A metallic sphere floats in an immiscible mixture of water `(rho_(w)=10^(3)kg//m^(3))` and a liquid `(rho_(L)=13.5xx10^(3)kg//m^(3))` such that its `(4)/(5)th` volume is in water and `(1)/(5)th` volume in the liquid. Find the density of metal.

Explanation

To solve the problem, we need to apply the principle of buoyancy, which states that the weight of the fluid displaced by the submerged part of the sphere is equal to the weight of the sphere itself.

### Step-by-Step Solution:

1. **Identify the given values:**
- Density of water, \( \rho_w = 10^3 \, \text{kg/m}^3 \)
- Density of liquid, \( \rho_L = 13.5 \times 10^3 \, \text{kg/m}^3 \)
- Volume of the sphere submerged in water, \( V_w = \frac{4}{5} V \)
- Volume of the sphere submerged in the liquid, \( V_L = \frac{1}{5} V \)

2. **Calculate the buoyant force:**
The buoyant force is equal to the weight of the fluid displaced by the submerged parts of the sphere. This can be calculated as:
\[ F_b = \text{Weight of water displaced} + \text{Weight of liquid displaced} \]
\[ F_b = \left( V_w \cdot \rho_w \cdot g \right) + \left( V_L \cdot \rho_L \cdot g \right) \]
Substituting the volumes:
\[ F_b = \left( \frac{4}{5} V \cdot \rho_w \cdot g \right) + \left( \frac{1}{5} V \cdot \rho_L \cdot g \right) \]

3. **Substitute the densities:**
\[ F_b = \left( \frac{4}{5} V \cdot 10^3 \cdot g \right) + \left( \frac{1}{5} V \cdot 13.5 \times 10^3 \cdot g \right) \]

4. **Factor out common terms:**
\[ F_b = V \cdot g \left( \frac{4}{5} \cdot 10^3 + \frac{1}{5} \cdot 13.5 \times 10^3 \right) \]

5. **Calculate the total buoyant force:**
\[ F_b = V \cdot g \left( \frac{4 \cdot 10^3 + 13.5 \times 10^3}{5} \right) \]
\[ F_b = V \cdot g \left( \frac{4 + 13.5}{5} \times 10^3 \right) = V \cdot g \left( \frac{17.5}{5} \times 10^3 \right) \]
\[ F_b = V \cdot g \left( 3.5 \times 10^3 \right) \]

6. **Weight of the sphere:**
The weight of the sphere is given by:
\[ W = V \cdot \rho_m \cdot g \]
where \( \rho_m \) is the density of the metal.

7. **Set the buoyant force equal to the weight of the sphere:**
\[ V \cdot g \left( 3.5 \times 10^3 \right) = V \cdot \rho_m \cdot g \]

8. **Cancel out common terms (V and g):**
\[ 3.5 \times 10^3 = \rho_m \]

9. **Final answer:**
\[ \rho_m = 3.5 \times 10^3 \, \text{kg/m}^3 \]

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