An increase in intensity level of 1 dB implies an increase in intensity of (given anti $l o g_{10} 0.1 = 1.2589)$

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Solution

The correct option is C
Intensity level is decibel is given by
$L = 10 l o g_{10} \frac{I}{I_{0}} L + 1 = 10 l o g_{10} \frac{I_{1}}{I_{0}} S u b t r a c t i n g, 1 = 10 l o g_{10} \frac{I_{1}}{I_{0}} - 10 l o g_{10} \frac{I}{I_{0}} o r \frac{1}{10} = l o g_{10} \frac{I_{1}}{I} o r 0.1 = l o g_{10} \frac{I_{1}}{I} o r \frac{I_{1}}{I} = 1.26$

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l o g_{10} 0.1 = 1.2589)

1 d B

(

a n t i l o g_{10} 0.1 = 1.2589)

{log}_{10} 2 = 0.3010

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