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A point source of power 50pi watts is pr...

A point source of power `50pi` watts is producint sound waves of frequency `1875Hz`. The velocity of sound is `330m//s`, atmospheric pressure is `1.0xx10^(5)Nm^(-2)` density of air is `1.0kgm^(-3)`. Then pressure amplitude at `r=sqrt(330)m` from the point source is (using `pi=22//7`:

A

`10Nm^(-2)`

B

`15 Nm^(-2)`

C

`20 Nm^(-2)`

D

`5Nm^(-2)`

Text Solution

Verified by Experts

The correct Answer is:
D
`I=(P_(0)^(2))/(2rhoV)=P/(4pir^(2))=(P_(0)^(2))/(2rhoV)` where `P_(1) P_(0), V` are power, pressure amplitude and velocity respectively `rArr P_(0)=sqrt((PrhoV)/(2pir^(2)))=sqrt((50pixx1xx330)/(2pixx330))=5`
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