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Page 1
1
FINAL JEE –MAIN EXAMINATION – AUGUST, 2021
(Held On Friday 27
th
August, 2021) TIME : 9 : 00 AM to 12 : 00 NOON
PHYSICS
TEST PAPER WITH SOLUTION
SECTION-A
1. A uniformly charged disc of radius R having
surface charge density ? is placed in the xy plane
with its center at the origin. Find the electric field
intensity along the z-axis at a distance Z from
origin :-
(1)
2 2 1/2
0
Z
E1
2 (Z R )
?? ?
??
??
??
??
(2)
2 2 1/2
0
Z
E1
2 (Z R )
?? ?
??
??
??
??
(3)
0
2 2 1/2
21
EZ
(Z R )
?? ?
??
??
??
??
(4)
2 2 2
0
11
E
2 (Z R ) Z
?? ?
??
??
??
??
Official Ans. by NTA (1)
Sol. Consider a small ring of radius r and thickness dr
on disc.
dr
z axis
Z
r
area of elemental ring on disc
dA = 2 ?rdr
charge on this ring dq = ?dA
dEz =
? ?
3/2
22
kdqz
zr ?
E =
R
z
22
0 0
z
dE 1 –
2
Rz
?? ?
?
??
?
? ??
?
2. There are 10
10
radioactive nuclei in a given
radioactive element, Its half-life time is 1 minute.
How many nuclei will remain after 30 seconds ?
? ?
2 1.414 ?
(1) 2 × 10
10
(2) 7 × 10
9
(3) 10
5
(4) 4 × 10
10
Official Ans. by NTA (2)
Sol.
0
N
N
=
1/2
t
t
1
2
??
??
??
10
N
10
=
30
60
1
2
??
??
??
? ? N = 10
10
×
1
2
1
2
??
??
??
=
10
10
2
? 7 × 10
9
3. Which of the following is not a dimensionless
quantity ?
(1) Relative magnetic permeability ( ?
r
)
(2) Power factor
(3) Permeability of free space ( ?
0
)
(4) Quality factor
Official Ans. by NTA (3)
Sol. [ ?
r
] = 1 as ?
r
=
m
?
?
[power factor (cos ?)] = 1
?
0
=
0
B
H
(unit = NA
–2
) : Not dimensionless
[ ?
0
] = [MLT
–2
A
–2
]
quality factor (Q) =
Energystored
Energy dissipated per cycle
So Q is unitless & dimensionless.
Page 2
1
FINAL JEE –MAIN EXAMINATION – AUGUST, 2021
(Held On Friday 27
th
August, 2021) TIME : 9 : 00 AM to 12 : 00 NOON
PHYSICS
TEST PAPER WITH SOLUTION
SECTION-A
1. A uniformly charged disc of radius R having
surface charge density ? is placed in the xy plane
with its center at the origin. Find the electric field
intensity along the z-axis at a distance Z from
origin :-
(1)
2 2 1/2
0
Z
E1
2 (Z R )
?? ?
??
??
??
??
(2)
2 2 1/2
0
Z
E1
2 (Z R )
?? ?
??
??
??
??
(3)
0
2 2 1/2
21
EZ
(Z R )
?? ?
??
??
??
??
(4)
2 2 2
0
11
E
2 (Z R ) Z
?? ?
??
??
??
??
Official Ans. by NTA (1)
Sol. Consider a small ring of radius r and thickness dr
on disc.
dr
z axis
Z
r
area of elemental ring on disc
dA = 2 ?rdr
charge on this ring dq = ?dA
dEz =
? ?
3/2
22
kdqz
zr ?
E =
R
z
22
0 0
z
dE 1 –
2
Rz
?? ?
?
??
?
? ??
?
2. There are 10
10
radioactive nuclei in a given
radioactive element, Its half-life time is 1 minute.
How many nuclei will remain after 30 seconds ?
? ?
2 1.414 ?
(1) 2 × 10
10
(2) 7 × 10
9
(3) 10
5
(4) 4 × 10
10
Official Ans. by NTA (2)
Sol.
0
N
N
=
1/2
t
t
1
2
??
??
??
10
N
10
=
30
60
1
2
??
??
??
? ? N = 10
10
×
1
2
1
2
??
??
??
=
10
10
2
? 7 × 10
9
3. Which of the following is not a dimensionless
quantity ?
(1) Relative magnetic permeability ( ?
r
)
(2) Power factor
(3) Permeability of free space ( ?
0
)
(4) Quality factor
Official Ans. by NTA (3)
Sol. [ ?
r
] = 1 as ?
r
=
m
?
?
[power factor (cos ?)] = 1
?
0
=
0
B
H
(unit = NA
–2
) : Not dimensionless
[ ?
0
] = [MLT
–2
A
–2
]
quality factor (Q) =
Energystored
Energy dissipated per cycle
So Q is unitless & dimensionless.
