JEE Main & Advanced JEE Main Paper (Held on 09-4-2019 Morning)

  • question_answer A uniform cable of mass 'M' and length 'L' is placed on a horizontal surface such that its \[{{\left( \frac{1}{n} \right)}^{th}}\]part is hanging below the edge of the surface. To lift the hanging part of the cable up to the surface, the work done should be :             [JEE Main 9-4-2019 Morning]

    A) \[\frac{MgL}{{{n}^{2}}}\]             

    B) \[\frac{MgL}{2{{n}^{2}}}\]

    C) \[\frac{2MgL}{{{n}^{2}}}\]                       

    D) \[nMgL\]

    Correct Answer: B

    Solution :

    Mass of the hanging part \[=\frac{M}{n}\]             \[{{h}_{COM}}=\frac{L}{2n}\] work done \[W=mg{{h}_{COM}}=\left( \frac{M}{n} \right)g\left( \frac{L}{2n} \right)=\frac{MgL}{2{{n}^{2}}}\] Option [b]

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