Question

The terminal velocity of a liquid drop of radius $r$ falling through air is $v$. If two such drops are combined to form a bigger drop, the terminal velocity with which the bigger drop falls through air is (Ignore any buoyant force due to air) :

• A. $\sqrt{2}v$
• B. $2v$
• C. $\sqrt[3]{34}v$
• D. $\sqrt[3]{32}v$

Answer 1

The correct option is C $\sqrt[3]{34}v$

On merging the two drops, the volume will become 2 times and since $V=\frac{4}{3}\pi r^3$, radius will become $2^{\frac{1}{3}}$ times.

The expression for terminal velocity as derived from Stoke's Law is

$v_t=\frac{2}{9}\frac{({\rho }_p-{\rho }_f)}{\mu }g{r}^2$

Since this is proportional to $r^2$, new terminal velocity will be $2^{\frac{2}{3}}=\sqrt[3]{34}$ times $v$