Two men `A` and `B` are carrying a uniform bar of length `L` on their shoulders. The bar is held horizontally such that A gets one-fourth load. If `A` is at one end of the bar, the distance of `B` from that end is

Options

A. `(L)/(3)`
B. `(L)/(2)`
C. `(2L)/(3)`
D. `(3L)/(4)`

Explanation

To solve the problem, we need to analyze the situation involving two men, A and B, carrying a uniform bar of length \( L \). The bar is held horizontally, and man A is at one end, carrying one-fourth of the total weight of the bar. We need to determine the distance of man B from man A.

### Step-by-Step Solution:

1. **Understanding the Load Distribution**:
- Let the total weight of the bar be \( W \).
- Man A carries \( \frac{W}{4} \) of the weight.
- Therefore, man B carries the remaining weight, which is \( W - \frac{W}{4} = \frac{3W}{4} \).

2. **Setting Up the Problem**:
- Since the bar is uniform, its center of mass is located at the midpoint, which is at a distance of \( \frac{L}{2} \) from either end.
- Let the distance of man B from man A be \( X \).

3. **Applying the Condition of Vertical Equilibrium**:
- The sum of the vertical forces must equal zero:
\[ \frac{W}{4} + \frac{3W}{4} = W \]
- This confirms that the forces are balanced.

4. **Applying the Condition of Rotational Equilibrium**:
- We will take moments about point A to ensure the bar does not rotate. The moment due to the weight at the center of the bar (acting at \( \frac{L}{2} \)) and the moment due to the weight carried by man B (acting at distance \( X \)) must balance.
- The clockwise moment due to the weight \( W \) at the center is:
\[ W \times \frac{L}{2} \]
- The anti-clockwise moment due to the weight carried by man B is:
\[ \frac{3W}{4} \times X \]

5. **Setting Up the Moment Equation**:
- Setting the sum of moments about point A to zero gives:
\[ W \times \frac{L}{2} = \frac{3W}{4} \times X \]

6. **Solving for \( X \)**:
- Cancel \( W \) from both sides (assuming \( W \neq 0 \)):
\[ \frac{L}{2} = \frac{3}{4} X \]
- Rearranging gives:
\[ X = \frac{L}{2} \times \frac{4}{3} = \frac{2L}{3} \]

7. **Conclusion**:
- The distance of man B from man A is \( \frac{2L}{3} \).

### Final Answer:
The distance of man B from man A is \( \frac{2L}{3} \).