A liquid of density `rho` is rotated by an angular speed `omega` as shown in figure, Using the concept of pressure equation, find a relation between `h_(1),h_(2),x_(1)` and `x_(2)`.

Explanation

Writing pressure equation between points `A` and `B`, we have
`p_(A)+rho gh_(1)-(rho omega^(2)x_(1)^(2))/(2)+(rho omega^(2)x_(2)^(2))/(2)-rho gh_(2)=p_(B)`
Points `A` and `B` are open to atmosphere.
Therefore,
`p_(A) = p_(B) = p_(0) =` atmospheric pressure.
Substituting the values in above equation we have,
`rho gh_(1) - (rho omega^(2)x_(1)^(2))/(2) +(rho omega^(2) x_(2)^(2))/(2) -rho g h_(2) = 0`
On simplifying we get,
`(h_(2)-h_(1))=(omega^(2))/(2g)(x_(2)^(2)-x_(1)^(2))`
This is the desired relation between `h_(1),h_(2),x_(1)` and `x_(2)`.