2
4. If E and H represents the intensity of electric field
and magnetising field respectively, then the unit of
E/H will be :
(1) ohm (2) mho
(3) joule (4) newton
Official Ans. by NTA (1)
Sol. Unit of
E
H
is
volt / metre
Ampere / metre
=
volt
Ampere
= ohm
5. The resultant of these forces OP, OQ, OR, OS and
OT is approximately ...... N.
[Take 3 1.7, 2 1.4 ?? Given
ˆ
i and
ˆ
j unit
vectors along x, y axis]
15N
20N
20N
15N
45°
45°
60°
30°
30°
O
R
S
T y
P
x
x'
y'
10N
Q
(1)
ˆˆ
9.25i 5j ? (2)
ˆˆ
3i 15j ?
(3)
ˆˆ
2.5i 14.5j ? (4)
ˆˆ
1.5i 15.5j ??
Official Ans. by NTA (1)
Sol.
15 cos 60º
20 cos 30º
10 sin 30º
15 sin 45º
20 sin 45º
O
10 cos 30º 20 sin 30º 20 cos 45º
X
15 COS 45º
15 Sin 60º
x
3 1 1 1 3
F 1 0 2 0 2 0 – 1 5 – 1 5 i
2 2 2 22
?? ??
? ? ? ? ??
? ? ? ??? ??
? ? ? ? ??
?? ??
?? ? ? ? ?
?? ??
= 9.25 i
y
F =
1 3 1 1 1
1 5 2 0 1 0 – 1 5 – 2 0 j
2 2 2 22
?? ??
? ? ? ? ? ? ? ?
???? ??
? ? ? ? ? ? ? ?
????
? ? ? ? ? ? ? ?
????
= 5 j
6. A balloon carries a total load of 185 kg at normal
pressure and temperature of 27°C. What load will
the balloon carry on rising to a height at which the
barometric pressure is 45 cm of Hg and the
temperature is –7°C. Assuming the volume
constant ?
(1) 181.46 kg (2) 214.15 kg.
(3) 219.07 kg (4) 123.54 kg
Official Ans. by NTA (4)
Sol. P
m
= ?RT
? ?
1 1 1
2 2 2
PT
PT
?
?
?
1 1 2
2 2 1
PT 76 266
P T 45 300
? ??
? ? ?
??
?
??
11
22
M 76 266
M 45 300
? ?
??
??
? ??M
2
??
45 300 185
76 266
??
?
? ????? ???kg?
7. An object is placed beyond the centre of curvature
C of the given concave mirror. If the distance of
the object is d
1
from C and the distance of the
image formed is d
2
from C, the radius of curvature
of this mirror is :
(1)
12
12
2d d
dd ?
(2)
12
12
2d d
dd ?
(3)
12
12
dd
dd ?
(4)
12
12
dd
d – d
Official Ans. by NTA (1)
Sol. Using Newton''s formula
? ? ? ?
2
12
f d f – d f ??
f
2
+ fd
1
– fd
2
– d
1
d
2
= f
2
f = ?
12
12
dd
d – d
? ??R = ?
12
12
2d d
d – d
?
Page 3
1
FINAL JEE –MAIN EXAMINATION – AUGUST, 2021
(Held On Friday 27
th
August, 2021) TIME : 9 : 00 AM to 12 : 00 NOON
PHYSICS
TEST PAPER WITH SOLUTION
SECTION-A
1. A uniformly charged disc of radius R having
surface charge density ? is placed in the xy plane
with its center at the origin. Find the electric field
intensity along the z-axis at a distance Z from
origin :-
(1)
2 2 1/2
0
Z
E1
2 (Z R )
?? ?
??
??
??
??
(2)
2 2 1/2
0
Z
E1
2 (Z R )
?? ?
??
??
??
??
(3)
0
2 2 1/2
21
EZ
(Z R )
?? ?
??
??
??
??
(4)
2 2 2
0
11
E
2 (Z R ) Z
?? ?
??
??
??
??
Official Ans. by NTA (1)
Sol. Consider a small ring of radius r and thickness dr
on disc.
dr
z axis
Z
r
area of elemental ring on disc
dA = 2 ?rdr
charge on this ring dq = ?dA
dEz =
? ?
3/2
22
kdqz
zr ?
E =
R
z
22
0 0
z
dE 1 –
2
Rz
?? ?
?
??
?
? ??
?
2. There are 10
10
radioactive nuclei in a given
radioactive element, Its half-life time is 1 minute.
How many nuclei will remain after 30 seconds ?
? ?
2 1.414 ?
(1) 2 × 10
10
(2) 7 × 10
9
(3) 10
5
(4) 4 × 10
10
Official Ans. by NTA (2)
Sol.
0
N
N
=
1/2
t
t
1
2
??
??
??
10
N
10
=
30
60
1
2
??
??
??
? ? N = 10
10
×
1
2
1
2
??
??
??
=
10
10
2
? 7 × 10
9
3. Which of the following is not a dimensionless
quantity ?
(1) Relative magnetic permeability ( ?
r
)
(2) Power factor
(3) Permeability of free space ( ?
0
)
(4) Quality factor
Official Ans. by NTA (3)
Sol. [ ?
r
] = 1 as ?
r
=
m
?
