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(A) Let $V_1$ be the volume of granite in water and $V_2$ be the volume of granite in mercury.According to the principle of floatation, the total buoy

Vedclass · Accepted Answer

$\frac{\rho_2-\rho}{\rho-\rho_1}$

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(A) Let $V_1$ be the volume of granite in water and $V_2$ be the volume of granite in mercury.According to the principle of floatation, the total buoyant force acting on the granite must be equal to its weight.The buoyant force from water is $F_{B1} = V_1 \rho_1 g$.The buoyant force from mercury is $F_{B2} = V_2 \rho_2 g$.The weight of the granite is $W = (V_1 + V_2) \rho g$.Equating the forces: $V_1 \rho_1 g + V_2 \rho_2 g = (V_1 + V_2) \rho g$.Dividing by $g$: $V_1 \rho_1 + V_2 \rho_2 = V_1 \rho + V_2 \rho$.Rearranging the terms to group $V_1$ and $V_2$: $V_1 \rho_1 - V_1 \rho = V_2 \rho - V_2 \rho_2$.$V_1(\rho_1 - \rho) = V_2(\rho - \rho_2)$.Therefore, the ratio of the volume of granite in water $(V_1)$ to the volume of granite in mercury $(V_2)$ is:$\frac{V_1}{V_2} = \frac{\rho - \rho_2}{\rho_1 - \rho} = \frac{\rho_2 - \rho}{\rho - \rho_1}$.

$A$ piece of granite floats at the interface of mercury and water contained in a beaker as shown in the figure. If the densities of granite, water, and mercury are $\rho, \rho_1$, and $\rho_2$ respectively, the ratio of the volume of granite in water to the volume of granite in mercury is:

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