Question

What will be the wavelength of sound waves having a frequency $262Hz$ in water? It has been already mentioned in the question that the bulk modulus of water is $\dfrac{{{10}^{11}}}{49}Pa$.
$\begin{aligned} & A.11m \\\ & B.22m \\\ & C.5.45m \\\ & D.2.75m \\\ \end{aligned}$

Answer 1

The frequency of a wave can be found by taking the ratio of the velocity of the wave to the wavelength of the wave. The velocity of the wave can be found by taking the square root of the bulk modulus of the water to the density of water. This will help you in answering this question.

Complete step by step solution:
The frequency of the sound wave has been mentioned to be as,
$\nu =262Hz$
As we all know, the frequency of a wave can be found by taking the ratio of the velocity of the wave to the wavelength of the wave. This can be expressed as an equation given as,
$\nu =\dfrac{v}{\lambda }$
Rearranging this equation in terms of velocity can be shown as,
$\lambda =\dfrac{v}{\nu }$
The velocity of the wave can be found by taking the square root of the bulk modulus of the water to the density of water. That is we can write that,
$v=\sqrt{\dfrac{B}{\rho }}$
Substituting this in the above obtained equation as,
$\lambda =\dfrac{\sqrt{\dfrac{B}{\rho }}}{\nu }$
It has been already mentioned in the question that the bulk modulus of the water be,
$B=\dfrac{{{10}^{11}}}{49}Pa$
The density of the water will be,
$\rho =1000kg{{m}^{-3}}$
Substituting these values in the equation will give,
$\lambda =\dfrac{\sqrt{\dfrac{{{10}^{11}}}{49\times 1000}}}{262}$
Simplifying the equation will give,
$\lambda =\dfrac{{{10}^{4}}}{7\times 262}=5.45m$
Therefore the wavelength of the wave mentioned in the question has been obtained.

So, the correct answer is “Option C”.

Note: The bulk modulus of a body can be defined as the measure of how much resistant the body will be to the compression. It can be defined as the ratio of the infinitesimal increase in the pressure to the net relative decrease of the volume.