Capillary Rise

A liquid of density $\rho$ and surface tension $\sigma$ rises in a capillary of inner radius $r$ to a height \begin{align} h=\frac{2\sigma\cos\theta}{\rho g r} \end{align} where $\theta$ is the contact angle made by the liquid meniscus with the capillary’s surface.

The liquid rises due to the forces of adhesion, cohesion, and surface tension. If adhesive force (liquid-capillary) is more than the cohesive force (liquid-liquid) then liquid rises as in case of water rise in a glass capillary. In this case, the contact angle is less than 90 deg and the meniscus is concave. If adhesive force is less than the cohesive force then liquid depresses as in case of mercury in a glass capillary. In this case, the contact angle is greater than 90 deg and the meniscus is convex.

The formula for capillary rise can be derived by balancing forces on the liquid column. The weight of the liquid ($\pi r^2 h \rho g$) is balanced by the upward force due to surface tension ($2\pi r \sigma \cos\theta$). This formula can also be derived using pressure balance.

The capillary rise experiment is used to measure the surface tension of a liquid.

Solved Problems on Capillary Rise

Problem from IIT JEE 2014

A glass capillary tube is of the shape of a truncated cone with an apex angle $\alpha$ so that its two ends have cross sections of different radii. When dipped in water vertically, water rises in it to a height $h$, where the radius of its cross-section is $b$. If the surface tension of water is $S$, its density is $\rho$, and its contact angle with glass is $\theta$, the value of $h$ will be (where $g$ is acceleration due to gravity),

1. $\frac{2S}{b\rho g}\cos(\theta-\alpha)$
2. $\frac{2S}{b\rho g}\cos(\theta+\alpha)$
3. $\frac{2S}{b\rho g}\cos(\theta-\alpha/2)$
4. $\frac{2S}{b\rho g}\cos(\theta+\alpha/2)$

Solution: Let $R$ be the radius of the meniscus formed with a contact angle $\theta$ (see figure). By geometry, this radius makes an angle $\theta+\frac{\alpha}{2}$ with the horizontal and, \begin{align} \label{oxb:eqn:1} \cos\left(\theta+\tfrac{\alpha}{2}\right)={b}/{R}. \end{align}

Let $P_0$ be the atmospheric pressure and $P_1$ be the pressure just below the meniscus. Excess pressure on the concave side of meniscus of radius $R$ is, \begin{align} \label{oxb:eqn:2} P_0-P_1={2S}/{R}. \end{align} The hydrostatic pressure gives, \begin{align} \label{oxb:eqn:3} P_0-P_1=h\rho g. \end{align} Eliminate $(P_0-P)$ from second and third equations and substitute $R$ from first equation to get, \begin{align} h=\frac{2S}{\rho g R}=\frac{2S}{b\rho g}\cos\left(\theta+\tfrac{\alpha}{2}\right).\nonumber \end{align}

Problem from IIT JEE 2018

A uniform capillary tube of inner radius $r$ is dipped vertically into a beaker filled with water. The water rises to a height $h$ in the capillary tube above the water surface in the beaker. The surface tension of water is $\sigma$. The angle of contact between water and the wall of the capillary tube is $\theta$. Ignore the mass of water in the meniscus. Which of the following statements is (are) true?

1. For a given material of the capillary tube, $h$ decreases with increase in $r$.
2. For a given material of the capillary tube, $h$ is independent of $\sigma$.
3. If this experiment is performed in a lift going up with a constant acceleration, then $h$ decreases.
4. $h$ is proportional to contact angle $\theta$.

Solution: Consider the portion of water that rises in the capillary tube. In equilibrium, the upward force due to surface tension, $F=2\pi r\sigma\cos\theta$, is balanced by the weight of the water, $W=\pi r^2 h \rho g$, i.e., $2\pi r\sigma\cos\theta=\pi r^2 h \rho g$ which gives \begin{align} \label{bcc:eqn:1} h=\frac{2 \sigma\cos\theta}{r\rho g}. \end{align} From this equation, $h$ decreases with increase in $r$, it depends on $\sigma$ and it is proportional to $\cos\theta$. The contact angle $\theta$ depends on the material of the capillary tube.

