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

Data: ρ = 0.9013 g/cm³, ρ_air = 0.0013 g/cm³, η = 1.8 × 10⁻⁴ P, v_t = 9.8 cm/s, g = 980 cm/s²

`v_t = (2r^2g(ρ-ρ_(air)))/(9η)`

∴ `r^2 = (9v_tη)/(2g(ρ-ρ_(air)))`

= `(9(9.8  cm//s)(1.8 xx 10^-4  P))/(2(980  cm//s^2)[(0.9013 - 0.0013  g//cm^3)])`

= `(0.9 xx 10^-4)/10`

= 9 × 10⁻⁶ cm²

∴ The radius of the oil drop, r = 3 × 10⁻³ cm.