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

Options

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

Answer

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

### Step 1: Understand the definitions of \(\epsilon_0\) and \(\mu_0\)
- \(\epsilon_0\) is the permittivity of free space, and \(\mu_0\) is the permeability of free space.

### Step 2: Use the relationship between force, charge, and permittivity
- From Coulomb's law, we know that the force \(F\) between two charges \(q_1\) and \(q_2\) separated by a distance \(r\) is given by:
\[ F = \frac{1}{4\pi \epsilon_0} \frac{q_1 q_2}{r^2} \]
- Rearranging gives us:
\[ \epsilon_0 = \frac{q_1 q_2}{4\pi F r^2} \]

### Step 3: Determine the dimensions of \(\epsilon_0\)
- The dimensions of force \(F\) are \([F] = M L T^{-2}\).
- The dimensions of charge \(q\) can be expressed as \([q] = I T\) (where \(I\) is current).
- Therefore, the dimensions of \(\epsilon_0\) can be derived as follows:
\[ [\epsilon_0] = \frac{[q]^2}{[F] \cdot [r]^2} = \frac{(I T)^2}{(M L T^{-2}) \cdot L^2} = \frac{I^2 T^2}{M L^3 T^{-4}} = M^{-1} L^{-3} I^2 T^4 \]

### Step 4: Use the relationship between speed of light, permittivity, and permeability
- The speed of light \(c\) is given by:
\[ c = \frac{1}{\sqrt{\mu_0 \epsilon_0}} \]
- Rearranging gives:
\[ \mu_0 = \frac{1}{c^2 \epsilon_0} \]

### Step 5: Determine the dimensions of \(\mu_0\)
- Rearranging the formula for \(c\):
\[ [\mu_0] = \frac{1}{[c]^2 [\epsilon_0]} = \frac{1}{(L T^{-1})^2 [\epsilon_0]} = \frac{1}{L^2 T^{-2} [\epsilon_0]} \]
- Substituting \([\epsilon_0]\):
\[ [\mu_0] = \frac{1}{L^2 T^{-2} (M^{-1} L^{-3} I^2 T^4)} = M L T^{-2} I^{-2} \]

### Step 6: Find the dimensions of \(\frac{\epsilon_0}{\mu_0}\)
- Now we can find:
\[ \frac{\epsilon_0}{\mu_0} = \frac{M^{-1} L^{-3} I^2 T^4}{M L T^{-2} I^{-2}} = M^{-2} L^{-2} I^4 T^6 \]

### Step 7: Find the dimensions of \(\sqrt{\frac{\epsilon_0}{\mu_0}}\)
- Taking the square root:
\[ \sqrt{\frac{\epsilon_0}{\mu_0}} = (M^{-2} L^{-2} I^4 T^6)^{1/2} = M^{-1} L^{-1} I^2 T^3 \]

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