A man places a chain (of mass `m` and length `l`) on a table slowly. Initially, the lower end of the chain just touches the table. The main brings down the chain by length `l//2`. Work done by the man in this process is

Options

A. (a) `-mg1/2`
B. (b) `-(mgl)/(4)`
C. (c) `(-3mgl)/(8)`
D. (d) `-(mgl)/(8)`

Explanation

To solve the problem, we need to calculate the work done by the man while lowering the chain by a length of \( \frac{l}{2} \). The work done can be determined by calculating the change in potential energy of the chain.

### Step-by-Step Solution:

1. **Understanding the Initial Position of the Chain:**
- Initially, the lower end of the chain just touches the table, and the entire chain is hanging vertically. The center of mass of the chain is at a height of \( \frac{l}{2} \) from the table.

2. **Calculating Initial Potential Energy (PE_initial):**
- The potential energy of the chain can be calculated using the formula:
\[ PE_{\text{initial}} = mgh \]
- Here, \( h = \frac{l}{2} \) (the height of the center of mass), so:
\[ PE_{\text{initial}} = mg \left(\frac{l}{2}\right) = \frac{mgl}{2} \]

3. **Understanding the Final Position of the Chain:**
- After lowering the chain by \( \frac{l}{2} \), half of the chain (length \( \frac{l}{2} \)) is on the table, and the other half is still hanging.
- The center of mass of the hanging part (which is now \( \frac{l}{2} \)) is at a height of \( \frac{l}{4} \) from the table.

4. **Calculating Final Potential Energy (PE_final):**
- The mass of the hanging part is \( \frac{m}{2} \) (since half of the chain is hanging). The potential energy of this hanging part is:
\[ PE_{\text{final}} = \left(\frac{m}{2}\right) g \left(\frac{l}{4}\right) = \frac{mgl}{8} \]

5. **Calculating the Change in Potential Energy:**
- The work done by the man is equal to the change in potential energy:
\[ \Delta PE = PE_{\text{final}} - PE_{\text{initial}} \]
- Substituting the values:
\[ \Delta PE = \frac{mgl}{8} - \frac{mgl}{2} \]
- To combine these, convert \( \frac{mgl}{2} \) to a fraction with a denominator of 8:
\[ \frac{mgl}{2} = \frac{4mgl}{8} \]
- Now substituting:
\[ \Delta PE = \frac{mgl}{8} - \frac{4mgl}{8} = -\frac{3mgl}{8} \]

6. **Conclusion:**
- The work done by the man in lowering the chain is:
\[ W = -\frac{3mgl}{8} \]
- The negative sign indicates that the work is done against the gravitational force.

### Final Answer:
The work done by the man in this process is \( -\frac{3mgl}{8} \).