?
[power factor (cos ?)] = 1
?
0
=
0
B
H
(unit = NA
–2
) : Not dimensionless
[ ?
0
] = [MLT
–2
A
–2
]
quality factor (Q) =
Energystored
Energy dissipated per cycle
So Q is unitless & dimensionless.
2
4. If E and H represents the intensity of electric field
and magnetising field respectively, then the unit of
E/H will be :
(1) ohm (2) mho
(3) joule (4) newton
Official Ans. by NTA (1)
Sol. Unit of
E
H
is
volt / metre
Ampere / metre
=
volt
Ampere
= ohm
5. The resultant of these forces OP, OQ, OR, OS and
OT is approximately ...... N.
[Take 3 1.7, 2 1.4 ?? Given
ˆ
i and
ˆ
j unit
vectors along x, y axis]
15N
20N
20N
15N
45°
45°
60°
30°
30°
O
R
S
T y
P
x
x'
y'
10N
Q
(1)
ˆˆ
9.25i 5j ? (2)
ˆˆ
3i 15j ?
(3)
ˆˆ
2.5i 14.5j ? (4)
ˆˆ
1.5i 15.5j ??
Official Ans. by NTA (1)
Sol.
15 cos 60º
20 cos 30º
10 sin 30º
15 sin 45º
20 sin 45º
O
10 cos 30º 20 sin 30º 20 cos 45º
X
15 COS 45º
15 Sin 60º
x
3 1 1 1 3
F 1 0 2 0 2 0 – 1 5 – 1 5 i
2 2 2 22
?? ??
? ? ? ? ??
? ? ? ??? ??
? ? ? ? ??
?? ??
?? ? ? ? ?
?? ??
= 9.25 i
y
F =
1 3 1 1 1
1 5 2 0 1 0 – 1 5 – 2 0 j
2 2 2 22
?? ??
? ? ? ? ? ? ? ?
???? ??
? ? ? ? ? ? ? ?
????
? ? ? ? ? ? ? ?
????
= 5 j
6. A balloon carries a total load of 185 kg at normal
pressure and temperature of 27°C. What load will
the balloon carry on rising to a height at which the
barometric pressure is 45 cm of Hg and the
temperature is –7°C. Assuming the volume
constant ?
(1) 181.46 kg (2) 214.15 kg.
(3) 219.07 kg (4) 123.54 kg
Official Ans. by NTA (4)
Sol. P
m
= ?RT
? ?
1 1 1
2 2 2
PT
PT
?
?
?
1 1 2
2 2 1
PT 76 266
P T 45 300
? ??
? ? ?
??
?
??
11
22
M 76 266
M 45 300
? ?
??
??
? ??M
2
??
45 300 185
76 266
??
?
? ????? ???kg?
7. An object is placed beyond the centre of curvature
C of the given concave mirror. If the distance of
the object is d
1
from C and the distance of the
image formed is d
2
from C, the radius of curvature
of this mirror is :
(1)
12
12
2d d
dd ?
(2)
12
12
2d d
dd ?
(3)
12
12
dd
dd ?
(4)
12
12
dd
d – d
Official Ans. by NTA (1)
Sol. Using Newton''s formula
? ? ? ?
2
12
f d f – d f ??
f
2
+ fd
1
– fd
2
– d
1
d
2
= f
2
f = ?
12
12
dd
d – d
? ??R = ?
12
12
2d d
d – d
?
3
8. A huge circular arc of length 4.4 ly subtends an
angle '4s' at the centre of the circle. How long it
would take for a body to complete 4 revolution if
its speed is 8 AU per second ?
Given : 1 ly = 9.46 × 10
15
m
1 AU = 1.5 × 10
11
m
(1) 4.1 × 10
8
s (2) 4.5 × 10
10
s
(3) 3.5 × 10
6
s (4) 7.2 × 10
8
s
Official Ans. by NTA (2)
Sol. R = ?
?
Time =
4 2 R
v
??
=
42
v
????
??
?
??
?
put ? = 4.4 × 9.46 × 10
15
v = 8 × 1.5 × 10
11
? =
4
3600
×
180
?
rad.
we get time = 4.5 × 10
10
sec
9. Calculate the amount of charge on capacitor of
4 ?F. The internal resistance of battery is 1 ? :
2 ?F
2 ?F
5V
4 ?
4 ?F
6 ?
(1) 8 ?C (2) zero
(3) 16 ?C (4) 4 ?C
Official Ans. by NTA (1)
Sol. On simplifying circuit we get
4 ?F 4 ?F
A B
4 ?
1 ?
5v
No current in upper wire.
? ? ? V
AB
=
5
41 ?
× 4 = 4 v.
? ? ? ? = (C
eq
)v
? 2 × 4 = 8 ?C
10. Moment of inertia of a square plate of side l about
the axis passing through one of the corner and
perpendicular to the plane of square plate is given
by :
(1)
2
M
6
l
(2) Ml
2
(3)
2
M
12
l
(4)
2
2
M
3
l
Official Ans. by NTA (4)
Sol. According to perpendicular Axis theorem.
