The line of action of the resultant of two like parallel forces shifts by one-fourth of the distance between the forces when the two forces are interchanged. The ratio of the two forces is:

Options

A. `1:2`
B. `2:3`
C. `3:4`
D. `3:5`

Explanation

To solve the problem, we need to analyze the situation involving two like parallel forces \( F_1 \) and \( F_2 \) acting at a distance \( L \) apart. The key information given is that the line of action of the resultant force shifts by one-fourth of the distance \( L \) between the forces when they are interchanged.

### Step-by-Step Solution:

1. **Define the Forces and Distance**:
- Let \( F_1 \) and \( F_2 \) be the two forces acting at points separated by a distance \( L \).
- Let the distance from \( F_1 \) to the resultant force \( F_R \) be \( x \) and the distance from \( F_2 \) to \( F_R \) be \( L - x \).

2. **Equilibrium Condition**:
- For the resultant force \( F_R \) to be in equilibrium, the torques about any point must be equal. We can set up the equation:
\[ F_1 \cdot x = F_2 \cdot (L - x) \]

3. **Interchanging the Forces**:
- When the forces are interchanged, the new distances become \( L - x \) for \( F_1 \) and \( x \) for \( F_2 \).
- The line of action shifts by \( \frac{L}{4} \), which means the new distance \( L - x - x = \frac{L}{4} \).

4. **Setting Up the Equation**:
- From the shift, we can write:
\[ L - 2x = \frac{L}{4} \]

5. **Solving for \( x \)**:
- Rearranging the equation gives:
\[ 2x = L - \frac{L}{4} \]
\[ 2x = \frac{4L}{4} - \frac{L}{4} = \frac{3L}{4} \]
\[ x = \frac{3L}{8} \]

6. **Finding the Ratio of the Forces**:
- Substitute \( x \) back into the torque equation:
\[ F_1 \cdot \frac{3L}{8} = F_2 \cdot \left(L - \frac{3L}{8}\right) \]
\[ F_1 \cdot \frac{3L}{8} = F_2 \cdot \frac{5L}{8} \]
- Dividing both sides by \( L \) and rearranging gives:
\[ \frac{F_1}{F_2} = \frac{5}{3} \]

7. **Final Ratio**:
- Thus, the ratio of the forces \( F_1 : F_2 \) is \( 5 : 3 \).

### Conclusion:
The ratio of the two forces is \( 5 : 3 \).