In a lift going up with a constant acceleration $a$, the force due to surface tension remains same but weight of the water increases i.e., $W_\text{lift}=\pi r^2 h \rho (g+a)$. Thus, the height of the water rise in a capillary becomes \begin{align} h=\frac{2 \sigma\cos\theta}{r\rho (g+a)}=\frac{2 \sigma\cos\theta}{r\rho g_\text{eff}},\nonumber \end{align} where $g_\text{eff}$ is the effective $g$ in the lift. Hence, $h$ decrease if the experiment is performed in a lift going up with a constant acceleration. Can you guess value of $h$ if the experiment is performed in a lift falling freely?

Questions on Capillary Rise

Question 1: In an experiment to measure the surface tension of water using capillary rise method, a student uses a capillary whose length is shorter than the height of capillary rise (obtained theoretically for this capillary). Which of the following statement is correct?

A. The water will rise to the top and then move out of the tube.
B. The water will rise to the top and then radius of curvature of the maniscus start increasing.

Question 2: In an experiment to measure the surface tension of a liquid using capillary rise method, a student kept the capillary inclined instead of the vertical. The vertical height of liquid rise in tilted capillary?

A. increases.
B. decreases.
C. remains the same.

Question 3: The capillary rise experiment was carried out in a space station under weightlessness conditions. The water will

A. not move inside the capillary.
B. move inside the capillary and move out from the other end.
C. move inside the capillary and stop after reaching the other end.
B. move inside the capillary upto a certain length and then stops.

Question 4: In capillary rise experiment, the work done by the force of surface tension is almost equal to the rise in gravitational potential energy of the liquid

A. true.
B. false.

Question 5: The meniscus surface in a fine capillary may be considered to be semispherical. In this case, the weight of the liquid above the lowest point of the meniscus is $\pi r^3\rho g/3$. Taking this into account, the formula for surface tension is

A. $\rho g r (h+r/3)/2$
B. $\rho g r (h+r/3)$
C. $\rho g r h/2$
D. $\rho g r h/2\cos\theta$

Question 6: Why is the meniscus of mercury convex?

A. A convex meniscus occurs when the particles of the liquid are more strongly attracted to the container than to each other, causing the liquid to climb the walls of the container.
B. A convex meniscus occurs when the particles in the liquid have a stronger attraction to each other than to the material of the container.

Question 7: Water rises to a height of 3 cm in capillary of radius 0.48 mm. If angle of contact is zero then surface tension of the water is

A. 0.0072 N/m.
B. 0.072 N/m.

Question 8: The inner radius of a barometer tube in 2.5 mm. The error introduced in the observed reading due to surface tension is (density of mercury is 13.6 g/cc, contact angle of mercury in glass is 135 deg)

A. 2.3 mm.
B. 0.23 mm

Question 9: The height of liquid rise in a capillary tube depends on the size of the tube but it is independent of the tube material.

A. true
B. false

Question 10: Two capillary tubes of diameter 6 mm and 3 mm are joined together to form a U-tube open at both ends. If the U-tube is filld with water then water levels on the two limbs of the tube are different.

A. true
B. false

Related Topics

1. Surface Enery and Surface Tension

References and External Links

1. IIT JEE Physics by Jitender Singh and Shraddhesh Chaturvedi
2. Concepts of Physics Part 1 by HC Verma (Link to Amazon)
3. Capillary Action (USGS website)
4. Cohesion, Adhesion and Capillary Action (opentextabc.ca website)
5. NCERT Experiment on Measurement of Surface Tension by Capillary Rise (pdf)
6. Origins of Wetting (detailed research paper, pdf)