A wire elongates by 9 mm when a load of 10 kg is suspended from it. What is the elongation when its radius is doubled, if all other quantities are same as before?

Answer

To solve the problem, we will use the relationship between elongation (ΔL), force (F), cross-sectional area (A), length (L), and Young's modulus (Y) of the material. The formula for elongation is given by:

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

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
- The initial elongation (ΔL₁) when a load of 10 kg is applied is given as 9 mm.
- The force (F) due to the load can be calculated using \( F = m \cdot g \), where \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).
- The initial radius of the wire is \( R \).

2. **Calculate the Initial Force**:
\[ F = 10 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 98.1 \, \text{N} \]

3. **Calculate the Initial Cross-Sectional Area**:
- The area \( A \) of the wire is given by \( A = \pi R^2 \).

4. **Set Up the Elongation Formula**:
- Using the elongation formula for the initial case:
\[ \Delta L_1 = \frac{F \cdot L}{A \cdot Y} \]
- We know \( \Delta L_1 = 9 \, \text{mm} \).

5. **Consider the Case When Radius is Doubled**:
- When the radius is doubled, the new radius becomes \( 2R \).
- The new cross-sectional area \( A' \) will be:
\[ A' = \pi (2R)^2 = 4\pi R^2 = 4A \]

6. **Set Up the Elongation Formula for the New Case**:
- The elongation for the new case (ΔL₂) can be expressed as:
\[ \Delta L_2 = \frac{F \cdot L}{A' \cdot Y} = \frac{F \cdot L}{4A \cdot Y} \]

7. **Relate the Two Elongations**:
- Since \( \Delta L_1 = \frac{F \cdot L}{A \cdot Y} \), we can express \( \Delta L_2 \) in terms of \( \Delta L_1 \):
\[ \Delta L_2 = \frac{1}{4} \Delta L_1 \]

8. **Calculate the New Elongation**:
- Substitute \( \Delta L_1 = 9 \, \text{mm} \):
\[ \Delta L_2 = \frac{1}{4} \cdot 9 \, \text{mm} = 2.25 \, \text{mm} \]

### Final Answer:
The elongation when the radius is doubled, while all other quantities remain the same, is **2.25 mm**.