In an experiment to verify Stokes law, a small spherical ball of radius $r$ and density $\rho$ 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)
Correct Option: 1
(1) Using, $v^{2}-u^{2}=2 g h$
$\Rightarrow v^{2}-0^{2}=2 g h \Rightarrow v=\sqrt{2 g h}$
$V_{T}=\frac{2}{9} \frac{r^{2}(\rho-\sigma) g}{\eta}$
After falling through $h$ the velocity should be equal to terminal velocity
$\therefore \sqrt{2 g h}=\frac{2}{9} \frac{r^{2}(\rho-\sigma) g}{\eta}$
$\Rightarrow 2 g h=\frac{4}{81} \frac{r^{4} g^{2}(\rho-\sigma)^{2}}{\eta^{2}}$
$\Rightarrow h=\frac{2 r^{4} g(\rho-\sigma)^{2}}{81 \eta^{2}} \Rightarrow h \alpha r^{4}$
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