Question

What is the sound level for a noise that has an intensity of 9.9 × 10-5 watts/m2?

Answer 1

Given: Sound Intensity ( I ) = $9.9 \times 10^{-5}\frac{watts}{m^2}$

We know,  the approximate threshold of human hearing is at 1kHz. In watts/m^2 it's value is $=\frac{9.9 \times 10^{-5}}{10^{-12}}$⁻¹² W/m².

So, we can say,  Reference sound intensity$(I_0)$ = 10⁻¹² W/m².

We have formula for "sound intensity level LI in dB" when entering sound intensity.

LI = 10×log (I / Io)  in dB

Plugging values of I and Io in formula.

LI = 10 × log ($\frac{9.9 \times 10^{-5}}{10^{-12}}$)

LI = 10 × log (99000000)

= 10 × 7.99564

LI = 79.96 dB.

Therefore,  79.96 dB is the sound level for a noise that has an intensity of 9.9 × 10-5 watts/m2.