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

Step 1: Find the mass of the hanging part of the chain.

The chain is uniform with total mass $M$ and total length $L$.

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

The length of the hanging portion is $l = \frac{L}{3}$.

$$m = \lambda \cdot \frac{L}{3}$$ $$m = \frac{M}{3}$$

Step 2: Find the center of mass of the hanging part.

For a uniform hanging segment of length $\frac{L}{3}$, the center of mass lies at its midpoint.

$$h = \frac{1}{2} \left(\frac{L}{3}\right)$$ $$h = \frac{L}{6}$$

Step 3: Calculate the work required using the center of mass.

The work done is equal to the potential energy gained by raising the mass of the hanging portion to the table level.

$$W = mgh$$ $$W = \left(\frac{M}{3}\right) g \left(\frac{L}{6}\right)$$ $$W = \frac{MgL}{18}$$

Step 4: Verify the result using integration.

Consider an infinitesimal element of length $dx$ at a depth $x$ below the table edge.

$$dm = \frac{M}{L} dx$$ $$dW = (dm)gx$$ $$dW = \frac{Mg}{L} x \, dx$$

Integrate from $x = 0$ to $x = \frac{L}{3}$.

$$W = \int_{0}^{L/3} \frac{Mg}{L} x \, dx$$ $$W = \frac{Mg}{L} \left[ \frac{x^2}{2} \right]_{0}^{L/3}$$ $$W = \frac{MgL}{18}$$

Answer: The work required to pull the hanging part onto the table is $\dfrac{MgL}{18}$.