Question

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

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{align}
  & {{v}_{m}}=2m{{s}^{-1}} \\
 & {{v}_{s}}=10m{{s}^{-1}} \\
\end{align} $
Now putting the values in the above formula, we get
 $ \begin{align}
  & \nu '=\nu \left[ 1445+2-0+2-10 \right] \\
 & \Rightarrow \nu '=\nu \times 1.007 \\
\end{align} $
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.