Two mercury droplets of radii 0.1 cm. and 0.2 cm. collapse into one single drop. What amount of energy is rele

NCERT Exemplar · Q18

T = 435.5 \times 10^{-3}

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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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Merging conserves volume, giving a drop of radius $R\approx 2.08\times10^{-3}$ m; the reduced surface area releases $\Delta E\approx 3.7\times10^{-6}$ J.

1. Conserve volume to find the combined radius ( $r_1=0.1$ cm $=1\times10^{-3}$ m, $r_2=0.2$ cm $=2\times10^{-3}$ m):

R^3=r_1^3+r_2^3=(1\times10^{-3})^3+(2\times10^{-3})^3=9\times10^{-9}\ \text{m}^3,

R=(9\times10^{-9})^{1/3}=2.08\times10^{-3}\ \text{m}.

2. Energy released = surface tension $\times$ decrease in area

\Delta E=T\big[4\pi(r_1^2+r_2^2)-4\pi R^2\big]=4\pi T\big(r_1^2+r_2^2-R^2\big).

3. Evaluate the area change …

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Q.Two mercury droplets of radii 0.1 cm. and 0.2 cm. collapse into one single drop. What amount of energy is released? The surface tension of mercury T=435.5×10−3T = 435.5 \times 10^{-3}T=435.5×10−3 N m−1^{-1}−1.

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Q.Two mercury droplets of radii 0.1 cm. and 0.2 cm. collapse into one single drop. What amount of energy is released? The surface tension of mercury $T = 435.5 \times 10^{-3}$ N m $^{-1}$ .

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