What is the dimensional formula of `(1)/(mu_(0)epsilon_(0))` where the symbols have their usual meanings?

Options

A. `M^(0)L^(2)T^(-2)`
B. `M^(0)L^(-2)T^(-2)`
C. `M^(0)L^(-2)T^(-2)`
D. `M^(0)L^(1)T^(-2)`

Answer

To find the dimensional formula of \(\frac{1}{\mu_0 \epsilon_0}\), where \(\mu_0\) is the magnetic permeability and \(\epsilon_0\) is the electric permittivity, we can use the relationship between these quantities and the speed of light \(c\).

### Step-by-Step Solution:

1. **Understanding the Relationship**:
We know from electromagnetic theory that the speed of light \(c\) is related to \(\mu_0\) and \(\epsilon_0\) by the equation:
\[ c^2 = \frac{1}{\mu_0 \epsilon_0} \]
This implies:
\[ \frac{1}{\mu_0 \epsilon_0} = c^2 \]

2. **Dimensional Formula of Speed of Light**:
The speed of light \(c\) has the dimensional formula of:
\[ [c] = \frac{L}{T} \]
Therefore, the dimensional formula of \(c^2\) is:
\[ [c^2] = \left(\frac{L}{T}\right)^2 = \frac{L^2}{T^2} \]

3. **Equating Dimensional Formulas**:
Since we have established that:
\[ \frac{1}{\mu_0 \epsilon_0} = c^2 \]
It follows that:
\[ \left[\frac{1}{\mu_0 \epsilon_0}\right] = [c^2] = \frac{L^2}{T^2} \]

4. **Final Dimensional Formula**:
Therefore, the dimensional formula of \(\frac{1}{\mu_0 \epsilon_0}\) is:
\[ \frac{L^2}{T^2} \]

### Conclusion:
The dimensional formula of \(\frac{1}{\mu_0 \epsilon_0}\) is:
\[ [M^0 L^2 T^{-2}] \]