A uniform chain of mass 4 kg and length 2 m is kept on table such that `3//10^("th")` of the chain hanges freely from the edge of the table. How much work has to be done in pulling the entire chain on the table?

Options

A. 7.2 J
B. 120 J
C. 1200 J
D. 3.6 J

Answer

To solve the problem of how much work has to be done in pulling the entire chain onto the table, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Mass of the chain, \( M = 4 \, \text{kg} \)
- Total length of the chain, \( L = 2 \, \text{m} \)
- Length of the chain hanging off the table, \( l = \frac{3}{10} \times L = \frac{3}{10} \times 2 = 0.6 \, \text{m} \)

2. **Calculate the Mass of the Hanging Part:**
- The mass of the hanging part of the chain can be calculated using the formula:
\[ m = M \times \frac{l}{L} \]
- Substituting the values:
\[ m = 4 \, \text{kg} \times \frac{0.6 \, \text{m}}{2 \, \text{m}} = 4 \times 0.3 = 1.2 \, \text{kg} \]

3. **Determine the Center of Mass of the Hanging Part:**
- The center of mass of the hanging part is located at half of its length from the edge of the table:
\[ h = \frac{l}{2} = \frac{0.6 \, \text{m}}{2} = 0.3 \, \text{m} \]

4. **Calculate the Work Done to Pull the Chain:**
- The work done in pulling the chain onto the table is equal to the change in potential energy of the hanging mass:
\[ W = m \cdot g \cdot h \]
- Assuming \( g = 10 \, \text{m/s}^2 \):
\[ W = 1.2 \, \text{kg} \times 10 \, \text{m/s}^2 \times 0.3 \, \text{m} \]
- Calculating this gives:
\[ W = 1.2 \times 10 \times 0.3 = 3.6 \, \text{J} \]

5. **Conclusion:**
- The total work done in pulling the entire chain onto the table is \( 3.6 \, \text{J} \).

### Final Answer:
The work done in pulling the entire chain on the table is **3.6 Joules**.

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