Physics - Fluid Question with Solution | TestHub

Let $r$ be the radius of a small drop and $R$ be the radius of the big drop.

Since volume remains constant when $125$ drops coalesce:

$\frac{4}{3}\pi R^3=125\times \frac{4}{3}\pi r^3$

$R^3=125r^3\Rightarrow R=5r$

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

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

Let $V$ be the terminal velocity of the big drop.

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

$\frac{V}{4}=(5{)}^2=25$

$V=100\text{ cm/s}=1\text{ m/s}$

Answer: $1\mathrm{m}/\mathrm{s}$