✒️SOUND INTENSITY LEVEL
[tex]\tt{\huge{\blue{Solution:}}}[/tex]
The problem asks to solve for the molecular weight of the solute. First, we need to identify the quantities to be used. These are
I = sound intensity
I₀ = reference sound intensity = [tex]10^{-12} \: \text{W/}\text{m}^{2}[/tex]
β = sound intensity level (in decibels)
Second, we need to identify the given values. These are
I = [tex]10^{-2} \: \text{W/}\text{m}^{2}[/tex]
Finally, we can now solve for the sound intensity level. Therefore, the sound intensity level in decibels is
[tex]\beta = 10 \: log (\frac{I}{I_{0}})[/tex]
[tex]\beta = 10 \: log (\frac{10^{-2} \: \text{W/}\text{m}^{2}}{10^{-12} \: \text{W/}\text{m}^{2}})[/tex]
[tex]\boxed{\beta = \text{100 dB}}[/tex]
[tex]\\[/tex]
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