Question

Two equal drops are falling through air with a steady velocity of $5\, cm /$ second. If two drops coalesce, then new terminal velocity will be

• A. $5 \times(4)^{1 / 3} cm / s$
• B. $5 \sqrt{2} cm / s$
• C. $\frac{5}{\sqrt{2}} cm / s$
• D. $5 \times 2\, cm / s$

Answer 1

Answer: (A)

$V_{\text {Terminal }} \propto r^{2}$
If initial radius $=r$, let new radius $=R$
Then $2 \times \frac{4}{3} \pi r^{3}=\frac{4}{3} \pi R^{3}$
$\Rightarrow (2)^{1 / 3} r=R$
$\Rightarrow V_{T} \propto R^{2}$
$\propto(2)^{2 / 3} r^{2}$ (For bigger drops)
$\frac{V_{T} \text { smaller drop }}{V_{T} \text { bigger drop }}=\frac{r^{2}}{(2)^{2 / 3} r^{2}}$
$\Rightarrow \frac{5}{x}=\frac{1}{(2)^{2 / 3}}$
$\Rightarrow 5 \times(2)^{2 / 3}=x$
$\Rightarrow 5 \times(4)^{1 / 3} cm / s =x$