Jurin's Law (Capillary Rise)

Equivalent forms

h = \frac{2 \gamma}{\rho g r} \text{ (perfect wetting)}

F_{cap} = 2\pi r \gamma \cos\theta

Surface tension lifts liquids — height scales inversely with tube radius.

Symbol	Name	SI unit	Dimension	Range
$\gamma$	Surface Tension Liquid-gas surface tension	N/m	M·T⁻²	0.001 – 1
$\theta$	Contact Angle Angle between liquid and tube wall	rad	1	0 – 3.14
$\rho$	Density Liquid density	kg/m³	M·L⁻³	0.1 – 14000
$r$	Tube Radius Inner radius of capillary tube	m	L	0.000001 – 0.01
$h$	Capillary Riseoutput Height liquid rises in tube	m	L	-1 – 1

Unit systems

Symbol	SI	CGS	Imperial
$\gamma$	N/m	dyn/cm	lbf/ft
$\theta$	rad	rad	rad
$\rho$	kg/m³	g/cm³	lb/ft³
$r$	m	cm	ft
$h$	m	cm	ft

Where it holds

Static equilibrium of a wetting liquid in a thin tube where tube radius is much smaller than the capillary length

(\sqrt{\gamma /\rho g} \approx 2.7\,\mathrm{mm}

for water). Excludes very wide tubes, evaporation-limited steady-state, and complex porous geometries.

Dimensional analysis

$[\gamma ]/([\rho ][g][r]) = (M\cdot T^{-2})/((M\cdot L^{-3})(L\cdot T^{-2})(L)) = L \checkmark$ (length)

Discovery

James Jurin · 1718

Jurin measured capillary rise in glass tubes of varying bore and discovered the inverse-radius relationship. The cos θ refinement was added later by Young and Laplace as wetting theory matured.

Try this

How does water climb 100 m up the trunk of a redwood without a pump?

A glass capillary tube of radius 0.0001 m is dipped in water (γ=0.072 N/m, ρ=1000 kg/m³, contact angle ≈0°). How high will water rise inside the tube?

Research status: stable

Real-world applications

Water uptake in plant xylem
Wicking in paper towels and fabrics
Capillary chromatography
Oil recovery in reservoir rocks

Common misconceptions

Capillary rise alone does NOT explain water transport in tall trees — transpiration and root pressure are also essential
h depends on contact angle, not just liquid properties — clean glass and water give $\theta \approx 0^{\circ}$
For mercury in glass $(\theta \approx 140^{\circ})$ , the liquid is DEPRESSED, not raised

Experimental verification

Direct measurement in glass capillaries of bore from

10 \mu m

to 1 mm confirms the formula across orders of magnitude. Modern X-ray microtomography in porous rocks confirms the inverse-radius scaling in heterogeneous capillaries.

Derivation

At the meniscus, the surface tension exerts a force

F = \gamma \cdot (2\pi r)\cdot \cos \theta

upward on the liquid column.

This must balance the weight of the column:

\rho \cdot g\cdot \pi r^{2}\cdot h

.

Equating and solving:

h = 2\gamma \cos \theta /(\rho gr)

.

Equivalently, the Young-Laplace pressure jump across the curved meniscus,

\Delta P = 2\gamma \cos \theta /r

, must equal the hydrostatic pressure

\rho gh

of the supported column.

Limiting cases

r \to 0

⟶

h \to \infty

Extremely thin tubes drive enormous rise — limited only by evaporation and tube length.

\theta = 90^{\circ}

⟶

h = 0

Non-wetting condition — no capillary force.

\theta

>

90^{\circ}

⟶ h < 0Liquid is depressed below the bulk level (e.g., mercury in glass).

What if…

What if you use a half-radius tube?

Capillary rise doubles. This is why thin paper tissues wick aggressively while thick cardboard does not.

What if the surface is hydrophobic

(\theta

>

90^{\circ})

?

$\cos \theta$ becomes negative and the liquid is pushed downward instead of rising. This is the principle behind waterproof coatings.

What if you reduce

\gamma

with detergent?

Rise drops proportionally. Detergents kill capillarity — a problem for soils and a feature for textiles.

1

Water in a 0.1 mm glass tube

Given ·

\gamma:: 0.072
\theta:: 0
\rho:: 1000
r:: 0.0001

Find · h

Steps

$h = 2\gamma \cos \theta /(\rho gr)$
Numerator: $2 \times 0.072 \times 1 = 0.144$
Denominator: $1000 \times 9.8 \times 0.0001 = 0.98$
$h \approx 0.147\,\mathrm{m} (14.7\,\mathrm{cm})$

Result ·

h = 2 \times 0.072 \times 1 / (1000 \times 9.8 \times 0.0001) \approx 0.147\,\mathrm{m}

2

Mercury depression in glass

Given ·

\gamma:: 0.485
\theta:: 2.44
\rho:: 13600
r:: 0.001

Find · h

Steps

$\cos (140^{\circ}) \approx -0.766$
$h = 2 \times 0.485 \times (-0.766) / (13600 \times 9.8 \times 0.001)$
$h \approx -5.6\,\mathrm{mm}$ (mercury is pushed DOWN)

Result ·

h = 2 \times 0.485 \times \cos (2.44) / (13600 \times 9.8 \times 0.001) \approx -0.0055\,\mathrm{m}

Young-Laplace Equation→Hydrostatic Pressure→