A mass `m` is suspended from a wire . Change in length of the wire is `Deltal`. Now the same wire is stretched to double its length and the same mass is suspended from the wire. The change in length in this case will become (it is suspended that elongation in the wire is within the proportional limit)

Options

A. `Deltal`
B. `2Deltal`
C. `4Deltal`
D. `8Deltal`

Explanation

To solve the problem, we will use the relationship between elongation (change in length) of a wire, the force applied, the original length of the wire, and Young's modulus.

### Step-by-Step Solution:

1. **Understanding the Initial Condition**:
When a mass \( m \) is suspended from a wire of original length \( L \), the change in length (elongation) of the wire is given as \( \Delta L_1 \).

2. **Using the Formula for Elongation**:
The elongation \( \Delta L \) of a wire can be expressed using the formula:
\[ \Delta L = \frac{F \cdot L}{A \cdot Y} \]
where:
- \( F \) is the force applied (which is equal to the weight of the mass, \( mg \)),
- \( L \) is the original length of the wire,
- \( A \) is the cross-sectional area of the wire,
- \( Y \) is Young's modulus of the material.

3. **Substituting the Force**:
For the initial case, the force \( F = mg \), so we can rewrite the elongation as:
\[ \Delta L_1 = \frac{mg \cdot L}{A \cdot Y} \]

4. **Changing the Length of the Wire**:
Now, the wire is stretched to double its original length, so the new length \( L_2 = 2L \).

5. **Calculating the New Elongation**:
The new elongation \( \Delta L_2 \) when the same mass \( m \) is suspended from the wire of length \( L_2 \) can be expressed as:
\[ \Delta L_2 = \frac{mg \cdot L_2}{A \cdot Y} \]
Substituting \( L_2 = 2L \):
\[ \Delta L_2 = \frac{mg \cdot (2L)}{A \cdot Y} = 2 \cdot \frac{mg \cdot L}{A \cdot Y} = 2 \Delta L_1 \]

6. **Considering the Proportional Limit**:
Since the elongation is within the proportional limit, we can use the relationship between elongation and length. The elongation is proportional to the square of the length:
\[ \frac{\Delta L_2}{\Delta L_1} = \left(\frac{L_2}{L_1}\right)^2 \]
Here, \( L_2 = 2L \) and \( L_1 = L \), hence:
\[ \frac{\Delta L_2}{\Delta L_1} = \left(\frac{2L}{L}\right)^2 = 4 \]
Therefore, we can conclude:
\[ \Delta L_2 = 4 \Delta L_1 \]

### Final Answer:
The change in length when the wire is stretched to double its length and the same mass is suspended from it will become \( 4 \Delta L_1 \).

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