A lead sphere of mass m falls in a viscous liquid with terminal velocity $v_0.$ Another lead sphere of mass M falls through the same viscous liquid with terminal velocity $4v_0.$The ratio $\frac{M}{m}$ is

$v_c=\frac{2}{9}{r}^2\frac{\left(\rho -{\rho }^{\mathrm{\prime }}\right)g}{\eta }$

Since two different lead balls of masses m and M are being compared, their radii must be different.

$\text{ Let }m=\rho \frac{4}{3}\pi r_1^3$

$M=\rho \cdot \frac{4}{3}\pi r_2^3\Rightarrow \frac{m}{M}={\left(\frac{r_1}{r_2}\right)}^3$

$\text{ Now }\frac{v_{c1}}{v_{c2}}=\frac{r_1^2}{r_2^2}\text{ or }\frac{v_0}{4v_0}={\left(\frac{r_1}{r_2}\right)}^2\Rightarrow \frac{r_1}{r_2}=\frac{1}{2}$

$\text{ Now }\frac{M}{m}={\left(\frac{r_2}{r_1}\right)}^3\Rightarrow \frac{M}{m}=(2{)}^3=8\text{. }$