Dimension of `(1)/(mu_(0)epsilon_(0))`, where symbols have usual meaning, are

Options

A. `L^(2)T^(-2)`
B. `LT^(-1)`
C. `L^(-1)T`
D. `L^(-2)T^(2)`

Answer

To find the dimension of \( \frac{1}{\mu_0 \epsilon_0} \), where \( \mu_0 \) is the permeability of free space and \( \epsilon_0 \) is the permittivity of free space, we can use the relationship between the speed of light \( c \), permeability, and permittivity.

### Step-by-Step Solution:

1. **Understand the Relationship**:
The speed of light \( c \) in a vacuum is given by the equation:
\[ c = \frac{1}{\sqrt{\mu_0 \epsilon_0}} \]
Squaring both sides gives:
\[ c^2 = \frac{1}{\mu_0 \epsilon_0} \]

2. **Express \( \frac{1}{\mu_0 \epsilon_0} \)**:
From the equation above, we can express \( \frac{1}{\mu_0 \epsilon_0} \) as:
\[ \frac{1}{\mu_0 \epsilon_0} = c^2 \]

3. **Determine the Dimensions of \( c^2 \)**:
The speed of light \( c \) has dimensions of length per time, which can be expressed as:
\[ [c] = [L][T^{-1}] \]
Therefore, squaring the dimensions gives:
\[ [c^2] = [L^2][T^{-2}] \]

4. **Conclusion**:
Thus, the dimensions of \( \frac{1}{\mu_0 \epsilon_0} \) are:
\[ \left[\frac{1}{\mu_0 \epsilon_0}\right] = [L^2][T^{-2}] \]

### Final Answer:
The dimensions of \( \frac{1}{\mu_0 \epsilon_0} \) are \( [L^2 T^{-2}] \).