Manipal Medical Manipal Medical Solved Paper-2005

  • question_answer
    An astronaut is approaching the moon. He sends out a radio signal of frequency 5000 MHz and the frequency of echo is different from that of the, original frequency by 100 kHz. His velocity of approach with respect to the moon is:

    A)  2 km/s      

    B)  3 km/s

    C)  4 km/s      

    D)  5 km/s

    Correct Answer: B

    Solution :

     Let the shift in frequency\[=\Delta n\] \[\Delta n=100\times {{10}^{3}}Hz\] \[\upsilon =\]relative speed Then,    \[\frac{\Delta n}{n}=2\frac{\upsilon }{c}\] \[\upsilon =\frac{\Delta n}{2n}c\] \[=\frac{100\times {{10}^{3}}\times 3\times {{10}^{8}}}{2\times 5000\times {{10}^{6}}}\] \[=3000\text{ }m/s\] \[=3km/s\]


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