An aluminium rod has a breaking strain `0.2%`. The minimum cross-sectional area of the rod in `m^(2)` in order to support a load of `10^(4)N` is fi (Young's modulus is `7xx10^(9) Nm^(-2)`)

Options

A. `1.7xx10^(-4)`
B. `1.7xx10^(-3)`
C. `7.1xx10^(-4)`
D. `1.4xx10^(-4)`

Explanation

To find the minimum cross-sectional area of the aluminium rod that can support a load of \(10^4 \, \text{N}\) given a breaking strain of \(0.2\%\) and Young's modulus \(Y = 7 \times 10^9 \, \text{N/m}^2\), we can follow these steps:

### Step 1: Convert Breaking Strain to Decimal
The breaking strain is given as \(0.2\%\). To convert this percentage to a decimal form, we divide by \(100\):
\[ \frac{\Delta L}{L} = \frac{0.2}{100} = 0.002 = 2 \times 10^{-3} \]

### Step 2: Use Young's Modulus Formula
Young's modulus \(Y\) is defined as:
\[ Y = \frac{F}{A} \cdot \frac{L}{\Delta L} \]
Rearranging this formula to find the cross-sectional area \(A\):
\[ A = \frac{F}{Y \cdot \frac{\Delta L}{L}} \]

### Step 3: Substitute Known Values
Now, we substitute the known values into the equation:
- Load \(F = 10^4 \, \text{N}\)
- Young's modulus \(Y = 7 \times 10^9 \, \text{N/m}^2\)
- Breaking strain \(\frac{\Delta L}{L} = 2 \times 10^{-3}\)

Substituting these values:
\[ A = \frac{10^4}{7 \times 10^9 \cdot 2 \times 10^{-3}} \]

### Step 4: Calculate the Area
Now, we calculate the area:
\[ A = \frac{10^4}{14 \times 10^6} = \frac{10^4}{1.4 \times 10^7} = \frac{10^4}{14 \times 10^6} = \frac{1}{14} \times 10^{-2} \]
Calculating \(14\):
\[ A = \frac{1}{14} \times 10^{-2} \approx 0.0714 \times 10^{-2} \, \text{m}^2 \]
This can be expressed as:
\[ A \approx 7.14 \times 10^{-4} \, \text{m}^2 \]

### Final Answer
The minimum cross-sectional area of the rod is approximately:
\[ A \approx 7 \times 10^{-4} \, \text{m}^2 \]

---