Question

A closed tube in the form of an equilateral triangle of side $l$ contains equal volumes of three liquids which do not mix and is placed vertically with its lowest side horizontal. Find $x$ in the figure if the densities of the liquids are in A.P. Fill 1/x in OMR sheet.

Answer 1

$\frac{2(\sqrt{3}+\sqrt{2})}{l}$

Answer 2

1. Geometric Setup: An equilateral triangle of side $l$ has a total height $H = \frac{\sqrt{3}}{2}l$. Since the tube is vertical with its base horizontal, the liquids form horizontal layers. Equal volumes imply equal cross-sectional areas for each liquid layer.

2. Height Calculation: Let the total area of the triangle be $A_{total}$. Each liquid layer has an area $A = A_{total}/3$.

   • The top liquid layer forms a smaller equilateral triangle. If its height is $h_1$, then its area is $A_1 = (\frac{h_1}{H})^2 A_{total}$. Since $A_1 = A_{total}/3$, we get $(\frac{h_1}{H})^2 = \frac{1}{3}$, so $h_1 = \frac{H}{\sqrt{3}}$.
   • The top two layers combined occupy $2A_{total}/3$. Let the height of this combined region from the top vertex be $h_{12}$. Then $(\frac{h_{12}}{H})^2 = \frac{2}{3}$, so $h_{12} = H\sqrt{\frac{2}{3}}$.
   • The height of the bottom liquid layer, denoted by $x$ in the figure, is $x = H - h_{12} = H(1 - \sqrt{\frac{2}{3}})$.

3. Expressing x in terms of l: Substitute $H = \frac{\sqrt{3}}{2}l$: $x = \frac{\sqrt{3}}{2}l \left(1 - \sqrt{\frac{2}{3}}\right) = \frac{\sqrt{3}}{2}l \left(1 - \frac{\sqrt{2}}{\sqrt{3}}\right) = \frac{\sqrt{3}}{2}l \left(\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3}}\right) = \frac{l}{2}(\sqrt{3} - \sqrt{2})$.

4. Calculating 1/x: The question asks to fill $1/x$ in the OMR sheet. $\frac{1}{x} = \frac{2}{l(\sqrt{3} - \sqrt{2})}$. Rationalizing the denominator: $\frac{1}{x} = \frac{2(\sqrt{3} + \sqrt{2})}{l(\sqrt{3} - \sqrt{2})(\sqrt{3} + \sqrt{2})} = \frac{2(\sqrt{3} + \sqrt{2})}{l(3 - 2)} = \frac{2(\sqrt{3} + \sqrt{2})}{l}$.

The condition that densities are in A.P. is not needed to determine the geometric distribution of the liquids, which depends only on equal volumes and the shape of the container.