when a weight of 10 kg is suspended from a copper wire of length 3m and diameter 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. 7.6 cm
B. 4.8 cm
C. 1.2 cm
D. 0.6 cm

Explanation

To solve the problem, we need to determine how the extension in the length of a copper wire changes when the diameter of the wire is doubled, while keeping the weight and length constant. Here's a step-by-step solution:

### Step 1: Understand the relationship between extension and diameter
The extension (ΔL) of a wire is given by the formula:
\[ \Delta L = \frac{F \cdot L}{A \cdot Y} \]
where:
- \(F\) is the force applied (weight),
- \(L\) is the original length of the wire,
- \(A\) is the cross-sectional area of the wire,
- \(Y\) is Young's modulus of the material.

### Step 2: Calculate the initial area
The area \(A\) of a cylindrical wire is calculated using the formula:
\[ A = \pi r^2 \]
Given the diameter \(D_1 = 0.4 \, \text{mm}\), the radius \(r_1\) is:
\[ r_1 = \frac{D_1}{2} = \frac{0.4 \, \text{mm}}{2} = 0.2 \, \text{mm} = 0.2 \times 10^{-3} \, \text{m} \]
Now, calculate the area \(A_1\):
\[ A_1 = \pi (0.2 \times 10^{-3})^2 = \pi (0.04 \times 10^{-6}) = 0.04\pi \times 10^{-6} \, \text{m}^2 \]

### Step 3: Relate the extensions
From the formula, we see that the extension is inversely proportional to the area:
\[ \Delta L \propto \frac{1}{A} \]
Thus, if we denote the initial extension as \(\Delta L_1\) and the new extension (when the diameter is doubled) as \(\Delta L_2\), we can write:
\[ \frac{\Delta L_2}{\Delta L_1} = \frac{A_1}{A_2} \]

### Step 4: Calculate the new area
When the diameter is doubled, \(D_2 = 2D_1 = 0.8 \, \text{mm}\), so the new radius \(r_2\) is:
\[ r_2 = \frac{D_2}{2} = \frac{0.8 \, \text{mm}}{2} = 0.4 \, \text{mm} = 0.4 \times 10^{-3} \, \text{m} \]
Now, calculate the new area \(A_2\):
\[ A_2 = \pi (0.4 \times 10^{-3})^2 = \pi (0.16 \times 10^{-6}) = 0.16\pi \times 10^{-6} \, \text{m}^2 \]

### Step 5: Set up the ratio of extensions
Now we can set up the ratio:
\[ \frac{\Delta L_2}{\Delta L_1} = \frac{A_1}{A_2} = \frac{0.04\pi \times 10^{-6}}{0.16\pi \times 10^{-6}} = \frac{0.04}{0.16} = \frac{1}{4} \]
Thus, we have:
\[ \Delta L_2 = \frac{1}{4} \Delta L_1 \]

### Step 6: Calculate the new extension
Given that \(\Delta L_1 = 2.4 \, \text{cm}\):
\[ \Delta L_2 = \frac{1}{4} \times 2.4 \, \text{cm} = 0.6 \, \text{cm} \]

### Final Answer
The extension in the length of the wire when the diameter is doubled is:
\[ \Delta L_2 = 0.6 \, \text{cm} \]