A) \[0.5\times {{10}^{3}}Hz\]
B) \[1\times {{10}^{3}}Hz\]
C) \[0.25\times {{10}^{3}}\]
D) \[0.25\times {{10}^{3}}Hz\]
Correct Answer: B
Solution :
[b] \[f=\left( \frac{v+{{v}_{0}}}{v} \right){{f}_{1}}={{f}_{1}}+{{f}_{1}}\frac{{{v}_{0}}}{v}\] \[{{v}_{0}}=gt\] So \[f={{f}_{1}}+\left( \frac{{{f}_{1}}g}{v} \right)t\] Slope of graph \[=\frac{{{f}_{1}}g}{v}\frac{2\times {{10}^{3}}-{{f}_{1}}}{30}=\frac{({{f}_{1}})(10)}{300}\] Or \[{{f}_{1}}={{10}^{3}}Hz\]You need to login to perform this action.
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