Water is filled in a capillary tube of radius R. If the surface of water is hemispherical `(theta = 0)`, then find pressure at a point 'A' which is at h depth below the surface.

Explanation

Draw normal (radial lines) at point `A` and `B` of periophery. The point `(C)` wherer radial lines meet is called centre of curvture. If contact angle is `theta`, from `Delta ACM, r_(C) = R sectheta`
So radius of curvature of the surface `r_(C) = R sectheta`
Point to remember :
If the liquid surface is hemispherical `(theta = 0)` then `r_(c) = R`
If liquid surface is not hemispherical `(theta ne 0)` then `r_(c) = R sec theta`

So pressure at `A` is `P_(0) - (2T)/(Rsectheta) + rhogh`