JEE Mains 27 August 2021 Question Paper Shift 1 | JEE Main & Advanced Previous Year Papers PDF Download

 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.