Equal volumes of two liquids A and B whose densities are in the ratio $$1 \colon 2$$ are mixed to form a homogenous mixture. When a solid body C is dropped in to this mixture, it floats with half of its volume submerged. If the ratio of the volumes liquids A and B is changed to $$2 \colon 1$$, find the fraction of the submerged volume of C.

Let the densities be $$\rho$$ and $$2\rho$$. The equal-volume mixture has density $$1.5\rho$$, and floating half submerged gives $$\rho_C = 0.75\rho$$. With volumes in the ratio $$2 \colon 1$$, the mixture density is $$\frac{2\rho + 2\rho}{3} = \frac{4\rho}{3}$$. The fraction submerged is $$\frac{0.75\rho}{4\rho/3} = \frac{9}{16} = 0.5625$$.