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Question

The sound intensity level at a point $$4m$$ from the point source is $$10dB$$, then the sound level at a distance $$2m$$  from the same source will be :


A
26dB
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B
16dB
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C
23dB
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D
32dB
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Solution

The correct option is B $$16dB$$
Sound level at $$A$$ is $$\bot o \bot B$$
If the intensity of sound at $$A\  4\ cm$$ away from the source is $$I_{A}$$
then. 
$$\bot o =10\log \dfrac{I_{A}}{I_{o}}$$            Where $$I_{o}=10^{-12} w\ m^{-2}$$
$$\Rightarrow \bot =\log \dfrac{I_{A}}{10^{-12}}$$
$$\Rightarrow \dfrac{I_{A}}{10^{-12}}=10$$
$$\Rightarrow I_{A}=10^{-11}w m^{-2}$$
If the source (s) is emmitting the sound energy of power $$P$$
$$I_{A}=\dfrac{P}{4\times (4)^{2}}$$
$$\Rightarrow P=10^{-11}\times 64 \pi$$   $$w$$
$$\Rightarrow$$ Intensity of sound $$(I_{B})$$ ar point $$B, 2\ m$$
$$I_{B}=\dfrac{P}{4\pi (2)^{2}}$$
$$I_{B}=\dfrac{64\pi\times 10^{-11}}{16\pi}10\ m^{-2}$$
$$\Rightarrow I_{B}=4\times 10^{-11}\ wm^{-2}$$
$$\Rightarrow$$ Sound level at $$2m=10\log \left(\dfrac{I_{B}}{I}\right)$$
$$=10 \log \left(\dfrac{4\times 10^{-11}}{10^{-12}}\right)$$
$$=10\log 40$$
$$=16.02$$
Hence Sound level at $$2m$$ from sourceis $$16\ dB$$

Physics

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