Question

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 $400\,m\,W$. The power increase in decibels as compared to the original power is :

• A. 13 dB
• B. 10 dB
• C. 20 dB
• D. 800 dB

Answer 1

Answer: (A)

We use
$L_{1} =10\, \log _{10}\left(\frac{I_{1}}{I_{0}}\right)$
and $L_{2}=10\, \log _{10}\left(\frac{I_{2}}{I_{0}}\right)$
So $L_{2}-L_{1} =10\, \log _{10}\left(\frac{I_{2}}{I_{1}}\right)$
As sound power $\propto$ intensity, we use
$L_{2}-L_{1} =10\, \log _{10}\left(\frac{P_{2}}{P_{1}}\right)$
$=10\, \log _{10}\left(\frac{400}{20}\right)$
$=10\, \log _{10} 20$
$=10\, \log (2 \times 10)$
$=10(0.301+1)=13\, dB$