The height to which a cylindrical vessel be filled with a homogeneous liquid, to make the average force with which the liquid presses the side of the vessel equal to the force exerted by the liquid on the bottom of the vessel, is equal to

To find the height to which a cylindrical vessel of radius ^$R$ must be filled with a homogeneous liquid of density ^$\rho$ such that the force on the side of the vessel equals the force on the bottom, we can calculate both forces step-by-step.

Step 1: Force exerted by the liquid on the bottom of the vessel
The pressure exerted by the liquid at the bottom of the vessel at depth $h$ is given by:
$P_{\text{bottom}}=\rho gh$
where $g$ is the acceleration due to gravity.
The area of the bottom of the cylindrical vessel is:
$A_{\text{bottom}}=\pi R^2$
Therefore, the force on the bottom of the vessel is:
$F_{\text{bottom}}=P_{\text{bottom}}\times A_{\text{bottom}}=\rho gh\times \pi R^2$

Step 2: Average force exerted by the liquid on the side of the vessel
The pressure of the liquid varies linearly with depth, from $0$ at the free surface to $\rho gh$ at the bottom. The average pressure on the vertical wall of the cylinder is:
$P_{\text{avg}}=\frac{0+\rho gh}{2}=\frac{1}{2}\rho gh$
The area of the cylindrical side wall in contact with the liquid is:
$A_{\text{side}}=2\pi Rh$
Therefore, the total force exerted by the liquid on the side wall is:
$F_{\text{side}}=P_{\text{avg}}\times A_{\text{side}}=\left(\frac{1}{2}\rho gh\right)\times \left(2\pi Rh\right)=\pi \rho gR{h}^2$

Step 3: Equating the two forces
According to the problem statement, the force on the side is equal to the force on the bottom:
$F_{\text{side}}=F_{\text{bottom}}$
Substituting the derived expressions:
$\pi \rho gR{h}^2=\pi \rho gh{R}^2$
Since $\pi$, $\rho$, $g$, $R$, and $h$ are non-zero, we can cancel $\pi \rho gRh$ from both sides:
$h=R$

Thus, the height to which the vessel must be filled is equal to the radius of the vessel.