A U-shaped wire is dipped in a soap solution, and removed. The thin soap film formed between the wire and the light slider supports a weight of $1.5 \times 10^{-2}\ \text{N}$ (which includes the small weight of the slider). The length of the slider is $30\ \text{cm}$. What is the surface tension of the film?

A U-shaped wire is dipped in a soap solution, and removed. The thin soap film formed between the wire and the

Exercises · 9.17

1.5 \times 10^{-2}\ \text{N}

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Concept understanding — Capillary Action

Capillary Action: The Physics of Water Defying Gravity

You've seen it happen a hundred times. Dip the corner of a paper towel into a spill, and watch the water climb upward into the towel — against gravity. Or take a thin glass tube (a capillary), put it in water, and the water rises inside it, higher than the outside surface. That's capillary action.

The name comes from capillus, Latin for "hair" — because the effect is strongest in tubes as thin as a hair.

The Intuition: Two Forces at War

Think of water molecules as tiny magnets. They attract each other strongly (cohesion), and they also stick to the walls of a container (adhesion). In a narrow tube, adhesion to the glass pulls the water upward along the walls. The water molecules, clinging to each other, drag the entire column up with them.

But gravity pulls down. The water rises only until the upward pull from adhesion is exactly balanced by the weight of the water column. That's the equilibrium height.

Note

If adhesion is stronger than cohesion (water-glass), the liquid rises. If cohesion is stronger (mercury-glass), the liquid is depressed — it falls below the outside level. Mercury doesn't wet glass.

The Precise Physics: What Determines the Height?

For a liquid that wets the tube (contact angle $\theta < 90^\circ$ ), the rise height $h$ is given by the Jurin's law:

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

Where:

$\gamma$ = surface tension of the liquid (N/m)
$\theta$ = contact angle between liquid and tube wall
$\rho$ = density of the liquid (kg/m³)
$g$ = acceleration due to gravity (9.8 m/s²)
$r$ = radius of the tube (m)

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

The key insight: $h$ is inversely proportional to $r$ . Halve the tube radius, and the water rises twice as high. That's why the effect is only noticeable in very narrow tubes — in a wide pipe, $h$ is negligible.

Why Does Surface Tension Pull Upward?

The surface tension $\gamma$ acts along the circumference of the water-air interface inside the tube. The total upward force is:

F_{\text{up}} = (2\pi r) \times \gamma \cos\theta

The weight of the water column (height $h$ , density $\rho$ ) is:

F_{\text{down}} = (\pi r^2 h) \times \rho g

Set them equal, cancel $\pi r$ , and you get Jurin's law.

Tip

For water in clean glass, $\theta \approx 0^\circ$ (perfect wetting), so $\cos\theta = 1$ . Then $h \approx \frac{2\gamma}{\rho g r}$ . For water at room temperature, $\gamma \approx 0.073$ N/m, so $h \approx \frac{0.015}{r}$ (with $r$ in metres). A tube of radius 0.1 mm gives a rise of about 15 cm.

Real-World Examples

Paper towels and sponges: The fibres form millions of tiny capillary channels. Water rises through them, soaking the towel. …

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The weight is supported by two film surfaces, each exerting a force $T L$ along the slider. Equating $2 T L = mg$ gives $T = \frac{1.5\times10^{-2}}{2\times0.30} = 2.5\times10^{-2}\ \text{N/m}$ .

The key insight here is that a soap film has two free surfaces — one on the front and one on the back of the film. Each surface contributes its own surface tension force along the length of the slider. So when you pull the slider, you're stretching two surfaces, not one.

Think of it this way: surface tension acts along the line where the film meets the slider, pulling inward on both sides. For a single surface, the force is $T L$ , where $L$ is the length of the slider. With two surfaces, the total upward force on the slider is $2 T L$ . That force balances the weight hanging from the slider.

Let's work through the numbers.

Write the force balance. The weight $mg = 1.5 \times 10^{-2}\ \text{N}$ is supported by the total surface tension force:

2 T L = mg

Convert length to SI units.

The slider length is $30\ \text{cm} = 0.30\ \text{m}$ .
Solve for $T$ .

T = \frac{mg}{2L} = \frac{1.5 \times 10^{-2}}{2 \times 0.30}

Compute the denominator: $2 \times 0.30 = 0.60$ .

Then: …

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