A chain of mass M is placed on a smooth table with 1/n of its length L hanging over the edge. The work done in pulling the hanging portion of the chain back to the surface of the table is:

Options

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

Answer

To solve the problem of calculating the work done in pulling the hanging portion of a chain back to the surface of the table, we can follow these steps:

### Step 1: Understand the Setup
We have a chain of mass \( M \) and length \( L \). A portion \( \frac{1}{n} \) of its length is hanging over the edge of a smooth table. Therefore, the length of the hanging part of the chain is:
\[ L_h = \frac{L}{n} \]

### Step 2: Calculate the Mass of the Hanging Portion
Since the mass of the chain is uniformly distributed, the mass of the hanging portion can be calculated as:
\[ M_h = \frac{M}{n} \]

### Step 3: Determine the Center of Mass of the Hanging Portion
The center of mass of the hanging portion (which is a uniform chain) is located at its midpoint. The distance from the table to the center of mass of the hanging portion is:
\[ h = \frac{L_h}{2} = \frac{L}{2n} \]

### Step 4: Calculate the Work Done Against Gravity
The work done in pulling the hanging portion back to the surface of the table is equal to the gravitational potential energy gained by the hanging mass when it is lifted to the table level. The work done \( W \) can be calculated using the formula:
\[ W = \text{Force} \times \text{Distance} \]
Here, the force is the weight of the hanging mass, which is given by:
\[ F = M_h \cdot g = \frac{M}{n} \cdot g \]
The distance the center of mass is raised is \( h \):
\[ h = \frac{L}{2n} \]
Thus, the work done is:
\[ W = F \cdot h = \left(\frac{M}{n} \cdot g\right) \cdot \left(\frac{L}{2n}\right) \]
\[ W = \frac{M \cdot g \cdot L}{2n^2} \]

### Final Answer
The work done in pulling the hanging portion of the chain back to the surface of the table is:
\[ W = \frac{M \cdot g \cdot L}{2n^2} \]

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