Surface tension

A water strider skates across a still pond on legs no thicker than a hair, and never gets wet. A sewing needle, laid flat with a little care, floats on the surface of a bowl of water even though steel is roughly eight times denser than water and should sink in an instant. Neither trick uses buoyancy in any ordinary sense — the needle does not displace enough water to float by Archimedes' principle. Something at the very skin of the liquid is holding both of them up, behaving less like a fluid and more like a taut rubber sheet stretched across a drum. That "skin" is surface tension, and it is not a metaphor — it is a real force, measurable in newtons per metre, that lives only in the outermost molecular layer of a liquid. It is the same effect that pulls a falling raindrop into a near-perfect sphere, that lets a soap film stretch across a wire loop without instantly collapsing, and that drags ink up through the fibres of a paper towel or sap up through the narrow vessels of a tree. This article builds surface tension from the molecular tug-of-war at a liquid's surface up to three formulas the CBSE Class 11 syllabus asks you to derive and apply: the excess pressure inside a drop, the doubled excess pressure inside a bubble, and the capillary-rise formula that lets you measure surface tension with nothing more than a thin glass tube and a ruler. By the end you should be able to explain why a narrower tube pulls water higher, not just recite the formula.

Key takeaways

Surface molecules feel a net inward pull. Molecules in the bulk of a liquid are tugged equally in all directions, but molecules at the surface have neighbours only below and beside them, so the surface behaves like a stretched membrane trying to minimise its own area.
Surface tension $S$ is force per unit length, $S = F/l$ , measured in N/m; numerically it equals surface energy per unit area, in J/m².
A curved surface always has higher pressure on its concave (inner) side. For a drop (one surface), $\Delta P = 2S/r$ ; for a soap bubble (two surfaces), $\Delta P = 4S/r$ .
Angle of contact $\theta$ decides wetting. Small $\theta$ (water on clean glass) means the liquid wets the surface and curves upward at the edges (concave meniscus); obtuse $\theta$ (mercury on glass) means it does not wet and curves downward (convex meniscus).
Capillary rise (or depression) follows $h = \dfrac{2S\cos\theta}{\rho g r}$ — narrower tubes give a larger rise, the opposite of most students' first guess.
Surface tension falls as temperature rises, because hotter molecules have weaker net attraction to their neighbours.
Key idea: unbalanced intermolecular attraction at a liquid's surface creates a tension S = F/l that raises the pressure inside any curved surface by 2S/r (drop) or 4S/r (bubble), and the same tension pulls liquid up a narrow tube to a height h = 2S cos θ/(ρ g r) that grows as the tube gets thinner.

The core insight

Unbalanced inward pulls on surface molecules make a liquid's surface act like a stretched membrane of tension $S = F/l$ , producing excess pressure $2S/r$ inside a drop or $4S/r$ inside a bubble, and driving capillary rise $h = 2S\cos\theta/(\rho g r)$ that is larger, not smaller, in a narrower tube.

Where it fits in the syllabus

Surface tension is the concluding topic of the CBSE Class 11 "Mechanical Properties of Fluids" unit, following pressure, Pascal's law, viscosity and Bernoulli's theorem. The board expects you to explain surface energy and surface tension at a molecular level, define the coefficient of surface tension $S = F/l$ , identify the angle of contact and what it tells you about wetting, derive and apply the excess-pressure formulas for a drop ( $2S/r$ ) and a bubble ( $4S/r$ ), and derive and apply the capillary-rise formula $h = 2S\cos\theta/(\rho g r)$ — all of it drawn on heavily in both conceptual (assertion-reason, "explain why") and numerical questions. Detailed treatments of surface energy as a distinct quantity, and the relation connecting surface energy to surface tension, are covered in the next two nodes of this unit; this section stays focused on the tension itself and its two headline consequences.

