Question 87 of 308

A small steel ball is dropped into a long cylinder containing glycerine. Which one of the following is the correct representation of the velocity time graph for the transit of the ball?

ABCD

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Answer: (B)

Consider a small steel ball as shown below:

As shown in diagram, buoyancy force ${F}_{B}=F$.
Applying Newton's second law, we get
$mg-{F}_{B}-{F}_{v}=ma$
$\Rightarrow \left(\rho \dfrac{4}{3}\pi {r}^{3}\right)g-\left({\rho }_{L}\dfrac{4}{3}\pi {r}^{3}\right)g-6\pi \eta rv=m\dfrac{dv}{dt}$
Let $\dfrac{4}{3m}\pi {R}^{3}g\left(\rho -{\rho }_{L}\right)={K}_{1}$ and $\dfrac{6\pi \eta r}{m}={K}_{2}$, then we get
$\Rightarrow \dfrac{dv}{dt}={K}_{1}-{K}_{2}v$
Integrating both sides, we get
$\Rightarrow {\int }_{0}^{v}\dfrac{dv}{{K}_{1}-{K}_{2}v}={\int }_{0}^{t}dt$
$\Rightarrow -\dfrac{1}{{K}_{2}}\ln {{\left[{K}_{1}-{K}_{2}v\right]}_{0}}^{v}=t$
$\Rightarrow \ln \left(\dfrac{{K}_{1}-{K}_{2}v}{{K}_{1}}\right)=-{K}_{2}t$
$\Rightarrow {K}_{1}-{K}_{2}v={K}_{1}{e}^{-{K}_{2}t}$
$\Rightarrow v=\dfrac{{K}_{1}}{{K}_{2}}\left[1-{e}^{-{K}_{2}t}\right]$
Clearly, option B matches with the given variation of velocity with time.