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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, work required to pull the hanging part on to the table is

A

MgL

B

`(MgL)/(3)`

C

`(MgL)/(9)`

D

`(MgL)/(18)`

Text Solution

Verified by Experts

The correct Answer is:
D
(d) Mass`(M)/(3)` has its centre of mass at `(L)/(6)` below the table surface.

`:. W=mgh=((M)/(3))(g)((L)/(6))=(MgL)/(18)`
`W_(1)=+W_(g)+W_(T)=0, W_(T)=0`
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