JEE Mains 20 July 2021 Question Paper Shift 2 | JEE Main & Advanced Previous Year Papers PDF Download

 Page 1


JEE-Main-20-07-2021-Shift-2 (Memory Based) 
 
PHYSICS 
 
Question: In a series LCR circuit, 5 , 0.5mH, 2.5µF RL C = ?= = . The RMS value of 
external voltage is 250 V. Find the power dissipated if circuit is in Resonance 
Answer: 12500 W 
Solution:  
Power
rms rms
R
VI
Z
? ?
=
? ?
? ?
 
Z = R (at resonance) 
250 5
Power 250
55
12500 W
= × ×
=
 
  
Question: The wavelength of sodium lamp is observed to be 2886Å from earth & original 
wavelength was 2880Å . Find speed of galaxy. 
Options: 
(a) 
51
3 10 ms
-
× 
(b) 
51
4 10 ms
-
× 
(c) 
51
6.25 10 ms
-
× 
(d) None 
Answer: (c) 
Solution: 2886 2880 6 ? ?= - = Å 
Using doppler shift, 
radial
V
C
?
?
?
-=- 
10
8
radial 10
6 10
3 10
2880 10
VC
?
?
-
-
?? ?× ? ?
? = = × ×
?? ? ?
×
? ?
??
 
51
6.25 10 ms
-
= × 
Hence, speed of galaxy 
51
6.25 10 ms
-
= × . 
 
Question: A body is under the influence of a force such that it delivers a constant power P. 
The variation of position with time of body as 
Page 2


JEE-Main-20-07-2021-Shift-2 (Memory Based) 
 
PHYSICS 
 
Question: In a series LCR circuit, 5 , 0.5mH, 2.5µF RL C = ?= = . The RMS value of 
external voltage is 250 V. Find the power dissipated if circuit is in Resonance 
Answer: 12500 W 
Solution:  
Power
rms rms
R
VI
Z
? ?
=
? ?
? ?
 
Z = R (at resonance) 
250 5
Power 250
55
12500 W
= × ×
=
 
  
Question: The wavelength of sodium lamp is observed to be 2886Å from earth & original 
wavelength was 2880Å . Find speed of galaxy. 
Options: 
(a) 
51
3 10 ms
-
× 
(b) 
51
4 10 ms
-
× 
(c) 
51
6.25 10 ms
-
× 
(d) None 
Answer: (c) 
Solution: 2886 2880 6 ? ?= - = Å 
Using doppler shift, 
radial
V
C
?
?
?
-=- 
10
8
radial 10
6 10
3 10
2880 10
VC
?
?
-
-
?? ?× ? ?
? = = × ×
?? ? ?
×
? ?
??
 
51
6.25 10 ms
-
= × 
Hence, speed of galaxy 
51
6.25 10 ms
-
= × . 
 
Question: A body is under the influence of a force such that it delivers a constant power P. 
The variation of position with time of body as 
Options: 
(a) 
1
2
t 
(b) 
3
2
t 
(c) 
5
2
t 
(d) None 
Answer: (b) 
Solution: Power = P 
F v = P 
dv
m vP
dt
??
·=
??
??
 
00
vt
P
vdv dt
m
? = ·
? ?
 
2P
v t
m
?= 
[ ]
2
assuming at 0, 0 & 0
dx P
t t x v
dt m
? = = = = 
00
2
xt
P
dx t dt
m
?= ·
??
 
3/2
22
3
P
tx t
m
?= = 
3/2
x t ??  
 
Question: When a metal is illuminated by light of wavelength ? ,the stopping potential is 
0
V
and for wavelength 2 ? , it is 
0
3V . Then the threshold wavelength is? 
Options: 
(a) 
2
3
?
 
(b) 
4
5
?
 
(c) 
3
?
 
(d) 
5
2
?
 
Answer: (b) 
Solution: 
( )
0
... 1
hc
eV f
?
= - 
Page 3


JEE-Main-20-07-2021-Shift-2 (Memory Based) 
 
PHYSICS 
 
Question: In a series LCR circuit, 5 , 0.5mH, 2.5µF RL C = ?= = . The RMS value of 
external voltage is 250 V. Find the power dissipated if circuit is in Resonance 
Answer: 12500 W 
Solution:  
Power
rms rms
R
VI
Z
? ?
=
? ?
? ?
 
