RAJASTHAN ­ PET Rajasthan PET Solved Paper-2008

  • question_answer
    The moment of inertia of a circular loop of radius R, at a distance of R/2 around a rotating axis parallel to horizontal diameter of loop is

    A)  \[M{{R}^{2}}\]               

    B)  \[\frac{1}{2}M{{R}^{2}}\]

    C)  \[2M{{R}^{2}}\]           

    D)  \[\frac{3}{4}M{{R}^{2}}\]

    Correct Answer: D

    Solution :

     According to 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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