
(where g is the ratio of specific heats):
(1)
-g
dV
V
(2)
-g
V
dV
(3)
-
g
1 dV
V
(4) 
dV
V
Official Ans. by NTA (1)
Sol. PV
g
 = constant
Differentiating
g
=-
dPP
dVV
g
=-
dP dV
PV
2
4. A proton and an a-particle, having kinetic
energies K
p 
and K
a
, respectively, enter into a
magnetic field at right angles.
The ratio of the radii of trajectory of proton to
that of a-particle is 2 : 1. The ratio of K
p
 : K
a
is :
(1) 1 : 8 (2) 8 : 1
(3) 1 : 4 (4) 4 : 1
Official Ans. by NTA (4)
Sol.
==
mvp
r
qB qB
a
=
p
m
4
m
a
aa
==
pp
p
rp
q 2
r qp1
aa
æö
==
ç÷
èø
pp
p 2q
1
2
p q2
a
=
p
p
1
p
a
a a
=
2
pp
p
p
Kp
m
Km p
 = (1) (4)
5. A plane electromagnetic wave propagating
along y-direction can have the following pair
of electric field ( )
E
r
 and magnetic field ( )
B
r
components.
(1) E
y
, B
y
 or E
z
, B
z
(2) E
y
, B
x
 or E
x
, B
y
(3) E
x
, B
z
 or E
z
, B
x
(4) E
x
, B
y
 or E
y
, B
x
Official Ans. by NTA (3)
Sol.
E
z
y
x
B
K
B
z
y
x
E
K
6. Consider a uniform wire of mass M and length
L. It is bent into a semicircle. Its moment of
inertia about a line perpendicular to the plane
of the wire passing through the centre is :
(1) 
p
2
2
1 ML
4
(2) 
p
2
2
2 ML
5
(3) 
p
2
2
ML
(4) 
p
2
2
1 ML
2
Official Ans. by NTA (3)
Sol.
p= Þ=
p
L
r Lr
==
p
2
2
2
ML
I Mr
7. The velocity-displacement graph of a particle
is shown in the figure.
x
0
x
v
v
0
O
The acceleration-displacement graph of the
same particle is represented by :
(1) 
x
a
O
(2) 
x
a
O
(3) 
x
a
O
(4) 
x
a
O
Official Ans. by NTA (3)
Sol.
æö
=-+
ç÷
èø
0
0
0
v
v xv
x
3
=
vdv
a
dx
éù æ ö éù
=- +-
êú ç÷ êú
êú è ø ëû ëû
00
0
00
vv
a xv
xx
æö
=-
ç÷
èø
2
2
00
00
vv
ax
xx
8. The correct relation between a (ratio of
collector current to emitter current) and b (ratio
of collector current to base current) of a
transistor is :
(1) 
a
b=
+a 1
(2) 
b
a=
-a 1
(3) 
b=
-a
1
1
(4) 
b
a=
+b 1
Official Ans. by NTA (4)
Sol.
a= b=
CC
EB
II
,
II
I
E
 = I
B
 + I
C
a==
+
+
C
B
BC
C
I 1
I
II
1
I
a=
+
b
1
1
1
b
a=
+b 1
9. Three rays of light, namely red (R), green (G)
and blue (B) are incident on the face PQ of a
right angled prism PQR as shown in figure.
P
R
Q
B
G
R
The refractive indices of the material of the
prism for red, green and blue wavelength are
1.27, 1.42 and 1.49 respectively. The colour
of the ray(s) emerging out of the face PR is :
(1) green (2) red
(3) blue and green (4) blue
Official Ans. by NTA (2)
Sol.
P
R
Q
45°
Assuming that the right angled prism is an
isoceles prism, so the other angles will be 45°
each.
Þ Each incident ray will make an angle of 45°
with the normal at face PR.
Þ The wavelength corresponding to which the
incidence angle is less than the critical angle,
will pass through PR.
Þ q
C
 = critical angle
Þ 
-
æö
q=
ç÷
m
èø
1
C
1
sin
Þ If q
C
 ³ 45°
the light ray will pass
Þ 
()
-
æö
q = =°
ç÷
èø
1
C
Red
1
sin 51.94
1.27
4
Red will pass.
