Sound WavesEasy
Question
A person speaking normally produces a sound intensity of 40 dB at a distance of 1m. If the threshold intensity for reasonable audibility is20 dB, the max. distance at which he can be heard clearly is :-
Options
A.4 m
B.5m
C.10 m
D.20 m
Solution
40 = 10 log10 
= 104
I0 = 10-12 wm-2
20 = 10 log10
= 102
⇒ r2 = 10 m (∵ r1 = 1m)]
= 104I0 = 10-12 wm-2
20 = 10 log10
= 102 ⇒ r2 = 10 m (∵ r1 = 1m)]Create a free account to view solution
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