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A ball of mass $m$ and radius $ r $ is gently released in a viscous liquid. The mass of the liquid displaced by it is $m' $ such that $m > m'$. The terminal velocity is proportional to
$\frac{{m - m'}}{r}$
$\frac{{m + m'}}{r}$
$\frac{{(m + m')}}{{{r^2}}}$
$(m - m') r^2$
Solution
The mass of ball is m and radius 8 .
The mass of the liquid displaced by it s The mass
$V_{t}=\frac{2}{9} \gamma^{2} \frac{(\text { foon }-\text { Peiquid })}{n} \times g$
$V_{+} \alpha \gamma^{2}\left(\rho_{\text {ball }}-\right.$ Seiquid $)$
$\rho_{\text {ball }}=\frac{m}{\frac{4 \pi r^{3}}{3}}=\frac{3 m}{4 \pi \gamma^{3}}$
$f_{\text {etquid }}=\rho_{l}$
$v \times f_{1} \times g=m g$
$\rho_{l}=\frac{m}{v}=\frac{3 m}{4 \pi r^{3}}$
Now $v_{t} \times \gamma^{2}\left(\frac{3 m}{4 \pi r^{3}}-\frac{3 m^{\prime}}{4 \pi r^{3}}\right)$
$v_{t} \times \frac{3}{4 \pi}\left(\frac{m}{\gamma}-\frac{-m^{\prime}}{\gamma}\right)$
$v_{t} \propto \frac{m-m^{\prime}}{\gamma}$
