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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)
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The correct answer is:
$\sqrt[3]{4} v$
Terminal velocity $v=\frac{2}{9} \frac{r^2(\rho-\sigma) g}{\eta}$
When, the two drops of same radius $r$ coalesce then radius of new drop is $R$.
$$
\begin{array}{llrl}
& \therefore & \frac{4}{3} \pi R^3 & =\frac{4}{3} \pi r^3+\frac{4}{3} \pi r^3 \\
& \Rightarrow & R & =2^{1 / 3} \cdot r \\
\text { Critical velocity } & \propto r^2 \\
& \therefore & \frac{v}{v_1} & =\frac{r^2}{2^{2 / 3} \cdot r^2} \\
& \Rightarrow & V_1 & =\sqrt[3]{4} \cdot v
\end{array}
$$
When, the two drops of same radius $r$ coalesce then radius of new drop is $R$.
$$
\begin{array}{llrl}
& \therefore & \frac{4}{3} \pi R^3 & =\frac{4}{3} \pi r^3+\frac{4}{3} \pi r^3 \\
& \Rightarrow & R & =2^{1 / 3} \cdot r \\
\text { Critical velocity } & \propto r^2 \\
& \therefore & \frac{v}{v_1} & =\frac{r^2}{2^{2 / 3} \cdot r^2} \\
& \Rightarrow & V_1 & =\sqrt[3]{4} \cdot v
\end{array}
$$
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