The dimensions of `(mu_(0)epsilon_(0))^(-1//2)` are

Options

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

Explanation

To find the dimensions of \((\mu_0 \epsilon_0)^{-1/2}\), we will follow these steps:

### Step 1: Understand the quantities involved

- \(\mu_0\) is the permeability of free space, and \(\epsilon_0\) is the permittivity of free space.
- The product \(\mu_0 \epsilon_0\) is related to the speed of light \(c\) in a vacuum, where \(c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}\).

### Step 2: Recall the dimensions of \(\mu_0\) and \(\epsilon_0\)

- The dimensions of permeability \(\mu_0\) are given by:
\[ [\mu_0] = \frac{\text{kg}}{\text{m} \cdot \text{s}^2} \cdot \text{m} = \frac{\text{kg}}{\text{m} \cdot \text{s}^2} \]
This can be simplified to:
\[ [\mu_0] = \text{M} \cdot \text{L}^{-1} \cdot \text{T}^{-2} \]

- The dimensions of permittivity \(\epsilon_0\) are given by:
\[ [\epsilon_0] = \frac{\text{C}^2}{\text{N} \cdot \text{m}^2} = \frac{\text{C}^2 \cdot \text{m}^{-2}}{\text{kg} \cdot \text{m} \cdot \text{s}^{-2}} = \frac{\text{C}^2}{\text{kg} \cdot \text{m} \cdot \text{s}^{-2}} \]
This can be simplified to:
\[ [\epsilon_0] = \text{M}^{-1} \cdot \text{L}^{-3} \cdot \text{T}^4 \cdot \text{I}^2 \]

### Step 3: Calculate the dimensions of \(\mu_0 \epsilon_0\)

Now, we multiply the dimensions of \(\mu_0\) and \(\epsilon_0\):
\[ [\mu_0 \epsilon_0] = [\mu_0] \cdot [\epsilon_0] = \left(\text{M} \cdot \text{L}^{-1} \cdot \text{T}^{-2}\right) \cdot \left(\text{M}^{-1} \cdot \text{L}^{-3} \cdot \text{T}^4 \cdot \text{I}^2\right) \]

### Step 4: Simplify the dimensions

When we multiply these, we get:
\[ [\mu_0 \epsilon_0] = \text{M}^{0} \cdot \text{L}^{-4} \cdot \text{T}^{2} \cdot \text{I}^2 \]

### Step 5: Find the dimensions of \((\mu_0 \epsilon_0)^{-1/2}\)

Now, we take the inverse square root:
\[ \left[\mu_0 \epsilon_0\right]^{-1/2} = \left(\text{M}^{0} \cdot \text{L}^{-4} \cdot \text{T}^{2} \cdot \text{I}^2\right)^{-1/2} = \text{M}^{0} \cdot \text{L}^{2} \cdot \text{T}^{-1} \cdot \text{I}^{-1} \]

### Final Answer:

Thus, the dimensions of \((\mu_0 \epsilon_0)^{-1/2}\) are:
\[ \text{L}^2 \cdot \text{T}^{-1} \cdot \text{I}^{-1} \]