A circular tube of uniform cross section is filled with two liquids densities `rho_(1)` and `rho_(2)` such that half of each liquid occupies a quarter to volume of the tube. If the line joining the free surfaces of the liquid makes an angle `theta` with horizontal find the value of `theta.`

Explanation

Let us find the pressure at the lowes point 1. Sicne the liquid has density `rho_(2)` and height of liquid column is `h_(2)^(')` on the right and side of point 1 we have
`P_(1)=rho_(1)gh_(1)`……i

Since two liquid coluns of heights `h_(1)` and `h_(2)` and densities `rho_(1)` and `rho_(2)` are situated above point 1, on the left hand side, we have
`P_(2)=rho_(1)gh_(2)+rho_(2)gh_(2)^(')`.............ii
Equating `P_(1)` and `P_(2)` from eqn i and ii we have
`rho_(1)h_(2) +rho_(2)h_(2)^(')=rho_(1)h_(1)`...........iii
Substituting `h_(2)^(')=Rsintheta+Rcostheta, h_(2)=R(1-costheta)` and `h_(1)=R(1-sintheta)` in eqn iii we have
`rho_(1) R(1-costheta)+rho_(2)R(sintheta+costheta)=rho_(1)R(1-sintheta)`
This gives
`(costheta+sintheta)/(costheta-sintheta)=(rho_(1))/(rho_(2))impliestantheta=(rho_(1)-rho_(2))/(rho_(1)+rho_(2))`
Then `theta=tan^(-1)((rho_(1)-rho_(2))/(rho_(1)-rho_(3)))`