Q. A ball of radius $r$ and density $\rho$ falls freely under gravity through a distance $h$ before entering water. Velocity of ball does not change even on entering wate r. If viscosity of water is $\eta$, the value of $h$ is given by
A) $\frac{2}{9}r^{2}\left( \frac{1 - \rho}{\eta} \right)g$
B) $\frac{2}{81}r^{2}\left( \frac{\rho - 1}{\eta} \right)g$
C) $\frac{2}{81}r^{4}\left( \frac{\rho - 1}{\eta} \right)^{2}g$
D) $\frac{2}{9}r^{4}\left( \frac{\rho - 1}{\eta} \right)^{2}g$
Explanation
(c) Velocity of ball when it strikes the water surface
$v = \sqrt{2gh}$ …(i)
Terminal velocity of ball inside the water
$v = \frac{2}{9}r^{2}g\left( \frac{\rho - 1}{\eta} \right)$ …(ii)
Equating (i) and (ii) we get $\sqrt{2gh} = \frac{2}{9}\frac{r^{2}g}{\eta}(\rho - 1)$
$\Rightarrow h = \frac{2}{81}r^{4}\left( \frac{\rho - 1}{\eta} \right)^{2}g$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.