JEE Advance 2022 Question Paper - 1 with Solutions | JEE Main & Advanced Mock Test Series

 Page 1


   
JEE-Advanced-28-08-2022 
[Paper-1]  
 
Physics 
 
Question: Two spherical stars A and B have densities 
A
? and ,
B
? respectively. A and B 
have the same radius, and their masses MA and MB are related by MB = 2MA. Due to an 
interaction process, star A loses some of its mass, so that its radius is halved, while its 
spherical shape is retained and its density remains .
A
? The entire mass lost by A is deposited 
as a thick spheric shell on B with the density the shell being .
A
? If 
A
v and 
B
v are the escape 
velocities from A and B after the interaction process, the ratio 
1/3
10
.
15
B
A
v n
v
= The value of n is 
________ 
Answer: 2.30 
Solution: 
0
2
2
8
A
A
v GM
v
R
A
= =
??
×
??
??
 
For, B, 
( )
33 3
4 47
3 38
rR R p p -= × 
1
3
15
8
rR
??
?=
??
??
 
Therefore, 
( ) ( )
0 11
33
7
22
23 2 8
15 8 15
2
AA
B
GM M
vv
R
? ?
× +
? ?
×
? ?
= = ×
×
 
Therefore,
( ) ( )
11
33
23 2.30 10
15 15
B
A
v
v
×
= = 
Therefore, n = 2.30 
 
Question: The minimum kinetic energy needed by an alpha particle to cause the nuclear 
reaction 
16 4 1 19
7 2 18
N He H + ?+ O in a laboratory frame is n (in MeV). Assume that 
16
7
N is at 
Page 2


   
JEE-Advanced-28-08-2022 
[Paper-1]  
 
Physics 
 
Question: Two spherical stars A and B have densities 
A
? and ,
B
? respectively. A and B 
have the same radius, and their masses MA and MB are related by MB = 2MA. Due to an 
interaction process, star A loses some of its mass, so that its radius is halved, while its 
spherical shape is retained and its density remains .
A
? The entire mass lost by A is deposited 
as a thick spheric shell on B with the density the shell being .
A
? If 
A
v and 
B
v are the escape 
velocities from A and B after the interaction process, the ratio 
1/3
10
.
15
B
A
v n
v
= The value of n is 
________ 
Answer: 2.30 
Solution: 
0
2
2
8
A
A
v GM
v
R
A
= =
??
×
??
??
 
For, B, 
( )
33 3
4 47
3 38
rR R p p -= × 
1
3
15
8
rR
??
?=
??
??
 
Therefore, 
( ) ( )
0 11
33
7
22
23 2 8
15 8 15
2
AA
B
GM M
vv
R
? ?
× +
? ?
×
? ?
= = ×
×
 
Therefore,
( ) ( )
11
33
23 2.30 10
15 15
B
A
v
v
×
= = 
Therefore, n = 2.30 
 
Question: The minimum kinetic energy needed by an alpha particle to cause the nuclear 
reaction 
16 4 1 19
7 2 18
N He H + ?+ O in a laboratory frame is n (in MeV). Assume that 
16
7
N is at 
   
rest in the laboratory frame. The masses of 
16
7
, N 
1 19
18
and HO can be taken to be 16.006 u, 
4.003 u, 1008 u and 19.003 u, respectively, where 1 u = 930 MeV c
-2
. The value of n is 
_______. 
Answer: 2.33 
Solution: 
( )
2
N He H O
Qm m m m c = + -- ×
 
( ) 16.006 4.003 1.008 19.003 930MeV = + -- ×
 
1.86MeV = -
 
1.86 energy absorbed. MeV =
 
And, 
2
14
max loss in kinetic energy
2 5
mm
v
m
×
× ×= 
2
15
24
mv Q ?=×
 
( )
5
1.86 2.325
4
MeV MeV =× =
 
Therefore, 2.33 n = 
 
Question: In the following circuit 
1 23
12 , 4 C FC C F µµ = = = and C4 = C5 = 2 µF. The charge 
stored in C3 is ______µC 
 
Answer: 8.00 
Solution: 
From circuit given, Potential difference across C3 is 2 V (constant) 
Therefore, 
3
24 8 Q C C µµ =×= 
 
Question: A rod of length 2 cm makes an angle 
2
3
p
rad with the principal axis of a thin 
convex lens. The lens has a focal length of 10 cm and placed at a distance of 
40
3
cm from the 
Page 3


   
JEE-Advanced-28-08-2022 
[Paper-1]  
 
Physics 
 
Question: Two spherical stars A and B have densities 
A
? and ,
B
? respectively. A and B 
have the same radius, and their masses MA and MB are related by MB = 2MA. Due to an 
interaction process, star A loses some of its mass, so that its radius is halved, while its 
spherical shape is retained and its density remains .
A
? The entire mass lost by A is deposited 
as a thick spheric shell on B with the density the shell being .
A
? If 
A
v and 
B
v are the escape 
velocities from A and B after the interaction process, the ratio 
1/3
10
.
15
B
A
v n
v
= The value of n is 
________ 
Answer: 2.30 
Solution: 
0
2
2
8
A
A
v GM
v
R
A
= =
??
×
??
??
 
