Question

A spherical oil drop falls at a steady rate of 2 cm/s in still air. Find diameter of the drop. Take g = 980 cm/s². The coefficient of viscosity of air is 1.8 x 10⁻⁴ Poise, Density of oil = 0.8 g/cm³, Density of air = I g/cm³.

Answer 1

given, terminal speed , v = 2cm/s = 0.02m/s

acceleration due to gravity, g = 9.8 m/s²

coefficient of viscosity, $\eta$ = 1.8 × 10^-4 Poise = 1.8 × 10^-5 Kg/ms [ because 1poise = 0.1 kg/ms ]

density of air , $\rho$ = 1000 kg/m³

density of oil, $\sigma$ = 800kg/m³


Let us assume that radius of spherical drop, r

use formula, r² = $\frac{9\eta v}{2g(\rho-\sigma)}$

r² = 9 × 1.8 × 10^-5 × 0.02/{2 × 9.8 × (1000-800)}

r² = 18 × 1.8 × 10^-7/{19.6 × 200}

r² = 18 × 18 × 10^-8/196 × 20

r = 1/√20 × 18 × 10^-4/14

r = 28.74 × 10^-6 m

hence, diameter of spherical drop = 57.48 × 10^-6m