Answer to Question #138571 in Electric Circuits for Tary

Question

A capillary tube is immersed in water of surface tension 70 x 10-2 Nm-1 and rises 6.2 cm. By what depth will mercury be depressed if the same capillary is immersed in it? (Surface tension of mercury = 0.54 Nm-1; angle of contact between mercury and glass = 140 degrees, density of mercury = 13600 kgm-3)

Answer 1

As the height of liquid rise in a capillary is equal to: $h=\frac{2\times σ\times \cos \theta}{p\times g\times R}$ , where: $σ$ -surface tension coefficient, $p$ is the fluid density, $g$ -acceleration of free fall, $R$ is the radius of the capillary.

Then: $R= \frac{p\times g\times h} {2\times σ\times \cos \theta }= \frac{10^3\times 10\times 0.062} {2\times 0.7\times \cos 0 }\approx442.86$ m.

Hence the height at mercury: $h=\frac{2\times σ\times \cos \theta}{p\times g\times R}=$ $\frac{2\times 0.54\times \cos {(180\degree-140\degree)}}{13600\times 10\times 442.86}\approx1. 37\times 10^{-8}$ m.