Wave Mechanics Question 641
Question: A person speaking normally produces a sound intensity of 40 dB at a distance of 1 m. If the threshold intensity for reasonable audibility is 20 dB, the maximum distance at which he can be heard clearly is
Options:
A) $ 4m $
B) $ ~5m $
C) $ 10m $
D) $ 20m $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ 40=10\log \frac{I _{1}}{I _{0}},\
Therefore \frac{I _{1}}{I _{2}}=10^{4} $ and $ 20=10\log \frac{I _{2}}{I _{0},}\
Therefore \frac{I _{2}}{I _{0}}=10^{2}\
Therefore \frac{I _{1}}{I _{2}}=100 $ Also $ \frac{I _{1}}{I _{2}}=100=\frac{r _{2}^{2}}{r _{1}^{2}}=\frac{r _{2}^{2}}{1^{2}},or,r _{2}=10m. $
BETA
