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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 accelelation due to gravity, the work done to pull the hanging part on to the table is?`
7 years ago

Ramesh V
70 Points
```										unit length of rope is M/L
given L/3 part of rope is hanging
we have to solve using integration methods, lets say 'x' part of rope is consided with an element 'dx' which is pulled up with limits from 0 to L/3
Potential energy = mgh
here dE= integral of ( Mx/L*g*dx)
E = integral of ( M/L*g*x.dx) limits from 0 to L/3
P.E = MgL/18
----
regards
Ramesh
```
7 years ago
Abhishek Gupta
44 Points
```										Work Done = mass of chain hanged * g * distance of chain from centre of mass of hanged part of chainSo, work done = M/3 * g * L/3*2                (This length is taken L/6 NOT L/3 becoz distance is taken from centre of mass)So, WORK DONE = MgL/18  (ANSWER)
```
11 months ago
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