Z = R (at resonance) 
250 5
Power 250
55
12500 W
= × ×
=
 
  
Question: The wavelength of sodium lamp is observed to be 2886Å from earth & original 
wavelength was 2880Å . Find speed of galaxy. 
Options: 
(a) 
51
3 10 ms
-
× 
(b) 
51
4 10 ms
-
× 
(c) 
51
6.25 10 ms
-
× 
(d) None 
Answer: (c) 
Solution: 2886 2880 6 ? ?= - = Å 
Using doppler shift, 
radial
V
C
?
?
?
-=- 
10
8
radial 10
6 10
3 10
2880 10
VC
?
?
-
-
?? ?× ? ?
? = = × ×
?? ? ?
×
? ?
??
 
51
6.25 10 ms
-
= × 
Hence, speed of galaxy 
51
6.25 10 ms
-
= × . 
 
Question: A body is under the influence of a force such that it delivers a constant power P. 
The variation of position with time of body as 
Options: 
(a) 
1
2
t 
(b) 
3
2
t 
(c) 
5
2
t 
(d) None 
Answer: (b) 
Solution: Power = P 
F v = P 
dv
m vP
dt
??
·=
??
??
 
00
vt
P
vdv dt
m
? = ·
? ?
 
2P
v t
m
?= 
[ ]
2
assuming at 0, 0 & 0
dx P
t t x v
dt m
? = = = = 
00
2
xt
P
dx t dt
m
?= ·
??
 
3/2
22
3
P
tx t
m
?= = 
3/2
x t ??  
 
Question: When a metal is illuminated by light of wavelength ? ,the stopping potential is 
0
V
and for wavelength 2 ? , it is 
0
3V . Then the threshold wavelength is? 
Options: 
(a) 
2
3
?
 
(b) 
4
5
?
 
(c) 
3
?
 
(d) 
5
2
?
 
Answer: (b) 
Solution: 
( )
0
... 1
hc
eV f
?
= - 
( )
0
3 ... 2
2
hc
eV f
?
= - 
Multiply by 3 in equation (1) 
( )
3
3 3 ... 3
hc
eV f
?
= - 
Equation (3) – (2) 
3
20
2
hc hc
f
? ?
- - = 
0
52
2
2
hc hc
f
??
= = 
0
5
4
hc hc
??
= 
0
4
5
?
? = 
 
Question: A gas is taken through an isothermal process as shown. Find the work done by the 
gas 
 
Options: 
(a) 240 J 
(b) 360 J 
(c) 560 J 
(d) None 
Answer: (c) 
Solution: 
2
1
Work ln
v
PV
v
= 
4
400 2ln
2
= × 
800ln 2 = 
800 0.7 = × 
Page 4


JEE-Main-20-07-2021-Shift-2 (Memory Based) 
 
PHYSICS 
 
Question: In a series LCR circuit, 5 , 0.5mH, 2.5µF RL C = ?= = . The RMS value of 
external voltage is 250 V. Find the power dissipated if circuit is in Resonance 
Answer: 12500 W 
Solution:  
Power
rms rms
R
VI
Z
? ?
=
? ?
? ?
 
Z = R (at resonance) 
250 5
Power 250
55
12500 W
= × ×
=
 
  
Question: The wavelength of sodium lamp is observed to be 2886Å from earth & original 
wavelength was 2880Å . Find speed of galaxy. 
Options: 
(a) 
51
3 10 ms
-
× 
(b) 
51
4 10 ms
-
× 
(c) 
51
6.25 10 ms
-
× 
(d) None 
Answer: (c) 
Solution: 2886 2880 6 ? ?= - = Å 
Using doppler shift, 
radial
V
C
?
?
?
-=- 
10
8
radial 10
6 10
3 10
2880 10
VC
?
?
-
-
?? ?× ? ?
? = = × ×
?? ? ?
×
? ?
??
 
51
6.25 10 ms
-
= × 
Hence, speed of galaxy 
51
6.25 10 ms
-
= × . 
 
