JEE Main 2024 · 6 April · evening · Single correct · +4 / −1

Pressure inside a soap bubble is greater than the pressure outside by an amount : (given : R= Radius of bubble, S= Surface tension of bubble)

  1. (a) S/R
  2. (b) (4 S)/R
  3. (c) (4 R)/S
  4. (d) (2 S)/R
Answer and solution

Answer (b) (4 S)/R

Solution

Physics sir · 4 steps

A soap bubble is a thin film with two surfaces, so the drop formula 2S/R gets counted twice.

  1. One curved surface. A single curved liquid surface of radius R pushes inward, so the pressure on its concave side is higher by 2S/R. A liquid drop has just this one surface.
  2. Two surfaces here. A soap bubble is a thin film of soap solution with air on both sides. It has an outer surface and an inner surface, both of radius almost exactly R because the film is so thin.
  3. Add both jumps. Going from outside to inside, the pressure rises by 2S/R across the outer surface and by another 2S/R across the inner surface.

    Pᵢₙ - Pₒᵤₜ = 2S/R + 2S/R = 4S/R

  4. So. The excess pressure inside the soap bubble is 4S/R, which is option (b).

The idea to keep. Count the surfaces first. Each curved surface adds 2S/R: a drop or an air bubble inside a liquid has one, giving 2S/R, and a soap bubble has two, giving 4S/R.

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

Study Surface Tension

Asked about 50 times in JEE Main since 2020. This question is from 2024.