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 is

The correct option is MgL/18.

Step-by-Step Derivation:

Let the total length of the uniform chain be L and its total mass be M.
The mass per unit length (linear mass density) of the chain, represented by $\lambda$, is given by:
$\lambda =\frac{M}{L}$

One-third of the chain's length hangs vertically over the edge of the table. Let us find the length and mass of this hanging section:
Length of the hanging part, $l=\frac{L}{3}$
Mass of the hanging part, $m=\lambda ·l=\left(\frac{M}{L}\right)·\left(\frac{L}{3}\right)=\frac{M}{3}$

Since the chain is uniform, the center of mass of the hanging part lies at its geometric midpoint. Therefore, the distance of the center of mass of the hanging portion below the edge of the table is:
$h=\frac{l}{2}=\frac{L}{3·2}=\frac{L}{6}$

The work required to pull the hanging part back onto the table is equal to the work done against gravity to lift its center of mass to the tabletop level. This is equal to the increase in potential energy of the hanging part:
$W=m·g·h$
Substitute the values of $m$ and $h$:
$W=\left(\frac{M}{3}\right)·g·\left(\frac{L}{6}\right)$
$W=\frac{MgL}{18}$

Thus, the work required to pull the hanging part on to the table is $\frac{MgL}{18}$.