Physics - Fluid Question with Solution | TestHub

The terminal velocity of a spherical drop of radius $r$ falling through a viscous medium is given by Stokes' law:

$V=\frac{2}{9}\frac{r^2(\rho -\sigma )g}{\eta }$

From this relation, the terminal velocity is directly proportional to the square of the radius of the drop:

$V\propto r^2$

For the smaller drop of radius $r$, the terminal velocity is $V$.

For the bigger drop of radius $R$, let the terminal velocity be $V^{\prime }$.

Taking the ratio of their terminal velocities:

$\frac{V^{\prime }}{V}=\frac{R^2}{r^2}$

$V^{\prime }=\frac{V{R}^2}{r^2}$

Answer: $\frac{V{R}^2}{r^2}$