Question

Two equal drops of water are falling through air with a steady velocity $v$. If the drops coalesce, the new velocity will be

• A. $2v$
• B. $\sqrt{2}v$
• C. $2^{2/3}v$
• D. $\frac{v}{\sqrt{2}}$

Answer 1

The correct option is C $2^{2/3}v$

Let $r$ be the radius of smaller drop and $R$ the radius of bigger drop. By equating the volume, we have,
$2\left(\frac{4}{3}\pi r^3\right)=\frac{4}{3}\pi R^3$
or $R=2^{1/3}.r$

We know that,
Terminal velocity, $v=\frac{2}{9}{r}^{2}g\left(\frac{\rho -\sigma }{\eta }\right)$

Since both the drops are of same density and both are moving in same fluid,

$\Rightarrow v\propto r^2$

$\Rightarrow v=k{r}^2\left[k\text{is constant}\right]$

$\therefore$ New velocity,

$v^{\prime }=k{R}^2=2^{2/3}v$