PhysicsCBSE Class 11 · Mechanical Properties of Fluids

Surface Tension, Excess Pressure and Capillary Rise Class 11

Updated 17 October 2026 · Fact-checked

Surface tension S is the force per unit length along a line in the surface (equivalently, surface energy per unit area). The excess pressure is 2S/r inside a liquid drop or an air bubble in a liquid and 4S/r inside a soap bubble (two surfaces). In a capillary tube of radius r, a liquid with contact angle θ rises to h = 2S cos θ/(ρgr); mercury, with an obtuse angle, is depressed.

Understand Surface Tension, Excess Pressure and Capillary Rise

Surface tension. Molecules in the surface layer are pulled inward, so a liquid surface behaves like a stretched membrane. The surface tension $S$ is the force per unit length acting perpendicular to a line drawn on the surface: $S = F / ℓ$ (N/m). It is also the surface energy per unit area (J/m²): increasing the surface area by $Δ A$ at constant temperature needs work $W = S Δ A$ . A soap film has two surfaces, so a film stretched on a frame of length $ℓ$ pulls with $F = 2 S ℓ$ and enlarging it by area $Δ A$ of film (each side) needs $W = 2 S Δ A$ . $S$ falls as temperature rises, and detergents reduce $S$ .

Angle of contact. At the line where liquid, solid and air meet, the angle $θ$ between the liquid surface and the solid (measured inside the liquid) is the angle of contact. Water on clean glass: $θ \approx 0$ (acute, liquid wets); mercury on glass: obtuse (about $14 0^{\circ}$ , liquid does not wet).

Excess pressure. Because the surface is curved, the pressure inside is higher:

liquid drop in air, or air bubble inside a liquid (one surface): $Δ P = \frac{2 S}{r}$ ;
soap bubble in air (two surfaces): $Δ P = \frac{4 S}{r}$ .

Smaller bubbles have larger excess pressure; when two bubbles are connected, air flows from the smaller to the larger one.

Capillary rise. In a narrow tube of radius $r$ dipped in a liquid of density $ρ$ , balancing the upward pull of surface tension, $2 π r S cos θ$ , with the weight of the raised column, $π r^{2} h ρ g$ :
$ $h = \frac{2 S c o s θ}{ρ g r} .$ $
For $θ < 9 0^{\circ}$ the liquid rises (water); for $θ > 9 0^{\circ}$ , $h$ is negative and the liquid is depressed (mercury). The height varies inversely with the tube radius. If the tube is shorter than $h$ , the liquid does not overflow; the meniscus flattens (its radius of curvature increases) so that $h^{'} R = constant$ .

Shape of drops. Surface energy is least for a sphere of given volume, which is why small free drops are spherical.

Surface Tension, Excess Pressure and Capillary Rise: formulas and key facts

Surface tension: $S = \frac{F}{ℓ} = \frac{W}{Δ A}$; N/m or J/m².
Soap film pull: $F = 2 S ℓ$; Two surfaces.
Excess pressure, drop or bubble in liquid: $Δ P = \frac{2 S}{r}$; One surface.
Excess pressure, soap bubble: $Δ P = \frac{4 S}{r}$; Two surfaces.
Capillary rise: $h = \frac{2 S cos θ}{ρ g r}$; Negative means depression.
Work in increasing surface area: $W = S Δ A$; Isothermal.

How to solve Surface Tension, Excess Pressure and Capillary Rise questions

Count the number of surfaces first (one for a drop, two for a soap film or bubble), then apply the formula for force, pressure or work.

1Decide how many free surfaces the liquid has (1 or 2).
2For force on a frame or slider, use F = (number of surfaces) × S × length.
3For pressure, use 2S/r for one surface and 4S/r for two.
4For work, find the change in surface area, then W = S × ΔA × (number of surfaces).
5For capillarity, use h = 2S cos θ/(ρgr) and keep the sign of cos θ.
6State if the liquid rises (θ < 90°) or is depressed (θ > 90°).

Surface Tension, Excess Pressure 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.

A drop on a leaf

Suppose a tiny water drop of radius 1.0 mm sits on a leaf (S = 0.07 N/m).

Find the excess pressure inside the drop.

Try it first, then show the solutionSolution

A drop has one surface: ΔP = 2S/r = 2 × 0.07/1.0 × 10⁻³ = 140 Pa.

Answer: 140 Pa.

Memory hook: Smaller drops have a larger excess pressure.

Water in a plant stem

Suppose a plant stem has fine vertical channels of radius 0.05 mm that behave like capillary tubes with θ = 0° for water (ρ = 1000 kg/m³, S = 0.07 N/m, g = 10 m/s²).