Question: A body is under the influence of a force such that it delivers a constant power P. 
The variation of position with time of body as 
Options: 
(a) 
1
2
t 
(b) 
3
2
t 
(c) 
5
2
t 
(d) None 
Answer: (b) 
Solution: Power = P 
F v = P 
dv
m vP
dt
??
·=
??
??
 
00
vt
P
vdv dt
m
? = ·
? ?
 
2P
v t
m
?= 
[ ]
2
assuming at 0, 0 & 0
dx P
t t x v
dt m
? = = = = 
00
2
xt
P
dx t dt
m
?= ·
??
 
3/2
22
3
P
tx t
m
?= = 
3/2
x t ??  
 
Question: When a metal is illuminated by light of wavelength ? ,the stopping potential is 
0
V
and for wavelength 2 ? , it is 
0
3V . Then the threshold wavelength is? 
Options: 
(a) 
2
3
?
 
(b) 
4
5
?
 
(c) 
3
?
 
(d) 
5
2
?
 
Answer: (b) 
Solution: 
( )
0
... 1
hc
eV f
?
= - 
( )
0
3 ... 2
2
hc
eV f
?
= - 
Multiply by 3 in equation (1) 
( )
3
3 3 ... 3
hc
eV f
?
= - 
Equation (3) – (2) 
3
20
2
hc hc
f
? ?
- - = 
0
52
2
2
hc hc
f
??
= = 
0
5
4
hc hc
??
= 
0
4
5
?
? = 
 
Question: A gas is taken through an isothermal process as shown. Find the work done by the 
gas 
 
Options: 
(a) 240 J 
(b) 360 J 
(c) 560 J 
(d) None 
Answer: (c) 
Solution: 
2
1
Work ln
v
PV
v
= 
4
400 2ln
2
= × 
800ln 2 = 
800 0.7 = × 
560J = 
 
 
  
Question: Two stars of masses 
1
m and 
2
m form a binary system, revolving around each other 
in circular orbits of radii 
12
and rr respectively. Time period of revolution for this system is  
 
Options: 
(a) 
( )
( )
3
12
12
2
rr
Gm m
p
+
+
 
(b) 
( )
( )
2
1 22
12
2
r rr
Gm m
p
+
+
 
(c) 
( )
( )
3
2
12
12
2 rr
Gm m
p +
+
 
(d) 
( )
( )
2
12 1
12
2 rr r
Gm m
p +
+
 
Answer: (c) 
Solution:  
( )
( )
( )
( )
( )
2
1 2 11
2
1
12
2 1 2 2 1
11 2
12
12
1 1 2 1
1 2
2 2
... 1
Gmm mv
r
rr
Gm r Gm r
vv
rr
rr
rr r r
T
v Gm
p p
=
+
= ?=
+
+
+
= =
  
By using COM concept. 
( )
21 2
1
12
mr r
r
mm
+
=
+
 
Put this value of r1 in eq
n
 (1) 
We get 
Page 5


JEE-Main-20-07-2021-Shift-2 (Memory Based) 
 
PHYSICS 
 
Question: In a series LCR circuit, 5 , 0.5mH, 2.5µF RL C = ?= = . The RMS value of 
external voltage is 250 V. Find the power dissipated if circuit is in Resonance 
Answer: 12500 W 
Solution:  
Power
rms rms
R
VI
Z
? ?
=
? ?
? ?
 
Z = R (at resonance) 
250 5
Power 250
55
12500 W
= × ×
=
 
  
Question: The wavelength of sodium lamp is observed to be 2886Å from earth & original 
wavelength was 2880Å . Find speed of galaxy. 
Options: 
(a) 
51
3 10 ms
-
× 
(b) 
51
4 10 ms
-
× 
(c) 
51
6.25 10 ms
-
× 
(d) None 
Answer: (c) 
Solution: 2886 2880 6 ? ?= - = Å 
Using doppler shift, 
radial
V
C
?
?
?
-=- 
10
8
radial 10
6 10
3 10
2880 10
VC
?
?
-
-
?? ?× ? ?
? = = × ×
?? ? ?
×
? ?
??
 
51
6.25 10 ms
-
= × 
Hence, speed of galaxy 
51
6.25 10 ms
-
= × . 
 
