Capillary rise

Dip the corner of a paper towel into a puddle and watch the water climb, silently, against gravity, staining the paper far above the puddle's surface. Push a fine glass tube into a beaker of water and the water inside the tube rises well above the level outside — sometimes visibly, sometimes by centimetres. No pump, no suction, nothing pushing from below: gravity is pulling down the whole time, and still the liquid climbs. This is capillary rise, and it is not a separate law of nature bolted onto fluid mechanics — it is the direct, inevitable consequence of two ideas already in this chapter, surface tension and the angle of contact, meeting inside a narrow tube. The story is short and satisfying to trace end to end. A wetting liquid inside a narrow tube curls its surface into a bowl-shaped meniscus because it prefers to hug the glass. A curved liquid surface, as the drops-and-bubbles article showed, always carries an excess pressure across it. That excess pressure quietly pulls liquid upward until the weight of the risen column pushes back just hard enough to balance it — and the tube stops rising exactly there. Everything in the final formula, $h = \dfrac{2S\cos\theta}{\rho g r}$ , is a bookkeeping of that one balance. This article rebuilds that balance from the excess-pressure formula for a curved surface, keeps the two Pass-1 worked numbers (7.35 cm for water in a 0.20 mm tube, and the 3.25 cm comparison with a second liquid) and extends them with board-level and competitive-level practice, the sign trap that costs the most marks (mercury falls, it does not rise), and the everyday examples — blotting paper, a lamp wick, tree sap movement, the soil around plant roots — that make this the single most visible surface-tension effect in daily life.

Key takeaways

Capillary rise is excess pressure in disguise. A curved meniscus inside a narrow tube has excess pressure across it, exactly like the excess pressure inside a drop or bubble, and this deficit pulls liquid up until hydrostatic pressure balances it.
The formula: $h = \dfrac{2S\cos\theta}{\rho g r}$ , where $S$ is surface tension, $\theta$ the angle of contact, $\rho$ the liquid density, $g$ gravity, and $r$ the inner radius of the tube.
Narrower tube, greater rise — $h \propto 1/r$ , which is why the effect is called capillary (Latin capillus, "hair") and is dramatic only in hair-thin tubes.
Wetting liquids rise, non-wetting liquids fall. Water in clean glass has $\theta \approx 0°$ so $\cos\theta \approx 1$ and $h$ is positive; mercury in glass has $\theta > 90°$ so $\cos\theta < 0$ and $h$ turns out negative — a real capillary depression, not an error.
The formula works both ways. Given $h$ , $r$ , and $\rho$ from an experiment, you can rearrange it to measure a liquid's surface tension $S$ — this is a standard practical.
It explains everyday phenomena — blotting paper soaking up ink, a cotton wick drawing up oil in a lamp, water moving through the fine capillaries of soil toward plant roots — all the same balance, different geometry.
Key idea: capillary rise balances the excess pressure of a curved meniscus, 2S cos theta over r, against the hydrostatic pressure rho g h of the risen column, giving h = 2S cos theta divided by rho g r, so rise grows as the tube narrows and reverses to depression whenever the angle of contact exceeds 90 degrees.

The core insight

A liquid rises (or falls) in a narrow tube until the hydrostatic pressure of the column, $\rho g h$ , exactly balances the excess pressure of the curved meniscus, $2S\cos\theta / r$ , giving $h = \dfrac{2S\cos\theta}{\rho g r}$ .

Where it fits in the syllabus

This topic closes the CBSE Class 11 Physics unit Mechanical Properties of Fluids, in the surface tension section, right after surface energy, surface tension, angle of contact, and excess pressure across drops and bubbles. Students are expected to derive the capillary rise formula from the excess-pressure idea, apply it numerically (finding $h$ , $r$ , or $S$ ), recognise capillary depression for non-wetting liquids from the sign of $\cos\theta$ , and connect it to everyday examples — all strictly at the algebraic level prescribed for boards; no calculus, no Navier–Stokes, no meniscus-shape derivations beyond the simple spherical-cap picture used here.

