Question:
A homogeneous solid cylinder of length L and cross-sectional area \(A/5\) is immersed such that it floats with its axis vertical at the liquid-liquid interface with length \(L/4\) in the denser liquid as shown in the figure. The lower density liquid is open to atmosphere having pressure \(p_{0}\). Then, density of the solid cylinder is
A homogeneous solid cylinder of length L and cross-sectional area \(A/5\) is immersed such that it floats with its axis vertical at the liquid-liquid interface with length \(L/4\) in the denser liquid as shown in the figure. The lower density liquid is open to atmosphere having pressure \(p_{0}\). Then, density of the solid cylinder is
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Equate the weight of the cylinder to the sum of the buoyant forces from the two liquid layers it displaces.
Updated On: Nov 12, 2026
- \(\frac{5d}{4}\)
- \(\frac{4d}{5}\)
- \(4d\)
- \(\frac{d}{5}\)
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept
The cylinder floats at the interface, so its weight equals the total upthrust from the two liquids. Atmospheric pressure acts equally on the top and bottom faces and adds no net force.
Step 2: Set up
From the figure the upper liquid has density \(d\) and the lower one \(2d\). The length \(\dfrac{L}{4}\) is in the lower liquid and \(\dfrac{3L}{4}\) in the upper one.
Step 3: Balance of forces
Let the cylinder density be \(\rho\) and its cross-section \(S\):
\(\rho LSg=\dfrac{3L}{4}S\,d\,g+\dfrac{L}{4}S\,(2d)\,g\)
Step 4: Solve
\(\rho=\dfrac{3d}{4}+\dfrac{2d}{4}=\dfrac{5d}{4}\), option (A).
Final Answer:
The density of the cylinder is \(\dfrac{5d}{4}\). \[ \boxed{\text{(A) }\dfrac{5d}{4}} \]
The cylinder floats at the interface, so its weight equals the total upthrust from the two liquids. Atmospheric pressure acts equally on the top and bottom faces and adds no net force.
Step 2: Set up
From the figure the upper liquid has density \(d\) and the lower one \(2d\). The length \(\dfrac{L}{4}\) is in the lower liquid and \(\dfrac{3L}{4}\) in the upper one.
Step 3: Balance of forces
Let the cylinder density be \(\rho\) and its cross-section \(S\):
\(\rho LSg=\dfrac{3L}{4}S\,d\,g+\dfrac{L}{4}S\,(2d)\,g\)
Step 4: Solve
\(\rho=\dfrac{3d}{4}+\dfrac{2d}{4}=\dfrac{5d}{4}\), option (A).
Final Answer:
The density of the cylinder is \(\dfrac{5d}{4}\). \[ \boxed{\text{(A) }\dfrac{5d}{4}} \]
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