Explanation

(c) If a sphere of mass M and radius R is dropped in a liquid, its weight $Mg\left( = \frac{4}{3}\pi\ R^{3}\rho g \right)$ acts vertically downwards. While upthrust $\frac{4}{3}\ \pi\ R^{3}\sigma g$ and viscous force $6\pi ɳRv$ acts vertically upwards. Initially the body will be accelerated down. At a certain instant when viscous force F will balance the net downward force, acceleration will become zero and the body will fall with constant velocity.
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$\therefore\ 6\pi ɳ\ Rv_{T} = \frac{4}{3}\pi\ R^{3}(\rho - \sigma)g$
$ie,\ \ v_{r} = \frac{2}{9}R^{2}\left( \frac{\rho - \sigma}{ɳ} \right)g$
$or\ \ \ \ \ v_{T} \propto R^{2}$