A uniform chain has a mass m and length l . It is held on a frictionless table with one - sixth of its length hanging over the edge. The work done in just pulling the hanging part back on the table is :

Options

A. `mgl/72`
B. `(mgl)/(36)`
C. `(mgl)/(12)`
D. `(mgl)/(6)`

Explanation

To find the work done in pulling the hanging part of the uniform chain back onto the table, we can follow these steps:

### Step 1: Understand the configuration of the chain
The uniform chain has a total length \( l \) and mass \( m \). One-sixth of its length is hanging over the edge of the table. Therefore, the length of the hanging part is:
\[ l_h = \frac{l}{6} \]
The length of the part on the table is:
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