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Question: A sound of intensity I is greater by 3.0103 dB from another sound of intensity \[10{\text{ }}nW\,c{m...

A sound of intensity I is greater by 3.0103 dB from another sound of intensity 10 nW cm−2.10{\text{ }}nW\,c{m^{ - 2}}. The absolute value of the intensity of sound level I in Wm−2is{\text{W}}{{\text{m}}^{ - 2}} is

A. 2.5×10−4 B. 2×10−4 C. 2.0×10−2 D. 2.5×10−2  A.\,2.5 \times {10^{ - 4}} \\\ B.\,2 \times {10^{ - 4}} \\\ C.\,2.0 \times {10^{ - 2}} \\\ D.\,2.5 \times {10^{ - 2}} \\\
Explanation

Solution

Loudness of sound with intensity is given by 10log⁡I10−1210\log \dfrac{I}{{{{10}^{ - 12}}}}

Formula used:
B=10log⁡I10−12B = 10\log \dfrac{I}{{{{10}^{ - 12}}}}
Where,

{B = {\text{ }}loudness{\text{ }}of{\text{ }}sound} \\\ {I = {\text{ }}Intensity{\text{ }}of{\text{ }}sound} \end{array}$$ **Complete step by step solution:** Given data, Intensity of another sound

{I_2} = {10_n},W,c{m^{ - 2}} \\
= 10 \times {10^{ - 9}}\dfrac{W}{{{{10}^{ - 4}}{m^2}}} \\
= {10^{ - 4}},10{m^{ - 2}} \\
\\

As we know, $B = 10\log \dfrac{I}{{{{10}^{ - 12}}}}$ If loudness of sound having intensity IEB, ${B_1} = 10\log \dfrac{I}{{{{10}^{ - 12}}}}$ And if soundness of another sound $ = {B_2}$ Then $ {B_2} = 10\log \dfrac{{{I_2}}}{{{{10}^{ - 12}}}} \\\ {B_2} = 10\log \dfrac{{{{10}^{ - 4}}}}{{{{10}^{ - 12}}}} \\\ {B_2} = 10\log {10^8} = 80 \\\ $ Since B, is greater by B2 by 3.0103 dB then,

\Rightarrow {B_1} = 3.0103 + {B_2} \\
\Rightarrow 10\log \dfrac{I}{{{{10}^{ - 12}}}} = 3.0103 + 80 \\
\Rightarrow \log \dfrac{I}{{{{10}^{ - 12}}}} = 8.30103 \\
\Rightarrow \dfrac{I}{{{{10}^{ - 12}}}} = {10^{8.30103}} \\
\Rightarrow I = {10^{ - 12}} \times {10^{8.30103}} \\
\Rightarrow I = 2 \times {10^{ - 4}}\dfrac{W}{{{m^2}}} \\

∗∗Hence,(B)iscorrectoption.∗∗∗∗AdditionalInformation:∗∗Amplitudeisthesizeofthevibrationandthisdetermineshowloudthesoundis.Soundintensity,alsoknownasacousticintensity,isdefinedasthepowercarriedbysoundwavesperunitareainadirectionperpendiculartothatarea.Sinceaudiblesoundconsistsofpressurewaves,oneofthewaystoquantifythesoundistostatetheamountofpressurevariationrelativetoatmosphericpressurecausedbythesound.Soundlevelmeasurementsindecibelsaregenerallyreferencedtothestandardthresholdofhearingat1000Hzforthehumanearwhichcanbestatedintermsofsoundintensity.DecibelsmeasuretheratioofagivenintensityItothethresholdofhearingintensitysothatthethresholdtakesthevalue0decibels.Toassesssoundloudness,asdistinctfromanobjectiveintensitymeasurement,thesensitivityoftheearmustbefactoredin.∗∗Note:∗∗Beforeapplicationoftheaboveformulaofloudness,theintensityofsecondshouldbeinaunitofW/m2. **Hence, (B) is correct option.** **Additional Information:** Amplitude is the size of the vibration and this determines how loud the sound is. Sound intensity, also known as acoustic intensity, is defined as the power carried by sound waves per unit area in a direction perpendicular to that area. Since audible sound consists of pressure waves, one of the ways to quantify the sound is to state the amount of pressure variation relative to atmospheric pressure caused by the sound. Sound level measurements in decibels are generally referenced to the standard threshold of hearing at 1000 Hz for the human ear which can be stated in terms of sound intensity. Decibels measure the ratio of a given intensity I to the threshold of hearing intensity so that the threshold takes the value 0 decibels. To assess sound loudness, as distinct from an objective intensity measurement, the sensitivity of the ear must be factored in. **Note:** Before application of the above formula of loudness, the intensity of second should be in a unit of W/m2.