A sound absorber attenuates the sound level b | vedclass

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Question

We have, $L_{1}=10 \log \left(\frac{I_{1}}{I_{0}}\right)$

$\mathrm{L}_{2}=10 \log \left(\frac{\mathrm{I}_{2}}{\mathrm{I}_{0}}\right)$

$\mathrm{L

Vedclass · Accepted Answer

$100$

Vedclass · Answer

We have, $L_{1}=10 \log \left(\frac{I_{1}}{I_{0}}ight)$

$\mathrm{L}_{2}=10 \log \left(\frac{\mathrm{I}_{2}}{\mathrm{I}_{0}}ight)$

$\mathrm{L}_{1}-\mathrm{L}_{2}=10 \log \left(\frac{\mathrm{I}_{1}}{\mathrm{I}_{0}}ight)-10 \log \left(\frac{\mathrm{I}_{2}}{\mathrm{I}_{0}}ight)$

or, $\quad \Delta \mathrm{L}=10 \log \left(\frac{\mathrm{I}_{1}}{\mathrm{I}_{0}} 	imes \frac{\mathrm{I}_{0}}{\mathrm{I}_{2}}ight)$

or, $\quad \Delta \mathrm{L}=10 \log \left(\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}ight)$

or, $\quad 20=10 \log \left(\frac{I_{1}}{I_{2}}ight)$

or, $\quad 2=\log \left(\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}ight)$

or, $\quad \frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=10^{2}$

or, $\quad \mathrm{I}_{2}=\frac{\mathrm{I}_{1}}{100}$

$\Rightarrow$ Intensity decreases by a factor $100 .$

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

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