A fire engine with its bell ringing with a frequency of 200 Hz is moving with a velocity of 54 kmph towards an observer at rest near a hut on fire. The apparent frequency of sound heard by the observer is (velocity of sound in air = 300 m/s)

210.5 Hz

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220.5 Hz

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230.5 Hz

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240.5 Hz

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Solution

The correct option is A 210.5 Hz
Given: $V_{0} = 0 m / s$
$V_{s} = 54 k m / h$ $= 54 \times \frac{1000}{3600}$ $= 15 m / s$
$V = 300 m / s$
$δ = 200 H z$
By Doppler effect formula:
$δ^{1} = δ (\frac{V - V_{0}}{V - V s})$
$= 200 (\frac{300 - 0}{300 - 15})$
$= \frac{200 \times 300}{285}$
$= 210.5 H z$

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A fire engine with its bell ringing with a frequency of 200 Hz is moving with a velocity of 54 kmph towards an observer at rest near a hut on fire. The apparent frequency of sound heard by the observer is (velocity of sound in air = 300 m/s)

The correct option is A 210.5 HzGiven: V0=0 m/sVs=54 km/h =54×10003600 =15 m/sV=300 m/sδ=200 HzBy Doppler effect formula:δ1=δ(V−V0V−Vs) =200(300−0300−15) =200×300285 =210.5 Hz