Question

A soap bubble of radius \(r = 1\,\text{mm}\) is rising in a liquid of density \( \rho = 2000\,\text{kg/m}^3 \). At the instant the bubble is rising upward with constant velocity \(v = \tfrac{1}{2}\,\text{cm/s}\). Find the coefficient of viscosity \( (\eta) \).

• A. \( \dfrac{2}{9}\ \text{N·s/m}^2 \)
• B. \( \dfrac{4}{9}\ \text{N·s/m}^2 \)
• C. \( \dfrac{2}{3}\ \text{N·s/m}^2 \)
• D. \( \dfrac{8}{9}\ \text{N·s/m}^2 \)

Hint:

For terminal velocity problems using Stokes' law: \[ v = \frac{2r^2(\rho-\rho_f)g}{9\eta} \] For bubbles, density of gas is negligible, so \( \rho_f \) dominates.

Answer 1

The Correct Option is A

Concept: For a sphere moving with terminal velocity in a viscous fluid (Stokes' law): \[ F_{\text{viscous}} = 6\pi \eta r v \] At constant velocity, net force is zero: \[ \text{Buoyant force} = \text{viscous force} \] For a bubble (negligible density): \[ \frac{4}{3}\pi r^3 \rho g = 6\pi \eta r v \] Step 1: Substitute the given values.} \[ r = 1\,\text{mm} = 10^{-3}\,\text{m} \] \[ v = \frac{1}{2}\,\text{cm/s} = 5\times10^{-3}\,\text{m/s} \] \[ \rho = 2000\,\text{kg/m}^3 \] Step 2: Solve for viscosity.} \[ \frac{4}{3}\pi r^3 \rho g = 6\pi \eta r v \] Cancel \( \pi \) and simplify: \[ \eta = \frac{2 r^2 \rho g}{9 v} \] Step 3: Substitute values.} \[ \eta = \frac{2(10^{-3})^2 (2000)(10)}{9(5\times10^{-3})} \] \[ \eta = \frac{2}{9}\ \text{N·s/m}^2 \]