In SI units, the dimensions of `sqrt((epsilon_(0))/(mu_(0)))` is

Options

A. `AT^(-3)ML^(3//2)`
B. `A^(-1)TML^(3)`
C. `AT^(2)M^(-1)L^(-1)`
D. `A^(2)T^(3)M^(-1)L^(-2)`

Explanation

To find the dimensions of the quantity \(\sqrt{\frac{\epsilon_0}{\mu_0}}\), we will follow these steps:

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

1. **Dimensions of \(\epsilon_0\)** (the permittivity of free space):
\[ \epsilon_0 = \frac{1}{\mu_0 c^2} \]
where \(c\) is the speed of light. The dimensions of \(c\) are \([L][T]^{-1}\).
The dimensions of \(\mu_0\) (the permeability of free space) are given as:
\[ \mu_0 = \frac{kg \cdot s^2}{A^2} \]
Thus, the dimensions of \(\epsilon_0\) can be derived as follows:
\[ [\epsilon_0] = \frac{1}{[\mu_0][c]^2} = \frac{1}{\left[M L T^{-2} A^{-2}\right]\left[L^2 T^{-2}\right]} = \frac{1}{M L^{-1} T^{-4} A^{-2}} = M^{-1} L^{-3} T^{4} A^{2} \]

2. **Dimensions of \(\mu_0\)**:
\[ [\mu_0] = M L T^{-2} A^{-2} \]

### Step 2: Write the expression for \(\sqrt{\frac{\epsilon_0}{\mu_0}}\)

Now we will write the expression for the dimensions of \(\sqrt{\frac{\epsilon_0}{\mu_0}}\):
\[ \sqrt{\frac{\epsilon_0}{\mu_0}} = \sqrt{\frac{M^{-1} L^{-3} T^{4} A^{2}}{M L T^{-2} A^{-2}}} \]

### Step 3: Simplify the expression

Now we simplify the fraction inside the square root:
\[ \frac{\epsilon_0}{\mu_0} = \frac{M^{-1} L^{-3} T^{4} A^{2}}{M L T^{-2} A^{-2}} = M^{-1-1} L^{-3-1} T^{4-(-2)} A^{2-(-2)} = M^{-2} L^{-4} T^{6} A^{4} \]

### Step 4: Take the square root

Taking the square root of the simplified expression:
\[ \sqrt{M^{-2} L^{-4} T^{6} A^{4}} = M^{-1} L^{-2} T^{3} A^{2} \]

### Final Answer

Thus, the dimensions of \(\sqrt{\frac{\epsilon_0}{\mu_0}}\) are:
\[ [M^{-1} L^{-2} T^{3} A^{2}] \]