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(C) Let $V$ be the outer volume of the sphere.Given that the inner volume is half the outer volume,the volume of the material of the sphere is $V_{mat

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$1.6 	imes 10^3 \ kg/m^3$

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(C) Let $V$ be the outer volume of the sphere.Given that the inner volume is half the outer volume,the volume of the material of the sphere is $V_{material} = V - V/2 = V/2$.According to the principle of floatation,the weight of the sphere is equal to the weight of the water displaced by the submerged part.Weight of sphere = $V_{material} 	imes ho 	imes g = (V/2) 	imes ho 	imes g$,where $ho$ is the density of the material.Weight of displaced water = $V_{submerged} 	imes \sigma 	imes g = (4/5)V 	imes \sigma 	imes g$,where $\sigma = 10^3 \ kg/m^3$ is the density of water.Equating the two: $(V/2) ho g = (4/5) V \sigma g$.$ho/2 = 4\sigma/5$.$ho = (8/5) \sigma = 1.6 	imes 10^3 \ kg/m^3$.

$A$ hollow sphere has an inner volume equal to half of its outer volume. If $4/5$ of its volume is submerged when placed in water,what is the density of the material of the sphere? (Density of water = $10^3 \ kg/m^3$)

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