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 __________.

Solution :

Terminal velocity of a spherical drop is given by :

$$v \propto r^2$$

Initial drop radius :

$$R$$

Initial terminal velocity :

$$v_1 \propto R^2$$

When the drop breaks into 64 identical droplets :

Volume is conserved.

Therefore,

$$\frac{4}{3}\pi R^3 = 64 \times \frac{4}{3}\pi r^3$$

$$R^3 = 64r^3$$

$$R = 4r$$

$$r = \frac{R}{4}$$

Terminal velocity of each small droplet :

$$v_2 \propto r^2$$

$$\propto \left(\frac{R}{4}\right)^2$$

$$= \frac{R^2}{16}$$

Therefore,

$$\frac{v_1}{v_2}=\frac{R^2}{R^2/16}$$

$$= 16$$

Final Answer :

$$16$$