The physical picture

Picture the liquid as a crowd of molecules held together by short-range attractive forces — the same forces responsible for a liquid not simply flying apart into gas. A molecule deep inside the liquid is surrounded on every side by neighbours pulling it inward, outward, sideways, in every direction at once; those pulls cancel, and the net force on it is zero. A molecule sitting right at the surface has neighbours below it and beside it, in the liquid, but essentially no neighbours pulling from above (only sparse vapour molecules there, exerting negligible attraction). The result is a net inward pull on every surface molecule, directed into the bulk of the liquid. That lopsided pull has two related consequences. First, dragging a molecule from the interior up to the surface costs energy, because you are working against the inward pull — this defines surface energy, the extra potential energy stored per unit area of surface, and it is why liquids minimise their surface area whenever nothing else constrains them (a small blob of mercury on a table pulls itself into a near-sphere, the shape with least surface area for a given volume). Second, and numerically identical for a given liquid, the surface behaves as though it is under a real mechanical tension, exactly like a stretched rubber sheet: any line drawn on the surface experiences a pull from the liquid on either side of it, perpendicular to the line and lying in the surface. This pull per unit length is the surface tension,

S = \frac{F}{l}

where: $F$ is the force exerted along the surface, perpendicular to an imaginary line of length $l$ drawn on that surface. The SI unit is N/m; because surface tension and surface energy per unit area turn out to be numerically equal for the same liquid, $S$ is equally often quoted in J/m². Surface tension is a property of the specific liquid-vapour (or liquid-liquid) interface, not of the bulk liquid alone — water against air has one value of $S$ , water against oil a different one. It decreases steadily as temperature rises, because thermal agitation weakens the net attractive pull that surface molecules feel from their neighbours, and it vanishes altogether at the critical temperature. When a liquid surface meets a solid — the wall of a glass tube, say — the geometry of the meniscus is fixed by the angle of contact, $\theta$ : the angle, measured inside the liquid, between the tangent to the liquid surface at the point of contact and the solid surface. Water in a clean glass tube has a small angle of contact (close to $0°$ in the idealised case); the liquid "wets" the glass, cohesive forces between water molecules being weaker than the adhesive pull of glass on water, and the meniscus curves upward at the edges into a concave shape. Mercury in glass has an obtuse angle of contact (around $140°$ ); mercury does not wet glass, cohesion among mercury atoms dominates over adhesion to glass, and the meniscus bulges downward into a convex shape.

Figure

A capillary tube dipped in water showing the liquid risen inside the tube above the outside level with a concave meniscus and the angle of contact marked where the curved surface meets the glass wall, next to a wider dish showing the flat outside water level

Because a curved liquid surface behaves like a stretched membrane, the pressure just inside a curved surface always exceeds the pressure just outside it — a flat surface, by contrast, has zero excess pressure across it, which is why the formulas below vanish as $r \to \infty$ . For a spherical liquid drop of radius $r$ , there is one liquid-air surface, and

\Delta P = \frac{2S}{r}

where: $\Delta P$ is the excess pressure, pressure inside minus pressure outside, $S$ is the surface tension of the liquid, and $r$ is the drop's radius. For a soap bubble of radius $r$ , there are two liquid-air surfaces — an inner film surface facing the trapped air and an outer film surface facing the atmosphere — and both contribute tension pulling inward, so the excess pressure doubles:

\Delta P = \frac{4S}{r}

Both formulas say the same physical thing in different multiples: a smaller drop or bubble has more sharply curved surface and needs a larger pressure difference to stay in mechanical equilibrium against its own surface tension — which is exactly why very small bubbles are hard to blow (they need a strong puff) and larger ones balloon out more easily once started. Combining the wetting geometry with the pressure balance across the curved meniscus gives capillary rise, the height $h$ to which a wetting liquid climbs inside a narrow tube above the level in the open container, or the depth to which a non-wetting liquid is pushed down (capillary depression):

h = \frac{2S\cos\theta}{\rho g r}

where: $\rho$ is the liquid's density, $g$ is the acceleration due to gravity, $r$ is the internal radius of the capillary tube, and $\theta$ is the angle of contact. For water in clean glass, $\theta$ is small and $\cos\theta$ is close to $+1$ , giving a genuine rise. For mercury in glass, $\theta$ is obtuse and $\cos\theta$ is negative, so $h$ comes out negative — read as a depression, the mercury level inside the tube sitting below the level outside.

