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# 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 acceleration due to gravity, the work required to pull the hanging part on to the table isa) MgLb) MgL/3c) MgL/9d) MgL/18

Arun Kumar IIT Delhi
7 years ago
Hi

See the centre of mass of the chain will be raised with the chain.
$\\depth of centre of mass=L/6 \\W=raise in PE=m_{hanging}*L/6 \\=>W=\frac{Mg}{3}*\frac{L}{6} \\=>W=\frac{MgL}{18}$

Thanks & Regards, Arun Kumar, Btech,IIT Delhi, Askiitians Faculty