The coefficient of real expansion of liquid is `gamma_(R)` and the coefficient of appareent expansion of the liquid is `gamma_(A)`. The coefficient of cubical expansion of the vessel is `gamma`. If `gamma_(R) : gamma_(A) = 4:1` then `gamma_(A) : gamma` is

Options

A. `3 : 1`
B. `1 : 3`
C. `4 : 1`
D. `1 : 4`

Explanation

To solve the problem, we need to use the relationship between the coefficients of real expansion, apparent expansion, and the cubical expansion of the vessel. Let's denote:

- \( \gamma_R \): Coefficient of real expansion of the liquid
- \( \gamma_A \): Coefficient of apparent expansion of the liquid
- \( \gamma \): Coefficient of cubical expansion of the vessel

Given that the ratio of the coefficients of real and apparent expansion is:

\[ \frac{\gamma_R}{\gamma_A} = 4:1 \]

This implies:

\[ \gamma_R = 4 \gamma_A \]

From the relationship between the coefficients, we have:

\[ \gamma_R = \gamma_A + \gamma \]

Substituting \( \gamma_R \) from the first equation into this relationship gives:

\[ 4 \gamma_A = \gamma_A + \gamma \]

Now, we can rearrange this equation to isolate \( \gamma \):

\[ 4 \gamma_A - \gamma_A = \gamma \]
\[ 3 \gamma_A = \gamma \]

Now, we need to find the ratio \( \frac{\gamma_A}{\gamma} \):

\[ \frac{\gamma_A}{\gamma} = \frac{\gamma_A}{3 \gamma_A} = \frac{1}{3} \]

Thus, the ratio \( \gamma_A : \gamma \) is:

\[ \gamma_A : \gamma = 1 : 3 \]

### Final Answer:
The ratio \( \gamma_A : \gamma \) is \( 1 : 3 \).