Consider a strip of width $d y$ at a depth $(h-y)$ below the surface.

'Torque on bottom length ( $O Z$ in diagram) due to force of water on strip of width $d y$ is

$d \tau=$ Force $\times$ Perpendicular distance

$=$ Pressure $\times$ Area $\times$ Distance

$d \tau=\rho g(h-y) \cdot(L \cdot d y) \cdot y$

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

$\tau_{1} =\int \limits_{0}^{h} \operatorname{L\rho g}(h-y) y \cdot d y $

$=L \rho g\left[\int \limits_{0}^{h}\left(h y-y^{2}\right) d y\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 $\tau_{1}=\frac{L \rho g h^{3}}{6}$ and torque due to water upto height $\frac{h}{2}$ and wall length $\frac{1}{2}$,

$\tau_{2}=\int \limits_{0}^{\frac{h}{2}} \frac{L}{2} \rho g\left(\frac{h}{2}-y\right) y d y$

$=\frac{L}{2} \rho g\left[\left(\frac{h y^{2}}{4}-\frac{h y^{3}}{3}\right)\right]_{0}^{\frac{h}{2}}$

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

So, $\frac{\tau_{1}}{\tau_{2}}=16$