An observer moves towards a stationary source of sound with a velocity of one-fifth of the velocity of sound. The percentage increase in the apparent frequency is (in $\%$)

When both source and listener are approaching each other,the observed frequency of sound is given by (where $V$ is the speed of sound,$V_L$ and $V_S$ are the velocities of the listener and source respectively,and $n_0$ is the radiated frequency):

$A$ source and an observer move away from each other with a velocity of $10\; m/s$ with respect to the ground. If the observer finds the frequency of sound coming from the source as $1950\; Hz$,then the actual frequency of the source is .... $Hz$ (velocity of sound in air = $340\; m/s$).

$A$ whistle sends out $256$ waves in a second. If the whistle approaches the observer with a velocity equal to $\frac{1}{3}$ of the velocity of sound in air,calculate the number of waves per second the observer will receive.

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:

$A$ and $B$ are two sources generating sound waves. $A$ listener is situated at $C$. The frequency of the source at $A$ is $500 \, Hz$. $A$ now moves towards $C$ with a speed of $4 \, m/s$. The number of beats heard at $C$ is $6$. When $A$ moves away from $C$ with a speed of $4 \, m/s$,the number of beats heard at $C$ is $18$. The speed of sound is $340 \, m/s$. The frequency of the source at $B$ is ..... $Hz$.