Question

A drop of water of radius 0.0015 mm is falling in air. If the coefficient of viscosity of air is $1.8\times {10}^{-5}kg/ms$, what will be the terminal velocity of the drop? (density of water = $1.0\times {10}^{3}kg/m^2$ and g = 9.8 N/kg). The density of air can be neglected

• A. $2.72\times {10}^{-4}m/s$
• B. $3.72\times {10}^{-4}m/s$
• C. $0.72\times {10}^{-4}m/s$
• D. $12.72\times {10}^{-4}m/s$

Answer 1

The correct option is A $2.72\times {10}^{-4}m/s$

By Stoke's law the terminal velocity of a water drop of radius r is given by
$v=\frac{2}{9}\frac{r^2(\rho -\sigma )g}{\eta }$
where $\rho$ is the density of water $\sigma$ is the density of air and $\eta$ the coefficient of viscosity of air Here $\sigma$ is negligible and r = 0.0015 mm = 1.5 x ${10}^{-3}$ = 1.5 x ${10}^{-6}$ m Substituting the values
$v=\frac{2}{9}\times \frac{(1.5\times {10}^{-6}{)}^2\times (1.0\times {10}^3)\times 9.8}{1.8\times {10}^{-5}}=2.72\times {10}^{-4}m/s$