Question:
A ship consumes 250 tonne of fuel when moving from sea to a river. On arrival, draught and trim (measured at rest) are the same as the corresponding values at sea.
The displacement of the ship in seawater is ______ tonne (answer in integer).
The density of seawater and river water are 1025 kg/m\(^3\) and 1000 kg/m\(^3\) respectively.
A ship consumes 250 tonne of fuel when moving from sea to a river. On arrival, draught and trim (measured at rest) are the same as the corresponding values at sea.
The displacement of the ship in seawater is ______ tonne (answer in integer).
The density of seawater and river water are 1025 kg/m\(^3\) and 1000 kg/m\(^3\) respectively.
The displacement of the ship in seawater is ______ tonne (answer in integer).
The density of seawater and river water are 1025 kg/m\(^3\) and 1000 kg/m\(^3\) respectively.
Show Hint
Same draught in both waters means the underwater volume is unchanged, only the density differs.
Updated On: Jul 28, 2026
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Verified By Collegedunia
Correct Answer: 10250
Solution and Explanation
Step 1: Note what stays the same.
Since the draught and trim on arrival match the values at sea, the underwater volume \(V\) is the same in both cases, even though the ship is now lighter and floating in less dense water.
Step 2: Write the equilibrium condition in each water.
At sea, weight equals buoyancy: \(W_{sea} = \rho_{sea} V\). On arrival in the river: \(W_{river} = \rho_{river} V\). The drop in weight between the two is exactly the fuel burned: \(W_{sea} - W_{river} = 250\) tonne.
Step 3: Solve for the common volume V.
\((\rho_{sea} - \rho_{river}) V = 250 \times 1000\) kg, so \((1025 - 1000) V = 250000\), giving \(V = 250000/25 = 10000\) m\(^3\).
Final Answer:
The displacement at sea is \(W_{sea} = \rho_{sea} V = 1025 \times 10000 / 1000 = 10250\) tonne. \[ \boxed{W_{sea} = 10250 \text{ tonne}} \]
Since the draught and trim on arrival match the values at sea, the underwater volume \(V\) is the same in both cases, even though the ship is now lighter and floating in less dense water.
Step 2: Write the equilibrium condition in each water.
At sea, weight equals buoyancy: \(W_{sea} = \rho_{sea} V\). On arrival in the river: \(W_{river} = \rho_{river} V\). The drop in weight between the two is exactly the fuel burned: \(W_{sea} - W_{river} = 250\) tonne.
Step 3: Solve for the common volume V.
\((\rho_{sea} - \rho_{river}) V = 250 \times 1000\) kg, so \((1025 - 1000) V = 250000\), giving \(V = 250000/25 = 10000\) m\(^3\).
Final Answer:
The displacement at sea is \(W_{sea} = \rho_{sea} V = 1025 \times 10000 / 1000 = 10250\) tonne. \[ \boxed{W_{sea} = 10250 \text{ tonne}} \]
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