A cubical block (side `l`) is in equilibrium at the interface of two liquids A and B. The spring balance reads W Newton. If the block is weighted in air, the reading of the spring balance will be :

Options

A. `W+(1)/(2)(rho_(1)+rho_(2))gl^(3)`
B. `W+(1)/(2)(rho_(2)-rho_(1))gl^(3)`
C. `W+(rho_(1)+rho_(2))gl^(3)`
D. `W`

Answer

A

Explanation

`W + F_(B) - mg =0`
`mg = W + F_(B)`
`F_(B)` = Net buoyany force
`= [rho_(1)((l)/(3))(l)^(2)+rho_(2)((l)/(2))l^(2)]g =(rho_(1)+rho_(2))(l^(3))/(2) g`
`mg = W + (1)/(2)(rho_(1)+rho_(2))gl^(3)` = weight in air