Question

By what factor is the intensity of sound at a rock concert louder than that of a whisper when the two intensity levels are 120 dB and 21 dB, respectively? c) 6.50x1010 a) 3.16x1010 d) 7.94x100 b) 1.58x1010 e) 3.98x109

Answer 1

Answer:

d) $7.94\times 10^{9}$

Explanation:

β₁ = sound level of sound at rock concert = 120 dB

β₂ = sound level of sound due to whisper = 21 dB

I₁ = Intensity of sound at rock concert

I₂ = Intensity of sound due to whisper

sound level of sound at rock concert is given as

$\beta _{1} = 10 log\left ( \frac{I_{1}}{10^{-12}} \right )$

$120 = 10 log\left ( \frac{I_{1}}{10^{-12}} \right )$

$12 = log\left ( \frac{I_{1}}{10^{-12}} \right )$               Eq-1

sound level due to whisper is given as

$\beta _{2} = 10 log\left ( \frac{I_{2}}{10^{-12}} \right )$

$21 = 10 log\left ( \frac{I_{2}}{10^{-12}} \right )$

$2.1 = log\left ( \frac{I_{2}}{10^{-12}} \right )$                          Eq-2

subtracting Eq-2 from Eq-1

$12 - 2.1 = log\left ( \frac{I_{1}}{10^{-12}} \right ) - log\left ( \frac{I_{2}}{10^{-12}} \right )$

$9.9 = log\left ( \frac{I_{1}}{I_{2}} \right )$

$\left ( \frac{I_{1}}{I_{2}} \right ) = 7.94\times 10^{9}$