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An air bubble of radius 0.1 cm in a liquid having surface tension 0.06N/m and density 1000kg/m3. The pressure inside the bubble is 1100Pa greater than the atmospheric pressure. At what depth below the surface of the liquid is the bubble ? (g=9.8m/s2)

An air bubble of radius 0.1 cm in a liquid having surface tension 0.06N/m and density 1000kg/m3. The pressure inside the bubble is 1100Pa greater than the atmospheric pressure. At what depth below the surface of the liquid is the bubble ? (g=9.8m/s2)

Grade:12

1 Answers

Sanchayan Ghosh
27 Points
10 years ago
The answer requires the usage of two concepts: 1. Gauge pressure. 2. Excess prssure in an air bubble. Imagine an air bubble at a depth h, having a radius r, in a liquid of density p. Using gauge pressure, we know, P - Pa = pgh (Pa = atmospheric pressure) So, P = pgh + Pa. Now, using excess pressure, Pi - Po = 4S/R [Two surfaces in contact). Pi = Pa + 1100. Po = Pa + pgh Pi - Po = 1100 - pgh so, 1100 - pgh = 4S/R. h = 1100/pg - 4S/Rpg. Solving, you get 8.7cm. Just check the answer, you get it like this.

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