Question

The velocity of water waves (v) may depend on their wavelength ${}^{\prime }\lambda {}^{\prime }$, density of water 'p' and acceleration due to gravity 'g'. The method of dimensions gives the relation between these quantities as

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

Answer 1

Answer: (C)

$v^2\propto g\lambda$

Dimension of Velocity is $L{T}^{-1}$
Dimension of wavelength $\lambda$ is $L$
Dimension of acceleration due to gravity $g$ is $L{T}^{-2}$
Dimension of Density of water $\rho$ is $M{L}^{-3}$
Let $v\propto {\lambda }^a{g}^b{\rho }^c$

Using Dimensional method:

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

$\left[L{T}^{-1}\right]=\left[L^{a+b-3c}{M}^{-3c}{T}^{-2b}\right]$
Comparing exponents of similar dimensions on both the sides,
$a+b-3c=1$ ( Length )
$-3c=0=>c=0$ ( Mass )
$-2b=-1=>b=0.5$ ( Time )
Hence, $a=0.5$
$\therefore v\propto \sqrt{g\lambda }$
$v^2\propto g\lambda$.