2. Solved Papers
  3. MGIMS WARDHA

MGIMS WARDHA MGIMS WARDHA Solved Paper-2005

  • question_answer
    A train approaches a stationary observer the velocity of train being\[\frac{1}{20}\]of the velocity of sound. A sharp blast is blown with the whistle of the engine at equal intervals of a second.         The interval between the successive blasts as heard by the observer?

    A) \[\frac{1}{20}\] sec                         

    B) \[\frac{1}{20}\] min

    C) \[\frac{19}{20}\] sec                                      

    D) \[\frac{19}{20}\] min

    Correct Answer: C

    Solution :

    The blast is blows at an interval of 1 sec,  so frequency\[=1\text{ }Hz\]. \[\therefore \] Frequency heard by the observer \[{{n}^{1}}=\frac{\upsilon }{\upsilon -{{\upsilon }_{s}}}.n\] \[=\frac{\upsilon }{\upsilon -\frac{\upsilon }{20}}\times 1\] \[=\frac{\upsilon }{19\frac{\upsilon }{20}}=\frac{20}{19}Hz\]       Therefore,   observed   time interval between two successive blasts \[=\frac{1}{20/19}=\frac{19}{20}\sec \] \[\therefore \]                  \[C\propto R\]


You need to login to perform this action.
You will be redirected in 3 sec spinner