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)

Options

A. `(mg(eta))/(r )`
B. `mg(eta)r`
C. `(mgr)/(eta)`
D. `(mg)/(reta)`

Answer

D

Explanation

Since the steedl ball is given to be small therefore Now, `6pietarv_0=mg` or `v_0pro(mg)/(etar)`