The sound level at a point \[5.0m\] away from a point source is \[40dB\]. What will be the level at a point \[50m\] away from the source?

Hint:We know that the intensity of sound decreases with an increase in distance. So, by using this relation to combine corresponding distances, we find the level at a point \[50m\] away from the source.

Formula used:
1.\[I \propto \dfrac{1}{{{r^2}}}\]
where \[r\] is the distance of the source and \[I\] is the sound intensity level
2. \[\beta = 10{\log _{10}}\left( {\dfrac{I}{{{I_0}}}} \right)\]
3. \[\log m - \log n = \log \left( {\dfrac{m}{n}} \right)\]

Complete step by step solution:
We know that the sound level depends on the distance from the source intensity level of sound (\[I\]) is inversely proportional to the square of the distance from the source (\[r\]) is
\[I \propto \dfrac{1}{{{r^2}}}\]. We are given that,
\[{r_1} = 5.0m \\
\Rightarrow {r_2} = 50m \\
\Rightarrow {\beta _1} = 40dB \]
Now let us assume that \[{I_1}\]and \[{I_2}\] be the intensity at distance \[{r_1}\]and \[{r_2}\] respectively. By using the above relation, we get
\[{I_1} \propto \dfrac{1}{{r_1^2}}...\left( 1 \right)\]
\[\Rightarrow {I_2} \propto \dfrac{1}{{{r_2}^2}}...\left( 2 \right)\]

By combining equations (1) and (2), we get
\[\dfrac{{{I_1}}}{{{I_2}}} = \dfrac{{{r_2}^2}}{{r_1^2}}\]
We know that the formula for sound level intensity is
\[\beta = 10{\log _{10}}\left( {\dfrac{I}{{{I_0}}}} \right)\]
By using the above formula, we assume that
\[{\beta _1} = 10{\log _{10}}\left( {\dfrac{{{I_1}}}{{{I_0}}}} \right)...\left( 3 \right)\]
\[\Rightarrow {\beta _2} = 10{\log _{10}}\left( {\dfrac{{{I_2}}}{{{I_0}}}} \right)...\left( 4 \right)\]
Now subtract equation (4) from (3):
\[{\beta _1} - {\beta _2} = 10{\log _{10}}\left( {\dfrac{{{I_1}}}{{{I_0}}}} \right) - 10{\log _{10}}\left( {\dfrac{{{I_2}}}{{{I_1}}}} \right)\]

We know that \[\log m - \log n = \log \left( {\dfrac{m}{n}} \right)\]
 \[{\beta _1} - {\beta _2} = 10{\log _{10}}\left( {\dfrac{{\dfrac{{{I_1}}}{{{I_0}}}}}{{\dfrac{{{I_2}}}{{{I_0}}}}}} \right) \\
\Rightarrow {\beta _1} - {\beta _2} = 10{\log _{10}}\left( {\dfrac{{{I_1}}}{{{I_0}}} \times \dfrac{{{I_0}}}{{{I_2}}}} \right) \\
\Rightarrow {\beta _1} - {\beta _2} = 10{\log _{10}}\left( {\dfrac{{{I_1}}}{{{I_2}}}} \right) \]
We know that \[\dfrac{{{I_1}}}{{{I_2}}} = \dfrac{{{r_2}^2}}{{r_1^2}}\]
On substituting the relation, we get
\[{\beta _1} - {\beta _2} = 10{\log _{10}}\left( {\dfrac{{{r_2}^2}}{{r_1^2}}} \right)\]

By substituting all the values in the above formula, we get
\[40 - {\beta _2} = 10{\log _{10}}\left( {\dfrac{{{{\left( {50} \right)}^2}}}{{{{\left( {5.0} \right)}^2}}}} \right) \\
\Rightarrow 40 - {\beta _2} = 10{\log _{10}}\left( {\dfrac{{2500}}{{25}}} \right) \\
\Rightarrow 40 - {\beta _2} = 10{\log _{10}}\left( {\dfrac{1}{{100}}} \right) \\
\Rightarrow 40 - {\beta _2} = 10 \times \left( 2 \right)\left( {\because {{\log }_{10}}\left( {100} \right) = 2} \right) \]
Further simplifying
\[40 - {\beta _2} = 20 \\
\Rightarrow 40 - 20 = {\beta _2} \\
\therefore {\beta _2} = 20 \]
Hence, the level at a point \[50m\] away from the source is \[20dB\].

Note: Students should ensure that the logarithm value is natural or to the base 10 and substitute the correct value and be careful while calculating the result as it is most confusing.