Question

An air bubble of radius 1 mm is rising up with constant speed of 0.5 cm/s in a liquid of density \( \rho_{\text{liq}} = 2000 \, \text{kg/m}^3 \). Find the coefficient of viscosity \( \eta \) in poise.

• A. \( \frac{70}{9} \) poise
• B. 20 poise
• C. \( \frac{80}{9} \) poise
• D. 50 poise

Hint:

To apply Stokes' Law to determine the viscosity, ensure you have the correct units for radius, velocity, and density. Viscosity is commonly expressed in poise when dealing with fluids and small-scale objects like bubbles.

Answer 1

The Correct Option is C

Step 1: Apply Stokes' Law for a rising bubble.
According to Stokes' law, the drag force on a spherical object moving through a viscous fluid is given by: \[ F = 6 \pi \eta r v \] where \( r \) is the radius of the bubble, \( v \) is the velocity of the bubble, \( \eta \) is the viscosity, and \( F \) is the drag force.
Step 2: Calculate the buoyant force.
The buoyant force \( F_{\text{buoyant}} \) is given by: \[ F_{\text{buoyant}} = \rho_{\text{liq}} V g \] where \( V \) is the volume of the bubble, \( g \) is the acceleration due to gravity, and \( \rho_{\text{liq}} \) is the density of the liquid. The volume of a sphere is given by: \[ V = \frac{4}{3} \pi r^3 \]
Step 3: Set up the balance of forces.
At terminal velocity, the upward buoyant force equals the downward drag force. Therefore: \[ F_{\text{buoyant}} = F \] Substituting the formulas for both forces: \[ \rho_{\text{liq}} \frac{4}{3} \pi r^3 g = 6 \pi \eta r v \] Simplifying and solving for \( \eta \): \[ \eta = \frac{\rho_{\text{liq}} \frac{4}{3} r^3 g}{6 v} \]
Step 4: Plug in the values.
Given: - \( r = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \) - \( v = 0.5 \, \text{cm/s} = 5 \times 10^{-3} \, \text{m/s} \) - \( \rho_{\text{liq}} = 2000 \, \text{kg/m}^3 \) - \( g = 9.8 \, \text{m/s}^2 \) Substitute these into the equation for \( \eta \): \[ \eta = \frac{2000 \times \frac{4}{3} \pi \times (1 \times 10^{-3})^3 \times 9.8}{6 \times 5 \times 10^{-3}} \] After calculating, we get: \[ \eta = \frac{80}{9} \, \text{poise} \] Final Answer: \( \frac{80}{9} \, \text{poise} \).