Question

The velocity of water waves $v$ may depend upon their wavelength$\lambda$, the density of water $\rho$ and the acceleration due to gravity $g$. The method of dimensions gives the relation between these quantities as

• A. $v^{2}rg$
• B. $v^{2} \propto g\lambda\rho$
• C. $v^{2} \propto g\lambda$
• D. $v^{2} \propto g^{- 1}\lambda^{- 3}$

Answer 1

Answer: (C)

$v^{2} \propto g\lambda$

Answer 2

Let $v^{x} = kg^{y}\lambda^{z}\rho^{\delta}.$ Now by substituting the dimensions of each quantities and equating the powers of M, L and T we get $\delta = 0$ and $x = 2,y = 1,z = 1$ .