CET Karnataka Medical CET - Karnataka Medical Solved Paper-2011

  • question_answer
      A wire under tension vibrates with a fundamental frequency of 600 Hz. If the length of the wire is doubled, the radius is halved and the wire is made to vibrate under one-ninth the tension. Then the fundamental frequency will become

    A)  200 Hz         

    B)  300 Hz

    C)  600 Hz         

    D)  400 Hz

    Correct Answer: A

    Solution :

    Frequency of wire, \[f=\frac{1}{2lr}\sqrt{\frac{T}{\pi \rho }}\] So here,        \[\frac{{{f}_{1}}}{{{f}_{2}}}=\frac{{{l}_{2}}}{{{l}_{1}}}\times \frac{{{r}_{2}}}{{{r}_{1}}}\times \sqrt{\frac{{{T}_{1}}}{{{T}_{2}}}}\] \[\frac{600}{{{f}_{2}}}=\frac{2}{1}\times \frac{1}{2}\times \sqrt{\frac{T}{T/9}}\] \[\frac{600}{{{f}_{2}}}=3\] \[\Rightarrow \] \[{{f}_{2}}=\frac{600}{3}\] \[{{f}_{2}}=200Hz\]


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