A heavy uniform chain lies on horizontal table top. If the co-efficient of friction between the chain and the table surface is 0.5, the maximum percentage of the length of the chain that can hang over one edge of the table is

Options

A. `20 %`
B. `33.3 %`
C. `75 %`
D. `50 %`

Answer

To solve the problem, we will analyze the forces acting on the chain and apply the principles of equilibrium and friction.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a heavy uniform chain lying on a horizontal table, with a portion hanging over the edge.
- Let the total length of the chain be \( l \).
- Let \( x \) be the length of the chain hanging over the edge of the table. Therefore, the length of the chain on the table is \( l - x \).

2. **Forces Acting on the Chain**:
- The weight of the hanging part of the chain is given by \( W_h = m \cdot g \cdot x \), where \( m \) is the mass per unit length of the chain, and \( g \) is the acceleration due to gravity.
- The normal force \( N \) acting on the part of the chain on the table is equal to the weight of the chain lying on the table: \( N = m \cdot g \cdot (l - x) \).

3. **Frictional Force**:
- The maximum static frictional force \( F_f \) that can act on the chain is given by:
\[ F_f = \mu \cdot N = \mu \cdot m \cdot g \cdot (l - x) \]
- Here, \( \mu \) is the coefficient of friction, which is given as 0.5.

4. **Equilibrium Condition**:
- For the chain to remain in equilibrium, the weight of the hanging part must be balanced by the frictional force:
\[ W_h = F_f \]
- Substituting the expressions for \( W_h \) and \( F_f \):
\[ m \cdot g \cdot x = \mu \cdot m \cdot g \cdot (l - x) \]
- We can cancel \( m \cdot g \) from both sides (assuming \( m \neq 0 \)):
\[ x = \mu \cdot (l - x) \]

5. **Solving for \( x \)**:
- Rearranging the equation:
\[ x + \mu x = \mu l \]
\[ x(1 + \mu) = \mu l \]
\[ x = \frac{\mu l}{1 + \mu} \]

6. **Substituting the Value of \( \mu \)**:
- Substituting \( \mu = 0.5 \):
\[ x = \frac{0.5 l}{1 + 0.5} = \frac{0.5 l}{1.5} = \frac{1}{3} l \]

7. **Calculating the Percentage**:
- The percentage of the chain that can hang over the edge is:
\[ \text{Percentage} = \left(\frac{x}{l}\right) \times 100 = \left(\frac{1/3 l}{l}\right) \times 100 = \frac{100}{3} \approx 33.33\% \]

### Final Answer:
The maximum percentage of the length of the chain that can hang over one edge of the table is approximately **33.33%**.