Question

A ball of radius r and density ρ falls freely under gravity through a distance h before entering water. Velocity of ball does not change even on entering water. If viscosity of water is η, 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. $\frac{2}{9}r^{4}\left( \frac{\rho - 1}{\eta} \right)^{2}g$

Answer 1

Answer: (C)

$\frac{2}{81}r^{4}\left( \frac{\rho - 1}{\eta} \right)^{2}g$

Answer 2

Velocity of ball when it strikes the water surface

$v = \sqrt{2gh}$ …(i)

Terminal velocity of ball inside the water

$v = \frac{2}{9}r^{2}g\frac{(\rho - 1)}{\eta}$ …(ii)

Equating (i) and (ii) we get

$\sqrt{2gh} = \frac{2}{9}\frac{r^{2}g}{\eta}(\rho - 1)$

⇒ $h = \frac{2}{81}r^{4}\left( \frac{\rho - 1}{\eta} \right)^{2}g$