Question

The velocity of water waves may depend on 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 (where $k$ is dimensionless constant):

• A. $v^2=k{\lambda }^{-1}{g}^{-1}{\rho }^{-1}$
• B. $v^2=kg\lambda$
• C. $v^2=kg\lambda \rho$
• D. $v^2=k{\lambda }^3{g}^{-1}{\rho }^{-1}$

Answer 1

Answer: (B)

$v^2=kg\lambda$

Here, $v\propto {\lambda }^a{\rho }^b{g}^c$ or $v=K{\lambda }^a{\rho }^b{g}^c$ where $K=$ proportionality constant

Using dimension analysis, $\left[L{T}^{-1}\right]=\left[L^a\right]\left[M^b{L}^{-3b}\right]\left[L^c{T}^{-2c}\right]$

or $\left[M^{0}L{T}^{-1}\right]=\left[M^b{L}^{a-3b+c}{T}^{-2c}\right]$

Equating power, $b=0$;

$1=a-3b+c$ and $-1=-2c$

So, $c=1/2$ and $a=1-c=1-1/2=1/2$

Thus, $v=K{\lambda }^{1/2}{\rho }^0{g}^{1/2}$

or $v^2=kg\lambda$ where $k=K^2$