A metallic sphere floats in an immiscible mixture of water `(rho_(W) = 10^(3)kg//m^(3))` and a liquid `(rho_(1) = 13.5 xx 10^(3) kg//m^(3))` such that is (4/5) th portion is in water and (1/5) th portion in the liquid. The density of metal is

Options

A. `4.5 xx 10^(3) kg//m^(3)`
B. `4.0 xx 10^(3) kg//m^(3)`
C. `3.5 xx 10^(3) kg//m^(3)`
D. `1.9 xx 10^(3) kg//m^(3)`

Explanation

To find the density of the metallic sphere that floats in an immiscible mixture of water and another liquid, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**:
The metallic sphere is floating in a mixture of two liquids: water and another liquid. The sphere is submerged such that \( \frac{4}{5} \) of it is in water and \( \frac{1}{5} \) is in the other liquid.

2. **Identify the Densities**:
- Density of water, \( \rho_W = 10^3 \, \text{kg/m}^3 \)
- Density of the other liquid, \( \rho_1 = 13.5 \times 10^3 \, \text{kg/m}^3 \)

3. **Calculate the Density of the Mixture**:
The density of the mixture \( \rho_m \) can be calculated using the weighted average based on the proportions of the sphere submerged in each liquid:
\[ \rho_m = \left(\frac{4}{5} \cdot \rho_W\right) + \left(\frac{1}{5} \cdot \rho_1\right) \]

4. **Substituting the Values**:
Substitute the known densities into the equation:
\[ \rho_m = \left(\frac{4}{5} \cdot 10^3\right) + \left(\frac{1}{5} \cdot 13.5 \times 10^3\right) \]

5. **Calculating Each Term**:
- Calculate \( \frac{4}{5} \cdot 10^3 = 800 \, \text{kg/m}^3 \)
- Calculate \( \frac{1}{5} \cdot 13.5 \times 10^3 = 2700 \, \text{kg/m}^3 \)

6. **Adding the Two Results**:
Now, add the two results to find the density of the mixture:
\[ \rho_m = 800 + 2700 = 3500 \, \text{kg/m}^3 \]

7. **Conclusion**:
The density of the metallic sphere must equal the density of the mixture for it to float. Thus, the density of the metal is:
\[ \rho_{metal} = 3.5 \times 10^3 \, \text{kg/m}^3 \]

### Final Answer:
The density of the metal is \( 3.5 \times 10^3 \, \text{kg/m}^3 \).