A chain is held on a frictionless talbe with `L//4` hanging over. Knowing total mass of the chain is `M` and total length is `L`, the work required to slowly pull hanging part back to the table is `:`

Options

A. `(MgL)/(16)`
B. `(MgL)/(8)`
C. `(MgL)/(32)`
D. `(MgL)/(24)`

Explanation

To solve the problem of calculating the work required to slowly pull the hanging part of the chain back to the table, we can follow these steps:

### Step 1: Identify the mass of the hanging part of the chain
The total mass of the chain is \( M \) and the total length of the chain is \( L \). Since \( \frac{L}{4} \) of the chain is hanging, the mass of the hanging part can be calculated as follows:

\[ \text{Mass of hanging part} = \frac{M}{L} \times \frac{L}{4} = \frac{M}{4} \]

### Step 2: Calculate the gravitational force on the hanging part
The gravitational force acting on the hanging part of the chain is given by the product of its mass and the acceleration due to gravity \( g \):

\[ \text{Gravitational force} = \text{Mass} \times g = \left(\frac{M}{4}\right) g = \frac{Mg}{4} \]

### Step 3: Determine the center of mass of the hanging part
The center of mass of the hanging part, which is \( \frac{L}{4} \) long, is located at its midpoint. Therefore, the distance from the table to the center of mass is:

\[ \text{Distance to center of mass} = \frac{L}{4} \div 2 = \frac{L}{8} \]

### Step 4: Calculate the work done to lift the hanging part
The work done \( W \) against gravity to lift the hanging part back to the table can be calculated using the formula:

\[ W = \text{Force} \times \text{Distance} \]

Since the force is acting in the opposite direction of the displacement (lifting the chain), we take the angle between force and displacement as 180 degrees, thus:

\[ W = \left(\frac{Mg}{4}\right) \times \left(\frac{L}{8}\right) \times \cos(180^\circ) \]

Since \( \cos(180^\circ) = -1 \):

\[ W = -\left(\frac{Mg}{4}\right) \times \left(\frac{L}{8}\right) = -\frac{MgL}{32} \]

However, we are interested in the magnitude of work done, so we take the positive value:

\[ W = \frac{MgL}{32} \]

### Final Answer
The work required to slowly pull the hanging part back to the table is:

\[ \boxed{\frac{MgL}{32}} \]