## Physics: Principles with Applications (7th Edition)

Let $V_I$ be the total volume of the iron. Let $V_M$ be the volume of the iron that is submerged in the mercury. Note that $V_M$ is the volume of mercury that is displaced. The buoyant force is equal to the weight of the iron. Therefore; $F_B = M_I~g$ $\rho_M~V_M~g = \rho_I~V_I~g$ $\frac{V_M}{V_I} = \frac{\rho_I}{\rho_M}$ $\frac{V_M}{V_I} = \frac{7.8\times 10^3~kg/m^3}{13.6\times 10^3~kg/m^3}$ $\frac{V_M}{V_I} = 0.57$ The fraction of the iron that is submerged is 0.57