For, B, 
( )
33 3
4 47
3 38
rR R p p -= × 
1
3
15
8
rR
??
?=
??
??
 
Therefore, 
( ) ( )
0 11
33
7
22
23 2 8
15 8 15
2
AA
B
GM M
vv
R
? ?
× +
? ?
×
? ?
= = ×
×
 
Therefore,
( ) ( )
11
33
23 2.30 10
15 15
B
A
v
v
×
= = 
Therefore, n = 2.30 
 
Question: The minimum kinetic energy needed by an alpha particle to cause the nuclear 
reaction 
16 4 1 19
7 2 18
N He H + ?+ O in a laboratory frame is n (in MeV). Assume that 
16
7
N is at 
   
rest in the laboratory frame. The masses of 
16
7
, N 
1 19
18
and HO can be taken to be 16.006 u, 
4.003 u, 1008 u and 19.003 u, respectively, where 1 u = 930 MeV c
-2
. The value of n is 
_______. 
Answer: 2.33 
Solution: 
( )
2
N He H O
Qm m m m c = + -- ×
 
( ) 16.006 4.003 1.008 19.003 930MeV = + -- ×
 
1.86MeV = -
 
1.86 energy absorbed. MeV =
 
And, 
2
14
max loss in kinetic energy
2 5
mm
v
m
×
× ×= 
2
15
24
mv Q ?=×
 
( )
5
1.86 2.325
4
MeV MeV =× =
 
Therefore, 2.33 n = 
 
Question: In the following circuit 
1 23
12 , 4 C FC C F µµ = = = and C4 = C5 = 2 µF. The charge 
stored in C3 is ______µC 
 
Answer: 8.00 
Solution: 
From circuit given, Potential difference across C3 is 2 V (constant) 
Therefore, 
3
24 8 Q C C µµ =×= 
 
Question: A rod of length 2 cm makes an angle 
2
3
p
rad with the principal axis of a thin 
convex lens. The lens has a focal length of 10 cm and placed at a distance of 
40
3
cm from the 
   
object as shown in the figure. The height of the image is 
30 3
13
cm and the angle made by it 
with respect to the principal axis is a rad. The value of a is 
n
p
rad, where n is ______ 
 
Answer: 6.00 
Solution: 
 
40
10
3
' 40
43
10
3
OA cm
×
= =
-
 
43
10
430
3
'
43
13
10
3
OB cm
×
= =
-
 
Therefore, 
430 90
' ' 40
13 13
A B cm = -= 
Therefore, 
30 3 1
tan
90
3
13
13
a = =
??
??
??
 
6
p
a ? =
 
Therefore, 6.00 n = 
 
Question: At time t = 0, a disk of radius 1 m starts to roll without slipping on a horizontal 
plane with an angular acceleration of 
2
2
.
3
rads a
-
= A small stone A stuck to the disk. At t = 
0, it is at the contact point of the disk and the plane. Later, at time , ts p = the stone detaches 
Page 4


   
JEE-Advanced-28-08-2022 
[Paper-1]  
 
Physics 
 
Question: Two spherical stars A and B have densities 
A
? and ,
B
? respectively. A and B 
have the same radius, and their masses MA and MB are related by MB = 2MA. Due to an 
interaction process, star A loses some of its mass, so that its radius is halved, while its 
spherical shape is retained and its density remains .
A
? The entire mass lost by A is deposited 
as a thick spheric shell on B with the density the shell being .
A
? If 
A
v and 
B
v are the escape 
velocities from A and B after the interaction process, the ratio 
1/3
10
.
15
B
A
v n
v
= The value of n is 
________ 
Answer: 2.30 
Solution: 
0
2
2
8
A
A
v GM
v
R
A
= =
??
×
??
??
 
For, B, 
( )
33 3
4 47
3 38
rR R p p -= × 
1
3
15
8
rR
??
?=
??
??
 
