A wire elongates by `l` units, when a load `w` is suspended from it . If the wire gets over a pulley ( equally on both the sides ) and two weights `w` each are hung at the two ends, the elongation of wire (in units ) will be

Options

A. `(a)Zero `
B. `(b)1/2`
C. `(c)l`
D. `(d)2l`

Explanation

To solve the problem step by step, let's analyze the situation described in the question.

### Step 1: Understand the first case
In the first case, a wire of length \( L \) elongates by \( l \) units when a load \( W \) is suspended from it. The relationship between elongation, force, length, area, and Young's modulus is given by:

\[ \Delta L = \frac{F \cdot L}{A \cdot Y} \]

Where:
- \( \Delta L \) is the elongation,
- \( F \) is the force applied (in this case, \( W \)),
- \( L \) is the original length of the wire,
- \( A \) is the cross-sectional area,
- \( Y \) is Young's modulus of the material.

From the problem, we know:

\[ l = \frac{W \cdot L}{A \cdot Y} \quad \text{(1)} \]

### Step 2: Analyze the second case
In the second case, the wire is passed over a pulley, and two weights \( W \) are hung at both ends. The effective length of the wire that is being stretched is now \( \frac{L}{2} \) on each side of the pulley.

### Step 3: Calculate the elongation in the second case
The total elongation \( \Delta L_2 \) in the second case will be the sum of the elongations at both ends of the wire.

Using the same formula for elongation, we can calculate the elongation at one end (let's say point A):

\[ \Delta L_A = \frac{W \cdot \frac{L}{2}}{A \cdot Y} \]

Similarly, the elongation at the other end (point B) is:

\[ \Delta L_B = \frac{W \cdot \frac{L}{2}}{A \cdot Y} \]

Thus, the total elongation \( \Delta L_2 \) is:

\[ \Delta L_2 = \Delta L_A + \Delta L_B = \frac{W \cdot \frac{L}{2}}{A \cdot Y} + \frac{W \cdot \frac{L}{2}}{A \cdot Y} \]

This simplifies to:

\[ \Delta L_2 = 2 \cdot \frac{W \cdot \frac{L}{2}}{A \cdot Y} = \frac{W \cdot L}{A \cdot Y} \]

### Step 4: Relate to the first case
From equation (1), we know that:

\[ \frac{W \cdot L}{A \cdot Y} = l \]

Therefore, we can conclude:

\[ \Delta L_2 = l \]

### Final Answer
The elongation of the wire when two weights \( W \) are hung at both ends is \( l \) units.

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