Relative permittivity of a medium is 27. If `epsi_m and epsi_0` care permittivity of medium and vacuum respectively then:

Options

A. `epsi_0=9epsi_m`
B. `epsi_0=27epsi_m`
C. `epsi_m=9epsi_0`
D. `epsi_m=27epsi_0`

Explanation

To solve the problem, we need to understand the relationship between the relative permittivity of a medium and the permittivity of free space (vacuum).

### Step-by-Step Solution:

1. **Understand Relative Permittivity**:
The relative permittivity (also known as the dielectric constant) of a medium is defined as the ratio of the permittivity of the medium (\( \epsilon_m \)) to the permittivity of free space (\( \epsilon_0 \)):
\[ \text{Relative Permittivity} = \frac{\epsilon_m}{\epsilon_0} \]

2. **Given Information**:
We are given that the relative permittivity of the medium is 27:
\[ \frac{\epsilon_m}{\epsilon_0} = 27 \]

3. **Rearranging the Equation**:
To find the relationship between \( \epsilon_m \) and \( \epsilon_0 \), we can rearrange the equation:
\[ \epsilon_m = 27 \cdot \epsilon_0 \]

4. **Identifying the Correct Option**:
Now, let's analyze the options provided:
- Option 1: \( \epsilon_0 = 9 \epsilon_m \) (Incorrect)
- Option 2: \( \epsilon_0 = 27 \epsilon_m \) (Incorrect)
- Option 3: \( \epsilon_m = 9 \epsilon_0 \) (Incorrect)
- Option 4: \( \epsilon_m = 27 \epsilon_0 \) (Correct)

5. **Conclusion**:
The correct expression that relates the permittivity of the medium to the permittivity of free space is:
\[ \epsilon_m = 27 \cdot \epsilon_0 \]
Therefore, the correct option is **Option 4**.