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 surface. To lift the hanging part of the cable upto the surface, the work done should be:

Options

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

Explanation

To solve the problem, we need to calculate the work done to lift the hanging part of a uniform cable of mass 'M' and length 'L' that is hanging off the edge of a horizontal surface. The hanging part is given as \( \frac{L}{n} \).

### Step-by-Step Solution:

1. **Identify the Length of the Hanging Part:**
The length of the hanging part of the cable is given by:
\[ L_h = \frac{L}{n} \]

2. **Determine the Mass of the Hanging Part:**
The mass of the hanging part can be calculated using the mass per unit length of the cable. The total mass of the cable is \( M \), and its total length is \( L \). Therefore, the mass per unit length \( \lambda \) is:
\[ \lambda = \frac{M}{L} \]
Thus, the mass of the hanging part \( M_h \) is:
\[ M_h = \lambda \cdot L_h = \frac{M}{L} \cdot \frac{L}{n} = \frac{M}{n} \]

3. **Calculate the Center of Mass of the Hanging Part:**
The center of mass of the hanging part is located at a distance of \( \frac{L_h}{2} \) from the edge of the surface. Therefore, the height of the center of mass from the surface is:
\[ h = \frac{L_h}{2} = \frac{1}{2} \cdot \frac{L}{n} = \frac{L}{2n} \]

4. **Calculate the Work Done to Lift the Hanging Part:**
The work done \( W \) to lift the hanging part to the surface is equal to the change in gravitational potential energy of the hanging mass. The change in potential energy \( \Delta PE \) is given by:
\[ W = M_h \cdot g \cdot h \]
Substituting the values we found:
\[ W = \left(\frac{M}{n}\right) \cdot g \cdot \left(\frac{L}{2n}\right) \]
Simplifying this expression gives:
\[ W = \frac{M g L}{2n^2} \]

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