A) 22.2% increase
B) 22.2% decrease
C) 18.2% decrease
D) 18.2% increase
Correct Answer: A
Solution :
Speed of sound = v (say) Speed of source = speed of observer\[=\frac{v}{10}\] (given) Let natural frequency of sound = n Apparent frequency of sound \[=n=n\left( \frac{v-{{v}_{o}}}{v-{{v}_{S}}} \right)\] Since, source is moving in the direction of sound and observer moving towards source then \[{{v}_{O}}=\frac{v}{10}\]and \[{{v}_{S}}=\frac{v}{10}\] \[\therefore \] \[n=n\frac{\left( v+\frac{v}{10} \right)}{\left( v-\frac{1}{10} \right)}=\left( \frac{\frac{11v}{10}}{\frac{9v}{10}} \right)\] \[n=\frac{11}{9}n\] \[\Rightarrow \] \[n-n=\frac{11}{9}n-n=\frac{2}{9}n\] \[\Rightarrow \] \[\frac{n-n}{n}=\frac{2}{9}\] \[\therefore \] \[\frac{n-n}{n}\times 100%=\frac{2}{9}\times 100%\] \[=\frac{200}{9}%=22.22\] \[\therefore \] Percentage increase in apparent change in frequency of sound = 22.2%.
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