The absolute coefficient of expansion of a liquid is 7 times that the volume coefficient of expansion of the vessel. Then the ratio of absolute and apparent expansion of the liquid is

Options

A. `(1)/(7)`
B. `(7)/(6)`
C. `(6)/(7)`
D. none of these

Explanation

To solve the problem, we need to find the ratio of the absolute expansion to the apparent expansion of a liquid when the absolute coefficient of expansion of the liquid is 7 times that of the volume coefficient of expansion of the vessel.

### Step-by-Step Solution:

1. **Define Coefficients**:
- Let the volume coefficient of expansion of the vessel be denoted as \( \gamma \).
- Then, the absolute coefficient of expansion of the liquid, denoted as \( \gamma_L \), is given by:
\[ \gamma_L = 7\gamma \]

2. **Initial Volume**:
- Assume the initial volume of the liquid and the vessel is \( V_0 \).

3. **Final Volume Calculation**:
- The final volume of the liquid after expansion can be expressed as:
\[ V_L = V_0 (1 + \gamma_L \Delta T) = V_0 (1 + 7\gamma \Delta T) \]
- The final volume of the vessel after expansion is:
\[ V_V = V_0 (1 + \gamma \Delta T) \]

4. **Apparent Expansion**:
- The apparent expansion of the liquid is the increase in volume of the liquid relative to the vessel. Thus, we can write:
\[ \text{Apparent Expansion} = V_L - V_V \]
- Substituting the expressions for \( V_L \) and \( V_V \):
\[ \text{Apparent Expansion} = V_0 (1 + 7\gamma \Delta T) - V_0 (1 + \gamma \Delta T) \]
- Simplifying this gives:
\[ \text{Apparent Expansion} = V_0 (7\gamma \Delta T - \gamma \Delta T) = V_0 (6\gamma \Delta T) \]

5. **Absolute Expansion**:
- The absolute expansion of the liquid is defined as:
\[ \text{Absolute Expansion} = V_L - V_0 = V_0 (1 + 7\gamma \Delta T) - V_0 = V_0 (7\gamma \Delta T) \]

6. **Ratio of Absolute to Apparent Expansion**:
- Now, we can find the ratio of the absolute expansion to the apparent expansion:
\[ \text{Ratio} = \frac{\text{Absolute Expansion}}{\text{Apparent Expansion}} = \frac{V_0 (7\gamma \Delta T)}{V_0 (6\gamma \Delta T)} \]
- The \( V_0 \) and \( \Delta T \) terms cancel out, leading to:
\[ \text{Ratio} = \frac{7\gamma}{6\gamma} = \frac{7}{6} \]

### Final Answer:
The ratio of absolute expansion to apparent expansion of the liquid is \( \frac{7}{6} \).