Question

The dimension of $\left(\mu_{0} \in_{0}\right)^{-1 / 2}$ are :

• A. $[L^{-1} T] $
• B. $[L T^{-1}] $
• C. $[L^{1/2} T^{1 /2}]$
• D. $[L^{1/2} T^{- 1/2}]$

Answer 1

The Correct Option is B

To determine the dimensions of the expression $\left(\mu_{0} \epsilon_{0}\right)^{-1 / 2}$, we need to understand the dimensions of each component involved: $\mu_{0}$ and $\epsilon_{0}$. Here, $\mu_{0}$ is the permeability of free space, and $\epsilon_{0}$ is the permittivity of free space.

1. The dimensional formula for the permeability of free space $\mu_{0}$ is $[M L T^{-2} A^{-2}]$.
2. The dimensional formula for the permittivity of free space $\epsilon_{0}$ is $[M^{-1} L^{-3} T^{4} A^{2}]$.
3. Now, we calculate the dimensional formula for their product:
   \[ \mu_{0} \epsilon_{0} = [M L T^{-2} A^{-2}] \cdot [M^{-1} L^{-3} T^{4} A^{2}] = [L^{-2} T^{2}] \]
4. Next, we find the dimensional formula for the expression \(\left(\mu_{0} \epsilon_{0}\right)^{-1 / 2}\):
   \[ \left(\mu_{0} \epsilon_{0}\right)^{-1 / 2} = [L^{-2} T^{2}]^{-1/2} = [L T^{-1}] \]

Therefore, the dimensional formula of $\left(\mu_{0} \epsilon_{0}\right)^{-1 / 2}$ is $[L T^{-1}]$.

This matches with the correct option: $[L T^{-1}]$, making it the right answer.