Question

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 water. If viscosity of water is $\eta$ the value of $h$ is given by

• A. $\frac{2}{9}{\left(\text{r}\right)}^2(\frac{1-\rho }{\eta })\text{g}$
• B. $\frac{2}{81}{\left(\text{r}\right)}^2(\frac{\rho -1}{\eta })\text{g}$
• 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)

Velocity of ball when the water surface
$v=\sqrt{2gh}\dots \dots \dots \dots$ (i)
Terminal velocity of ball inside the water
$v=\frac{2}{9}(r{)}^{2}g\frac{(\rho -1)}{\eta }\dots \dots \dots \dots$ (ii)
Equating (i) and (ii)we get
$\sqrt{2gh}=\frac{2}{9}\frac{r^{2}g}{\eta }(\rho -1)$
$\Rightarrow h=\frac{2}{81}{r}^4{\left(\frac{\rho -1}{\eta }\right)}^{2}g$