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 calculating 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, which is \( \frac{L}{3} \), is hanging vertically down from the edge of the table. We need to find the work done to pull this hanging part back onto the table.

### Step 2: Determine the mass per unit length
The mass per unit length of the chain can be calculated as:
\[ \text{Mass per unit length} = \frac{M}{L} \]
This means for a length \( x \), the mass \( m_x \) is given by:
\[ m_x = \frac{M}{L} \cdot x \]

### Step 3: Calculate the force acting on the hanging part
The force acting on the small length \( dx \) of the chain that is hanging is the weight of that length, which can be expressed as:
\[ F = m_x \cdot g = \left(\frac{M}{L} \cdot x\right) \cdot g \]
where \( g \) is the acceleration due to gravity.

### Step 4: Set up the work done for a small segment
The work done \( dW \) to lift a small segment \( dx \) of the chain from a height \( x \) to the table is given by:
\[ dW = F \cdot dx = \left(\frac{M}{L} \cdot x \cdot g\right) dx \]

### Step 5: Integrate to find the total work done
To find the total work done to lift the entire hanging part of the chain, we need to integrate from \( 0 \) to \( \frac{L}{3} \):
\[ W = \int_0^{\frac{L}{3}} \left(\frac{M}{L} \cdot x \cdot g\right) dx \]

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

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