Two equal drops are falling through air with a steady velocity of 5 ,cm / second . If two drops coal

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9-1.Fluid Mechanics

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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 ......... $cm / s$

A

$5 \times(4)^{13}$

B

$5 \sqrt{2}$

C

$\frac{5}{\sqrt{2}}$

D

$5 \times 2$

Solution

(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)

$\frac{V_{T \text { smaller drop }}}{V_{T \text { bigger drop }}}=\frac{r^2}{(2)^{2 / 3} r^2}$

$\Rightarrow \frac{5}{x}=\frac{1}{(2)^{2 / 3}}$

$\Rightarrow 5 \times(2)^{2 / 3}=x$

$\Rightarrow 5 \times(4)^{1 / 3} \,cm / s =x$

Standard 11

Physics

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