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JEE Main & Advanced Physics Wave Mechanics Question Bank Critical Thinking

  • question_answer
    A source producing sound of frequency 170 Hz is approaching a stationary observer with a velocity 17 ms?1. The apparent change in the wavelength of sound heard by the observer is (speed of sound in air = 340 ms?1) [EAMCET (Engg.) 2000]

    A)            0.1m                                        

    B)            0.2m

    C)            0.4m                                        

    D)            0.5m

    Correct Answer: A

    Solution :

                \[\lambda =\frac{v}{n}=\frac{340}{170}=2m,\,\,\,n'=\frac{340}{340-17}\times 170\Rightarrow n'=178.9Hz\]                    Now \[{\lambda }'=\frac{v}{{{n}'}}=\frac{340}{178.9}=1.9\]                    Þ \[\lambda -{\lambda }'=2-1.9=0.1\]


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