Dip one corner of a cotton handkerchief in a glass of water. Within minutes the wet patch climbs several centimetres up the cloth, against gravity. Look closely at the water in the glass. Right at the glass, the surface curves up a little, as if the water were trying to climb the wall. In a mercury barometer the surface does the opposite: it curves down at the glass, like the top of a dome.
Cohesion against adhesion
A molecule of liquid right next to a solid wall feels two kinds of pull. Its fellow molecules of the liquid pull it with cohesion. The molecules of the solid pull it with adhesion, the attraction between molecules of different kinds. Which of the two is stronger decides how the liquid meets the wall.
For water and glass, adhesion is stronger. The water near the wall is drawn up the glass, and the surface curves up to meet it: the surface is concave, hollow on top. For mercury and glass, cohesion is far stronger. The mercury pulls away from the glass, and its surface curves down at the wall: it is convex, domed on top.
The angle of contact \(\theta\) is measured where a liquid surface meets a solid, through the liquid. It is the angle between the solid surface and the tangent to the liquid surface.
A liquid with \(\theta < 90^\circ\) is said to wet the solid. Water on clean glass has \(\theta\) close to \(0^\circ\), and it spreads into a thin film. Water on a waxed plate has about \(105^\circ\), mercury on glass about \(135^\circ\), and water on a lotus leaf about \(160^\circ\). With \(\theta > 90^\circ\), the liquid does not wet the solid, and a drop stands up as a bead. Fig 5.1 shows all four with drops of one volume.
The angle can be changed. A little detergent lowers the surface tension of water and helps it wet glass, so \(\theta\) falls. Warming also lowers the surface tension, and with it the angle of contact. Grease or wax on the glass does the opposite, and water on greasy glass beads up.
Mercury on clean glass has an angle of contact of about \(135^\circ\). What does this tell you?
The climb up a narrow tube
Dip a narrow glass tube, a capillary, into water. The water inside rises above the level outside and stays there. This is capillary rise. Along the whole rim where the water surface meets the glass, the glass pulls the surface up. Each metre of rim pulls with \(T\), along the liquid surface, at the angle \(\theta\) to the wall, as in Fig 5.2.
The sideways parts of the pull cancel round the circle. The upward parts add: \(T\cos\theta\) for each metre of rim, over a rim of length \(2\pi r\). This lifts the column of water, of height \(h\) and volume \(\pi r^2 h\), whose weight is \(\pi r^2 h \rho g\).
\[ F_{\text{up}} = 2\pi r \cdot T\cos\theta \]
Why. the rim is a circle of length \(2\pi r\), and each metre of it pulls up with \(T\cos\theta\).
\[ W = \pi r^2 h\,\rho g \]
Why. the column above the outside level is a cylinder of radius \(r\) and height \(h\).
\[ 2\pi r T\cos\theta = \pi r^2 h\,\rho g \]
Why. the column is at rest, so the pull balances the weight.
\[ h = \dfrac{2T\cos\theta}{\rho g r} \]
Read it. divide both sides by \(\pi r^2 \rho g\).
For water in clean glass, \(\theta = 0\) and \(\cos\theta = 1\). In a tube of radius \(0.1\,\text{mm}\), \(h\) \(= 2 \times 0.07/(1000 \times 10 \times 10^{-4})\) \(= 0.14\,\text{m}\), a climb of \(14\,\text{cm}\). The radius is in the denominator, so \(h r\) is fixed for a given liquid and tube material: \(1.4 \times 10^{-5}\,\text{m}^2\) for water. A tube twice as wide lifts water half as high: \(7\,\text{cm}\) at \(0.2\,\text{mm}\), \(2.8\,\text{cm}\) at \(0.5\,\text{mm}\), \(1.4\,\text{cm}\) at \(1\,\text{mm}\).
The same result from pressure
The same law follows from the excess pressure of lesson 3. The meniscus of water in the tube is part of a sphere, curved with its hollow side up. Its radius \(R\) is set by the tube and the angle of contact: \(r = R\cos\theta\). Across a curved surface, the pressure is higher on the hollow side, by \(2T/R\). Here the hollow side is the air above, so the water just under the meniscus is at \(P_0 - 2T/R\), below atmospheric.
\[ P_{\text{under}} = P_0 - \dfrac{2T}{R} = P_0 - \dfrac{2T\cos\theta}{r} \]
Why. the air is on the hollow side of the meniscus, and \(R = r/\cos\theta\).
?
Your turn. which line fixes the height of the column?
\[ h = \dfrac{2T\cos\theta}{\rho g r} \]
Read it. substitute step 1 into step 2 and solve for \(h\).