I
y
I
x
I
x
+ I
y
= I
z
I
z
??
2
m
3
+
2
m
3
=
2
2m
3
11. For a transistor in CE mode to be used as an
amplifier, it must be operated in :
(1) Both cut-off and Saturation
(2) Saturation region only
(3) Cut-off region only
(4) The active region only
Official Ans. by NTA (4)
Sol. Active region of the CE transistor is linear region
and is best suited for its use as an amplifier
12. An ideal gas is expanding such that PT
3
=
constant. The coefficient of volume expansion of
the gas is :
(1)
1
T
(2)
2
T
(3)
4
T
(4)
3
T
Official Ans. by NTA (3)
Sol. PT
3
= constant
3
nRT
T
v
??
??
??
= constant
T
4
V
–1
= constant
T
4
= kV
? ??4
T
T
?
=
V
V
?
.........(1)
?V ?= V ??T.............(2)
comparing (1) and (2)
we get
? =
4
T
13. In a photoelectric experiment, increasing the
intensity of incident light :
(1) increases the number of photons incident and
also increases the K.E. of the ejected electrons
(2) increases the frequency of photons incident and
increases the K.E. of the ejected electrons.
(3) increases the frequency of photons incident and
the K.E. of the ejected electrons remains
unchanged
(4) increases the number of photons incident and the
K.E. of the ejected electrons remains unchanged
Official Ans. by NTA (4)
Sol. ? Increasing intensity means number of incident
photons are increased.
?Kinetic energy of ejected electrons depend on the
frequency of incident photons, not the intensity.
Page 4
1
FINAL JEE –MAIN EXAMINATION – AUGUST, 2021
(Held On Friday 27
th
August, 2021) TIME : 9 : 00 AM to 12 : 00 NOON
PHYSICS
TEST PAPER WITH SOLUTION
SECTION-A
1. A uniformly charged disc of radius R having
surface charge density ? is placed in the xy plane
with its center at the origin. Find the electric field
intensity along the z-axis at a distance Z from
origin :-
(1)
2 2 1/2
0
Z
E1
2 (Z R )
?? ?
??
??
??
??
(2)
2 2 1/2
0
Z
E1
2 (Z R )
?? ?
??
??
??
??
(3)
0
2 2 1/2
21
EZ
(Z R )
?? ?
??
??
??
??
(4)
2 2 2
0
11
E
2 (Z R ) Z
?? ?
??
??
??
??
Official Ans. by NTA (1)
Sol. Consider a small ring of radius r and thickness dr
on disc.
dr
z axis
Z
r
area of elemental ring on disc
dA = 2 ?rdr
charge on this ring dq = ?dA
dEz =
? ?
3/2
22
kdqz
zr ?
E =
R
z
22
0 0
z
dE 1 –
2
Rz
?? ?
?
??
?
? ??
?
2. There are 10
10
radioactive nuclei in a given
radioactive element, Its half-life time is 1 minute.
How many nuclei will remain after 30 seconds ?
? ?
2 1.414 ?
(1) 2 × 10
10
(2) 7 × 10
9
(3) 10
5
(4) 4 × 10
10
Official Ans. by NTA (2)
Sol.
0
N
N
=
1/2
t
t
1
2
??
??
??
10
N
10
=
30
60
1
2
??
??
??
? ? N = 10
10
×
1
2
1
2
??
??
??
=
10
10
2
? 7 × 10
9
3. Which of the following is not a dimensionless
quantity ?
(1) Relative magnetic permeability ( ?
r
)
(2) Power factor
(3) Permeability of free space ( ?
0
)
(4) Quality factor
Official Ans. by NTA (3)
Sol. [ ?
r
] = 1 as ?
r
=
m
?
?
[power factor (cos ?)] = 1
?
0
=
0
B
H
(unit = NA
–2
) : Not dimensionless
[ ?
0
] = [MLT
–2
A
–2
]
quality factor (Q) =
Energystored
Energy dissipated per cycle
So Q is unitless & dimensionless.
2
4. If E and H represents the intensity of electric field
and magnetising field respectively, then the unit of
E/H will be :
(1) ohm (2) mho
(3) joule (4) newton
Official Ans. by NTA (1)
Sol. Unit of
E
H
is
volt / metre
Ampere / metre
=
volt
Ampere
= ohm
5. The resultant of these forces OP, OQ, OR, OS and
OT is approximately ...... N.
[Take 3 1.7, 2 1.4 ?? Given
ˆ
i and
ˆ
j unit
vectors along x, y axis]
15N
20N
20N
15N
45°
45°
60°
30°
30°
O
R
S
T y
P
x
x'
y'
10N
Q
(1)
ˆˆ
9.25i 5j ? (2)
ˆˆ
3i 15j ?
(3)
ˆˆ
2.5i 14.5j ? (4)
ˆˆ
1.5i 15.5j ??
Official Ans. by NTA (1)
Sol.
