A) 4
B) 5
C) 6
D) 7
Correct Answer: C
Solution :
As B and listener are moving with the same velocity, frequency of the note from B remains unchanged. Frequency of the note from A, will be \[n=\left( \frac{\upsilon -{{\upsilon }_{1}}}{\upsilon -{{\upsilon }_{s}}} \right)n=\left( \frac{330-30}{330-0} \right)660=600Hz\] Hence, number of beats heard by the listener is \[600-594=6\]You need to login to perform this action.
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