Question 2490Deck Safety70% to pass
A shaft alley divides a vessel's cargo hold into two tanks, each 25 ft. wide by 50 ft. long. Each tank is filled with salt water below the level of the shaft alley. The vessel's displacement is 6,000 tons. What is the reduction in GM due to free surface effect?
AI Explanation
The correct answer is C) .62 foot.
The reduction in GM (metacentric height) due to the free surface effect can be calculated using the formula: Reduction in GM = IFL / (Volume of Displacement). In this case, the moment of inertia of the free surface (IFL) is calculated as (25 ft)^3 / 12 = 3,125 ft^4 for each tank. With two tanks, the total IFL is 6,250 ft^4. The volume of displacement is 6,000 tons, which is equivalent to 192,000 ft^3. Plugging these values into the formula, the reduction in GM is 6,250 ft^4 / 192,000 ft^3 = 0.62 ft.
The other answer choices are incorrect because they do not accurately reflect the reduction in GM based on the given information.
Related Questions
#2488 What is the reduction in metacentric height due to free surface when a tank 60 ft. wide and 60 ft. long is partially filled with saltwater? (The vessel's displacement is 10,000 tons.) #2489 A vessel carries three slack tanks of gasoline (SG .68 #2491 A shaft alley divides a vessel's cargo hold into two tanks, each 20 ft. wide by 60 ft. long. Each tank is filled with saltwater below the level of the shaft alley. The vessel's displacement is 7,000 tons. What is the reduction in GM due to free surface effect? #2492 On a vessel of 10,000 tons displacement, compute the reduction in metacentric height due to free surface in a hold having free water on tank tops. The hold is 50 feet long and 50 feet wide. What is the reduction in metacentric height? #2493 On a vessel of 12,000 tons displacement, what is the reduction in metacentric height due to free surface when a tank 60 feet long and 60 feet wide is partially filled with water?