State stokes’ law. By using it deduce the expression for :

$(i)$ initial acceleration of smooth sphere and

$(ii)$ equation of terminal velocity of sphere falling freely through the viscous medium.

$(iii)$ Explain : Upward motion of bubbles produced in fluid.

Scientist Stokes' said that, viscous force $\mathrm{F}_{(\mathrm{V})}$ on small spherical solid body of radius $r$ and moving with velocity $v$ through a viscous medium of large dimensions having coefficient of viscosity $\eta$ is $6 \pi \eta r v$.

As shown in figure a small spherical body of radius $r$, density $\rho$ falling in viscous medium of density $\sigma .$

Following forces acted on it.

$(i)$ Weight $\mathrm{F}_{1}=m g=$ (volume $\times$ density) $g$

$=\left(\frac{4}{3} \pi r^{3} \rho\right) g$

$\ldots$ $(1)$ (In downward)

Where $m=$ mass of sphere

$(ii)$ Buoyant force $\mathrm{F}_{2}=m_{\mathrm{o}} g$

$=\left(\frac{4}{3} \pi r^{3} \sigma\right) g$

$(2)$ (in upward)

where $m_{0}=$ mass of liquid having volume of sphere

$(iii)$ According to stokes' law

$\mathrm{F}_{(v)}=6 \pi \eta r v$

Resultant force acting on the sphere,

$\mathrm{F}_{\mathrm{R}}=\mathrm{F}_{1}-\mathrm{F}_{2}-\mathrm{F}_{(v)}$

$\mathrm{F}_{\mathrm{R}}=\left(\frac{4}{3} \pi r^{3} \rho\right) g-\left(\frac{4}{3} \pi r^{3} \sigma\right) g-6 \pi \eta r v$

$\mathrm{~F}_{\mathrm{R}}=\frac{4}{3} \pi r^{3} g(\rho-\sigma)-6 \pi \eta r v$