91Ó°ÊÓ

In this problem, the power of the sound wave is calculated by using the product of sound intensity and the area covered by sound.

The level of sound wave is \(\beta = 55\;{\rm{dB}}\).

The surface area of eardrum is \(A = 5.0 \times {10^{ - 5}}\;{{\rm{m}}^2}\).

The total energy received by eardrum is \(E = 1.0\;{\rm{J}}\).

The standard value of threshold intensity is \({I_0} = 1.0 \times {10^{ - 12}}\;{\rm{W/}}{{\rm{m}}^2}\).

The intensity of sound is calculated below:

\(\begin{aligned}{c}\beta = 10\log \left( {\frac{I}{{{I_0}}}} \right)\\I = {10^{\left( {\frac{\beta }{{10}}} \right)}} \times {I_0}\end{aligned}\)

Substitute the values in the above equation.

\(\begin{aligned}{c}I = {10^{\left( {\frac{{55\;{\rm{dB}}}}{{10}}} \right)}} \times \left( {1.0 \times {{10}^{ - 12}}\;{\rm{W/}}{{\rm{m}}^2}} \right)\\I = 3.16 \times {10^{ - 7}}\;{\rm{W/}}{{\rm{m}}^2}\end{aligned}\)

The energy received by the eardrum per second is calculated below:

\(P = IA\)

Substitute the values in the above equation.

\(\begin{aligned}{c}P = \left( {3.16 \times {{10}^{ - 7}}\;{\rm{W/}}{{\rm{m}}^2}} \right)\left( {5.0 \times {{10}^{ - 5}}\;{{\rm{m}}^2}} \right)\\P = 1.58 \times {10^{ - 11}}\;{\rm{J/s}}\end{aligned}\)

Hence, the energy received by the eardrum per second is \(1.58 \times {10^{ - 11}}\;{\rm{J/s}}\).

The time taken by the eardrum to take complete energy is calculated below:

\(t = \frac{E}{P}\)

Substitute the values in the above equation.

\(\begin{aligned}{c}t = \frac{{1.0\;{\rm{J}}}}{{1.58 \times {{10}^{ - 11}}\;{\rm{J/s}}}}\\t = 6.32 \times {10^{10}}\;{\rm{s}} \times \left( {\frac{{1\;{\rm{yr}}}}{{3.154 \times {{10}^7}\;{\rm{s}}}}} \right)\\t = 2005.25\;{\rm{yr}}\end{aligned}\)

Hence, the time taken by the eardrum to take complete energy is \(2005.25\;{\rm{yr}}\).