A) \[Dyne/c{{m}^{2}}\]
B) \[I=\frac{Q}{t}=\frac{[Q]}{[T]}=[{{M}^{0}}{{L}^{0}}{{T}^{-1}}Q]\]
C) \[Dyne/cm\]
D) \[Newton/{{m}^{2}}\]
Correct Answer: B
Solution :
Surface tension = \[\frac{\text{Force}}{\text{Length}}\] = \[Newtons/metre\]
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