Answer: Option C
:
C
The frequency that the observer receives directly from the source has frequency $n_1=500Hz$. As the observer and source both move towards the fixed wall with velocity u, the apparent frequency of the reflected wave coming from the wall to the observer will have frequency
$n_2=\left(\frac{V}{V-u}\right)500Hz$
where V is the velocity of sound wave in air. The apparent frequency of this reflected wave as heard by the observer will then be
$n_3=\left(\frac{V+u}{V}\right)n_2=\left(\frac{V+u}{V}\right)\left(\frac{V}{V-u}\right)500=\left(\frac{V+u}{V-u}\right)500$
It is given, that the number of beat per second in $n_3-n_1=10$
$\therefore (n_3-n_1)=10=\left(\frac{V+u}{V-u}\right)500-500=500[\frac{V+u}{V-u}-1]10=\frac{2\times u\times 500}{V-u}Hence,10V=1000u+10u=1010uPuttingu=4m/s,WehaveV=\frac{1}{10}\left[4040\right]=404m/s$