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JIPMER Jipmer Medical Solved Paper-2000

  • question_answer
    A uniform chain of length L and mass M is lying on a smooth table and one-third of its length is hanging vertically down over the edge of the table. If g is the acceleration due to gravity, the work required to pull the hanging part on to the table is:

    A) \[\frac{MgL}{18}\]                                          

    B) \[\frac{MgL}{9}\]                            

    C) \[\frac{MgL}{3}\]                            

    D)        \[MgL\]

    Correct Answer: A

    Solution :

    There will be the change in position of centre of gravity, which was \[L\text{/}6\] down now is shifted to the top of table hence, Work done\[=\frac{M}{3}\times g\times \frac{L}{6}=\frac{MgL}{18}\]


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