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Manipal Engineering Manipal Engineering Solved Paper-2014

  • question_answer
    A solid sphere of mass M, radius R and having moment of inertia about an axis passing through the centre of mass as I, is recast into a disc of thickness t, whose moment of inertia about an axis passing through its edge and perpendicular to its plane remains I. Then, radius of the disc will be

    A) \[\int_{0}^{\frac{3\pi }{2}}{\sin \left[ \frac{2x}{\pi } \right].dx,}\]                                             

    B) \[\frac{\pi }{2}(\sin 1+\cos 1)\]

    C)  4R                                         

    D) R

    Correct Answer: A

    Solution :

    \[\Delta {{S}_{1}}+\Delta {{S}_{2}}=\Delta {{S}_{3}}+\Delta {{S}_{4}}+\Delta {{S}_{5}}\]                             \[\text{J}\,\text{mo}{{\text{l}}^{\text{-}}}^{\text{1}}\]


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