Question
A spherical liquid drop of radius R acquires the terminal velocity $v_{1}$ when falls through a gas of viscosity $\eta$. Now the drop is broken into 64 identical droplets and each droplet acquires terminal velocity $v_{2}$ falling through the same gas. The ratio of terminal velocities $v_{1}/v_{2}$ is $\_\_\_\_$ .
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Solution
Sol. [Key Concept: Terminal velocity $\propto \ R^{2}$; volume conservation splits the drop into $64$ so $R_{2} = \frac{R_{1}}{4}$, giving $\frac{v_{1}}{v_{2}} = (\frac{R_{1}}{R_{2}})^{2} = 16$.]
Using volume conservation, $\frac{4}{3}\pi R_{1}^{3} = 64 \times \frac{4}{3}\pi R_{2}^{3} \Rightarrow R_{2} = \frac{R_{1}}{4}$ Terminal velocity $v \propto R^{2}$
$$\frac{v_{1}}{v_{2}} \Rightarrow \frac{R_{1}^{2}}{R_{2}^{2}} \Rightarrow (4)^{2} = 16 $$
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