Question

There is an air bubble of radius 1 mm in a liquid of surface tension 0.075 N/m and density 1000 kg/m3. The bubble is at a depth of 10 cm below the free surface. By what amount is the pressure inside the bubble greater than the atmospheric pressure is ___pa(g-10 ms^-2).

Answer 1

Correct Answer: 1150

1. Write the Expression for the Pressure Inside the Bubble:

The pressure inside the bubble \( P_{\text{in}} \) is given by the formula:

\( P_{\text{in}} = P_0 + \rho g h + \frac{2T}{r} \)

Where:

• \( P_0 \) is the atmospheric pressure.
• \( \rho \) is the density of the liquid.
• \( g \) is the acceleration due to gravity.
• \( h \) is the depth of the bubble.
• \( T \) is the surface tension of the liquid.
• \( r \) is the radius of the bubble.

2. Substitute the Given Values into the Formula:

Given:

\( P_0 = 0 \), \( \rho = 1000 \, \text{kg/m}^3 \), \( g = 10 \, \text{m/s}^2 \), \( h = 0.1 \, \text{m} \), \( T = 0.075 \, \text{N/m} \), \( r = 0.001 \, \text{m} \).

Substituting the values into the formula:

\( P_{\text{in}} = 0 + 1000 \times 10 \times 0.1 + \frac{2 \times 0.075}{0.001} \)

\( P_{\text{in}} = 1000 + \frac{0.15}{0.001} \)

\( P_{\text{in}} = 1000 + 150 \)

\( P_{\text{in}} = 1150 \, \text{Pa} \)

Final Answer

Thus, the pressure inside the bubble is greater than the atmospheric pressure by 1150 Pa.