If mass M, area A' and velocity V are chosen as fundamental units, then the dimension of coefficient of viscosity will be:

$P=\frac{\alpha }{\beta }lo{g}_e\left(\frac{kt}{\beta x}\right)$

$\frac{kt}{\beta x}$ = Dimensionless

$\beta =\frac{kt}{x}=\frac{\left[M{L}^2{T}^{-2}{k}^{-1}\right]\left[k\right]}{\left[L\right]}$

$\frac{\alpha }{\beta }=$ dimensionless

a = dimensionless of b

a = MLT^-2

Using F = MA = $m\frac{V}{T}$ $\frac{}{}$

m = FTV¹

Since, $\frac{x^2}{\alpha kT}$ should be dimensionless.

So, dimension of  $\alpha ,\left[\alpha \right]=\frac{L^2}{M{L}^2{T}^{-2}}=M^{-1}{T}^2$

Dimension of  $\alpha {\beta }^2$ should be that of W.

So, $\left[\alpha {\beta }^2\right]=M{L}^2{T}^{-2}$

$\Rightarrow \left[{\beta }^2\right]=\frac{M{L}^2{T}^{-2}}{M^{-1}{T}^2}=M^2{L}^2{T}^{-4}\Rightarrow \left[\beta \right]=ML{T}^{-2}$

The dimensional formula for energy E is [ML²T? ²].
The dimensional formula for the gravitational constant G is [M? ¹L³T? ²].
The ratio E/G has dimensions: [ML²T? ²] / [M? ¹L³T? ²] = [M²L? ¹T? ].

x = (IFV²)/ (WL? )
I = [ML²]
F = [MLT? ²]
V² = [L² T? ²]
W = [ML² T? ²]
Q? = [L? ]
X = [ML? ¹ T? ²]