JEE Mains 27 August 2021 Question Paper Shift 2 | 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 : 3 : 00 PM  to  6 : 00 PM 
PHYSICS 
 
TEST PAPER WITH SOLUTION 
 
 
 
 
 
 
 
SECTION-A 
1. Curved surfaces of a plano-convex lens of 
refractive index ?
1
 and a plano-concave lens of 
refractive index ?
2
 have equal radius of curvature 
as shown in figure. Find the ratio of radius of 
curvature to the focal length of the combined 
lenses. 
 
?
2
 ?
1
 
 
 (1) 
? ? ?
21
1
  (2) ?
1
 – ?
2
  
 (3) 
? ? ?
12
1
  (4) ?
2
 – ?
1
 
 Official Ans. by NTA (2) 
Sol. 
 
f
1 
f
2 
 
 
??
??
??
??
1
1
11
( µ 1 )
fR
 
 
??
? ? ?
??
??
2
2
11
( µ 1 )
fR
 
 
? ? ?
? ? ?
12
1 2 eq
( µ 1 ) ( µ 1 )
1 1 1
f f f R
 
 
?
?
12
eq
( µ µ ) 1
fR
 
 ??
12
eq
R
( µ µ )
f
 
2. The boxes of masses 2 kg and 8 kg are connected 
by a massless string passing over smooth pulleys. 
Calculate the time taken by box of mass 8 kg to 
strike the ground starting from rest. (use g = 10 m/s
2
) 
 
20cm 
2kg 
8kg 
 
 (1) 0.34 s  (2) 0.2 s  
 (3) 0.25 s  (4) 0.4 s  
 Official Ans. by NTA (4) 
Sol. 
 
m
1
g 
8kg
 
a
 
2T
 
m
2
g 
2kg
 
T
 
2a
 
 
 (m
1
g – 2T) = m
1
a – (1) 
 T – m
2
g = m
2
(2a) 
 2T – 2m
2
g = 4m
2
 a – (2) 
 m
1
g – 2m
2
g = (m
1
 + 4m
2
) a 
 a = 
?
??
?
(8 4)g 4 g
g
(8 8) 16 4
 
 a = 
10
4
 m/s
2
 
 S = 
2
1
at
2
 
 
??
?
2
0.2 2 4
t
10
 
 t = 0.4 sec 
Page 2


 
    1 
FINAL JEE –MAIN EXAMINATION – AUGUST, 2021 
(Held On Friday 27
th
 August, 2021)                         TIME : 3 : 00 PM  to  6 : 00 PM 
PHYSICS 
 
TEST PAPER WITH SOLUTION 
 
 
 
 
 
 
 
SECTION-A 
1. Curved surfaces of a plano-convex lens of 
refractive index ?
1
 and a plano-concave lens of 
refractive index ?
2
 have equal radius of curvature 
as shown in figure. Find the ratio of radius of 
curvature to the focal length of the combined 
lenses. 
 
?
2
 ?
1
 
 
 (1) 
? ? ?
21
1
  (2) ?
1
 – ?
2
  
 (3) 
? ? ?
12
1
  (4) ?
2
 – ?
1
 
 Official Ans. by NTA (2) 
Sol. 
 
f
1 
f
2 
 
 
??
??
??
??
1
1
11
( µ 1 )
fR
 
 
??
? ? ?
??
??
2
2
11
( µ 1 )
fR
 
 
? ? ?
? ? ?
12
1 2 eq
( µ 1 ) ( µ 1 )
1 1 1
f f f R
 
 
?
?
12
eq
( µ µ ) 1
fR
 
 ??
12
eq
R
( µ µ )
f
 
2. The boxes of masses 2 kg and 8 kg are connected 
by a massless string passing over smooth pulleys. 
Calculate the time taken by box of mass 8 kg to 
strike the ground starting from rest. (use g = 10 m/s
2
) 
 
20cm 
2kg 
8kg 
 
 (1) 0.34 s  (2) 0.2 s  
 (3) 0.25 s  (4) 0.4 s  
 Official Ans. by NTA (4) 
Sol. 
 
m
1
g 
8kg
 
a
 
2T
 
m
2
g 
2kg
 
T
 
2a
 
 
 (m
1
g – 2T) = m
1
a – (1) 
 T – m
2
g = m
2
(2a) 
 2T – 2m
2
g = 4m
2
 a – (2) 
 m
1
g – 2m
2
g = (m
1
 + 4m
2
) a 
 a = 
?
??
?
(8 4)g 4 g
g
(8 8) 16 4
 
 a = 
10
4
 m/s
2
 
 S = 
2
1
at
2
 
 
??
?
2
0.2 2 4
t
10
 
 t = 0.4 sec 
 
2 
 
 
3. For a transistor ? and ? are given as ? = 
C
E
I
I
 and  
? = 
C
B
I
I
. Then the correct relation between ? and ? 
will be : 
 (1) 
??
??
?
1
 
 (2) 
?
??
?? 1
 
 (3) ?? = 1  
 (4) 
?
??
?? 1
  
 Official Ans. by NTA (2) 
Sol. ??
C
E
I
,
I
 ??
C
B
I
I
; I
E
 = I
C
 + I
B
 
 
?
? ? ? ?
? ? ?
?
C C B
C
CB
B
I I / I
I
I I 1
1
I
+ 
 ??
??
11
1 
 
?
?
??
1 1 –
 
 ? = 
?
? 1–
   
4. Water drops are falling from a nozzle of a shower 
onto the floor, from a height of 9.8 m. The drops 
fall at a regular interval of time. When the first 
drop strikes the floor, at that instant, the third drop 
begins to fall. Locate the position of second drop 
from the floor when the first drop strikes the floor. 
 (1) 4.18 m  
 (2) 2.94 m  
 (3) 2.45 m  
 (4) 7.35 m  
 Official Ans. by NTA (4) 
Sol. 
 
