JEE Mains 20 July 2021 Question Paper Shift 1 | JEE Main & Advanced Previous Year Papers PDF Download

 Page 1


JEE-Main-20-07-2021-Shift-1 (Memory Based) 
 
PHYSICS 
 
Question: A butterfly in North East direction with a velocity of 42 / ms . Wind is blowing 
from North to South with a velocity of 1 m/s. Find the displacement of the bird in three 
seconds. 
Options: 
(a) 15 m, 37° North to East 
(b) 15 m, 37° East to North 
(c) 15 m, 37 ° North to West  
(d) None of these 
Answer: (a) 
Solution: 
 
ˆˆ
4/ 4/
butt
V m si m sj = +
?
 
ˆ
1/
wind
V m sj = -
?
 
,
ˆˆ
4 / 3/
butt wind
V m si m sj = +
?
 
Displacement of bird in three second. 
( ) ( )
22
4 3 33 D= × + × 
15 Dm = 
Direction 
1
3
tan
4
?
-
??
=
??
??
 
37 ? = ° 
?15m, 37 ° North to East. 
 
Question: A travelling wave moving with a velocity v. The equation of wave at two different 
instants t = 0 and t = 3 is given by 
2
1
1
y
x
=
+
and 
( )
2
1
11
y
x
=
++
respectively. Find the speed 
v of the wave. (y is in mm, x is in cm) 
Options: 
Page 2


JEE-Main-20-07-2021-Shift-1 (Memory Based) 
 
PHYSICS 
 
Question: A butterfly in North East direction with a velocity of 42 / ms . Wind is blowing 
from North to South with a velocity of 1 m/s. Find the displacement of the bird in three 
seconds. 
Options: 
(a) 15 m, 37° North to East 
(b) 15 m, 37° East to North 
(c) 15 m, 37 ° North to West  
(d) None of these 
Answer: (a) 
Solution: 
 
ˆˆ
4/ 4/
butt
V m si m sj = +
?
 
ˆ
1/
wind
V m sj = -
?
 
,
ˆˆ
4 / 3/
butt wind
V m si m sj = +
?
 
Displacement of bird in three second. 
( ) ( )
22
4 3 33 D= × + × 
15 Dm = 
Direction 
1
3
tan
4
?
-
??
=
??
??
 
37 ? = ° 
?15m, 37 ° North to East. 
 
Question: A travelling wave moving with a velocity v. The equation of wave at two different 
instants t = 0 and t = 3 is given by 
2
1
1
y
x
=
+
and 
( )
2
1
11
y
x
=
++
respectively. Find the speed 
v of the wave. (y is in mm, x is in cm) 
Options: 
(a) 
1
/
3
mm s 
(b) 3 mm/s 
(c) 
1
/
3
cm s 
(d) 3 cm/s 
Answer: (c) 
Solution: 
( )
2
1
,0
1
yx
x
=
+
 
( )
( )
2
1
,3
11
yx
x
=
++
 
If v is the velocity of the wave, then, 
( ) ( ) , ,0 yx t yx vt = + 
Therefore, 
( ) ( ) ,3 3 ,0 yx x v = + 
Replacing values we get 
( ) ( )
22
11
1 11 3 x xv
=
++ ++
 
13 x xv += + 
1
/
3
x cm s = + 
 
Question: A block of mass 20 kg is placed on a horizontal platform and three blocks each 
with mass 10kg are arranged as given in figure below. If platform accelerated downward with 
acceleration 
2
2/ ms , then the normal force between 10 kg and 20 kg block is- 
 
Options: 
(a) 100 N 
(b) 150 N 
(c) 120 N 
(d) 140 N 
Answer: (c) 
Solution: 
Page 3


JEE-Main-20-07-2021-Shift-1 (Memory Based) 
 
PHYSICS 
 
Question: A butterfly in North East direction with a velocity of 42 / ms . Wind is blowing 
from North to South with a velocity of 1 m/s. Find the displacement of the bird in three 
seconds. 
Options: 
(a) 15 m, 37° North to East 
(b) 15 m, 37° East to North 
(c) 15 m, 37 ° North to West  
(d) None of these 
Answer: (a) 
Solution: 
 
ˆˆ
4/ 4/
butt
V m si m sj = +
?
 
ˆ
1/
wind
V m sj = -
?
 
,
ˆˆ
4 / 3/
butt wind
V m si m sj = +
?
 
Displacement of bird in three second. 
( ) ( )
22
4 3 33 D= × + × 
15 Dm = 
Direction 
1
3
tan
4
?
-
??
=
??
??
 
37 ? = ° 
?15m, 37 ° North to East. 
 
