Fluid MechanicsMedium
Question
A spherical ball of radius r and relative density 0.5 is floating in equilibrium in water with half of it immersed in water. The work done in pushing the ball down so that whole of it is just immersed in water is [ρ is the density of water] -
Options
A.0.25 ρ rg
B.0.5 ρ rg
C.
πr3ρg
πr3ρgD.
πr3ρg
πr3ρgSolution
When the ball is pushed down, the water gains potential energy. Hence, gain in potential energy of water.
= (Vρ) rg -
(When half of the spherical ball is immersed in water, fuse of c.g. of displaced water =
)
= Vρrg
Loss in PE of ball = Vρ′rg -
πr4ρg
Work done =
= (Vρ) rg -
(When half of the spherical ball is immersed in water, fuse of c.g. of displaced water =
) = Vρrg
Loss in PE of ball = Vρ′rg -
πr4ρgWork done =
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