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JEE Advanced 2022 Paper 1
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© examsnet.com
Question : 19
Total: 54
Two spherical stars
A
and
B
have densities
ρ
A
and
ρ
B
, respectively.
A
and
B
have the same radius, and their masses
M
A
and
M
B
are related by
M
B
=
2
M
A
. 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
ρ
A
. The entire mass lost by
A
is deposited as a thick spherical shell on
B
with the density of the shell being
ρ
A
. If
v
A
and
v
B
are the escape velocities from
A
and
B
after the interaction process, the ratio
v
B
v
A
=
√
10
n
15
1
∕
3
. The value of
n
is
Your Answer:
Validate
Solution:
Given
R
A
=
R
B
=
R
M
B
=
2
M
A
Calculation of escape velocity for A:
Radius of remaining star
=
R
A
2
.
Mass of remaining star
=
ρ
A
4
3
π
R
A
3
8
=
M
A
8
−
GM
A
∕
B
R
A
∕
2
+
1
2
m
v
A
2
=
0
⇒
v
A
=
√
2
GM
A
∕
B
R
A
∕
2
=
√
GM
A
2
R
Calculation of escape velocity for B
Mass collectedover
B
=
7
8
M
A
Let the radius of
B
becomes
r
.
∴
4
3
π
(
r
3
−
R
B
3
)
ρ
A
=
7
8
ρ
A
4
3
π
R
A
3
⇒
π
3
=
7
8
R
A
3
+
R
B
3
=
(
15
)
1
∕
3
R
2
∴
V
B
2
2
=
23
GM
A
8
×
15
1
∕
3
R
2
=
23
GM
A
4
×
15
1
∕
3
R
∴
V
B
=
√
23
GM
A
2
×
15
1
∕
3
R
∴
V
B
V
A
=
√
23
15
1
∕
3
=
√
10
×
2.30
15
1
∕
3
n
=
2.30
© examsnet.com
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