First slide

Fluid statics

Question

A wooden ball of density D is immersed in water of density d to a depth h and then released . The height above the surface of water up to which the ball will jump out of water is ( neglect viscous effect )

Moderate

A

$\frac{d}{D} h$
B

$(\frac{d}{D} - 1) h$
C

h
D

zero

Solution

$Let the volume of the ball be V . Force on the ball due to upthrust = V d g Net upward force = V d g - V D g .' . Upward accelemtion is given bY VDa = Vdg - VDg a = (\frac{d - D}{D}) g velocity on reaching the surface, u = \sqrt{2 ah} Further u = \sqrt{2 g H} ∴ 2 ah = 2 gH or H = \frac{ah}{g} So, H = (\frac{d - D}{D}) h = (\frac{d}{D} - 1) h$

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