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A sound absorber attenuates the sound level by $20 \mathrm{~dB}$. The intensity decreases by a factor of
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Verified Answer
The correct answer is:
$100$
$100$
$\begin{aligned}
& \mathrm{B}_1=10 \log \left(\frac{\mathrm{I}}{\mathrm{I}_0}\right) \\
& \mathrm{B}_2=\log \left(\frac{\mathrm{I}^{\prime}}{\mathrm{I}_0}\right)
\end{aligned}$
given $B_2-B_1=20$
$\begin{aligned}
& 20=10 \log \left(\frac{\mathrm{I^{\prime}}}{\mathrm{I}}\right) \\
& \mathrm{I}^{\prime}=100 \mathrm{I}
\end{aligned}$
& \mathrm{B}_1=10 \log \left(\frac{\mathrm{I}}{\mathrm{I}_0}\right) \\
& \mathrm{B}_2=\log \left(\frac{\mathrm{I}^{\prime}}{\mathrm{I}_0}\right)
\end{aligned}$
given $B_2-B_1=20$
$\begin{aligned}
& 20=10 \log \left(\frac{\mathrm{I^{\prime}}}{\mathrm{I}}\right) \\
& \mathrm{I}^{\prime}=100 \mathrm{I}
\end{aligned}$
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