A Solid uniform ball of volume V floats on the interface of two immiscible liquids . [The specific gravity of the upper liquid is `gamma//2` and that of the lower liquid is `2 gamma`, where `gamma` is the specific gravity of the solid ball.] The fraction of the volume of the ball that will be in upper liquid is

Options

A. `2/3 V`
B. `1/3 V`
C. `V/4`
D. `(3V)/4`

Answer

To solve the problem, we need to find the fraction of the volume of a solid uniform ball that is submerged in the upper liquid when it floats at the interface of two immiscible liquids. Let's denote the specific gravity of the solid ball as \( \gamma \). The specific gravity of the upper liquid is \( \frac{\gamma}{2} \) and that of the lower liquid is \( 2\gamma \).

### Step-by-Step Solution:

1. **Identify the Densities**:
- The density of the solid ball, \( \rho_{ball} = \gamma \).
- The density of the upper liquid, \( \rho_{upper} = \frac{\gamma}{2} \).
- The density of the lower liquid, \( \rho_{lower} = 2\gamma \).

2. **Weight of the Ball**:
- The weight of the ball can be expressed as:
\[ W_{ball} = V \cdot \rho_{ball} \cdot g = V \cdot \gamma \cdot g \]
where \( V \) is the volume of the ball and \( g \) is the acceleration due to gravity.

3. **Displaced Weight of the Liquids**:
- Let \( V_1 \) be the volume of the ball submerged in the upper liquid, and \( V_2 \) be the volume submerged in the lower liquid.
- The total displaced weight due to the two liquids is:
\[ W_{displaced} = V_1 \cdot \rho_{upper} \cdot g + V_2 \cdot \rho_{lower} \cdot g \]
- Substituting the densities:
\[ W_{displaced} = V_1 \cdot \left(\frac{\gamma}{2}\right) \cdot g + V_2 \cdot (2\gamma) \cdot g \]

4. **Equating Weights**:
- Since the ball is floating, the weight of the ball equals the weight of the displaced liquid:
\[ V \cdot \gamma \cdot g = V_1 \cdot \left(\frac{\gamma}{2}\right) \cdot g + V_2 \cdot (2\gamma) \cdot g \]
- Dividing through by \( g \):
\[ V \cdot \gamma = V_1 \cdot \left(\frac{\gamma}{2}\right) + V_2 \cdot (2\gamma) \]

5. **Substituting \( V_2 \)**:
- Since the total volume of the ball is \( V \), we have:
\[ V_2 = V - V_1 \]
- Substitute \( V_2 \) into the weight equation:
\[ V \cdot \gamma = V_1 \cdot \left(\frac{\gamma}{2}\right) + (V - V_1) \cdot (2\gamma) \]

6. **Simplifying the Equation**:
- Expanding the equation:
\[ V \cdot \gamma = V_1 \cdot \left(\frac{\gamma}{2}\right) + 2\gamma V - 2\gamma V_1 \]
- Combine like terms:
\[ V \cdot \gamma = 2\gamma V - \frac{3\gamma}{2} V_1 \]

7. **Rearranging**:
- Rearranging gives:
\[ 2\gamma V - V \cdot \gamma = \frac{3\gamma}{2} V_1 \]
- This simplifies to:
\[ \gamma V = \frac{3\gamma}{2} V_1 \]

8. **Finding \( V_1 \)**:
- Dividing both sides by \( \gamma \):
\[ V = \frac{3}{2} V_1 \]
- Rearranging gives:
\[ V_1 = \frac{2}{3} V \]

9. **Finding the Fraction**:
- The fraction of the volume of the ball that is in the upper liquid is:
\[ \text{Fraction} = \frac{V_1}{V} = \frac{\frac{2}{3} V}{V} = \frac{2}{3} \]

### Final Answer:
The fraction of the volume of the ball that will be in the upper liquid is \( \frac{2}{3} \).