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

A point source of power `50pi` watts is producing sound waves of frequency `1875Hz`. The velocity of sound is `330m//s`, atmospheric pressure is `1.0xx10^(5)Nm^(-2)`, density of air is `(400)/(99pi)kgm^-3`. Then the displacement amplitude at `r=sqrt(330)`m from the point source is

A

`0.5 mum`

B

`0.2 mum`

C

`1 mum`

D

`2 mum`

Text Solution

Verified by Experts

`P_0=BKS_0 , k=(2pi)/lambda , lambda=v/f ,v=sqrt(B/rho)`
Using above, we get
`S_0=P_0/(2rhovpif)=5/(2xx1xx330xx3.14xx1875)`
`~=1mu` meter.
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