Let `[epsilon_(0)]` denote the dimensional formula of the permittivity of the vacuum, and `[mu_(0)]` that of the permeability of the vacuum. If `M = mass ,L = length, T = time and I = electric current,

Options

A. `[epsilon _(0) ] = [M^(-1)L^(-3)T^(2)I]`
B. ` [epsilon_(0) ] = [M^(-1)L^(-3)T^(4)I^(2)]`
C. ` [mu_(0)] = [MLT^(-2)I^(-2)] `
D. `[mu_(0)] = [ML^(2)T^(-1)I]`

Explanation

To find the dimensional formulas for the permittivity of vacuum \((\epsilon_0)\) and permeability of vacuum \((\mu_0)\), we will follow these steps:

### Step 1: Understanding the relationship between \(\epsilon_0\) and \(\mu_0\)
We start with the relationship between permittivity, permeability, and the speed of light \(c\):
\[ c = \frac{1}{\sqrt{\mu_0 \epsilon_0}} \]
From this, we can express \(\epsilon_0\) in terms of \(\mu_0\) and \(c\):
\[ \epsilon_0 = \frac{1}{\mu_0 c^2} \]

### Step 2: Finding the dimensional formula of \(c\)
The speed of light \(c\) has dimensions of length per time:
\[ [c] = L T^{-1} \]

### Step 3: Finding the dimensional formula of \(\epsilon_0\)
Using the expression for \(\epsilon_0\):
\[ \epsilon_0 = \frac{1}{\mu_0 c^2} \]
We need to find the dimensional formula of \(\mu_0\) first.

### Step 4: Using Coulomb's Law to find \(\epsilon_0\)
Coulomb's law states:
\[ F = \frac{1}{4\pi \epsilon_0} \frac{q_1 q_2}{r^2} \]
Rearranging gives:
\[ \epsilon_0 = \frac{q_1 q_2}{4\pi F r^2} \]
Where:
- \(F\) has dimensions of \(M L T^{-2}\) (force).
- \(q\) (charge) has dimensions of \(I T\) (current times time).

### Step 5: Substituting the dimensions into the formula for \(\epsilon_0\)
Substituting the dimensions into the expression for \(\epsilon_0\):
\[ \epsilon_0 = \frac{(I T)^2}{M L T^{-2} L^2} = \frac{I^2 T^2}{M L^2 T^{-2}} = \frac{I^2 T^4}{M L^2} \]
Thus, the dimensional formula for \(\epsilon_0\) is:
\[ [\epsilon_0] = M^{-1} L^{-3} T^{4} I^{2} \]

### Step 6: Finding the dimensional formula of \(\mu_0\)
Using the relationship:
\[ \mu_0 = \frac{1}{\epsilon_0 c^2} \]
Substituting the dimensional formula of \(\epsilon_0\) and \(c\):
\[ \mu_0 = \frac{1}{\left(M^{-1} L^{-3} T^{4} I^{2}\right) (L^2 T^{-2})} = \frac{M L^3 T^{-4} I^{-2}}{1} \]
Thus, the dimensional formula for \(\mu_0\) is:
\[ [\mu_0] = M^{-1} L^{3} T^{-4} I^{2} \]

### Final Answers
- The dimensional formula for \(\epsilon_0\) is:
\[ [\epsilon_0] = M^{-1} L^{-3} T^{4} I^{2} \]
- The dimensional formula for \(\mu_0\) is:
\[ [\mu_0] = M^{-1} L^{3} T^{-4} I^{-2} \]