Question:
The sound intensity increases from \(I_1=10^{-6}\,\mathrm{W/m^2}\) to \(I_2=10^{-4}\,\mathrm{W/m^2}\). The increase in sound level is
The sound intensity increases from \(I_1=10^{-6}\,\mathrm{W/m^2}\) to \(I_2=10^{-4}\,\mathrm{W/m^2}\). The increase in sound level is
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Sound level difference:
\[
\boxed{
\Delta L
=
10\log_{10}\left(\frac{I_2}{I_1}\right)
}
\]
A 100-fold increase in intensity corresponds to
\[
\boxed{20\,\mathrm{dB}.}
\]
Updated On: Jul 23, 2026
- \(10\,\mathrm{dB}\)
- \(20\,\mathrm{dB}\)
- \(30\,\mathrm{dB}\)
- \(40\,\mathrm{dB}\)
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The Correct Option is B
Solution and Explanation
Step 1: Use the sound level difference formula.
\[
\Delta L
=
10\log_{10}\left(\frac{I_2}{I_1}\right)
\]
Step 2: Substitute the given values. \[ \Delta L = 10\log_{10}\left(\frac{10^{-4}}{10^{-6}}\right) = 10\log_{10}(10^2) = 10\times2 = 20\,\mathrm{dB} \] Hence, \[ \boxed{20\,\mathrm{dB}} \] Therefore, \[ \boxed{(B)} \] is the correct answer.
Step 2: Substitute the given values. \[ \Delta L = 10\log_{10}\left(\frac{10^{-4}}{10^{-6}}\right) = 10\log_{10}(10^2) = 10\times2 = 20\,\mathrm{dB} \] Hence, \[ \boxed{20\,\mathrm{dB}} \] Therefore, \[ \boxed{(B)} \] is the correct answer.
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