A clean glass capillary tube with an inner radius of 0.20 mm is dipped vertically into water…

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Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A clean glass capillary tube with an inner radius of $0.20\ \text{mm}$ is dipped vertically into water. The surface tension of water is $0.072\ \text{N/m}$ , its contact angle with clean glass is zero, its density is $1000\ \text{kg/m}^3$ , and $g = 10\ \text{m/s}^2$ . The tube is long enough that the water never reaches its top.

How high does the water rise in the tube?
As the water rises, the surface tension at the rim of the meniscus pulls it up. How much work does this force do while the water rises? By how much does the gravitational potential energy of the raised water increase? Where does the rest of the work go, and how much is it?
The beaker and the tube are now carried in a lift that moves upwards with a steady acceleration of $2.5\ \text{m/s}^2$ . To what height does the water stand in the tube now?

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Step 1. The rise (a). Surface tension pulls up on the water with $F = 2\pi r T$ along the rim. It holds up the column of height $h$ : $2\pi r T = \pi r^2 h \rho g$ , so

\displaystyle h = \frac{2T}{\rho g r} = \frac{2 \times 0.072\ \text{N/m}}{1000\ \text{kg/m}^3 \times 10\ \text{m/s}^2 \times 2.0 \times 10^{-4}\ \text{m}} = 7.2\ \text{cm}

Step 2. The work of surface tension (b). The pull $2\pi r T$ stays the same while the meniscus climbs the height $h$ :

\displaystyle W = 2\pi r T h = 2\pi \times 2.0 \times 10^{-4}\ \text{m} \times 0.072\ \text{N/m} \times 0.072\ \text{m} = 6.514 \times 10^{-6}\ \text{J}

Step 3. Potential energy. The raised water has mass $m = \pi r^2 h \rho = 9.048 \times 10^{-6}\ \text{kg}$ , and its centre of mass is at $h/2$ :

\displaystyle \Delta U = mg\,\frac{h}{2} = 3.257 \times 10^{-6}\ \text{J}

exactly half of $W$ . The other half, $3.257 \times 10^{-6}\ \text{J}$ , is given out as heat. The water speeds up as it is pulled in, overshoots and slops up and down, and viscosity turns that motion into heat.

Step 4. In the lift (c). In the lift's frame the water feels $g + a = 12.5\ \text{m/s}^2$ downwards, while surface tension does not change:

\displaystyle h' = \frac{2T}{\rho (g + a) r} = 7.2\ \text{cm} \times \frac{10}{12.5} = 5.76\ \text{cm}

Answer: (a) $7.2\ \text{cm}$ . (b) $6.51 \times 10^{-6}\ \text{J}$ of work; $3.26 \times 10^{-6}\ \text{J}$ of potential energy; the other $3.26 \times 10^{-6}\ \text{J}$ becomes heat. (c) $5.76\ \text{cm}$ .

Check. The weight of the raised column is $mg = 9.048 \times 10^{-5}\ \text{N}$ , and the pull of surface tension is $2\pi r T = 9.048 \times 10^{-5}\ \text{N}$ : the same, as they must be when the water is at rest.

Watch out. It is tempting to set the work of surface tension equal to the gain in potential energy. That gives twice the true height, because it forgets the energy lost as heat. Balance the forces instead.

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