A body of density `rho` is released gently on the surface of a layer of liquid of depth d and density `rho'(rho' lt rho)` . Show that it will reach the bottom of the liquid after a time.
`[(2 d rho)/(g(rho-rho'))]^(1//2)` .

Explanation

Effective downward force on the body is
`F= mg -` upthrust = weight of balloon + tension in string
`=mg = (m/rho) rho' g = mg (1-(rho')/(rho))`
downward ac celerartion of body,
`a = F/m = g (1-(rho')/(rho)) = (g(rho-rho'))/(rho)`
Using the relation, `S= u t + 1/2 at^(2)`, we have
`S=d, u=0, a = (g (rho-rho'))/(rho)`, then `d= 0+ 1/2 (g (rho-rho'))/(rho) t^(2)`
or `t = [(2 d rho)/(g(rho-rho'))]^(1//2)` .