Question: A travelling wave moving with a velocity v. The equation of wave at two different 
instants t = 0 and t = 3 is given by 
2
1
1
y
x
=
+
and 
( )
2
1
11
y
x
=
++
respectively. Find the speed 
v of the wave. (y is in mm, x is in cm) 
Options: 
(a) 
1
/
3
mm s 
(b) 3 mm/s 
(c) 
1
/
3
cm s 
(d) 3 cm/s 
Answer: (c) 
Solution: 
( )
2
1
,0
1
yx
x
=
+
 
( )
( )
2
1
,3
11
yx
x
=
++
 
If v is the velocity of the wave, then, 
( ) ( ) , ,0 yx t yx vt = + 
Therefore, 
( ) ( ) ,3 3 ,0 yx x v = + 
Replacing values we get 
( ) ( )
22
11
1 11 3 x xv
=
++ ++
 
13 x xv += + 
1
/
3
x cm s = + 
 
Question: A block of mass 20 kg is placed on a horizontal platform and three blocks each 
with mass 10kg are arranged as given in figure below. If platform accelerated downward with 
acceleration 
2
2/ ms , then the normal force between 10 kg and 20 kg block is- 
 
Options: 
(a) 100 N 
(b) 150 N 
(c) 120 N 
(d) 140 N 
Answer: (c) 
Solution: 
 
20 kg is rest w.r.t platform. 
2 200 20 2 N= +× 
2 240 NN = 
120 NN = 
 
Question: A ball having charge to mass 8/ cg µ is placed at a distance of 10 cm from a wall. 
Suddenly an Electric field 
1
100Nm
-
is switched on. Assuming collisions to be Elastic. Find 
Time period of oscillations. 
 
Options: 
(a) 0.5 sec 
(b) 1 sec 
(c) 2 sec 
(d) 3 sec 
Answer: () 
Solution: 
8/
q
cg
m
µ = 
 
F qE = 
Page 4


JEE-Main-20-07-2021-Shift-1 (Memory Based) 
 
PHYSICS 
 
Question: A butterfly in North East direction with a velocity of 42 / ms . Wind is blowing 
from North to South with a velocity of 1 m/s. Find the displacement of the bird in three 
seconds. 
Options: 
(a) 15 m, 37° North to East 
(b) 15 m, 37° East to North 
(c) 15 m, 37 ° North to West  
(d) None of these 
Answer: (a) 
Solution: 
 
ˆˆ
4/ 4/
butt
V m si m sj = +
?
 
ˆ
1/
wind
V m sj = -
?
 
,
ˆˆ
4 / 3/
butt wind
V m si m sj = +
?
 
Displacement of bird in three second. 
( ) ( )
22
4 3 33 D= × + × 
15 Dm = 
Direction 
1
3
tan
4
?
-
??
=
??
??
 
37 ? = ° 
?15m, 37 ° North to East. 
 
Question: A travelling wave moving with a velocity v. The equation of wave at two different 
instants t = 0 and t = 3 is given by 
2
1
1
y
x
=
+
and 
( )
2
1
11
y
x
=
++
respectively. Find the speed 
v of the wave. (y is in mm, x is in cm) 
Options: 
(a) 
1
/
3
mm s 
(b) 3 mm/s 
(c) 
1
/
3
cm s 
(d) 3 cm/s 
Answer: (c) 
Solution: 
( )
2
1
,0
1
yx
x
=
+
 
( )
( )
2
1
,3
11
yx
x
=
++
 
If v is the velocity of the wave, then, 
( ) ( ) , ,0 yx t yx vt = + 
Therefore, 
( ) ( ) ,3 3 ,0 yx x v = + 
Replacing values we get 
( ) ( )
22
11
1 11 3 x xv
=
++ ++
 
13 x xv += + 
1
/
3
x cm s = + 
 
Question: A block of mass 20 kg is placed on a horizontal platform and three blocks each 
with mass 10kg are arranged as given in figure below. If platform accelerated downward with 
acceleration 
2
2/ ms , then the normal force between 10 kg and 20 kg block is- 
 
Options: 
(a) 100 N 
(b) 150 N 
(c) 120 N 
(d) 140 N 
Answer: (c) 
Solution: 
 
20 kg is rest w.r.t platform. 
2 200 20 2 N= +× 
2 240 NN = 
120 NN = 
 
Question: A ball having charge to mass 8/ cg µ is placed at a distance of 10 cm from a wall. 
Suddenly an Electric field 
1
100Nm
-
is switched on. Assuming collisions to be Elastic. Find 
Time period of oscillations. 
 
Options: 
(a) 0.5 sec 
(b) 1 sec 
(c) 2 sec 
(d) 3 sec 
Answer: () 
Solution: 
8/
q
cg
m
µ = 
 
F qE = 
qE
a
m
= 
6
3
8 10
100
1 10
a
-
-
×
= ×
×
 
2
0.8 / a ms = 
Time period (T) 
2
2
S
a
= 
2 0.1
2
8
×
= × 
T = 1 sec 
 
Question: If an ideal gas is taken through process AB then find work done on gas by external 
agent. Given curve AB is an ellipse. 
 
