2. Solved Papers
  3. BVP Medical

BVP Medical BVP Medical Solved Paper-2001

  • question_answer
    A tuning fork of frequency 90 Hz is sounded and moved towards an stationary observer with a velocity equal to one tenth of sound. The frequency of note heard by the observer is :

    A)  200 Hz                                 

    B)  100 Hz

    C)  50 Hz                                   

    D)  150 Hz

    Correct Answer: B

    Solution :

                               When source approaches to stationary observer then apparent frequency is \[n=\frac{\upsilon }{\upsilon -{{\upsilon }_{s}}}n\] (Given, \[{{\upsilon }_{s}}=\frac{\upsilon }{10},n=90Hz\]) So, \[n=\left( \frac{\upsilon }{\upsilon -\frac{\upsilon }{10}} \right)n=\frac{10\upsilon }{9\upsilon }90\] \[=100Hz\]


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