Question:
A ship floating in seawater has a TPC (Tonne Per Centimeter immersion) of 20.
If bilge keel of total mass 65.1 tonne and volume 44 \(\text{m}^3\) is added to the ship, then the change in the mean draft is ______ cm.
Assume that after bilge keel fitment, the TPC remains constant, and the ship undergoes parallel sinkage. Density of seawater is 1025 \(\text{kg/m}^3\).
A ship floating in seawater has a TPC (Tonne Per Centimeter immersion) of 20.
If bilge keel of total mass 65.1 tonne and volume 44 \(\text{m}^3\) is added to the ship, then the change in the mean draft is ______ cm.
Assume that after bilge keel fitment, the TPC remains constant, and the ship undergoes parallel sinkage. Density of seawater is 1025 \(\text{kg/m}^3\).
If bilge keel of total mass 65.1 tonne and volume 44 \(\text{m}^3\) is added to the ship, then the change in the mean draft is ______ cm.
Assume that after bilge keel fitment, the TPC remains constant, and the ship undergoes parallel sinkage. Density of seawater is 1025 \(\text{kg/m}^3\).
Show Hint
Subtract the buoyancy from the keel's own volume before dividing the rest by TPC.
Updated On: Jul 28, 2026
- 0.5
- 1
- 0.55
- 1.1
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The Correct Option is B
Solution and Explanation
Step 1: Split the added weight into two parts.
The bilge keel has mass 65.1 tonne and volume 44 \(\text{m}^3\), and it stays fully submerged, so its own volume displaces seawater and directly supports part of its weight.
The rest of its weight must be balanced by extra buoyancy from the ship sinking a little deeper.
Step 2: Find the buoyancy from the keel's own volume.
Buoyancy from the submerged keel volume \( = \rho_{sw} \times V = 1.025 \times 44 = 45.1 \) tonne.
Step 3: Find the weight that must be carried by parallel sinkage.
Remaining weight \( = 65.1 - 45.1 = 20.0 \) tonne, and this extra weight is what pushes the ship down further into the water.
Step 4: Use TPC to convert this weight into a draft change.
TPC gives the tonnes needed for 1 cm of parallel sinkage, so \( \text{sinkage} = \dfrac{20.0}{20} = 1.0 \) cm.
Final Answer:
The keel's own volume covers most of its weight, and only the leftover 20 tonne needs extra sinkage. \[ \boxed{\Delta d = 1.0 \text{ cm}} \]
The bilge keel has mass 65.1 tonne and volume 44 \(\text{m}^3\), and it stays fully submerged, so its own volume displaces seawater and directly supports part of its weight.
The rest of its weight must be balanced by extra buoyancy from the ship sinking a little deeper.
Step 2: Find the buoyancy from the keel's own volume.
Buoyancy from the submerged keel volume \( = \rho_{sw} \times V = 1.025 \times 44 = 45.1 \) tonne.
Step 3: Find the weight that must be carried by parallel sinkage.
Remaining weight \( = 65.1 - 45.1 = 20.0 \) tonne, and this extra weight is what pushes the ship down further into the water.
Step 4: Use TPC to convert this weight into a draft change.
TPC gives the tonnes needed for 1 cm of parallel sinkage, so \( \text{sinkage} = \dfrac{20.0}{20} = 1.0 \) cm.
Final Answer:
The keel's own volume covers most of its weight, and only the leftover 20 tonne needs extra sinkage. \[ \boxed{\Delta d = 1.0 \text{ cm}} \]
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