Q. A sound source is falling under gravity. At some time t=0, the detector lies vertically hallow sources at a depth $H\ $as shown in figure. If v is the velocity of sound and $f_{0}$ is frequency recorded after t=2s is

A) $f_{0}$
B) $\frac{f_{0}(v + 2g)}{v}$
C) $\frac{f_{0}(v - 2g)}{v}$
D) $f_{0}\left( \frac{v}{v - 2g} \right)$
Explanation

(d) From $v = u + at$
$v_{s} = 0 + g \times 2 = 2g$
As source is moving towards the observer.
$\therefore\ f = \frac{v}{v - v_{s}}f_{0} = \frac{vf_{0}}{v - 2g}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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