A) \[11890\text{ }cycles/sec.\]
B) \[10890\text{ }cycles/sec.\]
C) \[12890\text{ }cycles/sec.\]
D) none of these
Correct Answer: B
Solution :
The source of approaching the observer. Hence, the observed frequency is given by \[n'=\left( \frac{V}{V-{{V}_{s}}} \right)\,n\] ?..(i) Here, \[V=330m/s\] \[{{V}_{g}}=1080km\,h=1080\times \frac{5}{18}\] \[=300\,m/s\] \[n=990/\sec .\] from equation (i) we get \[n'=\left( \frac{330}{330-300} \right)\times 990\] \[=10890\text{ }cycles/sec\] Thus, the frequency beared by the observer is \[=10890\text{ }cycles/sec\].You need to login to perform this action.
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