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

### Step 1: Understand the Problem
We have a metallic sphere that is floating in a mixture of two liquids: water and another liquid. The density of water is given as \( \rho_w = 10^3 \, \text{kg/m}^3 \) and the density of the other liquid is \( \rho_L = 13.5 \times 10^3 \, \text{kg/m}^3 \). The sphere has \( \frac{4}{5} \) of its volume in water and \( \frac{1}{5} \) of its volume in the liquid.

### Step 2: Write the Equation for Buoyancy
According to Archimedes' principle, the buoyant force acting on the sphere is equal to the weight of the fluid displaced by the sphere. The total buoyant force \( F_b \) can be expressed as:
\[ F_b = \text{Weight of water displaced} + \text{Weight of liquid displaced} \]
This can be written mathematically as:
\[ F_b = \left(\frac{4}{5} V \cdot \rho_w \cdot g\right) + \left(\frac{1}{5} V \cdot \rho_L \cdot g\right) \]
where \( V \) is the total volume of the sphere and \( g \) is the acceleration due to gravity.

### Step 3: Write the Weight of the Sphere
The weight of the metallic sphere \( W \) can be expressed as:
\[ W = V \cdot \rho_m \cdot g \]
where \( \rho_m \) is the density of the metal.

### Step 4: Set Up the Equation
Since the sphere is floating, the buoyant force is equal to the weight of the sphere:
\[ \left(\frac{4}{5} V \cdot \rho_w \cdot g\right) + \left(\frac{1}{5} V \cdot \rho_L \cdot g\right) = V \cdot \rho_m \cdot g \]

### Step 5: Simplify the Equation
We can cancel \( V \) and \( g \) from both sides of the equation:
\[ \frac{4}{5} \rho_w + \frac{1}{5} \rho_L = \rho_m \]

### Step 6: Substitute the Values
Now, substitute the values of \( \rho_w \) and \( \rho_L \):
\[ \frac{4}{5} \cdot (10^3) + \frac{1}{5} \cdot (13.5 \times 10^3) = \rho_m \]

### Step 7: Calculate Each Term
Calculating the first term:
\[ \frac{4}{5} \cdot 10^3 = 800 \, \text{kg/m}^3 \]
Calculating the second term:
\[ \frac{1}{5} \cdot (13.5 \times 10^3) = 2700 \, \text{kg/m}^3 \]

### Step 8: Add the Two Results
Now, add the two results to find the density of the metal:
\[ \rho_m = 800 + 2700 = 3500 \, \text{kg/m}^3 \]

### Final Answer
Thus, the density of the metal is:
\[ \boxed{3500 \, \text{kg/m}^3} \]