Þ 
()
-
æö
q = =°
ç÷
èø
1
C
Green
1
sin 44.76
1.42
Green will not pass
Þ 
()
-
æö
q = =°
ç÷
èø
1
C
Blue
1
sin 42.15
1.49
Blue will not pass
Þ So only red will pass through PR.
10. If the angular velocity of earth's spin is
increased such that the bodies at the equator
start floating, the duration of the day would be
approximately :
(Take : g = 10 ms
–2
, the radius of earth,
R = 6400 × 10
3
 m, Take p = 3.14)
(1) 60 minutes
(2) does not change
(3) 1200 minutes
(4) 84 minutes
Official Ans. by NTA (4)
Sol. For objects to float
mg = mw
2
R
w = angular velocity of earth.
R = Radius of earth
w=
g
R
... (1)
Duration of day = T
p
=
w
2
T
... (2)
Þ =p
R
T2
g
´
=p
3
6400 10
2
10
Þ 
=
T
83.775
60
 minutes
;
 84 minutes
11. The decay of a proton to neutron is :
(1) not possible as proton mass is less than the
neutron mass
(2) possible only inside the nucleus
(3) not possible but neutron to proton
conversion is possible
(4) always possible as it is associated only with
b
+
 decay
Official Ans. by NTA (2)
Sol. It is possible only inside the nucleus and not
otherwise.
12. In a series LCR circuit, the inductive reactance
(X
L
) is 10 W and the capacitive reactance (X
C
)
is 4 W. The resistance (R) in the circuit is 6 W.
The power factor of the circuit is :
(1) 
1
2
(2) 
1
22
(3) 
1
2
(4) 
3
2
Official Ans. by NTA (3)
Sol.
X =10
L
W
X =4
C
W
C L
R = 6W
We know that power factor is cosf,
f=
R
cos
Z
... (1)
( ) = +-
2
2
LC
Z R XX ... (2)
f
Z
R
( L–1/ C) ww
( ) Þ = +-
2
2
Z 6 104
Þ= Z 62
 | f=
6
cos
62
f=
1
cos
2
Page 5


1
SECTION-A
1. Which of the following statements are correct?
(A) Electric monopoles do not exist whereas
magnetic monopoles exist.
(B) Magnetic field lines due to a solenoid at its
ends and outside cannot be completely
straight and confined.
(C) Magnetic field lines are completely confined
within a toroid.
(D) Magnetic field lines inside a bar magnet are
not parallel.
(E) c = – 1 is the condition for a perfect
diamagnetic material, where c is its
magnetic susceptibility.
Choose the correct answer from the options
given below :
(1) (C) and (E) only
(2) (B) and (D) only
(3) (A) and (B) only
(4) (B) and (C) only
Official Ans. by NTA (1)
Sol. Statement (C) is correct because, the magnetic
field outside the toroid is zero and they form
closed loops inside the toroid itself.
Statement (E) is correct because we know that
super conductors are materials inside which the
net magnetic field is always zero and they are
perfect diamagnetic.
µ
r
 = 1 + c
c = –1
µ
r
 = 0
For superconductors.
2. An object of mass m
1
 collides with another
object of mass m
2
, which is at rest. After the
collision the objects move with equal speeds
in opposite direction. The ratio of the masses
m
2
 : m
1
 is :
(1) 3 : 1 (2) 2 : 1
(3) 1 : 2 (4) 1 : 1
Official Ans. by NTA (1)
FINAL JEE–MAIN EXAMINATION – MARCH, 2021
(Held On Thursday 18
th
 March, 2021)    TIME : 3 : 00 PM   to  6 : 00 PM
PHYSICS TEST PAPER WITH ANSWER & SOLUTION
Sol.
m
1
m
2
v
1
v v
m
1
m
2
m
1
v
1
 = –m
1
v + m
2
v
=-+
2
1
1
m
v vv
m
( ) +
=
1
2
1
vv
m
vm
==
1
2v
e1
v
=
1
v
v
2
+
=
112
11
v v/2m
v/2m
=
2
1
m
3
m
3. For an adiabatic expansion of an ideal gas, the
fractional change in its pressure is equal to
(where g is the ratio of specific heats):
(1)
-g
dV
V
(2)
-g
V
dV
(3)
-
g
1 dV
V
(4) 
dV
V
Official Ans. by NTA (1)
Sol. PV
g
 = constant
Differentiating
g
=-
dPP
dVV
g
=-
dP dV
PV
2
4. A proton and an a-particle, having kinetic
energies K
p 
and K
a
, respectively, enter into a
magnetic field at right angles.