H=9.8 
3 
3 
2 
1 
h 
 
 H = 
2
1
gt
2
 
 
?
?
2
9.8 2
t
9.8
  
 t = 2 sec 
 ?t: time interval between drops  
 h = ??
2
1
g( 2 t)
2
 
 0 = ??
2
1
g( 2 2 t)
2
 
 ?t = 
1
2
 
 h = 
??
? ? ? ? ? ?
??
??
2
1 1 1 1 9.8
g 2 9.8 2.45m
2 2 2 4
2
 
 H – h = 9.8 – 2.45 
 = 7.35 m 
5. Two discs have moments of intertia I
1
 and I
2
 about 
their respective axes perpendicular to the plane and 
passing through the centre. They are rotating with 
angular speeds, ?
1
 and ?
2
 respectively and are 
brought into contact face to face with their axes of 
rotation coaxial. The loss in kinetic energy of the 
system in the process is given by : 
 (1) 
? ? ? ? ?
?
2
12
12
12
II
(I I )
  
 (2) 
? ? ?
?
2
1 2 1 2
12
(I I )
2(I I )
 
 (3) 
? ? ? ? ?
?
2 12
12
12
II
2(I I )
  
 (4) 
? ? ?
?
2
12
12
()
2(I I )
 
 Official Ans. by NTA (3) 
Page 3


 
    1 
FINAL JEE –MAIN EXAMINATION – AUGUST, 2021 
(Held On Friday 27
th
 August, 2021)                         TIME : 3 : 00 PM  to  6 : 00 PM 
PHYSICS 
 
TEST PAPER WITH SOLUTION 
 
 
 
 
 
 
 
SECTION-A 
1. Curved surfaces of a plano-convex lens of 
refractive index ?
1
 and a plano-concave lens of 
refractive index ?
2
 have equal radius of curvature 
as shown in figure. Find the ratio of radius of 
curvature to the focal length of the combined 
lenses. 
 
?
2
 ?
1
 
 
 (1) 
? ? ?
21
1
  (2) ?
1
 – ?
2
  
 (3) 
? ? ?
12
1
  (4) ?
2
 – ?
1
 
 Official Ans. by NTA (2) 
Sol. 
 
f
1 
f
2 
 
 
??
??
??
??
1
1
11
( µ 1 )
fR
 
 
??
? ? ?
??
??
2
2
11
( µ 1 )
fR
 
 
? ? ?
? ? ?
12
1 2 eq
( µ 1 ) ( µ 1 )
1 1 1
f f f R
 
 
?
?
12
eq
( µ µ ) 1
fR
 
 ??
12
eq
R
( µ µ )
f
 
2. The boxes of masses 2 kg and 8 kg are connected 
by a massless string passing over smooth pulleys. 
Calculate the time taken by box of mass 8 kg to 
strike the ground starting from rest. (use g = 10 m/s
2
) 
 
20cm 
2kg 
8kg 
 
 (1) 0.34 s  (2) 0.2 s  
 (3) 0.25 s  (4) 0.4 s  
 Official Ans. by NTA (4) 
Sol. 
 
m
1
g 
8kg
 
a
 
2T
 
m
2
g 
2kg
 
T
 
2a
 
 
 (m
1
g – 2T) = m
1
a – (1) 
 T – m
2
g = m
2
(2a) 
 2T – 2m
2
g = 4m
2
 a – (2) 
 m
1
g – 2m
2
g = (m
1
 + 4m
2
) a 
 a = 
?
??
?
(8 4)g 4 g
g
(8 8) 16 4
 
 a = 
10
4
 m/s
2
 
 S = 
2
1
at
2
 
 
??
?
2
0.2 2 4
t
10
 
 t = 0.4 sec 
 
2 
 
 
3. For a transistor ? and ? are given as ? = 
C
E
I
I
 and  
? = 
C
B
I
I
. Then the correct relation between ? and ? 
will be : 
 (1) 
??
??
?
1
 
 (2) 
?
??
?? 1
 
 (3) ?? = 1  
 (4) 
?
??
?? 1
  
 Official Ans. by NTA (2) 
Sol. ??
C
E
I
,
I
 ??
C
B
I
I
; I
E
 = I
C
 + I
B
 
 
?
? ? ? ?
? ? ?
?
C C B
C
CB
B
I I / I
I
I I 1
1
I
+ 
 ??
??
11
1 
 
?
?
??
1 1 –
 
 ? = 
?
? 1–
   
4. Water drops are falling from a nozzle of a shower 
onto the floor, from a height of 9.8 m. The drops 
fall at a regular interval of time. When the first 
drop strikes the floor, at that instant, the third drop 
begins to fall. Locate the position of second drop 
from the floor when the first drop strikes the floor. 
 (1) 4.18 m  
 (2) 2.94 m  
 (3) 2.45 m  
 (4) 7.35 m  
 Official Ans. by NTA (4) 
Sol. 
 
