PhysicsISC Class 11 · Mechanical Properties of Fluids

Surface Tension: Surface Energy, Excess Pressure and Capillary Rise

Updated 17 October 2026 · Fact-checked

Surface tension S is the force per unit length on a line in the surface, or equivalently the surface energy per unit area. Excess pressure is 2S/R inside a drop and 4S/R inside a soap bubble. A liquid rises or falls in a capillary of radius r by h = 2S cosθ/(rρg).

Understand Surface Tension and Capillary Rise

Definition. Surface tension acts along the surface, perpendicular to any line drawn on it: S = F/l, measured in N m⁻¹, dimensions [MT⁻²]. It is also the work needed to increase the area by one unit, S = W/ΔA, in J m⁻². A soap film has two surfaces, so a film on a frame pulls with F = 2Sl.

Surface energy. Increasing the surface area by ΔA at constant temperature needs work W = S ΔA. When small drops merge into a bigger one, the area falls, and the surface energy released appears as heat.

Angle of contact. The angle θ between the tangent to the liquid surface and the solid surface, measured inside the liquid. For water and glass θ is small (acute), the liquid wets the glass and the surface curves up; for mercury and glass θ is obtuse and the surface curves down.

Excess pressure. Across a curved surface the pressure is higher on the concave side. For a spherical drop, p − p₀ = 2S/R. For a soap bubble, which has two surfaces, p − p₀ = 4S/R. A smaller bubble has the greater pressure, so air flows from a smaller to a larger bubble when they are connected.

Capillary rise. In a narrow tube of radius r, the meniscus has radius r/cosθ, so the pressure difference 2S cosθ/r supports a column of height h: h = 2S cosθ/(rρg). It is raised for acute θ and depressed (a fall) for obtuse θ, and h ∝ 1/r.

Everyday links. Soap, detergents and heat lower the surface tension of water. Blotting paper, a candle wick and the rising of oil in a lamp wick are examples of capillarity.

Surface Tension and Capillary Rise: formulas and key facts

Surface tension: $S = F / l ∣ S = W /Δ A$; Soap film: F = 2Sl because it has two surfaces.
Excess pressure: $D r o p : 2 S / R ∣ B u bb l e : 4 S / R$; A soap bubble has two surfaces.
Capillary rise: $h = 2 S cos θ / (r ρ g)$; θ obtuse gives a negative h (depression).
Surface energy released on merging: $E = S \times (A_{i} ni t ia l - A_{f} ina l)$; Becomes heat.

Surface Tension and Capillary Rise in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Raindrops merging

Suppose 1000 tiny water droplets, each of radius 1 mm, merge into one large drop (S = 0.072 N m⁻¹).

Find the radius of the large drop and the surface energy released.

Try it first, then show the solutionSolution

Volume is conserved: R³ = 1000 r³, so R = 10 r = 10 mm.
Area before = 1000 × 4π(10⁻³)² = 4π × 10⁻³ m²; area after = 4π(10⁻²)² = 4π × 10⁻⁴ m².
E = S ΔA = 0.072 × 4π × 9 × 10⁻⁴ = 8.14 × 10⁻⁴ J.

Answer: R = 1 cm; energy released 8.1 × 10⁻⁴ J.

Memory hook: Merging reduces the total surface area, so energy is released.

Tricks to remember Surface Tension and Capillary Rise

Short memory aids. Say each one aloud once, then use it on a question the same day.

Bubble is twice a drop
A bubble has two surfaces, so 4S/R against 2S/R.
Use it when: Excess pressure questions.
Thin tube, high rise
h ∝ 1/r, so halving the radius doubles the rise.
Use it when: Comparing capillaries.

Surface Tension and Capillary Rise: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Units of surface tension · 1 mark · MCQBoard level

The dimensional formula of surface tension is:
(a) [M L T⁻²]
(b) [M T⁻²]
(c) [M L⁻¹ T⁻²]
(d) [M L² T⁻²]

Try it first, then show the solutionSolution

S = force/length = MLT⁻²/L = MT⁻².

