Two spherical stars \(A\) and \(B\) have densities \(\rho_{A}\) and \(\rho_{B}\), 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 \(\rho_{A}\). The entire mass lost by \(A\) is deposited as a thick spherical shell on \(B\) with the density of the shell being \(\rho_{A}\). If \(v_{A}\) and \(v_{B}\) are the escape velocities from \(A\) and \(B\) after the interaction process, the ratio \(\frac{v_{B}}{v_{A}}=\sqrt{\frac{10 n}{15^{1 / 3}}}\). The value of \(n\) is ____

JEE Advanced · 2022 · 28 Aug 2022 · Paper 1 · Gravitation

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

\frac{v _{B}}{v _{A}} = \frac{10 n}{1 5 ^{1/3}}

. The value of

n

is ____

Answer: 2.3

After the interaction, star A has radius

R_{A} /2

and mass

M_{A} /8

(since density remains

ρ_{A}

and volume becomes

1/8

).

Escape velocity from A:

v_{A} = \frac{2 G M _{A /8}}{R _{A} /2} = \frac{G M _{A}}{2 R}

Mass deposited on B:

7 M_{A} /8

. The new radius

r

of B satisfies:

\frac{4}{3} π (r^{3} - R^{3}) ρ_{A} = \frac{7}{8} M_{A}

Solving:

r^{3} = \frac{15 R ^{3}}{8}

, so

r = \frac{1 5 ^{1/3} R}{2}

Total mass of B:

2 M_{A} + 7 M_{A} /8 = 23 M_{A} /8

Escape velocity from B:

v_{B} = \frac{2 G ( 23 M _{A} /8 )}{1 5 ^{1/3} R /2} = \frac{23 G M _{A}}{2 \times 1 5 ^{1/3} R}

Ratio:

\frac{v _{B}}{v _{A}} = \frac{23}{1 5 ^{1/3}} = \frac{10 \times 2.30}{1 5 ^{1/3}}

Therefore,

n = 2.30

JEE Advanced · 2022 · 28 Aug 2022 · Paper 1 · Numerical answer