Question

A small steel ball of mass $m$ and radius $r$ is falling under gravity through a viscous liquid of coefficient of viscosity $\eta$. If $g$ is the value of acceleration due to gravity, then the terminal velocity of the ball is proportional to-

• A. $\frac{m g \eta}{r}$
• B. $m g \eta r$
• C. $\frac{m g r}{\eta}$
• D. $\frac{m g}{r \eta}$

Answer 1

Answer: (D)

As steel ball is small, we can neglect the buoyant force on it.
At critical velocity
$6 \pi \eta r v_{C}=m g \Rightarrow v_{C}=\frac{1}{6 \pi}\left(\frac{m g}{\eta r}\right)$
Thus, $v_{C} \propto \frac{m g}{\eta \cdot r}$