Free Study Material

NCERT Solutions

NCERT Books

Revision Notes

Practice Problems

This is second of our many quizzes on Sound intensity. Click on this to access more such competitive quizzes on SHM Waves and Optics.

Sound intensity Quiz 2

physics

Gupta Ji owns a cloth manufacturing company and earns a lot of profit, one day his manager suggested him about the modern machines for cloth manufacturing, but they produce a sound level of $\text{80 }dB$ each. The factory is in the commercial area so he wants to minimize the total sound level produced by $50$ such machines to avoid disturbances in nearby areas. The total sound level from all machines will be – 

Sound level will increase due to combined intensities of all the machines. 

 $\text{Sound intensity received from one machine-}\\  S=10\log \left( \dfrac{I'}{I} \right)\\  \Rightarrow 80=10\log \left( \dfrac{I'}{{{10}^{-12}}} \right)\\  \Rightarrow I'={{10}^{-4}}W/{{m}^{2}}\\  \text{Sound intensity received from 50 such machines will be-}\\  I=50 \times I'\\  \Rightarrow I=50 \times {{10}^{-4}}\,W/{{m}^{2}}\\  \Rightarrow S=10\log \left( \dfrac{I'}{I} \right)\\  \Rightarrow S =10\log \left( \dfrac{50\times {{10}^{-4}}}{{{10}^{-12}}} \right)\\  \Rightarrow S =10\log (50)-10\log ({{10}^{-8}})\\ \therefore S =17+80=97dB$

Mohit went to a huge party where he found a new branded music system, as we know he is very fond of music systems and love to try new things so he went $4\text{ }m$ away from the speaker and recorded an intensity $4.4\times {{10}^{-5}}W/{{m}^{2}}$, after that he further travels $4\text{ }m$ away and recorded an intensity. The value of new intensity will be- 

 Intensity is inversely proportional to the distance from the source to observer or we can say intensity will decrease by increasing distance between source and observer. 

  ${{I}_{1}}{{r}_{1}}^{2}={{I}_{2}}{{r}_{2}}^{2}\\  \Rightarrow 4.4\times {{10}^{-5}}\times {{4}^{2}}={{I}_{2}}\times {{8}^{2}}\\  \therefore {{I}_{2}}=1.1\times {{10}^{-5}}\,W/{{m}^{2}}$

Three friends $A$, $B$ and $C$ started walking from a house in three different directions North, south and east with a speed of $5\text{ }m{{s}^{-1}}$. A music system of average power output “P” is placed in the house and produces sound. The relation between intensity received by three friends will be- 

As the distance between source and friends increases, the intensity received by them will decrease. 

 Three friends starting from the same point with the same speed will be at equal distances from source so at any given point of time the intensity received by friends will be equal.

 In a laboratory two sound systems are present at a distance of $10\text{ }m$ from each other, and now a detector is placed in the middle of them. The intensity detected from each system is $5\times {{10}^{-5}}\,W/{{m}^{2}}$. Now the sound systems are moved away by some equal distance and intensity received by the detector from each system is $0.5\times {{10}^{-5}}W/{{m}^{2}}$. The percentage decrease in sound level detected by detector will be-<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/c8008d0f-84f8-4a3a-a13a-e8892794fb5d2958314483523949572.png"/>

Sound level is decreasing as the distance is increasing between source and detector. 

 $S=10\log \left( \dfrac{I'}{I} \right)\\  \text{In first case the total intensity received by detector will be-}\\ I'=5\times {{10}^{-5}}+5\times {{10}^{-5}}={{10}^{-4}}\,W/{{m}^{2}}\\  S=10\log \left( \dfrac{{{10}^{-4}}}{{{10}^{-12}}} \right)=80\text{ }dB\\  \text{In second case total intensity received by detector will be-}\\  I''=0.5\times {{10}^{-5}}+0.5\times {{10}^{-5}}\\  \Rightarrow I'' ={{10}^{-5}}W/{{m}^{2}}\\  \Rightarrow S=10\log \left( \dfrac{{{10}^{-5}}}{{{10}^{-12}}} \right)\\  \Rightarrow S =70\text{ }dB\\  \Rightarrow d=\dfrac{80-70}{80}\times 100\\  \therefore d=12.5\%$

Ram and Kishore went to a physics laboratory to perform experiments on sound intensities.First they have calculated the speed of sound as 350 m/s and density of air  $1.25\,Kg/{{m}^{3}}$. Now by different techniques they are calculating pressure amplitude from a source machine as $6.5\,N/{{m}^{2}}$. The intensity detected by their system from the source is equivalent to- 

Intensity is inversely proportional to the density of medium and velocity of sound in medium. 

 $  I=\dfrac{{{P}^{2}}}{2\rho v}\\  \Rightarrow I=\dfrac{{{(6.5)}^{2}}}{2\times 1.25\times 350}\\  \Rightarrow I =\dfrac{42.25}{2.5\times 350}\\  \therefore I =4.82\times {{10}^{-2}}\,W/{{m}^{2}}$