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There is an air bubble of radius 1.0 mm in a liquid of surface tension 0.072 N/m and density 103 kg/m3 . The bubble is at a depth of 10 cm below the free surface of the liquid. By what amount is the pressure inside the bubble greater than the’ atmospheric pressure? |
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Answer» Data : R = 10-3 m, T = 0.072 N/m, ρ = 103 kg/m3 , h = 0.1 m Let the atmospheric pressure be p0. Then, the absolute pressure within the liquid at a depth h is p = p0 + ρgh Hence, the pressure inside the bubble is pin = p0 + \(\frac{2T}R\) = p0 + pgh + \(\frac{2T}R\) The excess pressure inside the bubble over the atmospheric pressure is pin = p0 = pgh + \(\frac{2T}R\) = (103) (9.8) (0.1) + \(\frac{2(0.072)}{10^{-3}}\) = 980 + 144 = 1124 Pa |
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