Surface tension and viscosity of liquids

Aim

To find the surface tension of water by the rise in a capillary tube, and the viscosity of a liquid from the terminal speed of a falling ball.

Surface tension by capillary rise

Put the end of a very thin glass tube into water. The water climbs inside the tube, higher than the level outside. The rise comes from the surface tension $T$, the force along the surface of a liquid for each metre of a line drawn on it, in N/m.

The idea. The water touches the glass along a circle of length $2\pi r$. The surface pulls the water up along this circle. For a liquid that wets the glass, the angle of contact $\theta$ is nearly $0^\circ$, and the upward force is $2\pi r T \cos\theta$. It balances the weight of the lifted column, $\pi r^2 h \rho g$:

For clean glass and water, $\cos\theta = 1$. We want $T$, so $T = \dfrac{r h \rho g}{2}$. In the other form, $h = \dfrac{2T}{r\rho g}$, so the rise is in proportion to $1/r$.

A small correction is possible. The water in the curved top, the meniscus, adds a small volume equal to that of a column of height $\dfrac{r}{3}$. So the effective height is $h + \dfrac{r}{3}$. For $r = 0.5$ mm the change in $T$ is under one percent.

The steps.

1. Clean the tubes with dilute caustic soda or alcohol, and rinse with distilled water. The water must wet the glass fully.
2. Measure the bore of each tube. Use a travelling microscope, or the vernier calipers, at the broken end.
3. Clamp the tube vertically in a beaker of water. Fix a pointer whose tip just touches the water surface.
4. With the travelling microscope, find the lowest point of the meniscus inside the tube and the tip of the pointer. Their difference is $h$.
5. Repeat with five tubes of different radius, and measure the temperature.

The readings. For water at $20$ °C ($\rho = 998\ \text{kg/m}^3$):

The graph of $h$ against $1/r$ is a straight line through the origin. Its slope is $\dfrac{2T}{\rho g} = 14.8\ \text{mm}^2$, so $T = \dfrac{\rho g \times \text{slope}}{2}$. For one tube, $r = 0.50$ mm and $h = 29.9$ mm give

First, $r h = 0.50 \times 10^{-3}\ \text{m}$ $\times 29.9 \times 10^{-3}\ \text{m}$ $= 1.495 \times 10^{-5}\ \text{m}^2$. Next, $\rho g = 998\ \text{kg/m}^3 \times 9.8\ \text{m/s}^2$ $= 9780\ \text{N/m}^3$. So $T = \tfrac{1}{2} \times 1.495 \times 10^{-5}\ \text{m}^2$ $\times 9780\ \text{N/m}^3$ $= 0.0731\ \text{N/m}$

The standard value for water at $20$ °C is $0.0728$ N/m, and it is $0.0720$ N/m at $25$ °C. The surface tension falls as the temperature rises.

Halve the radius. What happens to $h$? Then raise the angle of contact to $60^\circ$. Why does the liquid rise less?

Precautions. Keep the tube vertical. Read the lowest point of the meniscus, not its edge. Do not touch the inside of the tube with a finger. Keep the temperature the same during the experiment.

Viscosity by Stokes' law

Honey is hard to stir, and water is easy. The difference is viscosity, the friction between layers of a liquid. A ball that moves slowly through a liquid feels a drag. For a small sphere of radius $r$ at a speed $v$, Stokes' law gives the viscous force

where $\eta$ is the coefficient of viscosity in Pa s.

The idea. Let a steel ball fall in the liquid. Three forces act on it: its weight $\tfrac{4}{3}\pi r^3\rho g$ downwards, the buoyant force $\tfrac{4}{3}\pi r^3\sigma g$ upwards, and the drag. The ball speeds up until the forces balance. The steady speed is the terminal speed $v_t$:

We want $\eta$, so $\eta = \dfrac{2r^2(\rho - \sigma)g}{9v_t}$. Here $\rho$ is the density of the ball and $\sigma$ is that of the liquid. The terminal speed is in proportion to $r^2$.

The steps.

