1. Given equation

   $\left[X+\frac{a}{Y^{2}}\right][Y-b]=RT$

   where:

   • $X$ = pressure
   • $Y$ = volume
   • $R$ = universal gas constant
   • $T$ = temperature

2. Find dimensions of $a$

   In the bracket $X + \frac{a}{Y^2}$ both terms must have the same dimensions.

   Since $X$ is pressure, $\left[\frac{a}{Y^2}\right] = [\text{pressure}]$

   Therefore, $[a] = [\text{pressure}]\,[Y^2]$

   Now, $[\text{pressure}] = M L^{-1} T^{-2}$ $[Y] = [\text{volume}] = L^3$

   So, $[a] = (M L^{-1} T^{-2})(L^3)^2 = (M L^{-1} T^{-2})(L^6) = M L^5 T^{-2}$

3. Find dimensions of $b$

   In the factor $Y-b$ both terms must have the same dimensions, so $[b] = [Y] = L^3$

4. Find dimensions of $\dfrac{a}{b}$

   $\left[\frac{a}{b}\right] = \frac{M L^5 T^{-2}}{L^3} = M L^2 T^{-2}$

5. Identify the physical quantity

   $M L^2 T^{-2}$ is the dimension of work/energy.

6. Check options

   • A: Impulse $= MLT^{-1}$ ❌
   • B: Energy $= ML^2T^{-2}$ ✅
   • C: Pressure gradient $= \dfrac{ML^{-1}T^{-2}}{L} = ML^{-2}T^{-2}$ ❌
   • D: Coefficient of viscosity $= ML^{-1}T^{-1}$ ❌

Therefore, the correct option is B: Energy.