H=9.8 
3 
3 
2 
1 
h 
 
 H = 
2
1
gt
2
 
 
?
?
2
9.8 2
t
9.8
  
 t = 2 sec 
 ?t: time interval between drops  
 h = ??
2
1
g( 2 t)
2
 
 0 = ??
2
1
g( 2 2 t)
2
 
 ?t = 
1
2
 
 h = 
??
? ? ? ? ? ?
??
??
2
1 1 1 1 9.8
g 2 9.8 2.45m
2 2 2 4
2
 
 H – h = 9.8 – 2.45 
 = 7.35 m 
5. Two discs have moments of intertia I
1
 and I
2
 about 
their respective axes perpendicular to the plane and 
passing through the centre. They are rotating with 
angular speeds, ?
1
 and ?
2
 respectively and are 
brought into contact face to face with their axes of 
rotation coaxial. The loss in kinetic energy of the 
system in the process is given by : 
 (1) 
? ? ? ? ?
?
2
12
12
12
II
(I I )
  
 (2) 
? ? ?
?
2
1 2 1 2
12
(I I )
2(I I )
 
 (3) 
? ? ? ? ?
?
2 12
12
12
II
2(I I )
  
 (4) 
? ? ?
?
2
12
12
()
2(I I )
 
 Official Ans. by NTA (3) 
 
 
3 
 
?  
Sol. From conservation of angular momentum we get 
 I
1
?
1
 + I
2
?
2
 = (I
1
 + I
2
) ? ?
 ? ????
? ? ?
?
1 1 2 2
12
II
II
 
 k
i
 = ? ? ?
22
1 1 2 2
11
II
22
 
 k
f
 = ??
2
12
1
(I I )
2
 
 
?? ? ? ?
? ? ? ? ? ?
??
?
??
2
22 1 1 2 2
i f 1 1 2 2
12
(I I ) 1
k k I I
2 I I
 
 Solving above we get 
 k
i
 – k
f
 = 
??
? ? ?
??
?
??
2 12
12
12
II 1
()
2 I I
 
6. Three capacitors C
1
 = 2 ?F, C
2
 = 6 ?F and  
C
3
 = 12 ?F are connected as shown in figure. Find 
the ratio of the charges on capacitors C
1
, C
2
 and C
3
 
respectively : 
 
A B 
C
1
 
D 
C
2
 C
3
 
v 
 
 (1) 2 : 1 : 1  (2) 2 : 3 : 3 
 (3) 1 : 2 : 2  (4) 3 : 4 : 4 
 Official Ans. by NTA (3) 
Sol.  
 
A B 
C
1
 
D 
C
2
 C
3
 
V 
0 
0 V 
V
D 
 
 
 (V
D
 – V) C
2
 + (V
D
 –0) C
3
 = 0 
 (V
D
 – V) 6 + (V
D
 –0) 12 = 0 
 V
D
–V
 
+ 2V
D
 = 0 
 V
D
 = 
V
3
 
 q
2
 = (V – V
D
) C
2
 = 
??
?
??
??
V
V
3
(6 µF) 
 q
2
 = (4V) µF 
 q
3
 = (V
D 
– 0) C
3
 = ??
V
1 2 µ F 4 V µ F
3
 
 q
1
 = (V – 0) C
1
 = V(2µF) 
 q
1
 : q
2
 : q
3 
 = 2 : 4 : 4 
 q
1
 : q
2
 : q
3 
= 1 : 2 : 2 
7. The colour coding on a carbon resistor is shown in 
the given figure. The resistance value of the given 
resistor is : 
 
Gold 
Red 
Green 
Violet 
 
 (1) (5700 ± 285) ? ?
 (2) (7500 ± 750) ? 
 (3) (5700 ± 375) ? ?
 (4) (7500 ± 375) ?  
 Official Ans. by NTA (4) 
Sol. R = 75 × 10
2
 ± 5% of 7500 
 R = (7500 ± 375) ? 
8. An antenna is mounted on a 400 m tall building. 
What will be the wavelength of signal of signal 
that can be radiated effectively by the transmission 
tower upto a range of 44 km? 
 (1) 37.8 m  
 (2) 605 m  
 (3) 75.6 m  
 (4) 302 m  
 Official Ans. by NTA (2) 
Sol. h : height of antenna 
 ? : wavelength of signal  
 h < ?  
 ? > h 
 ? > 400 m 
Page 4


 
    1 
FINAL JEE –MAIN EXAMINATION – AUGUST, 2021 
(Held On Friday 27
th
 August, 2021)                         TIME : 3 : 00 PM  to  6 : 00 PM 
PHYSICS 
 
TEST PAPER WITH SOLUTION 
 
 
 
 
 
 
 
SECTION-A 
1. Curved surfaces of a plano-convex lens of 
refractive index ?
1
 and a plano-concave lens of 
refractive index ?
2
 have equal radius of curvature 
as shown in figure. Find the ratio of radius of 
curvature to the focal length of the combined 
lenses. 
 
