Wave Mechanics Question 1096
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 mW. The power increase in decibels as compared to the original power is
Options:
A) 13 dB
B) 10 dB
C) 20 dB
D) 800 dB
Show Answer
Answer:
Correct Answer: A
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)=13dB $
BETA