Answer: (b) [M T⁻²]

Example 2: Soap film on a frame · 3 marks · Short answerBoard level

A film of soap solution (S = 0.03 N m⁻¹) is stretched on a U-shaped wire with a 10 cm slider. Find the force on the slider and the work done in pulling it 5 cm.

Try it first, then show the solutionSolution

The film has two surfaces: F = 2Sl = 2 × 0.03 × 0.10 = 6 × 10⁻³ N. (1 mark)
W = F d = 6 × 10⁻³ × 0.05. (1 mark)
= 3 × 10⁻⁴ J. Check: S × ΔA = 0.03 × 2 × (0.10 × 0.05) = 3 × 10⁻⁴ J. (1 mark)

Answer: F = 6 mN; W = 3 × 10⁻⁴ J.

Example 3: Soap bubble · 4 marks · Short answerHarder board level

Find the excess pressure inside a soap bubble of radius 2 cm (S = 0.03 N m⁻¹), and the work done in blowing it from radius 1 cm to 2 cm.

Try it first, then show the solutionSolution

Excess pressure = 4S/R = 4 × 0.03/0.02 = 6 Pa. (1 mark)
Increase in area = 2 × 4π(R₂² − R₁²) = 8π(4 × 10⁻⁴ − 10⁻⁴) = 7.54 × 10⁻³ m². (2 marks)
W = S ΔA = 0.03 × 7.54 × 10⁻³ = 2.26 × 10⁻⁴ J. (1 mark)

Answer: Excess pressure 6 Pa; work 2.3 × 10⁻⁴ J.

Example 4: Capillary rise and fall · 4 marks · Short answerHarder board level

Find the rise of water (S = 0.072 N m⁻¹, θ ≈ 0°, ρ = 1000 kg m⁻³) and the depression of mercury (S = 0.465 N m⁻¹, θ = 135°, ρ = 13 600 kg m⁻³) in glass tubes of radius 0.5 mm (g = 9.8 m s⁻²).

Try it first, then show the solutionSolution

Water: h = 2 × 0.072 × 1/(5 × 10⁻⁴ × 1000 × 9.8) = 0.144/4.9. (1 mark)
h = 2.94 × 10⁻² m = 2.94 cm. (1 mark)
Mercury: h = 2 × 0.465 × cos135°/(5 × 10⁻⁴ × 13 600 × 9.8) = −0.6576/66.64. (1 mark)
h = −9.9 × 10⁻³ m: depressed by about 9.9 mm. (1 mark)

Answer: Water rises 2.94 cm; mercury is depressed 9.9 mm.

Example 5: Merging drops · 5 marks · Long answerHOTS / competency

Suppose 1000 identical water droplets of radius 1 mm coalesce into one drop. Taking S = 0.072 N m⁻¹ and the specific heat of water as 4200 J kg⁻¹ K⁻¹, find the radius of the big drop, the energy released and the rise in temperature if all of it heats the water.

Try it first, then show the solutionSolution

R = 1000^(1/3) r = 10 r = 1 cm. (1 mark)
Area change = 4π[1000 × (10⁻³)² − (10⁻²)²] = 4π × 9 × 10⁻⁴ = 1.131 × 10⁻² m². (1 mark)
E = 0.072 × 1.131 × 10⁻² = 8.14 × 10⁻⁴ J. (1 mark)
Mass = ρ(4/3)πR³ = 1000 × 4.189 × 10⁻⁶ = 4.189 × 10⁻³ kg. (1 mark)
ΔT = E/(mc) = 8.14 × 10⁻⁴/(4.189 × 10⁻³ × 4200) = 4.6 × 10⁻⁵ K. (1 mark)

Answer: R = 1 cm; E = 8.1 × 10⁻⁴ J; ΔT ≈ 4.6 × 10⁻⁵ K (negligible).

Where students lose marks in Surface Tension and Capillary Rise

The slips examiners see most often, and what to write instead.

