A small steel ball is dropped into a long cylinder containing...

Consider a small steel ball as shown below:

As shown in diagram, buoyancy force ${F}_{B} = F$.

Applying Newton's second law, we get

$mg - {F}_{B} - {F}_{v} = ma$

$⇒ \left(ρ\dfrac{4}{3}\pi{r}^{3}\right)g - \left({ρ}_{L}\dfrac{4}{3}\pi{r}^{3}\right)g - 6\piηrv = m\dfrac{dv}{dt}$

Let $\dfrac{4}{3m}\pi{R}^{3}g\left(ρ - {ρ}_{L}\right) = {K}_{1}$ and $\dfrac{6\piηr}{m} = {K}_{2}$, then we get

$⇒ \dfrac{dv}{dt} = {K}_{1} - {K}_{2}v$

Integrating both sides, we get

$⇒ {∫}_{0}^{v}\dfrac{dv}{{K}_{1} - {K}_{2}v} = {∫}_{0}^{t}dt$

$⇒ - \dfrac{1}{{K}_{2}}ln{{\left[{K}_{1} - {K}_{2}v\right]}_{0}}^{v} = t$

$⇒ \ln \left(\dfrac{{K}_{1} - {K}_{2}v}{{K}_{1}}\right) = - {K}_{2}t$

$⇒ {K}_{1} - {K}_{2}v = {K}_{1}{e}^{- {K}_{2}t}$

$⇒ v = \dfrac{{K}_{1}}{{K}_{2}}\left[1 - {e}^{- {K}_{2}t}\right]$

Clearly, option B matches with the given variation of velocity with time.