Question

An air bubble of radius 1.0 mm is observed at a depth of 20 cm below the free surface of a liquid having surface tension \( 0.095 \, \text{J/m}^2 \) and density \( 10^3 \, \text{kg/m}^3 \). The difference between pressure inside the bubble and atmospheric pressure is: Given \( g = 10 \, \text{m/s}^2 \).

Hint:

The pressure difference inside a bubble is due to both the surface tension and the liquid pressure at the given depth. Use the given formulas to compute both contributions.

Answer 1

Correct Answer: 1

The pressure difference \( \Delta P \) between the inside and outside of the bubble due to the surface tension is given by: \[ \Delta P = \frac{4T}{r}, \] where \( T \) is the surface tension and \( r \) is the radius of the bubble. Also, the pressure at depth \( h \) due to the liquid is: \[ P_{\text{liquid}} = \rho g h. \] For the given values: - \( T = 0.095 \, \text{J/m}^2 \), - \( r = 1.0 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \), - \( \rho = 10^3 \, \text{kg/m}^3 \), - \( g = 10 \, \text{m/s}^2 \), - \( h = 20 \, \text{cm} = 0.2 \, \text{m} \). Now, calculating \( \Delta P \): \[ \Delta P = \frac{4 \times 0.095}{1 \times 10^{-3}} = 380 \, \text{N/m}^2. \] The pressure difference due to the liquid column is: \[ P_{\text{liquid}} = 10^3 \times 10 \times 0.2 = 2000 \, \text{N/m}^2. \] The total pressure difference inside the bubble is the sum of both effects: \[ \Delta P_{\text{total}} = 380 + 2000 = 2190 \, \text{N/m}^2. \]