The sound intensity level at a point 4 ,m fro | vedclass

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(B) The sound intensity $I$ at a distance $r$ from a point source is given by $I = \frac{P}{4\pi r^2}$,where $P$ is the power of the source. Thus,$I \

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$16$

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(B) The sound intensity $I$ at a distance $r$ from a point source is given by $I = \frac{P}{4\pi r^2}$,where $P$ is the power of the source. Thus,$I \propto \frac{1}{r^2}$.The sound intensity level $\beta$ in decibels $(dB)$ is defined as $\beta = 10 \log_{10} \left( \frac{I}{I_0} ight)$,where $I_0$ is the reference intensity.Let $\beta_1 = 10 \,dB$ at $r_1 = 4 \,m$ and $\beta_2$ be the level at $r_2 = 2 \,m$.$\beta_2 - \beta_1 = 10 \log_{10} \left( \frac{I_2}{I_1} ight) = 10 \log_{10} \left( \frac{r_1^2}{r_2^2} ight) = 20 \log_{10} \left( \frac{r_1}{r_2} ight)$.Substituting the values: $\beta_2 - 10 = 20 \log_{10} \left( \frac{4}{2} ight) = 20 \log_{10}(2)$.Using $\log_{10}(2) \approx 0.301$,we get $\beta_2 - 10 = 20 	imes 0.301 = 6.02$.Therefore,$\beta_2 = 10 + 6.02 = 16.02 \,dB \approx 16 \,dB$.

The sound intensity level at a point $4 \,m$ from the point source is $10 \,dB$. Then the sound level at a distance $2 \,m$ from the same source will be ........ $dB$.

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