JEE Mains 25 July 2021 Question Paper Shift 2 | JEE Main & Advanced Previous Year Papers PDF Download

 Page 1


JEE-Main-25-07-2021-Shift-2 (Memory Based) 
 
PHYSICS 
 
Question: A balloon is rising up with constant velocity of 10 m/s. At a height of 75m, a small 
stone is dropped from it. At what height from the ground would the balloon be, when the 
mass reaches the ground?  
Answer: 125 m 
Solution: Let the stone take time t in reaching the ground 
Initial velocity of stone 10m/s u = + 
Acceleration 
2
10m/s g =-=- 
Height h = -75 m 
2
2
2
1
75 10 10
2
15 2
5 3 15 0
5sec
tt
t t
t tt
t
- = -× ×
?- = -
? -+- =
?=
 
In this time balloon will move up by 10 5 50m × = 
Height of balloon from ground 75 50 125m += 
 
Question: For the force equation ( ) ( ) cos sin F A Bx C Dt = + . Find dimensions of ?
AD
B
 
Options: 
(a) 
11
MT
-
 
(b) 
12 3
M LT
-
 
(c) 
13
MT 
(d) 
13
MT
-
 
Answer: (b) 
Solution: ( ) ( ) cos sin F A Bx C Dt = + 
and Bx Dt will be dimensionless  
So, [ ]
0 10
1
[]
B M LT
x
-
?? = =
??
 and [ ]
00 1
1
[]
D M LT
t
-
?? = =
??
 
As cos and sin function is dimensionless, so A and C will have dimensions of force 
Page 2


JEE-Main-25-07-2021-Shift-2 (Memory Based) 
 
PHYSICS 
 
Question: A balloon is rising up with constant velocity of 10 m/s. At a height of 75m, a small 
stone is dropped from it. At what height from the ground would the balloon be, when the 
mass reaches the ground?  
Answer: 125 m 
Solution: Let the stone take time t in reaching the ground 
Initial velocity of stone 10m/s u = + 
Acceleration 
2
10m/s g =-=- 
Height h = -75 m 
2
2
2
1
75 10 10
2
15 2
5 3 15 0
5sec
tt
t t
t tt
t
- = -× ×
?- = -
? -+- =
?=
 
In this time balloon will move up by 10 5 50m × = 
Height of balloon from ground 75 50 125m += 
 
Question: For the force equation ( ) ( ) cos sin F A Bx C Dt = + . Find dimensions of ?
AD
B
 
Options: 
(a) 
11
MT
-
 
(b) 
12 3
M LT
-
 
(c) 
13
MT 
(d) 
13
MT
-
 
Answer: (b) 
Solution: ( ) ( ) cos sin F A Bx C Dt = + 
and Bx Dt will be dimensionless  
So, [ ]
0 10
1
[]
B M LT
x
-
?? = =
??
 and [ ]
00 1
1
[]
D M LT
t
-
?? = =
??
 
As cos and sin function is dimensionless, so A and C will have dimensions of force 
[ ] [ ]
2
A C MLT
-
?? = =
??
 
2 00 1
0 10
MLT M LT
AD
B M LT
--
-
? ? ? ?
? ?
? ? ? ?
=
? ?
??
? ?
??
 
23
MLT
-
?? =
??
 
Question: A particle starts from rest with acceleration 
2
a tt aß = + where and aß are 
constant. Find its displacement between t = 1 and t = 2 seconds. 
Options: 
(a) 
75
34
aß
+ 
(b) 
75
64
aß
+ 
(c) 7 5 aß + 
(d) None 
Answer: (b) 
Solution:  
Given, 
2
a tt aß = + 
2
dv
tt
dt
aß ? = + 
2
dv tdt t dt aß ?= +
?? ?
 
23
2 3
t t
vc
aß
?= + + 
At 0, 0 tv = = 
0 00 c ? = ++ 
0 c ?= 
23
2 3
dx t t
v
dt
aß
= = + 
2
1
22 23
11
23
x
x
tt
dx dt dt
aß
= +
?? ?
 
22
34
21
11
6 12
t t
xx
aß ? ?? ?
-= +
? ?? ?
? ?? ?
 
