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 viscous drag force acting on a small spherical bubble moving through a liquid is given by: \[ F = 6 \pi \eta r v \] where \( r \) is the radius of the bubble, \( v \) is its terminal velocity, \( \eta \) is the coefficient of viscosity of the liquid, and \( F \) represents the drag force.
Step 2: Determine the buoyant force.
The upward buoyant force acting on the bubble is given by: \[ F_{\text{buoyant}} = \rho_{\text{liq}} V g \] where \( \rho_{\text{liq}} \) is the density of the liquid, \( V \) is the volume of the bubble, and \( g \) is the acceleration due to gravity. The volume of the spherical bubble is: \[ V = \frac{4}{3} \pi r^3 \]
Step 3: Apply the condition for terminal velocity.
When the bubble rises with terminal velocity, the upward buoyant force balances the downward viscous drag force. Therefore: \[ F_{\text{buoyant}} = F \] Substituting the expressions for both forces: \[ \rho_{\text{liq}} \frac{4}{3} \pi r^3 g = 6 \pi \eta r v \] Cancelling the common terms and rearranging the equation to find viscosity \( \eta \): \[ \eta = \frac{\rho_{\text{liq}} \frac{4}{3} r^3 g}{6 v} \]
Step 4: Substitute the given values.
Given: - \( r = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \) - \( v = 0.5 \, \text{cm/s} = 5 \times 10^{-3} \, \text{m/s} \) - \( \rho_{\text{liq}} = 2000 \, \text{kg/m}^3 \) - \( g = 9.8 \, \text{m/s}^2 \) Substituting these values into the formula: \[ \eta = \frac{2000 \times \frac{4}{3} \pi \times (1 \times 10^{-3})^3 \times 9.8}{6 \times 5 \times 10^{-3}} \] On solving, the coefficient of viscosity of the liquid is obtained as: \[ \eta = \frac{80}{9} \, \text{poise} \]
Final Answer: \( \frac{80}{9} \, \text{poise} \).