$A$ siren emitting sound of frequency $800 \; Hz$ is going away from a static listener with a speed of $30 \; m/s$. The frequency of the sound heard by the listener is... $Hz$ (take velocity of sound as $330 \; m/s$).

Two sources $A$ and $B$ are sending notes of frequency $680 \ Hz$. $A$ listener moves from $A$ towards $B$ with a constant velocity $u$. If the speed of sound in air is $340 \ ms^{-1}$, what must be the value of $u$ so that he hears $10$ beats per second (in $ms^{-1}$)?

When the source of sound moves away from a stationary listener,we obtain $f_{L} < f_{S}$. Why?

Two stationary sources each emit waves of wavelength $\lambda$. An observer moves from one source to the other with velocity $u$. The number of beats heard by the observer is:

An engine moving away from a vertical cliff blows a horn at a frequency $f$. Its speed is $0.5 \%$ of the speed of sound in air. The frequency of the reflected sound received at the engine is ............ $f$.

When the observer moves towards a stationary source with velocity $V_{1}$,the apparent frequency of the emitted note is $F_{1}$. When the observer moves away from the source with velocity $V_{1}$,the apparent frequency is $F_{2}$. If $V$ is the velocity of sound in air and $F_{1} / F_{2} = 2$,then $V / V_{1}$ is equal to: