The dimensions of `[mu_(0)in_(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 \((\mu_0 \epsilon_0)^{1/2}\), where \(\mu_0\) is the permeability of free space and \(\epsilon_0\) is the permittivity of free space, we will follow these steps:

### Step 1: Identify the dimensions of \(\mu_0\) and \(\epsilon_0\)

1. **Permeability (\(\mu_0\))**: The permeability of free space has the dimension of magnetic permeability, which can be expressed in terms of the base units as:
\[ [\mu_0] = \frac{[M][L][T^{-2}]}{[I^2]} = \frac{kg \cdot m \cdot s^{-2}}{A^2} = [M L T^{-2} I^{-2}] \]

2. **Permittivity (\(\epsilon_0\))**: The permittivity of free space has the dimension of electric permittivity, which can be expressed as:
\[ [\epsilon_0] = \frac{[I^2][T^4]}{[M][L^3]} = \frac{A^2 \cdot s^4}{kg \cdot m^3} = [M^{-1} L^{-3} T^4 I^2] \]

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

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

### Step 3: Combine the dimensions

Combining the dimensions:
\[ [\mu_0 \epsilon_0] = [M^{1-1} L^{1-3} T^{-2+4} I^{-2+2}] = [M^{0} L^{-2} T^{2} I^{0}] \]
This simplifies to:
\[ [\mu_0 \epsilon_0] = [L^{-2} T^{2}] \]

### Step 4: Take the square root

Now, we need to find the dimensions of \((\mu_0 \epsilon_0)^{1/2}\):
\[ [(\mu_0 \epsilon_0)^{1/2}] = [L^{-2} T^{2}]^{1/2} = [L^{-1} T] \]

### Final Answer

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

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