Question

If mass \(M\), pressure \(P\) and velocity \(V\) are taken as fundamental quantities, then the dimensional formula of surface tension is

• A. \(\left[M^{\frac23}P^{\frac23}V^{\frac13}\right]\)
• B. \(\left[M^{\frac13}P^{\frac13}V^{\frac23}\right]\)
• C. \(\left[M^{\frac13}P^{\frac23}V^{\frac23}\right]\)
• D. \(\left[M^{\frac23}P^{\frac13}V^{\frac23}\right]\)

Hint:

When the fundamental quantities are changed, first write the dimensions of the new base quantities in terms of \(M,L,T\), assume unknown powers, and compare exponents.

Answer 1

The Correct Option is C

Step 1: Write the dimensional formula of surface tension. Surface tension is force per unit length. \[ [\text{Surface Tension}] = \frac{[MLT^{-2}]}{[L]} = [MT^{-2}]. \]

Step 2: Express it in terms of \(M\), \(P\) and \(V\). Let \[ [\text{Surface Tension}] = [M]^a[P]^b[V]^c. \] Now, \[ [P]=ML^{-1}T^{-2}, \] \[ [V]=LT^{-1}. \] Hence, \[ [M]^a[P]^b[V]^c = M^{a+b}L^{-b+c}T^{-2b-c}. \] Comparing with \[ [MT^{-2}], \] we obtain \[ a+b=1, \] \[ -b+c=0, \] \[ -2b-c=-2. \]

Step 3: Solve for the exponents. From \[ -b+c=0, \] \[ c=b. \] Substituting into \[ -2b-c=-2, \] gives \[ -3b=-2, \] \[ b=\frac23. \] Hence, \[ c=\frac23, \] and \[ a=1-\frac23=\frac13. \] Therefore, \[ \boxed{ [\text{Surface Tension}] = \left[M^{\frac13}P^{\frac23}V^{\frac23}\right]. } \] Thus, the correct option is \(\boxed{(C)}\).