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What is the excess pressure inside a bubble of soap solution of radius 5.00 mm, given that the surface tension of soap solution at the temperature (20°C) is 2.50 x 10-2 Nm-1? If an air bubble of the same dimension were formed at depth of 40.0 cm inside a container containing the soap solution (or relative density 1.20), what would be the pressure inside the bubble? (1 atmosphere pressure is 1.01 x 105 Pa) |
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Answer» Here, surface tension, T = 2.50 x 10-2 Nm-1 Radius, r = 5.00 mm = 5 x 10-3 m Excess pressure, p = 4T/r (For a bubble) = {4 x 2.50 x 10-2}/{5 x 10-3} = 20 Pa Now, p1 - p0 = p ⇒ p1 = p + p0 = 1.01 x 105 + 20 = 1.0102 x 105 Pa If the bubble of the same radius as above is at depth of 40 cm inside the soap solution, then excess of pressure is given by, p = 2T/r = {2 x 2.50 x 10-2}/{5 x 10-3} = 10 Pa Also, outside pressure is, p0 = Pressure of water above the bubble + atmosphere pressure = ρgh + 1.01 x 105 Here, ρ = 1.2 x 103 kg m-3 g = 9.8 ms-2 and h = 40 x 10-2 m p0 = 1.2 x 103 x 9.8 x 40 x 10-2 + 1.0 x 105 = 4.704 x 103 + 1.01 x 105 = 1.057 x 105 Pa If p1 is the inside pressure, then p1 - p0 = p ⇒ p1 = p + p0 = 1.057 x 105 + 10 = 1.057 x 105 |
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