A solid sphere having volume `V` and density `rho` floats at the interface of two immiscible liquids of densityes `rho_(1)` and `rho_(2)` respectively. If `rho_(1) lt rho lt rho_(2)`, then the ratio of volume of the parts of the sphere in upper and lower liquid is

Options

A. (rho-rho_(1))/(rho_(2)-rho)`
B. `(rho_(2)-rho)/(rho-rho_(1))`
C. (rho+rho_(1))/(rho+rho_(2))`
D. (rho+rho_(2))/(rho+rho_(1))`

Explanation

To solve the problem of a solid sphere floating at the interface of two immiscible liquids, we will follow these steps:

### Step 1: Understand the Problem
We have a solid sphere with volume \( V \) and density \( \rho \) floating at the interface of two immiscible liquids with densities \( \rho_1 \) (upper liquid) and \( \rho_2 \) (lower liquid). Given that \( \rho_1 < \rho < \rho_2 \), we need to find the ratio of the volumes of the parts of the sphere submerged in the upper and lower liquids.

### Step 2: Define Variables
Let:
- \( V_1 \) = Volume of the sphere submerged in the upper liquid (density \( \rho_1 \))
- \( V_2 \) = Volume of the sphere submerged in the lower liquid (density \( \rho_2 \))

### Step 3: Apply Archimedes' Principle
According to Archimedes' principle, the buoyant force (upthrust) acting on the sphere is equal to the weight of the liquid displaced by the submerged parts of the sphere. Therefore, we can write:

\[ U_1 + U_2 = mg \]

Where:
- \( U_1 = V_1 \cdot \rho_1 \cdot g \) (upthrust from the upper liquid)
- \( U_2 = V_2 \cdot \rho_2 \cdot g \) (upthrust from the lower liquid)
- \( mg = V \cdot \rho \cdot g \) (weight of the sphere)

### Step 4: Set Up the Equation
Substituting the expressions for \( U_1 \) and \( U_2 \) into the equation gives:

\[ V_1 \cdot \rho_1 \cdot g + V_2 \cdot \rho_2 \cdot g = V \cdot \rho \cdot g \]

We can cancel \( g \) from all terms (since \( g \neq 0 \)):

\[ V_1 \cdot \rho_1 + V_2 \cdot \rho_2 = V \cdot \rho \]

### Step 5: Relate Volumes
Since the total volume of the sphere is the sum of the submerged volumes:

\[ V = V_1 + V_2 \]

### Step 6: Solve for the Ratios
From the equations we have:

1. \( V_1 \cdot \rho_1 + V_2 \cdot \rho_2 = V \cdot \rho \)
2. \( V = V_1 + V_2 \)

We can express \( V_2 \) in terms of \( V_1 \):

\[ V_2 = V - V_1 \]

Substituting \( V_2 \) into the first equation:

\[ V_1 \cdot \rho_1 + (V - V_1) \cdot \rho_2 = V \cdot \rho \]

Expanding this gives:

\[ V_1 \cdot \rho_1 + V \cdot \rho_2 - V_1 \cdot \rho_2 = V \cdot \rho \]

Rearranging terms:

\[ V_1 (\rho_1 - \rho_2) = V \cdot \rho - V \cdot \rho_2 \]

Factoring out \( V \):

\[ V_1 (\rho_1 - \rho_2) = V (\rho - \rho_2) \]

Thus,

\[ V_1 = \frac{V (\rho - \rho_2)}{(\rho_1 - \rho_2)} \]

Now substituting \( V_1 \) back into the equation for \( V_2 \):

\[ V_2 = V - V_1 = V - \frac{V (\rho - \rho_2)}{(\rho_1 - \rho_2)} \]

### Step 7: Find the Ratio
Now we can find the ratio \( \frac{V_1}{V_2} \):

\[ \frac{V_1}{V_2} = \frac{\frac{V (\rho - \rho_2)}{(\rho_1 - \rho_2)}}{V - \frac{V (\rho - \rho_2)}{(\rho_1 - \rho_2)}} \]

This simplifies to:

\[ \frac{V_1}{V_2} = \frac{\rho_2 - \rho}{\rho - \rho_1} \]

### Final Answer
Thus, the ratio of the volume of the parts of the sphere in the upper and lower liquids is:

\[ \frac{V_1}{V_2} = \frac{\rho_2 - \rho}{\rho - \rho_1} \]