A) 1.4 m/s
B) 3.4 m/s
C) 1.7 m/s
D) 2.1 m/s
Correct Answer: B
Solution :
[b] \[{{f}_{1}}={{f}_{0}}\left( \frac{{{V}_{0}}}{{{V}_{0}}-V} \right){{f}_{2}}={{f}_{0}}\left( \frac{{{V}_{0}}}{{{V}_{0}}+V} \right)\] \[{{f}_{1}}-{{f}_{2}}={{f}_{0}}{{V}_{0}}\left( \frac{1}{{{V}_{0}}-V}-\frac{1}{{{V}_{0}}+V} \right)\] \[={{f}_{0}}{{V}_{0}}\left( \frac{{{V}_{0}}+V-{{V}_{0}}+V}{V_{0}^{2}-{{V}^{2}}} \right)\] \[={{f}_{0}}{{V}_{0}}\times \frac{2V}{V_{0}^{2}}={{f}_{0}}\frac{2V}{{{V}_{0}}}\] Given \[\frac{2V{{f}_{0}}}{{{V}_{0}}}=0.02\times {{f}_{0}}\Rightarrow V=0.01\,{{V}_{0}}=3.4\,m/s\]You need to login to perform this action.
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