A wire is loaded by 6 kg at its one end, the increase in length is 12 mm . If the radius of the wire is doubled and all other magnitudes are unchanged, then increase in length will be

Options

A. 6 mm
B. 3 mm
C. 24 mm
D. 48 mm

Answer

To solve the problem step by step, we will use the relationship between stress, strain, and the Young's modulus of the material.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions**:
- We have a wire loaded with a weight of 6 kg.
- The increase in length (ΔL1) of the wire due to this load is 12 mm.
- The radius of the wire is denoted as \( r \).

2. **Calculating the Force**:
- The force (F) applied to the wire due to the weight is given by:
\[ F = m \cdot g = 6 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 58.8 \, \text{N} \]

3. **Cross-Sectional Area of the Wire**:
- The cross-sectional area (A) of the wire is given by:
\[ A = \pi r^2 \]

4. **Using the Formula for Increase in Length**:
- The formula for the increase in length of the wire is:
\[ \Delta L = \frac{F \cdot L}{A \cdot Y} \]
- Rearranging gives:
\[ \Delta L \propto \frac{F \cdot L}{A} \]

5. **Expressing ΔL in Terms of Radius**:
- Since \( A = \pi r^2 \), we can express ΔL as:
\[ \Delta L \propto \frac{1}{r^2} \]

6. **Considering the New Conditions**:
- If the radius is doubled (i.e., \( r_2 = 2r \)), the new area becomes:
\[ A_2 = \pi (2r)^2 = 4\pi r^2 \]

7. **Finding the New Increase in Length (ΔL2)**:
- The new increase in length (ΔL2) can be expressed in terms of ΔL1:
\[ \frac{\Delta L_1}{\Delta L_2} = \frac{A_2}{A_1} = \frac{4\pi r^2}{\pi r^2} = 4 \]
- Thus, we can write:
\[ \Delta L_2 = \frac{\Delta L_1}{4} \]

8. **Substituting the Known Value**:
- We know that ΔL1 = 12 mm, so:
\[ \Delta L_2 = \frac{12 \, \text{mm}}{4} = 3 \, \text{mm} \]

### Final Answer:
The increase in length when the radius of the wire is doubled, while keeping all other magnitudes unchanged, will be **3 mm**.

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