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A solid floats with 2/3 of its volume submerged in one liquid and with 3/4 of its volume submerged in a second liquid. If the solid is now floated in a homogeneous mixture made of equal volumes of the two liquids, what fraction of its volume will be submerged?
- 6/7
- 8/11
- 11/16
- 12/17
Correct answer: 12/17
Solution
Let body density = d. In liquid 1: d = (2/3)*rho1 => rho1 = 3d/2. In liquid 2: d = (3/4)*rho2 => rho2 = 4d/3. Mixture (equal volumes): rhoₘ = (rho1 + rho2)/2 = (3d/2 + 4d/3)/2 = (9d/6 + 8d/6)/2 = (17d/6)/2 = 17d/12. Fraction submerged f = d/rhoₘ = d/(17d/12) = 12/17.
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