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((epsilon_(0))/(mu_(0)))`
B. `sqrt((mu_(0))/(epsilon_(0)))`
C. `(mu_(0))/(epsilon_(0))`
D. `(epsilon_(0))/(mu_(0))`

Explanation

To determine which combination has the dimension of electrical resistance, we need to analyze the dimensions of the permittivity of vacuum (ε₀) and the permeability of vacuum (μ₀), and then relate them to the dimensions of electrical resistance (R).

### Step-by-Step Solution:

1. **Identify the Dimensions of ε₀ and μ₀**:
- The dimension of permittivity of vacuum (ε₀) is given by:
\[ [\epsilon_0] = M^{-1} L^{-3} T^{4} A^{2} \]
- The dimension of permeability of vacuum (μ₀) is given by:
\[ [\mu_0] = M L T^{-2} A^{-2} \]

2. **Write the Expression for Resistance**:
- The dimension of electrical resistance (R) is given by:
\[ [R] = M L^{2} T^{-3} A^{-2} \]

3. **Combine ε₀ and μ₀**:
- We will consider the combination \( \sqrt{\frac{\mu_0}{\epsilon_0}} \) to find its dimensions.
- First, write the dimensions of \( \frac{\mu_0}{\epsilon_0} \):
\[ \frac{\mu_0}{\epsilon_0} = \frac{M L T^{-2} A^{-2}}{M^{-1} L^{-3} T^{4} A^{2}} = M^{1+1} L^{1+3} T^{-2-4} A^{-2-2} = M^{2} L^{4} T^{-6} A^{-4} \]

4. **Take the Square Root**:
- Now, take the square root of the dimensions obtained:
\[ \sqrt{\frac{\mu_0}{\epsilon_0}} = \sqrt{M^{2} L^{4} T^{-6} A^{-4}} = M^{1} L^{2} T^{-3} A^{-2} \]

5. **Conclusion**:
- The dimensions \( M^{1} L^{2} T^{-3} A^{-2} \) match exactly with the dimensions of electrical resistance (R).
- Therefore, the combination that has the dimension of electrical resistance is:
\[ \sqrt{\frac{\mu_0}{\epsilon_0}} \]

### Final Answer:
The correct option is \( \sqrt{\frac{\mu_0}{\epsilon_0}} \).