Question

A person speaking normally produces a sound intensity of \[40dB\] at a distance of \[1m\] . If the threshold intensity for reasonable audibility is \[20dB\] , the maximum distance at which he can be heard clearly is:
A. $4m$
B. $5m$
C. $10m$
D. $20m$

Answer 1

Hint: When we use a sound level metre to measure noise levels, we use decibel units to quantify the strength of the noise \[\left( {dB} \right)\] . A sound metre has a decibel range display with a resolution that is close to the range of the human ear. As a result, rather than the silent part, the upper range is used.

Complete answer:
Now, let us solve the question;
As we know that loudness is measured in decibels by: $\beta = 10{\log _{10}}\left( {\dfrac{I}{{{I_0}}}} \right)$
Here,
$\beta = $ Intensity in decibel
$I = $ Sound intensity and
${I_0} = $ Threshold intensity of sound \[ = {10^{ - 12}}W \cdot {m^2}\]
[The rate at which a sound wave carries energy per unit area is used to determine its intensity. It's the same as the average power per unit area. Watts per square metre is how we measure it.]
Now, we will put the given values in the above equation;
$40 = 10{\log _{10}}\left( {\dfrac{I}{{{I_0}}}} \right)\,\,\,\,\,.........\left( i \right)$
$20 = 10{\log _{10}}\left( {\dfrac{{{I_2}}}{{{I_0}}}} \right)\,\,\,\,...........\left( {ii} \right)$
Now, comparing both the equation $\left( i \right)$ and $\left( {ii} \right)$
$\dfrac{{\left( i \right)}}{{\left( {ii} \right)}} \Rightarrow \dfrac{{40}}{{20}} = {\log _{10}}\left( {\dfrac{{{I_1}}}{{{I_2}}}} \right)$
${I_0}$ in both the equations will cancel out each other.
$\Rightarrow 2 = {\log _{10}}\left( {\dfrac{{{I_1}}}{{{I_2}}}} \right)$
$\Rightarrow \dfrac{{{I_1}}}{{{I_2}}} = {10^2}$
Therefore, now we will compare the distance with the known values and find the unknown one.
$\therefore \dfrac{{r_1^2}}{{r_2^2}} = {10^2}\,\left( {since\,I\alpha \dfrac{1}{{{r^2}}}} \right)$
Here, ${r_1}\,and\,{r_2}$ are the distances where, ${r_1}$ is known and we will find out ${r_2}$
$ \Rightarrow r_2^2 = {10^2}r_1^2$
Square has been taken common on both the sides and cancelled out
$ \Rightarrow {r_2} = 10{r_1} = 10 \times 1 = 10m$
Hence, the maximum distance at which he can be heard clearly is: $10m$

So, the correct option is: C. $10m$

Note:
The decibel is helpful for displaying huge ratios and for representing multiplied effects like attenuation from several sources throughout a signal chain. It's less obvious how to use it in systems with additive effects. Because of the decibel's logarithmic scale, a wide range of ratios can be represented by a single number in a way akin to scientific notation. This enables for the visualisation of large changes in a quantity.