A boat floating in a water tank is carrying a number of large stones. If the stones are unloaded into water, what will happen to the water level?

Step 1: Analyze the initial state (stones inside the floating boat).

Let $m_b$ be the mass of the boat, $m_s$ be the mass of the stones, and $\rho_w$ be the density of water. Since the boat and stones float together, the total buoyant force equals their combined weight.

$$(m_b + m_s)g = \rho_w V_{\text{disp,1}} g$$ $$V_{\text{disp,1}} = \frac{m_b + m_s}{\rho_w}$$

Step 2: Analyze the final state (stones unloaded into the water).

Since the density of the stones is greater than the density of water ($\rho_s > \rho_w$), the stones sink to the bottom of the tank. The boat continues to float alone.

$$V_{\text{boat}} = \frac{m_b}{\rho_w}$$ $$V_{\text{stone}} = \frac{m_s}{\rho_s}$$

Since $\rho_s > \rho_w$, the volume of water displaced by the submerged stones is less than the volume of water equivalent to their mass.

$$V_{\text{stone}} < \frac{m_s}{\rho_w}$$

Step 3: Compare the total displaced volumes.

The total volume of water displaced in the second state is the sum of the volumes displaced by the floating boat and the submerged stones.

$$V_{\text{disp,2}} = \frac{m_b}{\rho_w} + \frac{m_s}{\rho_s}$$

Comparing the two displaced volumes shows that the total displaced volume decreases.

$$V_{\text{disp,2}} < V_{\text{disp,1}}$$

Step 4: Determine the change in water level.

The height $h$ of the water level in a tank of cross-sectional area $A$ depends on the constant volume of water $V_{\text{water}}$ and the displaced volume $V_{\text{disp}}$.

$$h = \frac{V_{\text{water}} + V_{\text{disp}}}{A}$$ $$\Delta h = \frac{V_{\text{disp,2}} - V_{\text{disp,1}}}{A} < 0$$

Since the total displaced volume decreases, the water level in the tank must decrease.

Answer: The water level falls.