A conical vessel wihout a bottom stands on a table. A liquid is poured with the vessel & as soon as the level reaches h, the pressure of the liquid raises the vessel. The radius of the base of the vessel is R and half angle of the cone `alpha` and the weight of the vessel is W. What is the density of the liquid ?

Explanation

Let.s assume that when the height of the liquid is h in the vessel , the vessel rises .
`tan alpha = (r)/(x) = (R)/(x + h)`
or `r = h - h tan alpha " " (14-31)`
The pressure on the bottom of the vessel is `rho gh` and the force with which the marked portion ( the truncated cone minus the cylinder volume) of the liquid (Fig ) presses on the table is

According to Newton.s third law an identical force acts on the liquid . At the time when the vessel starts rising , for equilibrium of the liquid , we have
`W + W_(1) = rho gh pi ) (2 R h tan alpha - h^(2) tan^(2) alpha)` where `W_(1)` is the weight of the shaded portion of the liquid
`W_(1) = (rho g h)/(3) [pi R^(2) + pi (R - h tan alpha)^(2) + pi R ( R- h tan alpha) ] - rho g h pi (R - h tan alpha)^(2)`
Therefore , `p = (W)/(pi g h^(2) tan alpha ( R - (1)/(3) h tan alpha))`