KEAM 2019 Physics and Chemistry Paper

Case I When train is moving towards a stationary observer. According to Doppler's effect of sound, apparent frequency heard, $n' = \frac{n v}{v - v_{s}}$ where, $n =$ frequency of source, $v =$ speed of sound and $v_{s} =$ speed of source. $\Rightarrow\: v_{s} = v\left[1 - \frac{n}{n'}\right]$.....(i) Case II When train is leaving the standing observer. Again, according to the Doppler's effect of sound, Apparent frequency heard, $n' = \frac{n v}{v + v_{s}} \Rightarrow v_{s} = v\left[\frac{n}{n'} - 1\right]$ .......(ii) From Eqs. (i) and (ii), we get $\frac{n}{n'} + \frac{n}{n'} = 2$ Given, $n' = 216\,\text{Hz}$ and $n' = 184\,\text{Hz}$ Substituting these values in the above equation, we get $n = \frac{2 \times 216 \times 184}{216 + 184} \approx 200\,\text{Hz}$