Capillary rise

Capillarity is the rise or fall of a liquid inside a narrow tube (a capillary) dipped into it, driven by surface tension and the curved meniscus that forms at the liquid's edge inside the tube. Water rises in a fine glass capillary because it wets glass; mercury is depressed because it does not.

Why it happens. In a uniform-bore tube of radius $r$ dipped into a wetting liquid (angle of contact $\theta < 90°$), the liquid inside curves into a concave meniscus, behaving like part of a sphere of radius $R = r/\cos\theta$. The excess-pressure result for a curved surface says the pressure just below this concave meniscus is lower than atmospheric by $2T/R$. That pressure deficit pulls liquid up from the reservoir, and the column keeps rising until its own weight exactly cancels the deficit — equilibrium.

Balancing pressure and weight. At the flat outside surface, pressure is atmospheric, $P_0$. Just below the meniscus inside the tube, pressure is $P_0 - 2T/R$. That point also sits a depth $h$ below the top of the risen column, so hydrostatically $P_0 = (P_0 - 2T/R) + \rho g h$, which rearranges to

where: $T$ is the liquid's surface tension, $\theta$ is the angle of contact with the tube material, $\rho$ is the liquid's density, $g$ is the acceleration due to gravity, and $r$ is the uniform radius of the capillary bore.

Key features. $h \propto 1/r$, so a narrower tube gives a greater rise — capillary effects matter only in fine tubes. If $\theta < 90°$ (wetting), $\cos\theta > 0$ and the liquid rises; if $\theta > 90°$ (non-wetting), $\cos\theta < 0$ and the liquid is depressed. For water in clean glass, $\theta \approx 0°$, so $\cos\theta \approx 1$ and the formula simplifies to $h = 2T/(\rho g r)$.

A clean glass capillary of radius $r = 0.20\ \text{mm}$ is dipped vertically in water, with $T_{\text{water}} = 0.072\ \text{N/m}$, $\rho_{\text{water}} = 1000\ \text{kg/m}^3$, $\theta \approx 0°$, and $g = 9.8\ \text{m/s}^2$.

Step 1. Substitute into $h = 2T\cos\theta/(\rho g r)$ with $\cos 0° = 1$: $h = \dfrac{2 \times 0.072 \times 1}{1000 \times 9.8 \times 0.20\times10^{-3}}$.

Step 2. Evaluate the denominator first: $1000 \times 9.8 \times 0.20\times10^{-3} = 1.96$.

Step 3. Divide: $h = \dfrac{0.144}{1.96} \approx 0.0735\ \text{m}$. Water rises about $7.35\ \text{cm}$ in this fine capillary.

Competitive stretch

The same tube is now dipped in mercury: $T_{\text{mercury}} = 0.465\ \text{N/m}$, $\rho_{\text{mercury}} = 13600\ \text{kg/m}^3$, $\theta = 140°$.

Step 1. $\cos 140° = -\cos 40° \approx -0.766$, negative because mercury is non-wetting.

Step 2. $h = \dfrac{2 \times 0.465 \times (-0.766)}{13600 \times 9.8 \times 0.20\times10^{-3}} = \dfrac{-0.7124}{26.656} \approx -0.0267\ \text{m}$.

Step 3. The negative sign confirms depression, not rise. Mercury is pushed down about $2.67\ \text{cm}$ below the outside level — consistent with mercury not wetting glass.

• Forgetting $\cos\theta$ — the simplified $h = 2T/(\rho g r)$ only holds for $\theta \approx 0°$ (pure water in very clean glass); other liquid-solid pairs need $\cos\theta$, which can flip the sign of $h$.
• Thinking a wider tube gives a bigger rise — $h$ is inversely proportional to $r$, so it is narrower tubes that rise higher.
• Mixing up rise and depression — wetting ($\theta < 90°$) rises, non-wetting ($\theta > 90°$) is depressed; mercury in glass is the standard depression example.
• Using the uniform-bore formula on a tube whose radius changes along its length — the derivation assumes constant $r$, so a tapering or irregular bore needs a different treatment.

Ask the tutor: Want to connect capillary rise back to excess pressure and angle of contact, or try a problem where the tube radius or liquid is unknown and you must invert the formula?