15 cos 60º
20 cos 30º
10 sin 30º
15 sin 45º
20 sin 45º
O
10 cos 30º 20 sin 30º 20 cos 45º
X
15 COS 45º
15 Sin 60º
x
3 1 1 1 3
F 1 0 2 0 2 0 – 1 5 – 1 5 i
2 2 2 22
?? ??
? ? ? ? ??
? ? ? ??? ??
? ? ? ? ??
?? ??
?? ? ? ? ?
?? ??
= 9.25 i
y
F =
1 3 1 1 1
1 5 2 0 1 0 – 1 5 – 2 0 j
2 2 2 22
?? ??
? ? ? ? ? ? ? ?
???? ??
? ? ? ? ? ? ? ?
????
? ? ? ? ? ? ? ?
????
= 5 j
6. A balloon carries a total load of 185 kg at normal
pressure and temperature of 27°C. What load will
the balloon carry on rising to a height at which the
barometric pressure is 45 cm of Hg and the
temperature is –7°C. Assuming the volume
constant ?
(1) 181.46 kg (2) 214.15 kg.
(3) 219.07 kg (4) 123.54 kg
Official Ans. by NTA (4)
Sol. P
m
= ?RT
? ?
1 1 1
2 2 2
PT
PT
?
?
?
1 1 2
2 2 1
PT 76 266
P T 45 300
? ??
? ? ?
??
?
??
11
22
M 76 266
M 45 300
? ?
??
??
? ??M
2
??
45 300 185
76 266
??
?
? ????? ???kg?
7. An object is placed beyond the centre of curvature
C of the given concave mirror. If the distance of
the object is d
1
from C and the distance of the
image formed is d
2
from C, the radius of curvature
of this mirror is :
(1)
12
12
2d d
dd ?
(2)
12
12
2d d
dd ?
(3)
12
12
dd
dd ?
(4)
12
12
dd
d – d
Official Ans. by NTA (1)
Sol. Using Newton''s formula
? ? ? ?
2
12
f d f – d f ??
f
2
+ fd
1
– fd
2
– d
1
d
2
= f
2
f = ?
12
12
dd
d – d
? ??R = ?
12
12
2d d
d – d
?
3
8. A huge circular arc of length 4.4 ly subtends an
angle '4s' at the centre of the circle. How long it
would take for a body to complete 4 revolution if
its speed is 8 AU per second ?
Given : 1 ly = 9.46 × 10
15
m
1 AU = 1.5 × 10
11
m
(1) 4.1 × 10
8
s (2) 4.5 × 10
10
s
(3) 3.5 × 10
6
s (4) 7.2 × 10
8
s
Official Ans. by NTA (2)
Sol. R = ?
?
Time =
4 2 R
v
??
=
42
v
????
??
?
??
?
put ? = 4.4 × 9.46 × 10
15
v = 8 × 1.5 × 10
11
? =
4
3600
×
180
?
rad.
we get time = 4.5 × 10
10
sec
9. Calculate the amount of charge on capacitor of
4 ?F. The internal resistance of battery is 1 ? :
2 ?F
2 ?F
5V
4 ?
4 ?F
6 ?
(1) 8 ?C (2) zero
(3) 16 ?C (4) 4 ?C
Official Ans. by NTA (1)
Sol. On simplifying circuit we get
4 ?F 4 ?F
A B
4 ?
1 ?
5v
No current in upper wire.
? ? ? V
AB
=
5
41 ?
× 4 = 4 v.
? ? ? ? = (C
eq
)v
? 2 × 4 = 8 ?C
10. Moment of inertia of a square plate of side l about
the axis passing through one of the corner and
perpendicular to the plane of square plate is given
by :
(1)
2
M
6
l
(2) Ml
2
(3)
2
M
12
l
(4)
2
2
M
3
l
Official Ans. by NTA (4)
Sol. According to perpendicular Axis theorem.
I
y
I
x
I
x
+ I
y
= I
z
I
z
??
2
m
3
+
2
m
3
=
2
2m
3
11. For a transistor in CE mode to be used as an
amplifier, it must be operated in :
(1) Both cut-off and Saturation
(2) Saturation region only
(3) Cut-off region only
(4) The active region only
Official Ans. by NTA (4)
Sol. Active region of the CE transistor is linear region
and is best suited for its use as an amplifier
12. An ideal gas is expanding such that PT
3
=
constant. The coefficient of volume expansion of
the gas is :
(1)
1
T
(2)
2
T
(3)
4
T
(4)
3
T
Official Ans. by NTA (3)
Sol. PT
3
= constant
3
nRT
T
v
??
??
??
= constant
T
4
V
–1
= constant
T
4
= kV
? ??4
T
T
?
=
V
V
?
.........(1)
?V ?= V ??T.............(2)
comparing (1) and (2)
we get
? =
4
T
13. In a photoelectric experiment, increasing the
intensity of incident light :
(1) increases the number of photons incident and
also increases the K.E. of the ejected electrons
(2) increases the frequency of photons incident and
increases the K.E. of the ejected electrons.
