How many order of magnitude more powerful is 90 dB sound than 40 dB sound?

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500

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10^{5}

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Solution

The correct option is D $10^{5}$
$90 = 10 l o g (\frac{I_{1}}{I})$

$40 = 10 l o g (\frac{I_{2}}{I})$

$90 - 40 = 10 l o g (\frac{I_{1}}{I_{2}})$

$\frac{I_{1}}{I_{2}} = 10^{5}$

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