A) 200 Hz
B) 100 Hz
C) 50 Hz
D) 150 Hz
Correct Answer: B
Solution :
When source approaches to stationary observer then apparent frequency is \[n=\frac{\upsilon }{\upsilon -{{\upsilon }_{s}}}n\] (Given, \[{{\upsilon }_{s}}=\frac{\upsilon }{10},n=90Hz\]) So, \[n=\left( \frac{\upsilon }{\upsilon -\frac{\upsilon }{10}} \right)n=\frac{10\upsilon }{9\upsilon }90\] \[=100Hz\]You need to login to perform this action.
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