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 = `2/9 (r^2(ρ-σ)g)/v`

r = `sqrt((9ηv)/(2(ρ-σ)g)`

r = `sqrt((9xx1.8xx10^-4xx9.8)/(2xx(0.9013-0.0013)xx980)`

r = `sqrt((1.5846xx10^-3)/(2xx0.9xx980))`

r = `sqrt((1.5846xx10^-3)/1764)`

r = `sqrt(8.98xx 10^-7)`

r = 0.000948