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A person speaking normally produces a sound intensity of 40 \mathrm{~dB} at a distance of 1 \mathrm{~m}. If the threshold intensity for reasonable audibility is \mathrm{20 \mathrm{~dB}}, the maximum distance at which person can be heard clearly is -
 

Option: 1

4 \mathrm{~m}




 


Option: 2

5 \mathrm{~m}


Option: 3

10 \mathrm{~m}


Option: 4

20 \mathrm{~m}


Answers (1)

best_answer

Sound level (\operatorname{in} \mathrm{dB})\mathrm{=10 \log 10\left(\frac{I}{I_0}\right)}, where

                                    \mathrm{I_0=10^{-12} \mathrm{w} / \mathrm{m}^2}

Since, \mathrm{40=10 \log 10\left(\frac{I_1}{I_0}\right) \Rightarrow \frac{I_1}{I_0}=10^{4}}_________(1)
Also, \mathrm{20=10 \log 10\left(\frac{I_2}{I_0}\right) \Rightarrow \frac{I_2}{I_0}=10^2}____________(2)

From equation (i) and (ii) we get -

\mathrm{ \frac{I_2}{I_1}=10^{-2}=\frac{\gamma_1^2}{\gamma_2^2} }

\mathrm{ \gamma_2^2=100 \gamma_1^2 }

\mathrm{ \gamma_2=10 \gamma_1 \Rightarrow \gamma_2=10 \mathrm{~m} \quad\left(\because \gamma_1=1 . \mathrm{m}\right)}

Hence option 3 is correct.





 

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