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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

`0.5mum`

B

`0.2mum`

C

`1mum`

D

`2mum`

Text Solution

Verified by Experts

The correct Answer is:
C
`P_0=BKS`, `k=(2pi)/(lamda)`, `lamda=(v)/(f)`, `v=sqrt((B)/(rho))`
Using above, we get
`S_0=(P_0)/(2rhovpif)=(5)/(2xx1xx330xx3.14xx1875)`
`=1mu` meter
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