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(B) Let $M_i$ be the mass of the ice and $M_m$ be the mass of the metal cube. When the ice is floating, the total weight is supported by the buoyant f

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Falls

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(B) Let $M_i$ be the mass of the ice and $M_m$ be the mass of the metal cube. When the ice is floating, the total weight is supported by the buoyant force: $F_B = (M_i + M_m)g$. The volume of water displaced by the floating system is $V_1 = \frac{M_i + M_m}{\rho_w}$, where $\rho_w$ is the density of water. After the ice melts, the ice becomes water with volume $V_{ice_melted} = \frac{M_i}{\rho_w}$. The metal cube sinks, displacing a volume of water equal to its own volume: $V_{metal} = \frac{M_m}{\rho_m}$, where $\rho_m$ is the density of the metal. The total volume of water displaced after melting is $V_2 = \frac{M_i}{\rho_w} + \frac{M_m}{\rho_m}$. Since the metal sinks, its density $\rho_m > \rho_w$, which implies $\frac{M_m}{\rho_m} Therefore, $V_2 < V_1$, meaning the water level falls.

$A$ metal cube is embedded inside a block of ice that is floating on the surface of water. The ice melts completely, and the metal cube sinks into the water. What happens to the water level in the container?

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