$A$ source of sound and a listener are approaching each other with a speed of $40 \, m/s$. The apparent frequency of the note produced by the source is $400 \, cps$. Then,its true frequency (in $cps$) is (velocity of sound in air $= 360 \, m/s$).

If a source emitting waves of frequency $f$ moves towards an observer with a velocity $v/4$ and the observer moves away from the source with a velocity $v/6$,the apparent frequency as heard by the observer will be ($v =$ velocity of sound).

$A$ source with a frequency of $90 \, \text{vibrations/sec}$ moves towards a stationary observer with a velocity of $1/10 \, V$, where $V$ is the speed of sound. What frequency will the observer hear in $\text{vibrations/sec}$?

$A$ stationary observer receives sound from two identical tuning forks, one of which approaches and the other one recedes with the same speed $v$ (much less than the speed of sound). The observer hears $2 \; \text{beats/sec}$. The oscillation frequency of each tuning fork is $\nu_{0} = 1400 \; \text{Hz}$ and the velocity of sound in air is $c = 350 \; \text{m/s}$. The speed of each tuning fork is close to:

$A$ source is stationary and the observer is in motion along a line joining the source and the observer. If the frequency heard by the observer is $1\%$ higher than the true frequency,the ratio of velocity of the observer and that of sound in air is:

Match Column-$A$ with Column-$B$.

Column-$A$	Column-$B$
$(a)$ When source of sound and listener move towards wall.	$(i)$ Listener would hear a sound with same frequency as that of source.
$(b)$ When source of sound and listener move away from each other.	$(ii)$ Listener would hear a sound with frequency smaller than that of source.
	$(iii)$ Frequency of echo would be greater than that of source.