Q.With what terminal velocity will an air bubble 0.4 mm in diameter rise in a liquid of viscosity 0.1 Ns/m² and specific gravity 0.9? Density of air is 1.29 kg/m³. [Ans. -0.782×10⁻³ m/s, The negative sign indicates that the bubble rises up]

Step 1. Diameter 0.4 mm gives $r = 0.2$ mm $= 0.2×10^{-3}$ m; the bubble is air ($\rho = 1.29$ kg/m³), rising through a liquid of specific gravity 0.9, i.e. $\sigma = 900$ kg/m³; $\eta = 0.1$ Ns/m²; $g = 9.8$ m/s².

Step 2. Apply $v = \dfrac{2r^2(\rho-\sigma)g}{9\eta}$ with the bubble (air) as the 'falling' object and the liquid as the medium: $v = \dfrac{2×(0.2×10^{-3})^2×(1.29-900)×9.8}{9×0.1}$.

Step 3. $2×(0.2×10^{-3})^2 = 8×10^{-8}$; $(1.29-900) = -898.71$; $8×10^{-8}×(-898.71) = -7.190×10^{-5}$; $×9.8 = -7.046×10^{-4}$; $÷0.9 = -7.829×10^{-4} \approx -0.782×10^{-3}$ m/s. …

Apply the terminal-velocity formula with ρ = density of the air bubble and σ = density of the surrounding liquid; …

Swapping ρ and σ (using the liquid's density as ρ and air's as σ), which would flip the sign and obsc …