A uniform plastic sphere of radius 31 - 3 meter is floating fully immersed in a fluid of density - as shown- If the force of the fluid pressure on the lower half of the sphere is - g- then find value of - Neglect atmospheric pressure-

F_buoyancy=F_Lower half F_upper half F_Lower=F_b+F_upper =F_buoyancy+weight of liquid above =43π r^3 ρ g+(π r^2 × 2r 23π r^3)ρ g =83πρ gr^3 =83 [ 3^1/3 ]^3πρ g= ...

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Standard XII

Physics

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A uniform plastic sphere of radius $3^{1 / 3}$ meter is floating fully immersed in a fluid of density $ρ$ as shown. If the force of the fluid pressure on the lower half of the sphere is $α π ρ g$ , then find value of $α$ . Neglect atmospheric pressure.

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Solution

$F_{buoyancy} = F_{Lower half} - F_{upper half}$
$F_{Lower} = F_{b} + F_{upper}$
$= F_{buoyancy} + weight of liquid above$
$= \frac{4}{3} π r^{3} ρ g + (π r^{2} \times 2 r - \frac{2}{3} π r^{3}) ρ g$
$= \frac{8}{3} π ρ g r^{3}$
$= \frac{8}{3} {[3^{\frac{1}{3}}]}^{3} π ρ g = 8 π ρ g$

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A uniform plastic sphere of radius 31/3 meter is floating fully immersed in a fluid of density ρ as shown. If the force of the fluid pressure on the lower half of the sphere is απρg, then find value of α. Neglect atmospheric pressure.

Fbuoyancy=FLower half−Fupper half FLower=Fb+Fupper =Fbuoyancy+weight of liquid above =43πr3ρg+(πr2×2r−23πr3)ρg =83πρgr3 =83[313]3πρg=8πρg

A uniform plastic sphere of radius $3^{1 / 3}$ meter is floating fully immersed in a fluid of density $ρ$ as shown. If the force of the fluid pressure on the lower half of the sphere is $α π ρ g$ , then find value of $α$ . Neglect atmospheric pressure.

$F_{buoyancy} = F_{Lower half} - F_{upper half}$
$F_{Lower} = F_{b} + F_{upper}$
$= F_{buoyancy} + weight of liquid above$
$= \frac{4}{3} π r^{3} ρ g + (π r^{2} \times 2 r - \frac{2}{3} π r^{3}) ρ g$
$= \frac{8}{3} π ρ g r^{3}$
$= \frac{8}{3} {[3^{\frac{1}{3}}]}^{3} π ρ g = 8 π ρ g$