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

Options

A. 10000
B. 10
C. 100
D. 1000

Answer

C

Explanation

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).