A thin uniform tube is bent into a circle of radius r in the...

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

{ρ}_{1}

and

{ρ}_{2} ( {ρ}_{1} > {ρ}_{2} ) ,

fill half the circle. The angle

\theta

between the radius vector passing through the common interface and the vertical is: A thin uniform tube is bent into a circle of radius r in the vertical plane. Equal volumes of two immiscible liquids,

A thin uniform tube is bent into a circle of radius r in the vertical plane. Equal volumes of two immiscible liquids,

Options

A $\theta = \tan^{- 1} \dfrac{\pi}{2}\left(\dfrac{{ρ}_{1 +}{ρ}_{2}}{{ρ}_{1} - {ρ}_{2}}\right)$
B $\theta = \tan^{- 1} \left[\left(\dfrac{{ρ}_{1 -}{ρ}_{2}}{{ρ}_{1} + {ρ}_{2}}\right)\right]$
C $\theta = \tan^{- 1} \dfrac{\pi}{2}\left(\dfrac{{ρ}_{2}}{{ρ}_{1}}\right)$
D $\theta = \tan^{- 1} \pi\left(\dfrac{{ρ}_{1}}{{ρ}_{2}}\right)$

Correct answer

B. $\theta = \tan^{- 1} \left[\left(\dfrac{{ρ}_{1 -}{ρ}_{2}}{{ρ}_{1} + {ρ}_{2}}\right)\right]$

Step-by-step solution

Pressure at the point

A

is the same on both sides of the tube,

{ρ}_{1} gr ( cos\theta - sin\theta ) = {ρ}_{2} gr ( sin\theta + cos\theta )

⇒ \dfrac{{ρ}_{1}}{{ρ}_{2}} = \dfrac{sin\theta + cos\theta}{cos\theta - sin\theta} = \dfrac{tan\theta + 1}{1 - tan\theta}

⇒ {ρ}_{1} - {ρ}_{1}tan\theta = {ρ}_{2} + {ρ}_{2}tan\theta

⇒ \left({ρ}_{1} + {ρ}_{2}\right)tan\theta = {ρ}_{1} - {ρ}_{2}

\theta = \tan^{- 1} \left(\dfrac{{ρ}_{1} - {ρ}_{2}}{{ρ}_{1} + {ρ}_{2}}\right)

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