8 16
6 6 12 12
x
aa ß ß
?= - + - 
Page 3


JEE-Main-25-07-2021-Shift-2 (Memory Based) 
 
PHYSICS 
 
Question: A balloon is rising up with constant velocity of 10 m/s. At a height of 75m, a small 
stone is dropped from it. At what height from the ground would the balloon be, when the 
mass reaches the ground?  
Answer: 125 m 
Solution: Let the stone take time t in reaching the ground 
Initial velocity of stone 10m/s u = + 
Acceleration 
2
10m/s g =-=- 
Height h = -75 m 
2
2
2
1
75 10 10
2
15 2
5 3 15 0
5sec
tt
t t
t tt
t
- = -× ×
?- = -
? -+- =
?=
 
In this time balloon will move up by 10 5 50m × = 
Height of balloon from ground 75 50 125m += 
 
Question: For the force equation ( ) ( ) cos sin F A Bx C Dt = + . Find dimensions of ?
AD
B
 
Options: 
(a) 
11
MT
-
 
(b) 
12 3
M LT
-
 
(c) 
13
MT 
(d) 
13
MT
-
 
Answer: (b) 
Solution: ( ) ( ) cos sin F A Bx C Dt = + 
and Bx Dt will be dimensionless  
So, [ ]
0 10
1
[]
B M LT
x
-
?? = =
??
 and [ ]
00 1
1
[]
D M LT
t
-
?? = =
??
 
As cos and sin function is dimensionless, so A and C will have dimensions of force 
[ ] [ ]
2
A C MLT
-
?? = =
??
 
2 00 1
0 10
MLT M LT
AD
B M LT
--
-
? ? ? ?
? ?
? ? ? ?
=
? ?
??
? ?
??
 
23
MLT
-
?? =
??
 
Question: A particle starts from rest with acceleration 
2
a tt aß = + where and aß are 
constant. Find its displacement between t = 1 and t = 2 seconds. 
Options: 
(a) 
75
34
aß
+ 
(b) 
75
64
aß
+ 
(c) 7 5 aß + 
(d) None 
Answer: (b) 
Solution:  
Given, 
2
a tt aß = + 
2
dv
tt
dt
aß ? = + 
2
dv tdt t dt aß ?= +
?? ?
 
23
2 3
t t
vc
aß
?= + + 
At 0, 0 tv = = 
0 00 c ? = ++ 
0 c ?= 
23
2 3
dx t t
v
dt
aß
= = + 
2
1
22 23
11
23
x
x
tt
dx dt dt
aß
= +
?? ?
 
22
34
21
11
6 12
t t
xx
aß ? ?? ?
-= +
? ?? ?
? ?? ?
 
8 16
6 6 12 12
x
aa ß ß
?= - + - 
7 15
6 12
x
aß
?= + 
75
64
x
aß
?= + 
 
Question: If velocity of photon is C and that of electron is v, then find the ratio of KE of 
electron to photon if their de-Broglie wavelength is same. 
Options: 
(a) 
C
v
 
(b) 
2C
v
 
(c) 
2
v
C
 
(d) 
v
C
 
Answer: (c) 
Solution: Given, 
Ph e
?? = 
( ) ... i
e
h
mv
? = 
Kinetic energy of photon ( ) ... ii
Ph e
hC hC
??
= = 
Kinetic energy of electron 
( )
( )
( )
2
2
2
... iii
22
e
mv
h
mm ?
= = 
( )
( )
( )
2
2
.. 2
..
e Ph
e
e
KE m
hC
KE h
?
?
= × 
( ) 2
e
mC
h
?
= 
From eq. (i) 
22 mC C
mv v
= = 
( )
( )
..
.. 2
E
Ph
KE
v
KE C
?= 
 
Page 4


JEE-Main-25-07-2021-Shift-2 (Memory Based) 
 
PHYSICS 
 
Question: A balloon is rising up with constant velocity of 10 m/s. At a height of 75m, a small 
stone is dropped from it. At what height from the ground would the balloon be, when the 
mass reaches the ground?  
Answer: 125 m 
Solution: Let the stone take time t in reaching the ground 
Initial velocity of stone 10m/s u = + 
Acceleration 
2
10m/s g =-=- 
Height h = -75 m 
2
2
2
1
75 10 10
2
15 2
5 3 15 0
5sec
tt
t t
t tt
t
- = -× ×
?- = -
? -+- =
?=
 
In this time balloon will move up by 10 5 50m × = 
Height of balloon from ground 75 50 125m += 
 
Question: For the force equation ( ) ( ) cos sin F A Bx C Dt = + . Find dimensions of ?
AD
B
 
Options: 
(a) 
11
MT
-
 
(b) 
12 3
M LT
-
 
(c) 
13
MT 
(d) 
13
MT
-
 
Answer: (b) 
Solution: ( ) ( ) cos sin F A Bx C Dt = + 
and Bx Dt will be dimensionless  
So, [ ]
0 10
1
[]
B M LT
x
-
?? = =
??
 and [ ]
00 1
1
[]
D M LT
t
-
?? = =
??
 
As cos and sin function is dimensionless, so A and C will have dimensions of force 
[ ] [ ]
2
A C MLT
-
?? = =
??
 
