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 (ignore buoyancy)

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

Answer 1

The correct option is A $\frac{mg}{r\eta }$


Viscous force, $F=6\pi \eta rv$

At terminal velocity, resistance due to liquid equals in magnitude the weight of the falling object.
$6\pi \eta r{v}_0=mg$

$v_0\propto \frac{mg}{\eta r}$
Final Answer: $\left(d\right)$