If energy (E), velocity [v] and time (T) are chosen as the fundamental quantities, the dimensional formula of surface tension will be

[E$v^{−2}T^{−1}$] [E$v^{−1}T^{−2}$] [E$v^{−2}T^{−2}$] [$E^{−2}v^{−1}T^{−3}$]

[E$v^{−2}T^{−2}$]

We know that Surface tension (S)=$\frac{Force[F]}{Lenght[L]}$ So,       [S] = $\frac{[MLT^{-2}]}{L}$ Energy (E) = Force *displacement          [E] = [M$LT^{-2}$][L]} = [M$L^{2}T^{-2}$] Velocity [V] = [L$T^{-1}$]          Since S ∝ $E^av^bT^c$ where, a, b, c are constants. From the principle of homogeneity,               Dimension of  [LHS] = Dimension of [RHS]                    $[ML^0T^{-2}]$ =  $[ML^{2}T^{-2}]^a[LT^{-1}]^b [T]^c$         =[$M^aL^{2a+b}T^{-2a-b+c}$] Equating the power on both sides, we get                           a=1 , 2a+b=0 ⇒ b=-2                             -2a-b+c= -2 ⇒ c = -2     So [S] = [E$V^{-2}T^{-2}$]