QJEEAdv17Ph583f: JEE Advanced 2017 Physics - Class 11 JEEAdv17Ph583f Solution: Consider an expanding sphere of instantaneous radius R whose…

Density $\rho = {{mass\,(m)} \over {volume\,(V)}}$

Volume of sphere $= {4 \over 3}\pi {R^3} \Rightarrow \rho = {m \over {{4 \over 3}\pi {R^3}}} = {{3m} \over {4\pi {R^3}}}$

Rearranging it we get

$\rho {R^2} = {3 \over {4\pi }}m$

It is given that mas m remains constant. Let the constant be k. Therefore,

$\rho {R^3} = k$

Differentiating it w.r.t. time, we get

${d \over {dt}}(\rho {R^3}) = 0$ (Since, differentiation of constant is zero)

${R^3}{{d\rho } \over {dt}} + 3{R^2}\rho {{dR} \over {dt}} = 0$

Now ${{dR} \over {dt}}$ , that is, rate of change of radius is equal to velocity. Therefore,

${R^3}{{d\rho } \over {dt}} + 3{R^2}\rho v = 0 \Rightarrow 3{R^2}\rho v = - {R^3}{{d\rho } \over {dt}}$

$\Rightarrow v = {{ - 1} \over 3}{{{R^3}} \over {{R^2}}}{1 \over \rho }{{d\rho } \over {dt}} = {{ - 1R} \over 3}{1 \over \rho }{{d\rho } \over {dt}}$

It is given that rate of fractional change in density $\left( {{1 \over \rho }{{d\rho } \over {dt}}} \right)$ is constant. Thus,

v $\propto$ R

Thus velocity of any point on surface of expanding sphere is proportional to R.