Question

The speed of propagation, \(c\), of a capillary wave depends on the density of the fluid, \(\rho\), the wavelength of the wave, \(\lambda\), and the surface tension, \(\sigma\). The dimensions are: \(\rho = ML^{-3}\), \(\lambda = L\), and \(\sigma = MT^{-2}\). Which expression is dimensionally correct?

• A. (c = \sqrt{\frac{\sigma}{\rho \lambda}}\)
• B. (c = \sqrt{\frac{\sigma}{\rho \lambda^2}}\)
• C. (c = \sqrt{\frac{\rho}{\sigma \lambda}}\)
• D. (c = \sqrt{\frac{\lambda}{\rho \sigma}}\)

Hint:

Dimensional analysis becomes easy when you separately balance the exponents of M, L, and T.

Answer 1

The Correct Option is C

Assume dependence: \[ c \propto \sigma^a \rho^b \lambda^c \] Dimensions: \[ [c] = LT^{-1},\quad [\sigma]=MT^{-2},\quad [\rho]=ML^{-3},\quad [\lambda]=L \] Writing dimensional equation: \[ LT^{-1} = (MT^{-2})^a (ML^{-3})^b (L)^c \] Matching M, L, T exponents:
Mass: \[ 0 = a + b \] Time: \[ -1 = -2a \Rightarrow a = \frac{1}{2} \] Then: \[ b = -\frac{1}{2} \] Length: \[ 1 = -3(-1/2) + c = \frac{3}{2} + c \Rightarrow c = -\frac{1}{2} \] Thus: \[ c \propto \sigma^{1/2} \rho^{-1/2} \lambda^{-1/2} \] \[ c = \sqrt{\frac{\sigma}{\rho \lambda}} \] Correct option is (A).