A train blowing its whistle moves with a constant velocity \[u\] away from an observer ground. The ratio of the natural frequency of the whistle to that measured by the observer is to be \[1.2\] . If the train is at rest and the observer moves away from it at the same velocity ratio would be given by :
${\text{A}}{\text{. 0}}{\text{.51}}$
${\text{B}}{\text{. 1}}{\text{.25}}$
${\text{C}}{\text{. 1}}{\text{.52}}$
${\text{D}}{\text{. 2}}{\text{.05}}$

Hint:
-Use the equation of the apparent frequency in terms of natural frequency for the condition of the train moving away from the observer.
-Calculate the ratio of the uniform velocity $(u)$ of the whistle of the train to the velocity of the source $({u_s})$ (when moving away) using the given ratio of the natural frequency of whistle is measured by its given observer.
-Use the equation of the apparent frequency in terms of natural frequency for the condition of the observer moving away. Find the required ratio for the source velocity and observer velocity is equal.

Formula used:
When the train is moving away, the apparent frequency, ${n_1} = \dfrac{u}{{u + {u_s}}}n$
Where,
 \[u\]is given by the uniform velocity of the train.
${u_s}$ = the source velocity.
$n = $ the natural frequency.
When the observer is moving away, the apparent frequency, ${n_2} = \dfrac{{u - {u_0}}}{u}n$, ${u_0}$ is the velocity of the observer.

Complete step by step answer:
Given, ${u_0} = {u_s}$.
The constant velocity of the train with blowing the whistle is given by $u$.
Now if the source velocity is ${u_s}$,
When the train is moving away, the apparent frequency, ${n_1} = \dfrac{u}{{u + {u_s}}}n$
$n = $ the natural frequency.
$ \Rightarrow \dfrac{{{n_1}}}{n} = \dfrac{u}{{u + {u_s}}}$
$ \Rightarrow \dfrac{n}{{{n_1}}} = 1 + \dfrac{{{u_s}}}{u}$
Given, $ \Rightarrow \dfrac{n}{{{n_1}}} = 1.2$
$\therefore 1 + \dfrac{{{u_s}}}{u} = 1.2$
_{On subtracting} \[1\] _{on both sides we get,}
$ \Rightarrow \dfrac{{{u_s}}}{u} = 0.2$
now, When the observer is moving away, the apparent frequency, ${n_2} = \dfrac{{u - {u_0}}}{u}n$, ${u_0}$ is the velocity of the observer.
The observer is moving with the same velocity as the source velocity, i.e. ${u_0} = {u_s}$
$\therefore {n_2} = \dfrac{{u - {u_s}}}{u}n$
$ \Rightarrow \dfrac{{{n_2}}}{n} = 1 - \dfrac{{{u_s}}}{u}$
Since, $\dfrac{{{u_s}}}{u} = 0.2$
$\therefore \dfrac{{{n_2}}}{n} = 1 - 0.2$
Let us subtract we get,
$ \Rightarrow \dfrac{{{n_2}}}{n} = 0.8$
$ \Rightarrow \dfrac{n}{{{n_2}}} = \dfrac{1}{{0.8}}$
$ \Rightarrow \dfrac{n}{{{n_2}}} = 1.25$
So the ratio is, $\dfrac{n}{{{n_2}}} = 1.25$

Hence the right answer is in option ${\text{B}}{\text{.}}$.

Note:In the first case, Here we take the equation of the apparent frequency by taking the velocity of the observer as zero, and since the train is moving away the source velocity is negative.
In the second case, we take the equation of the apparent frequency by taking the velocity of the source is zero.