Question

A sound absorber attenuates the sound level by $20\text{ dB}$. The intensity decreases by a factor of
A.$1000$
B.$10000$
C.$10$
D.$100$

Answer 1

Hint: In simple language, the question says that the sound absorber decreases the sound level by $20\text{ dB}$, therefore here we can use the formula for sound level intensity initially and then find the final sound level intensity when it is decreased by $20\text{ dB}$.
Formula used:
$L=10\log \left( \dfrac{I}{{{I}_{0}}} \right)$, here $I$ is the intensity of the sound and ${{I}_{0}}$ is the reference intensity and the unit of sound intensity level $L$ is $dB$. While the unit of intensity $I$ used in the formula is $\text{W}/{{\text{m}}^{2}}$.

Complete answer:
First let us calculate the sound intensity level initially. Sound intensity level compares the intensity of sound with a reference intensity. Therefore, the intensity level of the sound initially will be:
${{L}_{1}}=10\log \left( \dfrac{I}{{{I}_{0}}} \right)$
Now, when the intensity of the sound is decreased, then the final intensity level of sound will be:
${{L}_{2}}=10\log \left( \dfrac{I'}{{{I}_{0}}} \right)$
As it is given that the sound is decreased by $20\text{ dB}$ which means that the difference between initial sound intensity level and final sound intensity level is $20\text{ dB}$, therefore:
$\begin{align}
  & {{L}_{1}}-{{L}_{2}}=10\log \left( \dfrac{I}{{{I}_{0}}} \right)-10\log \left( \dfrac{I'}{{{I}_{0}}} \right) \\
 & \Rightarrow 20=10\log \left( \dfrac{I}{I'} \right) \\
 & \Rightarrow 2=\log \left( \dfrac{I}{I'} \right) \\
 & \Rightarrow -2=\log \left( \dfrac{I'}{I} \right) \\
 & \Rightarrow {{10}^{-2}}=\dfrac{I'}{I} \\
 & \Rightarrow I'={{10}^{-2}}I \\
 & \therefore I'=\dfrac{I}{100} \\
\end{align}$
Therefore, the intensity of sound decreases by a factor of $100$.

Hence option $D$ is correct.

Additional information:
The sound level intensity is also known as loudness. Loudness of sound is however subjective as if one person might find a particular sound waves as too loud, it is not necessary that every person would agree to his conclusion.

Note:
We must observe carefully that, the sound level is decreased, thus the initial sound level will be greater than the final sound level, thus when we write the difference as $20\text{ dB}$, then ${{L}_{1}}-{{L}_{2}}=20$ or it can be written as ${{L}_{2}}-{{L}_{1}}=-20$.