Explanation

(d) Let a nucleus of mass $M$ splits into two nuclear parts having masses $M_{1}$ and $M_{2}$ and radii $R_{1}$ and $R_{2}$ and densities $\rho_{1}$ and $\rho_{2}$
$\therefore M_{1} = \rho_{1}\frac{4}{3}\pi R_{1}^{3}$ and $M_{2} = \rho_{2}\frac{4}{3}\pi R_{2}^{3}$
Given: $\rho_{1} = \rho_{2}$
$\therefore\frac{M_{1}}{M_{2}} = \left( \frac{R_{1}}{R_{2}} \right)^{3}$
According to law conservation of linear momentum,
$M \times 0 = M_{1}v_{1} + M_{2}v_{2}$ or $\frac{M_{1}}{M_{2}} = - \frac{v_{2}}{v_{1}}$
$- ve$ sign show that both the parts are move in opposite direction in order to conserve the linear momentum
$\therefore\frac{v_{1}}{v_{2}} = \frac{M_{2}}{M_{1}}\ or\ \frac{v_{1}}{v_{2}} = \left( \frac{R_{2}}{R_{1}} \right)^{3}$
$\frac{v_{1}}{v_{2}} = \left( \frac{2}{1} \right)^{3} = \frac{8}{1}$ $\left\lbrack Given\frac{R_{1}}{R_{2}} = \frac{1}{2} \right\rbrack$