Options: 
(a) 200 J p 
(b) 200 J p - 
(c) 100 J p 
(d) 100 J p - 
Answer: (a) 
Solution: 
 
0
external gas
WW += 
external gas
WW = - 
20 20
2
J
p×× ??
=- -
??
??
 
200
external
WJ p = 
Page 5


JEE-Main-20-07-2021-Shift-1 (Memory Based) 
 
PHYSICS 
 
Question: A butterfly in North East direction with a velocity of 42 / ms . Wind is blowing 
from North to South with a velocity of 1 m/s. Find the displacement of the bird in three 
seconds. 
Options: 
(a) 15 m, 37° North to East 
(b) 15 m, 37° East to North 
(c) 15 m, 37 ° North to West  
(d) None of these 
Answer: (a) 
Solution: 
 
ˆˆ
4/ 4/
butt
V m si m sj = +
?
 
ˆ
1/
wind
V m sj = -
?
 
,
ˆˆ
4 / 3/
butt wind
V m si m sj = +
?
 
Displacement of bird in three second. 
( ) ( )
22
4 3 33 D= × + × 
15 Dm = 
Direction 
1
3
tan
4
?
-
??
=
??
??
 
37 ? = ° 
?15m, 37 ° North to East. 
 
Question: A travelling wave moving with a velocity v. The equation of wave at two different 
instants t = 0 and t = 3 is given by 
2
1
1
y
x
=
+
and 
( )
2
1
11
y
x
=
++
respectively. Find the speed 
v of the wave. (y is in mm, x is in cm) 
Options: 
(a) 
1
/
3
mm s 
(b) 3 mm/s 
(c) 
1
/
3
cm s 
(d) 3 cm/s 
Answer: (c) 
Solution: 
( )
2
1
,0
1
yx
x
=
+
 
( )
( )
2
1
,3
11
yx
x
=
++
 
If v is the velocity of the wave, then, 
( ) ( ) , ,0 yx t yx vt = + 
Therefore, 
( ) ( ) ,3 3 ,0 yx x v = + 
Replacing values we get 
( ) ( )
22
11
1 11 3 x xv
=
++ ++
 
13 x xv += + 
1
/
3
x cm s = + 
 
Question: A block of mass 20 kg is placed on a horizontal platform and three blocks each 
with mass 10kg are arranged as given in figure below. If platform accelerated downward with 
acceleration 
2
2/ ms , then the normal force between 10 kg and 20 kg block is- 
 
Options: 
(a) 100 N 
(b) 150 N 
(c) 120 N 
(d) 140 N 
Answer: (c) 
Solution: 
 
20 kg is rest w.r.t platform. 
2 200 20 2 N= +× 
2 240 NN = 
120 NN = 
 
Question: A ball having charge to mass 8/ cg µ is placed at a distance of 10 cm from a wall. 
Suddenly an Electric field 
1
100Nm
-
is switched on. Assuming collisions to be Elastic. Find 
Time period of oscillations. 
 
Options: 
(a) 0.5 sec 
(b) 1 sec 
(c) 2 sec 
(d) 3 sec 
Answer: () 
Solution: 
8/
q
cg
m
µ = 
 
F qE = 
qE
a
m
= 
6
3
8 10
100
1 10
a
-
-
×
= ×
×
 
2
0.8 / a ms = 
Time period (T) 
2
2
S
a
= 
2 0.1
2
8
×
= × 
T = 1 sec 
 
Question: If an ideal gas is taken through process AB then find work done on gas by external 
agent. Given curve AB is an ellipse. 
 
Options: 
(a) 200 J p 
(b) 200 J p - 
(c) 100 J p 
(d) 100 J p - 
Answer: (a) 
Solution: 
 
0
external gas
WW += 
external gas
WW = - 
20 20
2
J
p×× ??
=- -
??
??
 
200
external
WJ p = 
 
Question: If , A B A B ·= ×
?? ? ?
find AB -
? ?
 
Options: 
(a) 
22
2 A B A B ++ · 
(b) 
22
2 A B AB ++ 
(c) 
22
2 A B AB +- 
(d) 
22
AB + 
Answer: (c) 
Solution: 
A B A B ·= ×
?? ? ?
 
cos sin AB AB ?? = 
cos sin ?? = 
4
p
? ?= 
then 
22
2 cos 45 A B A B AB -= + - °
? ?
 
22
2 A B A B AB -= + -
? ?
 
 
Question: Two charges are kept at a fixed distance from each other. The sum of both charges 
is Q. What should be charge on each of them in order to maximize force between them. 
Options: 
(a) ,
22
QQ
 
(b) 
2
,
33
QQ
 
(c) 
2
,
33
QQ
 
(d) 
3
,
44
QQ
 
Answer: (a) 
Solution: 
Let one charge be 
1
q and other be 
1
q ? - 
( )
1 1
2
qQ q
FK
r
-
= 
For maximum force 
1
0
dF
dq
= 
and 
2
2
1
0
dF
dq
<