Question:
A vessel contains oil (Density = \(0.8 \, \text{g/cm}^3\)) over mercury (density = \(13.6 \, \text{g/cm}^3\)). A homogeneous sphere floats with half of its volume immersed in mercury and the other half in oil. The density of the material of the sphere is \( x \, \text{g/cm}^3 \). The value of \(x\) is
A vessel contains oil (Density = \(0.8 \, \text{g/cm}^3\)) over mercury (density = \(13.6 \, \text{g/cm}^3\)). A homogeneous sphere floats with half of its volume immersed in mercury and the other half in oil. The density of the material of the sphere is \( x \, \text{g/cm}^3 \). The value of \(x\) is
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If object is in multiple fluids $\longrightarrow$ take weighted average based on volume fractions.
Updated On: Apr 22, 2026
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Solution and Explanation
Concept:
Floating condition:
\[
\text{Weight} = \text{Buoyant force}
\]
Step 1: Apply buoyancy principle.
Half volume in oil and half in mercury: \[ \rho_{\text{sphere}} V g = \frac{V}{2} \rho_{\text{oil}} g + \frac{V}{2} \rho_{\text{Hg}} g \]
Step 2: Simplify.
\[ \rho_{\text{sphere}} = \frac{1}{2}(\rho_{\text{oil}} + \rho_{\text{Hg}}) \]
Step 3: Substitute values.
\[ = \frac{1}{2}(0.8 + 13.6) = \frac{14.4}{2} = 7.2 \] Thus, \(x = 7.2\).
Step 1: Apply buoyancy principle.
Half volume in oil and half in mercury: \[ \rho_{\text{sphere}} V g = \frac{V}{2} \rho_{\text{oil}} g + \frac{V}{2} \rho_{\text{Hg}} g \]
Step 2: Simplify.
\[ \rho_{\text{sphere}} = \frac{1}{2}(\rho_{\text{oil}} + \rho_{\text{Hg}}) \]
Step 3: Substitute values.
\[ = \frac{1}{2}(0.8 + 13.6) = \frac{14.4}{2} = 7.2 \] Thus, \(x = 7.2\).
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