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(C) The sound intensity level in decibels is given by $\beta = 10 \log_{10} \left( \frac{I}{I_0} \right)$.At $r_1 = 1 \ m$,$\beta_1 = 40 \ dB$. Thus,$

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$10$

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(C) The sound intensity level in decibels is given by $\beta = 10 \log_{10} \left( \frac{I}{I_0} \right)$.At $r_1 = 1 \ m$,$\beta_1 = 40 \ dB$. Thus,$40 = 10 \log_{10} \left( \frac{I_1}{I_0} \right) \implies \frac{I_1}{I_0} = 10^4$.At distance $r_2$,$\beta_2 = 20 \ dB$. Thus,$20 = 10 \log_{10} \left( \frac{I_2}{I_0} \right) \implies \frac{I_2}{I_0} = 10^2$.Dividing the two equations: $\frac{I_1}{I_2} = \frac{10^4}{10^2} = 10^2 = 100$.Since intensity $I \propto \frac{1}{r^2}$,we have $\frac{I_1}{I_2} = \frac{r_2^2}{r_1^2}$.Substituting the values: $100 = \frac{r_2^2}{(1)^2} \implies r_2^2 = 100 \implies r_2 = 10 \ m$.

$A$ person speaking normally produces a sound intensity of $40 \ dB$ at a distance of $1 \ m$. If the threshold intensity for reasonable audibility is $20 \ dB$,the maximum distance at which he can be heard clearly is ... $m$.

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