In terms of resistance R and time T, the dimension of ratio `(mu)/(varepsilon)` of the permeability `mu` and permittivity `varepsilon` is :

Options

A. `RT^(-2)`
B. `R^(2)T^(-2)`
C. `R^2`
D. `R^(2)T^(2)`

Explanation

To find the dimensions of the ratio \(\frac{\mu}{\varepsilon}\) in terms of resistance \(R\) and time \(T\), we will follow these steps:

### Step 1: Write down the dimensions of permittivity \(\varepsilon\) and permeability \(\mu\).

1. **Permittivity (\(\varepsilon\))**:
The dimensions of permittivity are given by:
\[ \varepsilon = M^{-1} L^{-3} T^{4} A^{2} \]

2. **Permeability (\(\mu\))**:
The dimensions of permeability are given by:
\[ \mu = M L T^{-2} A^{-2} \]

### Step 2: Set up the ratio \(\frac{\mu}{\varepsilon}\).

Now, we can express the ratio \(\frac{\mu}{\varepsilon}\):
\[ \frac{\mu}{\varepsilon} = \frac{M L T^{-2} A^{-2}}{M^{-1} L^{-3} T^{4} A^{2}} \]

### Step 3: Simplify the ratio.

When we simplify the ratio, we will divide the dimensions of \(\mu\) by the dimensions of \(\varepsilon\):
\[ \frac{\mu}{\varepsilon} = \frac{M L T^{-2} A^{-2}}{M^{-1} L^{-3} T^{4} A^{2}} = M^{1 - (-1)} L^{1 - (-3)} T^{-2 - 4} A^{-2 - 2} \]
This simplifies to:
\[ \frac{\mu}{\varepsilon} = M^{2} L^{4} T^{-6} A^{-4} \]

### Step 4: Express the dimensions in terms of resistance \(R\) and time \(T\).

The dimension of resistance \(R\) is given by:
\[ R = M L^{2} T^{-3} A^{-2} \]

Now, we need to express \(R^2\):
\[ R^2 = (M L^{2} T^{-3} A^{-2})^2 = M^{2} L^{4} T^{-6} A^{-4} \]

### Step 5: Relate \(\frac{\mu}{\varepsilon}\) to \(R^2\).

From our calculations, we see that:
\[ \frac{\mu}{\varepsilon} = R^2 \]

### Conclusion

Thus, the dimension of the ratio \(\frac{\mu}{\varepsilon}\) in terms of resistance \(R\) and time \(T\) is:
\[ \frac{\mu}{\varepsilon} = R^2 \]