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).
Correct Answer: 1150
Solution and Explanation
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.
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