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

Options

A. (a) `[L^(-1)T]`
B. (b) `[L^(-2)T^(2)]`
C. ( c ) `[L^(2)T^(-2)]`
D. (d) `[LT^(-1)]`

Explanation

To find the dimension of \( \frac{1}{\mu_0 \epsilon_0} \), we can follow these steps:

### Step 1: Understand the relationship between speed of light and \( \mu_0 \) and \( \epsilon_0 \)

The speed of light \( c \) is given by the equation:
\[ c = \frac{1}{\sqrt{\mu_0 \epsilon_0}} \]
This means that \( \sqrt{\mu_0 \epsilon_0} = \frac{1}{c} \).

### Step 2: Find the dimension of the speed of light

The speed of light \( c \) has dimensions of velocity, which can be expressed as:
\[ [c] = L^1 T^{-1} \]
where \( L \) represents length and \( T \) represents time.

### Step 3: Square both sides of the speed of light equation

Squaring both sides of the equation \( c = \frac{1}{\sqrt{\mu_0 \epsilon_0}} \) gives us:
\[ c^2 = \frac{1}{\mu_0 \epsilon_0} \]
Thus, we can express \( \frac{1}{\mu_0 \epsilon_0} \) as:
\[ \frac{1}{\mu_0 \epsilon_0} = c^2 \]

### Step 4: Substitute the dimension of \( c^2 \)

Now substituting the dimension of \( c \) into the equation, we have:
\[ [c^2] = (L^1 T^{-1})^2 = L^2 T^{-2} \]

### Step 5: Conclusion

Therefore, the dimension of \( \frac{1}{\mu_0 \epsilon_0} \) is:
\[ \frac{1}{\mu_0 \epsilon_0} \sim L^2 T^{-2} \]

### Final Answer

The dimension of \( \frac{1}{\mu_0 \epsilon_0} \) is \( L^2 T^{-2} \).

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