Two spherical raindrops of masses m and 8m fall through the air....

Let the radii of the two raindrops be $r_1$ and $r_2$.

Since the density $\rho$ is the same for both raindrops, mass $m = \dfrac{4}{3}\pi r^3 \rho$.

Therefore, $m \propto r^3$.

Given the ratio of masses is $m : 8m = 1 : 8$, the ratio of their radii is:

$\dfrac{r_1}{r_2} = \left( \dfrac{1}{8} \right)^{\dfrac{1}{3}} = \dfrac{1}{2}$

The terminal velocity $v$ of a spherical drop falling through a viscous medium is given by Stokes' law:

$v = \dfrac{2r^2(\rho - \sigma)g}{9\eta}$

Thus, terminal velocity $v \propto r^2$.

The ratio of their terminal velocities is:

$\dfrac{v_1}{v_2} = \left( \dfrac{r_1}{r_2} \right)^2 = \left( \dfrac{1}{2} \right)^2 = \dfrac{1}{4}$

Hence, the ratio is $1 : 4$.

Answer: $1 : 4$