Given equation of state

$\left(P+\frac{a}{V^2}\right)(V-b)=RT$

We need the dimensional formula of

$\frac{b^2}{a}$

and then match it with the correct physical quantity.

Find dimensions of $b$

In the factor $(V-b)$, both terms must have the same dimensions.

So,

$[b]=[V]$

Since volume has dimension

$[V]=L^3$

therefore,

$[b]=L^3$

Find dimensions of $a$

In the factor

$P+\frac{a}{V^2}$

both terms must have the same dimensions. Hence,

$\left[\frac{a}{V^2}\right]=[P]$

Therefore,

$[a]=[P][V^2]$

Now,

$[P]=ML^{-1}T^{-2}$ $[V^2]=(L^3)^2=L^6$

So,

$[a]=ML^{-1}T^{-2}\cdot L^6=ML^5T^{-2}$

Find dimensions of $\dfrac{b^2}{a}$

$[b^2]=(L^3)^2=L^6$

Thus,

$\left[\frac{b^2}{a}\right]=\frac{L^6}{ML^5T^{-2}}=M^{-1}LT^2$

Compare with given physical quantities

• Energy density

  $\frac{\text{energy}}{\text{volume}}=\frac{ML^2T^{-2}}{L^3}=ML^{-1}T^{-2}$

• Bulk modulus

  Same as pressure:

  $ML^{-1}T^{-2}$

• Modulus of rigidity

  Also same as pressure:

  $ML^{-1}T^{-2}$

• Compressibility

  Compressibility $= \dfrac{1}{\text{pressure}}$

  $[\text{compressibility}] = \left(ML^{-1}T^{-2}\right)^{-1}=M^{-1}LT^2$

Match

$\left[\frac{b^2}{a}\right]=M^{-1}LT^2$

This matches compressibility.

Final answer

The correct option is:

$\boxed{\text{D: Compressibility}}$