CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2000

  • question_answer
    The moment of inertia of a ring about of one its diameter is\[I\]. What will be its moment of inertia about a tangent parallel to the diameter?

    A)  \[4I\]                   

    B)         \[2I\]

    C)  \[\frac{3}{2}I\]                

    D)         \[3I\]

    E)  \[I\]

    Correct Answer: D

    Solution :

    \[I=\frac{1}{2}M{{R}^{2}}\] According to theorem of parallel axes. \[\therefore \]  \[I=\frac{1}{2}M{{R}^{2}}+M{{R}^{2}}=\frac{3}{2}M{{R}^{2}}=3I\]


You need to login to perform this action.
You will be redirected in 3 sec spinner