A small ball of density `rho` is placed from the bottom of a tank in which a non viscous liquid of density `2 rho` is filled . Find the time taken by the ball to reach the surface of liquid and maximum height reached above the bottom of container . (Size of the ball and any type of drag force acting on the ball are negligible )

Explanation

Acceleration of ball = `(F_(8) - W)/m uarr = (2 rho g - rhog)/(rho) = g uarr :` where `F_(B) ` & W are force of buoyancy & weight of the body respectively .
Now ` H = 0.1 + 1/2 "gt"^(2) rArr sqrt((2h)/g) = t `
Velocity at surface ` = 0+g sqrt((2H)/g) rArr v = sqrt(2Hg)`
Now maximum height above bottom
`= H + (v^(2))/(2g) = H + (2Hg)/(2g) = 2H`