An air bubble of radius 1.0 mm is observed at a depth 20 cm below the free surface of a liquid having surface tension 0.095 J / m² and density 10³ kg / m³. The difference between pressure inside the bubble and atmospheric pressure is __________ N / m². (Take g=10 m / s² )

Solution

Physics sir · 4 steps

Go down to the bubble's depth, then step across its one curved surface. Add both pressure jumps.

1. Liquid just outside. At depth h, the liquid touching the bubble is at P₀ + ρ g h, where P₀ is atmospheric pressure.
2. Inside the bubble. An air bubble inside a liquid has only one liquid-air surface, so the excess pressure across it is 2T/r, not 4T/r.

   Pᵢₙ - P₀ = ρ g h + 2T/r

3. Put in numbers. ρ g h = 10³ × 10 × 0.20 = 2000 N m⁻². 2T/r = (2 × 0.095)/(1.0 × 10⁻³) = 190 N m⁻².
4. So. Pᵢₙ - P₀ = 2000 + 190 = 2190 N m⁻².

The common slip: 2380. Using the soap-bubble formula 4T/r gives 380 + 2000 = 2380. An air bubble in a liquid has only one surface, so the term is 2T/r.

The idea to keep. Count the surfaces before you write the formula: an air bubble in a liquid or a liquid drop has one surface (2T/r), while a soap bubble in air has two (4T/r).

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