Question

A dog while barking delivers about 1 mW of power. Assume this power is uniformly distributed over a hemispherical area. Now, 5 dogs start barking at the same time from the same point, each delivering 1 mW of power. What would be the sound level at a distance of 5m? (Take, $lo{g}_{10}\left(6.37\right)=0.8$ and $lo{g}_{10}5=0.7$)

• A. 70 dB
• B. 75 dB
• C. 72 dB
• D. 80 dB

Answer 1

The correct option is B 75 dB

Due to one dog, $l_1=\frac{P}{2\pi r^2}=\frac{{10}^{-3}}{2\pi (5{)}^2}or{l}_1=6.37\times {10}^{-6}W{m}^{-2}$
Due to 5 dogs, $I_R=5l_1=31.85\times {10}^{-6}W{m}^{-2}$
Now, sound level is
$\beta =10lo{g}_{10}\frac{l_R}{l_0}=10lo{g}_{10}\left(\frac{31.85\times {10}^{-6}}{{10}^{-12}}\right)$
$Or\beta =10log{}_{10}(31.85\times {10}^6)$
$=10\left[lo{g}_{10}\right(6.37\times 5)+6lo{g}_{10}10]$