A solid cone of uniform density and height 2R and base radius R has a conical portion scooped out from its base with the same base radius but height R as shown in the figure. The solid cone is floatin

Question

A solid cone of uniform density and height 2R and base radius R has a conical portion scooped out from its base with the same base radius but height R as shown in the figure. The solid cone is floating in a liquid of density r with vertex A touching the fluid surface. If atmospheric pressure is $P_{0}$ , the force acting on the surface BCD due to liquid is:

TestHub · Accepted Answer

Pressure at depth h below the free surface: $P = P_{0} + ρ g h$

On the scooped surface BCD, a point at slant distance x has depth $h = R + x cos θ$

Force on element: $d F = P d A$ ; components along the axis add up: $(d F)_{n e t} = d F cos θ$

Integrating over the whole conical surface BCD:

$F = \frac{π R ^{3} ρ g}{3} [5 + \frac{3 P _{0}}{Rρ g}]$

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