JEE Advanced 2022 · Paper 1 · Physics Q1

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PhysicsNumerical+3Question 19 of 54

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 $M_B = 2M_A$ . 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 $\dfrac{v_B}{v_A} = \sqrt{\dfrac{10n}{15^{1/3}}}$ . The value of $n$ is _____.

Your answer

Answer and worked solution

Official answer: 2.2

Both stars start with radius $R$ , and $M_B=2M_A$ .

Star A. Its radius halves at the same density, so its mass becomes $\frac{M_A}8$ and its radius $\frac R2$ . It loses $\frac78M_A$ .

Star B. It gains a shell of density $\rho_A$ and volume $\frac78\cdot\frac43\pi R^3$ :

outer radius: $R'^3=R^3+\frac78R^3=\frac{15}8R^3$ , so $R'=\frac{15^{1/3}}2R$ ;
mass: $2M_A+\frac78M_A=\frac{23}8M_A$ .

Ratio. Since $v_e^2=\frac{2GM}{R}$ :
$\frac{v_B^2}{v_A^2}=\frac{(23M_A/8)/(15^{1/3}R/2)}{(M_A/8)/(R/2)}=\frac{23}{15^{1/3}}=\frac{10n}{15^{1/3}}.$

So $n=2.3$ .

Topics:Escape velocity Gravitational field and potential

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JEE Advanced 2022 · Paper 1, Physics question 1