(3) increases the frequency of photons incident and
the K.E. of the ejected electrons remains
unchanged
(4) increases the number of photons incident and the
K.E. of the ejected electrons remains unchanged
Official Ans. by NTA (4)
Sol. ? Increasing intensity means number of incident
photons are increased.
?Kinetic energy of ejected electrons depend on the
frequency of incident photons, not the intensity.
4
14. A bar magnet is passing through a conducting loop
of radius R with velocity ?. The radius of the bar
magnet is such that it just passes through the loop.
The induced e.m.f. in the loop can be represented
by the approximate curve :
R
loop
N S
l
? ?
(1)
l/ ?
emf
t
(2)
t
emf
l/ ?
(3)
t
emf
l/ ?
(4)
t
emf
l/ ?
Official Ans. by NTA (3)
Sol.
S N
? When magnet passes through centre region of
solenoid , no current / Emf is induced in loop.
??While entering flux increases so negative
induced emf
? ?While leaving flux decreases so positive
induced emf.
15. Two ions of masses 4 amu and 16 amu have
charges +2e and +3e respectively. These ions pass
through the region of constant perpendicular
magnetic field. The kinetic energy of both ions is
same. Then :
(1) lighter ion will be deflected less than heavier ion
(2) lighter ion will be deflected more than heavier ion
(3) both ions will be deflected equally
(4) no ion will be deflected.
Official Ans. by NTA (2)
Sol. r =
P 2mk
qB qB
?
Given they have same kinetic energy
r ?
m
q
1
2
r
r
=
4
2
×
3
16
=
3
4
1
2
4r
r
3
? (r
2
is for hearier ion and r
1
is for lighter ion)
x x x
x x x
x x x
x x x
d
? ? R
sin ? =
d
R
? ???Deflection
? ?
1
R
(R ? Radius of path)
?R
2
> R
1
? ??
2
< ?
1
16. Find the distance of the image from object O,
formed by the combination of lenses in the figure :
f =+10cm
5cm 10cm
f = –10cm
30cm
f =+30cm
O
(1) 75 cm (2) 10 cm
(3) 20 cm (4) infinity
Official Ans. by NTA (1)
Page 5
1
FINAL JEE –MAIN EXAMINATION – AUGUST, 2021
(Held On Friday 27
th
August, 2021) TIME : 9 : 00 AM to 12 : 00 NOON
PHYSICS
TEST PAPER WITH SOLUTION
SECTION-A
1. A uniformly charged disc of radius R having
surface charge density ? is placed in the xy plane
with its center at the origin. Find the electric field
intensity along the z-axis at a distance Z from
origin :-
(1)
2 2 1/2
0
Z
E1
2 (Z R )
?? ?
??
??
??
??
(2)
2 2 1/2
0
Z
E1
2 (Z R )
?? ?
??
??
??
??
(3)
0
2 2 1/2
21
EZ
(Z R )
?? ?
??
??
??
??
(4)
2 2 2
0
11
E
2 (Z R ) Z
?? ?
??
??
??
??
Official Ans. by NTA (1)
Sol. Consider a small ring of radius r and thickness dr
on disc.
dr
z axis
Z
r
area of elemental ring on disc
dA = 2 ?rdr
charge on this ring dq = ?dA
dEz =
? ?
3/2
22
kdqz
zr ?
E =
R
z
22
0 0
z
dE 1 –
2
Rz
?? ?
?
??
?
? ??
?
2. There are 10
10
radioactive nuclei in a given
radioactive element, Its half-life time is 1 minute.
How many nuclei will remain after 30 seconds ?
? ?
2 1.414 ?
(1) 2 × 10
10
(2) 7 × 10
9
(3) 10
5
(4) 4 × 10
10
Official Ans. by NTA (2)
Sol.
0
N
N
=
1/2
t
t
1
2
??
??
??
10
N
10
=
30
60
1
2
??
??
??
? ? N = 10
10
×
1
2
1
2
??
??
??
=
10
10
2
? 7 × 10
9
3. Which of the following is not a dimensionless
quantity ?
(1) Relative magnetic permeability ( ?
r
)
(2) Power factor
(3) Permeability of free space ( ?
0
)
(4) Quality factor
Official Ans. by NTA (3)
Sol. [ ?
r
] = 1 as ?
r
=
m
?
?
[power factor (cos ?)] = 1
?
0
=
0
B
H
(unit = NA
–2
) : Not dimensionless
[ ?
0
] = [MLT
–2
A
–2
]
quality factor (Q) =
Energystored
Energy dissipated per cycle
So Q is unitless & dimensionless.
2
4. If E and H represents the intensity of electric field
and magnetising field respectively, then the unit of
E/H will be :
(1) ohm (2) mho
(3) joule (4) newton
Official Ans. by NTA (1)
Sol. Unit of
E
H
is
volt / metre
Ampere / metre
=
volt
Ampere
= ohm
5. The resultant of these forces OP, OQ, OR, OS and
OT is approximately ...... N.
