Physics - Fluid Question with Solution | TestHub

Let the radius of a single tiny droplet be $r$ and its terminal velocity be $v$.

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

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

This implies $v\propto r^2$.

When $27$ such identical droplets coalesce to form a single larger drop of radius $R$, the total volume remains conserved:

$\frac{4}{3}\pi R^3=27\times \frac{4}{3}\pi r^3$

$R^3=27r^3$

$R=3r$

The new terminal velocity $v^{\prime }$ of the larger drop will be:

$v^{\prime }\propto R^2$

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

$\frac{v^{\prime }}{1}=(3{)}^2$

$v^{\prime }=9$ cm/s

Answer: $9$ cm/s