Question: A body is under the influence of a force such that it delivers a constant power P. 
The variation of position with time of body as 
Options: 
(a) 
1
2
t 
(b) 
3
2
t 
(c) 
5
2
t 
(d) None 
Answer: (b) 
Solution: Power = P 
F v = P 
dv
m vP
dt
??
·=
??
??
 
00
vt
P
vdv dt
m
? = ·
? ?
 
2P
v t
m
?= 
[ ]
2
assuming at 0, 0 & 0
dx P
t t x v
dt m
? = = = = 
00
2
xt
P
dx t dt
m
?= ·
??
 
3/2
22
3
P
tx t
m
?= = 
3/2
x t ??  
 
Question: When a metal is illuminated by light of wavelength ? ,the stopping potential is 
0
V
and for wavelength 2 ? , it is 
0
3V . Then the threshold wavelength is? 
Options: 
(a) 
2
3
?
 
(b) 
4
5
?
 
(c) 
3
?
 
(d) 
5
2
?
 
Answer: (b) 
Solution: 
( )
0
... 1
hc
eV f
?
= - 
( )
0
3 ... 2
2
hc
eV f
?
= - 
Multiply by 3 in equation (1) 
( )
3
3 3 ... 3
hc
eV f
?
= - 
Equation (3) – (2) 
3
20
2
hc hc
f
? ?
- - = 
0
52
2
2
hc hc
f
??
= = 
0
5
4
hc hc
??
= 
0
4
5
?
? = 
 
Question: A gas is taken through an isothermal process as shown. Find the work done by the 
gas 
 
Options: 
(a) 240 J 
(b) 360 J 
(c) 560 J 
(d) None 
Answer: (c) 
Solution: 
2
1
Work ln
v
PV
v
= 
4
400 2ln
2
= × 
800ln 2 = 
800 0.7 = × 
560J = 
 
 
  
Question: Two stars of masses 
1
m and 
2
m form a binary system, revolving around each other 
in circular orbits of radii 
12
and rr respectively. Time period of revolution for this system is  
 
Options: 
(a) 
( )
( )
3
12
12
2
rr
Gm m
p
+
+
 
(b) 
( )
( )
2
1 22
12
2
r rr
Gm m
p
+
+
 
(c) 
( )
( )
3
2
12
12
2 rr
Gm m
p +
+
 
(d) 
( )
( )
2
12 1
12
2 rr r
Gm m
p +
+
 
Answer: (c) 
Solution:  
( )
( )
( )
( )
( )
2
1 2 11
2
1
12
2 1 2 2 1
11 2
12
12
1 1 2 1
1 2
2 2
... 1
Gmm mv
r
rr
Gm r Gm r
vv
rr
rr
rr r r
T
v Gm
p p
=
+
= ?=
+
+
+
= =
  
By using COM concept. 
( )
21 2
1
12
mr r
r
mm
+
=
+
 
Put this value of r1 in eq
n
 (1) 
We get 
( )
( )
3/2
12
12
2 rr
T
Gm m
p +
=
+
 
 
Question: Tension in a spring is T1 when length of the spring is L1 and tension is T2 when its 
length is L2. The natural length of the spring is 
Options: 
(a) 
2 2 11
21
Tl Tl
TT
+
+
 
(b) 
2 2 11
21
Tl Tl
TT
-
-
 
(c) 
21 1 2
21
Tl Tl
TT
+
+
 
(d) 
21 1 2
21
Tl Tl
TT
-
-
 
Answer: (d) 
Solution: 
Let the natural length of wire be 
0
. l 
Using Hooke’s law, 
0
Tl
Y
A l
=
?
 
Where, 
0
l l l ?= - 
We get 
0
0
Tl
l l
AY
-= 
Case 1: Tension is T1 and length of wire 
1
l l = 
( )
10
10
... 1
Tl
ll
AY
?- = 
Case 2: Tension is T2 and length of wire 
2
l l = 
( )
20
20
... 2
Tl
l l
AY
?- = 
Dividing both equations  
10 1
20 2
12 2 2
0
21
ll T
l l T
lT l T
l
TT
-
=
-
-
=
-
 
 
Question: Find work done in the process AB ? (isothermal) by gas?