The physical picture

Dip a fine glass tube of inner radius $r$ into water. Because water wets clean glass, the angle of contact $\theta$ is small (close to $0°$ ), and the liquid surface inside the tube curls into a concave meniscus that looks like a small slice cut from a sphere. If that sphere has radius $R$ , simple geometry (the tube wall meets the meniscus at angle $\theta$ to the vertical) relates the two radii by

R = \frac{r}{\cos\theta}

where: $r$ is the tube's inner radius (fixed by the glassware) and $R$ is the radius of curvature of the meniscus itself, which is always the larger of the two whenever $\theta \neq 0$ (and $R \to r$ only in the limiting case $\theta \to 0$ ). The physical reading is that a wide angle of contact flattens the meniscus into a shallower bowl with a larger radius of curvature, while $\theta \approx 0°$ gives the tightest, most sharply curved bowl of radius essentially equal to the tube itself. A curved liquid surface, exactly as for the drop and bubble cases already met in this chapter, carries an excess pressure across it because surface tension pulls the surface toward flattening. For this concave meniscus, with air above and liquid below, the pressure just under the surface is lower than atmospheric by

\Delta P = \frac{2S}{R} = \frac{2S\cos\theta}{r}

where: $S$ is the surface tension of the liquid, and the substitution $R = r/\cos\theta$ turns a statement about the meniscus into a statement about the tube's own measurable radius. The physical reading: liquid immediately below this low-pressure zone gets pushed upward by the higher-pressure liquid surrounding it in the wider container, and it keeps rising — dragging the meniscus up with it — until the extra weight of the risen column supplies exactly the pressure needed to cancel this deficit. That balance point is the capillary rise, derived in full below.

Figure

A narrow glass tube dipped in water with a concave meniscus at the top of a risen liquid column of height h, the tube radius r and the meniscus curvature radius R marked, alongside a wider tube showing mercury with a convex meniscus that is depressed below the outer level

The physical picture, continued

Once the column of height $h$ has risen, it exerts its own hydrostatic pressure on the liquid just below the meniscus, given by the familiar $\rho g h$ from the pressure-with-depth article. Rise stops exactly when this hydrostatic term cancels the excess pressure across the meniscus:

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

where: $\rho$ is the liquid's density and $g = 9.8\ \text{m/s}^2$ . This is the balance condition, and solving it for $h$ gives the capillary rise formula derived in full in the next section. Two features fall out immediately without any further algebra. First, $h \propto 1/r$ : halve the tube's radius and the rise doubles, which is why capillary effects are dramatic only in genuinely fine tubes and negligible in a drinking glass. Second, the sign of $\cos\theta$ decides everything — an acute $\theta$ (wetting liquid) gives a positive $h$ and true rise, while an obtuse $\theta$ (non-wetting liquid, like mercury in glass) gives a negative $h$ , meaning the level inside the tube sits below the outer level: a capillary depression.

Derivation ADeriving the capillary rise formula from excess pressure across the meniscus

The goal here is to obtain $h = \dfrac{2S\cos\theta}{\rho g r}$ purely from two ideas already established elsewhere in this chapter — the excess pressure across a curved liquid surface (as used for drops and bubbles) and the hydrostatic pressure of a liquid column — combined through simple geometry relating the meniscus curvature to the tube's own radius and the angle of contact. No new physics is introduced; the whole derivation is bookkeeping the pressures on either side of the meniscus until they must be equal, since a liquid at rest cannot sustain an unbalanced pressure difference indefinitely.

Step 1. Picture the tube of inner radius $r$ dipped into the liquid, with the meniscus curving into a spherical cap of radius $R$ . Because the tube wall meets the liquid surface at the angle of contact $\theta$ , the geometry of a right triangle formed by the tube's centre line, the tube wall, and the line from the centre of curvature to the point of contact gives $r = R\cos\theta$ , so $R = r/\cos\theta$ . This step converts a hard-to-measure quantity (the curvature of a meniscus) into an easy-to-measure one (the tube's own bore, listed on the glassware or measured with a travelling microscope).

Step 2. Apply the excess-pressure-across-a-curved-surface result, the same one used for a spherical drop, to this meniscus: the pressure on the concave (lower-pressure) side is less than on the convex side by $\Delta P = 2S/R$ . A meniscus has only one liquid surface (unlike a soap bubble's two surfaces), so the factor is $2S/R$ , not $4S/R$ . Substituting $R = r/\cos\theta$ from Step 1 gives $\Delta P = \dfrac{2S\cos\theta}{r}$ — the pressure deficit expressed entirely in terms of the tube radius, the surface tension, and the angle of contact.

Step 3. Recognise that this pressure deficit is what pulls liquid up the tube. Immediately below the meniscus, inside the tube, the pressure is atmospheric pressure minus this deficit; immediately outside the tube, at the same horizontal level as the tube's meniscus but in the open liquid, the pressure is simply atmospheric (open to air, undisturbed surface). Liquid therefore flows from the higher-pressure outer region into the lower-pressure region under the meniscus, rising inside the tube. As the column rises to height $h$ above the outer free surface, it adds its own hydrostatic pressure $\rho g h$ at the level of the outer surface, which works to raise the pressure at the base back toward atmospheric.

Step 4. Set the condition for equilibrium: rise stops the moment the added hydrostatic pressure exactly cancels the excess-pressure deficit, so that pressure at any fixed horizontal level inside and outside the tube is once again equal, as it must be for a connected fluid at rest. This gives $\rho g h = \dfrac{2S\cos\theta}{r}$ , the balance equation reached independently in the physical picture above.

Step 5. Solve for $h$ by dividing both sides by $\rho g$ : $h = \dfrac{2S\cos\theta}{\rho g r}$ . This is the capillary rise formula. Every factor has a clear role — $S$ and $\cos\theta$ in the numerator drive the rise (stronger surface tension, more acute wetting angle, taller rise), while $\rho$ , $g$ , and $r$ in the denominator resist it (denser liquid, stronger gravity, or a wider tube all cut the rise down). Because $\cos\theta$ can be negative for $\theta > 90°$ , the formula automatically predicts depression for non-wetting liquids without needing a separate case.

Worked examples

Board level - rise of water in a fine capillary tube

A capillary tube of inner radius $r = 0.20\ \text{mm}$ is dipped vertically into water. Given surface tension of water $S = 7.2\times10^{-2}\ \text{N/m}$ , angle of contact $\theta \approx 0°$ , density of water $\rho = 1000\ \text{kg/m}^3$ , and $g = 9.8\ \text{m/s}^2$ , find the height to which water rises in the tube.

Step 1. Convert the radius to SI units: $r = 0.20\ \text{mm} = 2.0\times10^{-4}\ \text{m}$ . Working entirely in metres, kilograms, and seconds avoids unit-mismatch errors later.

Step 2. Write the capillary rise formula with $\cos 0° = 1$ , since water wets clean glass almost perfectly:

h = \frac{2S\cos\theta}{\rho g r} = \frac{2 \times 7.2\times10^{-2} \times 1}{1000 \times 9.8 \times 2.0\times10^{-4}}

Step 3. Compute the numerator and denominator separately. Numerator: $2 \times 7.2\times10^{-2} = 0.144$ . Denominator: $1000 \times 9.8 \times 2.0\times10^{-4} = 1.96$ .

Step 4. Divide: $h = \dfrac{0.144}{1.96} = 0.0735\ \text{m} = 7.35\ \text{cm}$ .

Water rises about $7.35\ \text{cm}$ in this fine tube — a substantial climb from a tube barely wider than a human hair, which is exactly the point of the name "capillary."