2 00 1
0 10
MLT M LT
AD
B M LT
--
-
? ? ? ?
? ?
? ? ? ?
=
? ?
??
? ?
??
 
23
MLT
-
?? =
??
 
Question: A particle starts from rest with acceleration 
2
a tt aß = + where and aß are 
constant. Find its displacement between t = 1 and t = 2 seconds. 
Options: 
(a) 
75
34
aß
+ 
(b) 
75
64
aß
+ 
(c) 7 5 aß + 
(d) None 
Answer: (b) 
Solution:  
Given, 
2
a tt aß = + 
2
dv
tt
dt
aß ? = + 
2
dv tdt t dt aß ?= +
?? ?
 
23
2 3
t t
vc
aß
?= + + 
At 0, 0 tv = = 
0 00 c ? = ++ 
0 c ?= 
23
2 3
dx t t
v
dt
aß
= = + 
2
1
22 23
11
23
x
x
tt
dx dt dt
aß
= +
?? ?
 
22
34
21
11
6 12
t t
xx
aß ? ?? ?
-= +
? ?? ?
? ?? ?
 
8 16
6 6 12 12
x
aa ß ß
?= - + - 
7 15
6 12
x
aß
?= + 
75
64
x
aß
?= + 
 
Question: If velocity of photon is C and that of electron is v, then find the ratio of KE of 
electron to photon if their de-Broglie wavelength is same. 
Options: 
(a) 
C
v
 
(b) 
2C
v
 
(c) 
2
v
C
 
(d) 
v
C
 
Answer: (c) 
Solution: Given, 
Ph e
?? = 
( ) ... i
e
h
mv
? = 
Kinetic energy of photon ( ) ... ii
Ph e
hC hC
??
= = 
Kinetic energy of electron 
( )
( )
( )
2
2
2
... iii
22
e
mv
h
mm ?
= = 
( )
( )
( )
2
2
.. 2
..
e Ph
e
e
KE m
hC
KE h
?
?
= × 
( ) 2
e
mC
h
?
= 
From eq. (i) 
22 mC C
mv v
= = 
( )
( )
..
.. 2
E
Ph
KE
v
KE C
?= 
 
Question: Two soap bubbles of radius R1 and R2 combine isothermally to form a new soap 
bubble. Find the radius of the new soap bubble formed 
Options: 
(a) 
12
2
RR +
 
(b) 
12
RR 
(c) 
12
12
RR
RR +
 
(d) 
22
12
RR + 
Answer: (d) 
Solution: Radius of first soap bubble is R1 
Radius of first soap bubble is R2 
Let, 
3
1 11
1
44
,
3
T
P VR
R
p = = , be the excess pressure inside first soap bubble and volume of first 
soap bubble respectively. 
3
2 22
2
44
,
3
T
P VR
R
p = = be the excess pressure inside second soap bubble and volume of second 
soap bubble respectively.  
3
4 4
,
3
T
P VR
R
p = = be the excess pressure inside new soap bubble, volume and radius of new 
soap bubble respectively. 
The two bubbles combine isothermally, hence  
11 2 2
3 3 3
12
12
4 4 4 4 4 4
3 33
PV PV PV
T TT
R RR
R RR
p pp
= +
?? ?? ??
= +
?? ?? ??
?? ?? ??
 
2 22
12
22
12
R RR
R RR
= +
= +
 
 
 
Question: Two dipoles p1 and p2 are perpendicular to each other, are placed in a uniform 
magnetic filed such that p1 makes an angle of 37° with field. Both dipoles experiences same 
torque. Find ratio of their dipole moment. 
Options: 
(a) 
4
3
 
(b) 
3
4
 
Page 5


JEE-Main-25-07-2021-Shift-2 (Memory Based) 
 
PHYSICS 
 
Question: A balloon is rising up with constant velocity of 10 m/s. At a height of 75m, a small 
stone is dropped from it. At what height from the ground would the balloon be, when the 
mass reaches the ground?  
Answer: 125 m 
Solution: Let the stone take time t in reaching the ground 
Initial velocity of stone 10m/s u = + 
Acceleration 
2
10m/s g =-=- 
Height h = -75 m 
2
2
2
1
75 10 10
2
15 2
5 3 15 0
5sec
tt
t t
t tt
t
- = -× ×
?- = -
? -+- =
?=
 
In this time balloon will move up by 10 5 50m × = 
Height of balloon from ground 75 50 125m += 
 
Question: For the force equation ( ) ( ) cos sin F A Bx C Dt = + . Find dimensions of ?
AD
B
 
Options: 
(a) 
11
MT
-
 
(b) 
12 3
M LT
-
 
(c) 
13
MT 
(d) 
13
MT
-
 
Answer: (b) 
Solution: ( ) ( ) cos sin F A Bx C Dt = + 
and Bx Dt will be dimensionless  
So, [ ]
0 10
1
[]
B M LT
x
-
?? = =
??
 and [ ]
00 1
1
[]
D M LT
t
-
?? = =
??
 
