Question

A spherical drop of oil falls at a constant speed of 9.8 cm/s in steady air. Calculate the radius of the drop. The density of oil is 0.9013 g/cm³, density of air is 0.0013 g/cm³ and the coefficient of viscosity of air is 1.8 × 10^-4 poise.

Answer 1

Given,

v = 9.8 cm/s

η = 1.8 × 10^-4 poise

ρ = 0.9013 g/cm³

σ = 0.0013 g/cm³

We have,

η = ${\displaystyle \frac{2}{9}\frac{r^2(\rho -\sigma )g}{v}}$

r = ${\displaystyle \sqrt{\frac{9\eta v}{2(\rho -\sigma )g}}}$

r = ${\displaystyle \sqrt{\frac{9\times 1.8\times {10}^{-4}\times 9.8}{2\times (0.9013-0.0013)\times 980}}}$

r = ${\displaystyle \sqrt{\frac{1.5846\times {10}^{-3}}{2\times 0.9\times 980}}}$

r = ${\displaystyle \sqrt{\frac{1.5846\times {10}^{-3}}{1764}}}$

r = ${\displaystyle \sqrt{8.98\times {10}^{-7}}}$

r = 0.000948