Question

A capillary tube of radius 0.50 mm is dipped vertically in a pot of water. Find the difference between the pressure of the water in the tube 5.0 cm below the surface and the atmospheric pressure. Surface tension of water = 0.075 N m⁻¹.

Answer 1

Given:
Radius of capillary tube r = 0.5 mm = 5 × 10⁻⁴ m
Depth (where pressure is to be found) h = 5.0 cm = 5 × 10⁻² m
Surface tension of water T = 0.075 N/m
Excess pressure at 5 cm before the surface:
P = ρhg = 1000 × (5 × 10⁻²) × 9.8 = 490 N/m²
Excess pressure at the surface is given by:
$$\begin{gathered} P_0=\frac{2T}{r}=\frac{2\times \left(0.75\right)}{\left(5\times {10}^{-4}\right)} \\ =300\mathrm{N}/{\mathrm{m}}^2 \end{gathered}$$
Difference in pressure: P₀ − P $=490-300=190\mathrm{N}/{\mathrm{m}}^2$

Hence, the required difference in pressure is 190 N/m².