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Question

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A sound absorber attenuates the sound level by 20 dB. The intensity decreases by a factor of

1000

10000

10

100

Solution

D.

100

Let intensity of sound be I and I'. Loudness of sound initially

$β_{1} = 10 \log (\frac{I}{I_{o}}) β_{2} = 10 \log (\frac{I'}{I_{o}})$

β₂ - β₁ = 20

$∴ 20 = 20 \log (\frac{I'}{I})$

∴ I' = 100 I

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