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

Question

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

TestHub · Accepted Answer

Let V be the volume of the ball.

New upward force $= V σ g - V ρ g$

Net upward acceleration, $a = \frac{V σ g - V ρ g}{V ρ} = \frac{(σ - ρ) g}{ρ}$

Velocity at the surface $= \sqrt{\frac{2 (σ - ρ) g h}{ρ}}$

If $h_{a}$ is the height in air to which the ball rises, then

$0 - \frac{2 (σ - ρ) g h}{ρ} = 2 (- g) h_{a}$

$∴ h_{a} = \frac{(σ - ρ) g h}{g ρ} = (\frac{σ}{ρ} - 1) h$

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