A cargo of 60 tons is to be loaded on deck 20 feet from the ship's centerline. The vessel's displacement including the 60 ton cargo will be 6,000 tons and the GM two feet. What would be the list of the vessel after loading this cargo?

Correct Answer: Option B — 5.72°

Applying standard transverse stability equations: $\tan(\text{List}) = \frac{60 \text{ tons} \times 20 \text{ ft}}{6000 \text{ tons} \times 2 \text{ ft}} = \frac{1200}{12000} = 0.1$. Taking the inverse tangent of 0.1 results in an equilibrium list angle of approximately 5.72 degrees.