A cylindrical block floats vertically in a li | vedclass

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Question

(A) Let $V$ be the total volume of the cylindrical block and $\rho$ be its density.In the first case, the block floats in a liquid of density $\rho_{1

Vedclass · Accepted Answer

$\frac{1-x_{2}}{x_{1}-x_{2}}$

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(A) Let $V$ be the total volume of the cylindrical block and $\rho$ be its density.In the first case, the block floats in a liquid of density $\rho_{1}$. According to the law of floatation, the weight of the block equals the buoyant force:$V \rho g = V x_{1} \rho_{1} g$$\Rightarrow \rho = x_{1} \rho_{1} \quad ... (1)$In the second case, the block is fully immersed in two immiscible liquids of densities $\rho_{1}$ and $\rho_{2}$. The volume fraction $x_{2}$ is in liquid $\rho_{1}$ and the remaining fraction $(1 - x_{2})$ is in liquid $\rho_{2}$. The total buoyant force equals the weight of the block:$V \rho g = V x_{2} \rho_{1} g + V (1 - x_{2}) \rho_{2} g$$\Rightarrow \rho = x_{2} \rho_{1} + (1 - x_{2}) \rho_{2} \quad ... (2)$Equating $(1)$ and $(2)$:$x_{1} \rho_{1} = x_{2} \rho_{1} + (1 - x_{2}) \rho_{2}$$x_{1} \rho_{1} - x_{2} \rho_{1} = (1 - x_{2}) \rho_{2}$$\rho_{1} (x_{1} - x_{2}) = \rho_{2} (1 - x_{2})$$\frac{\rho_{1}}{\rho_{2}} = \frac{1 - x_{2}}{x_{1} - x_{2}}$

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