A rubber ball of mass m and density $ρ$ is immersed in a liquid of density $3 ρ$ to depth h and released. To what height will the ball jump up above the surface due to buoyancy force of liquid on the ball is nh then n value is?

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Detailed Solution

Acceleration of ball moving in upward direction inside the liquid

$a = \frac{F_{net}}{m} = \frac{upthrust - weight}{m} = \frac{V_{tot} ρ_{l} g - mg}{m} = \frac{(\frac{m}{ρ}) (3 ρ) g - mg}{m} = 2 g$ $velocity = \sqrt{2 ah} = \sqrt{4 gh} H = \frac{V^{2}}{2 g} = \frac{4 gh}{2 g} = 2 h$

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