State Stokes' law. By using it,deduce the expression for:
$(i)$ Initial acceleration of a smooth sphere.
$(ii)$ Equation of terminal velocity of a sphere falling freely through a viscous medium.
$(iii)$ Explain: Upward motion of bubbles produced in a fluid.

(N/A) Stokes' Law states that the viscous drag force $F_v$ acting on a small spherical body of radius $r$ moving with velocity $v$ through a viscous medium of infinite extent with coefficient of viscosity $\eta$ is given by $F_v = 6 \pi \eta r v$.
Consider a small spherical body of radius $r$ and density $\rho$ falling in a viscous medium of density $\sigma$. The forces acting on it are:
$(i)$ Weight $F_1 = mg = \frac{4}{3} \pi r^3 \rho g$ (downward).
$(ii)$ Buoyant force $F_2 = \frac{4}{3} \pi r^3 \sigma g$ (upward).
$(iii)$ Viscous force $F_v = 6 \pi \eta r v$ (upward).
$(i)$ Initial acceleration: At $t=0$,$v=0$,so $F_v=0$. The net force $F_{net} = F_1 - F_2 = \frac{4}{3} \pi r^3 g(\rho - \sigma)$. Thus,acceleration $a = \frac{F_{net}}{m} = \frac{\frac{4}{3} \pi r^3 g(\rho - \sigma)}{\frac{4}{3} \pi r^3 \rho} = g(1 - \frac{\sigma}{\rho})$.
$(ii)$ Terminal velocity: At terminal velocity $v_t$,acceleration is zero,so $F_1 = F_2 + F_v$. Thus,$\frac{4}{3} \pi r^3 \rho g = \frac{4}{3} \pi r^3 \sigma g + 6 \pi \eta r v_t$. Solving for $v_t$,we get $v_t = \frac{2r^2 g(\rho - \sigma)}{9 \eta}$.
$(iii)$ Upward motion of bubbles: For a gas bubble,the density of gas $\rho$ is much less than the density of the liquid $\sigma$ $(\rho < \sigma)$. The buoyant force exceeds the weight,causing the bubble to accelerate upward. As it moves,the viscous force acts downward,eventually leading to a constant terminal velocity.