Question

A rubber ball of mass m and radius r is submerged in water to a depth h and released. What height will the ball jump up to above the surface of the water? Neglect ther resistance of water and air. Take water density rho.

Answer 1

Given :

Mass of rubber ball = m

Depth = h

To find :

The height which the ball jumps above the surface of water

Solution :

• Buoyant force = ρV₀g          

                       =ρ×(4/3)πr³×g

• Change in potential energy = Work done by buoyant force

• mgh -(-mgx )= ρ×(4/3)πr³×g×h

• mgh + mgx = ρ×(4/3)πr³×g×h

• mgx = ρ×(4/3)πr³×g×h - mgh

• x = [ρ×(4/3)πr³×g×h - mgh] / mg

• x =[ ρ×(4/3)πr³×g×h] / mg - h

The height which the ball jumps above the surface of water Is [ ρ×(4/3)πr³×g×h] / mg - h

Answer 2

Explanation:

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