A solid sphere of radius r is floating at the | vedclass

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(A) Let $F$ be the downward force exerted by the upper liquid on the upper hemisphere. The upward force exerted by the liquid at the interface on the

Vedclass · Accepted Answer

$p_0\pi r^2 + (h - 2/3r)\pi r^2\rho_1g$

Vedclass · Answer

(A) Let $F$ be the downward force exerted by the upper liquid on the upper hemisphere. The upward force exerted by the liquid at the interface on the lower surface of the upper hemisphere is $P \cdot A$,where $P$ is the pressure at the interface and $A = \pi r^2$ is the cross-sectional area of the sphere at the interface.The pressure at the interface is $P = p_0 + ho_1gh$.The buoyant force $F_B$ acting on the upper hemisphere due to the upper liquid is equal to the weight of the liquid displaced by the upper hemisphere:$F_B = V_{	ext{hemisphere}} \cdot ho_1 \cdot g = (\frac{2}{3} \pi r^3) ho_1 g$.Considering the equilibrium of the upper hemisphere,the net upward force is the difference between the upward pressure force and the downward force $F$ exerted by the liquid:$P \cdot A - F = F_B$$(p_0 + ho_1gh) \pi r^2 - F = \frac{2}{3} \pi r^3 ho_1 g$Rearranging to solve for $F$:$F = (p_0 + ho_1gh) \pi r^2 - \frac{2}{3} \pi r^3 ho_1 g$$F = p_0 \pi r^2 + ho_1 g \pi r^2 (h - \frac{2}{3} r)$.

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