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Intensity level of sound whose intensity is $10^{-8} \mathrm{Wm}^{-2}$ is
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Verified Answer
The correct answer is:
$40 \mathrm{~dB}$
Intensity level of sound $=10 \log \frac{\mathrm{I}}{\mathrm{I}_{0}} \mathrm{~dB}$
$$
\begin{aligned}
=10 \log \left(\frac{10^{-8}}{10^{-12}}\right) \mathrm{dB} &=40 \mathrm{~dB} \\
&\left.\because \mathrm{I}_{0}=10^{-12} \mathrm{~W} / \mathrm{m}^{2}\right)
\end{aligned}
$$
$$
\begin{aligned}
=10 \log \left(\frac{10^{-8}}{10^{-12}}\right) \mathrm{dB} &=40 \mathrm{~dB} \\
&\left.\because \mathrm{I}_{0}=10^{-12} \mathrm{~W} / \mathrm{m}^{2}\right)
\end{aligned}
$$
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