A cube of ice has an iron piece frozen inside | vedclass

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(B) Let $m_i$ be the mass of ice and $m_f$ be the mass of the iron piece.Before the ice melts, the total weight of the system is $(m_i + m_f)g$. Since

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level decreases

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(B) Let $m_i$ be the mass of ice and $m_f$ be the mass of the iron piece.Before the ice melts, the total weight of the system is $(m_i + m_f)g$. Since the system is floating, the buoyant force equals the weight of the system: $F_b = (m_i + m_f)g$.The volume of water displaced by the floating system is $V_{disp} = \frac{F_b}{\rho_w g} = \frac{m_i + m_f}{\rho_w} = \frac{m_i}{\rho_w} + \frac{m_f}{\rho_w}$.After the ice melts, the ice becomes water with volume $V_{water\_from\_ice} = \frac{m_i}{\rho_w}$. The iron piece sinks to the bottom, displacing its own volume $V_{iron} = \frac{m_f}{\rho_f}$.The total volume occupied by the water and the iron piece is $V_{total} = \frac{m_i}{\rho_w} + \frac{m_f}{\rho_f}$.Comparing the displaced volumes: $V_{disp} - V_{total} = (\frac{m_i}{\rho_w} + \frac{m_f}{\rho_w}) - (\frac{m_i}{\rho_w} + \frac{m_f}{\rho_f}) = m_f(\frac{1}{\rho_w} - \frac{1}{\rho_f})$.Since the density of iron $\rho_f$ is greater than the density of water $\rho_w$, we have $\frac{1}{\rho_w} > \frac{1}{\rho_f}$.Therefore, $V_{disp} > V_{total}$, which means the water level decreases.

$A$ cube of ice has an iron piece frozen inside it. The cube floats in a beaker filled with water. When the ice melts, the level of water in the beaker:

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