On a vessel of 6500 tons displacement, a...

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Question 1 of 2066636eec28f7522a1c51d0a0

On a vessel of 6500 tons displacement, a tank 30 ft. long, 32 ft. wide and 18 ft. deep is half filled with liquid cargo (S.G. 1.048) while the vessel is floating in saltwater (S.G. 1.026). What is the free surface constant for this tank?

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Question 1 of 20

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🔍 Key Concepts

• Free Surface Effect (FSE) and how it reduces a vessel's metacentric height (GM) • Formula for the Free Surface Constant (FSC) in terms of tank dimensions and displacement • Effect (or non‑effect) of different specific gravities (S.G.) of tank liquid vs. surrounding water on the FSC

💭 Think About

• Start by recalling the standard formula for free surface constant in feet-tons units and identify which tank dimensions are used in the numerator and how displacement appears in the denominator. • Decide whether the free surface constant depends only on the geometry of the tank and the ship’s displacement, or if you must also directly include the liquid’s specific gravity in the calculation. • Compute the second moment term for the tank’s free surface area, plug in the vessel displacement (6500 tons), and compare the resulting value to the closest multiple-choice option.

✅ Before You Answer

• Be sure you are using area moment of inertia about the correct axis for the free surface (i.e., which tank dimension is breadth vs. length for transverse stability). • Confirm your unit consistency: feet for length/ breadth/ depth and tons for displacement, and that your constant’s units match the choices. • Verify whether the different S.G. values (1.048 vs. 1.026) change the free surface constant itself or only affect the actual free surface correction (FSC × free surface correction factor).

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