A uniform cable of mass 'M' and length 'L' is placed on a horizontal surface such that its `((1)/(n))^(th)` part is hanging below the edge of the cable upto the surface, the work done should be :

Options

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

Answer

To solve the problem, we need to determine the work done to lift the hanging part of a uniform cable back onto a horizontal surface. Let's break down the solution step by step.

### Step-by-Step Solution:

1. **Identify the Length and Mass of the Hanging Part**:
- The total length of the cable is \( L \).
- The fraction of the cable that is hanging is \( \frac{1}{n} \), so the length of the hanging part is:
\[ L_h = \frac{L}{n} \]

2. **Calculate the Mass of the Hanging Part**:
- The mass of the entire cable is \( M \).
- The mass per unit length of the cable is:
\[ \text{mass per unit length} = \frac{M}{L} \]
- Therefore, the mass of the hanging part is:
\[ m_h = \frac{M}{L} \cdot L_h = \frac{M}{L} \cdot \frac{L}{n} = \frac{M}{n} \]

3. **Determine the Center of Mass of the Hanging Part**:
- The center of mass of the hanging part, which is a uniform cable, is located at half of its length from the top of the hanging part. Thus, the height of the center of mass from the table is:
\[ h = \frac{L_h}{2} = \frac{1}{2} \cdot \frac{L}{n} = \frac{L}{2n} \]

4. **Calculate the Initial Potential Energy of the Hanging Part**:
- The potential energy \( U \) of the hanging part when it is at height \( -h \) (below the reference level) is given by:
\[ U_i = -m_h \cdot g \cdot h = -\left(\frac{M}{n}\right) \cdot g \cdot \left(-\frac{L}{2n}\right) = \frac{M g L}{2n^2} \]

5. **Final Potential Energy**:
- When the entire cable is on the table, the potential energy \( U_f \) is zero since it is at the reference level:
\[ U_f = 0 \]

6. **Calculate the Work Done**:
- The work done \( W \) to lift the hanging part back onto the table is equal to the change in potential energy:
\[ W = U_f - U_i = 0 - \frac{M g L}{2n^2} = -\left(-\frac{M g L}{2n^2}\right) = \frac{M g L}{2n^2} \]

### Final Answer:
The work done to lift the hanging part of the cable back onto the table is:
\[ W = \frac{M g L}{2n^2} \]