Question

The velocity of water waves may depend on their wavelength $\lambda$ , the density of water $\rho$ and the acceleration due to gravity $g$. The method of dimensional analysis gives the relation between these quantities as:

• A. $V^2=K{\lambda }^{-1}{g}^{-1}{\rho }^{-1}$
• B. $V^2=K\lambda g$
• C. $V^2=K\lambda \rho g$
• D. $V^2=K{\lambda }^3{g}^{-1}{\rho }^{-1}$

Answer 1

Answer: (B)

$V^2=K\lambda g$

$V\alpha {\lambda }^a{\rho }^{-b}{g}^c$
$V=L{T}^{-1}$
$\lambda =L^1$
$\rho =M^1{L}^{-3}$
$g=L{T}^{-2}$
Equating LHS & RHS, we get, $b=0,a=\frac{1}{2},c=\frac{1}{2}$