1. Fill a tall glass jar with glycerine. Mark two lines A and B on it, the upper one at least $10$ cm below the surface, so that the ball has reached its steady speed. Measure the distance $s$ between them.
2. Measure the diameter of each steel ball with a screw gauge. Wet each ball with the liquid.
3. Drop a ball on the axis of the jar. Time it from A to B with a stopwatch. Repeat for balls of five sizes.
4. Find $v_t = s/t$ for each. Find the density $\sigma$ of the liquid with a hydrometer or a density bottle.

The readings. Glycerine, $s = 0.400$ m, steel balls ($\rho = 7800\ \text{kg/m}^3$), $\sigma = 1260\ \text{kg/m}^3$:

For $r = 1.50$ mm and $t = 12.0$ s, $v_t = \dfrac{0.400\ \text{m}}{12.0\ \text{s}} = 0.0333$ m/s. Then

The numerator is $2 \times (1.50 \times 10^{-3}\ \text{m})^2$ $\times 6540\ \text{kg/m}^3$ $\times 9.8\ \text{m/s}^2$ $= 0.288\ \text{kg/s}^2$. The denominator is $9 \times 0.0333\ \text{m/s}$ $= 0.3000\ \text{m/s}$. So $\eta = \dfrac{0.288\ \text{kg/s}^2}{0.3000\ \text{m/s}}$ $= 0.96\ \text{Pa s}$

The graph of $v_t$ against $r^2$ is a straight line through the origin. Its slope is $\dfrac{2(\rho - \sigma)g}{9\eta}$, so $\eta = \dfrac{2(\rho - \sigma)g}{9 \times \text{slope}}$. All five balls give $\eta = 0.95\ \text{Pa s}$, close to the value for glycerine near $25$ °C.

Double the radius. How does the terminal speed change? Which liquid, with a low or a high viscosity, gives a speed that is easy to time?

When is Stokes' law right? The law holds for a small, slow sphere in a wide container. The Reynolds number $Re = \dfrac{\sigma v_t (2r)}{\eta}$ must be less than about $1$. In the table, the greatest value is $0.32$. In a narrow jar the walls slow the ball. A correction factor $\left(1 + 2.4\,\dfrac{r}{R}\right)$ multiplies the speed, where $R$ is the radius of the jar.

Precautions. Release the ball at the axis of the jar. Wet it first, so that no air bubble sticks to it. Use the same temperature for all the readings, because the viscosity of a liquid falls quickly when it warms.

Practise with new numbers

Try it in Python

The program takes the table of the falling balls, fits the line of speed against $r^2$, and finds the viscosity. It also checks the Reynolds number.

Common mistakes

• Taking the diameter as the radius. Both formulas use $r$ and the bore of the tube is a diameter. Halve it.
• Leaving out the buoyant force. Use $\rho - \sigma$, not $\rho$.
• Reading the edge of the meniscus. Read the lowest point.
• Timing too early. The ball must reach its terminal speed between the marks.
• A dirty tube. Dirt changes the angle of contact, so $\cos\theta$ is not $1$.

Remember these

• $T = \dfrac{r h \rho g}{2\cos\theta}$. The graph of $h$ against $1/r$ is a line through the origin with the slope $\dfrac{2T}{\rho g}$.
• $F = 6\pi\eta r v$, and $\eta = \dfrac{2r^2(\rho - \sigma)g}{9v_t}$.
• The terminal speed is in proportion to $r^2$.
• Stokes' law needs $Re < 1$. Warm liquids are less viscous.
• Surface tension and viscosity both fall when the temperature rises.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. In the capillary experiment, why does a tube of half the radius give twice the rise?

2. Water rises $4.0$ cm in a tube of radius $0.37$ mm. What is the surface tension? Take $\rho = 1000\ \text{kg/m}^3$ and $\cos\theta = 1$.

3. A ball of radius $r$ falls at a terminal speed $v_t$ in a liquid. A second ball of the same material has a radius $2r$. What is its terminal speed?

4. Why is the first mark on the jar placed well below the surface in Stokes' experiment?

References

• NCERT, Physics Class XI, Chapter 9 (Mechanical Properties of Fluids), the sections on surface tension, viscosity and Stokes' law.
• NCERT, Laboratory Manual: Physics, Class XI, the experiments on surface tension by capillary rise and on viscosity by the terminal speed of a ball.
• OpenStax University Physics Volume 1, Chapter 14 (Fluid Mechanics), the section on viscosity.