?
2
 ?
1
 
 
 (1) 
? ? ?
21
1
  (2) ?
1
 – ?
2
  
 (3) 
? ? ?
12
1
  (4) ?
2
 – ?
1
 
 Official Ans. by NTA (2) 
Sol. 
 
f
1 
f
2 
 
 
??
??
??
??
1
1
11
( µ 1 )
fR
 
 
??
? ? ?
??
??
2
2
11
( µ 1 )
fR
 
 
? ? ?
? ? ?
12
1 2 eq
( µ 1 ) ( µ 1 )
1 1 1
f f f R
 
 
?
?
12
eq
( µ µ ) 1
fR
 
 ??
12
eq
R
( µ µ )
f
 
2. The boxes of masses 2 kg and 8 kg are connected 
by a massless string passing over smooth pulleys. 
Calculate the time taken by box of mass 8 kg to 
strike the ground starting from rest. (use g = 10 m/s
2
) 
 
20cm 
2kg 
8kg 
 
 (1) 0.34 s  (2) 0.2 s  
 (3) 0.25 s  (4) 0.4 s  
 Official Ans. by NTA (4) 
Sol. 
 
m
1
g 
8kg
 
a
 
2T
 
m
2
g 
2kg
 
T
 
2a
 
 
 (m
1
g – 2T) = m
1
a – (1) 
 T – m
2
g = m
2
(2a) 
 2T – 2m
2
g = 4m
2
 a – (2) 
 m
1
g – 2m
2
g = (m
1
 + 4m
2
) a 
 a = 
?
??
?
(8 4)g 4 g
g
(8 8) 16 4
 
 a = 
10
4
 m/s
2
 
 S = 
2
1
at
2
 
 
??
?
2
0.2 2 4
t
10
 
 t = 0.4 sec 
 
2 
 
 
3. For a transistor ? and ? are given as ? = 
C
E
I
I
 and  
? = 
C
B
I
I
. Then the correct relation between ? and ? 
will be : 
 (1) 
??
??
?
1
 
 (2) 
?
??
?? 1
 
 (3) ?? = 1  
 (4) 
?
??
?? 1
  
 Official Ans. by NTA (2) 
Sol. ??
C
E
I
,
I
 ??
C
B
I
I
; I
E
 = I
C
 + I
B
 
 
?
? ? ? ?
? ? ?
?
C C B
C
CB
B
I I / I
I
I I 1
1
I
+ 
 ??
??
11
1 
 
?
?
??
1 1 –
 
 ? = 
?
? 1–
   
4. Water drops are falling from a nozzle of a shower 
onto the floor, from a height of 9.8 m. The drops 
fall at a regular interval of time. When the first 
drop strikes the floor, at that instant, the third drop 
begins to fall. Locate the position of second drop 
from the floor when the first drop strikes the floor. 
 (1) 4.18 m  
 (2) 2.94 m  
 (3) 2.45 m  
 (4) 7.35 m  
 Official Ans. by NTA (4) 
Sol. 
 
H=9.8 
3 
3 
2 
1 
h 
 
 H = 
2
1
gt
2
 
 
?
?
2
9.8 2
t
9.8
  
 t = 2 sec 
 ?t: time interval between drops  
 h = ??
2
1
g( 2 t)
2
 
 0 = ??
2
1
g( 2 2 t)
2
 
 ?t = 
1
2
 
 h = 
??
? ? ? ? ? ?
??
??
2
1 1 1 1 9.8
g 2 9.8 2.45m
2 2 2 4
2
 
 H – h = 9.8 – 2.45 
 = 7.35 m 
5. Two discs have moments of intertia I
1
 and I
2
 about 
their respective axes perpendicular to the plane and 
passing through the centre. They are rotating with 
angular speeds, ?
1
 and ?
2
 respectively and are 
brought into contact face to face with their axes of 
rotation coaxial. The loss in kinetic energy of the 
system in the process is given by : 
 (1) 
? ? ? ? ?
?
2
12
12
12
II
(I I )
  
 (2) 
? ? ?
?
2
1 2 1 2
12
(I I )
2(I I )
 
 (3) 
? ? ? ? ?
?
2 12
12
12
II
2(I I )
  
 (4) 
? ? ?
?
2
12
12
()
2(I I )
 
 Official Ans. by NTA (3) 
 
 
3 
 
?  
Sol. From conservation of angular momentum we get 
 I
1
?
1
 + I
2
?
2
 = (I
1
 + I
2
) ? ?
 ? ????
? ? ?
?
1 1 2 2
12
II
II
 
 k
i
 = ? ? ?
22
1 1 2 2
11
II
22
 
 k
f
 = ??
2
12
1
(I I )
2
 
 
?? ? ? ?
? ? ? ? ? ?
??
?
??
2
22 1 1 2 2
i f 1 1 2 2
12
(I I ) 1
k k I I
2 I I
 
 Solving above we get 
 k
i
 – k
f
 = 
??
? ? ?
??
?
??
2 12
12
12
II 1
()
2 I I
 
6. Three capacitors C
1
 = 2 ?F, C
2
 = 6 ?F and  
C
3
 = 12 ?F are connected as shown in figure. Find 
the ratio of the charges on capacitors C
1
, C
2
 and C
3
 
respectively : 
 
A B 
C
1
 
D 
C
2
 C
3
 
v 
 
 (1) 2 : 1 : 1  (2) 2 : 3 : 3 
 (3) 1 : 2 : 2  (4) 3 : 4 : 4 
 Official Ans. by NTA (3) 
Sol.  
 
