In the previous question, suppose contact angle is not zero, but it is theta (the surface not hemispherical) now find pressure at point 'A'.

In the previous question, suppose contact angle is not zero, but it is

1.	In the previous question, suppose contact angle is not zero, but it is theta (the surface not hemispherical) now find pressure at point 'A'.
Answer» Solution : 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`