A solid uniform ball of volume `V` floats on the interface of two immiscible liquids (see the figure). The specific gravity of the upper liquid is `rho_(1)` and that of lower one is `rho_(2)` and the specific gravity of ball is `rho(rho_(1)gtrhogtrho_(2))` The fraction of the volume the ball in the upper liquid is

Answer

`0.8`

Explanation

Let V be the total volume of the ball and ` v ` be the volume of the ball in the upper liquid. Then ` V - v ` is the volume of the lower liquid displaced.

Using the law of floatation, we have
` V rho g = v rho _1 g + (V- v ) rho _2 g, V rho , v rho = v rho _1 + V rho _2 , V ( rho - rho _2 ) = v (rho _1 - rho _2 ) , (v)/(V) = ( rho - rho _2 ) /(rho _1 - rho _2 ) = (rho _2 - rho ) /( rho _2 - rho _1 ) `