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Surface tension and surface energy

Examines the molecular forces at liquid surfaces, including surface tension, surface energy, and related equations.

Why this shows up in the exam

You must understand these concepts to solve NEET questions on liquid drops, bubbles, and energy changes at surfaces.

How NEET tests this

Application · 3 QsNumericalDirect recall

Learn the idea

Surface tension is the force acting along a line on a liquid surface, and surface energy is the energy required to create a unit area of that surface; the key insight is that any change in surface area directly gives an energy change ΔE = T·ΔA.

🧠 Memory hook: Think of surface tension as a stretched rubber band: pull along a line (force) and the band’s stretch stores energy per square meter – just like a soap film.

Get this one clearly and it pays off every single time it shows up in the paper. 🎯

Formulas & facts to keep ready

  • Surface tension (T) = force per unit length (N m⁻¹) acting on a line of the liquid surface
  • Surface energy = energy per unit area (J m⁻²) and numerically equals T
  • Work done in forming a new surface = T ΔA
  • For a sphere, surface area = 4πr² and volume = 4⁄3πr³
  • Contact angle θ determines wetability via Young’s equation: γ_sv – γ_sl = γ_lv cosθ
  • Energy released when surface area decreases is taken as negative ΔE (the magnitude is released)

How to approach it

  1. 1Write the initial and final surface areas of the liquid configuration
  2. 2Find the change in area ΔA = A_final – A_initial
  3. 3Multiply by the surface tension T to obtain ΔE = T·ΔA (negative ΔE means energy released)
  4. 4If the problem involves droplets, use volume conservation (4⁄3πR³ = n·4⁄3πr³) to relate R, r and the number of drops

Worked example — watch it click

A certain number of spherical drops of a liquid of radius 'r' coalesce to form a single drop of radius 'R' and volume 'V'. If 'T' is the surface tension of the liquid, then

  • A)energy = 4VT (1/r − 1/R) is released.
  • B)energy = 3VT (1/r + 1/R) is absorbed.
  • ✅energy = 3VT (1/r – 1/R) is released.
  • D)energy is neither released nor absorbed.

The concept behind this problem

The example asks you to calculate the energy change when many small drops merge into one larger drop, so you must convert the change in total surface area into an energy change using ΔE = T·ΔA and use volume conservation to express everything in terms of V, r and R.

Step by step

  1. 1If n drops of radius r coalesce: V = (4/3)πR³ = n(4/3)πr³, so n = (R/r)³.
  2. 2Change in surface energy = T[4πR² - n·4πr²] = 4πT[R² - (R/r)³r²] = 4πTR²[1 - R/r] = -4πTR³/r = -3VT(1/r - 1/R).
  3. 3Energy is released (negative ΔE), so magnitude is 3VT(1/r - 1/R).

Watch out

Students often stop at ΔE = 4πT(R² – n r²) and forget to replace n r² using the volume relation, which drops the essential factor of 3.

Common slip-ups that cost marks

  • •Missing the factor 3 that appears when converting an area expression to volume using V = 4⁄3πr³
  • •Sign error: forgetting that a decrease in area gives a negative ΔE (energy is released)
  • •Using radius ratio R/r incorrectly; remember n = (R/r)³ for equal‑volume droplets

🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.

Practise it

These are real questions from past NEET papers that test this exact idea.

Question 1 of 4NEET 2013

The wetability of a surface by a liquid depends primarily on:

Push further

More challenging

5 harder questions built from the past papers above — a step up in difficulty, with distractors designed so you can't get there by elimination. Written and checked by our reviewers, not from a real paper.

Question 1 of 5

Two soap bubbles of radii R1 and R2 (R1 > R2) are connected by a narrow tube. What happens to the bubbles after connection?

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