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. $\frac{mg\eta }{r}$
• B. $mg\eta r$
• C. $\frac{mgr}{\eta }$
• D. $\frac{mg}{r\eta }$

Answer 1

Answer: (D)

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