Two similar wires under the same load yield elongation of `0.1 mm` and `0.05 mm` respectively. If the area of cross-section of the first wire is `4 mm^(2)`, then the area of cross-section of the second wire is

Options

A. `6 mm^(2)`
B. `8 mm^(2)`
C. `10 mm^(2)`
D. `12 mm^(2)`

Explanation

To solve the problem, we need to use the relationship between the elongation of the wires, their cross-sectional areas, and the load applied. The elongation of a wire under a load is given by the formula:

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

where:
- \(\Delta L\) is the elongation,
- \(F\) is the force (load) applied,
- \(L\) is the original length of the wire,
- \(A\) is the cross-sectional area,
- \(Y\) is the Young's modulus of the material.

Since both wires are made of the same material and are under the same load, we can set up a ratio of their elongations and areas:

\[ \frac{\Delta L_1}{\Delta L_2} = \frac{A_2}{A_1} \]

Given:
- \(\Delta L_1 = 0.1 \, \text{mm}\)
- \(\Delta L_2 = 0.05 \, \text{mm}\)
- \(A_1 = 4 \, \text{mm}^2\)

Now, substituting the values into the ratio:

1. **Step 1: Write the ratio of elongations.**
\[ \frac{\Delta L_1}{\Delta L_2} = \frac{0.1 \, \text{mm}}{0.05 \, \text{mm}} = 2 \]

**Hint:** Remember that the elongation is directly proportional to the load and inversely proportional to the area.

2. **Step 2: Set up the equation using the ratio of areas.**
\[ \frac{A_2}{A_1} = \frac{\Delta L_1}{\Delta L_2} = 2 \]

3. **Step 3: Substitute the known area of the first wire.**
\[ A_2 = 2 \cdot A_1 = 2 \cdot 4 \, \text{mm}^2 \]

4. **Step 4: Calculate the area of the second wire.**
\[ A_2 = 8 \, \text{mm}^2 \]

Thus, the area of cross-section of the second wire is \(8 \, \text{mm}^2\).

### Summary of Steps:
1. Calculate the ratio of elongations.
2. Set up the equation relating the areas using the ratio.
3. Substitute the known area of the first wire.
4. Calculate the area of the second wire.