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A uniform long tube is bent into a circl...

A uniform long tube is bent into a circle of radius R and it lies in vertical plane. Two liquids of same volume but densities `rho " and " delta` fill half the tube. The angle `theta` is

A

`tan^(-1) ( (rho -del)/(rho+ g))`

B

`tan^(-1) rho/g`

C

`tan^(-1) del/rho`

D

`tan^(-1) ((rho+ delta)/(rho+ delta))`

Text Solution

Verified by Experts

The correct Answer is:
A

Vertical height of the liquid in portion AC
`h_1 = DO + OE = Rsin theta + Rcos theta = R(sin theta + cos theta)`
Vertical height of the liquid in portion CP
`h_2 =R-Rcos theta = R(1-cos theta)`
Vertical height of the liquid in portion PB
`h_3 = R -R sin theta = R ( 1- sin theta)`
In equilibrium, the pressure due to liquid on the both sides must be equal at the lowest point P
`delgh_1 + rho gh_2 + rhogh_3 ` [ as pressure `= h rho g`]
`del g R (sin theta + cos theta) + rho g R(1- cos theta) = rho g R (1 - sin theta)`
`del(sin theta + cos theta) + rho (1 - cos theta) = rho ( 1- sin theta)`
`(rho + del) sin theta = (rho - del) cos theta`
`tan theta = ((rho - delta))/((rho- delta)) rArr theta = tan^(-1) ((rho - g)/(rho + g))`
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