The ratio of the radii of trajectory of proton to
that of a-particle is 2 : 1. The ratio of K
p
 : K
a
is :
(1) 1 : 8 (2) 8 : 1
(3) 1 : 4 (4) 4 : 1
Official Ans. by NTA (4)
Sol.
==
mvp
r
qB qB
a
=
p
m
4
m
a
aa
==
pp
p
rp
q 2
r qp1
aa
æö
==
ç÷
èø
pp
p 2q
1
2
p q2
a
=
p
p
1
p
a
a a
=
2
pp
p
p
Kp
m
Km p
 = (1) (4)
5. A plane electromagnetic wave propagating
along y-direction can have the following pair
of electric field ( )
E
r
 and magnetic field ( )
B
r
components.
(1) E
y
, B
y
 or E
z
, B
z
(2) E
y
, B
x
 or E
x
, B
y
(3) E
x
, B
z
 or E
z
, B
x
(4) E
x
, B
y
 or E
y
, B
x
Official Ans. by NTA (3)
Sol.
E
z
y
x
B
K
B
z
y
x
E
K
6. Consider a uniform wire of mass M and length
L. It is bent into a semicircle. Its moment of
inertia about a line perpendicular to the plane
of the wire passing through the centre is :
(1) 
p
2
2
1 ML
4
(2) 
p
2
2
2 ML
5
(3) 
p
2
2
ML
(4) 
p
2
2
1 ML
2
Official Ans. by NTA (3)
Sol.
p= Þ=
p
L
r Lr
==
p
2
2
2
ML
I Mr
7. The velocity-displacement graph of a particle
is shown in the figure.
x
0
x
v
v
0
O
The acceleration-displacement graph of the
same particle is represented by :
(1) 
x
a
O
(2) 
x
a
O
(3) 
x
a
O
(4) 
x
a
O
Official Ans. by NTA (3)
Sol.
æö
=-+
ç÷
èø
0
0
0
v
v xv
x
3
=
vdv
a
dx
éù æ ö éù
=- +-
êú ç÷ êú
êú è ø ëû ëû
00
0
00
vv
a xv
xx
æö
=-
ç÷
èø
2
2
00
00
vv
ax
xx
8. The correct relation between a (ratio of
collector current to emitter current) and b (ratio
of collector current to base current) of a
transistor is :
(1) 
a
b=
+a 1
(2) 
b
a=
-a 1
(3) 
b=
-a
1
1
(4) 
b
a=
+b 1
Official Ans. by NTA (4)
Sol.
a= b=
CC
EB
II
,
II
I
E
 = I
B
 + I
C
a==
+
+
C
B
BC
C
I 1
I
II
1
I
a=
+
b
1
1
1
b
a=
+b 1
9. Three rays of light, namely red (R), green (G)
and blue (B) are incident on the face PQ of a
right angled prism PQR as shown in figure.
P
R
Q
B
G
R
The refractive indices of the material of the
prism for red, green and blue wavelength are
1.27, 1.42 and 1.49 respectively. The colour
of the ray(s) emerging out of the face PR is :
(1) green (2) red
(3) blue and green (4) blue
Official Ans. by NTA (2)
Sol.
P
R
Q
45°
Assuming that the right angled prism is an
isoceles prism, so the other angles will be 45°
each.
Þ Each incident ray will make an angle of 45°
with the normal at face PR.
Þ The wavelength corresponding to which the
incidence angle is less than the critical angle,
will pass through PR.
