A chain is held on a frictionless table with `1//n` th of its length hanging over the edge. If the chain has a length L and a mass M, how much work is required to pull the hanging part back on the table?

Options

A. `(MgL)/(5)`
B. `(MgL)/(50)`
C. `(MgL)/(18)`
D. `(MgL)/(10)`

Answer

To solve the problem of how much work is required to pull the hanging part of the chain back on the table, we can follow these steps:

### Step 1: Understand the Problem
We have a chain of total length \( L \) and mass \( M \), with \( \frac{1}{n} \) of its length hanging over the edge of a frictionless table. We need to find the work done to pull the hanging part back onto the table.

### Step 2: Determine the Length of the Hanging Part
The length of the hanging part of the chain is given by:
\[ L_h = \frac{L}{n} \]

### Step 3: Calculate the Mass of the Hanging Part
The mass of the hanging part can be calculated using the mass per unit length of the chain. The mass per unit length \( \lambda \) is:
\[ \lambda = \frac{M}{L} \]
Therefore, the mass of the hanging part \( M_h \) is:
\[ M_h = \lambda \cdot L_h = \frac{M}{L} \cdot \frac{L}{n} = \frac{M}{n} \]

### Step 4: Find 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 table. Thus, the height \( h \) of the center of mass from the table is:
\[ h = \frac{L_h}{2} = \frac{1}{2} \cdot \frac{L}{n} = \frac{L}{2n} \]

### Step 5: Calculate the Work Done Against Gravity
The work done \( W \) to pull the hanging part back onto the table is equal to the gravitational potential energy gained by the hanging part when it is lifted to the height of the table. This can be calculated using 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 gives:
\[ W = \frac{M \cdot g \cdot L}{2n^2} \]

### Final Answer
Thus, the work required to pull the hanging part back on the table is:
\[ W = \frac{M \cdot g \cdot L}{2n^2} \]

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