A tube filled with water and closed at both ends uniformly rotates in a horizontal plane about the `OO'` axis. The manometers `A` and `B` fixed in the tube at distances `r_(1)` and `r_(2)` from rotational axis indicate pressure `p_(1)` and `p_(2)` respectively. Determine the angular velocity `omega` of rotation of the tube.

Options

A. `omega=sqrt((2(p_(2)-p_(1)))/(rho(r_(2)^(2)-r_(1)^(2))))`
B. `omega=sqrt((2(p_(2)-p_(1)))/(rho(r_(2)^(2)+r_(1)^(2))))`
C. `omega=sqrt((2(r_(2)^(2)-r_(1)^(2)))/(p_(1)-p_(2)))`
D. `omega=sqrt((2(p_(2)-p_(1)))/(rhor_(1)r_(2)))`

Answer

A

Explanation

Consider the condition of equilibrium for the mass of water contained between cross section separated by `x` and `x+dx` from rotation axis.
`dp=rhoomega^(2)dxdx, int_(p_(1))^(p_(2))dp=int_(r_(1))^(r_(2))rhoomega^(2) x dx `

`p_(2)-p_(1)=rhoomega^(2)[(x^(2))/2]_(r_(1))^(r_(2))=rho omega^(2)((r_(2)^(2)-r_(1)^(2))/2)`
`omega=sqrt((2(p_(2)-p_(1)))/(rho(r_(2)^(2)-r_(1)^(2))))`