Oil of density 0.5 g/cm`""^(3)` is contained over mercury of density 13.6 g/cm`""^(3)` in a vessel. A homogeneous body is floating with its half of volume immersed inside mercury and half inside oil. The density of homogeneous material will be

Options

A. `6g//cm^(3)`
B. `7.05 g//cm^(3)`
C. 8.05 g/cc
D. 10 g/cc

Explanation

To find the density of the homogeneous body floating in the oil and mercury, we can follow these steps:

### Step-by-step Solution:

1. **Identify Given Values:**
- Density of oil (\( \rho_{oil} \)) = 0.5 g/cm³
- Density of mercury (\( \rho_{mercury} \)) = 13.6 g/cm³

2. **Understand the Problem:**
- The body is floating with half of its volume (\( V/2 \)) immersed in mercury and half in oil.
- We need to find the density of the homogeneous material (\( \rho_{body} \)).

3. **Apply the Principle of Buoyancy:**
- According to Archimedes' principle, the weight of the fluid displaced by the body is equal to the weight of the body.
- The buoyant force acting on the body due to the mercury and oil must balance the weight of the body.

4. **Calculate the Buoyant Forces:**
- Buoyant force due to mercury (\( F_{b, mercury} \)):
\[ F_{b, mercury} = \text{Volume submerged in mercury} \times \text{Density of mercury} \times g = \frac{V}{2} \times \rho_{mercury} \times g \]
- Buoyant force due to oil (\( F_{b, oil} \)):
\[ F_{b, oil} = \text{Volume submerged in oil} \times \text{Density of oil} \times g = \frac{V}{2} \times \rho_{oil} \times g \]

5. **Set Up the Equation:**
- The total buoyant force (\( F_b \)) is the sum of the buoyant forces from both fluids:
\[ F_b = F_{b, mercury} + F_{b, oil} = \left(\frac{V}{2} \times \rho_{mercury} + \frac{V}{2} \times \rho_{oil}\right) \times g \]
- The weight of the body (\( F_w \)):
\[ F_w = \text{Density of body} \times \text{Volume} \times g = \rho_{body} \times V \times g \]

6. **Equate Buoyant Force and Weight:**
\[ \rho_{body} \times V \times g = \left(\frac{V}{2} \times \rho_{mercury} + \frac{V}{2} \times \rho_{oil}\right) \times g \]
- Cancel \( V \) and \( g \) from both sides:
\[ \rho_{body} = \frac{1}{2} \left( \rho_{mercury} + \rho_{oil} \right) \]

7. **Substitute the Values:**
\[ \rho_{body} = \frac{1}{2} \left( 13.6 + 0.5 \right) = \frac{1}{2} \times 14.1 = 7.05 \, \text{g/cm}^3 \]

### Final Answer:
The density of the homogeneous material is \( 7.05 \, \text{g/cm}^3 \).

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