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: [where K is a 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

The correct option is B $v^2=Kg\lambda$

By dimension analysis:
$v^2=k{\lambda }^a{\rho }^b{g}^c$
$\Rightarrow [L{T}^{-1}{]}^2=L^a[M{L}^{-3}{]}^b[L{T}^{-2}{]}^c$
By solving it
$a=1,b=0andc=1$
$\therefore v^2=K\lambda g$