Two spherical stars A and B have densities _A and _B, respectivel… - Vidyadip

MCQ

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 $\rho_4$. 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. . . . .

✓
$2.30$
B
$2.35$
C
$2.40$
D
$2.45$

✓

Answer

Correct option: A.

$2.30$

a
Given $R_A=R_B=R$

$M _{ B }=2 M _{ A }$

Calculation of escape velocity for $A$ :

Radius of remaining star $=\frac{R_A}{2}$.

Mass of remaining star $=\rho_A \frac{4}{3} \pi \frac{R_A^3}{8}=\frac{M_A}{8}$

$\frac{- GM _{ A / B }}{ R _{ A / 2}}+\frac{1}{2} mv _{ A }^2=0 \Rightarrow v _{ A }=\sqrt{\frac{2 GM _{A / B}}{ R _{A / 2}}}=\sqrt{\frac{ GM _A}{2 R }}$

Calculation of escape velocity for $B$

Mass collected over $B =\frac{7}{8} M _{\text {A }}$

Let the radius of $B$ becomes $r$.

$\therefore \frac{4}{3} \pi\left( r ^3- R _B^3\right) \rho_A=\frac{7}{8} \rho_A \frac{4}{3} \pi R _A^3 \Rightarrow \pi^3=\frac{7}{8} R _A^3+ R _{ B }^3=\frac{(15)^{1 / 3} R }{2}$

$\therefore \frac{ V _{ B }^2}{2}=\frac{23 GM _A}{8 \times 15^{1 / 3} \frac{ R }{2}}=\frac{23 GM _A}{4 \times 15^{1 / 3} R }$

$\therefore V _{ B }=\sqrt{\frac{23 GM _A}{2 \times 15^{1 / 3} R }}$

$\therefore \frac{ V _{ B }}{ V _A}=\sqrt{\frac{23}{15^{1 / 3}}}=\sqrt{\frac{10 \times 2.30}{15^{1 / 3}}}$

$n =2.30$

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