If mu_0$\mu_0$ and varepsilon_0$\varepsilon_0$ are the permeability and permittivity of free space, respectively, then the dimension of left(frac1mu_0varepsilon_0right)$\left(\frac{1}{\mu_0\varepsilon_0}\right)$ is:

### Related Formula Maxwell's relation for the speed of light (c$c$) in vacuum: c = frac1sqrtmu_0 varepsilon_0$$c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$$ ### Core Logic Square both sides of the speed of light formula: c^2 = frac1mu_0 varepsilon_0$$c^2 = \frac{1}{\mu_0 \varepsilon_0}$$ Since c$c$ represents the speed of light (velocity), its dimensional formula is: [c] = [L T^-1]$$[c] = [L T^{-1}]$$ Therefore, the dimensions of c^2$c^2$ are: [c^2] = [L T^-1]^2 = [L^2 T^-2]$$[c^2] = [L T^{-1}]^2 = [L^2 T^{-2}]$$ ### Step 1: Match options Express the dimensions in terms of the given variable ratios: [c^2] = fracL^2T^2$$[c^2] = \frac{L^2}{T^2}$$ This perfectly matches Option (2). ### Pattern Recognition Sees: Dimensions of vacuum permittivity-permeability inverse product. Trap: Trying to find the separate dimensions of both mu_0$\mu_0$ and varepsilon_0$\varepsilon_0$, then conducting manual division. This is extremely slow and prone to algebraic error. Shortcut: Directly identify the term as c^2$c^2$. The velocity squared has units of mathrmm^2mathrms^-2$\mathrm{m}^2\mathrm{s}^{-2}$, giving dimensions of mathrmL^2mathrmT^-2$\mathrm{L}^2\mathrm{T}^{-2}$ immediately. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Waves Class 11 Physics: Units and Measurements