A light semi cylindrical gate of radius $R$ is piovted at its mid point $O$, of the diameter as shown in the figure holding liquid of density $\rho $. The force $F$ required to prevent the rotation of the gate is equal to
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The pressure at any point in a liquid is a function of the density of the liquid and the height of the point from some base reference.
Therefore, $P=\rho g h$
Let's consider the topmost point as the base reference to be at zero pressure.
As we move below, consider a semi-circular ring of radius $r$ and thickness $dr.$
The height of this ring from the topmost reference point $=\mathrm{R}-\mathrm{r}$
Hence, the pressure outside any such ring, $\mathrm{dP}=\rho g(R-r)$
The force exerted by the fluid on any such ring, $\mathrm{dF}=d P \times A$
Hence, dF $=\rho g(R-r) \times 2 \pi r d r$
The counteracting force required to keep the gate stable is the integration of this pressure exerted force.
$\mathrm{F}=\int_{0}^{R} \rho g(R-r) \times 2 \pi r d r$
$F=\frac{\pi \rho g R^{3}}{3}$
Hence, the answer is none of these.
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