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Q. A small speaker delivers $3\, W$ of audio output. The distance from the speaker where one can detect $120\, dB$ intensity sound is $\alpha$. Write the value of $100 \alpha$.
[Given reference intensity of sound as $10^{-12} W m ^{-2}($ in m) Take $\frac{3}{\pi}=0.96, \sqrt{0.96}=0.98$

NTA AbhyasNTA Abhyas 2022

Solution:

Given, audio output $=3W$
Intensity, $I=120dB$
$I_{0}=10^{- 12}W/m^{2}$
From loudness relation,
$dB=10log\frac{I}{I_{0}}$
$\Rightarrow 120=10log\left(\frac{I}{\left(10\right)^{- 12}}\right)$
$\Rightarrow 10I=10$
$\Rightarrow I=1W/m^{2}$
Final intensity $=\frac{P}{4 \pi r^{2}}$
$\Rightarrow I=\frac{P}{4 \pi r^{2}}$
$\Rightarrow r=\sqrt{\frac{3}{4 \pi \left(\right. 1 \left.\right)}}=\frac{0 . 98}{2}=0.49m=\alpha $