When both the source of sound and the observer approach each other with a speed equal to $10 \%$ of the speed of sound,then the percentage change in frequency heard by the observer is nearly (in $\%$)

An observer moves towards a stationary source of sound,with a velocity one-fifth of the velocity of sound. What is the percentage increase in the apparent frequency?

$A$ 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 ..... $Hz$.

An observer is standing $500 \,m$ away from a vertical hill. Starting between the observer and the hill, a police van sounding a siren of frequency $1000 \,Hz$ moves towards the hill with a uniform speed. If the frequency of the sound heard directly from the siren is $970 \,Hz$, the frequency of the sound heard after reflection from the hill (in $Hz$) is about, (velocity of sound $= 330 \,m/s$):

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

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: