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An air bubble of radius 1.0 mm lies at a depth of 10 cm below the free surface of a liquid having surface tension 0.075 N/m and density 1000 kg/m³. By how much does the pressure inside the bubble exceed the atmospheric pressure? Take g = 9.8 m/s².
- 1130 Pa
- 952 Pa
- 890 Pa
- 1208 Pa
Correct answer: 1130 Pa
Solution
P_inside - P_atm = rho g h + 2T/r. Hydrostatic: 1000 x 9.8 x 0.10 = 980 Pa. Surface tension (single interface): 2T/r = 2 x 0.075 / 1.0e-3 = 150 Pa. Sum = 1130 Pa.
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