Question:medium
A ship of 4000 m\(^3\) displacement has a block coefficient of 0.8.
If the beam is 0.1 times the length and twice the draught, then the length of the ship is ______ m (answer in integer).
A ship of 4000 m\(^3\) displacement has a block coefficient of 0.8.
If the beam is 0.1 times the length and twice the draught, then the length of the ship is ______ m (answer in integer).
If the beam is 0.1 times the length and twice the draught, then the length of the ship is ______ m (answer in integer).
Show Hint
Write beam and draught as fractions of length, then use the block coefficient volume formula.
Updated On: Jul 28, 2026
Show Solution
Correct Answer: 100
Solution and Explanation
Step 1: Set up a single unknown scale factor.
Let the length be $L$. Then the beam is $B = 0.1L$ and, since the beam is twice the draught, the draught is $T = 0.05L$. All three dimensions are now written as multiples of $L$.
Step 2: Express the underwater volume and match it to the given displacement.
The block coefficient links the actual underwater volume to the box volume $L \times B \times T$, so $\nabla = C_b L B T$. Multiplying the three length terms gives $L \times 0.1L \times 0.05L = 0.005L^3$, so $\nabla = 0.8 \times 0.005 L^3 = 0.004L^3$.
Step 3: Solve the cubic.
Setting this equal to the given displacement, $0.004L^3 = 4000$, so $L^3 = 10^6$ and taking the cube root gives $L = 100$ m. Checking: $B = 10$ m, $T = 5$ m, and $0.8 \times 100 \times 10 \times 5 = 4000$ m$^3$, which confirms the result.
Final Answer:
The ship length works out to 100 m. \[ \boxed{L = 100 \text{ m}} \]
Let the length be $L$. Then the beam is $B = 0.1L$ and, since the beam is twice the draught, the draught is $T = 0.05L$. All three dimensions are now written as multiples of $L$.
Step 2: Express the underwater volume and match it to the given displacement.
The block coefficient links the actual underwater volume to the box volume $L \times B \times T$, so $\nabla = C_b L B T$. Multiplying the three length terms gives $L \times 0.1L \times 0.05L = 0.005L^3$, so $\nabla = 0.8 \times 0.005 L^3 = 0.004L^3$.
Step 3: Solve the cubic.
Setting this equal to the given displacement, $0.004L^3 = 4000$, so $L^3 = 10^6$ and taking the cube root gives $L = 100$ m. Checking: $B = 10$ m, $T = 5$ m, and $0.8 \times 100 \times 10 \times 5 = 4000$ m$^3$, which confirms the result.
Final Answer:
The ship length works out to 100 m. \[ \boxed{L = 100 \text{ m}} \]
Was this answer helpful?
0
