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 $R=(2{)}^{1/3}.r$
Now, terminal velocity $v\propto (\text{radius}{)}^2$
$\therefore$ New velocity, $v^{\prime }=(2{)}^{2/3}v$