Identify the standard dimensions: Energy ($E$) = $[ML^2T^{-2}]$ Velocity ($v$) = $[LT^{-1}]$ Time ($T$) = $[T]$ Surface Tension ($S$) = Force / Length = $[MLT^{-2}] / [L] = [MT^{-2}]$

Set up the dimensional equation: Let Surface Tension $S \propto E^a v^b T^c$ $S = k E^a v^b T^c$ (where $k$ is a dimensionless constant) Substituting the dimensions into the equation: $[M^1L^0T^{-2}] = [ML^2T^{-2}]^a [LT^{-1}]^b [T]^c$ $[M^1L^0T^{-2}] = [M^a L^{2a+b} T^{-2a-b+c}]$

Equate the powers of $M$, $L$, and $T$: For $M$: $a = 1$ For $L$: $2a + b = 0$ For $T$: $-2a - b + c = -2$

Solve for $a$, $b$, and $c$: Since $a = 1$, substituting in the $L$ equation: $2(1) + b = 0 \Rightarrow b = -2$ Substituting $a$ and $b$ into the $T$ equation: $-2(1) - (-2) + c = -2 \Rightarrow -2 + 2 + c = -2 \Rightarrow c = -2$

Final Dimensional Formula: Substituting the values of $a$, $b$, and $c$ back into the assumed relation gives $S = [E^1 v^{-2} T^{-2}] = [E v^{-2} T^{-2}]$.