Force-balance derivation of capillary rise

Derivation AWhy a narrow tube pulls liquid up to a definite height

The goal is to find the equilibrium height $h$ to which a wetting liquid rises inside a thin capillary tube of internal radius $r$ , dipped vertically into a wide open container of the same liquid, using nothing beyond the definition of surface tension as a force per unit length and the ordinary hydrostatic idea that a column of liquid has weight $\rho g \times \text{volume}$ . The physical picture is that the curved meniscus pulls the liquid column upward along the wetted rim of the tube, and the column rises until that upward pull is exactly balanced by the weight of the extra liquid it has lifted above the outside level — an equilibrium condition that fixes $h$ uniquely once $S$ , $\theta$ , $\rho$ and $r$ are known.

Step 1. Identify the upward pulling force. Surface tension acts along the circular line where the liquid surface meets the glass wall, a circle of circumference $2\pi r$ . The liquid surface pulls on the tube wall (and, by Newton's third law, the wall pulls the liquid surface) with a force of magnitude $S$ per unit length of that contact line, directed along the liquid surface at the angle of contact $\theta$ to the tube wall. Only the vertical component of this pull actually lifts the liquid column; the vertical component of the tension per unit length is $S\cos\theta$ .

Step 2. Total the upward force. Multiplying the vertical tension per unit length by the length of the contact circle gives the total upward force supporting the raised column: $F_{\text{up}} = S\cos\theta \times 2\pi r = 2\pi r S\cos\theta$ . This is the entire mechanical support holding the water column above the level of the surrounding liquid — there is nothing else pushing it up.

Step 3. Write the weight of the raised column. Treat the raised liquid inside the tube, of height $h$ above the outside level, as an ordinary cylinder of liquid of radius $r$ and height $h$ (this ignores the small volume in the curved meniscus cap itself, an approximation valid for a narrow tube, which is exactly the regime the formula is meant for). Its volume is $\pi r^2 h$ , its mass is $\rho \pi r^2 h$ , and its weight is $W = \rho \pi r^2 h g$ .

Step 4. Balance the two forces. At equilibrium the column neither keeps rising nor falls back, so the upward pull exactly supports the weight of liquid lifted: $2\pi r S\cos\theta = \rho \pi r^2 h g$ .

Step 5. Solve for $h$ . Cancel one factor of $\pi r$ from both sides: $2 S\cos\theta = \rho r h g$ , and divide through by $\rho g r$ to isolate $h$ :

h = \frac{2S\cos\theta}{\rho g r}

Step 6. Read off the radius dependence, which is the single most counter-intuitive result in this chapter. The upward force in Step 2 scales with the circumference $2\pi r$ (one power of $r$ ), while the weight to be lifted in Step 3 scales with the cross-sectional area $\pi r^2$ (two powers of $r$ ) for a given height $h$ . Because the pulling force grows more slowly with $r$ than the cross-section available to fill does, equilibrium can only be reached at a smaller height when the tube is wide, and a larger height when the tube is narrow — hence $h \propto 1/r$ , confirmed directly by the $r$ in the denominator of the final formula. A hair-thin capillary can lift liquid many centimetres; a wide test tube shows almost no visible rise at all.

Worked examples

Board level - excess pressure inside a soap bubble

A soap bubble of radius $1.0\ \text{cm}$ is formed by blowing air through a straw. Find the excess pressure inside the bubble compared to the outside atmosphere. Take the surface tension of the soap solution $S = 0.030\ \text{N/m}$ .

Step 1. Recognise that a soap bubble has two liquid-air surfaces (inner and outer film), so the excess-pressure formula to use is $\Delta P = 4S/r$ , not the single-surface drop formula $2S/r$ .

Step 2. Convert the radius to SI units: $r = 1.0\ \text{cm} = 0.01\ \text{m}$ .

Step 3. Substitute the values into the formula.

\Delta P = \frac{4 \times 0.030}{0.01} = \frac{0.12}{0.01} = 12\ \text{Pa}

The pressure inside the soap bubble is 12 Pa greater than the surrounding atmospheric pressure — this small excess is exactly what keeps the thin soap film inflated and taut against its own surface tension, and it is why blowing a bubble needs a brief burst of extra air pressure from your lungs to get it started, after which the bubble holds its own shape.

Competitive level - finding surface tension from a capillary-rise measurement

Water rises to a height of $3.0\ \text{cm}$ in a capillary tube of radius $0.50\ \text{mm}$ dipped into a wide dish of water. Assuming the angle of contact between water and clean glass is $0°$ (so $\cos\theta = 1$ ), find the surface tension of water. Take $\rho = 1000\ \text{kg/m}^3$ and $g = 9.8\ \text{m/s}^2$ . As a follow-up, find the new rise if an identical tube of half the radius is used instead, with the same liquid.

Step 1. Start from the capillary-rise formula and rearrange it to make $S$ the subject.

h = \frac{2S\cos\theta}{\rho g r} \quad\Rightarrow\quad S = \frac{h\rho g r}{2\cos\theta}

Step 2. Convert every given quantity to SI base units: $h = 3.0\ \text{cm} = 0.03\ \text{m}$ , $r = 0.50\ \text{mm} = 5.0\times10^{-4}\ \text{m}$ .

Step 3. Substitute into the rearranged formula, with $\cos\theta = 1$ .

S = \frac{0.03 \times 1000 \times 9.8 \times 5.0\times10^{-4}}{2 \times 1}

Step 4. Evaluate the numerator first: $0.03 \times 1000 \times 9.8 \times 5.0\times10^{-4} = 0.147$ .

Step 5. Divide by 2 to get the final value: $S = 0.147 / 2 = 0.0735\ \text{N/m}$ . This gives a surface tension of about 0.0735 N/m, satisfyingly close to the accepted textbook value for water at room temperature, about $0.072\ \text{N/m}$ , confirming that the capillary-rise method is a genuine, classroom-feasible way to measure surface tension.

Step 6. Now use $h \propto 1/r$ for the follow-up, holding $S$ , $\rho$ , $\theta$ fixed for the same liquid and tube material. Halving the radius from $0.50\ \text{mm}$ to $0.25\ \text{mm}$ doubles the rise: $h' = 2 \times 3.0\ \text{cm} = 6.0\ \text{cm}$ . The thinner tube pulls the water up to 6.0 cm — no need to redo the full calculation once you recognise the inverse proportionality. Timing note: In a competitive exam, once you have identified $h \propto 1/r$ (with $S$ , $\rho$ , $\theta$ unchanged), a radius-ratio question like Step 6 should take under 15 seconds — do not re-derive $S$ and resubstitute; scale $h$ directly by the inverse of the radius ratio.

Common traps

Using $\Delta P = 2S/r$ for a soap bubble instead of $\Delta P = 4S/r$ — forgetting the bubble has two surfaces (inner and outer film), unlike a simple liquid drop, which has only one liquid-air surface.
Forgetting the $\cos\theta$ factor in the capillary-rise formula, or getting the sign wrong for a liquid such as mercury that shows capillary depression rather than rise.
Believing surface tension is just "stickiness" or the same thing as viscosity — it is a surface-specific effect arising from unbalanced intermolecular forces right at the liquid's boundary, and it is unrelated to viscosity, which is an internal-friction effect throughout the bulk of a flowing fluid.
Assuming a narrower capillary gives a smaller rise because "less space for water to climb" — actually $h \propto 1/r$ , so a narrower tube gives a larger rise, the single most commonly misremembered relation in this topic.
Treating surface tension as constant with temperature — it decreases as temperature rises, because weaker thermal-agitated attraction between surface molecules means a weaker net inward pull.

Common trapA narrower capillary tube gives a larger rise, not a smaller one

The instinct many students carry into this topic is that a thinner tube should struggle to lift liquid, the way a narrow straw feels harder to drink through — so a smaller radius must mean a smaller rise. The force-balance derivation shows the opposite is true, and the reason is a mismatch in how the two competing effects scale with radius. The upward pull from surface tension acts only along the circumference of the contact circle, which grows as one power of $r$ ; the weight that has to be lifted to reach a given height grows as the cross-sectional area, two powers of $r$ . Because the lifting force is "weaker" in its scaling than the load it must support, the equilibrium height has to fall as $r$ increases and rise as $r$ shrinks — giving $h \propto 1/r$ , with $r$ sitting in the denominator of the formula, not the numerator. A capillary tube fine enough to be called a hair can lift water several centimetres, which is precisely the mechanism (scaled up through many fine vessels in wood and paper fibres) behind capillary action in plant stems and paper towels.

