Castor oil at $20^{\circ}\;C$ has a coefficient of viscosity $2.42\;N\,s\, /m^2$ and density $940\; kg/m^3$. Calculate the terminal velocity of steel ball of radius $2\;mm$ falling under gravity in the oil, taking density of steel as $7800\;kg/m^3$.

Given,
Here coefficient of viscosity of castor oil $(\eta) = 2.42\; Ns/m^2$
Density of the oil $(\sigma) = 940\;kg/m^3$
Radius of the steel ball $(r) = 2\;mm = 0.002\;m$
Density of the steel $(\rho) = 7800\;kg/m^3$
Terminal velocity $(v) = ?$

When the steel ball falls downward with terminal velocity $(v)$, then
Total upward force = Total downward force
Viscous force $(F)$ + Upthrust $(U)$ = Weight of the steel ball $(W)$

$F = W  -  U$

$6\; \pi \; \eta \; r \; v = \frac{4}{3} \pi \; r^3 \; \rho \; g - \frac{4}{3} \pi r^3 \sigma g$

By Solving, we get

$v = \frac{2\;r^2\;(\rho - \sigma)\;g}{9\; \eta} = \frac{2 \, (0.002)^2 \; * \; (7800 \; - \; 940)\;*\;10}{9\; * \; 2.42}$ $= 0.025\;m/s$

$\therefore$ Hence, the required terminal velocity of the steel ball is $0.025\;m/s$

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