A B 
C
1
 
D 
C
2
 C
3
 
V 
0 
0 V 
V
D 
 
 
 (V
D
 – V) C
2
 + (V
D
 –0) C
3
 = 0 
 (V
D
 – V) 6 + (V
D
 –0) 12 = 0 
 V
D
–V
 
+ 2V
D
 = 0 
 V
D
 = 
V
3
 
 q
2
 = (V – V
D
) C
2
 = 
??
?
??
??
V
V
3
(6 µF) 
 q
2
 = (4V) µF 
 q
3
 = (V
D 
– 0) C
3
 = ??
V
1 2 µ F 4 V µ F
3
 
 q
1
 = (V – 0) C
1
 = V(2µF) 
 q
1
 : q
2
 : q
3 
 = 2 : 4 : 4 
 q
1
 : q
2
 : q
3 
= 1 : 2 : 2 
7. The colour coding on a carbon resistor is shown in 
the given figure. The resistance value of the given 
resistor is : 
 
Gold 
Red 
Green 
Violet 
 
 (1) (5700 ± 285) ? ?
 (2) (7500 ± 750) ? 
 (3) (5700 ± 375) ? ?
 (4) (7500 ± 375) ?  
 Official Ans. by NTA (4) 
Sol. R = 75 × 10
2
 ± 5% of 7500 
 R = (7500 ± 375) ? 
8. An antenna is mounted on a 400 m tall building. 
What will be the wavelength of signal of signal 
that can be radiated effectively by the transmission 
tower upto a range of 44 km? 
 (1) 37.8 m  
 (2) 605 m  
 (3) 75.6 m  
 (4) 302 m  
 Official Ans. by NTA (2) 
Sol. h : height of antenna 
 ? : wavelength of signal  
 h < ?  
 ? > h 
 ? > 400 m 
 
4 
 
 
9. If the rms speed of oxygen molecules at 0°C is  
160 m/s, find the rms speed of hydrogen molecules 
at 0°C. 
 (1) 640 m/s  (2) 40 m/s 
 (3) 80 m/s  (4) 332 m/s  
 Official Ans. by NTA (1) 
Sol. V
rms
 = 
3KT
M
 
 ??
22
22
rms O H
rms H O
(V ) M
2
(V ) M 32
 
 
? ? ??
2
2
rms H rms
O
(V ) 4 V 
             = 4 × 160 
             = 640 m/s 
10. A constant magnetic field of 1 T is applied in the  
x > 0 region. A metallic circular ring of radius 1m 
is moving with a constant velocity of 1 m/s along 
the x-axis. At t = 0s, the centre of O of the ring is 
at x = – 1m.  What will be the value of the induced 
emf in the ring at t = 1s? (Assume the velocity of 
the ring does not change.) 
 
x=0 
x 
 
O 
 
 (1) 1 V  (2) 2 ? ?V  
 (3) 2 V  (4) 0 V  
 Official Ans. by NTA (3) 
Sol. emf = ?LV 
       = 1.(2R) .1 
       = 2 V 
11. A mass of 50 kg is placed at the centre of a 
uniform spherical shell of mass 100 kg and radius 
50 m. If the gravitational potential at a point, 25 m 
from the centre is V kg/m. The value of V is : 
 (1) – 60 G  (2) + 2 G 
 (3) – 20 G  (4) – 4 G  
 Official Ans. by NTA (4) 
Sol. 
 
A 
r 
50 kg 
100 kg 
50 m 
 
 V
A
 = 
??
?
??
??
12
GM GM
–
rR
 
      = 
??
?
??
??
50 100
– G G
25 50
 
       = –4G  
12. For full scale deflection of total 50 divisions,  
50 mV voltage is required in galvanometer. The 
resistance of galvanometer if its current sensitivity 
is 2 div/mA will be : 
 (1) 1 ? (2) 5 ? (3) 4 ? (4) 2 ?  
 Official Ans. by NTA (4) 
Sol. I
max
 = ?
50
25mA
2
 
