Question

What will be the frequency detected by a receiver kept in the river downstream?
$\begin{aligned} & \left( A \right)500696Hz \\\ & \left( B \right)100696Hz \\\ & \left( C \right)200696Hz \\\ & \left( D \right)300696Hz \\\ \end{aligned}$

Answer 1

Hint : In order to solve the question, we need to first calculate the velocity of sound in water and then the frequency received by detector can be calculated by using a formula for it where in the value of actual frequency of sound waves in water is required along with the various other velocity values.
For velocity of sound in water
${{v}_{w}}=\sqrt{\dfrac{B}{\rho }}$
Where, $B$ is coefficient of stiffness and its value is $2.088\times {{10}^{9}}$ and $\rho$ is density for water equal to ${{10}^{3}}$
For frequency received by detector
$\nu '=\nu \left[ {{v}_{w}}+{{v}_{m}}-\dfrac{{{v}_{0}}}{c}+{{v}_{m}}-{{v}_{s}} \right]$
Here, $\nu '$ is frequency detected
$\nu$ is actual frequency, ${{v}_{0}}$ is zero, ${{v}_{w}}$ is speed of sound in water.

Complete Step By Step Answer:
In order to solve the question, we must know the velocity of sound in water. The velocity of sound in water is given by the formula
${{v}_{w}}=\sqrt{\dfrac{B}{\rho }}$
Where, $B$ is coefficient of stiffness and its value is $2.088\times {{10}^{9}}$ and $\rho$ is density for water equal to ${{10}^{3}}$
${{v}_{w}}=\sqrt{\dfrac{2.088\times {{10}^{9}}}{{{10}^{3}}}}=1445m{{s}^{-1}}$
Therefore, sound travels in water with the speed $1445m{{s}^{-1}}$
We know that the frequency is given by the formula, i.e.,
$\nu '=\nu \left[ {{v}_{w}}+{{v}_{m}}-\dfrac{{{v}_{0}}}{c}+{{v}_{m}}-{{v}_{s}} \right]$
Here, $\nu '$ is frequency detected
$\nu$ is actual frequency, ${{v}_{0}}$ is zero, ${{v}_{w}}$ is speed of sound in water
$\begin{aligned} & {{v}_{m}}=2m{{s}^{-1}} \\\ & {{v}_{s}}=10m{{s}^{-1}} \\\ \end{aligned}$
Now putting the values in the above formula, we get
$\begin{aligned} & \nu '=\nu \left[ 1445+2-0+2-10 \right] \\\ & \Rightarrow \nu '=\nu \times 1.007 \\\ \end{aligned}$
The frequency $\nu$ can be found from the speed of sound in water and its wavelength
So,
$\nu =\dfrac{{{v}_{w}}}{\lambda }=\dfrac{1445}{14.45\times {{10}^{-3}}}={{10}^{5}}$
Therefore, the frequency detected by the receiver at the downstream is
$\nu '=\nu \times 1.007={{10}^{5}}\times 1.007=100700Hz$
Hence, if approximation is neglected, then, option $\left( B \right)100696Hz$ is correct.

Note :
It is very important to note that the sound is a special case for the velocities and hence all the values that are taken , are constants and hence to be memorized only, however, the velocity of sound has been calculated here by using the formula , but the value can also be memorized to make it simpler.