How many times more intense is $90 dB$ sound than $40 dB$ sound?

5

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50

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500

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10^{5}

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Solution

The correct option is D $10^{5}$
Loudness of the sound is diven by
$β_{2} - β_{1} = 10 l o g_{10} \frac{I_{2}}{I_{1}}$
$\Rightarrow 50 = 10 l o g_{10} \frac{I_{2}}{I_{1}} \Rightarrow \frac{I_{2}}{I_{1}} = 10^{5}$

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