Sol. $$ \begin{aligned} & \mathrm{mg}-\mathrm{F}_{\mathrm{B}}-\mathrm{F}_{\mathrm{v}}=\mathrm{ma} \\ & \left(\rho \frac{4}{3} \pi \mathrm{r}^3\right) g-\left(\rho_{\mathrm{L}} \frac{4}{3} \pi \mathrm{r}^3\right) \mathrm{g}-6 \pi \eta \mathrm{rv}=\mathrm{m} \frac{\mathrm{dv}}{\mathrm{dt}} \\ & \text { Let } \frac{4}{3 \mathrm{~m}} \pi \mathrm{R}^3 \mathrm{~g}\left(\rho-\rho_{\mathrm{L}}\right)=\mathrm{K}_1 \text { and } \frac{6 \pi \eta \mathrm{r}}{\mathrm{~m}}=\mathrm{K}_2 \\ & \frac{\mathrm{dv}}{\mathrm{dt}}=\mathrm{K}_1-\mathrm{K}_2 \mathrm{~V} \\ & \int_0^v \frac{d v}{K_1-K_2 v}=\int_0^{\mathrm{t}} d t \\ & -\frac{1}{\mathrm{~K}_2} \ln \left[\mathrm{~K}_1-\mathrm{K}_2 \mathrm{v}\right]_0^{\mathrm{v}}=\mathrm{t} \end{aligned} $$ $\begin{aligned} & \ell\left(\frac{K_1-K_2 V}{K_1}\right)=-K_2 t \\ & K_1-K_2 v=K_1 e^{-K_2 t} \\ & v=\frac{K_1}{K_2}\left[1-e^{-K_2 t}\right]\end{aligned}$