A solid whose volume does not change with temperature floats in a liquid. For two different temperatures `t_1` and `t_2` of the liqiud, fraction `f_1` and `f_2` of the volume of the solid remain submerged in the liquid. The coefficient of volume expansion of the liquid is equal to

Options

A. `(f_1-f_2)/(f_2t_1-f_1t_2)`
B. `(f_1-f_2)/(f_1t_1-f_2t_2)`
C. `(f_1+f_2)/(f_2t_1+f_1t_2)`
D. `(f_1+f_2)/(f_1t_1+f_2t_2)`

Explanation

To solve the problem, we need to find the coefficient of volume expansion (γ) of the liquid based on the fractions of the solid submerged in the liquid at two different temperatures.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a solid that does not change its volume with temperature.
- At two different temperatures \( t_1 \) and \( t_2 \), the fractions of the solid submerged in the liquid are \( f_1 \) and \( f_2 \), respectively.

2. **Volume Expansion of the Liquid**:
- The volume of the liquid at temperature \( t_1 \) can be expressed as:
\[ V_{L1} = V_{L0} (1 + \gamma t_1) \]
- The volume of the liquid at temperature \( t_2 \) can be expressed as:
\[ V_{L2} = V_{L0} (1 + \gamma t_2) \]
- Here, \( V_{L0} \) is the initial volume of the liquid, and \( \gamma \) is the coefficient of volume expansion.

3. **Relating Submerged Volume to Liquid Volume**:
- The fraction of the solid submerged in the liquid at temperature \( t_1 \) is given by:
\[ f_1 = \frac{V_{S}}{V_{L1}} = \frac{V_{S}}{V_{L0} (1 + \gamma t_1)} \]
- The fraction of the solid submerged in the liquid at temperature \( t_2 \) is given by:
\[ f_2 = \frac{V_{S}}{V_{L2}} = \frac{V_{S}}{V_{L0} (1 + \gamma t_2)} \]

4. **Setting Up the Equations**:
- Rearranging the equations for \( f_1 \) and \( f_2 \):
\[ f_1 (1 + \gamma t_1) = \frac{V_{S}}{V_{L0}} \quad \text{(1)} \]
\[ f_2 (1 + \gamma t_2) = \frac{V_{S}}{V_{L0}} \quad \text{(2)} \]

5. **Dividing the Equations**:
- Dividing equation (1) by equation (2):
\[ \frac{f_1 (1 + \gamma t_1)}{f_2 (1 + \gamma t_2)} = 1 \]

6. **Rearranging the Equation**:
- Rearranging gives us:
\[ f_1 (1 + \gamma t_1) = f_2 (1 + \gamma t_2) \]
- Expanding this:
\[ f_1 + f_1 \gamma t_1 = f_2 + f_2 \gamma t_2 \]

7. **Isolating γ**:
- Rearranging to isolate \( \gamma \):
\[ f_1 - f_2 = \gamma (f_2 t_2 - f_1 t_1) \]
- Thus, we can express \( \gamma \) as:
\[ \gamma = \frac{f_1 - f_2}{f_2 t_2 - f_1 t_1} \]

### Final Answer:
The coefficient of volume expansion of the liquid is given by:
\[ \gamma = \frac{f_1 - f_2}{f_2 t_2 - f_1 t_1} \]