Which of the following combinations has the dimension of electrical resistance ($$\varepsilon_0$$ is the permittivity of vacuum and $$\mu_0$$ is the permeability of vacuum)?

We have electrical resistance denoted by $$R$$ and, from Ohm’s law, the defining relation is $$V = IR.$$ So $$R = \dfrac{V}{I}.$$

First we find the dimensional formula of the potential difference $$V.$$ Potential is work done per unit charge and work (energy) has the same dimensions as joule.

Energy $$E$$ has the dimensions $$[E] = M L^{2} T^{-2}.$$ Charge $$q$$ has the dimensions $$[q] = I\,T.$$ Therefore $$[V] = \dfrac{[E]}{[q]} = \dfrac{M L^{2} T^{-2}}{I T}=M L^{2} T^{-3} I^{-1}.$$

Substituting this in $$R = V/I,$$ we get $$[R]=\dfrac{M L^{2} T^{-3} I^{-1}}{I}=M L^{2} T^{-3} I^{-2}.$$

Now we need the dimensions of the permittivity of free space $$\varepsilon_0.$$ We recall Coulomb’s law: $$F=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{r^{2}}.$$ Re-writing, $$\varepsilon_0=\dfrac{q^{2}}{4\pi F r^{2}}$$ (constants like $$4\pi$$ do not affect dimensions).

Hence, $$[\varepsilon_0]=\dfrac{[q]^2}{[F]\,[r]^2} =\dfrac{(I\,T)^2}{(M L T^{-2})\,L^{2}} =\dfrac{I^{2} T^{2}}{M L^{3} T^{-2}} = M^{-1} L^{-3} T^{4} I^{2}.$$

Next we find the dimensions of the permeability of free space $$\mu_0.$$ We use the formula for the magnetic force per unit length between two long parallel currents: $$\dfrac{F}{\ell}=\dfrac{\mu_0}{2\pi}\dfrac{I_1 I_2}{r},$$ so that $$\mu_0=\dfrac{F r}{I^{2}}\;(\text{again the constant }2\pi\text{ is ignored}).$$

Therefore, $$[\mu_0]=\dfrac{[F]\,[r]}{I^{2}} =\dfrac{(M L T^{-2})\,L}{I^{2}} = M L T^{-2} I^{-2}.$$

We now examine the four given combinations.

First, $$\left[\dfrac{\mu_0}{\varepsilon_0}\right] =\dfrac{M L T^{-2} I^{-2}}{M^{-1} L^{-3} T^{4} I^{2}} = M^{1-(-1)} L^{1-(-3)} T^{-2-4} I^{-2-2} = M^{2} L^{4} T^{-6} I^{-4}.$$

The square root of this ratio is $$\left[\sqrt{\dfrac{\mu_0}{\varepsilon_0}}\right] =\left(M^{2} L^{4} T^{-6} I^{-4}\right)^{1/2} = M^{1} L^{2} T^{-3} I^{-2}.$$

Comparing with $$[R]=M L^{2} T^{-3} I^{-2},$$ we see an exact match.

For completeness, observe that $$\left[\dfrac{\varepsilon_0}{\mu_0}\right] =\left(M^{-1} L^{-3} T^{4} I^{2}\right)\big/\left(M L T^{-2} I^{-2}\right) = M^{-2} L^{-4} T^{6} I^{4},$$ and $$\left[\sqrt{\dfrac{\varepsilon_0}{\mu_0}}\right] = M^{-1} L^{-2} T^{3} I^{2},$$ neither of which matches the dimensional formula of resistance.

So only the combination $$\sqrt{\dfrac{\mu_0}{\varepsilon_0}}$$ possesses the dimensions $$M L^{2} T^{-3} I^{-2}$$ required for electrical resistance.

Hence, the correct answer is Option D.