Question

When a person wears a hearing aid, the sound intensity level increases by $30\,{\text{dB}}$. The sound intensity increases by
A. ${e^3}$
B. ${10^3}$
C. $30$
D. ${10^2}$

Answer 1

Use the formula for loudness of the sound heard in decibel. This formula gives the relation between the loudness of the sound in decibel, intensity of the sound heard and minimum intensity of the sound that can be heard by humans. Write this formula for initial and increased loudness and take subtraction of them to determine the relation between increased intensity and original intensity.

Formula used:
The loudness $\beta$ of the sound is given by
$\beta = 10{\log _{10}}\left( {\dfrac{I}{{{I_0}}}} \right)$ ….. (1)
Here, $I$ is the intensity of the sound heard and ${I_0}$ is the minimum intensity of the sound heard by humans.

Complete step by step answer:
We have given that when the person wears a hearing aid, the loudness of the sound heard by the person increases by $30\,{\text{dB}}$.
Let the intensity of the sound heard by the person without wearing a hearing aid is $I$. The equation (1) for loudness $\beta$ of the sound heard by the person becomes
$\beta = 10{\log _{10}}\left( {\dfrac{I}{{{I_0}}}} \right)$

Let the intensity of the sound heard by the person without wearing a hearing aid is $I'$. The equation (1) for loudness $\beta '$ of the sound heard by the person becomes
$\beta ' = 10{\log _{10}}\left( {\dfrac{{I'}}{{{I_0}}}} \right)$
The loudness of the sound is increased by $30\,{\text{dB}}$.
$\beta ' - \beta = 30\,{\text{dB}}$

Substitute $10{\log _{10}}\left( {\dfrac{I}{{{I_0}}}} \right)$ for $\beta$ and $10{\log _{10}}\left( {\dfrac{{I'}}{{{I_0}}}} \right)$ for $\beta '$ in the above equation.
$10{\log _{10}}\left( {\dfrac{{I'}}{{{I_0}}}} \right) - 10{\log _{10}}\left( {\dfrac{I}{{{I_0}}}} \right) = 30\,{\text{dB}}$
$\Rightarrow 10\left( {{{\log }_{10}}I' - {{\log }_{10}}{I_0}} \right) - 10\left( {{{\log }_{10}}I - {{\log }_{10}}{I_0}} \right) = 30\,{\text{dB}}$
$\Rightarrow 10{\log _{10}}I' - 10{\log _{10}}I = 30\,{\text{dB}}$
$\Rightarrow 10{\log _{10}}\left( {\dfrac{{I'}}{I}} \right) = 30\,{\text{dB}}$
$\Rightarrow {\log _{10}}\left( {\dfrac{{I'}}{I}} \right) = 3$

Take antilog on both sides of the above equation.
$\Rightarrow \left( {\dfrac{{I'}}{I}} \right) = {\text{Antilog}}\left( 3 \right)$
$\therefore I' = {10^3}I$
Therefore, the intensity of the sound is increased by a factor ${10^3}$.

Hence, the correct option is B.

Note: The students should keep in mind that the intensity of the sound has increased by 30 decibel means the difference between the loudness of the sound heard by the person is 30 decibel for two different intensities of the sound heard by the person by wearing the hearing aid and without the hearing aid and not the changed intensity of the sound is 30 decibel.