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If a source emitting waves of frequency $f$ moves towards an observer with a velocity $\frac{v}{4}$ and the observer moves away from the source with a velocity $v / 6$, the apparent frequency as heard by the observer will be $(v=$ velocity of sound)
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Verified Answer
The correct answer is:
$\frac{10}{9} f$
When source and observer both are moving in the same direction and observer is ahead of source, then apparent frequency
$$
f^{\prime}=\frac{v-v_{o}}{v-v_{s}} f
$$
$$
=\frac{v-\frac{v}{6}}{v-\frac{v}{4}}=\frac{\frac{5 v}{6}}{\frac{3 v}{4}} f=\frac{10}{9} f
$$
$$
f^{\prime}=\frac{v-v_{o}}{v-v_{s}} f
$$
$$
=\frac{v-\frac{v}{6}}{v-\frac{v}{4}}=\frac{\frac{5 v}{6}}{\frac{3 v}{4}} f=\frac{10}{9} f
$$
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