Physics - Fluid Question with Solution | TestHub

Let $r$ be the radius of each small drop and $v$ be its terminal velocity.

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

Thus, $v\propto r^2$.

When two such drops coalesce to form a single big drop of radius $R$, the total volume remains conserved.

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

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

Let $V$ be the terminal velocity of the big drop. Since $V\propto R^2$, we have:

$\frac{V}{v}={\left(\frac{R}{r}\right)}^2={\left(2^{1/3}\right)}^2=2^{2/3}=4^{1/3}$

$V=v\times 4^{1/3}$

Given $v=5\text{ cm/s}$, we get:

$V=5\times (4{)}^{1/3}\text{ cm/s}$

Answer: $5\times (4{)}^{1/3}\mathrm{cm}/\mathrm{s}$