A vertical `U`-tube is spinned about the vertical axis passing through the left hand limb, with an angular speed `omega` such that the liquid columns in the limbs have heights `h` and `3h`, as shown in the figure. Find the value of to `omega`.

Answer

`sqrt((5g)/(41h))`

Explanation

Let pressure at `P` be `p_(1)` Now at a distance `x`

`p-p_(1)=(rhoomega^(2))/2[x^(2)]_(0)^(4h)`
`p-p_(1)=(rhoomega^(2))/2xx16h^(2)`…..i
`p_(2)-p=(2rhoomega^(2))/2[x^(2)]_(4h)^(7h)`
`p^(2)-p=rhoomega^(2)xx(33h^(2))`………ii
Adding eqn i and ii `p_(2)-p_(1)=41rhoomega^(2)h^(2)`
`5rhogh=41rhoh^(2)omega^(2)impliessqrt((5g)/(41h))=omega`