Explanation

(d) The terminal velocity of the spherical body of radius $R$, density $\rho$ falling through a liquid of density $\sigma$ is given by
$v_{T} = \frac{2}{9}\frac{R^{2}(\rho - \sigma)g}{\eta}$
Where $\eta$ is the coefficient of viscosity of the liquid
$\therefore v_{T_{1}} = \frac{2R_{1}^{2}\left( \sigma_{2} - \sigma \right)g}{9\eta}$ and $v_{T_{2}} = \frac{2R_{2}^{2}\left( \sigma_{2} - \sigma \right)g}{9\eta}$
According to the given problem, $v_{T_{1}} = v_{T_{2}}$
$R_{1}^{2}\left( \rho_{1} - \sigma \right) = R_{2}^{2}(\rho_{2} - \sigma)$ or $\frac{R_{1}^{2}}{R_{2}^{2}} = \frac{\rho_{2} - \sigma}{\rho_{1} - \sigma}$
Substituting the given values, we get
$\frac{R_{1}^{2}}{R_{2}^{2}} = \frac{(11 \times 10^{3} - 2 \times 10^{3})}{(8 \times 10^{3} - 2 \times 10^{3})} = \frac{9}{6} = \frac{3}{2}$
$\frac{R_{1}}{R_{2}} = \sqrt{\frac{3}{2}}$