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  3. CEE Kerala Engineering

CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2009

  • question_answer
    From a circular ring of mass M and radius R, an arc corresponding to a\[{{90}^{o}}\]sector is removed. The moment of inertia of the remaining part of the ring about an axis passing through the centre of the ring and perpendicular to the plane of the ring is k times\[M{{R}^{2}}\]. Then the value of k is

    A)  \[\frac{3}{4}\]                  

    B)         \[\frac{7}{8}\]

    C)  \[\frac{1}{4}\]                  

    D)         \[1\]

    E)  \[\frac{1}{8}\]

    Correct Answer: A

    Solution :

    The moment of inertia of circular ring \[=M{{R}^{2}}\] The moment of inertia of removed sector                 \[=\frac{1}{4}M{{R}^{2}}\] The moment of inertia of remaining part                 \[=M{{R}^{2}}-\frac{1}{4}M{{R}^{2}}\]                 \[=\frac{3}{4}M{{R}^{2}}\] According to question, the moment of inertia of the remaining part                 \[=kM{{R}^{2}}\] then          \[k=\frac{3}{4}\]


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