Estimate how high water would rise by capillarity alone.

Try it first, then show the solutionSolution

h = 2S/(ρgr) = 0.14/(1000 × 10 × 5 × 10⁻⁵) = 0.14/0.5 = 0.28 m.

Answer: About 0.28 m.

Memory hook: Narrower channels lift water higher.

Tricks to remember Surface Tension, Excess Pressure and Capillary Rise

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

One, two
Drop or bubble in liquid: 2S/r. Soap bubble: 4S/r.
Use it when: Excess pressure.
Narrow tube, high rise
h ∝ 1/r.
Use it when: Capillary comparisons.
Negative cos, depression
Mercury θ is obtuse so h is negative.
Use it when: Mercury questions.
Work = S × new area
Two surfaces on a film, so double it.
Use it when: Blowing and stretching.

Surface Tension, Excess Pressure and Capillary Rise: solved examples

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

Example 1: Force on a soap film · 2 marks · Very short answerBoard level

A soap film is stretched on a rectangular frame with a movable wire 5 cm long (S = 0.07 N/m). Find the force the film exerts on the wire.

Try it first, then show the solutionSolution

A film has two surfaces, so F = 2Sℓ. (1)
F = 2 × 0.07 × 0.05 = 7.0 × 10⁻³ N. (1)

Answer: 7.0 × 10⁻³ N.

Example 2: Capillary rise of water · 3 marks · Short answerBoard level

Find the height to which water rises in a capillary tube of radius 0.4 mm (S = 0.07 N/m, θ = 0°, ρ = 1000 kg/m³, g = 10 m/s²). What will it be for a tube of radius 0.8 mm, and check the first answer by comparing the force of surface tension with the weight of the column.

Try it first, then show the solutionSolution

h = 2S/(ρgr) = 0.14/(1000 × 10 × 4 × 10⁻⁴) = 0.035 m = 3.5 cm. (1)
Doubling the radius halves the rise: 1.75 cm. (1)
Check: 2πrS = 2π × 4 × 10⁻⁴ × 0.07 = 1.76 × 10⁻⁴ N; weight = πr²hρg = π × 1.6 × 10⁻⁷ × 0.035 × 10⁴ = 1.76 × 10⁻⁴ N. (1)

Answer: 3.5 cm; 1.75 cm; the two forces balance.

Example 3: Depression of mercury · 3 marks · Short answerHarder board level

Mercury in a glass capillary of radius 1.0 mm makes an angle of contact of 140° (S = 0.47 N/m, ρ = 13600 kg/m³, g = 10 m/s²). Find the change of level of mercury in the tube relative to the outside.

Try it first, then show the solutionSolution

h = 2S cos θ/(ρgr) with cos 140° = −0.766. (1)
h = 2 × 0.47 × (−0.766)/(13600 × 10 × 10⁻³) = −0.720/136. (1)
h ≈ −5.3 × 10⁻³ m: the mercury is depressed by about 5.3 mm. (1)

Answer: Depressed by about 5.3 mm.

Example 4: Excess pressure in three cases · 4 marks · Long answerHarder board level

Compare the excess pressure in (a) a water drop of radius 2 mm in air (S = 0.07 N/m), (b) an air bubble of radius 2 mm inside water, and (c) a soap bubble of radius 2 mm in air (S of soap solution = 0.03 N/m).

Try it first, then show the solutionSolution

(a) A drop has one surface: ΔP = 2S/r = 2 × 0.07/2 × 10⁻³ = 70 Pa. (1)
(b) An air bubble in water also has one surface: ΔP = 2S/r = 70 Pa. (1)
(c) A soap bubble has two surfaces: ΔP = 4S/r = 4 × 0.03/2 × 10⁻³ = 60 Pa. (1)
The soap bubble has the smaller excess pressure although it has two surfaces, because S of the soap solution is lower. (1)

Answer: (a) 70 Pa; (b) 70 Pa; (c) 60 Pa.

Example 5: Tube longer than the rise · 4 marks · Long answerHOTS / competency

A capillary tube of radius 0.5 mm is dipped in water (S = 0.07 N/m, ρ = 1000 kg/m³, g = 10 m/s²). (a) Find the height of rise. (b) The tube is now raised so that only 1.5 cm projects above the water surface. Will the water overflow? Find the radius of curvature of the meniscus and the angle of contact it implies. (c) Explain.

