1.

A sound absorber attenuates the sound level by 20 dB. The intensity decreases by a factor of :

Answer»

10000
10
100
1000

Solution :LET initial SOUND LOUD of sound wave intensity `I_(1)` be 100 dB.
` THEREFORE L = 10 log_(10)" " (I_(1))/(I_(0)) "" rArr "" 100 = 10 log_(10)" "(I_(1))/(I_(0))`
`rArr "" (I_(1))/(I_(0)) = 10^(10)"" `.... (i)
Since sound absorber attenuates the sound level by 20 dB.
`therefore` (100 -20) dB= 10 log `(I_(2))/(I_(0))`
`rArr "" 80= 10"log" (I_(2))/(I_(0)) rArr (I_(2))/(I_(0)) = 10^(8)` .... (ii)
Dividing (ii) by (i)
`(I_(2))/(I_(1)) = (10^(8))/(10^(10)) = (1)/(100) `
`rArr "" I_(2) = (I_(1))/(100) ` . so correct choice is (c).


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