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JEE Main & Advanced JEE Main Paper (Held On 12 April 2014)

  • question_answer
    A source of sound A emitting waves of frequency 1800 Hz is falling towards ground with a terminal speed v. The observer B on the ground directly beneath the source receives waves of frequency 2150 Hz. The source A receives waves, reflected from ground of frequency nearly: (Speed of sound  = 343 m/s)   [JEE Main Online Paper ( Held On 12 Apirl  2014 )

    A) 2150 Hz                                

    B) 2500 Hz

    C) 1800 Hz                                

    D) 2400 Hz

    Correct Answer: B

    Solution :

                    Given \[{{f}_{A}}=1800Hz\] \[{{v}_{t}}=v\] \[{{f}_{B}}=2150Hz\] Reflected wave frequency received by A, \[{{f}_{A}}'=?\] Applying doppler?s effect of sound, \[f'=\frac{{{v}_{s}}f}{{{v}_{s}}-{{v}_{t}}}\]here, \[{{v}_{t}}={{v}_{s}}\left( 1-\frac{{{f}_{A}}}{{{f}_{B}}} \right)\] \[=343\left( 1-\frac{1800}{2150} \right)\]\[{{v}_{t}}=55.8372m/s\] Now, for the reflected wave, \[\therefore \]\[f_{A}^{'}=\left( \frac{{{v}_{s}}+{{v}_{t}}}{{{v}_{s}}-{{v}_{t}}} \right){{f}_{A}}\]\[=\left( \frac{343+55.83}{343-55.83} \right)\times 1800\] \[=2499.44\approx 2500Hz\]


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