The dimension of $$\sqrt{\frac{\mu_0}{\epsilon_0}}$$ is equal to that of : ($$\mu_0$$ = Vacuum permeability and $$\epsilon_0$$ = Vacuum permittivity)

The expression is $$\sqrt{\dfrac{\mu_0}{\epsilon_0}}$$.
We must find its dimensional formula and compare it with the dimensions of the given physical quantities.

Step 1 : Dimensions of vacuum permeability $$\mu_0$$
In SI, $$\mu_0 = 4\pi \times 10^{-7}\;N\,A^{-2}$$, where $$N = kg \; m \; s^{-2}$$.
Hence, $$\mu_0$$ has dimensions
$$[\,\mu_0\,] = M^{1}\,L^{1}\,T^{-2}\,I^{-2}$$.

Step 2 : Dimensions of vacuum permittivity $$\epsilon_0$$
From Coulomb’s law $$F = \dfrac{1}{4\pi\epsilon_0}\dfrac{q^{\,2}}{r^{\,2}}$$, we get
$$\epsilon_0 = \dfrac{q^{\,2}}{F\,r^{\,2}}$$.
Using $$q = I\,T$$ and $$F = M\,L\,T^{-2}$$:
$$[\,\epsilon_0\,] = \dfrac{(I\,T)^2}{M\,L\,T^{-2}\;L^{2}} = M^{-1}\,L^{-3}\,T^{4}\,I^{2}$$.

Step 3 : Dimensions of $$\dfrac{\mu_0}{\epsilon_0}$$
$$\left[\dfrac{\mu_0}{\epsilon_0}\right] = M^{1-(-1)}\,L^{1-(-3)}\,T^{-2-4}\,I^{-2-2}$$
$$= M^{2}\,L^{4}\,T^{-6}\,I^{-4}$$.

Step 4 : Taking the square root
$$\left[\sqrt{\dfrac{\mu_0}{\epsilon_0}}\right] = M^{1}\,L^{2}\,T^{-3}\,I^{-2}$$.

Step 5 : Dimensions of resistance
Resistance $$R = \dfrac{V}{I}$$.
Electrical power $$P = V\,I$$ has dimensions $$M\,L^{2}\,T^{-3}$$.
Therefore voltage $$V = \dfrac{P}{I}$$ has dimensions $$M\,L^{2}\,T^{-3}\,I^{-1}$$.
Thus,
$$[\,R\,] = \dfrac{[\,V\,]}{[\,I\,]} = M\,L^{2}\,T^{-3}\,I^{-2}$$.

Step 6 : Comparison
The dimension $$M\,L^{2}\,T^{-3}\,I^{-2}$$ obtained for $$\sqrt{\dfrac{\mu_0}{\epsilon_0}}$$ is exactly the same as that of resistance.

Hence, $$\sqrt{\dfrac{\mu_0}{\epsilon_0}}$$ has the dimensions of resistance.
Correct choice: Option D (Resistance).