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(B) For a long linear source of sound,the intensity $I$ at a distance $r$ is given by $I = \frac{P}{2 \pi r L}$.The sound level $\beta$ in decibels is

Vedclass · Accepted Answer

$70$

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(B) For a long linear source of sound,the intensity $I$ at a distance $r$ is given by $I = \frac{P}{2 \pi r L}$.The sound level $\beta$ in decibels is given by $\beta = 10 \log_{10} \left( \frac{I}{I_0} \right) = 10 \log_{10} \left( \frac{P}{2 \pi L I_0 r} \right)$.Let $\beta_A$ be the sound level at distance $r$ and $\beta_B$ be the sound level at distance $10r$.$\beta_A - \beta_B = 10 \log_{10} \left( \frac{P}{2 \pi L I_0 r} \right) - 10 \log_{10} \left( \frac{P}{2 \pi L I_0 (10r)} \right)$.$\beta_A - \beta_B = 10 \log_{10} \left( \frac{P / (2 \pi L I_0 r)}{P / (2 \pi L I_0 10r)} \right) = 10 \log_{10} (10) = 10 \ dB$.Given $\beta_A = 80 \ dB$,we have $80 \ dB - \beta_B = 10 \ dB$.Therefore,$\beta_B = 80 \ dB - 10 \ dB = 70 \ dB$.

There is a long linear source of sound whose power is $P$ and length is $L$. At point $A$,the sound level is $80 \ dB$. Then at point $B$,the sound level will be .... $dB$.

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