At a distance of 20m from a point source, the loudness is 30 db. After a minimum distance of (x×5010) metre, the sound is inaudible. The value of ‘x’ is _______.

I=p4πr2 Iα1r2 B= 10log10II0 30=10 log10II0⇒II0=103 r02r2=103 r0=r×103 r0=20×1010=20010 r0=4×5010 X=4.

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At a distance of 20m from a point source, the loudness is 30 db. After a minimum distance of $(x \times 50 \sqrt{10})$ metre, the sound is inaudible. The value of ‘x’ is _______.

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answer is 4.

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Detailed Solution

$I = \frac{p}{4 π r^{2}} I α \frac{1}{r^{2}} B = 10 \log_{10} \frac{I}{I_{0}} 30 = 10 \log_{10} \frac{I}{I_{0}} \Rightarrow \frac{I}{I_{0}} = 10^{3} \frac{r_{0}^{2}}{r^{2}} = 10^{3} r_{0} = r \times \sqrt{10^{3}} r_{0} = 20 \times 10 \sqrt{10} = 200 \sqrt{10} r_{0} = 4 \times 50 \sqrt{10} X = 4 .$