A block of wood floats in fresh water with 0.715 of its volume V submerged and i

Question

A block of wood floats in fresh water with 0.715 of its volume V submerged and in oil with 0.939 V submerged. Find the density of (a) the wood and (b) the oil.

Explanation / Answer

let p termed as density the block displaces 0.715V of fluid in water: ma = (weight of block) - (buoyant force) = p(wood)Vg - 0.715p(water)Vg. Since the block floats, a = 0, and: 0 = p(wood)Vg - 0.715p(water)Vg ==> p(wood) = 0.715p(water). Applying Archimede's Principle to the block in oil: ma = (weight of block) - (buoyant force) = p(wood)Vg - 0.939p(oil)Vg. a = 0, so: 0 = p(wood)Vg - 0.939p(water)Vg p(wood) = 0.939p(oil). By comparing the values of p(wood): 0.731p(water) = 0.858p(oil) ==> p(oil) = (0.731/0.858)p(water) = (0.715/0.939)(1000 kg/m^3) = 761 kg/m^3. Then, since p(wood) = 0.731p(water): p(wood) = (0.715)(1000 kg/m^3) = 715 kg/m^3.

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