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

Hint:

If object is in multiple fluids $\longrightarrow$ take weighted average based on volume fractions.

Answer 1

Step 1: Understanding the Concept:
Archimedes' Principle states that the buoyant force on an object is equal to the weight of the fluid it displaces. For a floating object, the total buoyant force from all layers of fluid must balance the weight of the object.
Step 2: Key Formula or Approach:
Equilibrium condition: \( \text{Weight of Sphere} = \text{Buoyant Force from Oil} + \text{Buoyant Force from Mercury} \).
\[ V \rho_s g = V_{oil} \rho_{oil} g + V_{Hg} \rho_{Hg} g \]
Step 3: Detailed Explanation:
Let the total volume of the sphere be \( V \).
The problem states half is in oil and half in mercury:
\( V_{oil} = \frac{V}{2} \) and \( V_{Hg} = \frac{V}{2} \).
Weight of sphere: \( W = V \rho_s g \).
Buoyant force: \( F_B = \left( \frac{V}{2} \right) \rho_{oil} g + \left( \frac{V}{2} \right) \rho_{Hg} g \).
Setting \( W = F_B \):
\[ V \rho_s g = \frac{V}{2} (\rho_{oil} + \rho_{Hg}) g \]
Dividing by \( Vg \):
\[ \rho_s = \frac{\rho_{oil} + \rho_{Hg}}{2} \]
\[ \rho_s = \frac{0.8 + 13.6}{2} = \frac{14.4}{2} = 7.2 \text{ g/cm}^3 \]
Thus, \( x = 7.2 \).
Step 4: Final Answer:
The value of x is 7.2.