There is a circular tube in a vertical plane. Two liquids are filled in the tube whose densities are given by $373 {kg/m}^{3}$ and $100 {kg/m}^{3}$ respectively . It is observed that the liquids do not mix and each liquid subtends an angle of $90^{\circ}$ at the centre. Radius joining their interfaces makes an angle $α$ with the vertical as shown in figure. Find the value of $α$ (in degrees) is

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Solution

From the data given in the question

At equilibrium, pressure at point A on left side and right side will be equal
$p_{A} (left) = p_{A} (right)$
$373 r g (1 - sin α) = 100 r g sin α + 100 r g cos α + 373 r g (1 - cos α)$
$[∵ p = ρ g h]$
$\Rightarrow 373 r g (cos α - sin α) = 100 r g (sin α + cos α)$
Dividing by $cos α$ on both sides,
$\Rightarrow 373 r g - 373 r g tan α = 100 r g tan α + 100 r g$
$\Rightarrow 473 r g tan α = 273 r g$
$\Rightarrow tan α = \frac{273}{473} = 0.577 = \frac{1}{\sqrt{3}}$
$\Rightarrow α = 30^{\circ}$

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Similar questions

Q. There is a circular tube in a vertical plane. Two liquids are filled in the tube whose densities are given by

373 {kg/m}^{3}

and

100 {kg/m}^{3}

respectively . It is observed that the liquids do not mix and each liquid subtends an angle of

90^{\circ}

at the centre. Radius joining their interfaces makes an angle

α

with the vertical as shown in figure. Find the value of

α

(in degrees) is

There is a circular tube in a vertical plane. Two liquids which do not mix and of densities $d_{1}$ and $d_{2}$ are filled in the tube. Each liquid subtends $90^{\circ}$ angle at centre. Radius joining their interface makes an angle $α$ with vertical. Ratio of $\frac{d_{1}}{d_{2}}$ is