A right circular solid cone of radius R, height R and density $2 ρ$ suspended by a string is just immersed completely in a fluid of density $ρ .$ Neglect atmospheric pressure then

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Net horizontal force due to hydrostatic pressure on the half part of the cone ACB is $\frac{8}{3} ρ g R^{3} .$

The horizontal force exerted by the half of the cone ACB on the other half of the cone is $\frac{2}{3} ρ g π R^{3} .$

Tension in the string is $\frac{2}{3} ρ g π R^{3} .$

Net vertical force due to hydrostatic pressure on the curved surface of the half part of the cone is $\frac{8}{3} ρ g R^{3} .$

answer is B, D.

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Detailed Solution

Buoyant force on the cone = $v ρ g$
$= \frac{1}{3} π R^{2} 2 R ρ g = \frac{2}{3} ρ g π R^{3}$
If $F_{y}$ be the vertical component of the force due to fluid pressure,
$ρ g 2 R \times π R^{2} - F_{y} = \frac{2}{3} ρ g π R^{3}$
$F_{y} = \frac{4}{3} ρ g π R^{3}$
$T + \frac{2}{3} ρ g π R^{3} = \frac{1}{3} π R^{2} 2 R \times 2 ρ g$
$T = \frac{2}{3} ρ g π R^{3}$
Horizontal force exerted by the fluid on the curved surface ABC = Horizontal force exerted by imaginary vertical surface AB by the fluid.
Question Image
$F_{x} = \int d F_{x} = \int_{0}^{2 R} (ρ g y) 2 x d y$
$= \int_{0}^{2 R} ρ g y^{2} d y = \frac{8 ρ g R^{3}}{3}$