Physics - Unit & Dimension Question with Solution | TestHub

According to the principle of dimensional homogeneity, the dimensions of each  terms of a dimensional equation on both sides are the same.

So, dimension of $b$ and $V$ will be same,  $\left[b\right]=\left[V\right]$ 

And dimension of $\frac{a}{V^2}$ and $P$ will be same, $\left[\frac{a}{V^2}\right]=\left[P\right]$

From above two relations, we have $\left[\frac{a}{b^2}\right]=\left[P\right]$

$\therefore \left[\frac{b^2}{a}\right]=\frac{1}{\left[P\right]}={\left[M{L}^{-1}{T}^{-2}\right]}^{-1}$

Thus, $\left[M^{-1}{L}^1{T}^2\right]$  is the dimension of compressibility.