Question

The velocity $v$ of waves produced in water depends on their wavelength $\lambda$, the density of water $\rho$, and acceleration due gravity $g$. The square of velocity is proportional to:-

• A. $K\sqrt{\left(g\lambda \right)}$
• B. ${\lambda }^{-1}{g}^{-1}{\rho }^{-1}$
• C. $\lambda \rho g$
• D. ${\lambda }^2{g}^{-2}{\rho }^{-1}$

Answer 1

The correct option is A $K\sqrt{\left(g\lambda \right)}$

Let$,$ $K\times {\lambda }^a{\rho }^b{g}^c$

$V=\left[L\right]\left[T^{-1}\right]$$

$\lambda =\left[L\right]$

$\rho =density=\left[M\right]{\left[L\right]}^{-3}$

$g=\left[L\right]{\left[T\right]}^{-2}$

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

Comparing the dimensions$:$

$b=0$ $,$ $c=1/2$ $,$ $a=1/2$

So$,$ $V=K\sqrt{\left(g\lambda \right)}$

Hence,

option $\left(A\right)$ is correct answer.