Explanation

(d) Law of conservation of momentum gives
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ m_{1}v_{1} = m_{2}v_{2}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{m_{1}}{m_{2}} = \frac{v_{2}}{v_{1}}$
But $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ m = \frac{4}{3}\pi r^{3}\rho$
$\ or\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ m \propto r^{3}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{m_{1}}{m_{2}} = \frac{r_{1}^{3}}{r_{2}^{3}} = \frac{v_{2}}{v_{1}}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{r_{1}}{r_{2}} = \left( \frac{1}{2} \right)^{1/3}$
$\therefore$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ r_{1}:r_{2} = 1:2^{1/3}$