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 sound source producing waves of frequency 300 Hz and wavelength 1 m observer is stationary, while source is going away with the velocity $30 \mathrm{~m} / \mathrm{s}$, then apparent frequency heared by the observer is
PhysicsWaves and SoundJIPMERJIPMER 2014
Options:
  • A $270\mathrm{~Hz}$
  • B $273\mathrm{~Hz}$
  • C $383\mathrm{~Hz}$
  • D $300\mathrm{~Hz}$
Solution:
2621 Upvotes Verified Answer
The correct answer is: $273\mathrm{~Hz}$
Apparent frequency $n^{\prime}=n\left(\frac{v}{v+v_s}\right)$
$\begin{aligned} v & =\text { velocity of sound, } \\ v_s & =\text { velocity of source }\end{aligned}$
$\therefore \quad n^{\prime}=300\left(\frac{332}{332+30}\right)=273 \mathrm{~Hz}$

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.