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To determine how the decibel level is affected when the intensity of a sound is increased by a factor of 100, we can utilize the formula for calculating the intensity level in decibels (dB):
[tex]\[ L = 10 \log_{10} \left(\frac{I}{I_0}\right) \][/tex]
where:
- [tex]\( L \)[/tex] is the intensity level in decibels.
- [tex]\( I \)[/tex] is the intensity of the sound.
- [tex]\( I_0 \)[/tex] is the reference intensity.
When the intensity is increased by a factor of 100, the new intensity ([tex]\( I_{\text{new}} \)[/tex]) is expressed as:
[tex]\[ I_{\text{new}} = 100 \times I \][/tex]
We want to find the change in decibel level due to this increase. The change in decibel level [tex]\( \Delta L \)[/tex] can be found by calculating the difference between the new decibel level and the old decibel level.
First, let's express the change in logarithmic terms:
[tex]\[ \Delta L = 10 \log_{10} \left(\frac{I_{\text{new}}}{I_0}\right) - 10 \log_{10} \left(\frac{I}{I_0}\right) \][/tex]
Since we know that:
[tex]\[ I_{\text{new}} = 100 \times I \][/tex]
Substitute [tex]\( I_{\text{new}} \)[/tex] into the change in decibel level formula:
[tex]\[ \Delta L = 10 \log_{10} \left(\frac{100 \times I}{I_0}\right) - 10 \log_{10} \left(\frac{I}{I_0}\right) \][/tex]
Simplify the expression:
[tex]\[ \Delta L = 10 \left[ \log_{10} \left(\frac{100 \times I}{I_0}\right) - \log_{10} \left(\frac{I}{I_0}\right) \right] \][/tex]
Using the properties of logarithms, we get:
[tex]\[ \Delta L = 10 \left[ \log_{10} (100) + \log_{10} \left(\frac{I}{I_0}\right) - \log_{10} \left(\frac{I}{I_0}\right) \right] \][/tex]
Notice that [tex]\( \log_{10} \left(\frac{I}{I_0}\right) \)[/tex] cancels out:
[tex]\[ \Delta L = 10 \log_{10} (100) \][/tex]
We know that:
[tex]\[ \log_{10} (100) = 2 \][/tex]
Therefore:
[tex]\[ \Delta L = 10 \times 2 = 20 \][/tex]
Thus, the decibel level increases by 20 dB.
Therefore, the correct answer is:
Increase of 20 dB
[tex]\[ L = 10 \log_{10} \left(\frac{I}{I_0}\right) \][/tex]
where:
- [tex]\( L \)[/tex] is the intensity level in decibels.
- [tex]\( I \)[/tex] is the intensity of the sound.
- [tex]\( I_0 \)[/tex] is the reference intensity.
When the intensity is increased by a factor of 100, the new intensity ([tex]\( I_{\text{new}} \)[/tex]) is expressed as:
[tex]\[ I_{\text{new}} = 100 \times I \][/tex]
We want to find the change in decibel level due to this increase. The change in decibel level [tex]\( \Delta L \)[/tex] can be found by calculating the difference between the new decibel level and the old decibel level.
First, let's express the change in logarithmic terms:
[tex]\[ \Delta L = 10 \log_{10} \left(\frac{I_{\text{new}}}{I_0}\right) - 10 \log_{10} \left(\frac{I}{I_0}\right) \][/tex]
Since we know that:
[tex]\[ I_{\text{new}} = 100 \times I \][/tex]
Substitute [tex]\( I_{\text{new}} \)[/tex] into the change in decibel level formula:
[tex]\[ \Delta L = 10 \log_{10} \left(\frac{100 \times I}{I_0}\right) - 10 \log_{10} \left(\frac{I}{I_0}\right) \][/tex]
Simplify the expression:
[tex]\[ \Delta L = 10 \left[ \log_{10} \left(\frac{100 \times I}{I_0}\right) - \log_{10} \left(\frac{I}{I_0}\right) \right] \][/tex]
Using the properties of logarithms, we get:
[tex]\[ \Delta L = 10 \left[ \log_{10} (100) + \log_{10} \left(\frac{I}{I_0}\right) - \log_{10} \left(\frac{I}{I_0}\right) \right] \][/tex]
Notice that [tex]\( \log_{10} \left(\frac{I}{I_0}\right) \)[/tex] cancels out:
[tex]\[ \Delta L = 10 \log_{10} (100) \][/tex]
We know that:
[tex]\[ \log_{10} (100) = 2 \][/tex]
Therefore:
[tex]\[ \Delta L = 10 \times 2 = 20 \][/tex]
Thus, the decibel level increases by 20 dB.
Therefore, the correct answer is:
Increase of 20 dB
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