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Equal volumes of two miscible liquids of relative densities 6 and 2 are mixed to form a homogenous mixture. A cube floats on this liquid mixture with half its volume submerged. If instead equal masses of these liquids were mixed, find the fraction of the volume of the cube that would submerge in the mixture.
Correct Answer: 0.67
The equal-volume mixture has relative density $$\frac{6 + 2}{2} = 4$$, and floating half submerged gives the cube a relative density of 2. The equal-mass mixture has relative density $$\frac{2 \times 6 \times 2}{6 + 2} = 3$$. So the fraction submerged is $$\frac{2}{3} \approx 0.67$$.
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