Question

The velocity of water wave ($v$) 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

• 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

The correct option is 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 $c=\frac{1}{2}$
$\therefore a=\frac{1}{2}$
$\therefore v\propto {\lambda }^{\frac{1}{2}}{\rho }^0{g}^{\frac{1}{2}}$ or $v^2\propto \lambda g$