Answer to Question #138567 in Electric Circuits for Jay

Question

If water has a surface tension of 7 x 10^2 Nm and an angle of contact with water of zero. At what height does it rise in a capillary of diameter 0.5 mm?
What are the dimensions of surface tension?

Answer 1

As the height of rise of liquid in capillary is equal to: $h=\frac{2\times σ\times \cos \theta}{(p\times g\times R)}$

Where: $σ$ -surface tension coefficient, $σ = 7\times10^2$ H/m, p is the fluid density, $p$ (water) $= 10^3$ kg/m$^3$ , $g$ -acceleration of free fall, $g = 10$ m/S$^2$ , $R$ is the radius of the capillary $R =\frac{1}{2}\times 5\times10^{-4}$ m.

Then: $h=\frac{2\times σ\times \cos \theta }{(p\times g\times R)}=\frac{2\times7\times10^2 \times \cos 0}{(103\times 10 \times \frac{1}{2}\times5\times10^{-4})}=5600$ m

The size of the surface tension is

$S= h\times \pi \times R^2 = 5600\times$ $\pi \times (2.5\times 10^{-4})^2\approx 0.001$ m^2.