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 ____ .

Question

Accepted Answer

D. 1. A falling sphere reaches terminal velocity when the viscous drag balances the net downward force from gravity and buoyancy, so Stokes' Law and the force balance apply.
$$F_g = \frac{4}{3}\pi r^3 ho g, \quad F_b = \frac{4}{3}\pi r^3 \sigma g, \quad F_v = 6\pi \eta r v_t$$

2. Setting the viscous drag equal to the net weight at terminal velocity gives the terminal velocity formula.
$$6\pi \eta r v_t = \frac{4}{3}\pi r^3(ho - \sigma)g$$
So,
$$v_t = \frac{2r^2(ho - \sigma)g}{9\eta}$$

3. Since $ho$, $\sigma$, $g$, and $\eta$ are the same for the big drop and the droplets, the terminal velocity depends only on the square of the radius.
$$v \propto r^2$$

4. Because the total volume is conserved when the drop breaks into 64 droplets, we can relate the radius of a droplet to $R$.
$$\frac{4}{3}\pi R^3 = 64 	imes \frac{4}{3}\pi r^3$$
Therefore,
$$R = 4r \Rightarrow r = \frac{R}{4}$$

5. Using $v \propto r^2$ for the big drop ($v_1$, radius $R$) and small droplet ($v_2$, radius $r$):
$$\frac{v_1}{v_2} = \frac{R^2}{r^2} = \frac{R^2}{(R/4)^2} = 16$$

Hence, the answer is D.