A uniform chain of mass m & length L is kept on a smooth horizontal table such that `(L)/(n) `portion of the chaing hangs from the table. The work dione required to slowly bringsthe chain completely on the table is

Options

A. `(mgL)/(n)`
B. `(mgL)/(2)`
C. `(mgL)/(n^(2))`
D. `(mgL)/(2n^(2))`

Answer

To solve the problem of finding the work done required to slowly bring a uniform chain of mass \( m \) and length \( L \) completely onto a smooth horizontal table, where \( \frac{L}{n} \) portion of the chain hangs from the table, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Length of the Hanging Portion:**
The length of the chain that is hanging from the table is given by:
\[ L_h = \frac{L}{n} \]

2. **Determine the Mass of the Hanging Portion:**
Since the chain is uniform, the mass per unit length is:
\[ \text{Mass per unit length} = \frac{m}{L} \]
Therefore, the mass of the hanging portion is:
\[ m_h = \frac{m}{L} \cdot L_h = \frac{m}{L} \cdot \frac{L}{n} = \frac{m}{n} \]

3. **Calculate the Center of Mass of the Hanging Portion:**
The center of mass of the hanging portion is located at a distance of \( \frac{L_h}{2} \) from the edge of the table. Thus:
\[ \text{Distance of center of mass from the table} = \frac{L_h}{2} = \frac{1}{2} \cdot \frac{L}{n} = \frac{L}{2n} \]

4. **Determine the Height the Center of Mass is Raised:**
When the hanging portion is lifted onto the table, the center of mass of the hanging portion is raised by a height of \( \frac{L}{2n} \).

5. **Calculate the Work Done:**
The work done \( W \) to lift the mass \( m_h \) by a height \( h \) is given by the formula:
\[ W = m_h \cdot g \cdot h \]
Substituting the values we found:
\[ W = \left(\frac{m}{n}\right) \cdot g \cdot \left(\frac{L}{2n}\right) \]
Simplifying this expression gives:
\[ W = \frac{m \cdot g \cdot L}{2n^2} \]

### Final Answer:
The work done required to slowly bring the chain completely onto the table is:
\[ W = \frac{m \cdot g \cdot L}{2n^2} \]