 R = ??
V 50mV
I 25mA
2? 
13. A monochromatic neon lamp with wavelength of 
670.5 nm illuminates a photo-sensitive material 
which has a stopping voltage of 0.48 V. What will 
be the stopping voltage if the source light is 
changed with another source of wavelength of 
474.6 nm? 
 (1) 0.96 V (2) 1.25 V (3) 0.24 V (4) 1.5 V 
 Official Ans. by NTA (2) 
Sol.  kE
max
 = ??
?
i
hc
 
 or  eV
o
 = ??
?
i
hc
 
  when ?
i
 = 670.5 nm ;   V
o
 = 0.48 
  when ?
i
 = 474.6 nm ;   V
o
 = ? 
 So, e(0.48) = 
1240
670.5
+ ?  ...(1) 
  e(V
o
) = ??
1240
474.6
  ...(2) 
 (2) – (1) 
  e(V
o
 – 0.48) = 1240 
??
?
??
??
11
474.6 670.5
eV 
  V
o
 = 0.48 + 1240 
??
??
?
??
6 7 0 . 5 – 4 7 4 . 6
474.6 670.5
 Volts 
 V
o
 = 0.48 + 0.76 
 V
o
 = 1.24 V  1.25 V 
Page 5


 
    1 
FINAL JEE –MAIN EXAMINATION – AUGUST, 2021 
(Held On Friday 27
th
 August, 2021)                         TIME : 3 : 00 PM  to  6 : 00 PM 
PHYSICS 
 
TEST PAPER WITH SOLUTION 
 
 
 
 
 
 
 
SECTION-A 
1. Curved surfaces of a plano-convex lens of 
refractive index ?
1
 and a plano-concave lens of 
refractive index ?
2
 have equal radius of curvature 
as shown in figure. Find the ratio of radius of 
curvature to the focal length of the combined 
lenses. 
 
?
2
 ?
1
 
 
 (1) 
? ? ?
21
1
  (2) ?
1
 – ?
2
  
 (3) 
? ? ?
12
1
  (4) ?
2
 – ?
1
 
 Official Ans. by NTA (2) 
Sol. 
 
f
1 
f
2 
 
 
??
??
??
??
1
1
11
( µ 1 )
fR
 
 
??
? ? ?
??
??
2
2
11
( µ 1 )
fR
 
 
? ? ?
? ? ?
12
1 2 eq
( µ 1 ) ( µ 1 )
1 1 1
f f f R
 
 
?
?
12
eq
( µ µ ) 1
fR
 
 ??
12
eq
R
( µ µ )
f
 
2. The boxes of masses 2 kg and 8 kg are connected 
by a massless string passing over smooth pulleys. 
Calculate the time taken by box of mass 8 kg to 
strike the ground starting from rest. (use g = 10 m/s
2
) 
 
20cm 
2kg 
8kg 
 
 (1) 0.34 s  (2) 0.2 s  
 (3) 0.25 s  (4) 0.4 s  
 Official Ans. by NTA (4) 
Sol. 
 
m
1
g 
8kg
 
a
 
2T
 
m
2
g 
2kg
 
T
 
2a
 
 
 (m
1
g – 2T) = m
1
a – (1) 
 T – m
2
g = m
2
(2a) 
 2T – 2m
2
g = 4m
2
 a – (2) 
 m
1
g – 2m
2
g = (m
1
 + 4m
2
) a 
 a = 
?
??
?
(8 4)g 4 g
g
(8 8) 16 4
 
 a = 
10
4
 m/s
2
 
 S = 
2
1
at
2
 
 
??
?
2
0.2 2 4
t
10
 
 t = 0.4 sec 
 
2 
 
 
3. For a transistor ? and ? are given as ? = 
C
E
I
I
 and  
? = 
C
B
I
I
. Then the correct relation between ? and ? 
will be : 
 (1) 
??
??
?
1
 
 (2) 
?
??
?? 1
 
 (3) ?? = 1  
 (4) 
?
??
?? 1
  
 Official Ans. by NTA (2) 
Sol. ??
C
E
I
,
I
 ??
C
B
I
I
; I
E
 = I
C
 + I
B
 
 
?
? ? ? ?
? ? ?
?
C C B
C
CB
B
I I / I
I
I I 1
1
I
+ 
 ??
??
11
1 
 
?
?
??
1 1 –
 
 ? = 
?
? 1–
   
4. Water drops are falling from a nozzle of a shower 
onto the floor, from a height of 9.8 m. The drops 
fall at a regular interval of time. When the first 
drop strikes the floor, at that instant, the third drop 
begins to fall. Locate the position of second drop 
from the floor when the first drop strikes the floor. 
 (1) 4.18 m  
 (2) 2.94 m  
 (3) 2.45 m  
 (4) 7.35 m  
 Official Ans. by NTA (4) 
Sol. 
 
