Match List-I with List-II, where List-II gives the dimensional...

Let us express the standard fundamental quantities (Mass, Length, Time) in terms of Energy ($E$), Momentum ($P$), and Force ($F$).

The standard dimensional formulas are:
$[E] = [ML^2 T^{-2}]$
$[P] = [MLT^{-1}]$
$[F] = [MLT^{-2}]$

To find Length ($L$):
$\dfrac{[E]}{[F]} = \dfrac{[ML^2 T^{-2}]}{[MLT^{-2}]} = [L]$
So, $[L] = [E F^{-1}]$. This matches (C) with (I).

To find Time ($T$):
$\dfrac{[P]}{[F]} = \dfrac{[MLT^{-1}]}{[MLT^{-2}]} = [T]$
So, $[T] = [P F^{-1}]$. This matches (B) with (IV).

To find Mass ($M$):
Since kinetic energy $E = \dfrac{P^2}{2M}$, we have $M \propto \dfrac{P^2}{E}$.
Dimensionally, $\dfrac{[P]^2}{[E]} = \dfrac{[M^2 L^2 T^{-2}]}{[ML^2 T^{-2}]} = [M]$.
So, $[M] = [P^2 E^{-1}]$. This matches (A) with (II).

To find Surface Tension:
Surface Tension = $\dfrac{\text{Force}}{\text{Length}} = \dfrac{[F]}{[L]}$
Substituting $[L] = [E F^{-1}]$:
$[\text{Surface Tension}] = \dfrac{[F]}{[E F^{-1}]} = [E^{-1} F^2]$. This matches (D) with (III).

Thus, the correct matching is (A)-(II), (B)-(IV), (C)-(I), (D)-(III).

Answer: (A)-(II), (B)-(IV), (C)-(I), (D)-(III)