Question

If velocity $V$, energy $E$ and time $T$ are chosen as fundamental quantities then dimensional representation of surface tension in this system will be

• A. $E^{1}V^{-2}T^{-2}$
• B. $E^{1}V^{-1}T^{-2}$
• C. $E^{-2}V^{-1}T^{-3}$
• D. $E^{1}V^{-2}T^{-1}$

Hint:

Surface tension can also be thought of as Energy per unit Area ($E/L^2$). Since $V=L/T$, then $L^2 = V^2T^2$, making $S = E/(V^2T^2) = EV^{-2}T^{-2}$.

Answer 1

The Correct Option is A

Step 1: Concept
Express surface tension ($S$) as $S \propto E^{a}V^{b}T^{c}$ and equate the dimensions of $M, L, T$ on both sides.

Step 2: Meaning
Surface tension is Force/Length, so $[S] = [MT^{-2}]$. Fundamental units are $V = [LT^{-1}]$, $E = [ML^{2}T^{-2}]$, and $T = [T]$.

Step 3: Analysis $[MT^{-2}] = [ML^{2}T^{-2}]^{a} [LT^{-1}]^{b} [T]^{c} = M^{a} L^{2a+b} T^{-2a-b+c}$. Equating powers: $a=1$; $2a+b=0 \implies b=-2$; $-2a-b+c = -2 \implies -2(1)-(-2)+c = -2 \implies c=-2$.

Step 4: Conclusion
Substituting the values gives $E^{1}V^{-2}T^{-2}$.
Final Answer: (A)