A small speaker delivers 2, W of audio output | vedclass

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(C) The loudness level in decibels $(dB)$ is given by the formula: $L = 10 \log_{10} \left( \frac{I}{I_0} \right)$,where $I$ is the intensity of sound

Vedclass · Accepted Answer

$40$

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(C) The loudness level in decibels $(dB)$ is given by the formula: $L = 10 \log_{10} \left( \frac{I}{I_0} ight)$,where $I$ is the intensity of sound and $I_0 = 10^{-12}\, W/m^2$ is the reference intensity.Given $L = 120\, dB$,we have: $120 = 10 \log_{10} \left( \frac{I}{10^{-12}} ight)$.Dividing by $10$,we get $12 = \log_{10} \left( \frac{I}{10^{-12}} ight)$,which implies $\frac{I}{10^{-12}} = 10^{12}$.Thus,$I = 10^{12} 	imes 10^{-12} = 1\, W/m^2$.The intensity $I$ at a distance $r$ from a point source is given by $I = \frac{P}{4 \pi r^2}$,where $P = 2\, W$ is the power.Substituting the values: $1 = \frac{2}{4 \pi r^2}$.Solving for $r^2$: $r^2 = \frac{2}{4 \pi} = \frac{1}{2 \pi} \approx \frac{1}{6.28} \approx 0.159\, m^2$.Taking the square root: $r = \sqrt{0.159} \approx 0.399\, m$.Converting to centimeters: $r \approx 0.399 	imes 100 = 39.9\, cm \approx 40\, cm$.

$A$ small speaker delivers $2\, W$ of audio output. At what distance from the speaker will one detect $120\, dB$ intensity sound in $cm$?

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