If dimensions of critical velocity, v c of a liquid flowing through a tube are expressed as η x ρ y r z ,…

Dimensions of velocity, coefficient of viscosity and density are,

$\left[v\right]=\left[{\mathrm{LT}}^{-1}\right]$

$\left[\eta \right]=\frac{\left[F\right]}{\left[A{\displaystyle \frac{dv}{dx}}\right]}=\frac{\left[{\mathrm{MLT}}^{-2}\right]}{\left[{\mathrm{L}}^2\right]\left[{\mathrm{T}}^{-1}\right]}=\left[{\mathrm{ML}}^{-1}{\mathrm{T}}^{-1}\right]$

$\left[\rho \right]=\left[{\mathrm{ML}}^{-3}\right]$

Now, $\left[v\right]={\left[\eta \right]}^x{\left[\rho \right]}^y{\left[r\right]}^z$

 $\Rightarrow \left[{\mathrm{LT}}^{-1}\right]={\left[{\mathrm{ML}}^{-1}{\mathrm{T}}^{-1}\right]}^{\mathrm{x}}{\left[{\mathrm{ML}}^{-3}\right]}^{\mathrm{y}}{\left[\mathrm{L}\right]}^{\mathrm{z}}$

Now, by using dimensional homogeneity, we will get,

${\mathrm{M}}^0={\mathrm{M}}^{\left(x+y\right)}$

$\Rightarrow y=-x$

For $T$, 

$x=1$

$\Rightarrow y=-x=-1$

For $L$,

$1=-x-3y+z$

$\Rightarrow z=-1$