If dimensions of critical velocity ${\nu }_c$ of a liquid flowing through a tube are expressed as $\left[{\eta }^x{\rho }^y{r}^z\right]$ 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

$\left[v_c\right]=\left[{\eta }^x{\rho }^y{r}^2\right] \text{ (given) . . . .(i)}$

$Writing the dimensions of various quantities in eqn. \left(i\right), we get$

$\left[{\mathrm{M}}^0{\mathrm{LT}}^{-1}\right]={\left[{\mathrm{ML}}^{-1}{ \mathrm{T}}^{-1}\right]}^x{\left[{\mathrm{ML}}^{-3}{ \mathrm{T}}^0\right]}^y{\left[{\mathrm{M}}^0{\mathrm{LT}}^0\right]}^z$

$=\left[{\mathbf{M}}^{x+y}{ \mathrm{L}}^{-x-3y+z}{ \mathrm{T}}^{-x}\right]$

$Applying the principle of homogeneity of dimensions, we get$

$x+y=0 ;-x-3 y+z=1 ;-x=-1$

$\text{ On solving, we get }$

$x=1, y=-1, z=-1$