# Excess Pressure in Bubbles and Drops > Soap bubble 4T/r, liquid drop 2T/r, air bubble 2T/r — why surfaces count and how bubble energy problems are set in JEE Physics. - Canonical URL: https://prepelephant.com/topics/jee/physics/excess-pressure-bubble - Exam / course: JEE · Subject: Physics - Publisher: PrepElephant (https://prepelephant.com) — Prepared and reviewed by the PrepElephant Academic Review Team - First published: 2026-10-02 - Last updated: 2026-10-02 - How to cite: "Excess Pressure in Bubbles and Drops", PrepElephant, https://prepelephant.com/topics/jee/physics/excess-pressure-bubble ## Direct answer The pressure inside a curved liquid surface exceeds the outside pressure by an amount fixed by surface tension and the geometry. A liquid drop has one surface and excess pressure 2T/r; a soap bubble has two surfaces (inner and outer) and excess pressure 4T/r; an air bubble submerged in liquid has one liquid–gas surface and behaves like a drop at 2T/r. The number of surfaces, not the substance, is what students must count. Blowing a soap bubble from radius r1 to r2 demands work equal to the rise in surface energy, W = 8πT(r2² − r1²), because both surfaces grow. When two bubbles connect, air flows from the smaller (higher excess pressure) into the larger, and the common interface adopts a radius set by 1/r = 1/r1 − 1/r2. ## What you must remember - **Drop (one surface):** excess pressure 2T/r, derived by balancing surface tension 2πr T around the rim against the pressure difference on the great circle πr². - **Soap bubble (two surfaces):** 4T/r; the film has an inner and an outer surface, each pulling with 2T/r — the single most tested distinction in this chapter. - **Air bubble in liquid:** 2T/r, one interface; strictly the radius used is of the gas pocket, and the pressure outside it is the liquid pressure at that depth, P0 + hρg. - **Cylindrical film (soap film on a wire loop):** excess pressure T/r, one curved surface of a cylinder. - **Surface energy:** a drop carries T × 4πr²; a soap bubble carries 2 × T × 4πr² = 8πr²T, so doubling a drop's radius costs the difference in these energies — work that must come from outside. - **Coalescing bubbles:** two soap bubbles of radii r1 < r2 connected by a tube share air until the interface radius is r = r1r2/(r2 − r1); the smaller bubble shrinks because 4T/r1 > 4T/r2. - **Capillary connection:** the same 2T/r curvature logic drives capillary rise h = 2T cos θ/(rρg), with θ the angle of contact — JEE frequently mixes the two in one question. ## Blowing a bubble, in joules A soap solution with T = 0.025 N/m is blown from radius 2 cm to 4 cm. The bubble's surface energy is 8πr²T at each stage, so the work done is 8πT(r2² − r1²) = 8π × 0.025 × (16 − 4) × 10⁻⁴ = 8π × 0.025 × 1.2 × 10⁻³ ≈ 7.5 × 10⁻⁴ J. Part of this went into creating fresh surface; the rest pushed the atmosphere back as the bubble grew, which is why the full thermodynamic accounting includes a P dV term at the more advanced level, though JEE's standard treatment equates work with the change in surface energy. Now the coalescence classic. Bubbles of radii 1 cm and 3 cm are connected. Excess pressures are 4T/1 and 4T/3, so the smaller is at higher internal pressure and pumps air into the larger until the pressure difference across their common wall balances: 4T/r = 4T/r1 − 4T/r2 gives 1/r = 1/1 − 1/3 = 2/3, so the interface curves with r = 1.5 cm. The smaller bubble need not vanish — it stops shrinking when its shrinking radius and the shared-wall geometry satisfy this relation, and JEE Advanced has asked precisely for the interface radius rather than the naive "smaller disappears" answer. ## Where students slip The dominant error is the surface count: quoting 4T/r for a mercury drop in vacuum (it is 2T/r — drops have one surface) or 2T/r for a soap bubble. The second slip is absolute versus excess pressure: a bubble at depth h in water has internal pressure P0 + hρg + 4T/r, and problems that ask "pressure inside" expect all three terms. Watch also the radius arithmetic in energy problems — 8πr²T has radius squared, so a factor-of-two radius error costs a factor of four, and options are spaced to catch exactly that. Finally, in the two-bubble problem, students who compare radii instead of pressures conclude the large bubble feeds the small one; pressure, not size, drives the flow. ## Frequently asked questions ### Why is a soap bubble's excess pressure double a drop's? Because the film has two free surfaces, inner and outer, each contributing 2T/r, giving 4T/r in total. ### What is the excess pressure inside an air bubble in water? 2T/r above the surrounding liquid pressure at that depth, since a submerged air bubble presents only one liquid–gas interface. ### How much work is needed to blow a soap bubble to radius r? The surface energy is 8πr²T (two surfaces), so blowing from nothing at constant surface tension requires that much work; expanding r1 to r2 costs 8πT(r2² − r1²). ### What happens when two soap bubbles of different radii are connected? Air flows from the smaller bubble to the larger because its excess pressure 4T/r is higher, until the common interface settles at radius r1r2/(r2 − r1). ### Does atmospheric pressure affect the excess pressure formula? No — the excess depends only on surface tension and radius; atmospheric pressure sets the baseline inside and outside equally and cancels in the difference.