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Two bodies of equal volume connected by a string are floating with one of them completely submerged and the other with half of its volume outside the liquid as in the figure (Below, Left). Find the densities of the materials of the bodies if the densities are in the ratio $$1 \colon 3$$. If the volume of the floating body is doubled, find the fraction of the floating body that now lies submerged in the liquid. (Density of liquid $$= 800\text{ kg m}^{-3}$$)
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Correct Answer: 0
PLACEHOLDER ANSWER - the question has two answers. With densities $$\rho$$ and $$3\rho$$ and each volume $$V$$, total weight equals buoyancy, $$4\rho V = 800 \times 1.5V$$, so $$\rho = 300\text{ kg m}^{-3}$$ and the other body is $$900\text{ kg m}^{-3}$$. If the upper body's volume becomes $$2V$$, the weight is $$300 \times 2V + 900V = 1500V$$. With a fraction $$f$$ of it submerged, $$800(V + 2fV) = 1500V$$, giving $$f = \frac{7}{16}$$.
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