A cone of radius R and height H is hanging in | vedclass

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(C) The force exerted by the liquid on the base of the cone is $F_{base} = P \cdot A = (\rho g H) \cdot (\pi R^2) = \pi \rho g H R^2$.The buoyant forc

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$\frac{4}{3} \pi \rho gHR^2$

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(C) The force exerted by the liquid on the base of the cone is $F_{base} = P \cdot A = (\rho g H) \cdot (\pi R^2) = \pi \rho g H R^2$.The buoyant force $F_B$ acting on the cone is equal to the weight of the displaced liquid,which is $F_B = V \rho g = \frac{1}{3} \pi R^2 H \rho g$.The buoyant force is the net upward force exerted by the liquid on the entire surface of the cone. It is the difference between the upward force on the slant surface $(F_{slant})$ and the downward force on the base $(F_{base})$,or vice versa depending on orientation. In this configuration,the liquid exerts a downward force on the base and an upward force on the slant surface.Thus,$F_B = F_{slant} - F_{base}$.$F_{slant} = F_B + F_{base} = \frac{1}{3} \pi R^2 H \rho g + \pi R^2 H \rho g = \frac{4}{3} \pi \rho g H R^2$.

$A$ cone of radius $R$ and height $H$ is hanging inside a liquid of density $\rho$ by means of a string,as shown in the figure. The force due to the liquid acting on the slant surface of the cone is:

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