As cos and sin function is dimensionless, so A and C will have dimensions of force 
[ ] [ ]
2
A C MLT
-
?? = =
??
 
2 00 1
0 10
MLT M LT
AD
B M LT
--
-
? ? ? ?
? ?
? ? ? ?
=
? ?
??
? ?
??
 
23
MLT
-
?? =
??
 
Question: A particle starts from rest with acceleration 
2
a tt aß = + where and aß are 
constant. Find its displacement between t = 1 and t = 2 seconds. 
Options: 
(a) 
75
34
aß
+ 
(b) 
75
64
aß
+ 
(c) 7 5 aß + 
(d) None 
Answer: (b) 
Solution:  
Given, 
2
a tt aß = + 
2
dv
tt
dt
aß ? = + 
2
dv tdt t dt aß ?= +
?? ?
 
23
2 3
t t
vc
aß
?= + + 
At 0, 0 tv = = 
0 00 c ? = ++ 
0 c ?= 
23
2 3
dx t t
v
dt
aß
= = + 
2
1
22 23
11
23
x
x
tt
dx dt dt
aß
= +
?? ?
 
22
34
21
11
6 12
t t
xx
aß ? ?? ?
-= +
? ?? ?
? ?? ?
 
8 16
6 6 12 12
x
aa ß ß
?= - + - 
7 15
6 12
x
aß
?= + 
75
64
x
aß
?= + 
 
Question: If velocity of photon is C and that of electron is v, then find the ratio of KE of 
electron to photon if their de-Broglie wavelength is same. 
Options: 
(a) 
C
v
 
(b) 
2C
v
 
(c) 
2
v
C
 
(d) 
v
C
 
Answer: (c) 
Solution: Given, 
Ph e
?? = 
( ) ... i
e
h
mv
? = 
Kinetic energy of photon ( ) ... ii
Ph e
hC hC
??
= = 
Kinetic energy of electron 
( )
( )
( )
2
2
2
... iii
22
e
mv
h
mm ?
= = 
( )
( )
( )
2
2
.. 2
..
e Ph
e
e
KE m
hC
KE h
?
?
= × 
( ) 2
e
mC
h
?
= 
From eq. (i) 
22 mC C
mv v
= = 
( )
( )
..
.. 2
E
Ph
KE
v
KE C
?= 
 
Question: Two soap bubbles of radius R1 and R2 combine isothermally to form a new soap 
bubble. Find the radius of the new soap bubble formed 
Options: 
(a) 
12
2
RR +
 
(b) 
12
RR 
(c) 
12
12
RR
RR +
 
(d) 
22
12
RR + 
Answer: (d) 
Solution: Radius of first soap bubble is R1 
Radius of first soap bubble is R2 
Let, 
3
1 11
1
44
,
3
T
P VR
R
p = = , be the excess pressure inside first soap bubble and volume of first 
soap bubble respectively. 
3
2 22
2
44
,
3
T
P VR
R
p = = be the excess pressure inside second soap bubble and volume of second 
soap bubble respectively.  
3
4 4
,
3
T
P VR
R
p = = be the excess pressure inside new soap bubble, volume and radius of new 
soap bubble respectively. 
The two bubbles combine isothermally, hence  
11 2 2
3 3 3
12
12
4 4 4 4 4 4
3 33
PV PV PV
T TT
R RR
R RR
p pp
= +
?? ?? ??
= +
?? ?? ??
?? ?? ??
 
2 22
12
22
12
R RR
R RR
= +
= +
 
 
 
Question: Two dipoles p1 and p2 are perpendicular to each other, are placed in a uniform 
magnetic filed such that p1 makes an angle of 37° with field. Both dipoles experiences same 
torque. Find ratio of their dipole moment. 
Options: 
(a) 
4
3
 
(b) 
3
4
 
(c) 
3
5
 
(d) 
4
5
 
Answer: (a) 
Solution:  
 
Torque sin P E PE ? = × =
??
 
Torque on dipole one 
11
PE t = × 
1
1
sin 37
3
5
PE
PE
= °
= ×
 
Similarly 
22
PE t = × 
( )
2
2
2
12
12
1
2
sin 90 37
cos37
4
5
34
55
4
3
PE
PE
PE
PE PE
P
P
tt
= +°
= °
??
=
??
??
=
×=
=
 
 
 
Question: If and X Y
??? ? ?
 are two vectors, such that & 10 , X Y XY X Y = -= +
? ? ?? ? ?
then. Find 
the angle between and X Y
??? ? ?
 
Options: