Surface Tension and Capillary Rise

Surface tension is the reason a paperclip can float on water, mosquitoes can walk across a pond, and water rises up a thin straw without anyone sucking on it. It’s a small but high-yield chapter in Class 11 — appears every year in CBSE and gives 1–2 questions in NEET. JEE Main occasionally pulls a tricky numerical involving capillary rise in a non-vertical tube or two-liquid system.

We’ll build up from the molecular picture, get the formulas down with the right sign conventions, and then work through problems graded easy to JEE-Advanced.

What Surface Tension Actually Is

Inside a liquid, every molecule is pulled equally in all directions by its neighbours — the net force is zero. At the surface, molecules have neighbours below and beside them but none above, so the net pull is downward. To pull a molecule from the bulk to the surface requires work — that work shows up as surface energy, and the same effect manifests as a tangential force per unit length at the surface, called surface tension $T$ (or $\sigma$ ).

Surface tension $T$ is defined as the force per unit length acting tangent to the surface, perpendicular to any line drawn on the surface. SI unit: N/m. Equivalently, it equals the surface energy per unit area in J/m².

For water at 20°C, $T = 0.0728$ N/m. For mercury, $T = 0.465$ N/m. The number is small, but pressure differences and capillary heights are very sensitive to it.

Key Terms & Definitions

Cohesion — Attraction between molecules of the same substance (water-water).

Adhesion — Attraction between molecules of different substances (water-glass, water-wax).

Angle of contact $\theta$ — The angle between the liquid surface and the solid wall, measured inside the liquid. For water-glass, $\theta \approx 0$ (water wets glass). For mercury-glass, $\theta \approx 140^\circ$ (mercury doesn’t wet glass).

Meniscus — The curved liquid surface at the wall. Concave when adhesion > cohesion (water in glass tube), convex when cohesion > adhesion (mercury in glass).

Capillarity — The rise or fall of a liquid in a narrow tube due to surface tension and the angle of contact.

Core Formulas

Force on a wire of length $L$ pulled across a film: $F = 2TL$ (factor of 2 because film has two surfaces).

Excess pressure inside a liquid drop: $\Delta P = \dfrac{2T}{R}$ .

Excess pressure inside a soap bubble: $\Delta P = \dfrac{4T}{R}$ (two surfaces — inner and outer).

Capillary rise height: $h = \dfrac{2T\cos\theta}{r\rho g}$ .

Where $r$ is the tube radius, $\rho$ is the liquid density, $g$ is gravitational acceleration.

The capillary formula is the workhorse — almost every problem reduces to it.

Methods & Concepts

Deriving the Capillary Rise Formula

The vertical component of surface-tension force around the circumference of the meniscus must support the weight of the lifted column of liquid.

Force pulling liquid up: $F_{\text{up}} = T \cos\theta \times 2\pi r$ .

Weight of lifted column: $W = (\pi r^2 h)\rho g$ .

Setting $F_{\text{up}} = W$ :

2\pi r T \cos\theta = \pi r^2 h \rho g \Rightarrow h = \frac{2T\cos\theta}{r\rho g}

Two takeaways: (1) $h$ is inversely proportional to $r$ — narrower tubes give higher rise, (2) when $\theta > 90^\circ$ (mercury), $\cos\theta < 0$ , giving negative $h$ , i.e., depression.

Excess Pressure — Drops vs Bubbles

A liquid drop has one surface (water-air boundary) — excess pressure is $2T/R$ .

A soap bubble in air has two surfaces (inner and outer of the soap film) — excess pressure is $4T/R$ .

A bubble inside a liquid (like an air bubble in water) again has one surface — back to $2T/R$ . Students mix these up constantly. Always count the number of liquid-gas interfaces.

Surface Energy and Work

When a soap film of area $A$ is stretched by $dA$ , work done = $2T \, dA$ (factor of 2 because two surfaces). Equivalently, the surface energy per unit area is $T$ .

This explains why small drops merge into a big drop: a single big drop has less total surface area than many small ones, so combining lowers surface energy.

Solved Examples

Easy (CBSE level) — Capillary rise

Q. A capillary tube of internal diameter 0.4 mm is dipped vertically in water. To what height does water rise? Take $T = 0.073$ N/m, $\theta = 0$ , $\rho = 1000$ kg/m³, $g = 9.8$ m/s².

Solution. Radius $r = 0.2$ mm $= 2 \times 10^{-4}$ m.

h = \frac{2 \times 0.073 \times 1}{2 \times 10^{-4} \times 1000 \times 9.8} = \frac{0.146}{1.96} \approx 0.0745 \text{ m} \approx 7.45 \text{ cm}

Medium (JEE Main level) — Combining drops

Q. Eight identical small water droplets each of radius 1 mm coalesce into a single big drop. Find the ratio of (a) total surface area before to after, (b) total surface energy before to after. Surface tension of water $= 0.072$ N/m.

Solution. By volume conservation, $8 \cdot \tfrac{4}{3}\pi r^3 = \tfrac{4}{3}\pi R^3 \Rightarrow R = 2r = 2$ mm.

Initial total area: $8 \times 4\pi r^2 = 32\pi r^2$ . Final: $4\pi R^2 = 16\pi r^2$ . Ratio = $2 : 1$ .

Energy ratio is the same as area ratio (energy = $T \times$ area). Surface energy is halved when 8 drops merge into 1 — and the released energy heats the drop slightly.

Hard (JEE Advanced level) — Capillary in a tilted tube

Q. A capillary tube of radius 0.5 mm is held inclined at 60° to the horizontal in water (T = 0.073 N/m, $\theta = 0$ ). Find the length of the water column in the tube, and the vertical rise of the water.

