A rectangular tube of uniform cross section has three liquids of densities `rho_(1), rho_(2)` and `rho_(3)`. Each liquid column has length `l` equal to length of sides of the equilateral triangle. Find the length `x` of the liquid of density `rho_(1)` in the horizontal limb of the tube, if the triangular tube is kept in the vertical plane.

Answer

`x=((rho_(2)-rho_(1)l)/(2rho_(2)=(rho_(1)+rho_(3)))`

Explanation

Let us consider two points `1` and `2` in the horizontal limb. Pressure at `1` is `P_(1)` and `rho_(1)gh_(1)+rho_(2)gh_(2)`

pressure at `2` is `P_(2)=rhogh_(2)^(')=rho_(3)gh_(3)`
Since `P_(1)=P_(2)` for non accelerating liquid, we have
`rho_(1)h_(1)+rho_(2)h_(2)=rho_(2)h_(2)^(')+rho_(3)h_(3)`
Substituting `h_(1)=h_(2)^(')(l-x)sin60^@ ` and `h_(2)-h_(3)=xsin60^@,` we have
`x=((rho_(2)-rho_(1))l)/(2rho_(2)-(rho_(1)+rho_(3)))`