A uniform chain of mass `M` and length `L` is lying on a frictionless table in such a way that its `1//3` parts is hanging vertically down. The work done in pulling the chain up the table is

Options

A. `(MgL)/(18)`
B. `(MgL)/(9)`
C. `(MgL)/(6)`
D. `(MgL)/(3)`

Answer

To solve the problem of calculating the work done in pulling a uniform chain of mass \( M \) and length \( L \) such that \( \frac{1}{3} \) of it is hanging vertically down from a frictionless table, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the mass of the hanging part of the chain:**
Since \( \frac{1}{3} \) of the chain is hanging, the length of the hanging part is:
\[ L_h = \frac{L}{3} \]
The mass of the hanging part can be calculated as:
\[ M_h = \frac{M}{3} \]

2. **Determine the gravitational force acting on the hanging part:**
The force due to gravity acting on the hanging mass is:
\[ F = M_h \cdot g = \frac{M}{3} \cdot g \]

3. **Consider an infinitesimal element of the hanging chain:**
Let’s consider an infinitesimal length \( dx \) of the chain that is at a distance \( x \) below the edge of the table. The mass of this infinitesimal element is:
\[ dm = \frac{M}{L} \cdot dx \]

4. **Calculate the force acting on this infinitesimal element:**
The gravitational force acting on this infinitesimal mass is:
\[ dF = dm \cdot g = \frac{M}{L} \cdot g \cdot dx \]

5. **Determine the work done to lift this infinitesimal element:**
The work done \( dW \) in lifting this element from a height \( x \) to the table (which is at height 0) is given by:
\[ dW = -dF \cdot x = -\left(\frac{M}{L} \cdot g \cdot dx\right) \cdot x \]
The negative sign indicates that the force and displacement are in opposite directions.

6. **Integrate to find the total work done:**
To find the total work done \( W \) in pulling the entire hanging part of the chain back onto the table, we integrate from \( x = 0 \) to \( x = \frac{L}{3} \):
\[ W = -\int_0^{\frac{L}{3}} \left(\frac{M}{L} \cdot g \cdot x\right) dx \]

7. **Perform the integration:**
\[ W = -\frac{M}{L} \cdot g \cdot \int_0^{\frac{L}{3}} x \, dx \]
The integral of \( x \) is:
\[ \int x \, dx = \frac{x^2}{2} \]
Evaluating from 0 to \( \frac{L}{3} \):
\[ \int_0^{\frac{L}{3}} x \, dx = \left[\frac{x^2}{2}\right]_0^{\frac{L}{3}} = \frac{1}{2} \left(\frac{L}{3}\right)^2 = \frac{L^2}{18} \]

8. **Substituting back into the work equation:**
\[ W = -\frac{M}{L} \cdot g \cdot \frac{L^2}{18} = -\frac{MgL}{18} \]
Since work done is a scalar quantity, we can take the absolute value:
\[ W = \frac{MgL}{18} \]

### Final Answer:
The work done in pulling the chain up onto the table is:
\[ W = \frac{MgL}{18} \]