H=9.8 
3 
3 
2 
1 
h 
 
 H = 
2
1
gt
2
 
 
?
?
2
9.8 2
t
9.8
  
 t = 2 sec 
 ?t: time interval between drops  
 h = ??
2
1
g( 2 t)
2
 
 0 = ??
2
1
g( 2 2 t)
2
 
 ?t = 
1
2
 
 h = 
??
? ? ? ? ? ?
??
??
2
1 1 1 1 9.8
g 2 9.8 2.45m
2 2 2 4
2
 
 H – h = 9.8 – 2.45 
 = 7.35 m 
5. Two discs have moments of intertia I
1
 and I
2
 about 
their respective axes perpendicular to the plane and 
passing through the centre. They are rotating with 
angular speeds, ?
1
 and ?
2
 respectively and are 
brought into contact face to face with their axes of 
rotation coaxial. The loss in kinetic energy of the 
system in the process is given by : 
 (1) 
? ? ? ? ?
?
2
12
12
12
II
(I I )
  
 (2) 
? ? ?
?
2
1 2 1 2
12
(I I )
2(I I )
 
 (3) 
? ? ? ? ?
?
2 12
12
12
II
2(I I )
  
 (4) 
? ? ?
?
2
12
12
()
2(I I )
 
 Official Ans. by NTA (3) 
 
 
3 
 
?  
Sol. From conservation of angular momentum we get 
 I
1
?
1
 + I
2
?
2
 = (I
1
 + I
2
) ? ?
 ? ????
? ? ?
?
1 1 2 2
12
II
II
 
 k
i
 = ? ? ?
22
1 1 2 2
11
II
22
 
 k
f
 = ??
2
12
1
(I I )
2
 
 
?? ? ? ?
? ? ? ? ? ?
??
?
??
2
22 1 1 2 2
i f 1 1 2 2
12
(I I ) 1
k k I I
2 I I
 
 Solving above we get 
 k
i
 – k
f
 = 
??
? ? ?
??
?
??
2 12
12
12
II 1
()
2 I I
 
6. Three capacitors C
1
 = 2 ?F, C
2
 = 6 ?F and  
C
3
 = 12 ?F are connected as shown in figure. Find 
the ratio of the charges on capacitors C
1
, C
2
 and C
3
 
respectively : 
 
A B 
C
1
 
D 
C
2
 C
3
 
v 
 
 (1) 2 : 1 : 1  (2) 2 : 3 : 3 
 (3) 1 : 2 : 2  (4) 3 : 4 : 4 
 Official Ans. by NTA (3) 
Sol.  
 
A B 
C
1
 
D 
C
2
 C
3
 
V 
0 
0 V 
V
D 
 
 
 (V
D
 – V) C
2
 + (V
D
 –0) C
3
 = 0 
 (V
D
 – V) 6 + (V
D
 –0) 12 = 0 
 V
D
–V
 
+ 2V
D
 = 0 
 V
D
 = 
V
3
 
 q
2
 = (V – V
D
) C
2
 = 
??
?
??
??
V
V
3
(6 µF) 
 q
2
 = (4V) µF 
 q
3
 = (V
D 
– 0) C
3
 = ??
V
1 2 µ F 4 V µ F
3
 
 q
1
 = (V – 0) C
1
 = V(2µF) 
 q
1
 : q
2
 : q
3 
 = 2 : 4 : 4 
 q
1
 : q
2
 : q
3 
= 1 : 2 : 2 
7. The colour coding on a carbon resistor is shown in 
the given figure. The resistance value of the given 
resistor is : 
 
Gold 
Red 
Green 
Violet 
 
 (1) (5700 ± 285) ? ?
 (2) (7500 ± 750) ? 
 (3) (5700 ± 375) ? ?
 (4) (7500 ± 375) ?  
 Official Ans. by NTA (4) 
Sol. R = 75 × 10
2
 ± 5% of 7500 
 R = (7500 ± 375) ? 
8. An antenna is mounted on a 400 m tall building. 
What will be the wavelength of signal of signal 
that can be radiated effectively by the transmission 
tower upto a range of 44 km? 
 (1) 37.8 m  
 (2) 605 m  
 (3) 75.6 m  
 (4) 302 m  
 Official Ans. by NTA (2) 
Sol. h : height of antenna 
 ? : wavelength of signal  
 h < ?  
 ? > h 
 ? > 400 m 
 
4 
 
 
9. If the rms speed of oxygen molecules at 0°C is  
160 m/s, find the rms speed of hydrogen molecules 
at 0°C. 
 (1) 640 m/s  (2) 40 m/s 
 (3) 80 m/s  (4) 332 m/s  
 Official Ans. by NTA (1) 
Sol. V
rms
 = 
3KT
M
 
 ??
22
22
rms O H
rms H O
(V ) M
2
(V ) M 32
 
 
? ? ??
2
2
rms H rms
O
(V ) 4 V 
             = 4 × 160 
             = 640 m/s 
10. A constant magnetic field of 1 T is applied in the  
x > 0 region. A metallic circular ring of radius 1m 
is moving with a constant velocity of 1 m/s along 
the x-axis. At t = 0s, the centre of O of the ring is 
at x = – 1m.  What will be the value of the induced 
emf in the ring at t = 1s? (Assume the velocity of 
the ring does not change.) 
 
x=0 
x 
 
O 
 
 (1) 1 V  (2) 2 ? ?V  
 (3) 2 V  (4) 0 V  
 Official Ans. by NTA (3) 
Sol. emf = ?LV 
       = 1.(2R) .1 
       = 2 V 
11. A mass of 50 kg is placed at the centre of a 
uniform spherical shell of mass 100 kg and radius 
50 m. If the gravitational potential at a point, 25 m 
from the centre is V kg/m. The value of V is : 
 (1) – 60 G  (2) + 2 G 
 (3) – 20 G  (4) – 4 G  
 Official Ans. by NTA (4) 
Sol. 
 
