A uniform chain of mass m and length l overhangs a table with its two third part on the table. Fine the work to be done by a person to put the hanging part back on the table.

Explanation

To solve the problem of finding the work done by a person to put the hanging part of a uniform chain back on the table, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a uniform chain of mass \( m \) and length \( l \).
- Two-thirds of the chain is on the table, and one-third is hanging off the edge.
- We need to find the work done to lift the hanging part back onto the table.

2. **Identify the Length of the Hanging Part**:
- The length of the hanging part of the chain is \( \frac{1}{3}l \).

3. **Consider an Element of the Chain**:
- Let’s consider a small element of the chain of length \( dx \) at a distance \( x \) from the edge of the table.
- The mass of this small element \( dx \) can be expressed as:
\[ dm = \frac{m}{l} \cdot dx \]
- Here, \( \frac{m}{l} \) is the mass per unit length of the chain.

4. **Determine the Height to Lift**:
- The height \( h \) from which this small element \( dx \) needs to be lifted is equal to \( x \) (the distance from the table).

5. **Calculate the Work Done on the Element**:
- The work done \( dW \) to lift this small element \( dx \) is given by:
\[ dW = dm \cdot g \cdot h = \left(\frac{m}{l} \cdot dx\right) \cdot g \cdot x \]
- Therefore, we have:
\[ dW = \frac{mg}{l} \cdot x \cdot dx \]

6. **Integrate to Find Total Work Done**:
- To find the total work done \( W \), we integrate \( dW \) from \( x = 0 \) to \( x = \frac{1}{3}l \):
\[ W = \int_0^{\frac{1}{3}l} \frac{mg}{l} \cdot x \, dx \]
- This simplifies to:
\[ W = \frac{mg}{l} \int_0^{\frac{1}{3}l} x \, dx \]

7. **Evaluate the Integral**:
- The integral \( \int x \, dx \) evaluates to \( \frac{x^2}{2} \):
\[ W = \frac{mg}{l} \left[ \frac{x^2}{2} \right]_0^{\frac{1}{3}l} \]
- Substituting the limits gives:
\[ W = \frac{mg}{l} \cdot \frac{1}{2} \left(\frac{1}{3}l\right)^2 = \frac{mg}{l} \cdot \frac{1}{2} \cdot \frac{1}{9}l^2 \]

8. **Final Simplification**:
- Simplifying this expression results in:
\[ W = \frac{mg l}{18} \]

### Final Answer:
The work done by a person to put the hanging part back on the table is:
\[ W = \frac{mg l}{18} \]