A) Zero
B) 0.5%
C) 5%
D) 20%
Correct Answer: D
Solution :
Given, \[{{v}_{o}}=\frac{v}{5}\] When observer moves towards the stationary source, then Hence, percentage increase \[=\left( \frac{64}{320}\times 100 \right)%=20%\]You need to login to perform this action.
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