A 
r 
50 kg 
100 kg 
50 m 
 
 V
A
 = 
??
?
??
??
12
GM GM
–
rR
 
      = 
??
?
??
??
50 100
– G G
25 50
 
       = –4G  
12. For full scale deflection of total 50 divisions,  
50 mV voltage is required in galvanometer. The 
resistance of galvanometer if its current sensitivity 
is 2 div/mA will be : 
 (1) 1 ? (2) 5 ? (3) 4 ? (4) 2 ?  
 Official Ans. by NTA (4) 
Sol. I
max
 = ?
50
25mA
2
 
 R = ??
V 50mV
I 25mA
2? 
13. A monochromatic neon lamp with wavelength of 
670.5 nm illuminates a photo-sensitive material 
which has a stopping voltage of 0.48 V. What will 
be the stopping voltage if the source light is 
changed with another source of wavelength of 
474.6 nm? 
 (1) 0.96 V (2) 1.25 V (3) 0.24 V (4) 1.5 V 
 Official Ans. by NTA (2) 
Sol.  kE
max
 = ??
?
i
hc
 
 or  eV
o
 = ??
?
i
hc
 
  when ?
i
 = 670.5 nm ;   V
o
 = 0.48 
  when ?
i
 = 474.6 nm ;   V
o
 = ? 
 So, e(0.48) = 
1240
670.5
+ ?  ...(1) 
  e(V
o
) = ??
1240
474.6
  ...(2) 
 (2) – (1) 
  e(V
o
 – 0.48) = 1240 
??
?
??
??
11
474.6 670.5
eV 
  V
o
 = 0.48 + 1240 
??
??
?
??
6 7 0 . 5 – 4 7 4 . 6
474.6 670.5
 Volts 
 V
o
 = 0.48 + 0.76 
 V
o
 = 1.24 V  1.25 V 
 
 
5 
 
?  
14. Match List-I with List-II.  
           List-I             List-II  
 (a) R
H
 (Rydberg constant)        (i) kg m
–1 
s
–1
  
 (b) h(Planck's constant)       (ii) kg m
2
 s
–1 
 
(c) ?
B
 (Magnetic field        (iii) m
–1 
          
energy density)
 
 (d) ?(coefficient of viscocity)       (iv) kg m
–1
 s
–2
  
 Choose the most appropriate answer from the 
options given below :  
 (1) (a) –(ii), (b) –(iii), (c) –(iv), (d) –(i) 
 (2) (a) –(iii), (b) –(ii), (c) –(iv), (d) –(i) 
 (3) (a) –(iv), (b) –(ii), (c) –(i), (d) –(iii) 
 (4) (a) –(iii), (b) –(ii), (c) –(i), (d) –(iv) 
 Official Ans. by NTA (2) 
Sol. SI unit of Rydberg const. = m
–1
 
 SI unit of Plank's const. = kg m
2
s
–1
 
 SI unit of Magnetic field energy density= kg m
–1
s
–2
 
 SI unit of coeff. of viscosity = kg m
–1
s
–1
 
15. If force (F), length (L) and time (T) are taken as 
the fundamental quantities. Then what will be the 
dimension of density :  
 (1)[FL
–4 
T
2
]  
 (2) [FL
–3 
T
2
]  
 (3) [FL
–5 
T
2
]  
 (4) [FL
–3 
T
3
]  
 Official Ans. by NTA (1) 
Sol. Density = [F
a
L
b
T
c
] 
 [ML
–3
] = [M
a
L
a
T
–2a
L
b
T
c
] 
 [M
1
L
–3
] = [M
a
L
a+b 
T
–2a+c
] 
 a = 1   ;   a + b = –3   ;  –2a + c = 0 
     1 + b = –3 c = 2a 
     b = – 4 c = 2 
 So, density = [F
1
L
–4
T
2
] 
16. A coaxial cable consists of an inner wire of radius 
'a' surrounded by an outer shell of inner and outer 
radii 'b' and 'c' respectively. The inner wire carries 
an electric current i
0
, which is distributed 
uniformly across cross-sectional area. The outer 
shell carries an equal current in opposite direction 
and distributed uniformly. What will be the ratio of 
the magnetic field at a distance x from the axis 
when (i) x < a and (ii) a < x < b ? 
 (1) 
2
2
x
a
  (2) 
2
2
a
x
  
 (3)
2
22
x
b – a
  (4) 
22
2
b – a
x
 
 Official Ans. by NTA (1) 
Sol. 
 
i
o 
a
 
i
o 
b
 
c
 
 
when x < a 
 B
1
 (2 ?x) = 
??
?
??
?
??
2 o
o 2
i
µx
a
 
 B(2 ?x) = 
2
oo
2
µ i x
a
 
 B
1
 = 
?
oo
2
µ i x
2a
  ...(1) 
 when a < x < b 
 B
2
(2?x) = µ
o
i
o
 
 B
2
 = 
?
oo
µi
2x
  ...(2) 
 
?
?
?
oo 2
1
oo
2
x
µi
B
2a
µi
B
2x
 = 
2
2
x
a