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JEE Main & Advanced JEE Main Paper (Held On 12-Jan-2019 Evening)

  • question_answer
    The moment of inertia of a solid sphere, about an axis parallel to its diameter and at a distance of x from it, is I(x). Which one of the graphs represents the variation of I(x) with x correctly? [JEE Main Online Paper Held On 12-Jan-2019 Evening]

    A)                    

    B)     

    C)                               

    D)

    Correct Answer: B

    Solution :

    The moment of inertia I(x) at distance x is \[I(x)=\frac{2}{5}M{{R}^{2}}+M{{x}^{2}}\] \[{{x}^{2}}=\frac{1}{M}(I(x)-\frac{2}{5}M{{R}^{2}})\] This equation resembles the standard equation of parabola i.e., \[{{(x-h)}^{2}}=4p(y-k)\] Hence the curve will be parabolic.


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