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.
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,
$\mathrm{F}_{\mathrm{R}}=\mathrm{F}_{1}-\mathrm{F}_{2}-\mathrm{F}_{(v)}$
$\mathrm{F}_{\mathrm{R}}=\left(\frac{4}{3} \pi r^{3} \rho\right) g-\left(\frac{4}{3} \pi r^{3} \sigma\right) g-6 \pi \eta r v$
$\mathrm{~F}_{\mathrm{R}}=\frac{4}{3} \pi r^{3} g(\rho-\sigma)-6 \pi \eta r v$
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