A block of ice with a lead shot embedded in it is floating on water contained in a vessel. The temperature of the system is maintained at $0^\circ\text{C}$ as the ice melts. When the ice melts completely the level of water in the vessel rises.

Step 1: Set up the equilibrium condition for the floating block.

The buoyant force balances the total weight of the ice block and the embedded lead shot.

$$\rho_w V_{\text{disp}} g = (m_i + m_l) g$$

Step 2: Express the initial displaced volume.

Solve for the volume of water displaced by the floating system, $V_{\text{disp}}$.

$$V_{\text{disp}} = \frac{m_i + m_l}{\rho_w}$$ $$V_{\text{disp}} = \frac{m_i}{\rho_w} + \frac{m_l}{\rho_w}$$

Step 3: Determine the final volume after the ice melts.

The melted ice becomes water of volume $V_{\text{melt}}$, while the lead shot sinks and displaces its own volume $V_{\text{lead}}$.

$$V_{\text{melt}} = \frac{m_i}{\rho_w}$$ $$V_{\text{lead}} = \frac{m_l}{\rho_l}$$ $$V_f = \frac{m_i}{\rho_w} + \frac{m_l}{\rho_l}$$

Step 4: Compare the initial and final volumes.

Calculate the difference between the initial displaced volume and the final volume.

$$V_{\text{disp}} - V_f = \left(\frac{m_i}{\rho_w} + \frac{m_l}{\rho_w}\right) - \left(\frac{m_i}{\rho_w} + \frac{m_l}{\rho_l}\right)$$ $$V_{\text{disp}} - V_f = m_l \left(\frac{1}{\rho_w} - \frac{1}{\rho_l}\right)$$

Since the density of lead is greater than the density of water ($\rho_l > \rho_w$), the difference is positive.

$$\frac{1}{\rho_w} - \frac{1}{\rho_l} > 0$$ $$V_{\text{disp}} > V_f$$

Since the total volume occupied decreases, the water level in the vessel falls.

Answer: The water level falls (the statement that the water level rises is FALSE).