Intensity level 200 cm from a source of sound is 80 dB. If there is no loss of acoustic power in air and intensity of threshold hearing is 10−12 Wm−2 then, what is the intensity level at a distance of 4000 cm from source
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a
Zero
b
54 dB
c
64 dB
d
44 dB
answer is B.
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Detailed Solution
I∝1r2⇒I2I1=r12r22=22(40)2=1400⇒I1=400I2Intensity level at point 1, L1=10log10I1I0and intensity at point 2, L2=10log10I2I0∴L1−L2=10logI1I2=10log10(400)⇒L1−L2=10×2.602=26L2=L1−26=80−26=54 dB