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: Understanding the Concept:
We express the target quantity (Surface Tension) as a product of the new fundamental quantities raised to unknown powers: \( S = k E^a V^b T^c \).
Step 2: Key Formula or Approach:
1. Surface Tension (\(S\)) = Force/Length = \( [MT^{-2}] \).
2. Energy (\(E\)) = \( [ML^2T^{-2}] \).
3. Velocity (\(V\)) = \( [LT^{-1}] \).
Step 3: Detailed Explanation:
\[ [MT^{-2}] = [ML^2T^{-2}]^a [LT^{-1}]^b [T]^c \] \[ M^1L^0T^{-2} = M^a L^{2a+b} T^{-2a-b+c} \] Equating powers: For M: \( a = 1 \). For L: \( 2a + b = 0 \implies 2(1) + b = 0 \implies b = -2 \). For T: \( -2a - b + c = -2 \implies -2(1) - (-2) + c = -2 \implies 0 + c = -2 \implies c = -2 \). Substituting values: \( S = E^1 V^{-2} T^{-2} \).
Step 4: Final Answer:
The dimensional representation is \( E^1V^{-2}T^{-2} \).