Explanation:
It is given that,
Sound intensity level, [tex]\beta =90\ dB[/tex]
Area of eardrum, [tex]A=1.56\times 10^{-4}\ m^2[/tex]
Time taken, t = 9 hours = 32400 s
Sound intensity level is given by :
[tex]\beta =10dB\ log(\dfrac{I}{I_0})[/tex]
[tex]9=log(\dfrac{I}{10^{-12}})[/tex]
[tex]10^9=\dfrac{I}{10^{-12}}[/tex]
[tex]I=10^{-3}\ W/m^2[/tex]
Sound intensity is given by :
[tex]I=\dfrac{P}{A}[/tex]
And P = E/t
[tex]I=\dfrac{E}{At}[/tex]
[tex]E=I\times A\times t[/tex]
[tex]E=10^{-3}\times 1.56\times 10^{-4}\times 32400[/tex]
E = 0.00505 Joules
or
[tex]E=5.05\times 10^{-3}\ J[/tex]
So, the incident energy on the eardrum during this time is [tex]5.05\times 10^{-3}\ J[/tex]. Hence, this is the required solution.
