The dimensional formula `mu_(0)epsilon_0` is

Options

A. `M^0L^(-2)T^2`
B. `M^0 L^2T^(-2)`
C. `M^(0) L^(1) T^(-1)`
D. `M^(0) L^(-1) T^(1)`

Answer

To find the dimensional formula of the product \( \mu_0 \epsilon_0 \), we start with the relationship involving the speed of light \( c \):

1. **Understanding the relationship**: The speed of light \( c \) is given by the equation:
\[ c = \frac{1}{\sqrt{\mu_0 \epsilon_0}} \]
Squaring both sides gives:
\[ c^2 = \frac{1}{\mu_0 \epsilon_0} \]
Therefore, we can express \( \mu_0 \epsilon_0 \) as:
\[ \mu_0 \epsilon_0 = \frac{1}{c^2} \]

2. **Finding the dimensional formula of \( c^2 \)**: The speed of light \( c \) has dimensions of length over time:
\[ [c] = \frac{[L]}{[T]} \quad \text{(where [L] is length and [T] is time)} \]
Thus, squaring \( c \) gives:
\[ [c^2] = \left(\frac{[L]}{[T]}\right)^2 = \frac{[L]^2}{[T]^2} \]

3. **Finding the dimensional formula of \( \mu_0 \epsilon_0 \)**: Since we have \( \mu_0 \epsilon_0 = \frac{1}{c^2} \), we can write:
\[ [\mu_0 \epsilon_0] = \frac{1}{[c^2]} = \frac{1}{\frac{[L]^2}{[T]^2}} = \frac{[T]^2}{[L]^2} \]

Thus, the dimensional formula of \( \mu_0 \epsilon_0 \) is:
\[ [\mu_0 \epsilon_0] = [T]^2 [L]^{-2} \]