Question

If the velocity of water wave $v$ depends upon the wavelength $\lambda$, the density of water $\rho$ and the acceleration due to gravity $g$, then which of the following options depicts correct relation dimensionally?

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

Answer 1

The correct option is C $v^2\propto g\lambda$

Dimensional formula of velocity, $v$
$\left[v\right]=\left[M^0{L}^1{T}^{-1}\right]$
Wavelength, $\lambda$
$\left[\lambda \right]=\left[M^0{L}^1{T}^0\right]$
Density, $\rho$
$\left[\rho \right]=\left[M^1{L}^{-3}{T}^0\right]$
Acceleration due to gravity, $g$
$\left[g\right]=\left[M^0{L}^1{T}^{-2}\right]$
Let us suppose the relation is
$v^2\propto {\lambda }^a{g}^b{\rho }^c$
$\Rightarrow \left[v^2\right]=\left[{\lambda }^a{g}^b{\rho }^c\right]$
$\Rightarrow \left[\right(M^0{L}^1{T}^{-1}{)}^2]=[M^0{L}^1{T}^0{]}^a\times [M^0{L}^1{T}^{-2}{]}^b\times [M^1{L}^{-3}{T}^0{]}^c$
$\Rightarrow \left[M^0{L}^2{T}^{-2}\right]=\left[M^{\left(c\right)}{L}^{(a+b-3c)}{T}^{(-2b)}\right]$
On comparing, we get,
$\begin{array}{}c=0& \text{(1)}\end{array}$
$\begin{array}{}a+b-3c=2& \text{(2)}\end{array}$
$\begin{array}{}-2b=-2& \text{(3)}\end{array}$
On solving these equations, we get,
$a=1,b=1$ and $c=0$
Thus,
$v^2\propto {\lambda }^a{g}^b{\rho }^c\Rightarrow v^2\propto g\lambda$