The sufrace tension and vapour pressure of water at 20°C is $7.28\times 10^{-2}$ Nm$^{-1}$ and $2.33\times 10^{3}$ Pa, respectively. What is the radius of the smallest spherical water droplet which can form without evaporating at 20°C?

The sufrace tension and vapour pressure of water at 20°C is 7.28× 10⁻² Nm⁻¹ and 2.33× 10³ Pa, respectively. Wh

NCERT Exemplar · Q20

7.28\times 10^{-2}

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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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Surface tension squeezes a spherical droplet, creating an excess internal pressure of $2T/r$ . A droplet can only exist without immediately evaporating if this excess pressure does not exceed the surrounding vapour pressure — setting these equal gives the smallest stable radius, $r=2T/P\approx6.25\times10^{-5}$ m (about $62.5\ \mu$ m).

Why smaller droplets are more prone to evaporating

A curved liquid surface, due to surface tension, squeezes the liquid inside it, raising the pressure inside relative to outside — this is the Laplace excess pressure. For a spherical droplet of radius $r$ :

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

The smaller the droplet, the larger this excess pressure. A very small droplet is squeezed so hard that its effective vapour pressure exceeds the ambient vapour pressure, and it evaporates rapidly. A droplet can persist (not evaporate away) only if this excess pressure doesn't exceed what the surroundings can sustain — i.e. the ambient vapour pressure $P$ .

Setting up the boundary condition …

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Q.The sufrace tension and vapour pressure of water at 20°C is 7.28×10−27.28\times 10^{-2}7.28×10−2 Nm−1^{-1}−1 and 2.33×1032.33\times 10^{3}2.33×103 Pa, respectively. What is the radius of the smallest spherical water droplet which can form without evaporating at 20°C?

Why smaller droplets are more prone to evaporating

Setting up the boundary condition …

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Q.The sufrace tension and vapour pressure of water at 20°C is $7.28\times 10^{-2}$ Nm $^{-1}$ and $2.33\times 10^{3}$ Pa, respectively. What is the radius of the smallest spherical water droplet which can form without evaporating at 20°C?

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