Q.

A small ball of density r is immersed in a liquid of density σ(> ρ) to a depth h and released. The height above the surface of water upto which the ball will jump is

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a

(σρ-1)h

b

(ρσ-1)h

c

(ρσ+1)h

d

(σρ+1)h

answer is A.

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Detailed Solution

Let V be the volume of the ball.Net upward force = Vσg-VρgNet upward acceleration,a = Vσg-VρgVρ = (σ-ρ)gρVelocity at the surface = 2(σ-ρ)ghρIf ha is the height in air to which the ball rises, then 0 -2(σ-ρ)ghρ = 2(-g)haha = (σ-r)ghgρ = (σρ-1)h
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