$\text{H}=\frac{\text{B}}{\mu_0}$
$\frac{\text{E}_0}{\text{H}_0}=\frac{\frac{\text{B}_0}{(\mu_0\in_0\text{C})}}{\frac{\text{B}_0}{\mu_0}}=\frac{1}{\in_0\text{C}}$
$=\frac{1}{8.85\times10^{-12}\times3\times10^8}$
$=376.6\Omega=377\Omega$
$\text{Dimension}\frac{1}{\in_0\text{C}}=\frac{1}{\big[\text{LT}^{-1}\big]\big[\text{M}^{-1}\text{L}^{-3}\text{T}^{4}\text{A}^2\big]}$
$=\frac{1}{\text{M}^{-1}\text{L}^{-2}\text{T}^{3}\text{A}^2}=\text{M}^1\text{L}^2\text{T}^{-3}\text{A}^{-2}=[\text{R}]$
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$\text{v}=\frac{\pi}{8}\frac{\text{Pr}^4}{\eta\text{l}}$
where P is the pressure difference between the two ends of the pipe and $\eta$ is coefficent of viscosity of the liquid having dimensional formula ML-1 T-1. Check whether the equation is dimensionally correct.