RAJASTHAN PMT Rajasthan - PMT Solved Paper-1996

  • question_answer
    A rocket is receding from earth with a velocity\[0.2c\]. It emits a signal of frequency\[4\times {{10}^{7}}Hz\]. The frequency as observed on earth will be:

    A)  \[4\times {{10}^{6}}Hz\]                             

    B)         \[3.3\times {{10}^{7}}Hz\]

    C)         \[3\times {{10}^{6}}\,Hz\]                          

    D)         \[5\times {{10}^{7}}Hz\]

    Correct Answer: B

    Solution :

    Apparent frequency \[v=v\left( 1-\frac{\upsilon }{c} \right)\] \[=4\times {{10}^{7}}\left( 1-\frac{0.2c}{c} \right)=3.2\times {{10}^{7}}Hz\]


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