Which of the following combinations has the dimension of electrical resistance (`in_(0)` is the permittivity of vacuum and `mu_(0)` is the permeability of vacuum)?

Options

A. `sqrt((in_(0))/(mu_(0))`
B. `sqrt((mu_(0))/(in_(0)))`
C. `(mu_(0))/(in_(0))`
D. `(in_(0))/(mu_(0))`

Explanation

To determine which combination has the dimension of electrical resistance, we will follow these steps:

### Step 1: Understand the dimensions of electrical resistance
Electrical resistance (R) is defined as the ratio of voltage (V) to current (I):
\[ R = \frac{V}{I} \]
The dimensions of voltage (V) can be expressed in terms of work done (W) and charge (Q):
\[ V = \frac{W}{Q} \]
Where:
- Work (W) has dimensions of \( [M][L^2][T^{-2}] \)
- Charge (Q) can be expressed in terms of current (I) and time (T): \( Q = I \cdot T \)

Thus, the dimensions of voltage become:
\[ V = \frac{[M][L^2][T^{-2}]}{[I][T]} = [M][L^2][T^{-3}][I^{-1}] \]

Now substituting this into the resistance formula:
\[ R = \frac{[M][L^2][T^{-3}][I^{-1}]}{[I]} = [M][L^2][T^{-3}][I^{-2}] \]

### Step 2: Determine the dimensions of permittivity (\( \epsilon_0 \))
The permittivity of free space (\( \epsilon_0 \)) can be derived from the electrostatic force equation:
\[ F = \frac{1}{4\pi \epsilon_0} \frac{q_1 q_2}{r^2} \]
Rearranging gives:
\[ \epsilon_0 = \frac{q_1 q_2}{4\pi F r^2} \]
Substituting charge \( q = I \cdot T \) and force \( F = [M][L][T^{-2}] \):
\[ \epsilon_0 = \frac{(I \cdot T)^2}{[M][L][T^{-2}] \cdot [L^2]} \]
Calculating the dimensions:
\[ \epsilon_0 = \frac{[I^2][T^2]}{[M][L^3][T^{-2}]} = [M^{-1}][L^{-3}][T^4][I^2] \]

### Step 3: Determine the dimensions of permeability (\( \mu_0 \))
Using the relationship between speed of light \( c \), permittivity, and permeability:
\[ c^2 = \frac{1}{\mu_0 \epsilon_0} \]
Thus:
\[ \mu_0 = \frac{1}{c^2 \epsilon_0} \]
Where the dimensions of \( c \) are \( [L][T^{-1}] \):
\[ c^2 = [L^2][T^{-2}] \]
Substituting this into the equation for \( \mu_0 \):
\[ \mu_0 = \frac{[M][L^3][T^{-4}][I^{-2}]}{[L^2][T^{-2}]} = [M][L^{-1}][T^{-2}][I^{-2}] \]

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

### Conclusion
The dimensions of electrical resistance \( R \) are:
\[ [M][L^2][T^{-3}][I^{-2}] \]
And we find that:
\[ \sqrt{\frac{\mu_0}{\epsilon_0}} \]
has the same dimensions as resistance.

Thus, the correct option is:
**Option B: \( \sqrt{\frac{\mu_0}{\epsilon_0}} \)**