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 `2gamma`, where `gamma` is th especific gracity 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)`

Explanation

To solve the problem of determining the fraction of the volume of a solid uniform ball that floats on the interface of two immiscible liquids, we can follow these steps:

### Step 1: Define Variables
Let:
- \( V \) = total volume of the ball
- \( V' \) = volume of the ball submerged in the upper liquid
- \( V'' \) = volume of the ball submerged in the lower liquid
- The specific gravity of the solid ball = \( \gamma \)
- The specific gravity of the upper liquid = \( \frac{\gamma}{2} \)
- The specific gravity of the lower liquid = \( 2\gamma \)

### Step 2: Relate Volumes
Since the ball is floating, the total volume of the ball can be expressed as:
\[ V = V' + V'' \]

### Step 3: Express Weight of the Ball
The weight of the ball \( W \) can be expressed in terms of its specific gravity:
\[ W = \text{Specific Gravity} \times \text{Volume} \times g = \gamma V g \]

### Step 4: Calculate Buoyant Forces
The buoyant force from the upper liquid \( F_1 \) is given by:
\[ F_1 = \text{Density of upper liquid} \times V' \times g = \left(\frac{\gamma}{2}\right) V' g \]

The buoyant force from the lower liquid \( F_2 \) is given by:
\[ F_2 = \text{Density of lower liquid} \times V'' \times g = (2\gamma) V'' g \]

### Step 5: Set Up the Equation for Equilibrium
For the ball to float, the total buoyant force must equal the weight of the ball:
\[ F_1 + F_2 = W \]
Substituting the expressions we derived:
\[ \left(\frac{\gamma}{2}\right) V' g + (2\gamma) V'' g = \gamma V g \]

### Step 6: Simplify the Equation
Cancelling \( g \) from both sides, we have:
\[ \frac{\gamma}{2} V' + 2\gamma V'' = \gamma V \]
Dividing through by \( \gamma \):
\[ \frac{1}{2} V' + 2 V'' = V \]

### Step 7: Substitute \( V'' \)
From \( V = V' + V'' \), we can express \( V'' \) as:
\[ V'' = V - V' \]
Substituting this into the buoyancy equation:
\[ \frac{1}{2} V' + 2(V - V') = V \]
Expanding and simplifying:
\[ \frac{1}{2} V' + 2V - 2V' = V \]
\[ 2V - \frac{3}{2} V' = V \]
Rearranging gives:
\[ 2V - V = \frac{3}{2} V' \]
\[ V = \frac{3}{2} V' \]

### Step 8: Solve for \( V' \)
Now, solving for \( V' \):
\[ V' = \frac{2}{3} V \]

### Step 9: Find the Fraction of Volume in Upper Liquid
The fraction of the volume of the ball that is in the upper liquid is:
\[ \text{Fraction} = \frac{V'}{V} = \frac{\frac{2}{3} V}{V} = \frac{2}{3} \]

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

---