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. `Delta l`
B. `2 Delta l`
C. `4 Delta l`
D. `8 Delta l`

Answer

To solve the problem, we need to analyze how the elongation of the wire changes when the length of the wire is doubled while the same mass is suspended from it.

### Step-by-Step Solution:

1. **Understanding the Original Situation**:
- When a mass \( m \) is suspended from a wire of original length \( L \), the change in length (elongation) of the wire is given as \( \Delta l \).
- The formula for elongation \( \Delta l \) is given by:
\[ \Delta l = \frac{F L}{A Y} \]
where \( F \) is the force (weight of the mass), \( A \) is the cross-sectional area of the wire, \( Y \) is the Young's modulus of the material, and \( L \) is the original length of the wire.

2. **Doubling the Length of the Wire**:
- Now, the wire is stretched to double its original length, so the new length \( L' = 2L \).
- The force \( F \) acting on the wire remains the same since the same mass \( m \) is suspended.

3. **Calculating the New Elongation**:
- The new elongation \( \Delta l' \) for the wire of length \( L' \) can be expressed as:
\[ \Delta l' = \frac{F L'}{A Y} \]
- Substituting \( L' = 2L \) into the equation gives:
\[ \Delta l' = \frac{F (2L)}{A Y} = 2 \cdot \frac{F L}{A Y} = 2 \Delta l \]

4. **Considering the Proportionality**:
- The elongation is proportional to the square of the length of the wire. Therefore, if the length is doubled, the elongation will change as follows:
\[ \Delta l' = \frac{F (2L)^2}{A Y} = \frac{F \cdot 4L^2}{A Y} = 4 \cdot \frac{F L^2}{A Y} = 4 \Delta l \]

5. **Conclusion**:
- Hence, when the wire is stretched to double its length, the new change in length \( \Delta l' \) will be:
\[ \Delta l' = 4 \Delta l \]

### Final Answer:
The change in length when the wire is stretched to double its length is \( 4 \Delta l \).