A person speaking normally produces a sound of intensity 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

4 m

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5 m

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10 m

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20 m

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Solution

The correct option is C 10 m
$40 = 10 l o g_{10} (\frac{I_{1}}{I_{0}}) ∴ \frac{I_{1}}{I_{0}} = 10^{4} (i) A l s o, 20 = 10 l o g_{10} (\frac{I_{2}}{I_{0}}) \Rightarrow \frac{I_{2}}{I_{0}} = 10^{2} (i i) ∴ \frac{I_{2}}{I_{0}} = 10^{- 2} = \frac{r_{1}^{2}}{r_{2}^{2}} ∴ r_{2} = 100 r_{1}^{2} \Rightarrow r_{2} = 10 m (∴ r_{1} = 1 m)$

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A person speaking normally produces a sound of intensity 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

The correct option is C 10 m40=10log10(I1I0)∴I1I0=104(i)Also,20=10log10(I2I0)⇒I2I0=102(ii)∴I2I0=10−2=r21r22∴r2=100r21⇒r2=10m(∴r1=1m)