Q. A whistle giving out 450 Hz approaches a stationary observer at a speed of $33{ms}^{- 1}$. The frequency heard by the observer in Hz is [velocity of sound in air=$333ms^{- 1}$]
A) 409
B) 429
C) 517
D) 500
Explanation
(d) As source is moving towards observer,
$\therefore v^{'} = \frac{uv}{u - v_{1}} = \frac{333 \times 450}{333 - 30} = 499.5 = 500$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.