1.

The minimum intensity of audibility of sound is 10^(-2) watt/m^(2) intensity of sound is 10^(-9) watt/m^(2) the intensity level of this sound in decibels is :

Answer»

1000
100
30
300

Solution :`I_(0) = 10^(-12) "watt"//m^(2) "" I = 10^(-9)" watt"//m^(2)` In decibels.
L = 10 `log_(10)"" (I)/(I_(0)) = 10 log_(10)" "(10^(-9))/(10^(-12)) = 10 XX 3 `
L = 30 dB.


Discussion

No Comment Found

Related InterviewSolutions