Question

Two equal drops are falling through air with a steady velocity of $5cm/$ 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. $5\times 2cm/s$

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)
$\frac{V_T\text{ smaller drop }}{V_T\text{ bigger drop }}=\frac{r^2}{(2{)}^{2/3}{r}^2}$
$\Rightarrow \frac{5}{x}=\frac{1}{(2{)}^{2/3}}$
$\Rightarrow 5\times (2{)}^{2/3}=x$
$\Rightarrow 5\times (4{)}^{1/3}cm/s=x$