Question:
If an air bubble of diameter 2 mm rises steadily through a liquid of density 2000 kg/m³ at a rate of 0.5 cm/s, then the coefficient of viscosity of the liquid is _______ Poise.
If an air bubble of diameter 2 mm rises steadily through a liquid of density 2000 kg/m³ at a rate of 0.5 cm/s, then the coefficient of viscosity of the liquid is _______ Poise.
Updated On: Apr 10, 2026
- 0.88
- 8.8
- 88.8
- 0.088
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The Correct Option is A
Solution and Explanation
Step 1: Apply Stokes' Law.
The formula for the velocity of a rising bubble in a liquid is given by: \[ v = \frac{2r^2 (\rho_l - \rho_g) g}{9 \eta} \] where \( v = 0.5 \, \text{cm/s} = 0.005 \, \text{m/s} \), \( r = 1 \, \text{mm} = 0.001 \, \text{m} \), \( \rho_l = 2000 \, \text{kg/m}^3 \), and \( g = 10 \, \text{m/s}^2 \).
Step 2: Solve for the viscosity \( \eta \).
Rearranging the formula: \[ \eta = \frac{2r^2 (\rho_l - \rho_g) g}{9v} \] Substitute the values: \[ \eta = \frac{2 \times (0.001)^2 \times 2000 \times 10}{9 \times 0.005} = 0.88 \, \text{Poise} \]
Final Answer: 0.88
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