A uniform chain of mass `M` and length `L` is lying on a frictionless table in such a way that its `1//3` parts is hanging vertically down. The work done in pulling the chain up the table is

Options

A. `(MgL)/(9)`
B. `(MgL)/(18)`
C. `(2MgL)/(27)`
D. `(2MgL)/(3)`

Answer

B

Explanation

If length `x` of the chain is pulled up on the table. Then the length of hanging part of the chain would be `((L)/(3)-x)` and its weight would be `(M)/(L)((L)/(3)-x)g`, If it is pulled up further by a distance `dx`, the work done in pulling up.
`rArrdW=(m)/(L)((L)/(3)-x)gdxrArr`
`W=int_0^(L/3)(M)/(L)((L)/(3)-x)gdx=(MgL)/(18)`
OR Required work done `=((M)/(3))g((L)/(6))=(MgL)/(18)`