Water and liquid is filled up behind a square wall of side `l`
Find out (a). Pressure at A, B and C
(b). Forces in part AB and BC
(c). Total force and point of application of force. (Neglect atmosphere pressure in every calculation.)

Explanation

(a). As there is no liquid above 'A'
So pressure at `A,p_(A)=0`
pressure at `B,p_(B)=rhogh_(1)`
Pressure at `C,p_(C)=rhogh_(1)+2rhogh_(2)`
(b). Force at A=0
Take strip of width dx at a depth x in part AB
Pressure is equal to `rhogx`
force on strip=pressure`xx`area
`dF=rhogxldx`
Total force upto `B:Funderset(0)overset(h_(1))intrhogxldx=(rhogxlh_(1)^(2))/(2)=(1000xx10xx10xx5xx5)/(2)=1.25xx10^(6)N`
In part BC for force take a elementary strip of width dx in portion BC.
pressure is equal to `=rhogh_(1)+2rhog(x-h_(1))`
force on elementary strip=pressure`xx`area `dF=[rhogh_(1)+2rhog(x-h_(1))]ldx`
Total force on part BC `F=int_(h_(1))^(l)[rhogh_(1)+2rhog(x-h_(1))]ldx`
`=[rhogh_(1)x+2rhog[(x^(2))/(2)-h_(1)x]]_(h_(1))^(l)l`
`=rhogh_(1)h_(2)l+2rhol[(l^(2)-h_(1)^(2))/(2)-h_(1)l+h_(1)^(2)]`
`=rhogh_(1)h_(2)l+(2rhogl)/(2)[l^(2)+h_(1)^(2)-2h_(1)l]=rhogh_(1)h_(2)l+rhogl(l-h_(1))^(2)`
`=rhogh_(2)l[h_(1)+h_(2)]=rhogh_(2)l^(2)=1000xx10xx5xx10xx10=5xx10^(6)N`
(c). Total force
Taking torque about A
Total torque of force in `AB=intdF*x=int_(0)^(h_(1))rhogxldx.x`
`=[(rhoghlx^(3))/(3)]_(0)^(h_(1))=(rhoglh_(1)^(3))/(3)=(1000xx10xx10xx125)/(3)=(1.25xx10^(7))/(3)N-m`
Total torque of force in BC `=intdF*x`
On solving we get `=rhogh_(1)h_(2)l[h_(1)+(h_(2))/(2)]+rhogh_(2)^(2)l[h_(1)+(2h_(2))/(3)]`
`=1000xx10xx5xx5xx10[5+2.5]+1000xx10xx25xx10[5+(10)/(3)]`
`=2.5xx7.5xx10^(6)+(62.5)/(3)xx10^(6)=(118.75)/(3)xx10^(6)`
total torque `=(11.875xx10^(7))/(3)+(1.25xx10^(7))/(3)=(13.125xx10^(7))/(3)`
total torque =total `xx` distance of point of application of force from top `=F.x_(p)`
`6.25xx10^(6)x_(p)=(13.125xx10^(7))/(3)`
`x_(p)=7m`