[Take 3 1.7, 2 1.4 ?? Given
ˆ
i and
ˆ
j unit
vectors along x, y axis]
15N
20N
20N
15N
45°
45°
60°
30°
30°
O
R
S
T y
P
x
x'
y'
10N
Q
(1)
ˆˆ
9.25i 5j ? (2)
ˆˆ
3i 15j ?
(3)
ˆˆ
2.5i 14.5j ? (4)
ˆˆ
1.5i 15.5j ??
Official Ans. by NTA (1)
Sol.
15 cos 60º
20 cos 30º
10 sin 30º
15 sin 45º
20 sin 45º
O
10 cos 30º 20 sin 30º 20 cos 45º
X
15 COS 45º
15 Sin 60º
x
3 1 1 1 3
F 1 0 2 0 2 0 – 1 5 – 1 5 i
2 2 2 22
?? ??
? ? ? ? ??
? ? ? ??? ??
? ? ? ? ??
?? ??
?? ? ? ? ?
?? ??
= 9.25 i
y
F =
1 3 1 1 1
1 5 2 0 1 0 – 1 5 – 2 0 j
2 2 2 22
?? ??
? ? ? ? ? ? ? ?
???? ??
? ? ? ? ? ? ? ?
????
? ? ? ? ? ? ? ?
????
= 5 j
6. A balloon carries a total load of 185 kg at normal
pressure and temperature of 27°C. What load will
the balloon carry on rising to a height at which the
barometric pressure is 45 cm of Hg and the
temperature is –7°C. Assuming the volume
constant ?
(1) 181.46 kg (2) 214.15 kg.
(3) 219.07 kg (4) 123.54 kg
Official Ans. by NTA (4)
Sol. P
m
= ?RT
? ?
1 1 1
2 2 2
PT
PT
?
?
?
1 1 2
2 2 1
PT 76 266
P T 45 300
? ??
? ? ?
??
?
??
11
22
M 76 266
M 45 300
? ?
??
??
? ??M
2
??
45 300 185
76 266
??
?
? ????? ???kg?
7. An object is placed beyond the centre of curvature
C of the given concave mirror. If the distance of
the object is d
1
from C and the distance of the
image formed is d
2
from C, the radius of curvature
of this mirror is :
(1)
12
12
2d d
dd ?
(2)
12
12
2d d
dd ?
(3)
12
12
dd
dd ?
(4)
12
12
dd
d – d
Official Ans. by NTA (1)
Sol. Using Newton''s formula
? ? ? ?
2
12
f d f – d f ??
f
2
+ fd
1
– fd
2
– d
1
d
2
= f
2
f = ?
12
12
dd
d – d
? ??R = ?
12
12
2d d
d – d
?
3
8. A huge circular arc of length 4.4 ly subtends an
angle '4s' at the centre of the circle. How long it
would take for a body to complete 4 revolution if
its speed is 8 AU per second ?
Given : 1 ly = 9.46 × 10
15
m
1 AU = 1.5 × 10
11
m
(1) 4.1 × 10
8
s (2) 4.5 × 10
10
s
(3) 3.5 × 10
6
s (4) 7.2 × 10
8
s
Official Ans. by NTA (2)
Sol. R = ?
?
Time =
4 2 R
v
??
=
42
v
????
??
?
??
?
put ? = 4.4 × 9.46 × 10
15
v = 8 × 1.5 × 10
11
? =
4
3600
×
180
?
rad.
we get time = 4.5 × 10
10
sec
9. Calculate the amount of charge on capacitor of
4 ?F. The internal resistance of battery is 1 ? :
2 ?F
2 ?F
5V
4 ?
4 ?F
6 ?
(1) 8 ?C (2) zero
(3) 16 ?C (4) 4 ?C
Official Ans. by NTA (1)
Sol. On simplifying circuit we get
4 ?F 4 ?F
A B
4 ?
1 ?
5v
No current in upper wire.
? ? ? V
AB
=
5
41 ?
× 4 = 4 v.
? ? ? ? = (C
eq
)v
? 2 × 4 = 8 ?C
10. Moment of inertia of a square plate of side l about
the axis passing through one of the corner and
perpendicular to the plane of square plate is given
by :
(1)
2
M
6
l
(2) Ml
2
(3)
2
M
12
l
(4)
2
2
M
3
l
Official Ans. by NTA (4)
Sol. According to perpendicular Axis theorem.
I
y
I
x
I
x
+ I
y
= I
z
I
z
??
2
m
3
+
2
m
3
=
2
2m
3
11. For a transistor in CE mode to be used as an
amplifier, it must be operated in :
(1) Both cut-off and Saturation
(2) Saturation region only
(3) Cut-off region only
(4) The active region only
Official Ans. by NTA (4)
Sol. Active region of the CE transistor is linear region
and is best suited for its use as an amplifier
12. An ideal gas is expanding such that PT
3
=
constant. The coefficient of volume expansion of
the gas is :
(1)
1
T
(2)
2
T
(3)
4
T
(4)
3
T
Official Ans. by NTA (3)
Sol. PT
3
= constant
3
nRT
T
v
??
??
??
