How many times more intense 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 $10^{5}$
Let $I$ represents the intensity of sound.
Loudness of sound $L = 10 l o g_{10} \frac{I}{I_{o}}$
$⟹ L_{2} - L_{1} = 10 l o g_{10} \frac{I_{2}}{I_{1}}$
$90 - 40 = 10 {log}_{10} \frac{I_{2}}{I_{1}}$
$5 = l o g_{10} \frac{I_{2}}{I_{1}} ⟹ \frac{I_{2}}{I_{1}} = 10^{5}$

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