A uniform chain of length `l` is placed on a rough table, with length `l//3` hanging over the edge. If the chain just begins to slide off the table by itself from this portion, the coefficient of friction between the chain and the table is 1/K. Then find the value of K.

Answer

To solve the problem, we need to analyze the forces acting on the chain when it is in the state just before it starts to slide off the table.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a uniform chain of length `l` on a rough table.
- A portion of the chain, specifically `l/3`, is hanging off the edge of the table.
- The remaining portion, `2l/3`, is on the table.

2. **Calculating the Mass of the Chain**:
- Let the total mass of the chain be `M`.
- The mass per unit length of the chain is \( \frac{M}{l} \).
- The mass of the hanging portion (length \( l/3 \)) is:
\[ m = \frac{M}{l} \cdot \frac{l}{3} = \frac{M}{3} \]

3. **Weight of the Hanging Portion**:
- The weight of the hanging portion is:
\[ W_h = m \cdot g = \frac{M}{3} \cdot g \]

4. **Calculating the Mass on the Table**:
- The mass of the portion on the table (length \( 2l/3 \)) is:
\[ m' = \frac{M}{l} \cdot \frac{2l}{3} = \frac{2M}{3} \]

5. **Weight of the Portion on the Table**:
- The weight of the portion on the table is:
\[ W_t = m' \cdot g = \frac{2M}{3} \cdot g \]

6. **Normal Force**:
- The normal force \( N \) acting on the chain from the table is equal to the weight of the portion of the chain that is on the table:
\[ N = W_t = \frac{2M}{3} \cdot g \]

7. **Frictional Force**:
- The maximum static frictional force \( F_s \) that can act on the chain is given by:
\[ F_s = \mu N = \frac{1}{K} \cdot \frac{2M}{3} \cdot g \]

8. **Equilibrium Condition**:
- At the moment the chain just begins to slide, the tension \( T \) in the chain (due to the hanging mass) is equal to the maximum static frictional force:
\[ T = F_s \]
- The tension \( T \) is equal to the weight of the hanging portion:
\[ T = W_h = \frac{M}{3} \cdot g \]

9. **Setting Up the Equation**:
- Equating the two expressions for tension:
\[ \frac{M}{3} \cdot g = \frac{1}{K} \cdot \frac{2M}{3} \cdot g \]

10. **Solving for K**:
- Cancel \( g \) and \( M \) from both sides (assuming \( M \neq 0 \) and \( g \neq 0 \)):
\[ \frac{1}{3} = \frac{2}{3K} \]
- Cross-multiplying gives:
\[ K = 2 \]

### Final Answer:
The value of \( K \) is \( 2 \).