You should be able to…

Explain surface tension at the molecular level (unbalanced inward pull on surface molecules) and connect it to surface energy.
Define angle of contact and explain wetting versus non-wetting behaviour, contrasting water on glass with mercury on glass.
Apply $\Delta P = 2S/r$ (drop) and $\Delta P = 4S/r$ (bubble) correctly, and explain in words why a bubble's factor is double a drop's.
Derive the capillary-rise formula $h = 2S\cos\theta/(\rho g r)$ from a force balance between the surface-tension pull around the wetted circumference and the weight of the raised liquid column.
Apply and rearrange $h = 2S\cos\theta/(\rho g r)$ to solve for any one unknown, including using it to measure $S$ experimentally.
Explain why capillary rise increases as tube radius decreases, using the different scaling of force ( $\propto r$ ) and weight ( $\propto r^2$ ) with radius.

Ask the tutor: Want a walkthrough of why the surface-tension force in the derivation only counts its vertical component, or help predicting the exact capillary rise for a liquid and tube you haven't seen before? Ask the AI tutor to work through it with you.

Why does surface tension make drops spherical instead of some other shape?

Surface tension makes a liquid surface behave as though it is trying to shrink to the smallest possible area, because increasing surface area costs energy (surface energy) that the system has no reason to spend unless something forces it to. For a fixed volume of liquid, the shape with the least possible surface area is a sphere — this is a purely geometric fact, true for any blob of any material. So in the absence of other forces strong enough to distort it (gravity flattens larger drops, air resistance distorts falling raindrops), a small liquid drop settles into a sphere simply because that is the lowest-surface-area, lowest-surface-energy shape it can take.

Is surface tension the same thing as adhesion, or as viscosity?

No, and mixing these up costs marks. Surface tension is a cohesive effect — it comes from liquid molecules attracting other liquid molecules at the surface, and it exists even for a liquid sitting in a vacuum with nothing to touch. Adhesion is different: it is the attraction between a liquid's molecules and a different material's molecules (like water and glass), and it is what decides the angle of contact and whether a liquid wets a surface at all. Viscosity is different again — it is an internal-friction effect that resists relative motion between layers of a flowing liquid, has nothing to do with the surface specifically, and is governed by Stokes' law and terminal velocity, not by $S = F/l$ . All three matter in this chapter, but they are three distinct physical effects.

Why does surface tension decrease when you heat a liquid?

Surface tension exists because surface molecules feel a net inward pull from their neighbours below and beside them. Heating a liquid gives every molecule more kinetic energy, so each one jostles about more vigorously and spends, on average, more time slightly farther from its neighbours — weakening the net attractive pull that produces the tension in the first place. This is a completely general trend: nearly every liquid's surface tension falls smoothly as temperature rises, right up to the critical temperature, where the distinction between liquid and vapour disappears altogether and surface tension drops to zero. It is also the practical reason warm soapy water blows bubbles more easily and washes greasy dishes better than cold water — lower surface tension lets the liquid spread and wet surfaces more readily.

Why does the exam ask us to derive the capillary-rise formula using force balance instead of just quoting it?

Because the derivation is exactly where the physical reasoning of this whole chapter gets tested, not just formula recall. Quoting $h = 2S\cos\theta/(\rho g r)$ proves you memorised a result; reconstructing it from "the vertical component of the surface-tension pull around the wetted circumference supports the weight of the raised column" proves you understand why a curved meniscus can lift liquid at all, and it is exactly this reasoning — force per unit length times a length equals a force, set equal to a weight — that lets you predict, without any new formula, that $h \propto 1/r$ or that mercury shows depression instead of rise. Board exams frequently award full marks only when each force-balance step (tension force, its vertical component, the weight of the column, and the final algebra) is shown explicitly.