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(C) The sound intensity level $\beta$ is given by $\beta = 10 \log_{10} \left( \frac{I}{I_0} \right)$.At distance $r_1 = 1\, m$,the intensity level $\

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$10$

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(C) The sound intensity level $\beta$ is given by $\beta = 10 \log_{10} \left( \frac{I}{I_0} \right)$.At distance $r_1 = 1\, m$,the intensity level $\beta_1 = 40\, dB$.At distance $r_2$,the intensity level $\beta_2 = 20\, dB$.We know that intensity $I \propto \frac{1}{r^2}$,so $\frac{I_1}{I_2} = \frac{r_2^2}{r_1^2}$.From the intensity level formula: $\beta_1 - \beta_2 = 10 \log_{10} \left( \frac{I_1}{I_2} \right)$.Substituting the values: $40 - 20 = 10 \log_{10} \left( \frac{r_2^2}{r_1^2} \right)$.$20 = 10 \log_{10} \left( \frac{r_2^2}{1^2} \right) \Rightarrow 2 = \log_{10} (r_2^2)$.Taking antilog on both sides: $10^2 = r_2^2$.$r_2 = 10\, m$.

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

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