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JEE Mains Physics · Mechanical Properties of Fluids

Surface Energy of Drops and Bubbles

A liquid surface stores energy T for every unit of area, so making new surface costs T × ΔA: splitting a drop needs work, merging drops releases energy, and blowing a soap bubble pays for two surfaces.

All Mechanical Properties of Fluids notesJEE Mains Physics strategy

Why this matters

Twenty PYQs, seven of them numeric, and five from 2026. Fourteen are drops that merge or split: the energy released or needed, the ratio of the surface energies before and after, and the heat this gives; six are the work to blow a soap bubble bigger, where both surfaces of the film count.

Concept 1 of 2: Drops that merge or split

Volume is kept, surface is not. Split one drop into n equal droplets: each has radius R/n1/3R/n^{1/3}, and together they have n1/3n^{1/3} times the surface of the drop, so work must be done. Merge n droplets into one and the surface shrinks by the same factor, so energy is released, mostly as heat.

Definition

  • n droplets of radius r and one drop of radius R hold the same volume: R=n1/3rR = n^{1/3}r.
  • A drop has one surface: E=4πr2TE = 4\pi r^{2}T.
  • EdropletsEdrop=n1/3\dfrac{E_{\text{droplets}}}{E_{\text{drop}}} = n^{1/3}. For 1000 droplets this is 10.
  • Splitting one drop of radius R into n: W=4πR2T(n1/3−1)W = 4\pi R^{2}T\left(n^{1/3} - 1\right).
  • Merging n droplets of radius r: energy released =4πr2T(n−n2/3)= 4\pi r^{2}T\left(n - n^{2/3}\right).
  • Heat per unit volume on merging: 3TJ(1r−1R)\dfrac{3T}{J}\left(\dfrac{1}{r} - \dfrac{1}{R}\right), J the mechanical equivalent of heat.

Splitting a drop into n

W=4πR2T(n1/3−1)W = 4\pi R^{2}T\left(n^{1/3} - 1\right)

Worked example

A water drop of radius 3 mm splits into 27 equal droplets. The surface tension is 0.07 N/m. Find the work needed. (π=22/7\pi = 22/7)
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 2 Apr 2026 Shift 1 · Q20Moderate

Example 1 · Mechanical Properties of Fluids · Surface Energy of Drops and Bubbles

A liquid drop of diameter 2 mm breaks into 512 droplets. The change in surface energy is α×10−6 J\alpha \times10^{- 6}\text{ }J. The value of α\alpha is ____\_\_\_\_ . (Take surface tension of liquid =0.08 N/m= 0.08\text{ }N/m )

A diameter in the stem

A drop of diameter 2 mm has radius 1 mm. Squaring the diameter makes every energy four times too big.

The ratio of energies is n^(1/3), not n

1000 droplets have 10 times the surface energy of the drop they form, not 1000 times. Each droplet is smaller, so the surfaces do not simply add up n-fold.

Comparing two splittings uses n^(1/3) − 1

Splitting one drop into 64 instead of 27 needs (4 − 1)/(3 − 1) = 3/2 times the work, not 64/27 times.
Drill 13 more on drops that merge or split

Concept 2 of 2: Work to blow a soap bubble

A soap film has two faces, the inside and the outside, and each has surface tension T. Blowing a bubble from radius r1r_1 to r2r_2 adds 4π(r22−r12)4\pi(r_2^{2} - r_1^{2}) of area to EACH face, so the work is twice T times that.

Definition

  • Soap bubble (two surfaces): W=2T×4π(r22−r12)=8πT(r22−r12)W = 2T \times 4\pi\left(r_2^{2} - r_1^{2}\right) = 8\pi T\left(r_2^{2} - r_1^{2}\right).
  • A drop, or an air bubble inside a liquid, has ONE surface: W=4πT(r22−r12)W = 4\pi T\left(r_2^{2} - r_1^{2}\right).
  • Read the stem for diameter or radius: diameters of 2 cm and 6 cm are radii of 1 cm and 3 cm.
  • CGS units: 1 dyne/cm = 10−310^{-3} N/m, and 1 erg = 10−710^{-7} J.

Soap bubble

W=8πT(r22−r12)W = 8\pi T\left(r_2^{2} - r_1^{2}\right)

Worked example

A soap solution has surface tension 0.025 N/m. Find the work to blow a bubble of it from radius 2 cm to radius 4 cm. (π=3.14\pi = 3.14)
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 2 Apr 2026 Shift 2 · Q8Moderate

Example 2 · Mechanical Properties of Fluids · Surface Energy of Drops and Bubbles

The surface tension of a soap bubble is 0.03 N/m0.03\text{ }N/m. The work done in increasing the diameter of bubble from 2 cm to 6 cm is απ×10−4 J\alpha\pi\times10^{- 4}\text{ }J. The value of α\alpha is ____\_\_\_\_ . (Take π=3.14\pi= 3.14 )

Count both faces of a soap film

A soap bubble needs 8πT(r₂² − r₁²). The 4π form is for a single surface and gives half the answer, which is always one of the options.

Diameters in the stem

Halve diameters before squaring. Using diameters as radii makes the work four times too big.
Drill 5 more on work to blow a soap bubble

Summary — formulas & gotchas at a glance

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Formulas (2)

Watch out for (5)

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