Solution. Vertical rise $h$ doesn’t depend on tube orientation:

h = \frac{2T}{r\rho g} = \frac{2 \times 0.073}{5 \times 10^{-4} \times 1000 \times 9.8} \approx 0.0298 \text{ m} \approx 2.98 \text{ cm}

Length along the tube: $L = h/\sin 60^\circ = 2.98/0.866 \approx 3.44$ cm. The vertical rise is the same as in a vertical tube — the tube just provides a longer path for the same lift.

Exam-Specific Tips

CBSE: Memorise $h = 2T\cos\theta/(r\rho g)$ . The 2-mark question is almost always plug-and-chug. The 3-mark question typically asks you to derive this formula starting from the force-balance argument.

JEE Main: Watch for soap bubble vs water drop distinction. The factor of 2 in $\Delta P = 4T/R$ for bubbles vs $2T/R$ for drops is the most asked single fact.

NEET: Conceptual MCQs on angle of contact and meniscus shape. Know that water + clean glass gives $\theta \approx 0$ , water + wax gives $\theta > 90^\circ$ , mercury + glass gives $\theta \approx 140^\circ$ .

Common Mistakes to Avoid

1. Wrong factor in bubble vs drop. Always check: how many liquid-gas surfaces does this object have?

2. Forgetting $\cos\theta$ . When $\theta = 0$ , $\cos\theta = 1$ and the term disappears — but for mercury problems it’s essential.

3. Confusing radius and diameter. Capillary problems often state diameter; you need radius. Halve before plugging in.

4. Wrong sign for mercury depression. Mercury falls in a glass tube. The formula gives negative $h$ when $\theta > 90^\circ$ — keep the sign or report magnitude with “depression.”

5. Treating surface energy as ordinary energy. Surface energy is reversible work to create a surface — it’s not heat or kinetic. Combining drops releases surface energy as heat, but that’s a separate calculation.

Practice Questions

Q1. A tube of radius 0.4 mm rises water by 3.6 cm. Find surface tension. ( $\rho = 1000$ , $g = 10$ , $\theta = 0$ .)

$T = h r \rho g / 2 = (0.036)(4 \times 10^{-4})(1000)(10)/2 = 0.072$ N/m.

Q2. Excess pressure inside a 4-mm-diameter water drop. ( $T = 0.073$ N/m.)

$\Delta P = 2T/R = 2(0.073)/0.002 = 73$ Pa.

Q3. Why does a soap bubble eventually burst?

The soap film thins as liquid drains under gravity, eventually reaching molecular thickness and rupturing. Surface tension itself doesn’t cause bursting — it actually holds the bubble together until evaporation/drainage thins the film below stability.

Q4. Two soap bubbles of radii 3 cm and 4 cm coalesce keeping a common interface. Find radius of the common surface.

For the common film between two bubbles: $\frac{1}{R_c} = \frac{1}{r_1} - \frac{1}{r_2} = \frac{1}{3} - \frac{1}{4} = \frac{1}{12}$ . So $R_c = 12$ cm.

Q5. A capillary tube is broken at half its rise height. What happens to the water?

Water rises only to the broken end (it cannot rise above the tube). The angle of contact adjusts so that surface tension still balances the column weight — meniscus becomes flatter. No water overflows.

Q6. Find work needed to break a 2-mm spherical drop into 1000 droplets of equal size. ( $T = 0.072$ N/m.)

Volume conservation: $\tfrac{4}{3}\pi R^3 = 1000 \cdot \tfrac{4}{3}\pi r^3 \Rightarrow r = R/10 = 0.1$ mm. Increase in area = $1000 \cdot 4\pi r^2 - 4\pi R^2 = 4\pi(10R^2 - R^2) = 36\pi R^2 = 36\pi (10^{-3})^2 \approx 1.13 \times 10^{-4}$ m². Work = $T \cdot \Delta A \approx 8.14 \times 10^{-6}$ J.

Q7. What’s the angle of contact for water on a perfectly clean glass surface?

$\theta = 0$ . Water completely wets clean glass.

Q8. Why does mercury form a convex meniscus in glass?

Mercury-mercury cohesion is much stronger than mercury-glass adhesion. The liquid pulls itself away from the glass wall, giving angle of contact > 90° and a convex shape.

FAQs

Why does a needle float on water?

The water surface acts like a stretched elastic membrane. The needle’s weight depresses the surface, and the upward component of surface tension around the contact line balances gravity — as long as the needle is light and the surface is undisturbed.

Does surface tension change with temperature?

Yes — it decreases with rising temperature. Hotter molecules have more thermal motion, which weakens the inward cohesive pull at the surface. At the critical temperature of any liquid, surface tension drops to zero.

Why are raindrops spherical?

A sphere has the smallest surface area for a given volume — so a freely falling drop minimises surface energy by becoming spherical. Air drag distorts very large drops into a flattened “hamburger” shape.

Can capillary rise be infinite for a very thin tube?

In theory $h \propto 1/r$ predicts unbounded rise as $r \to 0$ . In practice, the assumption “tube radius is much greater than molecular size” breaks down. Below a few nanometres, continuum surface tension stops being meaningful.

Why don’t insects sink while heavier ducks float?

Insects rely on surface tension — their tiny weight is supported by the curvature of the water surface around their feet. Ducks rely on buoyancy — their average density is less than water. Two completely different effects.

Is capillary action affected by gravity?

Yes — $h \propto 1/g$ . In zero gravity, capillary action would lift liquid until surface tension is balanced by something else (like inertia or container limits). NASA experiments confirm this.