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JEE Main & Advanced Sample Paper JEE Main Sample Paper-20

  • question_answer
    A ring of mass M and radius R lies in x-y plane with its centre at origin as shown. The mass distribution of ring is nun-uniform such that at any point P on the ring, the mass per unit length is given by \[\lambda ={{\lambda }_{0}}{{\cos }^{2}}\theta \] (where \[{{\lambda }_{0}}\] is a positive constant). Then the moment of inertia of the ring about z-axis is -            

    A)  \[M{{R}^{2}}\]                      

    B)  \[\frac{1}{2}M{{R}^{2}}\]

    C)  \[\frac{1}{2}\frac{M}{{{\lambda }_{0}}}R\]                             

    D)  \[\frac{1}{\pi }\frac{M}{{{\lambda }_{0}}}R\]

    Correct Answer: A

    Solution :

     Divide the ring into infinitely small lengths of mass \[d{{m}_{1}}.\] Even though mass distribution is non-uniform, each mass \[d{{m}_{1}}.\] is at same distance R from origin. \[\therefore \] MI o fringe bout z-axis is \[=d{{m}_{1}}{{R}^{2}}+d{{m}_{2}}{{R}^{2}}+........+d{{m}_{n}}{{R}^{2}}=M{{R}^{2}}\]


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