Question

A spherical drop of oil falls at a constant speed of 6 cm/s in steady air. Calculate the radius of the drop. The density of the oil is 0.9 g/cm', density of air is 1.0 g/cm' and the coefficient of viscosity of air is 1.8 × 10“ poise. (g = 980 cm/s³)​

Answer 1

Given :-

• Terminal velocity of oil = v = 6 cms-¹ = 0.06 ms-¹

• Acceleration due to gravity = g = 980 cms-² = 9.8 ms-²

• Density of air = ρ = 1000 Kgm-³

• Density of oil = σ = 800 Kgm-³

• Viscosity of air = η = 1.8 × 10-⁴ Poise

• η = 1.8 × 10^(-5) Kg/ms

Now, As per formula,

r² = 9vη/2g(ρ - σ)

r² = [9 × 0.06 × 1.8 × 10^(-5)]/2 × 9.8(1000 - 800)

r² = [9 × 6 × 10-² × 18 × 10-¹ × 10^(-5)]/(19.6 × 200)

r² = [972 × 10^(-8)× 10-²]/(19.6 × 2)

r² = [972 × 10^(-10)]/39.2

r² = [24.60 × 10^-(10)]

r = √[24.6 × 10^(-10)]

r = 4.9 × 10^(-5) m

Hence,

The radius of oil drop = r = 4.9 × 10^(-5) m.