A capillary tube with a bore diameter of 0.5 mm is dipped upright in a liquid having surface tension 0.03 N. `m^-1`. The specific gravity of the liquid is `0.8` and the liquid wets the surface of the tube completely. Up to what height will the liquid rise in the tube?

Explanation

If a liquid wets the wall of tube completely, then the angle of contact becomes approximately zero.
`therefore T=(rh rho g)/(2)` [ As `theta` is very small, `costheta~=1`] `=(2xx0.03)/(0.25xx10^-3xx0.8xx10^3xx9.8)=0.0306 m.`