GATE Mechanical Engineering (ME) 2020 Shift 1 Solved Paper

Hydrostatic force in reservoir $=\mathit{\rho }\mathit{g}\mathit{A}\overline{\mathit{x}}$
$\mathit{A}=\mathit{h}\times 1$ as width is unity
$\mathit{A}=\mathit{h}$
$\overline{\mathit{x}}\to$ centroid of centre of gravity
$=\frac{\mathit{h}}{2}$
${\mathit{F}}_1={\mathit{F}}_2=\mathit{\rho }\mathit{g}\mathit{A}\frac{\mathit{h}}{2}=\frac{\mathit{\rho }\mathit{g}{\mathit{h}}^2}{2}$
${\mathit{F}}_1=\frac{\mathit{\rho }\mathit{g}{\mathit{h}}_1^2}{2}$
${\mathit{F}}_2=\frac{\mathit{\rho }\mathit{g}{\mathit{h}}_2^2}{2}$
Moment around $\mathit{A}={\mathit{F}}_1\times {\mathit{h}}^{*}-{\mathit{F}}_2{\mathit{h}}^{*}$
Centre of pressure $\left({\mathit{h}}^{*}\right)=\frac{{\mathit{I}}_{\mathit{G}}}{\mathit{A}\mathit{X}}+\overline{\mathit{X}}$
${\mathit{I}}_{\mathit{G}}=\frac{\mathit{b}{\mathit{h}}^3}{12}=\frac{{\mathit{h}}^3}{12}$${\mathit{h}}^{*}=\frac{\frac{{\mathit{h}}^3}{12}}{\mathit{h}\times 1\times \frac{\mathit{h}}{2}}+\frac{\mathit{h}}{2}=\frac{2\mathit{h}}{3}$
Now this centre of pressure is from top
from bottom distance of centre of pressure $=\frac{\mathit{h}}{3}$
So, Net moment around $\mathit{A}$
$=\frac{1}{2}\mathit{\rho }\mathit{g}{\mathit{h}}_1^2\times \frac{{\mathit{h}}_1}{3}-\frac{1}{2}\mathit{\rho }\mathit{g}{\mathit{h}}_2^2\times \frac{{\mathit{h}}_2}{3}$
$=\frac{1}{2}\mathit{\rho }\mathit{g}\left[\frac{{\mathit{h}}_1^3}{3}-\frac{{\mathit{h}}_2^3}{3}\right]$
$=\frac{1000\times 10}{2\times \mathrm{3}}\left[4^3-1^3\right]$
$=105\mathit{K}\mathit{N}\mathit{m}$