The pitch of a whistle of an engine appears to drop by \(30 \%\) of original value when it passes a stationary observer. If the speed of sound in air is \(350 \mathrm{~ms}^{-1}\), then the speed of engine in \(\mathrm{ms}^{-1}\) is

Speed of sound \(=350 \mathrm{~m} / \mathrm{s}\)
Also,'
....(given)
\(\mathrm{n}^{\prime}=\mathrm{n}-\frac{30}{100} \mathrm{n}=\frac{70}{100} \mathrm{n}=0.7 \mathrm{n}\)
If \(n\) is the original (actual) frequency of the whistle and \(\mathrm{n}^{\prime}\) the frequency as the engine passes the stationary observer, then
\(\begin{array}{ll} & \frac{\mathrm{n}^{\prime}}{\mathrm{n}}=\left(\frac{\mathrm{v}}{\mathrm{v}+\mathrm{v}_{\text {cngine }}}\right) \\ \therefore \quad & 0.7=\frac{350}{350+\mathrm{v}_{\text {engine }}} \\ & 0.7 \mathrm{y}_{\text {engine }}+245=350 \\ \therefore \quad & \mathrm{v}_{\text {engine }}=150 \mathrm{~m} / \mathrm{s}\end{array}\)