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