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**Hint:**To solve this question, we need to use the Doppler Effect formula for the apparent frequency as received by a receiver. We have to substitute the speeds with the proper sign convention for the two cases. On solving the two equations formed, we will get the required speed of the aeroplane.

**Formula used:**The formula used to solve this question is given by

\[\nu ' = \left( {\dfrac{{v - {v_L}}}{{v - {v_S}}}} \right){\nu _0}\], here $\nu '$ is the apparent frequency as received by a receiver moving with a speed of ${v_L}$ due to a source of sound moving with a speed of ${v_S}$ and emitting a sound of frequency ${\nu _0}$. The velocity of the wave is $v$.

**Complete step-by-step solution:**

In the given question, the direction of the velocity of the aeroplane is not given to us. Therefore we assume that it is moving away from the radar.

From the Doppler’s Effect formula, we know that the apparent frequency received by an observer is given by

\[\nu ' = \left( {\dfrac{{v - {v_L}}}{{v - {v_S}}}} \right){\nu _0}\]

According to the question, the radio wave has a frequency of $\nu $. Also we know that radio waves are a type of electromagnetic wave which has a speed of $c$. Therefore substituting $v = c$, and ${\nu _0} = \nu $ in the above equation, we get

\[\nu ' = \left( {\dfrac{{c - {v_L}}}{{c - {v_S}}}} \right)\nu \]................(1)

In the first case, the radar is the source and the aeroplane is the receiver. Since the radar is stationary, so we have${v_S} = 0$. Also, since the aeroplane is moving with a speed of ${v_a}$ parallel to the signal, we have \[{v_L} = {v_a}\]. Substituting these in (1), we get the frequency received by the aeroplane as

\[\nu ' = \left( {\dfrac{{c - {v_a}}}{{c - 0}}} \right)\nu \]

$ \Rightarrow \nu ' = \left( {\dfrac{{c - {v_a}}}{c}} \right)\nu $ …………………..(2)

Now, this frequency will be reflected by the aeroplane. So this time the aeroplane becomes the source and the radar becomes the receiver. Now, the source is moving opposite to the signal with a speed of ${v_a}$, so we have \[{v_S} = - {v_a}\]. Also, the receiver, the radar is stationary. So we have ${v_L} = 0$. Substituting these in (1) we get the frequency of the reflected radio wave received by the radar as

$\nu '' = \left( {\dfrac{{c - 0}}{{c + {v_a}}}} \right)\nu '$

$ \Rightarrow \nu '' = \left( {\dfrac{c}{{c + {v_a}}}} \right)\nu '$

Putting (2) in the above, we get

$\nu '' = \left( {\dfrac{c}{{c + {v_a}}}} \right)\left( {\dfrac{{c - {v_a}}}{c}} \right)\nu $

$ \Rightarrow \nu '' = \left( {\dfrac{{c - {v_a}}}{{c + {v_a}}}} \right)\nu $

According to the question, the frequency of the reflected wave is greater than the original frequency, so $\nu '' > \nu $. But from the above equation, we have $\nu '' < \nu $. This means that our assumption of the direction of velocity of the aeroplane is incorrect. So the aeroplane must be moving towards the radar. So we replace \[{v_a}\] by $ - {v_a}$ in the above equation to get

$\nu '' = \left( {\dfrac{{c + {v_a}}}{{c - {v_a}}}} \right)\nu $

Now, we have

$\Delta \nu = \nu '' - \nu $

$ \Rightarrow \Delta \nu = \left( {\dfrac{{c + {v_a}}}{{c - {v_a}}}} \right)\nu - \nu $

Dividing by $\nu $ both the sides, we get

$\dfrac{{\Delta \nu }}{\nu } = \left( {\dfrac{{c + {v_a}}}{{c - {v_a}}}} \right) - 1$

$ \Rightarrow \dfrac{{\Delta \nu }}{\nu } = \dfrac{{2{v_a}}}{{c - {v_a}}}$

According to the question, $\left( {{v_a} < < c} \right)$. So we approximate $c - {v_a} \approx c$ in the above equation to get

$\dfrac{{\Delta \nu }}{\nu } = \dfrac{{2{v_a}}}{c}$

\[ \Rightarrow {v_a} = \dfrac{{c\Delta \nu }}{{2\nu }}\]

Thus, the velocity of the aeroplane is equal to $\dfrac{{c\Delta \nu }}{{2\nu }}$.

**Hence, the correct answer is option C.**

**Note:**The direction of velocity is from the source to the receiver. According to the sign convention, all the speeds which are parallel to the direction from the source to the receiver are taken to be positive and those in the opposite direction are taken to be negative. Do not apply the concept of relative velocity while substituting the speeds of the source and the receiver.

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