A siren placed at a railway platform is emitting sound of frequency $5 kHz$ . A passenger sitting in a moving train $A$ records a frequency of $5.5 kHz$ , while the train approaches the siren. During his return journey in a different train $B$ he records a frequency of $6.0 kHz$ while approaching the same siren.The ratio of the velocity of train $B$ to that train $A$ is

\frac{242}{252}

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2

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\frac{5}{6}

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\frac{11}{6}

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Solution

The correct option is B $2$
Using the formula $f^{'} = f (\frac{v + v_{0}}{v})$

we get,
$5.5 = 5 (\frac{v + v_{A}}{v}) . . . (i)$ and
$6.0 = 5 (\frac{v + v_{B}}{v}) . . . (i i)$

From $(i)$ and $(i i)$
$\frac{v_{B}}{v_{A}} = \frac{0.2 v}{0.1 v} = 2$

$∴ (b)$

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Similar questions

Q. A siren placed at a railway platform is emitting sound of frequency

5 k H z

. A passenger sitting in a moving train A records a frequency of

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while the train approaches the siren. During his return journey in a different train

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he records a frequency of

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while approaches the same siren. The ratio of the velocity of train

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A siren placed at a railway platform is emitting sound of frequency 5 kHz. A passenger sitting in a moving train A, records a frequency of 5.5 kHz while the train approaches the siren. During his return journey in a different train B, he records a frequency of 6.0 kHz while approaching the same siren. The ratio of the velocity of train B to that of train A is

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. The apparent frequency of sound as heard by a passenger at the rear end of the train is

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The correct option is B 2Using the formula f′=f(v+v0v) we get, 5.5=5(v+vAv) ...(i) and 6.0=5(v+vBv) ...(ii) From (i) and (ii) vBvA=0.2v0.1v=2 ∴ (b)

The correct option is B $2$
Using the formula $f^{'} = f (\frac{v + v_{0}}{v})$

we get,
$5.5 = 5 (\frac{v + v_{A}}{v}) . . . (i)$ and
$6.0 = 5 (\frac{v + v_{B}}{v}) . . . (i i)$

From $(i)$ and $(i i)$
$\frac{v_{B}}{v_{A}} = \frac{0.2 v}{0.1 v} = 2$

$∴ (b)$