A spherical metal ball of radius ' $r$ ' falls through viscous liquid with velocity ' $\mathrm{V}$ '.…
A spherical metal ball of radius ' $r$ ' falls through viscous liquid with velocity ' $\mathrm{V}$ '. Another metal ball of same material but of radius $\left(\frac{\mathrm{r}}{3}\right)$ falls through same liquid, then its terminal velocity will be
$\frac{\mathrm{V}}{3}$
$\frac{\mathrm{V}}{4}$
$\frac{\mathrm{V}}{6}$
$\frac{\mathrm{V}}{9}$
Solution
$\mathrm{v}=\frac{2 \mathrm{r}^2(\rho-\sigma) \mathrm{g}}{9 \Delta}$
$\rho, \sigma$ and $\eta$ are constant
$\Rightarrow \mathrm{v} \propto \mathrm{r}^2$
The ratio of terminal velocities is
$\begin{aligned}
\frac{\mathrm{v}_1}{\mathrm{v}_2} & =\frac{\mathrm{r}_1^2}{\mathrm{r}_2^2} \\
\frac{\mathrm{v}_1}{\mathrm{v}_2} & =\frac{\mathrm{r}^2}{\left(\frac{\mathrm{r}}{3}\right)^2} \\
\frac{\mathrm{v}_1}{\mathrm{v}_2} & =9 \\
\therefore \quad \mathrm{v}_2 & =\frac{\mathrm{v}}{9}
\end{aligned}$