A small uniform tube is bent into a circle of radius `r` whose plane is vertical. Equal volumes of two fluids whose densities are`rho` and `sigma(rhogtsigma)` fill half the circle. Find the angle that the radius passing through the interface makes with the vertical.

Options

A. `tan^-1((rho+sigma)/(rho-sigma))`
B. `tan^-1((rho-sigma)/(rho-sigma))`
C. `tan^-1((rho+sigma)/(rho+sigma))`
D. `tan^-1((rho-sigma)/(rho+sigma))`

Answer

D

Explanation

`h_(AB)=r-r cos(90^(@)- theta)=r-r sin theta`

`h_(BC)=r-r costheta`
`h_(CD)=r sin(90^(@)-theta)=r costheta`
`h_(DE)=r sin theta`
Writing pressure equation between points `A`and `E` we have
`p_(A)+(r-r sin theta) rho g-(r-r costheta) -rho g-(r cos theta)(sigma)g-(r sin theta) sigma g=p_(E)`
But `p_(A)=p_(E)=p_(gas)`
Solving this equation, we get
`tan theta =((rho-sigma)/(rho-sigma))`