How many times more intense is a 60 dB sound than a 30 dB sound?

1000

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100

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Solution

The correct option is A 1000
Loudness of sound is given by $L = 10 {log}_{10} \frac{I}{I_{o}}$
So, we get $L_{2} - L_{1} = 10 {log}_{10} \frac{I_{2}}{I_{1}}$
Given : $L_{1} = 30 d B$ and $L_{2} = 60 d B$
$∴$ $60 - 30 = 10 {log}_{10} \frac{I_{2}}{I_{1}}$
Or ${log}_{10} \frac{I_{2}}{I_{1}} = 3$
Or $\frac{I_{2}}{I_{1}} = 10^{3} = 1000$
$⟹ I_{2} = 1000 I_{1}$

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