When an observer moves towards a stationary source with velocity ' $\mathrm{V}_1$ ', the appearent frequency of emitted note is ' $\mathrm{F}_1$ '. When observer moves away from stationary source with velocity ' $\mathrm{V}_1$ ' the appearent frequency is ' $F_2$ '. If ' $v$ ' is velocity of sound in air and $\frac{F_1}{F_2}=2$, then $\frac{V}{V_1}$ is equal to

The Doppler effect for sound waves gives the apparent frequency $F'$ when an observer moves relative to a stationary source:

$F' = F \left( \frac{v \pm V_o}{v} \right)$

where $F$ is the source frequency, $v$ the speed of sound, and $V_o$ the observer's speed.

When the observer approaches the source, the apparent frequency is $F_1 = F \left( \frac{v + V_1}{v} \right)$.

When moving away, $F_2 = F \left( \frac{v - V_1}{v} \right)$.

Given $\frac{F_1}{F_2} = 2$, substitute the expressions:

$\frac{v + V_1}{v - V_1} = 2$

Solving for $\frac{v}{V_1}$:

$2(v - V_1) = v + V_1$
$2v - 2V_1 = v + V_1$
$v = 3V_1$

Thus $\frac{v}{V_1} = 3$, corresponding to option C.