
A normal conversation involves sound intensities of about 3.0 x10-6 Wm-2
What is the decibel level for this intensity? What is the intensity of the sound for 100 dB?
A. 65
B. 70
C. 71
D. 24
Answer
587.1k+ views
Hint: The threshold intensity of hearing is ${I_o} = {10^{ - 12}}W/{m^2}$
Formula: $\beta (dB) = 10{\log _{10}}\left( {\dfrac{I}{{{I_o}}}} \right)$
Complete step by step solution:
Given,
Sound intensity $\left( I \right) = 30\times{10^{ - 6}}W/{m^2}$
The threshold intensity of hearing is ${I_o} = {10^{ - 12}}W/{m^2}$
The sound intensity level β in decibels of a sound having an intensity I in watts per meter squared is defined to be
$\beta (dB) = 10{\log _{10}}\left( {\dfrac{I}{{{I_o}}}} \right)$
∴Sound level=$\beta (dB) = 10{\log _{10}}\left( {\dfrac{{30\times{{10}^{ - 6}}}}{{{{10}^{ - 12}}}}} \right) = 10\left( {6.48} \right) = 64.8$
Thus, the decibel level = 65 dB (approx.)
Correct answer: A) 65
Note: The decibel (dB) is the preferred unit for measuring sound.
Formula: $\beta (dB) = 10{\log _{10}}\left( {\dfrac{I}{{{I_o}}}} \right)$
Complete step by step solution:
Given,
Sound intensity $\left( I \right) = 30\times{10^{ - 6}}W/{m^2}$
The threshold intensity of hearing is ${I_o} = {10^{ - 12}}W/{m^2}$
The sound intensity level β in decibels of a sound having an intensity I in watts per meter squared is defined to be
$\beta (dB) = 10{\log _{10}}\left( {\dfrac{I}{{{I_o}}}} \right)$
∴Sound level=$\beta (dB) = 10{\log _{10}}\left( {\dfrac{{30\times{{10}^{ - 6}}}}{{{{10}^{ - 12}}}}} \right) = 10\left( {6.48} \right) = 64.8$
Thus, the decibel level = 65 dB (approx.)
Correct answer: A) 65
Note: The decibel (dB) is the preferred unit for measuring sound.
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