Twenty-five hundred (2500) tons of iron ore with a stowage factor of 17 is stowed in a cargo hold. The dimensions of the hold are 55 feet long and 45 feet wide and 35 feet high. What is the height of the center of gravity of the ore above the bottom of the hold?

Correct Answer: Option B — 8.6 feet

The volume of the ore is 2500 tons multiplied by the 17 cubic feet per ton stowage factor, totaling 42,500 cubic feet. Divided by the hold footprint (55x45 ft), the cargo height is 17.15 feet. Its center of gravity is halfway up, at 8.6 feet.