A) 5.7 km/hour
B) 7.2 km/hour
C) 10.3 km/hour
D) 44.8 km/hour
Correct Answer: A
Solution :
: When source is approaching observer \[n=\frac{vn}{(v-{{v}_{s}})}\] When source is receding from observer, \[n=\frac{vn}{(v+{{v}_{s}})}\] \[\therefore \]\[n-n=\frac{2nv{{v}_{s}}}{{{v}^{2}}}=\frac{2n{{v}_{s}}}{v}=\frac{n-n}{n}=\frac{1}{100}\] \[\therefore \]\[\frac{2n{{v}_{s}}}{nv}=\frac{1}{100}\] Or \[2{{v}_{s}}=\frac{v}{100}=\frac{320}{100}=3.2\]or\[{{v}_{s}}=\frac{3.2}{2}=1.6\,m/s\] Or \[{{v}_{s}}=\frac{1.6\times 3600}{1000}\frac{km}{h}=5.76\frac{km}{h}\] Since\[v>>{{v}_{s}},v_{s}^{2}\]can be ignored \[\therefore \] \[n-n=\frac{2nv{{v}_{s}}}{{{v}^{2}}}=\frac{2n{{v}_{s}}}{v}\frac{n-n}{n}=\frac{1}{100}\] \[\therefore \] \[\frac{2n{{v}_{s}}}{nv}=\frac{1}{100}\] Or \[2{{v}_{s}}=\frac{v}{100}=\frac{320}{100}=3.2\]or \[{{v}_{s}}=\frac{3.2}{2}=1.6\,m/s\] Or \[{{v}_{s}}=\frac{1.6\times 3600}{1000}\frac{km}{h}=5.76\frac{km}{h}\]
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