(a) How high will water rise in a glass capillary tube with a 0.500-mm

radius?

(b) How much gravitational potential energy does the water gain?

(c) Discuss possible sources of this energy.

Capillary action is the mechanism by which a liquid moves through a small opening defying external forces like gravity, or even working against them.

The radius of the capillary tube is\(r = 0.500 \times 1{0^{ - 3}}m\)

\(h = \frac{{2\gamma cos\theta }}{{\rho gr}}\)

Here,

θ is contact angle = \({0^0}\) for water-glass surface

\(\gamma \)isthe surface tension of water at \(2{0^0}C = 0.07275 J/{m^2}\) .

ρ is the density of the water is\(997 kg/{m^3}\)

substituting these values in the above equation, we get

\(\begin{array}{l}h = \frac{{2*0.07275*cos0}}{{997*9.8*0.5 \times 1{0^{ - 3}}}}\\h = \frac{{0.1455}}{{4.8853}}\\h = 0.02978 m\\h = 2.978 cm\end{array}\)

Therefore, the water will rise: \(h = 2.978 cm\).

\({E_P} = mgh\)

Here, \(g\)and ‘\(h\)’ is known.

The mass of the water in the cylindrical capillary tube is calculated as:

\(\begin{array}{l}m = \rho \times V\\m = \rho \times \pi {r^2}h\\m = 997 \times 3.14 \times {(0.5 \times 1{0^{ - 3}})^2} \times 0.02978\\m = 2.3307 \times 1{0^{ - 5}}kg\end{array}\)

The water molecules inside the capillary tube are at different heights, hence, we are going to consider their average height.

Therefore, the gain in gravitational potential energy is given by:

\(\begin{array}{l}{E_P} = mg(h/2)\\{E_P} = 2.3307 \times 1{0^{ - 5}}*9.8*(0.02978/2)\\{E_P} = 3.4 \times 1{0^{ - 6}}J\end{array}\)

Hence, the gravitational potential energy is:\({E_P} = 2.17 \times 1{0^{ - 6}}J\).

Therefore, energy is provided by surface tension.