A thin uniform tube is bent into a circle of radius r in the vertical plane. Equal volumes of two immiscible liquids, whose densities are `rho_(1)` and `rho_(2) (rho_(1) gt rho_(2))` fill half the circle. The angle `theta` between the radius vector passing though the common interface and the vertical is :

Options

A. `theta=tan^(-1) ((p_(1)+p_(2))/(p_(1)-p_(2))`
B. `theta=tan^(-1) ((p_(1)-p_(2))/p_(1)+p_(2))`
C. `theta=tan^(-1) ((p_(2))/p_(1))`
D. `theta=tan^(-1) ((p_(1))/p_(2))`

Answer

B

Explanation

Let r be the radius of the ciruclar tube
Let `P_(A)=P_(C)=P_(0)`

`P_(B)=P_(0)+P_(1) (r-rsin theta)g .......(i)`
`P_(B)=P_(0)+p_(2) (r sin theta+r cos theta) g +p_(1) (r-r cos theta) g .......(ii)`
Equation (i) and (ii)
`p_(1) (r-rsin theta) =p_(2) (r sin theta+r cos theta)+p_(1) (r-r cos theta)`
`p_(1) (r-r sin theta -r+r cos theta)=p_(2) (r sin theta+r cos theta)`
`p_(1) (r cos theta-r sin theta)=p_(2) (r sin theta+r cos theta)`
`p_(1)/p_(2)=(sin theta+cos theta)/(cos theta-sin theta)=(tan theta+1)/(1-tan theta)`
`p_(1)-p_(1) tan theta=p_(2)+p_(2) tan theta`
`(p_(1)+p_(2)) tan theta=p_(1)-p_(2)`
`theta=tan^(-1) ((p_(1)-p_(2))/(p_(1)+p_(2)))`