A small spherical ball of radius $r$ falls freely under gravity through a distance $h$ before entering a tank of water. If, after entering the water, the velocity of the ball does not change, then $h$ is proportional to 

As indicated in the question as the ball enters the water it has already attained the terminal velocity. Terminal velocity is reached when the viscous force (acting upwards) balances the weight $mg$ of the ball. Therefore, 

                                 $6\mathrm{\pi \eta r}v=mg$

where $6\pi \eta rv$ is the viscous force on the ball. Here $v$ its terminal velocity and $\eta$, the coefficient of viscosity of water. If $\rho$ is the density of the material of the ball, 

we have     $m=\frac{4}{3}{\mathrm{\pi r}}^3\mathrm{\rho } \therefore 6\mathrm{\pi \eta rv}=\frac{4}{3}{\mathrm{\pi r}}^3\mathrm{\rho g}$

or              $v=\frac{2r^2\rho g}{9\eta } \left(1\right)$

Now         $v^2-0=2gh \Rightarrow h=\frac{v^2}{2g} \left(2\right)$

From (1) and (2) we find that h$\propto$r⁴, which is choice (3).