= constant
T
4
V
–1
= constant
T
4
= kV
? ??4
T
T
?
=
V
V
?
.........(1)
?V ?= V ??T.............(2)
comparing (1) and (2)
we get
? =
4
T
13. In a photoelectric experiment, increasing the
intensity of incident light :
(1) increases the number of photons incident and
also increases the K.E. of the ejected electrons
(2) increases the frequency of photons incident and
increases the K.E. of the ejected electrons.
(3) increases the frequency of photons incident and
the K.E. of the ejected electrons remains
unchanged
(4) increases the number of photons incident and the
K.E. of the ejected electrons remains unchanged
Official Ans. by NTA (4)
Sol. ? Increasing intensity means number of incident
photons are increased.
?Kinetic energy of ejected electrons depend on the
frequency of incident photons, not the intensity.
4
14. A bar magnet is passing through a conducting loop
of radius R with velocity ?. The radius of the bar
magnet is such that it just passes through the loop.
The induced e.m.f. in the loop can be represented
by the approximate curve :
R
loop
N S
l
? ?
(1)
l/ ?
emf
t
(2)
t
emf
l/ ?
(3)
t
emf
l/ ?
(4)
t
emf
l/ ?
Official Ans. by NTA (3)
Sol.
S N
? When magnet passes through centre region of
solenoid , no current / Emf is induced in loop.
??While entering flux increases so negative
induced emf
? ?While leaving flux decreases so positive
induced emf.
15. Two ions of masses 4 amu and 16 amu have
charges +2e and +3e respectively. These ions pass
through the region of constant perpendicular
magnetic field. The kinetic energy of both ions is
same. Then :
(1) lighter ion will be deflected less than heavier ion
(2) lighter ion will be deflected more than heavier ion
(3) both ions will be deflected equally
(4) no ion will be deflected.
Official Ans. by NTA (2)
Sol. r =
P 2mk
qB qB
?
Given they have same kinetic energy
r ?
m
q
1
2
r
r
=
4
2
×
3
16
=
3
4
1
2
4r
r
3
? (r
2
is for hearier ion and r
1
is for lighter ion)
x x x
x x x
x x x
x x x
d
? ? R
sin ? =
d
R
? ???Deflection
? ?
1
R
(R ? Radius of path)
?R
2
> R
1
? ??
2
< ?
1
16. Find the distance of the image from object O,
formed by the combination of lenses in the figure :
f =+10cm
5cm 10cm
f = –10cm
30cm
f =+30cm
O
(1) 75 cm (2) 10 cm
(3) 20 cm (4) infinity
Official Ans. by NTA (1)
5
Sol.
1
1 1 1
V 30 10
??
1
12
V 30
? ? V
1
= 15 cm
2
11
–
V 10
= –
1
10
2
1
V
= 0 V
2
= ? ?
? ? V
3
= 30 cm
OV
3
= 75 cm ?
17. In Millikan's oil drop experiment, what is viscous
force acting on an uncharged drop of radius
2.0 × 10
–5
m and density 1.2 × 10
3
kgm
–3
? Take
viscosity of liquid = 1.8 × 10
–5
Nsm
–2
. (Neglect
buoyancy due to air).
(1) 3.8 × 10
–11
N (2) 3.9 × 10
–10
N
(3) 1.8 × 10
–10
N (4) 5.8 × 10
–10
N
Official Ans. by NTA (2)
Sol. Viscous force = Weight
= ? ×
3
4
rg
3
??
?
??
??
= 3.9 × 10
–10
18. Electric field in a plane electromagnetic wave is
given by E = 50 sin(500x – 10 × 10
10
t) V/m
The velocity of electromagnetic wave in this
medium is :
(Given C = speed of light in vacuum)
(1)
3
C
2
(2) C (3)
2
C
3
(4)
C
2
Official Ans. by NTA (3)
Sol. V =
K
?
=
10
10 10
500
?
= 2 × 10
8
V =
C
3
?
.
19. Five identical cells each of internal resistance 1 ?
and emf 5V are connected in series and in parallel
with an external resistance 'R'. For what value of
'R', current in series and parallel combination will
remain the same ?
(1) 1 ? (2) 25 ?
(3) 5 ? (4) 10 ?
Official Ans. by NTA (1)
Sol. i
1
=
25
5R ?
i
2
=
5
1
R
5
?
i
1
= i
2
?
1
5 R 5 R
5
??
? ? ?
??
??
4R = 4
R =1 ?
20. The variation of displacement with time of a
particle executing free simple harmonic motion is
shown in the figure.
x
O
A B
C
t
The potential energy U(x) versus time (t) plot of
the particle is correctly shown in figure :
(1)
U(x)
O
A B C
t
(2)
U(x)
O
A B
C
t
(3)
U(x)
O
A B
C
t
(4)
U(x)
O
A B
C
t
Official Ans. by NTA (4)
Sol. Potential energy is maximum at maximum distance
from mean.
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their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
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students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of JEE Mains 27 August 2021 Question Paper Shift 1 is prepared as per the latest JEE syllabus.
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