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 the acceleration due to gravity, the work required to pull the hanging part on to the table is

Options

A. `MgL`
B. `MgL//3`
C. `MgL//9`
D. `MgL//18`

Explanation

To solve the problem of finding the work required to pull the hanging part of a uniform chain onto the table, we can follow these steps:

### Step 1: Understand the Problem
We have a uniform chain of length \( L \) and mass \( M \). One-third of the chain's length is hanging vertically down over the edge of the table. We need to calculate the work done to pull this hanging part back onto the table.

### Step 2: Determine the Length and Mass of the Hanging Part
The length of the hanging part of the chain is:
\[ \text{Length of hanging part} = \frac{1}{3}L \]
The mass of the hanging part can be calculated using the mass per unit length:
\[ \text{Mass per unit length} = \frac{M}{L} \]
Thus, the mass of the hanging part is:
\[ \text{Mass of hanging part} = \frac{M}{L} \times \frac{1}{3}L = \frac{M}{3} \]

### Step 3: Calculate the Force Acting on the Hanging Part
The force acting on the hanging part due to gravity is given by:
\[ F = \text{mass} \times g = \frac{M}{3} \times g \]

### Step 4: Calculate the Work Done to Lift the Hanging Part
To find the work done, we need to consider the varying distance each infinitesimal segment of the hanging part is lifted.

Let’s consider a small segment of the chain of length \( dx \) at a distance \( x \) from the top of the hanging part. The distance this segment needs to be lifted to reach the table is \( x \).

The small work done \( dW \) to lift this segment is:
\[ dW = \text{Force} \times \text{Distance} = \left(\frac{M}{L} \cdot g \cdot x\right) dx \]

### Step 5: Set Up the Integral for Total Work Done
We need to integrate \( dW \) from \( x = 0 \) to \( x = \frac{1}{3}L \):
\[ W = \int_0^{\frac{1}{3}L} \left(\frac{M}{L} \cdot g \cdot x\right) dx \]

### Step 6: Perform the Integration
Calculating the integral:
\[ W = \frac{Mg}{L} \int_0^{\frac{1}{3}L} x \, dx \]
The integral of \( x \) is:
\[ \int x \, dx = \frac{x^2}{2} \]
Thus,
\[ W = \frac{Mg}{L} \left[ \frac{x^2}{2} \right]_0^{\frac{1}{3}L} = \frac{Mg}{L} \cdot \frac{1}{2} \left(\frac{1}{3}L\right)^2 \]
Calculating this gives:
\[ W = \frac{Mg}{L} \cdot \frac{1}{2} \cdot \frac{1}{9}L^2 = \frac{MgL}{18} \]

### Final Answer
The work required to pull the hanging part of the chain onto the table is:
\[ W = \frac{MgL}{18} \]