In an experiment to verify Stoke's law, a small spherical ball of radius $r$ and density $\rho$ fall

Question

In an experiment to verify Stoke's law, a small spherical ball of radius $r$ and density $ho$ falls under gravity through a distance $h$ in air before entering a tank of water. If the terminal velocity of the ball inside water is same as its velocity just before entering the water surface, then the value of $h$ is proportional to (ignore viscosity of air)\

Professor PC Thomas Classes · Answer

(a) : Velocity attained by the spherical ball while falling under gravity through a distance $h, v=\sqrt{2 g h}$.\ Velocity attained by the spherical ball after entering into the water tank,\ $v_{T}=\frac{2}{9} \frac{r^{2}(ho-\sigma) g}{\eta}$\ where $\quad ho=$ density of the spherical ball $r=$ radius of the spherical ball\ $\sigma=$ density of fluid (water)\ and $\eta=$ coefficient of viscosity\ According to the question, $v=v_{T}$\ i.e., $\sqrt{2 g h}=\frac{2}{9} \frac{r^{2}(ho-\sigma) g}{\eta}$\ or $h=\frac{2 r^{4}(ho-\sigma)^{2} g}{81 \eta^{2}}$\ $	herefore \quad h$ is proportional to $r^{4}$.\