There is an air bubble of radius 1.0 mm in a liquid of surface tension 0.075 Nm⁻¹ and density 1000 kg m⁻³ at a depth of 10 cm below the free surface. The amount by which the pressure inside the bubble is greater than the atmospheric pressure is _________ Pa(g=10 ms⁻²)

Solution

Physics sir · 4 steps

Stack two pressures on top of the atmosphere: the water above the bubble, then its one curved surface.

1. Pressure at that depth. Just outside the bubble, the liquid pressure is P₀ + ρ g h. Here ρ g h = 1000 × 10 × 0.1 = 1000 Pa.
2. One surface only. An air bubble inside a liquid has a single liquid-air surface. A soap bubble in air has two. So the excess pressure inside is 2T/r, not 4T/r.

   Pᵢₙ - P₀ = ρ g h + 2T/r

3. Put in numbers. 2T/r = (2 × 0.075)/(1.0 × 10⁻³) = 150 Pa. Adding the depth term gives 1000 + 150 = 1150 Pa.
4. So. The pressure inside the bubble exceeds atmospheric pressure by 1150 Pa.

The common slip: 1300. Using 4T/r as if it were a soap bubble gives 300 Pa for the surface term, which makes the total 1300 instead of 1150.

The idea to keep. Count the free surfaces first. A drop, or an air bubble inside a liquid, has one surface and gives 2T/r. A soap bubble in air has two surfaces and gives 4T/r. Then add ρ g h if the bubble is under the surface.

Answer from the NTA final answer key. Solution by PrepWiser.