Answer to Question #138571 in Electric Circuits for Tary

Question #138571
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)
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Expert's answer
2020-10-23T11:54:32-0400

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

Then: R=p×g×h2×σ×cosθ=103×10×0.0622×0.7×cos0442.86R= \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=2×σ×cosθp×g×R=h=\frac{2\times σ\times \cos \theta}{p\times g\times R}= 2×0.54×cos(180°140°)13600×10×442.861.37×108\frac{2\times 0.54\times \cos {(180\degree-140\degree)}}{13600\times 10\times 442.86}\approx1. 37\times 10^{-8} m.


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