A) 340
B) 330
C) 360
D) 320
Correct Answer: C
Solution :
When listener is moving towards the source then apparent frequency \[n=\frac{\upsilon +{{\upsilon }_{o}}}{\upsilon }\times n\Rightarrow \,\,\,200=\frac{\upsilon +40}{\upsilon }\times n\] ... (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 \[=\frac{200}{160}=\frac{\upsilon +40}{\upsilon -40}\] \[=5\,\upsilon -200=4\upsilon +160\] \[\therefore \] \[\upsilon =360\,m/s\]You need to login to perform this action.
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