A) \[1.5m{{s}^{-1}}\]
B) \[3\text{ }m{{s}^{-1}}\]
C) \[6m{{s}^{-1}}\]
D) \[~0.25\text{ }km{{s}^{-1}}\]
Correct Answer: C
Solution :
The apparent frequency when source approaches the stationary observer. \[{{v}_{1}}=v\left( \frac{\upsilon }{\upsilon -{{\upsilon }_{s}}} \right)\] The apparent frequency when source recesses the stationary observer \[{{v}_{2}}=v\left( \frac{\upsilon }{\upsilon +{{\upsilon }_{s}}} \right)\] Given: \[{{v}_{1}}-{{v}_{2}}=\frac{2}{100}{{v}_{1}}\] \[v\left( \frac{\upsilon }{\upsilon -{{\upsilon }_{s}}} \right)-v\left( \frac{\upsilon }{\upsilon +{{\upsilon }_{s}}} \right)=0.2v\left( \frac{\upsilon }{\upsilon -{{\upsilon }_{s}}} \right)\] or \[\frac{1}{\upsilon -{{\upsilon }_{s}}}-\frac{0.02}{\upsilon -{{\upsilon }_{s}}}=\frac{1}{\upsilon +{{\upsilon }_{s}}}\] or \[\frac{98}{\upsilon -{{\upsilon }_{s}}}=\frac{100}{\upsilon +{{\upsilon }_{s}}}\] \[\therefore \] \[{{\upsilon }_{s}}=\frac{2\times 300}{198}\approx 3m/s\]
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