A liquid occupies half of a vessel at a perticular temperature. The volume of the unoccupied part remains constant at all temperatures. If `alpha` and `gamma` are the coefficients of linear and real expansions of a vessel and liquid, then `gamma` is

Options

A. `3alpha`
B. `3alpha//2`
C. `6alpha`
D. `9alpha`

Explanation

To solve the problem, we need to analyze the relationship between the coefficients of linear expansion of the vessel and the coefficient of real expansion of the liquid. Here's a step-by-step solution:

### Step 1: Define the initial conditions
Let the volume of the vessel be \( V_V \) and the volume of the liquid be \( V_L \). According to the problem, at a particular temperature \( T_1 \):
\[ V_L = \frac{1}{2} V_V \]

### Step 2: Understand the expansion of the vessel and liquid
When the temperature increases to \( T_2 \), both the vessel and the liquid will expand. The volume of the vessel after expansion can be expressed as:
\[ V_V' = V_V (1 + 3\alpha \Delta T) \]
where \( \alpha \) is the coefficient of linear expansion of the vessel and \( \Delta T = T_2 - T_1 \).

The volume of the liquid after expansion can be expressed as:
\[ V_L' = V_L (1 + \gamma \Delta T) \]
where \( \gamma \) is the coefficient of real expansion of the liquid.

### Step 3: Set up the equation for constant unoccupied volume
The problem states that the volume of the unoccupied part remains constant at all temperatures. Therefore, we can write:
\[ V_V - V_L = V_V' - V_L' \]

Substituting the expressions for \( V_V' \) and \( V_L' \):
\[ V_V - V_L = V_V (1 + 3\alpha \Delta T) - V_L (1 + \gamma \Delta T) \]

### Step 4: Simplify the equation
Expanding both sides gives us:
\[ V_V - V_L = V_V + 3\alpha V_V \Delta T - V_L - \gamma V_L \Delta T \]

Rearranging the equation leads to:
\[ 0 = 3\alpha V_V \Delta T - \gamma V_L \Delta T \]

### Step 5: Factor out \( \Delta T \)
Since \( \Delta T \) is not zero, we can factor it out:
\[ 0 = 3\alpha V_V - \gamma V_L \]

### Step 6: Substitute \( V_L \) in terms of \( V_V \)
From our initial condition, we know \( V_L = \frac{1}{2} V_V \). Substituting this into the equation gives:
\[ 0 = 3\alpha V_V - \gamma \left(\frac{1}{2} V_V\right) \]

### Step 7: Solve for \( \gamma \)
We can simplify this equation:
\[ 3\alpha V_V = \frac{1}{2} \gamma V_V \]

Dividing both sides by \( V_V \) (assuming \( V_V \neq 0 \)):
\[ 3\alpha = \frac{1}{2} \gamma \]

Multiplying both sides by 2 gives:
\[ 6\alpha = \gamma \]

### Conclusion
Thus, the relationship between the coefficient of real expansion of the liquid (\( \gamma \)) and the coefficient of linear expansion of the vessel (\( \alpha \)) is:
\[ \gamma = 6\alpha \]