A) 340
B) 330
C) 360
D) 320
Correct Answer: C
Solution :
When listener is moving towards the source then apparent frequency \[n=\frac{\upsilon +{{\upsilon }_{0}}}{\upsilon }\times n\Rightarrow 200=\frac{\upsilon +40}{\upsilon }\times n\] ?(i) where v = velocity of sound in air n = actual frequency of sound source Similarly, when listener is moving away, then \[160=\frac{\upsilon -40}{\upsilon }\times n\] ...(ii) From Eqs. (i) and (ii), we have \[\frac{200}{160}=\frac{\upsilon +40}{\upsilon -40}\] \[5\upsilon -200=4\upsilon +160\] \[\therefore \] \[\upsilon =360m/s\]You need to login to perform this action.
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