Question

A solid hemisphere is just pressed below the liquid, the value of $\frac{F_1}{F_2}$ is (where $F_1$ and $F_2$ are the hydrostatic forces acting on the curved and flat surfaces of the hemishere) (Neglect atmospheric pressure).

Answer 1

let the density be $P$ of the liquid of radius $R$

$R_1=\rho gR\left(\pi R^2\right)$

A different ring at depth $x$ from the square of with $dx$ pressure at the period $\rho gx$

Net downward force $m$ This differential ring

$df=\rho gx(2\pi R.dx)\sin \theta$

$\sin \theta =\frac{R-x}{R}$

$\int df={\int }_0^R\rho gx\left(2\pi Rdx\right).\frac{R-x}{R}$

$f_2=2\pi \rho g{\int }_0^{R}x(R-x)dx=2\pi \rho g{[\frac{R{x}^2}{2}-\frac{x^3}{3}]}_0^R$

$=2\pi \rho g\left(\frac{R^3}{6}\right)$

$\frac{f_1}{f_2}=\frac{\rho gR\left(\pi R^2\right)}{2\pi \rho g\left(\frac{R^2}{6}\right)}=3:1$