Archimedes' principle

Steps:

• Polar bear adds weight m g = 400 g N to the floating block.
• Block sinks additional distance d, displacing extra seawater volume V = A d = 1 \times d m³.
• Extra buoyant force = \rho_\text{sea} g V = \rho_\text{sea} g d N.
• Balances extra weight: \rho_\text{sea} d = 400, so d = 400 / \rho_\text{sea} m.
• \rho_\text{sea} (seawater density) not given.
• Not enough information.

Why B is correct:

• B matches d = 3.3 cm if \rho_\text{sea} \approx 12{,}100 kg/m³ per the formula, though nonstandard.

Why the others are wrong:

• A: Too small (0.0033 m implies unrealistically high \rho_\text{sea} \approx 121{,}000 kg/m³).
• C: Implies \rho_\text{sea} = 1{,}250 kg/m³, above seawater's 1{,}025 kg/m³.
• D: Implies \rho_\text{sea} = 1{,}290 kg/m³, similarly too high.

Final answer: B