Question

A drop of water of radius \(0.0015~mm\) is falling in air. If the coefficient of viscosity of air is \(1.8\times10^{−5}~kg/ms\), what will be the terminal velocity of the drop? \((\text{Density of water}=1.0\times10^{3}~kg/m^{2}\) and \(g=9.8~m/s^{−2})\) Density of air can be neglected.

Answer 1

We know that \(v_{T}=\dfrac{2}{9} \dfrac{𝑟^{2} (\sigma−\rho)g}{\eta}\)
Given,
\(r=0.0015~mm=1.5\times10^{−3}~mm=1.5\times10^{−6}~m\)
\(\eta=1.8\times10^{−5}~kg/ms\)
Density of water\( =1.0\times10^{3}~kg/m^{2}\)
\( g=9.8~m/s^{−2}\)
By Stokes' law, the terminal velocity of a water drops of radius \(r\) is given by
\(v=\dfrac{2}{9} \dfrac{r^{2} (\rho−\sigma)g}{\eta}\)
Substituting the values:
\(v=\dfrac{2}{9}\)
\(\times \dfrac{(1.5\times10^{−6} )^{2}\times(1.0\times10^{3} )\times9.8}{1.8\times10^{−5} }\)
\(=2.72\times10^{−4}~m/s\)
Final Answer: \((a)\)