The dimensional formula of $\mu_{0} \varepsilon_{0}$ is . . . . . . .

If dimensions of critical velocity $v_c$ of a liquid flowing through a tube are expressed as $[\eta^x \rho^y r^z]$,where $\eta, \rho,$ and $r$ are the coefficient of viscosity of the liquid,density of the liquid,and radius of the tube respectively,then the values of $x, y,$ and $z$ are given by:

If $\varepsilon_0$ is the permittivity of free space,$e$ is the charge of a proton,$G$ is the universal gravitational constant,and $m_p$ is the mass of a proton,then the dimensional formula for $\frac{e^2}{4 \pi \varepsilon_0 G m_p^2}$ is:

If force $(F)$,velocity $(V)$,and time $(T)$ are considered as fundamental physical quantities,then the dimensional formula of density will be:

In the relation $P = \frac{\alpha}{\beta} e^{-\frac{\alpha Z}{k\theta}}$,$P$ is pressure,$Z$ is distance,$k$ is Boltzmann constant,and $\theta$ is temperature. The dimensional formula of $\beta$ is:

$E, m, l$ and $G$ denote energy,mass,angular momentum,and gravitational constant respectively. Then the dimensions of $\frac{El^2}{m^5G^2}$ are: