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JEE Advance - Physics (2022 - Paper 1 Online - No. 1)

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 $\rho_{A}$. 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

Explanation

JEE Advanced 2022 Paper 1 Online Physics - Gravitation Question 6 English Explanation 1

Due to an interaction process, star A losses some of it's mass and radius becomes $${R \over 2}$$. Let new mass of star A is M'A. Here in both cases density of star A remains same $${\rho _A}$$.

$$\therefore$$ Initially $$({\rho _A}) = {{{M_A}} \over {{4 \over 3}\pi {R^3}}}$$

Finally $$({\rho _A}) = {{M{'_A}} \over {{4 \over 3}\pi {{\left( {{R \over 2}} \right)}^3}}}$$

Density remains same,

$$\therefore$$ $${{{M_A}} \over {{4 \over 3}\pi {R^3}}} = {{M{'_A}} \over {{4 \over 3}\pi {{\left( {{R \over 2}} \right)}^3}}}$$

$$ \Rightarrow 8M{'_A} = {M_A}$$

$$ \Rightarrow M{'_A} = {{{M_A}} \over 8}$$

$$\therefore$$ Lost mass by $$A = {M_A} - M{'_A} = {M_A} - {{{M_A}} \over 8} = {{7{M_A}} \over 8}$$

This lost mass $${{7{M_A}} \over 8}$$ is attached on the star B and density of the attached mass stay $${\rho _A}$$. So new radius of star B is $${R_2}$$.

JEE Advanced 2022 Paper 1 Online Physics - Gravitation Question 6 English Explanation 2

Density of the removed part from star A is,

$${\rho _A} = {{{{7{M_A}} \over 8}} \over {{4 \over 3}\pi \left( {{R^3} - {{\left( {{R \over 2}} \right)}^3}} \right)}}$$

Density of the added part in star B stay's same as $${\rho _A}$$,

$$\therefore$$ $${\rho _A} = {{{{7{M_A}} \over 8}} \over {{4 \over 3}\pi \left( {R_2^3 - {R^3}} \right)}}$$

$$\therefore$$ $${{{{7{M_A}} \over 8}} \over {{4 \over 3}\pi \left( {{R^3} - {{\left( {{R \over 2}} \right)}^3}} \right)}} = {{{{7{M_A}} \over 8}} \over {{4 \over 3}\pi \left( {R_2^3 - {R^3}} \right)}}$$

$$ \Rightarrow {R^3} - {{{R^3}} \over 8} = R_2^3 - {R^3}$$

$$ \Rightarrow R_2^3 = 2{R^3} - {{{R^3}} \over 8}$$

$$ \Rightarrow R_2^3 = {{15{R^3}} \over 8}$$

$$ \Rightarrow {R_2} = {(15)^{{1 \over 3}}} \times {R \over 2}$$

Escape velocity from star A after interaction process,

$${V_A} = \sqrt {{{2G\left( {{{{M_A}} \over 8}} \right)} \over {{R \over 2}}}} $$

And escape velocity from star B after interaction process,

$${V_B} = \sqrt {{{2G\left( {{{23{M_A}} \over 8}} \right)} \over {{{(15)}^{{1 \over 3}}} \times {R \over 2}}}} $$

$$\therefore$$ $${{{V_B}} \over {{V_A}}} = \sqrt {{{{{23{M_A}} \over 8}} \over {{{(15)}^{{1 \over 3}}} \times {R \over 2}}} \times {{{R \over 2}} \over {{{{M_A}} \over 8}}}} $$

$$ = \sqrt {{{23} \over {{{(15)}^{{1 \over 3}}}}}} $$ ..... (1)

Given,

$${{{V_B}} \over {{V_A}}} = \sqrt {{{10n} \over {{{(15)}^{{1 \over 3}}}}}} $$ ..... (2)

Comparing equation (1) and (2), we get,

$$10n = 23$$

$$ \Rightarrow n = 2.3$$

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