QJEEAdv22Phfcf1: JEE Advanced 2022 Physics - Class 11 JEEAdv22Phfcf1 Solution: Two spherical stars and have densities and , respectively. and…

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 ${R_2}$ .

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,

$10n = 23$

$\Rightarrow n = 2.3$