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The pressure inside a soap bubble is greater than the pressure outside by an amount (given: R= radius of the bubble, S= surface tension of the bubble)

Asked in JEE Main 6th April 2nd Shift 2024 · Excess pressure

• (1) (4S)/R
• (2) (4R)/S
• (3) S/R
• (4) (2S)/R

Step-by-step solution

Why the other options are wrong

• (2) comes from inverts the fraction; (4R)/S has the units m/(N m⁻¹)=m² N⁻¹, which is not a pressure at all, and it would make a bigger bubble harder to blow.
• (3) comes from loses both factors of 2: the 2 that comes from balancing S(2π R) against Δ p(π R²) for one surface, and the further doubling for the film's second surface.
• (4) (2S)/R is the excess pressure for a liquid drop, and also for a gas bubble inside a liquid — each has a single interface. A soap bubble in air has an inner film surface and an outer one, so its excess is doubled to (4S)/R.

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