A small ball of density `rho` is immersed in a liquid of density `sigma(gtrho)` to a depth `h` and released. The height above the surface of water up to which the ball will jump is

Explanation

Let V be the volume of the ball.
Weight of the ball = `V rho g`
Upward thrust on ball = `V sigma g`
Effective upward thrust = `V sigma g - V rho g`
Upward ac celeration
`a = (V sigma g - V rho g)/(V rho ) = ((sigma - rho )/(rho)) g`
Let `upsilon` be the velocity of ball while reaching the surface of water, after being released at depth h in water.
then `upsilon^(2) = 2 aS = 2 xx ((sigma - rho )/(rho)) gh`
for the motion of ball outside the water, K.E. at the surface of water = P.E. at height H.
`1.2 upsilon^(2) = mg H`
or `H=(upsilon^2)/(2g) = ((sigma - rho )/(rho)) h = ((sigma)/(rho) - 1)h`.