Question

A vessel contains oil (density \(= 0.8\text{ gm}/\text{cm}^3\)) over mercury (density \(= 13.6\text{ gm}/\text{cm}^3\)). 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 in \(\text{gm}/\text{cm}^3\) is

• A. \(12.8\)
• B. \(7.2\)
• C. \(6.4\)
• D. \(3.3\)

Hint:

Weight of the sphere equals the sum of upthrusts from the two liquids, each acting on half the volume. That gives the mean of the two densities.

Answer 1

The Correct Option is B

Step 1: Understanding the Figure:
A beaker holds mercury at the bottom and oil above it. A homogeneous sphere floats at the boundary, with half its volume in mercury and half in oil.

Step 2: Key Formula or Approach:
A floating body is in equilibrium: its weight equals the total buoyant force. The buoyant force from each liquid is (volume immersed in that liquid) \(\times\) (density of the liquid) \(\times\, g\).

Step 3: Set up:
Let the sphere's volume be \(V\) and density \(\rho\). \[ V\rho g = \frac V2 \rho_{Hg}\, g + \frac V2 \rho_{oil}\, g \]

Step 4: Solve:
Cancel \(Vg\) from every term: \[ \rho = \frac{\rho_{Hg} + \rho_{oil}}{2} = \frac{13.6 + 0.8}{2} = \frac{14.4}{2} = 7.2\ \text{g/cm}^3 \]

Step 5: Why the other options are wrong:
The value 12.8 is \(13.6 - 0.8\), which subtracts instead of averaging. The value 6.4 is half of 12.8. The value 3.3 does not come from any correct step. Only the average of the two densities is right, because the sphere is split half and half.

Final Answer:
The density of the sphere is 7.2 g/cm\(^3\), option (B). \[ \boxed{7.2\ \text{g/cm}^3} \]