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Question: A closed tube in the form of an equilateral triangle of side l contains equal volumes of three liqui...

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.

A

3+1l\frac{\sqrt{3}+1}{l}

B

3−1l\frac{\sqrt{3}-1}{l}

C

2(3+1)l\frac{2(\sqrt{3}+1)}{l}

D

2(3−1)l\frac{2(\sqrt{3}-1)}{l}

Answer

3+1l\frac{\sqrt{3}+1}{l}

Explanation

Solution

Let the height of the equilateral triangle be H=32lH = \frac{\sqrt{3}}{2}l. Since the three liquids have equal volumes, they occupy equal areas in the 2D cross-section of the triangle, each being A/3A/3, where AA is the total area. Let the densities of the liquids be ρ1,ρ2,ρ3\rho_1, \rho_2, \rho_3 in arithmetic progression. Due to gravity, the liquids will arrange themselves in order of density, with the densest at the bottom and the least dense at the top. Let the densities from top to bottom be ρtop<ρmiddle<ρbottom\rho_{top} < \rho_{middle} < \rho_{bottom}.

The area of a similar triangle with height h′h' from the vertex is proportional to (h′)2(h')^2. If the height from the vertex to the top surface of the first liquid (topmost) is h1′h'_1, then the area occupied by this liquid is A(h1′/H)2A(h'_1/H)^2. Given that this area is A/3A/3, we have A(h1′/H)2=A/3A(h'_1/H)^2 = A/3, which implies h1′=H/3h'_1 = H/\sqrt{3}. This is the height of the top liquid layer.

For the top two liquids, the combined area is 2A/32A/3. If h2′h'_2 is the height from the vertex to the interface between the second and third liquid (from the top), then A(h2′/H)2=2A/3A(h'_2/H)^2 = 2A/3, which implies h2′=H2/3h'_2 = H\sqrt{2/3}.

The height of the top liquid layer is h1=h1′=H/3h_1 = h'_1 = H/\sqrt{3}. The height of the middle liquid layer is h2=h2′−h1′=H2/3−H/3=H(2−13)h_2 = h'_2 - h'_1 = H\sqrt{2/3} - H/\sqrt{3} = H(\frac{\sqrt{2}-1}{\sqrt{3}}). The height of the bottom liquid layer is h3=H−h2′=H−H2/3=H(1−23)h_3 = H - h'_2 = H - H\sqrt{2/3} = H(1-\frac{\sqrt{2}}{\sqrt{3}}).

The figure shows xx as the vertical height of the interface between the middle and top liquid, measured from the base of the triangle. This interface is the top surface of the middle liquid. Therefore, xx is the sum of the heights of the bottom and middle liquid layers: x=h3+h2=H(1−23)+H(2−13)=H(1−23+2−13)=H(1+−2+2−13)=H(1−13)x = h_3 + h_2 = H(1-\frac{\sqrt{2}}{\sqrt{3}}) + H(\frac{\sqrt{2}-1}{\sqrt{3}}) = H(1 - \frac{\sqrt{2}}{\sqrt{3}} + \frac{\sqrt{2}-1}{\sqrt{3}}) = H(1 + \frac{- \sqrt{2} + \sqrt{2} - 1}{\sqrt{3}}) = H(1 - \frac{1}{\sqrt{3}}).

Substituting H=32lH = \frac{\sqrt{3}}{2}l: x=32l(1−13)=32l−l2=l2(3−1)x = \frac{\sqrt{3}}{2}l (1 - \frac{1}{\sqrt{3}}) = \frac{\sqrt{3}}{2}l - \frac{l}{2} = \frac{l}{2}(\sqrt{3}-1).

The question asks to fill 1/x1/x in the OMR sheet. 1x=2l(3−1)\frac{1}{x} = \frac{2}{l(\sqrt{3}-1)}. To rationalize the denominator, multiply the numerator and denominator by (3+1)(\sqrt{3}+1): 1x=2(3+1)l(3−1)(3+1)=2(3+1)l(3−1)=2(3+1)2l=3+1l\frac{1}{x} = \frac{2(\sqrt{3}+1)}{l(\sqrt{3}-1)(\sqrt{3}+1)} = \frac{2(\sqrt{3}+1)}{l(3-1)} = \frac{2(\sqrt{3}+1)}{2l} = \frac{\sqrt{3}+1}{l}.