Physics - Fluid Question with Solution | TestHub

The terminal velocity of a spherical drop of radius $r$ falling through a viscous medium is given by $V=\frac{2}{9}\frac{r^2(\rho -\sigma )g}{\eta }$.

Thus, the terminal velocity is directly proportional to the square of the radius, $V\propto r^2$.

When two identical drops of radius $r$ combine to form a single drop of radius $R$, the total volume remains conserved.

$\frac{4}{3}\pi R^3=2\times \frac{4}{3}\pi r^3$

$R^3=2r^3\Rightarrow R=2^{1/3}r$

The new terminal velocity $V^{\prime }$ of the combined drop is proportional to $R^2$.

$\frac{V^{\prime }}{V}={\left(\frac{R}{r}\right)}^2$

$\frac{V^{\prime }}{V}={\left(2^{1/3}\right)}^2=2^{2/3}$

$V^{\prime }=2^{2/3}V$

Answer: $(2{)}^{2/3}\mathrm{V}$