A chain of uniform mass `m` and length `L` is held on a frictionless table in such a way that its `(1)/(n)`th part is hanging below the edge of table. The work done to pull the hanging part of chain is `:-`

Options

A. `sqrtn`
B. `n`
C. `n^(-3)`
D. `n^(-2)`

Answer

To find the work done to pull the hanging part of the chain back onto the table, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Parameters**:
- Let the total mass of the chain be \( m \).
- The total length of the chain is \( L \).
- The part of the chain hanging below the edge of the table is \( \frac{L}{n} \).

2. **Determine the Mass of the Hanging Part**:
- The mass of the hanging part of the chain can be calculated using the ratio of the lengths:
\[ m_{\text{hanging}} = \frac{m}{L} \cdot \frac{L}{n} = \frac{m}{n} \]

3. **Find the Center of Mass of the Hanging Part**:
- The center of mass of the hanging part (which is a uniform chain) is located at its midpoint. Therefore, the distance from the edge of the table to the center of mass of the hanging part is:
\[ h = \frac{L}{2n} \]

4. **Calculate the Initial Potential Energy**:
- The potential energy (PE) of the hanging part when it is hanging is given by:
\[ PE_{\text{initial}} = -m_{\text{hanging}} \cdot g \cdot h \]
Substituting the values:
\[ PE_{\text{initial}} = -\left(\frac{m}{n}\right) \cdot g \cdot \left(\frac{L}{2n}\right) = -\frac{m g L}{2n^2} \]

5. **Calculate the Final Potential Energy**:
- When the entire chain is pulled onto the table, the potential energy of the chain becomes zero:
\[ PE_{\text{final}} = 0 \]

6. **Calculate the Work Done**:
- The work done to pull the hanging part of the chain is equal to the change in potential energy:
\[ W = PE_{\text{final}} - PE_{\text{initial}} = 0 - \left(-\frac{m g L}{2n^2}\right) = \frac{m g L}{2n^2} \]

### Final Answer:
The work done to pull the hanging part of the chain is:
\[ W = \frac{m g L}{2n^2} \]