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(A) Let the vertical line passing through the center be the reference. The interface between the two liquids is at an angle $	heta$ from the vertical

Vedclass · Accepted Answer

$	an ^{-1}\left(\frac{ho-\delta}{ho+\delta}ight)$

Vedclass · Answer

(A) Let the vertical line passing through the center be the reference. The interface between the two liquids is at an angle $	heta$ from the vertical.Since the tube is in a vertical plane,the pressure at the lowest point must be the same from both sides.Let the density of the liquid on the left be $\delta$ and on the right be $ho$.The vertical height of the center of mass of the liquid column of density $\delta$ from the lowest point is $h_1 = R(1 - \cos 	heta)$.The vertical height of the center of mass of the liquid column of density $ho$ from the lowest point is $h_2 = R(1 - \cos 	heta)$.For equilibrium,the pressure at the lowest point of the tube must be equal from both sides.The pressure exerted by a liquid column in a circular tube is proportional to the vertical depth of its center of mass.The condition for equilibrium is given by:$\delta g R(\cos 	heta + \sin 	heta) = ho g R(\cos 	heta - \sin 	heta)$Dividing both sides by $gR$:$\delta(\cos 	heta + \sin 	heta) = ho(\cos 	heta - \sin 	heta)$$\delta \cos 	heta + \delta \sin 	heta = ho \cos 	heta - ho \sin 	heta$$\sin 	heta(ho + \delta) = \cos 	heta(ho - \delta)$$	an 	heta = \frac{ho - \delta}{ho + \delta}$$	heta = 	an ^{-1}\left(\frac{ho - \delta}{ho + \delta}ight)$

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

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