A) \[1.5m{{s}^{-1}}\]
B) \[12m{{s}^{-1}}\]
C) \[6m{{s}^{-1}}\]
D) \[3m{{s}^{-1}}\]
Correct Answer: D
Solution :
When source approaches the observer, the apparent frequency heard by observer is \[n=n\left( \frac{v}{v-{{v}_{s}}} \right)\] ? (i) \[{{v}_{s}}=\]speed of source of sound. During it recession, apparent frequency \[n\,\,=n\left( \frac{v}{v-{{v}_{s}}} \right)\] ? (ii) Accordingly \[n=n\,\,=\frac{2}{100}n\] (given) \[\therefore \] \[n\left( \frac{v}{v-{{v}_{s}}} \right)=n\left( \frac{v}{v+{{v}_{s}}} \right)=\frac{2}{100}n\] or \[v\left[ \frac{v+{{v}_{s}}-v+{{v}_{s}}}{(v-{{v}_{s}})v+{{v}_{s}}} \right]=\frac{2}{100}\] or \[\frac{2v{{v}_{s}}}{(v-{{v}_{s}})(v+{{v}_{s}})}=\frac{2}{100}\] or \[100v{{v}_{s}}={{v}^{2}}-v_{s}^{2}\] But speed of sound in air\[v=300\text{ }m/s\] \[\therefore \] \[30000{{v}_{s}}={{(300)}^{2}}-v_{s}^{2}\] \[\Rightarrow \] \[v_{s}^{2}+30000{{v}_{s}}-90000=0\] \[\therefore \] \[{{v}_{s}}=\frac{-30000\pm \sqrt{{{(30000)}^{2}}+4\times 90000}}{2}\] \[=\frac{6}{2}=3m{{s}^{-1}}\] (taking \[+ve\] sign only)
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