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Uttarakhand PMT Uttarakhand PMT Solved Paper-2008

  • question_answer
    Moment of inertia of circular loop of radius R about the axis of rotation parallel to horizontal diameter at a distance R/2 from it is

    A)  \[M{{R}^{2}}\]        

    B)  \[\frac{1}{2}M{{R}^{2}}\]MR

    C)  \[2M{{R}^{2}}\]        

    D)  \[\frac{3}{4}M{{R}^{2}}\]

    Correct Answer: D

    Solution :

     Applying theorem of parallel axis \[I={{I}_{CM}}+M{{\left( \frac{R}{2} \right)}^{2}}\] \[I=\frac{1}{2}M{{R}^{2}}+\frac{M{{R}^{2}}}{4}\] \[I=\frac{3}{4}M{{R}^{2}}\]


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