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Pascal's Law

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Question

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 $\frac{d_{1}}{d_{2}}$ is:

\frac{1 + tan α}{1 - tan α}

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\frac{1 + sin α}{1 - cos α}

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\frac{1 + sin α}{1 - sin α}

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\frac{1 + cos α}{1 - cos α}

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Solution

The correct option is A $\frac{1 + tan α}{1 - tan α}$
Let the point be O of tube touching the ground. Pressure due to liquid of density $d_{1}$ on the left side of O = pressure due to liquid of density $d_{2}$ on the right side of O.

$d_{1} (R - R sin α) = d_{1} (R - R cos α) + d_{2} (R sin α + R cos α)$
$d_{1} (1 - sin α) = d_{1} (1 - cos α) + d_{2} (sin α + cos α)$
$d_{1} - d_{1} sin α = d_{1} - d_{1} cos α + d_{2} (sin α + cos α)$
$d_{1} (cos α - sin α) = d_{2} (sin α + cos α)$
$\frac{d_{1}}{d_{2}} = \frac{sin α + cos α}{cos α - sin α} = \frac{1 + tan α}{1 - tan α}$

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There is a circular tube in a vertical plane. Two liquids which do not mix and of densities d1 and d2 are filled in the tube. Each liquid subtends 90∘ angle at centre. Radius joining their interface makes an angle α with vertical. Ratio d1d2 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 $\frac{d_{1}}{d_{2}}$ is: