A man is sitting in a boat which is floating in a pond. If the man drinks some water from the pond, the level of the water in the pond decreases.

Step 1: Analyze the initial equilibrium of the floating boat.

For a floating boat of mass $M_{\text{boat}}$ containing a man of mass $M_{\text{man}}$, the buoyant force equals the total weight of the system.

$$\rho V_d g = (M_{\text{man}} + M_{\text{boat}}) g$$

The initial volume of water displaced by the boat, $V_d$, is:

$$V_d = \frac{M_{\text{man}} + M_{\text{boat}}}{\rho}$$

Step 2: Express the initial water level of the pond.

Let $V_{\text{water}}$ be the volume of the water in the pond, and $A$ be the horizontal cross-sectional area of the pond. The initial water level $h$ is given by:

$$h = \frac{V_{\text{water}} + V_d}{A}$$

Step 3: Determine the changes after the man drinks water.

When the man drinks a mass $m$ of water from the pond, two changes occur:

1. The volume of the liquid water in the pond decreases:

$$V_{\text{water}} \to V_{\text{water}} - \frac{m}{\rho}$$

2. The total mass of the boat-man system increases by $m$, which increases the displaced volume:

$$V_d \to \frac{M_{\text{man}} + M_{\text{boat}} + m}{\rho} = V_d + \frac{m}{\rho}$$

Step 4: Calculate the new water level.

Substitute the new values into the expression for the new water level $h'$.

$$h' = \frac{\left(V_{\text{water}} - \frac{m}{\rho}\right) + \left(V_d + \frac{m}{\rho}\right)}{A}$$ $$h' = \frac{V_{\text{water}} + V_d}{A} = h$$

The change in water level is:

$$\Delta h = 0$$

The water level remains unchanged.

Answer: The given statement is False (the level of the water in the pond remains the same).