2. Solved Papers
  3. J & K CET Medical

J & K CET Medical J & K - CET Medical Solved Paper-2010

  • question_answer
    A source of sound is approaching an observer with speed of 30 ms-1 and the observer is approaching the source with a speed of 60 ms-1. Then the fractional change in the frequency of sound (speed of sound in air\[=330\text{ }m{{s}^{-1}}\]) is

    A) \[\frac{1}{3}\]

    B) \[\frac{3}{10}\]

    C) \[\frac{2}{5}\]

    D) \[\frac{2}{3}\]

    Correct Answer: B

    Solution :

                     Given,  \[{{v}_{s}}=30\text{ }m{{s}^{-1}}\] \[{{V}_{o}}=60\text{ }m{{s}^{-1}}\] Apparent frequency\[v=\left[ \frac{v+{{v}_{o}}}{v-{{v}_{s}}} \right]\]                 \[v=v\left[ \frac{330+60}{330-30} \right]\]       \[v=v\left[ \frac{390}{300} \right]\]\[\Rightarrow \]\[v=\frac{39}{30}v\] \[\therefore \]Fractional change in frequency                 \[\frac{\frac{39}{30}v-v}{v}=\frac{9}{30}=\frac{3}{10}\]


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