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# For a sound source of intensity $I = (W/{m^2})$, corresponding sound level is ${B_o}$ decibel. If the intensity is increased to 4I, a new sound level becomes approximately?1) $2{B_o}dB$2) $({B_o} + 3)dB$3) $({B_o} + 6)dB$4) $4{B_o}dB$

Last updated date: 23rd May 2024
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Hint: Generally one might think that the sound intensity and loudness are the same but they are very different from each other. The sound intensity (I) is defined as power (p) upon Area (A) while the loudness is the ratio of intensity of a given sound upon the intensity at the threshold of hearing. Human ear can only hear sounds in a particular range of intensities. The threshold of human hearing has an intensity of about 120dB. Now, Define the formula for sound level and then establish a proper relation between the current sound level and the final sound level while applying logarithmic properties, Use: (log (a.b) = loga+logb).

Formula Used:
Here we have used the formula
${B_o} = 10\log (\dfrac{I}{{{I_o}}})$
Where
${B_o}$= Sound level
I = Intensity of sound
${I_o}$= ${I_o} = {10^{ - 12}}W/{m^2}$
Given:
I=4I
Find:
Sound level (B) = ??

Find out the intensity and solve for the unknown
Here we know the formula of sound wave which is
${B_o} = 10\log (\dfrac{I}{{{I_o}}})$
Here ${I_o}$ is given by
${I_o} = {10^{ - 12}}W/{m^2}$
Now we have been given
I=4I
So now, it becomes
$B = 10\log (\dfrac{{4I}}{{{I_o}}})$
Apply trigonometric property (log (a.b) =loga+logb)
B =10(log$\dfrac{I}{{{I_o}}}$) +10log4
Since${B_o} = 10\log (\dfrac{I}{{{I_o}}})$, So
B = ${B_o}$+$10 \times 0.60$
B = (${B_o}$+6) dB

$\therefore$The new sound level becomes option (3) approximately B = (${B_o}$+6) dB.