In the expression $P = El^2m^{-5}G^{-2}$, $E$, $l$, $m$ and $G$ denote energy, mass, angular momentum and gravitational constant respectively. Show that $P$ is a dimensionless quantity.
Given, expression is, $\quad P=E L^{2} m^{-5} G^{-2}$
where $\mathrm{E}$ is energy
$[\mathrm{E}]=\left[\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-2}\right]$
$\mathrm{m}$ is mass
$[\mathrm{m}]=\left[\mathrm{M}^{1}\right]$
$\mathrm{L}$ is angular momentum $[\mathrm{L}]=\left[\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-1}\right]$
$\mathrm{G}$ is gravitational constant $[\mathrm{G}]=\left[\mathrm{M}^{-1} \mathrm{~L}^{3} \mathrm{~T}^{-2}\right]$
Substituting dimensions of each term in the given expression,
$[\mathrm{P}] =\left[\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-2}\right] \times\left[\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-1}\right]^{2} \times\left[\mathrm{M}^{1}\right]^{-5} \times\left[\mathrm{M}^{-1} \mathrm{~L}^{3} \mathrm{~T}^{-2}\right]^{-2}$
$=\left[\mathrm{M}^{1+2-5+2} \mathrm{~L}^{2+4-6} \mathrm{~T}^{-2-2+4}\right]=\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}\right]$
Hence, $P$ is a dimensionless quantity.
The foundations of dimensional analysis were laid down by
If dimensions of critical velocity $v_c$ of a liquid flowing through a tube are expressed as$ [\eta ^x \rho ^yr^z]$ where $\eta ,\rho $ and $r $ are the coefficient of viscosity of liquid, density of liquid and radius of the tube respectively, then the values of $x, y$ and $z$ are given by
The quantity $X = \frac{{{\varepsilon _0}LV}}{t}$: ${\varepsilon _0}$ is the permittivity of free space, $L$ is length, $V$ is potential difference and $t$ is time. The dimensions of $X$ are same as that of
In the relation $P = \frac{\alpha }{\beta }{e^{ - \frac{{\alpha Z}}{{k\theta }}}}$ $P$ is pressure, $Z$ is the distance, $k$ is Boltzmann constant and $\theta$ is the temperature. The dimensional formula of $\beta$ will be