Question 63 ONC01 - Master/Chief Mate Unlimited Tonnage

The Correct Answer is B ### Explanation for Option B (35 cubic feet) In maritime engineering and naval architecture, particularly when using Imperial units (tons and cubic feet), the standard calculation for displacement volume relies on the density of seawater and the definition of a ton. 1. **Standard Seawater Density:** Seawater is typically assumed to weigh 64 pounds per cubic foot (64 $lb/ft^3$). 2. **Standard Maritime Ton:** When discussing ship displacement, a "ton" usually refers to the **long ton**, which equals 2,240 pounds. 3. **Calculation:** To find the volume (in cubic feet) that weighs one long ton, divide the weight of the ton by the density of the water: $$\text{Volume} = \frac{\text{Weight}}{\text{Density}} = \frac{2,240 \text{ lbs}}{64 \text{ lbs/ft}^3} = 35 \text{ cubic feet}$$ Therefore, 35 cubic feet of seawater will increase the weight displacement of the ship by one long ton. *** ### Why Other Options Are Incorrect **A) 64 cubic feet** This volume is often confused with the density (64 $lb/ft^3$). If you had 64 cubic feet of seawater, it would weigh $64 \times 64 = 4,096$ pounds, which is approximately 1.8 long tons. **C) 100 gallons** The weight of 100 gallons of seawater is significantly less than one ton. A US gallon of seawater weighs about 8.5 pounds. 100 gallons would weigh only 850 pounds, far short of the 2,240 pounds required for a long ton. **D) 500 liters** This is a metric measure. Since one liter of seawater weighs approximately 1.02 kg, 500 liters weighs about 510 kg. A metric tonne (the standard metric equivalent of a ton) is 1,000 kg. Therefore, 500 liters is only about half a metric tonne.