Therefore, 
( ) ( )
0 11
33
7
22
23 2 8
15 8 15
2
AA
B
GM M
vv
R
? ?
× +
? ?
×
? ?
= = ×
×
 
Therefore,
( ) ( )
11
33
23 2.30 10
15 15
B
A
v
v
×
= = 
Therefore, n = 2.30 
 
Question: The minimum kinetic energy needed by an alpha particle to cause the nuclear 
reaction 
16 4 1 19
7 2 18
N He H + ?+ O in a laboratory frame is n (in MeV). Assume that 
16
7
N is at 
   
rest in the laboratory frame. The masses of 
16
7
, N 
1 19
18
and HO can be taken to be 16.006 u, 
4.003 u, 1008 u and 19.003 u, respectively, where 1 u = 930 MeV c
-2
. The value of n is 
_______. 
Answer: 2.33 
Solution: 
( )
2
N He H O
Qm m m m c = + -- ×
 
( ) 16.006 4.003 1.008 19.003 930MeV = + -- ×
 
1.86MeV = -
 
1.86 energy absorbed. MeV =
 
And, 
2
14
max loss in kinetic energy
2 5
mm
v
m
×
× ×= 
2
15
24
mv Q ?=×
 
( )
5
1.86 2.325
4
MeV MeV =× =
 
Therefore, 2.33 n = 
 
Question: In the following circuit 
1 23
12 , 4 C FC C F µµ = = = and C4 = C5 = 2 µF. The charge 
stored in C3 is ______µC 
 
Answer: 8.00 
Solution: 
From circuit given, Potential difference across C3 is 2 V (constant) 
Therefore, 
3
24 8 Q C C µµ =×= 
 
Question: A rod of length 2 cm makes an angle 
2
3
p
rad with the principal axis of a thin 
convex lens. The lens has a focal length of 10 cm and placed at a distance of 
40
3
cm from the 
   
object as shown in the figure. The height of the image is 
30 3
13
cm and the angle made by it 
with respect to the principal axis is a rad. The value of a is 
n
p
rad, where n is ______ 
 
Answer: 6.00 
Solution: 
 
40
10
3
' 40
43
10
3
OA cm
×
= =
-
 
43
10
430
3
'
43
13
10
3
OB cm
×
= =
-
 
Therefore, 
430 90
' ' 40
13 13
A B cm = -= 
Therefore, 
30 3 1
tan
90
3
13
13
a = =
??
??
??
 
6
p
a ? =
 
Therefore, 6.00 n = 
 
Question: At time t = 0, a disk of radius 1 m starts to roll without slipping on a horizontal 
plane with an angular acceleration of 
2
2
.
3
rads a
-
= A small stone A stuck to the disk. At t = 
0, it is at the contact point of the disk and the plane. Later, at time , ts p = the stone detaches 
   
itself and flies off tangentially from the disk. The maximum height (in m) reached by the 
stone measured from the plane is 
1
.
2 10
x
+ The value of x is ___________       [Take g = 10 
ms
-2
] 
Answer: 0.52 
Solution: 
The angle rotated by disc in ts p = is 
( )
2
0
2
1
2
12
2 3
3
tt
rad
?? a
?p
p
= +
?= ×
=
 
and the angular velocity of disc is 
0
2
/
3
t
rad s
? ? a
p
= +
=
 
And 
2
1
3
cm
vR
p
? = = × 
2
/
3
ms
p
= 
So, at the moment it detaches the situation is 
 
( ) ( )
2
2
2 cos120
cm cm
v R v Rv ?? = ++ ° 
2
/
3
sin120
and tan
cos120
cm
cm
v ms
R
vR
p
?
?
?
= =
°
=
+°
 
Page 5


   
JEE-Advanced-28-08-2022 
[Paper-1]  
 
Physics 
 
Question: Two spherical stars A and B have densities 
A
? and ,
B
? respectively. A and B 
have the same radius, and their masses MA and MB are related by MB = 2MA. Due to an 
interaction process, star A loses some of its mass, so that its radius is halved, while its 
spherical shape is retained and its density remains .
A
? The entire mass lost by A is deposited 
as a thick spheric shell on B with the density the shell being .
A
? If 
A
v and 
B
v are the escape 
velocities from A and B after the interaction process, the ratio 
1/3
10
.
15
B
A
v n
v
= The value of n is 
________ 
Answer: 2.30 
Solution: 
0
2
2
8
A
A
v GM
v
R
A
= =
??
×
??
??
 
For, B, 
( )
33 3
4 47
3 38
rR R p p -= × 
1
3
15
8
rR
??
?=
??
??
 
