A) Zero
B) 0.5%
C) 5%
D) 20%
Correct Answer: D
Solution :
Given \[{{v}_{o}}=\frac{v}{5}\Rightarrow {{v}_{o}}=\frac{320}{5}=64\,m/s\] When observer moves towards the stationary source, then \[n'=\left( \frac{v+{{v}_{o}}}{v} \right)n\] \[n'=\left( \frac{320+64}{320} \right)\] \[n'=\left( \frac{384}{320} \right)n\] \[\frac{n'}{n}=\frac{384}{320}\] Hence, percentage increase \[\left( \frac{n'-n}{n} \right)=\left( \frac{384-320}{320}\times 100 \right)%\] \[=\left( \frac{64}{320}\times 100 \right)%=20%\]
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