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. \(E^1V^{-2}T^{-2}\)
• B. \(E^1V^{-1}T^{-2}\)
• C. \(E^{-2}V^{-1}T^{-3}\)
• D. \(E^1V^{-2}T^{-1}\)

Hint:

In dimensional analysis with new fundamental quantities, assume the required quantity as \(E^aV^bT^c\), then compare powers.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
This is a problem of change of units where we express one quantity in terms of non-standard fundamental quantities.
We need to express Surface Tension (\(S\)) in terms of Velocity (\(V\)), Energy (\(E\)), and Time (\(T\)).
Step 2: Key Formula or Approach:
Surface Tension \(S = \frac{\text{Force}}{\text{Length}}\). Its dimensions are \([MT^{-2}]\).
Energy \(E = [ML^2T^{-2}]\).
Velocity \(V = [LT^{-1}]\).
Time \(T = [T]\).
Assume \(S = E^a V^b T^c\) and solve for \(a, b, c\) using dimensional analysis.
Step 3: Detailed Explanation:

Dimensional formulas in standard SI units:
\[ [S] = [MT^{-2}] \]
\[ [E] = [ML^2T^{-2}] \]
\[ [V] = [LT^{-1}] \]
\[ [T] = [T] \]

Equating dimensions:
\[ [MT^{-2}] = [ML^2T^{-2}]^a \cdot [LT^{-1}]^b \cdot [T]^c \]
\[ M^1 L^0 T^{-2} = M^a L^{2a+b} T^{-2a-b+c} \]

Comparing powers:
1) For \(M\): \(a = 1\)
2) For \(L\): \(2a + b = 0 \implies 2(1) + b = 0 \implies b = -2\)
3) For \(T\): \(-2a - b + c = -2\)
Substitute \(a=1, b=-2\):
\(-2(1) - (-2) + c = -2\)
\(-2 + 2 + c = -2 \implies c = -2\)

Thus, the dimensional formula is \(E^1 V^{-2} T^{-2}\).

Step 4: Final Answer:
By solving the system of equations for the powers of energy, velocity, and time, we found the relation to be \(E^1 V^{-2} T^{-2}\).