There is a circular tube in a vertical plane. | vedclass

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(B) Let $R$ be the radius of the circular tube. The interface between the two liquids is at an angle $\alpha$ with the vertical.For the liquid of dens

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$\frac{1 + 	an\alpha}{1 - 	an\alpha}$

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(B) Let $R$ be the radius of the circular tube. The interface between the two liquids is at an angle $\alpha$ with the vertical.For the liquid of density $d_1$,the vertical height of the column is $h_1 = R \cos \alpha - R \sin \alpha$.For the liquid of density $d_2$,the vertical height of the column is $h_2 = R \cos \alpha + R \sin \alpha$.Since the pressure at the same horizontal level in a continuous static fluid is the same,we equate the pressures at the lowest point of the interface:$P_1 = P_2$$d_1 g h_1 = d_2 g h_2$$d_1 (R \cos \alpha - R \sin \alpha) = d_2 (R \cos \alpha + R \sin \alpha)$$\frac{d_1}{d_2} = \frac{R \cos \alpha + R \sin \alpha}{R \cos \alpha - R \sin \alpha}$Dividing the numerator and denominator by $R \cos \alpha$:$\frac{d_1}{d_2} = \frac{1 + 	an \alpha}{1 - 	an \alpha}$

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 a $90^o$ angle at the centre. The radius joining their interface makes an angle $\alpha$ with the vertical. The ratio $\frac{d_1}{d_2}$ is

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