Competitive level - comparing the rise of two liquids in the same tube

Water rises to a height of $6.0\ \text{cm}$ in a certain capillary tube. The same tube is now dipped into a second liquid of density $\rho_2 = 800\ \text{kg/m}^3$ , surface tension $S_2 = 3.6\times10^{-2}\ \text{N/m}$ , and angle of contact $30°$ . Take surface tension of water $S_1 = 7.2\times10^{-2}\ \text{N/m}$ , angle of contact of water $0°$ , and density of water $\rho_1 = 1000\ \text{kg/m}^3$ . Find the height to which the second liquid rises.

Step 1. Since the same tube is used for both liquids, the radius $r$ is common to both and cancels if the two capillary-rise equations are divided. Form the ratio:

\frac{h_2}{h_1} = \frac{S_2 \cos\theta_2}{\rho_2} \times \frac{\rho_1}{S_1\cos\theta_1}

Step 2. Substitute the values, using $\cos 30° = 0.866$ and $\cos 0° = 1$ :

\frac{h_2}{h_1} = \frac{(3.6\times10^{-2})(0.866)}{800} \times \frac{1000}{(7.2\times10^{-2})(1)}

Step 3. Compute in stages to avoid arithmetic slips. First, $(3.6\times10^{-2})(0.866) = 3.118\times10^{-2}$ ; dividing by $800$ gives $3.897\times10^{-5}$ . Next, $\dfrac{1000}{7.2\times10^{-2}} = 1.389\times10^{4}$ . Multiplying the two: $\dfrac{h_2}{h_1} = 3.897\times10^{-5} \times 1.389\times10^{4} = 0.5414$ .

Step 4. Recover $h_2$ using $h_1 = 6.0\ \text{cm}$ : $h_2 = 0.5414 \times 6.0\ \text{cm} \approx 3.25\ \text{cm}$ .

The second liquid rises to about $3.25\ \text{cm}$ , noticeably less than water — mainly because of its lower surface tension and the reduced $\cos\theta$ factor from the larger contact angle, even though its lower density alone would have tended to push the rise up. Timing note: this ratio method is the fast, exam-safe path whenever the same tube is reused for a second liquid — it avoids computing $r$ altogether and turns a two-formula problem into one substitution, well suited to a 90-second competitive-exam slot.

Common traps

Forgetting $\cos\theta$ entirely and writing $h = 2S/(\rho g r)$ — this simplification is only valid in the special case $\theta \approx 0°$ (water on clean glass); dropping the $\cos\theta$ silently is wrong whenever the angle of contact is not zero.
Using the outer radius of the tube instead of the inner radius — the formula depends on the inner radius, because that is what shapes the meniscus and encloses the risen liquid.
Treating a negative $h$ (from an obtuse $\theta$ ) as a mistake to fix rather than as the correct physical answer — it means capillary depression, a genuine result for non-wetting liquids like mercury in glass.
Assuming capillary rise depends on how tall the tube is — a tube taller than the calculated $h$ still shows exactly the same rise; the extra length above the meniscus plays no role once equilibrium is reached.
Confusing the meniscus radius $R$ with the tube radius $r$ when applying the excess-pressure formula directly — always substitute $R = r/\cos\theta$ first, or use the combined form $2S\cos\theta/r$ directly.

Common trapA negative h means depression, not a wrong answer, and mercury does not rise in a capillary tube

The single biggest mark-loser on this topic is panicking when a calculation returns a negative rise and "fixing" it by dropping the minus sign or reworking the problem. For mercury in a glass tube, the angle of contact is around $138°$ , well past $90°$ , so $\cos\theta$ is genuinely negative, and the formula $h = 2S\cos\theta/(\rho g r)$ correctly predicts that mercury sits lower inside the tube than the level outside it — a capillary depression, not a rise. This is the exact mirror image of water's behaviour, and both outcomes come from the very same formula; only the sign of $\cos\theta$ , set by whether the liquid wets or does not wet the tube material, decides which way the liquid moves. Reporting "depression of magnitude such-and-such centimetres" for a non-wetting liquid is the fully correct, fully expected answer.

