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 lt rho lt rho _2 ) ` The fraction of the volume of 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 ) `