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Question

The sound level at a point 5.0 m away from a point source is 40 dB. What will be the level at a point 50 m away from the source?


Solution

Let βA be the sound level at a point 5 m (= r1) away from the point source and βB be the sound level at a distance of 50 m (= r2) away from the point source.
∴​ βA = 40 dB
Sound level is given by:
β=10log10II0
According to the question,
 βA=10 log10 IAI0. IAI0=10βA10   .....1βB=10 log10IBIo IBI0=10βB10   .....2From 1 and 2, we get:     IAIB=10βA-βB10   ....3Also, IAIB=rB2rA2=5052 = 102   .....4From 3 and 4, we get:102=10βA-βB10 βA-βB10=2  βA-βB=20 βB=40-20=20 dB

Thus, the sound level of a point 50 m away from the point source is 20 dB.

Physics
HC Verma - I
Standard XI

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