A) 18%
B) 24%
C) 6%
D) 12%
Correct Answer: D
Solution :
\[v'=v\left[ \frac{V}{V\pm {{V}_{s}}} \right]\] Frequency increases while approacting and decreases while receding. \[100\frac{\Delta v}{v}=2\left( {}^{{{V}_{s}}}/{}_{V} \right)(100)\] \[=200\left( \frac{20}{320} \right)=12\]You need to login to perform this action.
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