The two ways agree, as they must. For the \(0.1\,\text{mm}\) tube, the step under the hemispherical meniscus is \(2 \times 0.07/10^{-4} = 1400\,\text{Pa}\). A column of water \(14\,\text{cm}\) tall adds \(1000 \times 10 \times 0.14 = 1400\,\text{Pa}\). The column rises exactly until its weight makes up the pressure lost across the curved surface.
The mass that is lifted
A narrow tube lifts water higher, but a wide tube lifts more water. The mass of the column is \(m = \pi r^2 h \rho = \pi r^2 \rho \times 2T\cos\theta/(\rho g r)\), which simplifies to \(m = 2\pi r T\cos\theta/g\). It grows with the radius, because the lifting force \(2\pi r T\cos\theta\) grows with the rim. The \(0.1\,\text{mm}\) tube holds up \(2\pi \times 10^{-4} \times 0.07/10\) \(\approx 4.4 \times 10^{-6}\,\text{kg}\) of water, about \(4\,\text{mg}\).
Water rises \(14\,\text{cm}\) in a glass tube of radius \(0.1\,\text{mm}\). Find the rise in the same tube for two other liquids. Liquid (a) has the same \(T\) and density, but \(\theta = 60^\circ\). Liquid (b) is soap solution with \(T = 0.03\,\text{N/m}\), the density of water and \(\theta = 0\).
(a) \(h \propto \cos\theta\), and \(\cos 60^\circ = 0.5\), so \(h = 14 \times 0.5 = 7\,\text{cm}\).
(b) \(h \propto T\), so \(h = 14 \times 0.03/0.07 = 6\,\text{cm}\).
A clean glass tube of radius \(0.25\,\text{mm}\) dips into water. Take \(T = 0.07\,\text{N/m}\), \(\theta = 0\), \(\rho = 1000\,\text{kg/m}^3\) and \(g = 10\,\text{m/s}^2\). How high does the water rise, in centimetres?
A liquid of density \(800\,\text{kg/m}^3\) rises \(7\,\text{cm}\) in a capillary of radius \(0.1\,\text{mm}\), with an angle of contact of zero. Take \(g = 10\,\text{m/s}^2\). Find its surface tension, in N/m.
Water rises to a height \(h\) in a clean glass capillary. Which statements are true?
Mercury goes down
Now dip a glass tube into mercury. Its angle of contact is about \(135^\circ\), and \(\cos 135^\circ \approx -0.71\) is negative. Jurin's law then gives a negative \(h\): the mercury in the tube stands below the level outside. This is capillary depression. The meniscus is convex, domed on top. Its hollow side now faces down, into the mercury. So the mercury just under it is above atmospheric pressure, and the column sinks until the pressures match.
For a glass tube of radius \(0.5\,\text{mm}\), with \(T = 0.5\,\text{N/m}\) and \(\rho = 13\,600\,\text{kg/m}^3\) for mercury, the top of Jurin's law is \(2T\cos\theta = 2 \times 0.5 \times (-0.707)\) \(= -0.707\,\text{N/m}\). The bottom is \(\rho g r = 13\,600 \times 10 \times 5 \times 10^{-4}\) \(= 68\,\text{N/m}^2\). So \(h = -0.707/68\) \(\approx -0.0104\,\text{m}\). The mercury stands about \(1.04\,\text{cm}\) below the outside level. The second chip of the sim above shows it.
A glass tube of radius \(1\,\text{mm}\) dips into mercury. Take \(T = 0.5\,\text{N/m}\), \(\theta = 135^\circ\), \(\rho = 13\,600\,\text{kg/m}^3\) and \(g = 10\,\text{m/s}^2\). How far below the outside level does the mercury in the tube stand, in centimetres?
The column of a mercury barometer is always a little lower than it should be, because of capillary depression. In a tube of radius \(5\,\text{mm}\), the error is about \(1\,\text{mm}\) of mercury, already a noticeable part of the \(760\,\text{mm}\) column. In a narrow tube it would be far worse. So barometer tubes are made wide, and careful readings add a small correction for the depression.
James Jurin was a London physician and a Fellow of the Royal Society, where he also served as its secretary. Others had seen water climb in thin glass tubes, but no one had found the rule. In 1718 Jurin dipped tubes of different bores into water and measured the heights. He showed that the height is inversely proportional to the width of the tube. He argued that the lift comes from the ring of glass at the top of the column. The law \(h r = \text{constant}\) still carries his name.
- Measuring the angle of contact outside the liquid.Measure \(\theta\) through the liquid, from the solid to the tangent of the liquid surface.
- Using the diameter for \(r\) in Jurin's law.\(r\) is the inner radius of the tube.
- Thinking a wider tube lifts less water.It lifts water less high, but more of it: \(m = 2\pi r T\cos\theta/g\).
- Forgetting the sign of \(\cos\theta\) for mercury.For \(\theta > 90^\circ\), \(\cos\theta < 0\) and the column is depressed.