When a weight of `10 kg` is suspended from a copper wire of length 3 metres and diameters `0.4 mm`, its length increases by `2.4 cm`. If the diameter of the wire is doubled, then the extension in its length will be

Options

A. `9.6 cm`
B. `4.8 cm`
C. `1.2 cm`
D. `0.6 cm`

Answer

To solve the problem, we will use the relationship between stress, strain, and Young's modulus of elasticity. We will derive the extension in length when the diameter of the wire is doubled.

### Step-by-Step Solution:

1. **Identify Given Data:**
- Weight (Force, F) = 10 kg
- Length of wire (L) = 3 m = 300 cm
- Initial diameter (D1) = 0.4 mm = 0.04 cm
- Initial extension (ΔL1) = 2.4 cm

2. **Convert Weight to Force:**
- The weight of 10 kg can be converted to force using \( F = mg \), where \( g \approx 9.81 \, \text{m/s}^2 \).
- Thus, \( F = 10 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 98.1 \, \text{N} \).

3. **Calculate the Area of the Wire:**
- The area \( A \) of the wire can be calculated using the formula for the area of a circle:
\[ A = \frac{\pi D^2}{4} \]
- For the initial diameter \( D1 = 0.4 \, \text{mm} = 0.0004 \, \text{m} \):
\[ A1 = \frac{\pi (0.0004)^2}{4} \approx 5.0265 \times 10^{-8} \, \text{m}^2 \]

4. **Use Young's Modulus Relation:**
- Young's modulus \( Y \) is defined as:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L} \]
- Rearranging gives:
\[ \Delta L = \frac{F \cdot L}{Y \cdot A} \]

5. **Relate the Extension with Diameter:**
- When the diameter is doubled, the new diameter \( D2 = 2D1 = 0.8 \, \text{mm} = 0.0008 \, \text{m} \).
- The new area \( A2 \) becomes:
\[ A2 = \frac{\pi (0.0008)^2}{4} = \frac{\pi (0.0004)^2 \cdot 4}{4} = 4A1 \]

6. **Calculate the New Extension:**
- Since the force and the original length remain the same, we can express the new extension \( \Delta L2 \) as:
\[ \Delta L2 = \frac{F \cdot L}{Y \cdot A2} = \frac{F \cdot L}{Y \cdot 4A1} = \frac{\Delta L1}{4} \]
- Given \( \Delta L1 = 2.4 \, \text{cm} \):
\[ \Delta L2 = \frac{2.4 \, \text{cm}}{4} = 0.6 \, \text{cm} \]

### Final Answer:
The extension in the length of the wire when its diameter is doubled is **0.6 cm**.