The terminal velocity $$v_t$$ of the spherical rain drop depends on the radius $$r$$ of the spherical rain drop as

We need to determine how the terminal velocity $$v_t$$ of a spherical raindrop depends on its radius $$r$$.

When a spherical raindrop falls through air at terminal velocity, three forces act on it: the weight (downward) $$W = \frac{4}{3}\pi r^3 \rho g$$, where $$\rho$$ is the density of water; the buoyant force (upward) $$F_b = \frac{4}{3}\pi r^3 \sigma g$$, where $$\sigma$$ is the density of air; and the viscous drag (upward) given by Stokes’ law as $$F_d = 6\pi \eta r v_t$$, where $$\eta$$ is the viscosity of air.

At terminal velocity, the net force is zero, so $$W = F_b + F_d$$ which implies $$\frac{4}{3}\pi r^3 \rho g = \frac{4}{3}\pi r^3 \sigma g + 6\pi \eta r v_t$$.

Rearranging to solve for the terminal velocity yields $$6\pi \eta r v_t = \frac{4}{3}\pi r^3 (\rho - \sigma) g$$ and hence $$v_t = \frac{2r^2(\rho - \sigma)g}{9\eta}$$.

Since $$\rho$$, $$\sigma$$, $$g$$, and $$\eta$$ are constants for a given system, it follows that $$v_t \propto r^2$$. The correct answer is Option B: $$r^2$$.