A heavy uniform chain lies on horizontal table top. If the coefficient of friction between the chain and the table surface is 0.25, then the maximum fraction of the length of the chain that can hang over one edge of the table is

Options

A. 0.2
B. 0.25
C. 0.35
D. 0.15

Explanation

To solve the problem, we need to determine the maximum fraction of the length of a heavy uniform chain that can hang over the edge of a table without slipping, given that the coefficient of friction between the chain and the table is 0.25.

### Step-by-Step Solution:

1. **Define Variables:**
- Let the total length of the chain be \( L \).
- Let the length of the chain hanging over the edge of the table be \( l \).
- The mass density of the chain is given by \( \lambda = \frac{m}{L} \), where \( m \) is the total mass of the chain.

2. **Calculate Mass of Hanging Length:**
- The mass of the hanging part of the chain (length \( l \)) is:
\[ m_h = \lambda \cdot l = \frac{m}{L} \cdot l \]

3. **Calculate Weight of Hanging Part:**
- The weight of the hanging part is:
\[ W_h = m_h \cdot g = \left(\frac{m}{L} \cdot l\right) g \]

4. **Calculate Mass of the Remaining Part:**
- The mass of the remaining part of the chain (length \( L - l \)) is:
\[ m_r = \lambda \cdot (L - l) = \frac{m}{L} \cdot (L - l) \]

5. **Calculate Normal Force:**
- The normal force \( N \) acting on the chain due to the part lying on the table is equal to the weight of the remaining part:
\[ N = m_r \cdot g = \left(\frac{m}{L} \cdot (L - l)\right) g \]

6. **Calculate Frictional Force:**
- The maximum static frictional force \( F_f \) that can act on the chain is given by:
\[ F_f = \mu \cdot N = \mu \cdot \left(\frac{m}{L} \cdot (L - l)\right) g \]
- Substituting \( \mu = 0.25 \):
\[ F_f = 0.25 \cdot \left(\frac{m}{L} \cdot (L - l)\right) g \]

7. **Set Up the Equation for Equilibrium:**
- For the chain to be in equilibrium, the frictional force must balance the weight of the hanging part:
\[ F_f = W_h \]
- Thus:
\[ 0.25 \cdot \left(\frac{m}{L} \cdot (L - l)\right) g = \left(\frac{m}{L} \cdot l\right) g \]

8. **Cancel Common Terms:**
- Cancel \( \frac{m}{L} \) and \( g \) from both sides:
\[ 0.25 \cdot (L - l) = l \]

9. **Rearrange the Equation:**
- Rearranging gives:
\[ 0.25L - 0.25l = l \]
- Combine like terms:
\[ 0.25L = l + 0.25l = 1.25l \]

10. **Solve for \( l \):**
- Thus:
\[ l = \frac{0.25L}{1.25} = \frac{0.25}{1.25}L = \frac{1}{5}L \]

11. **Calculate the Fraction:**
- The fraction of the length of the chain that can hang over the edge is:
\[ \frac{l}{L} = \frac{1}{5} = 0.2 \]

### Final Answer:
The maximum fraction of the length of the chain that can hang over one edge of the table is **0.2**.