Archimedes’ principle

pressure=Hρg
slant surface area=Ï€R√(R²+H²)
force=(Ï€R√(R²+H²))*Hρg

We can calculate the area of the slant height using Ï€R√(R²+H²). Then, find out the area of the triangle ABC. Area= RH/2. Then, we multiply the area with the area of slant height = RH/2 * Ï€R√(R²+H²). Let this value be k. D=M/V. So, M=DV. Density is ρ and volume is k. We multiply them and get mass=ρk. To find out the force, we multiply it using g. So, final answer=
ρ RH Ï€R√(R²+H²) g2

Sir,

Is this height the height of the cylinder? If yes, may I know at how much depth this cylinder is kept?

Also, do we have to find out how much force is exerted on the lateral surface of the cylinder or throughout the entire surface of the cylinder?

Thanx...

\dpi{150} \fn_cs \textit{Let us consider a small segment of radius r at a depth x below the surface}

\dpi{150} \fn_cs \frac{r}{x} = \frac{R}{H}\Rightarrow r = \frac{x\cdot R}{H}

\dpi{150} \fn_cs dA = 2\pi rdx = \frac{(2\pi Rx )dx}{H}

\dpi{150} \fn_cs \textit{Pressure by the fluid at a depth x = }x\rho g

\dpi{150} \fn_cs dF = P\cdot dA = \frac{2\pi R\rho gx^{2}dx}{H}

\dpi{150} \fn_cs F = \int \frac{2\pi R\rho gx^{2}dx}{H}

\dpi{150} \fn_cs F = \frac{2\pi H^{2}R\rho g}{3}

B=pVg=TTR²pgH3=pressure on curved part+pressure on base
pressure on base=pgHTTR²
Pressure on curved surface=B-pressure on base=TTR²pgH3-pgHTTR²