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CMC Medical CMC-Medical VELLORE Solved Paper-2015

  • question_answer
    For a wire, the value of \[\mu \] is \[5\times {{10}^{-3}}kg\,{{m}^{-1}}.\] The wire is streched between two rigid supports with a tension of 450 N. The wire resonates at a frequency of 420 Hz. The next higer frequency at which the same wire resonates is 490 Hz, then the length of the wire will be

    A)  1.9 m                                   

    B)  1.8 m   

    C)  2.0 m                                   

    D)  2.1 m

    E)  3.2m

    Correct Answer: D

    Solution :

                    Suppose, nth harmonic of frequency 420 Hz equals to (n + 1) th harmonic of 490 Hz. then \[420=\frac{n}{2l}\sqrt{\frac{T}{\mu }}\]                               ?(i) and        \[490=\frac{n+1}{2l}\sqrt{T/\mu }\]        ...(ii) On dividing Eq. (ii) by Eq. (i), we get                 \[\frac{490}{420}=\frac{n+1}{n}\] \[\Rightarrow \]               \[n=6\] On putting \[n=6\]in Eq (i), we get \[420=\frac{6}{2l}\sqrt{\frac{450}{5\times {{10}^{-3}}}}\Rightarrow L=\frac{900}{420}=2.1\,m\]


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