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

Concept: Surface tension is force per unit length: \[ S=\frac{F}{L} \] Its dimensional formula is: \[ [S]=MT^{-2} \]

Step 1: Write dimensions of energy and velocity. \[ [E]=ML^2T^{-2} \] \[ [V]=LT^{-1} \]

Step 2: Suppose surface tension is represented as: \[ [S]=E^aV^bT^c \] Substitute dimensions: \[ MT^{-2}=(ML^2T^{-2})^a(LT^{-1})^bT^c \] \[ MT^{-2}=M^aL^{2a+b}T^{-2a-b+c} \]

Step 3: Compare powers of \(M,L,T\). For \(M\): \[ a=1 \] For \(L\): \[ 2a+b=0 \] Put \(a=1\): \[ 2+b=0 \] \[ b=-2 \] For \(T\): \[ -2a-b+c=-2 \] Put \(a=1,\ b=-2\): \[ -2(1)-(-2)+c=-2 \] \[ -2+2+c=-2 \] \[ c=-2 \]

Step 4: Hence: \[ [S]=E^1V^{-2}T^{-2} \] Therefore, \[ \boxed{E^1V^{-2}T^{-2}} \]