Mistake: Forgetting two surfaces in a soap film
Using F = Sl or 2S/R for a bubble.
Do this: A film or bubble has two surfaces: F = 2Sl and 4S/R.
Usually costs: 1 to 2 marks.
Mistake: Sign of cosθ for mercury
Reporting a rise for mercury in glass.
Do this: θ > 90° makes cosθ negative, so the level is depressed.
Usually costs: 1 mark.
Mistake: Not converting mm to metre
Substituting the radius in mm.
Do this: Use SI units: r in m, S in N m⁻¹.
Usually costs: 1 to 2 marks.

Exam tips

Say whether the surface is a drop (one surface) or bubble (two surfaces) before choosing the formula.
For capillary questions give the sign convention: positive h is a rise.
For merging drops, conserve volume first to get the radius.
State units: N m⁻¹ for S and J m⁻² for surface energy.

Practice questions on Surface Tension and Capillary Rise

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The excess pressure inside a soap bubble of radius R is p. For a bubble of radius 2R blown from the same solution it is:
(a) p/4
(b) p/2
(c) p
(d) 2p
Hint
Excess pressure = 4S/R.
Answer
(b) p/2
Q2 · 2 marks · Short answerBoard level
A wire 0.2 m long is held horizontally in a soap film (S = 0.03 N m⁻¹), lifting it out of the solution. What force pulls the wire down due to the film?
Hint
The film has two faces.
Answer
F = 2Sl = 2 × 0.03 × 0.2 = 0.012 N.
Q3 · 3 marks · Short answerHarder board level
Water rises 1.5 cm in a capillary of radius 1 mm. How high does it rise in a capillary of radius 0.5 mm of the same material?
Hint
h ∝ 1/r.
Answer
h₂ = 1.5 × 1/0.5 = 3.0 cm.
Q4 · 3 marks · Short answerHarder board level
Compare the excess pressures in a water drop (S = 0.072 N m⁻¹) and a soap bubble (S = 0.03 N m⁻¹), each of radius 2 mm.
Hint
Drop: 2S/R; bubble: 4S/R.
Answer
Drop: 2 × 0.072/0.002 = 72 Pa. Bubble: 4 × 0.03/0.002 = 60 Pa. The drop has the larger excess pressure.
Q5 · 3 marks · Short answerHarder board level
A liquid (S = 0.07 N m⁻¹, ρ = 1000 kg m⁻³) rises 1.2 cm in a tube of radius 0.4 mm (g = 9.8 m s⁻²). Find the angle of contact.
Hint
cosθ = h r ρ g/(2S).
Answer
cosθ = 0.012 × 4 × 10⁻⁴ × 1000 × 9.8/(2 × 0.07) = 0.04704/0.14 = 0.336, so θ = 70.4°.
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose two soap bubbles of radii 3 cm and 6 cm (S = 0.03 N m⁻¹) are joined by a thin tube.
(i) Find the excess pressure in each. (1 mark)
(ii) In which direction does the air flow? (1 mark)
(iii) What happens to the sizes of the bubbles? (1 mark)
(iv) Find the work done in blowing the 3 cm bubble from nothing. (1 mark)
Hint
Pressure is higher in the smaller bubble.
Answer
(i) 4 × 0.03/0.03 = 4 Pa and 4 × 0.03/0.06 = 2 Pa. (ii) From the smaller bubble (higher pressure) to the larger one. (iii) The smaller shrinks and the larger grows. (iv) W = S × 2 × 4πR² = 0.03 × 2 × 4π × 9 × 10⁻⁴ = 6.8 × 10⁻⁴ J.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Define surface tension.Show answer
Force per unit length acting in the surface, perpendicular to a line in it.
2.Excess pressure in a drop?Show answer
2S/R.
3.Capillary rise formula?Show answer
h = 2S cosθ/(rρg).
4.What is the angle of contact for mercury and glass?Show answer
Obtuse (about 135°).

Surface Tension and Capillary Rise: frequently asked questions

Is surface tension a vector?

No. Although it is defined as a force per unit length, it is a scalar property of the surface; the force it produces acts perpendicular to the line you choose.

Why do small drops tend to be spherical?

A sphere has the least surface area for a given volume, so the surface energy is least.

How does temperature affect surface tension?

The surface tension of a liquid decreases as the temperature rises, and falls to zero at the critical temperature.