JEE Advanced 2022 — Paper 1 — PHY-I solution

JEE Advanced2022Paper 1PHY-I

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 ______.

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Solution

Star A after mass loss: radius becomes $R /2$ , density unchanged at $ρ_{A}$ . The new mass is $M_{A}^{'} = ρ_{A} \cdot \frac{4}{3} π (R /2)^{3} = M_{A} /8$ .

Escape velocity from A: $v_{A} = \frac{2 G M _{A}^{'}}{R /2} = \frac{2 G M _{A}}{8 \cdot ( R /2 )} = \frac{v _{0}}{2}$ , where $v_{0} = \frac{2 G M _{A}}{R}$ .

Star B after accretion: mass lost by A equals $\frac{7 M _{A}}{8}$ , deposited as a shell. Using $\frac{4}{3} π (r^{3} - R^{3}) ρ_{A} = \frac{7 M _{A}}{8}$ with $M_{A} = \frac{4}{3} π R^{3} ρ_{A}$ , we get $r = (\frac{15}{8})^{1/3} R$ .

New mass of B is $M_{B}^{'} = 2 M_{A} + \frac{7 M _{A}}{8} = \frac{23 M _{A}}{8}$ .

$v_{B} = \frac{2 G \cdot \frac{23 M _{A}}{8}}{( 15/8 ) ^{1/3} R} = v_{0} \frac{23 \cdot 2}{8 \cdot 1 5 ^{1/3}}$ .

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

$∴ n = 2.30$ .

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