Question

The waterplane area of a ship floating in seawater is \(2000 \, \text{m}^2\). The density of seawater is \(1025 \, \text{kg/m}^3\). If a mass of \(246 \, \text{tonnes}\) is added to the ship, then the TPC (Tonnes Per Centimeter immersion) and increase in draft (in cm) respectively are:

• A. 20.50 and 12
• B. 20 and 12.3
• C. 20.50 and 24
• D. 10.25 and 24.6

Hint:

1. TPC is directly proportional to the waterplane area and the density of water.
2. Ensure all units are consistent when calculating draft changes.
3. For quick approximations, use \(\Delta d = \text{Mass}/\text{TPC}\).

Answer 1

The Correct Option is A

Step 1: Calculate TPC (Tonnes Per Centimeter immersion). The TPC is calculated using the formula: \[ \text{TPC} = \frac{\text{Waterplane Area} \times \text{Density of Water}}{100} \] Where: - Waterplane Area \( = 2000 \, \text{m}^2 \), - Density of seawater \( = 1025 \, \text{kg/m}^3 = 1.025 \, \text{tonnes/m}^3\). Substitute the values: \[ \text{TPC} = \frac{2000 \times 1.025}{100} = 20.50 \, \text{tonnes per cm}. \] Step 2: Calculate the increase in draft. The increase in draft (\(\Delta d\)) is calculated as: \[ \Delta d = \frac{\text{Added Mass}}{\text{TPC}} \] Where: - Added Mass \( = 246 \, \text{tonnes} \), - TPC \( = 20.50 \, \text{tonnes/cm} \). Substitute the values: \[ \Delta d = \frac{246}{20.50} = 12 \, \text{cm}. \] Conclusion: The TPC is \( \mathbf{20.50 \, \text{tonnes per cm}} \) and the increase in draft is \( \mathbf{12 \, \text{cm}} \). The correct option is \( \mathbf{(A)} \).