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

Given, critical velocity of liquid flowing through tube is expressed as, ${\mathrm{v}}_{\mathrm{c}}\propto {\mathrm{\eta }}_{\mathrm{x}}{\mathrm{\rho }}^{\mathrm{y}}{\mathrm{r}}^{\mathrm{z}}$
Coefficient of viscosity of liquid, $\mathrm{\eta }=\left[{\mathrm{ML}}^{-1}{\mathrm{T}}^{-1}\right]$
Density of liquid, $\mathrm{\rho }=\left[{\mathrm{ML}}^{-3}\right]$
Radius of a tube, r = [L]
Critical velocity of liquid, ${\mathrm{v}}_{\mathrm{c}}=\left[{\mathrm{M}}^0{\mathrm{LT}}^{-1}\right]$
$$\begin{gathered} \Rightarrow \left[{\mathrm{M}}^0{\mathrm{LT}}^{-1}\right]={\left[{\mathrm{ML}}^{-1}{\mathrm{T}}^{-1}\right]}^{\mathrm{x}}{\left[{\mathrm{ML}}^{-3}\right]}^{\mathrm{y}}{\left[\mathrm{L}\right]}^{\mathrm{z}} \\ \left[{\mathrm{M}}^0{\mathrm{LT}}^{-1}\right]=\left[{\mathrm{M}}^{\mathrm{x}+\mathrm{y}}{\mathrm{L}}^{-\mathrm{x}-3\mathrm{y}}{\mathrm{T}}^{-\mathrm{x}}\right] \end{gathered}$$
Comparing powers of M, L and T, we get
x+y = 0, -x-3y+z = 1, -x = -1
On solving above equations, we get
x=1, y=-1, z=-1