which one of the following represents correct dimensional formula of ` 1 //mu_0 epsi_0` ?

Options

A. `[M^0 LT^(-1)]`
B. `[M^(1) Lt^(-1)]`
C. `[m^0 L^(2 ) T^(-2)]`
D. `[M^2 L^2 T^(-2)]`

Explanation

To find the correct dimensional formula for \( \frac{1}{\mu_0 \epsilon_0} \), we can follow these steps:

### Step 1: Understand the relationship between \( c \), \( \mu_0 \), and \( \epsilon_0 \)
The speed of light \( c \) in a vacuum is given by the equation:

\[ c = \frac{1}{\sqrt{\mu_0 \epsilon_0}} \]

### Step 2: Rearranging the equation
From the equation above, we can express \( \frac{1}{\mu_0 \epsilon_0} \) as:

\[ \frac{1}{\mu_0 \epsilon_0} = c^2 \]

### Step 3: Identify the dimensional formula for speed of light \( c \)
The dimensional formula for speed (or velocity) is:

\[ [c] = [L T^{-1}] \]

### Step 4: Calculate the dimensional formula for \( c^2 \)
Since \( c^2 \) is the square of the speed of light, we can find its dimensional formula by squaring the dimensional formula of \( c \):

\[ [c^2] = [L T^{-1}]^2 = [L^2 T^{-2}] \]

### Step 5: Conclusion
Thus, the dimensional formula for \( \frac{1}{\mu_0 \epsilon_0} \) is:

\[ \frac{1}{\mu_0 \epsilon_0} = [L^2 T^{-2}] \]

### Final Answer
The correct dimensional formula for \( \frac{1}{\mu_0 \epsilon_0} \) is:

\[ [L^2 T^{-2}] \]

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