An air bubble of radius 0.1 cm lies at a depth of 20 cm below the free surface of a liquid of density 1000 kg / m³. If the pressure inside the bubble is 2100 N / m² greater than the atmospheric pressure, then the surface tension of the liquid in SI unit is (use g=10 m / s² )

Solution

Physics sir · 4 steps

Inside the bubble you have atmosphere, plus the liquid above it, plus 2T/r. Peel off the liquid part first.

1. Pressure inside. An air bubble inside a liquid has only one liquid surface, so its excess pressure is 2T/r. The liquid just outside the bubble is at P₀ + ρ g h, so the pressure inside is that plus 2T/r.

   Pᵢₙ - P₀ = ρ g h + 2T/r

2. Liquid column. With h = 0.2 m, ρ g h = 1000 × 10 × 0.2 = 2000 N m⁻². So 2T/r = 2100 - 2000 = 100 N m⁻².
3. Solve for T. With r = 0.1 cm = 1 × 10⁻³ m, T = (100 × 10⁻³)/2 = 0.05 N m⁻¹.
4. So. The surface tension is 0.05 N m⁻¹, which is option (b).

The idea to keep. Air bubble in a liquid has one surface, so use 2T/r. A soap bubble in air has two surfaces, so use 4T/r. If the bubble is under a liquid, add ρ g h too.

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