Step-by-Step Solution

1. Initial State of Stars A and B

Let the initial radius of both stars be $R$. So, $R_A = R_B = R$. Let their densities be $ho_A$ and $ho_B$. The initial masses are given by: $M_A = \rho_A \times \frac{4}{3}\pi R^3$ $M_B = \rho_B \times \frac{4}{3}\pi R^3$

We are given the relation $M_B = 2 M_A$. This implies: $\rho_B \frac{4}{3}\pi R^3 = 2 \left(\rho_A \frac{4}{3}\pi R^3\right) \implies \rho_B = 2\rho_A$.

2. Final State of Star A

After the interaction, star A loses some mass. Its new radius is halved: $R'_A = \frac{R}{2}$. The density of star A remains $\rho_A$. Its new mass, $M'_A$, is: $M'_A = \rho_A \times \frac{4}{3}\pi (R'_A)^3 = \rho_A \frac{4}{3}\pi \left(\frac{R}{2}\right)^3 = \rho_A \frac{4}{3}\pi \frac{R^3}{8}$ $M'_A = \frac{1}{8} \left(\rho_A \frac{4}{3}\pi R^3\right) = \frac{1}{8} M_A$.

3. Mass Transferred

The mass lost by star A is: $\Delta M_A = M_A - M'_A = M_A - \frac{1}{8}M_A = \frac{7}{8}M_A$. This mass is deposited as a shell on star B.

4. Final State of Star B

The mass of the shell added to star B is $M_{shell} = \Delta M_A = \frac{7}{8}M_A$. The new total mass of star B, $M'_B$, is: $M'_B = M_B + M_{shell} = 2M_A + \frac{7}{8}M_A = \left(2 + \frac{7}{8}\right)M_A = \frac{23}{8}M_A$.

The density of the shell is given as $\rho_A$. Let the new outer radius of star B be $R'_B$. The volume of the shell, $V_{shell}$, is: $V_{shell} = \frac{4}{3}\pi ((R'_B)^3 - R^3)$. Also, $M_{shell} = \rho_A V_{shell}$. $\frac{7}{8}M_A = \rho_A \left(\frac{4}{3}\pi ((R'_B)^3 - R^3)\right)$. Substitute $M_A = \rho_A \frac{4}{3}\pi R^3$: $\frac{7}{8}\left(\rho_A \frac{4}{3}\pi R^3\right) = \rho_A \left(\frac{4}{3}\pi ((R'_B)^3 - R^3)\right)$. Canceling common terms $(\rho_A \frac{4}{3}\pi)$: $\frac{7}{8}R^3 = (R'_B)^3 - R^3$ $(R'_B)^3 = R^3 + \frac{7}{8}R^3 = \frac{15}{8}R^3$ $R'_B = \left(\frac{15}{8}\right)^{1/3} R = \frac{15^{1/3}}{2} R$.

5. Escape Velocities

The escape velocity from the surface of a spherical body of mass $M$ and radius $R$ is $v_{esc} = \sqrt{\frac{2GM}{R}}$.

For star A after the process: $v_A = \sqrt{\frac{2GM'_A}{R'_A}} = \sqrt{\frac{2G(\frac{1}{8}M_A)}{\frac{R}{2}}} = \sqrt{\frac{\frac{1}{4}GM_A}{\frac{R}{2}}} = \sqrt{\frac{GM_A}{2R}}$.

For star B after the process: $v_B = \sqrt{\frac{2GM'_B}{R'_B}} = \sqrt{\frac{2G(\frac{23}{8}M_A)}{\frac{15^{1/3}}{2}R}} = \sqrt{\frac{\frac{23}{4}GM_A}{\frac{15^{1/3}}{2}R}} = \sqrt{\frac{23 \cdot 2 \cdot GM_A}{4 \cdot 15^{1/3} \cdot R}} = \sqrt{\frac{23GM_A}{2R \cdot 15^{1/3}}}$.

6. Ratio of Escape Velocities

Now, we find the ratio $\frac{v_B}{v_A}$: $\frac{v_B}{v_A} = \frac{\sqrt{\frac{23GM_A}{2R \cdot 15^{1/3}}}}{\sqrt{\frac{GM_A}{2R}}} = \sqrt{\frac{\frac{23GM_A}{2R \cdot 15^{1/3}}}{\frac{GM_A}{2R}}} = \sqrt{\frac{23GM_A}{2R \cdot 15^{1/3}} \times \frac{2R}{GM_A}}$. $\frac{v_B}{v_A} = \sqrt{\frac{23}{15^{1/3}}}$.

7. Finding the value of n

We are given the expression for the ratio: $\frac{v_B}{v_A} = \sqrt{\frac{10n}{15^{1/3}}}$.

Equating our derived expression with the given one: $\sqrt{\frac{23}{15^{1/3}}} = \sqrt{\frac{10n}{15^{1/3}}}$.

Squaring both sides: $\frac{23}{15^{1/3}} = \frac{10n}{15^{1/3}}$.

$23 = 10n$ $n = \frac{23}{10} = 2.3$.

The value of n is 2.3.