1. Given data

• Total length of chain: $L = 2\,\text{m}$
• Hanging length: $x = 60\,\text{cm} = 0.6\,\text{m}$
• Total mass: $M = 4\,\text{kg}$

We need the work done to pull the hanging part completely onto the table.

2. Mass per unit length of the chain

Since the chain is uniform,

$\lambda = \frac{M}{L} = \frac{4}{2} = 2\,\text{kg/m}$

So the mass of the hanging part is

$m_h = \lambda x = 2 \times 0.6 = 1.2\,\text{kg}$

3. Idea of work done

Only the hanging part is raised. The part already on the table does not change its height.

The hanging segment has length $0.6\,\text{m}$, so its center of mass is at a distance

$\frac{x}{2} = \frac{0.6}{2} = 0.3\,\text{m}$

below the table level.

To bring the entire hanging part onto the table, its center of mass must be raised by $0.3\,\text{m}$.

Hence work done = increase in gravitational potential energy:

$W = m_h g \left(\frac{x}{2}\right)$

Substitute values:

$W = 1.2 \times 10 \times 0.3 = 3.6\,\text{J}$

4. Alternative compact formula

Using $m_h = \lambda x$,

$W = \lambda x g \cdot \frac{x}{2} = \frac{\lambda g x^2}{2}$

$W = \frac{2 \times 10 \times (0.6)^2}{2} = 10 \times 0.36 = 3.6\,\text{J}$

5. Checking options

• A: $12\,\text{J}$ — incorrect
• B: $3.6\,\text{J}$ — correct
• C: $7.2\,\text{J}$ — incorrect
• D: $1200\,\text{J}$ — incorrect

6. Comparison with stored answer

Stored correct answer: B

Our derived answer: B ($3.6\,\text{J}$)

So, the stored answer is correct.