If dimensions of the coefficient of viscosity $\eta$ of a liquid flowing through a tube are expresse

Question

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

Professor PC Thomas Classes · Answer

(a) : $[\eta]=\left[v_{c}^{x} ho^{y} r^{z}ight]$\\ Writing the dimensions of various quantities, we get $\left[\mathrm{ML}^{-1} \mathrm{~T}^{-1}ight]=\left[\mathrm{LT}^{-1}ight]^{x}\left[\mathrm{ML}^{-3}ight]^{y}[\mathrm{~L}]^{z}$ \\ $=\left[\mathrm{M}^{y} \mathrm{~L}^{x-3 y+z} \mathrm{~T}^{-x}ight]$ \\ Applying the principle of homogeneity of dimensions, we get \\ $y=1 ;-x=-1 \Rightarrow x=1$ \\ $x-3 y+z=-1 \Rightarrow 1-3(1)+z=-1 \Rightarrow z=1$ \\ On solving, we get $x=1, y=1, z=1$