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JCECE Medical JCECE Medical Solved Paper-2004

  • question_answer
    A stretched string is 1 m long. Its mass per unit length is 0.5 g/m. It is stretched with a force of  20 N. It is plucked at a distance of 25 cm from one end. The frequency of note emitted by it will be:  

    A)  400 m       

    B)  300 Hz

    C)  200 Hz       

    D)  100 Hz

    Correct Answer: C

    Solution :

     As stretched string of 1m long is plucked at a distance of 25 cm from end, so there will be 2 loops. Frequency,    \[n=\frac{p}{2l}\sqrt{\frac{T}{m}}\] Given, \[p=2,l=1\,m,T=20N,\] \[m=0.5\times {{10}^{-3}}kg/m\] \[\therefore \] \[n=\frac{2}{2\times 1}\sqrt{\frac{20}{0.5\times {{10}^{-3}}}}\] \[\Rightarrow \] \[n=200\,Hz\]


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