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Haryana PMT Haryana PMT Solved Paper-2003

  • 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 plucked at a distance of 25 cm from one end. The frequency of note emitted by it will be :

    A)  400 Hz                                 

    B)  300 Hz

    C)  200 Hz                                 

    D)  100 Hz

    Correct Answer: C

    Solution :

                    Using relation \[f=\frac{p}{2l}\sqrt{\frac{T}{m}}\] (Since, string is plucked at 25 cm hence, it will vibrate in two segments\[p=2\])                 \[f=\frac{2}{2\times 1}\sqrt{\frac{20}{0.5\times {{10}^{-3}}}}=200Hz\]


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