You should be able to…

Derive the capillary rise formula $h = \dfrac{2S\cos\theta}{\rho g r}$ starting from the excess pressure across a curved meniscus, using $R = r/\cos\theta$ to connect meniscus curvature to tube radius.
Explain why $h \propto 1/r$ and why the effect is strongest, and most visibly named "capillary," in the narrowest tubes.
Predict capillary depression (not rise) for non-wetting liquids purely from the sign of $\cos\theta$ , and state that mercury in glass is the standard example.
Solve numerically for $h$ , $r$ , or $S$ given the other quantities, including comparative ratio problems between two liquids sharing the same tube.
Rearrange the formula to find a liquid's surface tension from an experimentally measured rise, tube radius, and density.
Connect the formula qualitatively to everyday capillary-rise examples — water in a thin glass tube, blotting paper, a lamp wick, water movement through soil.

Ask the tutor: want to practice rearranging the capillary rise formula to measure surface tension from an experimental rise, or work through a mercury-depression numerical? Ask the AI tutor to set up a practice problem for you.

Why does a narrower tube give a bigger rise? It feels backwards.

It only feels backwards if you think of the tube as "helping" the liquid up by size — it is the opposite. The upward pull comes from the curved meniscus, and a narrow tube forces that meniscus into a tighter, more sharply curved bowl (a small radius of curvature $R \approx r$ for a wetting liquid), and a more sharply curved surface has a larger excess pressure across it, from $\Delta P = 2S/R$ . A wide tube lets the meniscus flatten out into a much gentler curve, so the pressure deficit driving the rise is small, and the liquid barely climbs at all. So it is not that a narrow tube pushes harder from below — it is that a narrow tube forces a sharper meniscus, and a sharper meniscus means a bigger pressure imbalance to correct.

Does the water in the capillary tube weigh less than an equal column outside, since it is somehow being held up by surface tension?

No — the risen water weighs exactly what any column of that height and cross-section would weigh; gravity acts on it the same as on any water. What is different is the pressure at its base. The meniscus's curvature keeps the pressure just under it slightly below atmospheric, so the liquid does not need atmospheric pressure alone to support it — the deficit itself, transmitted through the connected liquid, is what lets a taller column sit in mechanical equilibrium than atmospheric pressure could support on its own in a flat-surfaced container. It is a pressure-balance effect, not the liquid magically becoming lighter.

Is capillary rise the same effect as osmosis moving water into plant roots?

No, though both eventually help move water in plants and both are commonly mentioned together. Capillary rise, the topic here, works purely through surface tension and the curved meniscus in narrow channels, and needs no membrane or concentration difference — it happens in a plain glass tube. Osmosis is a completely different mechanism involving the movement of water across a semi-permeable membrane from a region of lower solute concentration to higher, driven by osmotic pressure, and belongs to biology and chemistry rather than this chapter. In a real plant, capillary action in the fine xylem tubes, transpiration pull from evaporating leaves, and osmotic pressure in root cells all contribute together, but only the capillary piece is examined in this Class 11 Physics chapter.

Can I use the capillary rise formula to actually measure a liquid's surface tension in the lab?

Yes, and this is one of the standard Class 11 practicals. Rearrange the formula to $S = \dfrac{h \rho g r}{2\cos\theta}$ , then measure the tube's inner radius $r$ (with a travelling microscope or by weighing a known length of mercury thread), dip the tube in the liquid and measure the rise $h$ with a scale or travelling microscope, and use the liquid's known density $\rho$ . For a liquid that wets the tube well, $\theta$ is taken as approximately $0°$ so $\cos\theta \approx 1$ , simplifying the formula to $S = h\rho g r/2$ . This reverse use of the same equation is exactly why the formula is worth memorising both ways round, not just for finding $h$ .