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, it is clear that the terminal velocity is directly proportional to the square of the radius of the drop.

$V\propto r^2$

For the small drop of radius $r$, the terminal velocity is $V\propto r^2$.

For the big drop of radius $R$, let the terminal velocity be $V^{\prime }$. Then, $V^{\prime }\propto R^2$.

Taking the ratio of the two terminal velocities, we get:

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

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

Answer: $\frac{{\mathrm{VR}}^2}{{\mathrm{r}}^2}$