2. Solved Papers
  3. JIPMER

JIPMER Jipmer Medical Solved Paper-2006

  • question_answer
    A person is observing two trains one coming towards him and other leaving with the same velocity 4 m/s. If their whistling frequencies are 240 Hz each, fen the number of beats per second heard by the person will be : (if velocity of sound is 320 m/s)

    A) 3 

    B)                        6                             

    C)        9                             

    D)        zero

    Correct Answer: B

    Solution :

    Observed frequency of first train \[{{n}_{1}}=\frac{v}{v-{{v}_{s}}}\times n\]     \[=\frac{320}{320-4}\times 240\]     \[=\frac{320}{316}\times 240\]     \[=243\,Hz.\] Observed frequency of second train \[{{n}_{2}}=\frac{v}{v+{{v}_{s}}}n\]      \[=\frac{320}{320+4}\times 240\]     \[=237\,Hz.\] \[\therefore \] Number of beats\[=243-237=6\]


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