Question

Calculate the height to which water will rise in a capillary tube of 0.4 mm radius. If the capillary is inclined at an angle of 60° with the vertical, then what will be the position of the water in the tube? Surface tension of water = 7 x 10^-2 N/m.

Answer 1

r = 0.4 mm = 0.4 x 10^-3 m; h = ?; d = 10³ kg/m³; g = 9.8 m/s²; T = 7 x 10^-2 N/m

∵ \(T = \frac{rhdg}{2}\Rightarrow h = \frac{2T}{rdg}\)

\(\therefore\ h = \frac{2\times7\times10^{-2}}{0.4\times10^{-3}\times10^3\times9.8} = \frac{14\times10^{-2}}{0.4\times9.8}\)

= 3.57 x 10^-2 m

h = 3.57 cm

If capillary tube is inclined to ß angle from vertical, then

l = \( \frac{h}{\cos\beta}\)

Here ß = 60° ∴ cos ß = cos 60° = \(\frac{1}{2}\)

∴ l = \(\frac{3.57}{1/2}\) = 2 x 3.57

or l = 7.14 cm