A metallic sphere floats in immiscible mixture of water (density `10^(3) kgm^(-3))` and a liquid (density `8xx10^(3) kgm^(-3))` such that its `(2//3)` part is in water and `(1//3)` part in the liquid. The density of the metal is

Options

A. `(5000)/(3)kgm^(-3)`
B. `(10000)/(3)kgm^(-3)`
C. `5000 kgm^(-3)`
D. `2000 kgm^(-3)`

Explanation

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

### Step 1: Understand the problem
The sphere is floating, which means the weight of the sphere is balanced by the buoyant forces acting on it from both the water and the other liquid. The sphere is partially submerged in both liquids: \( \frac{2}{3} \) in water and \( \frac{1}{3} \) in the liquid.

### Step 2: Set up the equation for buoyancy
The weight of the sphere (W) can be expressed as:
\[ W = V \cdot \rho_m \cdot g \]
where:
- \( V \) = total volume of the sphere,
- \( \rho_m \) = density of the metallic sphere,
- \( g \) = acceleration due to gravity.

The buoyant force from the water (B_w) and the liquid (B_l) can be expressed as:
\[ B_w = \left( \frac{2}{3} V \right) \cdot \rho_w \cdot g \]
\[ B_l = \left( \frac{1}{3} V \right) \cdot \rho_l \cdot g \]
where:
- \( \rho_w = 10^3 \, \text{kg/m}^3 \) (density of water),
- \( \rho_l = 8 \times 10^3 \, \text{kg/m}^3 \) (density of the other liquid).

### Step 3: Write the equilibrium condition
Since the sphere is floating, the total buoyant force is equal to the weight of the sphere:
\[ W = B_w + B_l \]

Substituting the expressions we have:
\[ V \cdot \rho_m \cdot g = \left( \frac{2}{3} V \cdot \rho_w \cdot g \right) + \left( \frac{1}{3} V \cdot \rho_l \cdot g \right) \]

### Step 4: Simplify the equation
We can cancel \( V \) and \( g \) from both sides (assuming \( V \neq 0 \) and \( g \neq 0 \)):
\[ \rho_m = \frac{2}{3} \rho_w + \frac{1}{3} \rho_l \]

### Step 5: Substitute the values
Now, substituting the known values of \( \rho_w \) and \( \rho_l \):
\[ \rho_m = \frac{2}{3} (10^3) + \frac{1}{3} (8 \times 10^3) \]
\[ \rho_m = \frac{2 \times 10^3}{3} + \frac{8 \times 10^3}{3} \]
\[ \rho_m = \frac{(2 + 8) \times 10^3}{3} \]
\[ \rho_m = \frac{10 \times 10^3}{3} \]
\[ \rho_m = \frac{10,000}{3} \, \text{kg/m}^3 \]

### Final Answer
The density of the metallic sphere is:
\[ \rho_m \approx 3333.33 \, \text{kg/m}^3 \]

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