A chain is lying on a rough table with a fraction 1/n of its length hanging down from the edge of the table. if it is just on the point of sliding down from the table, then the coefficient of friction between the table and the chain is -

Options

A. `(1)/(n)`
B. `(1)/((n-1))`
C. `(1)/((n+1))`
D. `(n-1)/((n+1))`

Explanation

To solve the problem, we need to analyze the forces acting on the chain that is lying on the rough table with a fraction \( \frac{1}{n} \) of its length hanging down. We will derive the coefficient of friction step by step.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- Let the total length of the chain be \( L \).
- The length of the chain hanging down from the table is \( \frac{L}{n} \).
- Therefore, the length of the chain on the table is \( L - \frac{L}{n} = L\left(1 - \frac{1}{n}\right) \).

2. **Mass of the Chain**:
- Let the mass of the entire chain be \( M \).
- The mass of the portion hanging down (length \( \frac{L}{n} \)) is \( \frac{M}{n} \).
- The mass of the portion on the table (length \( L\left(1 - \frac{1}{n}\right) \)) is \( M - \frac{M}{n} = M\left(1 - \frac{1}{n}\right) \).

3. **Weight of the Hanging Portion**:
- The weight of the hanging portion is given by:
\[ W_{\text{hanging}} = \text{mass} \times g = \frac{M}{n} \cdot g \]

4. **Normal Force on the Chain**:
- The normal force acting on the portion of the chain on the table is equal to the weight of that portion:
\[ N = \left(M\left(1 - \frac{1}{n}\right)\right)g \]

5. **Frictional Force**:
- The frictional force \( F_f \) that opposes the sliding is given by:
\[ F_f = \mu \cdot N \]
- Substituting for \( N \):
\[ F_f = \mu \cdot \left(M\left(1 - \frac{1}{n}\right)g\right) \]

6. **Condition for the Chain on the Verge of Sliding**:
- At the verge of sliding, the frictional force equals the weight of the hanging portion:
\[ F_f = W_{\text{hanging}} = \frac{M}{n} \cdot g \]

7. **Setting the Forces Equal**:
- Equating the two expressions for forces:
\[ \mu \cdot \left(M\left(1 - \frac{1}{n}\right)g\right) = \frac{M}{n} \cdot g \]

8. **Canceling Common Terms**:
- We can cancel \( M \) and \( g \) from both sides (assuming \( M \neq 0 \) and \( g \neq 0 \)):
\[ \mu \cdot \left(1 - \frac{1}{n}\right) = \frac{1}{n} \]

9. **Solving for the Coefficient of Friction \( \mu \)**:
- Rearranging gives:
\[ \mu = \frac{1/n}{1 - 1/n} = \frac{1}{n - 1} \]

### Final Answer:
The coefficient of friction \( \mu \) between the table and the chain is:
\[ \mu = \frac{1}{n - 1} \]