Capillary rise formula h = 2T cosθ/(ρgr) — derive and solve numerical

Question

Derive the formula for capillary rise: $h = \frac{2T\cos\theta}{\rho g r}$ . Then calculate the rise of water in a glass capillary tube of radius 0.5 mm. Given: surface tension of water $T = 0.07$ N/m, contact angle $\theta = 0°$ , density $\rho = 1000$ kg/m $^3$ .

(JEE Main 2022, similar pattern)

Solution — Step by Step

Identify the upward force — surface tension

At the meniscus, the liquid surface meets the tube wall at the contact angle $\theta$ . Surface tension $T$ acts along the liquid surface, tangent to the meniscus. The vertical component of this force pulls the liquid up.

The contact perimeter is $2\pi r$ . The upward component of surface tension along this perimeter:

$F_{up} = T \cdot 2\pi r \cdot \cos\theta$

Balance with the weight of liquid column

The liquid rises to height $h$ , forming a column of volume $\pi r^2 h$ and weight:

$F_{down} = \rho g \cdot \pi r^2 h$

Equate and solve for h

At equilibrium:

$T \cdot 2\pi r \cos\theta = \rho g \pi r^2 h$

Cancel $\pi r$ from both sides:

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

This is the Jurin's law of capillary rise.

Plug in the numbers

$r = 0.5$ mm $= 5 \times 10^{-4}$ m, $\theta = 0°$ so $\cos\theta = 1$ .

$h = \frac{2 \times 0.07 \times 1}{1000 \times 10 \times 5 \times 10^{-4}} = \frac{0.14}{5} = 0.028 \text{ m}$

$\boxed{h = 2.8 \text{ cm}}$

Why This Works

Surface tension at the contact line creates an upward pull on the liquid. The liquid keeps rising until this pull is balanced by the weight of the liquid column. Narrower tubes mean less liquid weight per unit circumference, so the liquid rises higher — this is why capillary rise is inversely proportional to the tube radius.

For mercury in glass ( $\theta > 90°$ ), $\cos\theta$ is negative, meaning the mercury is pushed down (capillary depression). The same formula works — you just get a negative $h$ .

Alternative Method

You can also derive this using pressure. At the curved meniscus, the excess pressure due to surface tension is $\Delta P = 2T\cos\theta / r$ . This excess pressure supports a liquid column of height $h$ where $\Delta P = \rho g h$ . Equating gives the same result.

💡 Expert Tip

For NEET, remember the inverse proportionality: $h \propto 1/r$ . If the tube radius is halved, the rise doubles. This is a common MCQ pattern — no calculation needed, just the proportionality.

Common Mistake

⚠️ Common Mistake

Many students forget the $\cos\theta$ factor, writing $h = 2T/(\rho g r)$ . This only works when $\theta = 0°$ . For mercury in glass ( $\theta \approx 140°$ ), omitting the cosine gives the wrong sign and magnitude. Always include $\cos\theta$ in the formula.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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