Question

The power of a sound from the speaker of a radio is $20mW$. By turning the knob of the volume control, the power of the sound is increased to $400mW$. 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
$L_2-L_1=10{\log }_{10}\left(\frac{P_2}{P_1}\right)$
$=10{\log }_{10}\left(\frac{400}{20}\right)$
$=10{\log }_{10}20$
$=10\log (2\times 10)$
$=10(0.301+1)=13dB$