Therefore, 
( ) ( )
0 11
33
7
22
23 2 8
15 8 15
2
AA
B
GM M
vv
R
? ?
× +
? ?
×
? ?
= = ×
×
 
Therefore,
( ) ( )
11
33
23 2.30 10
15 15
B
A
v
v
×
= = 
Therefore, n = 2.30 
 
Question: The minimum kinetic energy needed by an alpha particle to cause the nuclear 
reaction 
16 4 1 19
7 2 18
N He H + ?+ O in a laboratory frame is n (in MeV). Assume that 
16
7
N is at 
   
rest in the laboratory frame. The masses of 
16
7
, N 
1 19
18
and HO can be taken to be 16.006 u, 
4.003 u, 1008 u and 19.003 u, respectively, where 1 u = 930 MeV c
-2
. The value of n is 
_______. 
Answer: 2.33 
Solution: 
( )
2
N He H O
Qm m m m c = + -- ×
 
( ) 16.006 4.003 1.008 19.003 930MeV = + -- ×
 
1.86MeV = -
 
1.86 energy absorbed. MeV =
 
And, 
2
14
max loss in kinetic energy
2 5
mm
v
m
×
× ×= 
2
15
24
mv Q ?=×
 
( )
5
1.86 2.325
4
MeV MeV =× =
 
Therefore, 2.33 n = 
 
Question: In the following circuit 
1 23
12 , 4 C FC C F µµ = = = and C4 = C5 = 2 µF. The charge 
stored in C3 is ______µC 
 
Answer: 8.00 
Solution: 
From circuit given, Potential difference across C3 is 2 V (constant) 
Therefore, 
3
24 8 Q C C µµ =×= 
 
Question: A rod of length 2 cm makes an angle 
2
3
p
rad with the principal axis of a thin 
convex lens. The lens has a focal length of 10 cm and placed at a distance of 
40
3
cm from the 
   
object as shown in the figure. The height of the image is 
30 3
13
cm and the angle made by it 
with respect to the principal axis is a rad. The value of a is 
n
p
rad, where n is ______ 
 
Answer: 6.00 
Solution: 
 
40
10
3
' 40
43
10
3
OA cm
×
= =
-
 
43
10
430
3
'
43
13
10
3
OB cm
×
= =
-
 
Therefore, 
430 90
' ' 40
13 13
A B cm = -= 
Therefore, 
30 3 1
tan
90
3
13
13
a = =
??
??
??
 
6
p
a ? =
 
Therefore, 6.00 n = 
 
Question: At time t = 0, a disk of radius 1 m starts to roll without slipping on a horizontal 
plane with an angular acceleration of 
2
2
.
3
rads a
-
= A small stone A stuck to the disk. At t = 
0, it is at the contact point of the disk and the plane. Later, at time , ts p = the stone detaches 
   
itself and flies off tangentially from the disk. The maximum height (in m) reached by the 
stone measured from the plane is 
1
.
2 10
x
+ The value of x is ___________       [Take g = 10 
ms
-2
] 
Answer: 0.52 
Solution: 
The angle rotated by disc in ts p = is 
( )
2
0
2
1
2
12
2 3
3
tt
rad
?? a
?p
p
= +
?= ×
=
 
and the angular velocity of disc is 
0
2
/
3
t
rad s
? ? a
p
= +
=
 
And 
2
1
3
cm
vR
p
? = = × 
2
/
3
ms
p
= 
So, at the moment it detaches the situation is 
 
( ) ( )
2
2
2 cos120
cm cm
v R v Rv ?? = ++ ° 
2
/
3
sin120
and tan
cos120
cm
cm
v ms
R
vR
p
?
?
?
= =
°
=
+°
 
   
tan 3
3
rad
?
p
?
?=
?=
 
So, 
2 2
max
sin
2
u
H
g
?
= 
2
2
2
sin 60
3
2 10
p
? ?
×°
? ?
? ?
=
×
 
43
92 104
60
m
p
p
×
=
×× ×
=
 
So, height from ground will be 
( )
1
1 cos 60
60 2 10
0.52
6
x
R
x
p
p
- °+ = +
?= =
 
Question: A solid sphere of mass I kg and radius 1 m rolls without slipping on a fixed 
inclined plane with an angle of inclination ? = 30
o
 from the horizontal. Two forces of 
magnitude 1 N each, parallel to the incline, act on the sphere, both at distance r = 0.5 m from 
the center of the sphere, as shown in the figure. The acceleration of the sphere down the plane 
is _____ms
-2
 (Take g = 10 ms
-2
)  
 
Answer: 0.286 
Solution: