Question:
A small speaker delivers \(2\ \mathrm{W}\) of audio output. At what distance from the speaker will one detect \(120\ \mathrm{dB}\) intensity sound ? [Given reference intensity of sound as \(10^{-12}\ \mathrm{W/m^2}\)]
A small speaker delivers \(2\ \mathrm{W}\) of audio output. At what distance from the speaker will one detect \(120\ \mathrm{dB}\) intensity sound ? [Given reference intensity of sound as \(10^{-12}\ \mathrm{W/m^2}\)]
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Convert 120 dB to an intensity using \(\beta = 10\log_{10}(I/I_0)\), then equate it to \(P/(4\pi r^2)\).
Updated On: Nov 11, 2026
- \(20\ \mathrm{cm}\)
- \(10\ \mathrm{cm}\)
- \(40\ \mathrm{cm}\)
- \(30\ \mathrm{cm}\)
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The Correct Option is C
Solution and Explanation
\(120 = 10\log_{10}\frac{I}{10^{-12}}\Rightarrow \frac{I}{10^{-12}} = 10^{12}\Rightarrow I = 1\ \mathrm{W/m^2}\).
For a point source, \(I = \frac{P}{4\pi r^2}\), so \(r^2 = \frac{P}{4\pi I} = \frac{2}{4\pi\times1} = \frac{1}{2\pi} = 0.159\ \mathrm{m^2}\).
\(r \approx 0.40\ \mathrm{m} = 40\ \mathrm{cm}\).
Answer: (3) \(40\ \mathrm{cm}\)
For a point source, \(I = \frac{P}{4\pi r^2}\), so \(r^2 = \frac{P}{4\pi I} = \frac{2}{4\pi\times1} = \frac{1}{2\pi} = 0.159\ \mathrm{m^2}\).
\(r \approx 0.40\ \mathrm{m} = 40\ \mathrm{cm}\).
Answer: (3) \(40\ \mathrm{cm}\)
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