Question

A liquid of density $\rho$ is filled in a conical vessel as shown in fig. Force exerted by liquid on side wall is

Answer 1

$\frac{\pi R h \sqrt{R^2 + h^2} \rho g}{3}$

Answer 2

To find the force exerted by the liquid on the side wall of the conical vessel, we consider the pressure exerted by the liquid at different depths and integrate it over the area of the side wall.

Let the conical vessel have height $h$ and base radius $R$. The density of the liquid is $\rho$. Let $L$ be the slant height of the cone, so $L = \sqrt{R^2 + h^2}$.

Consider a horizontal ring on the side wall at a vertical depth $y$ from the free surface of the liquid. The radius of the cone at this depth is $r = R(1 - y/h)$. The pressure at this depth is $P = \rho g y$.

Consider a small strip of the side wall of vertical height $dy$ at depth $y$. The slant height of this strip is $ds$. The relationship between $ds$ and $dy$ is $ds = \frac{L}{h} dy$. The area of this elemental ring on the side wall is $dA = 2\pi r ds = 2\pi R(1 - y/h) \frac{L}{h} dy$.

The force $dF$ exerted by the liquid on this elemental ring is $dF = P dA$. This force is perpendicular to the side wall and directed outwards. $dF = (\rho g y) \left( 2\pi R \frac{L}{h} (1 - y/h) dy \right)$ $dF = \frac{2\pi RL \rho g}{h} y \left(1 - \frac{y}{h}\right) dy$

The total force $F$ exerted by the liquid on the side wall is the integral of $dF$ from the surface ($y=0$) to the base ($y=h$): $F = \int_{0}^{h} dF = \int_{0}^{h} \frac{2\pi RL \rho g}{h} \left(y - \frac{y^2}{h}\right) dy$ $F = \frac{2\pi RL \rho g}{h} \int_{0}^{h} \left(y - \frac{y^2}{h}\right) dy$ $F = \frac{2\pi RL \rho g}{h} \left[\frac{y^2}{2} - \frac{y^3}{3h}\right]_{0}^{h}$ $F = \frac{2\pi RL \rho g}{h} \left(\frac{h^2}{2} - \frac{h^3}{3h}\right)$ $F = \frac{2\pi RL \rho g}{h} \left(\frac{h^2}{2} - \frac{h^2}{3}\right)$ $F = \frac{2\pi RL \rho g}{h} \left(\frac{3h^2 - 2h^2}{6}\right)$ $F = \frac{2\pi RL \rho g}{h} \left(\frac{h^2}{6}\right)$ $F = \frac{\pi RL \rho g h}{3}$

Substituting $L = \sqrt{R^2 + h^2}$: $F = \frac{\pi R \sqrt{R^2 + h^2} \rho g h}{3}$