If momentum P , area A and time T are taken as fundamental quantities, then the dimensional formula for…

We know, Viscosity$=$Pascal$‐$second

Let $\left[\eta \right]=$$P^x{A}^y{T}^z=\left[{\mathrm{M}}^1{\mathrm{L}}^{-1}{\mathrm{T}}^{-1}\right]$

$\Rightarrow {\left[{\mathrm{M}}^1{\mathrm{L}}^{+1}{\mathrm{T}}^{-1}\right]}^{\mathrm{x}}{\left[{\mathrm{L}}^2\right]}^y{\left[{ \mathrm{T}}^1\right]}^z={\mathrm{M}}^1{\mathrm{L}}^{-1}{\mathrm{T}}^{-1}$

$\Rightarrow {\mathrm{M}}^{\mathrm{x}}{\mathrm{L}}^{+\mathrm{x}+2\mathrm{y}}{\mathrm{T}}^{-\mathrm{x}+z}={\mathrm{M}}^1{\mathrm{L}}^{-1}{\mathrm{T}}^{-1}$

From above, we have 

$$\begin{gathered} x=1 \\ x+2y=-1 \\ -x+z=-1 \end{gathered}$$

Solving above three equations, we get 

$y=-1$ and $\mathrm{z}=0$

So, Viscosity $=\left[{\mathrm{P}}^1{\mathrm{A}}^{-1}{\mathrm{T}}^0\right]$