A sound absorber attenuates the sound level b | vedclass

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

Vedclass · Accepted Answer

$100$

Vedclass · Answer

(C) The sound level $L$ in decibels $(dB)$ is given by $L = 10 \log_{10} \left( \frac{I}{I_0} \right)$,where $I$ is the intensity and $I_0$ is the reference intensity.The change in sound level is given by $\Delta L = L_1 - L_2 = 20 \ dB$.Substituting the formula,we get $\Delta L = 10 \log_{10} \left( \frac{I_1}{I_0} \right) - 10 \log_{10} \left( \frac{I_2}{I_0} \right) = 10 \log_{10} \left( \frac{I_1}{I_2} \right)$.Given $\Delta L = 20$,we have $20 = 10 \log_{10} \left( \frac{I_1}{I_2} \right)$.Dividing by $10$,we get $2 = \log_{10} \left( \frac{I_1}{I_2} \right)$.Taking the antilog,we get $\frac{I_1}{I_2} = 10^2 = 100$.Therefore,the intensity decreases by a factor of $100$.

$A$ sound absorber attenuates the sound level by $20 \ dB$. The intensity decreases by a factor of

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