A ball of mass m and density `rho` is immersed in a liquid of density `3 rho` at a depth `h` and released. To what height will the ball jump up above the surface of liqud ? (neglect the reistance of water and air).

Explanation

Volume of ball `V=(m)/(rho)`
Acceleration of ball moving in upward direction inside the liquid
`a=(F_("net"))/(m)=("upthrust"-"weight")/(m)=(V_("total")rho_(l)g-mg)/(m)`
`a=(((m)/(rho))(3rho)(g)-mg)/(m)=2g" (upwards)"`
`therefore" velocity of ball while crossing the surface "`
`v=sqrt(2gh)=sqrt(4gh)`
`therefore" The ball will jump to a height "`
`H=(v^(2))/(2g)=(4gh)/(2g)=2h`