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 =sqrt frac 10 n 15 1 / 3 . The value of n is

Physics · Gravitation

JEE Advanced 2022 — Paper 1 — Question 19

Two spherical stars A and B have densities $\rho_{\mathrm{A}}$ and $\rho_{\mathrm{B}}$ , 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 $\rho_{\mathrm{A}}$ . 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

Answer: 2.3

Numerical answer — enter this value.

Step-by-step solution

$M_{B}=2 M_{A} \quad \Rightarrow \quad \rho_{B}=2 \rho_{A}$ Now after the mass transfer, $\mathrm{M}_{\mathrm{A}}^{\prime}=\rho_{\mathrm{A}} \frac{4}{3} \times \frac{R^{3}}{8}$ and

$\mathrm{M}_{\mathrm{B}}=\mathrm{G} \frac{4}{3} \pi\left(2 \rho_{\mathrm{A}} \mathrm{R}^{3}+\frac{7}{8} \rho_{\mathrm{A}} \mathrm{R}^{3}\right)$ and outer radius $\mathrm{R}^{\prime}==\frac{(15)^{1 / 3}}{2} R$

So, $\quad \mathrm{v}_{\mathrm{A}}=\sqrt{\frac{2 G M_{A}^{\prime}}{R}}=\sqrt{\frac{2}{3} G \rho_{A} \pi R^{2}}$ \end{enumerate} $\begin{aligned} \mathrm{v}_{\mathrm{B}} & =\sqrt{\frac{2 G(4 / 3) \pi\left(2 \rho_{A} R^{3}+\frac{7}{8} \rho_{A} R^{3}\right)}{R^{\prime}}} \frac{v_{B}}{v_{A}} & =\sqrt{\frac{23}{(15)^{1 / 3}}}=\sqrt{\frac{(2.3)(10)}{(15)^{1 / 3}}} \end{aligned}$

So, $\quad \mathrm{n}=2.30$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2022
Paper: Paper 1
Subject: Physics
Chapter: Gravitation
Topic: Motion of Satellites and Escape Speed