A wire elongates by 8 mm when a load of 9 kg is suspended from it. What is the elongation when its radius is doubled, if all other quantites are the same as before ?

Explanation

To solve the problem, we will use the concept of Young's modulus and the relationship between elongation, force, and the dimensions of the wire.

### Step-by-Step Solution:

1. **Understand the Given Information**:
- Initial elongation (\( \Delta L_1 \)) = 8 mm
- Load (\( F \)) = 9 kg
- Radius of the wire = \( r \)
- New radius of the wire = \( 2r \)

2. **Convert the Load to Force**:
- The force due to the load can be calculated using the formula \( F = mg \), where \( g \) (acceleration due to gravity) is approximately \( 9.81 \, \text{m/s}^2 \).
- Therefore, \( F = 9 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 88.29 \, \text{N} \).

3. **Use the Formula for Young's Modulus**:
- Young's modulus (\( Y \)) is defined as:
\[ Y = \frac{F \cdot L}{A \cdot \Delta L} \]
- Here, \( A \) (cross-sectional area) for a circular wire is given by \( A = \pi r^2 \).

4. **Relate Elongation to Radius**:
- From the formula, we can express the elongation in terms of the radius:
\[ \Delta L = \frac{F \cdot L}{Y \cdot A} = \frac{F \cdot L}{Y \cdot \pi r^2} \]
- This means that \( \Delta L \) is inversely proportional to the square of the radius.

5. **Set Up the Relationship for New Radius**:
- When the radius is doubled (\( r_2 = 2r \)), the new area becomes:
\[ A_2 = \pi (2r)^2 = 4\pi r^2 \]
- The new elongation (\( \Delta L_2 \)) can be expressed as:
\[ \Delta L_2 = \frac{F \cdot L}{Y \cdot A_2} = \frac{F \cdot L}{Y \cdot 4\pi r^2} \]

6. **Relate the Two Elongations**:
- Since \( \Delta L_1 \) is known, we can relate \( \Delta L_2 \) to \( \Delta L_1 \):
\[ \Delta L_2 = \frac{\Delta L_1}{4} \]
- Substituting \( \Delta L_1 = 8 \, \text{mm} \):
\[ \Delta L_2 = \frac{8 \, \text{mm}}{4} = 2 \, \text{mm} \]

### Final Answer:
The elongation when the radius is doubled is **2 mm**.