Question

The velocity of water waves $\left(v\right)$ 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 is

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

Answer 1

Answer: (B)

$v^2\propto g\lambda$

Let $v=k{\lambda }^a{\rho }^b{g}^c$
So, $\left[M^{0}L{T}^{-1}\right]=L^a\left(M{L}^{-3}{)}^b\right(L{T}^{-2}{)}^c$$=M^b{L}^{a-3b+c}{T}^{-2c}$
Applying principle of homogeneity,

$b=0,a-3b+c=1$ and $-2c=-1$
On solving, we get $c=1/2$

$\therefore a=\frac{1}{2}$
$\therefore v\propto {\lambda }^{1/2}{\rho }^0{g}^{1/2}$ or $v^2\propto \lambda g$