Try it first, then show the solutionSolution

(a) h = 2S/(ρgr) = 0.14/(1000 × 10 × 5 × 10⁻⁴) = 0.028 m = 2.8 cm. (1)
(b) The water cannot rise above 1.5 cm, so it does not overflow. The meniscus radius R satisfies h'R = 2S/ρg, so R = 0.14/(10⁴ × 0.015) = 0.93 mm. (1)
cos θ = r/R = 0.5/0.933 = 0.536, so θ ≈ 58°. (1)
(c) The meniscus becomes flatter (larger radius of curvature) so that the pull of surface tension matches the lower column's weight; the water simply stays at the top. (1)

Answer: (a) 2.8 cm; (b) no overflow; R ≈ 0.93 mm; θ ≈ 58°.

Where students lose marks in Surface Tension, Excess Pressure and Capillary Rise

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

Mistake: Using 2S/r for a soap bubble.
Forgetting the two surfaces.
Do this: A soap bubble has two surfaces: 4S/r.
Usually costs: The answer mark.
Mistake: Forgetting the factor of 2 in the force on a soap film.
One surface assumed.
Do this: F = 2Sℓ for a film.
Usually costs: The answer mark.
Mistake: Writing cos 140° as positive for mercury.
Dropping the sign.
Do this: Obtuse angle, negative cosine, depression.
Usually costs: Reasoning mark.
Mistake: Using diameter in place of radius in the capillary formula.
Formula mismatch.
Do this: r is the radius of the tube.
Usually costs: The answer mark.

Exam tips

Open every answer with "number of surfaces = ...".
State the direction (rise or fall) and the sign of cos θ.
For work questions, say whether the new area is on one side or two.
Quote S, ρ and g at the start.
Express heights in cm and pressures in Pa at the end.

Practice questions on Surface Tension, Excess Pressure 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 angle of contact for water on clean glass is
(a) zero or small (b) 90° (c) about 140° (d) 180°
Hint
Water wets glass.
Answer
(a).
Q2 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Detergents help to clean clothes.
Reason (R): Detergents lower the surface tension of water so that it wets the cloth more easily and penetrates the fibres.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Lower S, better wetting.
Answer
(a).
Q3 · 3 marks · Short answerHarder board level
The radius of a soap bubble is increased from 1 cm to 2 cm (S = 0.03 N/m). Find the ratio of the excess pressures inside, and the work done against surface tension.
Hint
Two surfaces: ΔP = 4S/r, W = 2S × 4π(r₂² − r₁²).
Answer
ΔP₁ = 12 Pa, ΔP₂ = 6 Pa, ratio 2 : 1. W = 2 × 0.03 × 4π(4 × 10⁻⁴ − 10⁻⁴) = 2.26 × 10⁻⁴ J.
Q4 · 3 marks · Short answerHarder board level
Two soap bubbles of radii 2 cm and 5 cm are connected by a narrow tube (S = 0.03 N/m). Find the excess pressure inside each and say in which direction air flows.
Hint
Smaller bubble, larger pressure.
Answer
4S/r: 6 Pa and 2.4 Pa. Air flows from the smaller bubble to the larger; the smaller one shrinks and the larger grows. The pressure difference is 3.6 Pa.
Q5 · 4 marks · Long answerHOTS / competency
A soap film on a rectangular frame has a movable slider of length 8 cm (S = 0.03 N/m). (a) Find the force needed to hold the slider in place. (b) Find the work done in pulling the slider out 5 cm. (c) State the change in surface energy and what happens to this energy if the slider is released.
Hint
Two surfaces: F = 2Sℓ.
Answer
(a) F = 2 × 0.03 × 0.08 = 4.8 × 10⁻³ N. (b) W = F x = 4.8 × 10⁻³ × 0.05 = 2.4 × 10⁻⁴ J. (c) Area added = 0.08 × 0.05 on each of two sides, ΔE = S × 2 × 4 × 10⁻³ = 2.4 × 10⁻⁴ J, equal to the work. On release, the film contracts and the energy becomes kinetic energy of the slider, then heat.

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.Excess pressure in a soap bubble?Show answer
4S/r.
2.Excess pressure in a liquid drop?Show answer
2S/r.
3.Capillary rise?Show answer
2S cos θ/(ρgr).
4.SI unit of surface tension?Show answer
N/m.
5.Angle of contact of mercury with glass?Show answer
Obtuse, about 140°.
6.Surface energy per unit area?Show answer
Equal to S.

Surface Tension, Excess Pressure and Capillary Rise: frequently asked questions

Why are small drops spherical?

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

Does surface tension depend on temperature?

Yes, it decreases as temperature rises.

Why does mercury get depressed in a glass tube?

Its angle of contact is obtuse, so cos θ is negative and the meniscus is convex.

Why does a soap bubble have a pressure of 4S/r?

It has two surfaces, inside and outside, each adding 2S/r.

Why does a towel absorb water?

Its fine pores act as capillaries and the water rises into them.