Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
A car sounding a horn of frequency $1000 \mathrm{~Hz}$ passes a stationary observer. The ratio of frequencies of the horn noted by the observer before and after passing the car is $11: 9$. If the speed of sound is ' $v$ ', the speed of the car is
PhysicsWaves and SoundMHT CETMHT CET 2023 (11 May Shift 1)
Options:
  • A v
  • B $\frac{\mathrm{v}}{2}$
  • C $\frac{\mathrm{v}}{5}$
  • D $\frac{\mathrm{v}}{10}$
Solution:
2858 Upvotes Verified Answer
The correct answer is: $\frac{\mathrm{v}}{10}$
Frequency of a source moving towards a stationary listener is $\mathrm{n}_{\mathrm{b}}=\left(\frac{\mathrm{v}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}\right) \mathrm{n}$

Frequency of a source moving towards a stationary listener is $\mathrm{n}_{\mathrm{a}}=\left(\frac{\mathrm{v}}{\mathrm{v}+\mathrm{v}_{\mathrm{s}}}\right) \mathrm{n}$

Taking the ratio
$$
\frac{\mathrm{n}_{\mathrm{b}}}{\mathrm{n}_{\mathrm{a}}}=\left(\frac{\mathrm{v}+\mathrm{v}_{\mathrm{s}}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}\right)
$$
$\begin{aligned} & \frac{11}{9}=\left(\frac{\mathrm{v}+\mathrm{v}_{\mathrm{s}}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}\right) \\ & 11 \mathrm{v}+11 \mathrm{v}_{\mathrm{s}}=9 \mathrm{v}-9 \mathrm{v}_{\mathrm{s}} \\ & 2 \mathrm{v}=20 \mathrm{v}_{\mathrm{s}} \\ & \mathrm{v}_{\mathrm{s}}=\frac{1}{10} \mathrm{v}\end{aligned}$

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.