In expressing sound intensity, we take $10^{-12} \mathrm{W} / \mathrm{m}^{2}$ as the reference level. For ordinary conversation, the intensity level is about $10^{-6} \mathrm{W} / \mathrm{m}^{2}$. Expressed in decibel, this is
A. \[{{10}^{6}}\]
B \[6\]
C.$60$
D.$\log _{n}\left(10^{5}\right)$
Answer
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Hint: In decibels (dB), the loudness of the sound is measured. Actually, this is a measure of intensity, which relates to how much energy the wave of pressure has. A relative measurement is decibels. They relate the pressure wave intensity to a normal or standard pressure wave.The sound waves' amplitude and how far they have travelled from the sound source. Calculate Intensity of the sound I and Io and then put it in the formula of the Loudness.
Formula used:
IL $=10 \log _{10}\left[\dfrac{\mathrm{I}}{\mathrm{Io}}\right]$
Complete solution Step-by-Step:
Sound intensity, also known as acoustic intensity, is defined in a direction perpendicular to that area as the power carried by sound waves per unit area. The SI intensity unit, which includes the intensity of sound, is the watt per square meter.
We know that:
IL $=10 \log _{10}\left[\dfrac{\mathrm{I}}{\mathrm{Io}}\right]$
where, $\mathrm{I}=10^{-6}$ and $\mathrm{I}_{0}=10^{-12}$
Substituting the values,
$\therefore \mathrm{IL}=10 \log _{10}\left[\dfrac{10^{-6}}{10^{-12}}\right]$
$=10 \log _{10}\left[10^{6}\right]$
$=6 \times 10 \log _{10} 10$
$\therefore \text{dB}=60\text{dB}$
$60\text{dB}$ Expressed in decibel.
Hence, the correct option is (C).
Note:
In Watts, the intensity of a sound is the power of the sound divided by the area covered by the sound in square meters. A sound's loudness relates the intensity of any given sound to the intensity at the hearing threshold. It is measured in (dB) decibels. A sound's pitch depends on the frequency, while a sound's loudness depends on the sound waves' amplitude. The amount of energy transmitted (as by acoustic or electromagnetic radiation); "the intensity of the sound was adjusted"; "they measured the signal strength of the station" intensity, strength.
Formula used:
IL $=10 \log _{10}\left[\dfrac{\mathrm{I}}{\mathrm{Io}}\right]$
Complete solution Step-by-Step:
Sound intensity, also known as acoustic intensity, is defined in a direction perpendicular to that area as the power carried by sound waves per unit area. The SI intensity unit, which includes the intensity of sound, is the watt per square meter.
We know that:
IL $=10 \log _{10}\left[\dfrac{\mathrm{I}}{\mathrm{Io}}\right]$
where, $\mathrm{I}=10^{-6}$ and $\mathrm{I}_{0}=10^{-12}$
Substituting the values,
$\therefore \mathrm{IL}=10 \log _{10}\left[\dfrac{10^{-6}}{10^{-12}}\right]$
$=10 \log _{10}\left[10^{6}\right]$
$=6 \times 10 \log _{10} 10$
$\therefore \text{dB}=60\text{dB}$
$60\text{dB}$ Expressed in decibel.
Hence, the correct option is (C).
Note:
In Watts, the intensity of a sound is the power of the sound divided by the area covered by the sound in square meters. A sound's loudness relates the intensity of any given sound to the intensity at the hearing threshold. It is measured in (dB) decibels. A sound's pitch depends on the frequency, while a sound's loudness depends on the sound waves' amplitude. The amount of energy transmitted (as by acoustic or electromagnetic radiation); "the intensity of the sound was adjusted"; "they measured the signal strength of the station" intensity, strength.
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