Solution:
Applying the work-energy theorem, work done by the buoyant force over depth \(h\) equals total work done against gravity: \(V \sigma g h = V \rho g (h + H)\). Solving for height \(H\) gives \(H = \left(\frac{\sigma}{\rho} - 1\right)h\).
A small ball of density \(rho\) is immersed in a liquid of density \(sigma (\sigma > \rho)\) to a depth h and released. The height above the surface of water upto which the ball jumps is:
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