Answer:

Correct Answer: C

Solution:

We know that Surface tension $ \text{(S)=}\frac{\text{Force }[\text{ F }]}{\text{Lenght }[\text{ L }]} $

So, $ [S]=\frac{[MK{T^{-2}}]}{[L]}=[ML^{0}{T^{-2}}] $ Energy (E) = Force $ \times $ displacement

$ \Rightarrow ,[E]=[ML^{2}T^{2}] $ Velocity (v) $ \text{=}\frac{displacement}{time} $

$ \Rightarrow ,[v]=[L{T^{-1}}] $ A.s, $ S\propto E^{a}v^{b}T^{c} $ where, a, b, c are constants. From the principle of homogeneity, [LHS] = [RHS]

$ \Rightarrow ,[ML^{0}{T^{-2}}]={{[ML^{2}{T^{-2}}]}^{a}}{{[L{T^{-1}}]}^{b}}{{[T]}^{c}} $

$ \Rightarrow ,[ML^{0}{T^{-2}}]=[M^{a}{L^{2a+b}}{T^{-2a-b+c}}] $

Equating the power on both sides, we get $ a=1,2a+b=0,b=-2 $

$ \Rightarrow -2a-b+c=-2 $

$ \Rightarrow \ c=(2a+b)-2=0-2=-2 $

So $ [S],=[E{v^{-2}}{T^{-2}}],=[E{v^{-2}}{T^{-2}}] $