CET Karnataka Engineering CET - Karnataka Engineering 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,               \[\frac{{{f}_{1}}}{{{f}_{2}}}=\frac{{{l}_{2}}}{{{l}_{1}}}\times \frac{{{r}_{2}}}{{{r}_{1}}}\times \sqrt{\frac{{{T}_{1}}}{{{T}_{2}}}}\]                                 \[=\frac{600}{{{f}_{2}}}=\frac{2}{1}\times \frac{1}{2}\times \sqrt{\frac{{{T}_{1}}}{T/9}}\]                                 \[\frac{600}{{{f}_{2}}}=3\] \[\Rightarrow \]                               \[{{f}_{2}}=\frac{600}{3}\]                                 \[{{f}_{2}}=200\,Hz\]


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