A car sounding a horn of frequency $1000 \mathrm{~Hz}$ passes an observer. The ratio of frequencies of the…

A car sounding a horn of frequency $1000 \mathrm{~Hz}$ passes an observer. The ratio of frequencies of the both noted by the observer before and after passing of the car is 11:9. If the speed of sound is $V$, the speed of the car is
  1. V
  2. $\frac{V}{10}$
  3. $\frac{V}{100}$
  4. $\frac{V}{5}$

Solution

$n_{\text {Before }}=\left(\frac{v}{v+v_c}\right) n$ And $n_{\text {After }}=\left(\frac{v}{v+v_c}\right) \cdot n$ $\frac{n_{\text {Before }}}{n_{\text {After }}}=\frac{11}{9}=\left(\frac{v+v_c}{v-v_c}\right)$ $\Rightarrow v_c \Rightarrow \frac{v}{10}$

Asked in: MHT CET 2022 (06 Aug Shift 2)

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