How many times more intense is a 60 dB sound | vedclass

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(A) The sound intensity level $L$ in decibels $(dB)$ is given by the formula $L = 10 \log_{10} \left( \frac{I}{I_0} \right)$,where $I$ is the intensit

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$1000$

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(A) The sound intensity level $L$ in decibels $(dB)$ is given by the formula $L = 10 \log_{10} \left( \frac{I}{I_0} \right)$,where $I$ is the intensity of the sound and $I_0$ is the reference intensity.For a $60 \ dB$ sound: $60 = 10 \log_{10} \left( \frac{I_1}{I_0} \right) \Rightarrow 6 = \log_{10} \left( \frac{I_1}{I_0} \right) \Rightarrow \frac{I_1}{I_0} = 10^6$.For a $30 \ dB$ sound: $30 = 10 \log_{10} \left( \frac{I_2}{I_0} \right) \Rightarrow 3 = \log_{10} \left( \frac{I_2}{I_0} \right) \Rightarrow \frac{I_2}{I_0} = 10^3$.To find how many times more intense the $60 \ dB$ sound is compared to the $30 \ dB$ sound,we calculate the ratio $\frac{I_1}{I_2}$:$\frac{I_1}{I_2} = \frac{I_1 / I_0}{I_2 / I_0} = \frac{10^6}{10^3} = 10^3 = 1000$.Thus,a $60 \ dB$ sound is $1000$ times more intense than a $30 \ dB$ sound.

How many times more intense is a $60 \ dB$ sound than a $30 \ dB$ sound?

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