SaitechAI — Capillarity (Surface Tension in Tubes)

Concept

When a fine tube touches a liquid surface, surface tension and wetting cause the liquid to rise (water in clean glass) or fall (mercury in glass). For a vertical tube of radius $R$ at equilibrium, the height $h$ is

Core formula

$$h=\frac{2S\cos\theta}{\rho g R}$$

$S$ = surface tension (N m⁻¹), $\rho$ = density (kg m⁻³), $g$ = $9.8$–$9.81$ m s⁻², $R$ = tube radius (m), $\theta$ = contact angle.
Rise if $\theta<90^\circ$ (wetting, $\cos\theta>0$); fall if $\theta>90^\circ$ (non-wetting, $\cos\theta<0$).
Formula assumes a circular tube and a static meniscus; $R\!\ll$ capillary length.

Derivation sketch. Vertical force balance on the meniscus: upward component of surface tension around the rim $= 2\pi R S\cos\theta$; weight of the liquid column $= \rho g(\pi R^2 h)$. Equate and simplify to get the formula above.

Common conversions

Quantity	Convert
Radius	$\;1~\text{cm}=10^{-2}\text{ m},\;1~\text{mm}=10^{-3}\text{ m}$
Density	$\rho_{\text{water}}\approx 10^3\ \text{kg m}^{-3}$, $\rho_{\text{Hg}}=13.6\times10^3$
Angle	$\cos 0^\circ=1,\;\cos 60^\circ=\tfrac12$

Proportionalities

$h\propto S$ (more surface tension → more rise).
$h\propto \dfrac{1}{\rho}$ and $h\propto \dfrac{1}{R}$ (heavier liquid or wider tube → smaller rise).
Sign of $h$ given by $\cos\theta$ (rise $+$, fall $-$).

Example 1

Water rise in a narrow tube

“A capillary of radius $0.05$ cm is dipped in water. Find the rise if $S=0.073$ N/m and the contact angle is $0^\circ$.”

Solution steps

Convert $R=0.05\ \text{cm}=5\times10^{-4}\ \text{m}$.
Use $\rho=10^3\ \text{kg m}^{-3}$, $g=9.8\ \text{m s}^{-2}$, $\cos0^\circ=1$.
$$h=\frac{2(0.073)(1)}{(10^3)(9.8)(5\times10^{-4})}$$
$$h\approx 0.0298\ \text{m} = 2.98\ \text{cm}$$

Answer: $h \approx 0.0298\ \text{m}$ (≈ 3.0 cm rise).

Example 2

Ratio of surface tensions: mercury vs water

“In the same glass capillary, water rises to $10.0$ cm while mercury falls by $5.0$ cm. If $\theta_{\text{water}}=0^\circ$ and $\theta_{\text{Hg}}=60^\circ$, find $S_{\text{Hg}}:S_{\text{water}}$.”

Solution steps

For a fixed tube, $S=\dfrac{h\,\rho\,g\,R}{2\cos\theta} \ \Rightarrow \ \dfrac{S_{\text{Hg}}}{S_{\text{water}}}=\dfrac{h_{\text{Hg}}\rho_{\text{Hg}}\cos\theta_{\text{water}}}{h_{\text{water}}\rho_{\text{water}}\cos\theta_{\text{Hg}}}$.

$h_{\text{water}} = 10.0\ \text{cm}=0.10\ \text{m}$ (rise)
$h_{\text{Hg}} = 5.0\ \text{cm}=0.05\ \text{m}$ (use magnitude)
$\rho_{\text{water}}=10^3$, $\rho_{\text{Hg}}=13.6\times10^3$ kg m⁻³
$\cos 0^\circ=1$, $\cos 60^\circ=\tfrac12$

$$\frac{S_{\text{Hg}}}{S_{\text{water}}}=\frac{0.05\times 13.6\times10^3\times 1}{0.10\times 10^3\times \tfrac12}=13.6$$

Answer: $S_{\text{Hg}}:S_{\text{water}} = \boxed{13.6:1}$.

Try-it-yourself (guided template)

Use these mini-prompts to practice. Reveal steps after attempting.

Practice ARise with different radius

An ethanol column in a glass tube ($\rho=790\ \text{kg m}^{-3}$, $S=0.022\ \text{N m}^{-1}$, $\theta=0^\circ$) has $R=0.30$ mm. Compute $h$.

Show steps

Convert $R=3.0\times10^{-4}$ m. Use $h=\dfrac{2S}{\rho g R}$.

$$h=\frac{2(0.022)}{(790)(9.8)(3.0\times10^{-4})}\approx 0.019\ \text{m}=1.9\ \text{cm}$$

Practice BEffect of contact angle

Water in a tube of $R=0.25$ mm has a contact angle $\theta=30^\circ$. With $S=0.073$ N/m, find $h$.

Show steps

$\cos30^\circ=\sqrt{3}/2$. Put in $h=\dfrac{2S\cos\theta}{\rho g R}$ with $\rho=10^3$.

$$h=\frac{2(0.073)\left(\frac{\sqrt3}{2}\right)}{(10^3)(9.8)(2.5\times10^{-4})}\approx 0.0206\ \text{m}=2.06\ \text{cm}$$

Quick checks

If the same liquid is used in two tubes with radii $R_1$ and $R_2$, show that $\dfrac{h_1}{h_2}=\dfrac{R_2}{R_1}$.
Why does mercury depress in a clean glass tube?
How would impurities that reduce $S$ affect capillary rise?