1. Given data

• Total mass of cable $= M$
• Total length of cable $= L$
• Hanging length $= \dfrac{L}{n}$

Since the cable is uniform, its mass per unit length is

$\lambda = \frac{M}{L}$

2. Mass of the hanging part

The hanging part has length $\dfrac{L}{n}$, so its mass is

$m = \lambda \cdot \frac{L}{n} = \frac{M}{L}\cdot \frac{L}{n} = \frac{M}{n}$

3. How much is this part to be lifted?

The hanging part extends vertically below the table edge from $0$ to $\dfrac{L}{n}$.

To bring it fully onto the horizontal surface, each small element at depth $x$ must be lifted upward by distance $x$.

So the work done equals the increase in gravitational potential energy of the hanging part.

4. Using center of mass method

The hanging part is a uniform segment of length $\dfrac{L}{n}$, so its center of mass is at

$\frac{1}{2}\cdot \frac{L}{n} = \frac{L}{2n}$

below the surface.

To bring the whole hanging part onto the surface, its center of mass rises by

$\frac{L}{2n}$

Hence the increase in potential energy is

$W = m g \left(\frac{L}{2n}\right)$

Substitute $m = \dfrac{M}{n}$:

$W = \frac{M}{n} g \cdot \frac{L}{2n}$

$W = \frac{MgL}{2n^2}$

5. Match with the options

$\boxed{W = \frac{MgL}{2n^2}}$

This matches Option C.

6. Comparison with stored correct answer

Stored correct answer: C

My derived answer: C

So they agree.