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Q.

The power of a sound from the speaker of a radio is 20 mW. By turning the knob of the volume control, the power of the sound is increased to 400 mW. The power increase in decibels as compared to the original power is

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a

10 dB

b

800 dB

c

13 dB

d

20 dB

answer is A.

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Detailed Solution

$P \propto I$

$L_{1} = 10 \log_{10} (\frac{I_{1}}{I_{0}}) and L_{2} = 10 \log_{10} (\frac{I_{2}}{I_{0}})$

So $L_{2} - L_{1} = 10 \log_{10} (\frac{I_{2}}{I_{1}})$

$= 10 \log_{10} (\frac{P_{2}}{P_{1}}) = 10 \log_{10} (\frac{400}{20}) = 10 \log_{10} 20$

$= 10 \log (2 \times 10) = 10 (0.301 + 1) = 13 dB$

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