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Question

$\mathrm{mg}-\mathrm{F}_{\mathrm{B}}-\mathrm{F}_{\mathrm{v}}=\mathrm{ma}$

$\left(\rho \frac{4}{3} \pi \mathrm{r}^3\right) \mathrm{g}-\left(\rho_{\m

Vedclass · Accepted Answer

Vedclass · Answer

$\mathrm{mg}-\mathrm{F}_{\mathrm{B}}-\mathrm{F}_{\mathrm{v}}=\mathrm{ma}$

$\left(ho \frac{4}{3} \pi \mathrm{r}^3ight) \mathrm{g}-\left(ho_{\mathrm{L}} \frac{4}{3} \pi \mathrm{r}^3ight) \mathrm{g}-6 \pi \eta \mathrm{r}=\mathrm{m} \frac{\mathrm{dv}}{\mathrm{dt}}$

$	ext { Let } \frac{4}{3 \mathrm{~m}} \pi \mathrm{R}^3 \mathrm{~g}\left(ho-ho_{\mathrm{L}}ight)=\mathrm{K}_1 	ext { and } \frac{6 \pi \eta \mathrm{r}}{\mathrm{m}}=\mathrm{K}_2$

$\frac{\mathrm{dv}}{\mathrm{dt}}=\mathrm{K}_1-\mathrm{K}_2 \mathrm{v}$

$\int_0^{\mathrm{v}} \frac{\mathrm{dv}}{\mathrm{K}_1-\mathrm{K}_2 \mathrm{v}}=\int_0^{\mathrm{t}} \mathrm{dt}$

$-\frac{1}{\mathrm{~K}_2} \ln \left[\mathrm{K}_1-\mathrm{K}_2 \mathrm{v}ight]_0^{\mathrm{v}}=\mathrm{t}$

$\ln \left(\frac{\mathrm{K}_1-\mathrm{K}_2 \mathrm{v}}{\mathrm{K}_1}ight)=-\mathrm{K}_2 \mathrm{t}$

$\mathrm{K}_1-\mathrm{K}_2 \mathrm{v}=\mathrm{K}_1 \mathrm{e}^{-\mathrm{K}_2 \mathrm{t}} $

$\mathrm{v}=\frac{\mathrm{K}_1}{\mathrm{~K}_2}\left[1-\mathrm{e}^{-\mathrm{K}_2 \mathrm{t}}ight]$

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