The coefficient of real expansion `gamma_(R)` of a liquid is `5` times the coefficient of linear expansion of the material of the container in which the liquid is present. The ratio is of the cofficient of apparent expansion and real expansion of the liquid is

Options

A. `5:2`
B. `1:5`
C. `2:5`
D. `5:1`

Explanation

To solve the problem, we need to find the ratio of the coefficient of apparent expansion (γ_a) to the coefficient of real expansion (γ_r) of a liquid, given that the coefficient of real expansion of the liquid is 5 times the coefficient of linear expansion (α) of the container material.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Coefficient of real expansion of the liquid: γ_r = 5α
- Coefficient of linear expansion of the container material: α

2. **Understand the Relationship:**
- The relationship between the coefficients of apparent expansion (γ_a), real expansion (γ_r), and linear expansion (α) is given by:
\[ γ_r = γ_a + 3α \]

3. **Substitute the Known Values:**
- Substitute γ_r = 5α into the equation:
\[ 5α = γ_a + 3α \]

4. **Rearrange the Equation:**
- Rearranging gives:
\[ γ_a = 5α - 3α \]
- Simplifying this, we find:
\[ γ_a = 2α \]

5. **Calculate the Ratio:**
- Now, we need to find the ratio of γ_a to γ_r:
\[ \frac{γ_a}{γ_r} = \frac{2α}{5α} \]
- The α terms cancel out:
\[ \frac{γ_a}{γ_r} = \frac{2}{5} \]

6. **Final Answer:**
- The ratio of the coefficient of apparent expansion to the coefficient of real expansion is:
\[ \text{Ratio} = 2 : 5 \]