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. (a) l
B. (b) 2l
C. (c) zero
D. (d) `l/2`

Explanation

To solve the problem of how much a wire elongates when two weights \( W \) are hung from both ends over a pulley, we can follow these steps:

### Step 1: Understand the Initial Condition
When a single load \( W \) is hung from the wire, it elongates by \( l \) mm. This elongation can be expressed using Hooke's Law, which states that the elongation \( \Delta L \) is proportional to the force applied and the original length of the wire.

### Step 2: Analyze the New Condition
In the new scenario, the wire goes over a pulley, and two weights \( W \) are hung at both ends. The total force acting on the wire now is \( 2W \) (since each side has a weight \( W \)).

### Step 3: Determine the Effective Length
Since the wire is now effectively divided into two segments (each of length \( L/2 \)), we need to consider the elongation for each segment.

### Step 4: Apply Hooke's Law
Using Hooke's Law, the elongation \( \Delta L' \) for the wire under the new load can be expressed as:
\[ \Delta L' = \frac{F \cdot L}{A \cdot Y} \]
where:
- \( F = 2W \) (the total force),
- \( L = L/2 \) (the effective length of each segment),
- \( A \) is the cross-sectional area,
- \( Y \) is the Young's modulus of the material.

### Step 5: Substitute Values
Substituting the values into the equation gives:
\[ \Delta L' = \frac{2W \cdot (L/2)}{A \cdot Y} = \frac{W \cdot L}{A \cdot Y} \]

### Step 6: Relate to Original Elongation
From the initial condition, we know that:
\[ \Delta L = \frac{W \cdot L}{A \cdot Y} = l \text{ mm} \]
Thus, we can conclude that:
\[ \Delta L' = l \text{ mm} \]

### Conclusion
The elongation of the wire when two weights \( W \) are hung from both ends over a pulley is \( l \) mm.