91Ó°ÊÓ

The intensity of sound varies with an alteration in the distance among the source and receiver.

The vertical distance of person 1 from explosion is \({r_1} = 100\;{\rm{m}}\).

The horizontal distance of person 2 from explosion is \({r_2}' = 200\;{\rm{m}}\).

The distance of person 2 from explosion is calculated below:

\({r_2} = \sqrt {{{\left( {{r_1}} \right)}^2} + {{\left( {{r_2}'} \right)}^2}} \)

Substitute the values in the above equation.

\(\begin{aligned}{c}{r_2} = \sqrt {{{\left( {100\;{\rm{m}}} \right)}^2} + {{\left( {200\;{\rm{m}}} \right)}^2}} \\{r_2} = 223.60\;{\rm{m}}\end{aligned}\)

The intensity ratio of sound between both persons is calculated below:

\(\frac{{{I_1}}}{{{I_2}}} = \frac{{{{\left( {{r_2}} \right)}^2}}}{{{{\left( {{r_1}} \right)}^2}}}\)

Substitute the values in the above equation.

\(\begin{aligned}{c}\frac{{{I_1}}}{{{I_2}}} = \frac{{{{\left( {223.60\;{\rm{m}}} \right)}^2}}}{{{{\left( {100\;{\rm{m}}} \right)}^2}}}\\\frac{{{I_1}}}{{{I_2}}} = 5\end{aligned}\)

The sound level of the explosion is calculated below:

\(\beta = 10\log \left( {\frac{{{I_1}}}{{{I_2}}}} \right)\)

Substitute the values in the above equation.

\(\begin{aligned}{c}\beta = 10\log \left( 5 \right)\\\beta = 6.99\;{\rm{dB}}\end{aligned}\)

Hence, the sound level of the explosion is \(6.99\;{\rm{dB}}\).