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$90 \mathrm{~dB}$ sound is $x$ times more intense than $40 \mathrm{~dB}$ sound, then $x$ is
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Verified Answer
The correct answer is:
$10^{5}$
The level of sound is given by
$L=10 \log \left(\frac{I}{I_{0}}\right) \mathrm{dB}$
Let for $40 \mathrm{~dB}$ sound intensity be $I$ and for $90 \mathrm{~dB}$ sound intensity be $I_{x}$, then
$40=10 \log \left(\frac{I}{I_{0}}\right)$
$\Rightarrow \quad \log \left(\frac{I}{I_{0}}\right)=4 ...(i)$
and $\quad 90=10 \log \left(\frac{I_{x}}{I_{0}}\right)$
$\Rightarrow \quad \log \left(\frac{I_{x}}{I_{0}}\right)=9$
Given, $\quad I_{x}=x I$
$\Rightarrow \quad \log \left(\frac{x I}{I_{0}}\right)=9$
or $\quad \log x+\log \left(\frac{I}{I_{0}}\right)=9$
$\Rightarrow \quad \log x+4=9 \quad$ [from Eq. (i)]
$\Rightarrow \quad \log x=5 \quad \Rightarrow \quad x=10^{5}$
$L=10 \log \left(\frac{I}{I_{0}}\right) \mathrm{dB}$
Let for $40 \mathrm{~dB}$ sound intensity be $I$ and for $90 \mathrm{~dB}$ sound intensity be $I_{x}$, then
$40=10 \log \left(\frac{I}{I_{0}}\right)$
$\Rightarrow \quad \log \left(\frac{I}{I_{0}}\right)=4 ...(i)$
and $\quad 90=10 \log \left(\frac{I_{x}}{I_{0}}\right)$
$\Rightarrow \quad \log \left(\frac{I_{x}}{I_{0}}\right)=9$
Given, $\quad I_{x}=x I$
$\Rightarrow \quad \log \left(\frac{x I}{I_{0}}\right)=9$
or $\quad \log x+\log \left(\frac{I}{I_{0}}\right)=9$
$\Rightarrow \quad \log x+4=9 \quad$ [from Eq. (i)]
$\Rightarrow \quad \log x=5 \quad \Rightarrow \quad x=10^{5}$
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