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
B
16dB
C
23dB
D
32dB

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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