Consider the wall of a dam to be straight with height $H$ and length $L$. It holds a lake of water of height $h\left(h<H\right)$ on one side. Let the density of water be ${\rho }_w$ Denote the torque about the axis along the bottom length of the wall by ${\tau }_1$. Denote also a similar torque due to the water up to height $h/2$ and wall length $L/2$ by ${\tau }_2$. Then, ${\tau }_1/{\tau }_2$ (ignore atmospheric pressure) is

Consider a strip of width $dy$ at a depth $\left(h-y\right)$ below the surface. 

Torque on bottom length ( $OZ$ in diagram) due to force of water on strip of width $dy$ is 

$d\tau =$ Force $\times$ Perpendicular distance $=$ Pressure $\times$ Area $\times$ Distance

$d\tau =\rho g\left(h-y\right)·\left(L·dy\right)·y$

So, torque on bottom length due to complete volume of water, 

${\tau }_1={\int }_0^{h}L\rho g\left(h-y\right)y·dy$

$=L\rho g\left[{\int }_0^h\left(hy-y^2\right)dy\right]$

$=L\rho g{\left[\frac{h{y}^2}{2}-\frac{y^3}{3}\right]}_0^h$

$=L\rho g\left[\frac{h^3}{2}-\frac{h^3}{3}\right]$

or ${\tau }_1=\frac{L\rho g{h}^3}{6}$  $...\left(\mathrm{i}\right)$

and torque due to water upto height $\frac{h}{2}$ and wall length $\frac{1}{2}$, ${\tau }_2={\int }_0^{\frac{h}{2}}\frac{L}{2}\rho g\left(\frac{h}{2}-y\right)ydy$

$=\frac{L}{2}\rho g{\left[\left(\frac{h{y}^2}{4}-\frac{h{y}^3}{3}\right)\right]}_0^h$

$=\frac{L}{2}\rho g\left[\frac{h^3}{16}-\frac{h^3}{24}\right]=\frac{L\rho g{h}^3}{16\times 6}$

So, $\frac{{\tau }_1}{{\tau }_2}=16$