JEE Advanced 2021 Paper 2 Physics: Paragraph 1

The observer is behind the source, so for the direct sound the source recedes from him: $f_1=\dfrac{f_0v}{v+u}=\dfrac{850\times340}{360}=802.78\ \mathrm{Hz}$. The wall is a stationary receiver towards which the source approaches, so it receives $f_w=\dfrac{f_0v}{v-u}=\dfrac{850\times340}{320}=903.13\ \mathrm{Hz}$. A stationary wall re-radiates the same frequency, and the stationary observer hears the reflected sound at $f_2=903.13\ \mathrm{Hz}$. The beat frequency is $f_2-f_1=f_0v\left(\dfrac1{v-u}-\dfrac1{v+u}\right)=\dfrac{2f_0vu}{v^2-u^2}=\dfrac{2\times850\times340\times20}{115200}=100.35\ \mathrm{Hz}$. Note that the speed of the air is zero, so the wave speed relative to the ground is always $v$ and only the source motion enters.

The source and wall are unchanged, so the frequency of the wave travelling away from the source towards the observer is still $f_0v/(v+u)$ in the ground frame, with wavelength $(v+u)/f_0$, while the reflected wave has wavelength $(v-u)/f_0$ and travels from the wall towards the observer. The observer moves towards the wall; both waves travel away from the wall, so he moves against the direction of propagation in both cases, and relative to the air each wave reaches him with speed $v+w$, so each frequency is multiplied by $(v+w)/v$: $f_1'=\dfrac{f_0(v+w)}{v+u}$ and $f_2'=\dfrac{f_0(v+w)}{v-u}$. The beat frequency is therefore $f_2'-f_1'=\dfrac{v+w}{v}\times100.347=\dfrac{350}{340}\times100.347=103.30\ \mathrm{Hz}$.