A container contains mercury (rho = 13.6 text | vedclass

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(C) Let $V$ be the total volume of the sphere and $\rho$ be the density of the sphere.According to the principle of flotation,the weight of the sphere

Vedclass · Accepted Answer

$7.2$

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(C) Let $V$ be the total volume of the sphere and $ho$ be the density of the sphere.According to the principle of flotation,the weight of the sphere is equal to the total buoyant force exerted by the two liquids.Weight of the sphere = $V ho g$Buoyant force due to oil = $(\frac{V}{2}) ho_{	ext{oil}} g$Buoyant force due to mercury = $(\frac{V}{2}) ho_{	ext{Hg}} g$Equating the forces: $V ho g = \frac{V}{2} ho_{	ext{oil}} g + \frac{V}{2} ho_{	ext{Hg}} g$Dividing by $Vg$: $ho = \frac{ho_{	ext{oil}} + ho_{	ext{Hg}}}{2}$Substituting the values: $ho = \frac{0.8 + 13.6}{2} = \frac{14.4}{2} = 7.2 	ext{ g/cm}^3$.

$A$ container contains mercury $(\rho = 13.6 \text{ g/cm}^3)$ and oil $(\rho = 0.8 \text{ g/cm}^3)$. $A$ sphere floats such that half of its volume is in mercury and half of its volume is in oil. What is the density of the material of the sphere in $\text{g/cm}^3$?

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