The quantities x = 1 μ _ 0 ∈ _ 0 , y = (E)/(B) and z = (l)/(CR) are...

(a) Dimensions of quantity are given bellow

${\mu}_{0} = \left[Tm{A}^{- 1}\right] = \left[Wb{m}^{- 1}{A}^{- 1}\right] = \left[Vs{m}^{- 1}{C}^{- 1}s\right] = \left[N{C}^{- 2}{s}^{2}\right]{ε}_{0} = \left[{C}^{2}{N}^{- 1}{m}^{- 2}\right]$

Now the dimension of $x$ is given by

\left[x\right] = \dfrac{1}{\left[\sqrt{{\mu}_{0}{\in}_{0}}\right]} = \sqrt{\dfrac{1}{\left[N{C}^{- 2}{s}^{2}\right]\left[{C}^{2}{N}^{- 1}{m}^{- 2}\right]}}\left[x\right] = m{s}^{- 1}$</p><p>(b) The dimension of electric and magnetic field is given by,</p><p>$B = \left[T\right] = \left[Wb{m}^{- 2}\right] = \left[Vs{m}^{- 2}\right]E = \left[V{m}^{- 1}\right]$</p><p>Now the dimension of$y$is given by,</p><p> \left[y\right] = \dfrac{\left[E\right]}{\left[B\right]} = \dfrac{\left[V{m}^{- 1}\right]}{\left[Vs{m}^{- 2}\right]} = m{s}^{- 1}$

(c) The dimension of capacitance and resistance is given by,

$C = \left[C{V}^{- 1}\right]R = \left[V{C}^{- 1}s\right]$

Now the dimension of $z$ is given by

\left[z\right] = \dfrac{\left[l\right]}{\left[CR\right]} = \dfrac{\left[m\right]}{\left[C{V}^{- 1}\right]\left[V{C}^{- 1}s\right]} = m{s}^{- 1}

All quantities are having the dimensions of velocity.