A galss capillary length `l=11`cm and inside diameter `d=20mum` is submerged vertically into water. The upper end of the capillary is sealed. The outside pressure is considered to be `1xx10^(5)(N)/(m^(2))` . To what length has the capillary to be submerged to make the water levels inside and outside the capillary coincide?

Explanation

Initially the pressure of air in the cppiliary is `p_0` and it's length is `l`. When submerged under water, the pressure of air in the portion above water must be `p_0 + 4(alpha)/(d)`, since the level of water inside the capillary is the same as the level outside. Thus by Boyle's law
`(p_0 + (4 alpha)/(d))(l - x) = p_0 l`
or `(4 alpha)/(d) (l -x) = p_0 x` or `x = (l)/(1 + (p_0 d)/(4 alpha))`.