Þ q
C
 = critical angle
Þ 
-
æö
q=
ç÷
m
èø
1
C
1
sin
Þ If q
C
 ³ 45°
the light ray will pass
Þ 
()
-
æö
q = =°
ç÷
èø
1
C
Red
1
sin 51.94
1.27
4
Red will pass.
Þ 
()
-
æö
q = =°
ç÷
èø
1
C
Green
1
sin 44.76
1.42
Green will not pass
Þ 
()
-
æö
q = =°
ç÷
èø
1
C
Blue
1
sin 42.15
1.49
Blue will not pass
Þ So only red will pass through PR.
10. If the angular velocity of earth's spin is
increased such that the bodies at the equator
start floating, the duration of the day would be
approximately :
(Take : g = 10 ms
–2
, the radius of earth,
R = 6400 × 10
3
 m, Take p = 3.14)
(1) 60 minutes
(2) does not change
(3) 1200 minutes
(4) 84 minutes
Official Ans. by NTA (4)
Sol. For objects to float
mg = mw
2
R
w = angular velocity of earth.
R = Radius of earth
w=
g
R
... (1)
Duration of day = T
p
=
w
2
T
... (2)
Þ =p
R
T2
g
´
=p
3
6400 10
2
10
Þ 
=
T
83.775
60
 minutes
;
 84 minutes
11. The decay of a proton to neutron is :
(1) not possible as proton mass is less than the
neutron mass
(2) possible only inside the nucleus
(3) not possible but neutron to proton
conversion is possible
(4) always possible as it is associated only with
b
+
 decay
Official Ans. by NTA (2)
Sol. It is possible only inside the nucleus and not
otherwise.
12. In a series LCR circuit, the inductive reactance
(X
L
) is 10 W and the capacitive reactance (X
C
)
is 4 W. The resistance (R) in the circuit is 6 W.
The power factor of the circuit is :
(1) 
1
2
(2) 
1
22
(3) 
1
2
(4) 
3
2
Official Ans. by NTA (3)
Sol.
X =10
L
W
X =4
C
W
C L
R = 6W
We know that power factor is cosf,
f=
R
cos
Z
... (1)
( ) = +-
2
2
LC
Z R XX ... (2)
f
Z
R
( L–1/ C) ww
( ) Þ = +-
2
2
Z 6 104
Þ= Z 62
 | f=
6
cos
62
f=
1
cos
2
5
13. The angular momentum of a planet of mass M
moving around the sun in an elliptical orbit is
L
r
. The magnitude of the areal velocity of the
planet is :
(1) 
4L
M
(2) 
L
M
(3) 
2L
M
(4) 
L
2M
Official Ans. by NTA (4)
Sol.
q r
S
ds
For small displacement ds of the planet its area
can be written as
q
r
ds
d l
=
1
dA rd
2
l
= 
q
1
r dssin
2
A.vel = 
q
= q=
dA 1 ds Vrsin
rsin
dt 2 dt 2
q
==
dA1mVrsinL
dt 2 m 2m
14. The function of time representing a simple
harmonic motion with a period of 
p
w
 is :
(1) sin(wt) + cos (wt)
(2) cos(wt) + cos (2wt) + cos (3wt)
(3) sin
2
(wt)
(4) 
pæö
-w
ç÷
èø
3cos 2t
4
Official Ans. by NTA (4)
Sol. Time period 
p
=
w
2
T
'
pp
=
ww
2
'
w' = 2w ® Angular frequency of SHM
Option (3)
( ) ( ) w= w= -w
22
11
sint 2sint 1 cos2t
22
Angular frequency of 
æö
-w
ç÷
èø
11
cos2t
22
 is 2w
Option (4)
Angular frequency of SHM
pæö
-w
ç÷
èø
3cos 2t
4
 is 2w.
So option (3) & (4) both have angular
frequency 2w but option (4) is direct answer.
15. A solid cylinder of mass m is wrapped with an
inextensible light string and, is placed on a
rough inclined plane as shown in the figure.
The frictional force acting between the cylinder
and the inclined plane is :
60°
[The coefficient of static friction, µ
s
, is 0.4]
(1) 
7
mg
2
(2) 5 mg
(3) 
mg
5
(4) 0
Official Ans. by NTA (3)