A wire elongates by l mm when a load W is hanged from it. If the wire goes over a pulley and two weights W each are hung at the two ends, the elongation of the wire will be (in mm)

Options

A. l
B. 2l
C. zero
D. `(l)/(2)`

Answer

To solve the problem, we will analyze the elongation of the wire in two different scenarios:

### Step-by-Step Solution:

1. **Understanding the First Case:**
- When a single load \( W \) is hung from the wire, it elongates by \( l \) mm.
- The tension \( T \) in the wire due to the load is equal to \( W \).
- The elongation \( l \) can be expressed using Young's modulus \( Y \):
\[ Y = \frac{F/A}{\Delta L/L} \quad \text{(where \( F \) is the force, \( A \) is the cross-sectional area, \( \Delta L \) is the elongation, and \( L \) is the original length)} \]
- For this case, we have:
\[ Y = \frac{W/A}{l/L} \]

2. **Understanding the Second Case:**
- In this scenario, the wire goes over a pulley, and two weights \( W \) are hung at both ends.
- The tension \( T \) in the wire remains equal to \( W \) since both ends are experiencing the same force.
- The total length of the wire remains the same, but the effective length of the wire under tension is halved for each segment of the wire, meaning each segment will elongate by \( \frac{l}{2} \) mm.

3. **Applying Young's Modulus for the Second Case:**
- The elongation for each half of the wire can be expressed as:
\[ Y = \frac{W/A}{(l/2)/L} \]
- Rearranging gives:
\[ Y = \frac{2W/A}{l/L} \]
- Since the Young's modulus \( Y \) must be the same in both cases, we can equate the two expressions for \( Y \):
\[ \frac{W/A}{l/L} = \frac{2W/A}{l/L} \]

4. **Conclusion:**
- Since both cases have the same Young's modulus, the elongation in the wire for the second case is also \( l \) mm.
- Therefore, the elongation of the wire when two weights \( W \) are hung at both ends is \( l \) mm.

### Final Answer:
The elongation of the wire will be \( l \) mm.

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