Correct answer is (B)

$\sqrt{\frac{{\in }_0}{{\mu }_0}}$ = $\frac{{\in }_0}{\sqrt{{\mu }_0{\in }_0}}$ = c $\times$ ${\in }_0$

$\therefore$ $\left[\sqrt{\frac{{\in }_0}{{\mu }_0}}\right]$ = $\left[L{T}^{-1}\right]\times \left[{\in }_0\right]$

We know, F = $\frac{1}{4\pi {\in }_0}\frac{q^2}{r^2}$

$\therefore$ ${\in }_0=\frac{q^2}{4\pi r^{2}F}$

$\Rightarrow$ $\left[{\in }_0\right]=\frac{{\left[AT\right]}^2}{\left[ML{T}^{-2}\right]\times \left[L^2\right]}$ = $\left[A^2{M}^{-1}{L}^{-3}{T}^4\right]$

$\therefore$ $\left[\sqrt{\frac{{\in }_0}{{\mu }_0}}\right]$ = $\left[L{T}^{-1}\right]$ $\times$ $\left[A^2{M}^{-1}{L}^{-3}{T}^4\right]$

                 = [A²T³M^–1L^–2]

Correct answer is (A)

$\left[\frac{\ell }{r}\right]=$ T

[CV] $=$ AT

So,   $\left[\frac{\ell }{rCV}\right]$ = $\frac{T}{AT}$ = [A^$-$1]

Correct answer is (C)

We know. resistivity ($\rho$) = $\frac{RA}{L}$

and conductivity = $\frac{1}{\rho }$ = $\frac{1}{RA}$

As   R = $\frac{V}{\mathrm{I}}$

$\therefore$   conductivity = $\frac{L\mathrm{I}}{VA}$

Also  V = $\frac{\omega }{q}$ = $\frac{\omega }{it}$ = $\frac{\left[M{L}^2{T}^{-2}\right]}{\left[\mathrm{I}\right]\left[T\right]}$ = $\left[M{L}^2{T}^{-3}{\mathrm{I}}^{-1}\right]$

$\therefore$   Conductivity = $\frac{\left[\mathrm{L}\right]\left[\mathrm{I}\right]}{\left[M{L}^2{T}^{-3}{\mathrm{I}}^{-1}\right]\left[L^2\right]}$

=   $\left[M^{-1}{L}^{-3}{T}^3{\mathrm{I}}^2\right]$

Correct answer is (B)

From Coulomb's law we know,

$F=\frac{1}{4\pi {\in }_0}\frac{q_1{q}_2}{r^2}$

$\therefore$ ${\in }_0=\frac{1}{4\pi }\frac{q_1{q}_2}{F{r}^2}$

Hence, $\left[{\in }_0\right]=\frac{\left[AT\right]\left[AT\right]}{\left[ML{T}^{-2}\right]\left[L^2\right]}$

= $\left[M^{-1}{L}^{-3}{T}^4{A}^2\right]$

Correct answer is (C)

We know that,
Lorentz force $\left|\overrightarrow{F}\right|=\left|q\overrightarrow{v}\times \overrightarrow{B}\right|$

$\therefore [B]$ = $\frac{[F]}{[q][v]}$ = $\frac{[ML{T}^{-2}]}{[C]\times [L{T}^{-1}]}$ = [$M{T}^{-1}{C}^{-1}$]