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Two rain drops of same radius $r$ falling with terminal velocity $V$ merge and form bigger drop with radius $R$, terminal velocity is
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Verified Answer
The correct answer is:
$\frac{V R^2}{r^2}$
Expression for terminal velocity can be obtained by balancing weight of the droplet with viscous retardation force and the buoyancy force.
Terminal velocity is given by $v=\frac{2}{9}\left(\frac{r^2(\rho-\sigma)}{\eta}\right)$
$\therefore v \propto r^2$
Now, let $V^{\prime}$ be the terminal velocity of the merged droplet
$\begin{aligned}
& \therefore \frac{V}{V^{\prime}}=\frac{r^2}{R^2} \\
& \Rightarrow V^{\prime}=\frac{V R^2}{r^2}
\end{aligned}$
Terminal velocity is given by $v=\frac{2}{9}\left(\frac{r^2(\rho-\sigma)}{\eta}\right)$
$\therefore v \propto r^2$
Now, let $V^{\prime}$ be the terminal velocity of the merged droplet
$\begin{aligned}
& \therefore \frac{V}{V^{\prime}}=\frac{r^2}{R^2} \\
& \Rightarrow V^{\prime}=\frac{V R^2}{r^2}
\end{aligned}$
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