The diameter of a brass rod is 4 mm and Young's modulus of brass is `9xx10^(10)" N/m"^(2)`. The force required to stretch by `0.1%` of its length

Options

A. `360 piN`
B. 36 N
C. `144pixx10^(3)N`
D. `36 pixx10^(5)N`

Explanation

To solve the problem, we need to find the force required to stretch a brass rod by 0.1% of its length. We will use the formula derived from Young's modulus.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Diameter of the brass rod (d) = 4 mm = 4 × 10^(-3) m
- Young's modulus of brass (Y) = 9 × 10^(10) N/m²
- Percentage elongation (ΔL/L) = 0.1% = 0.1/100 = 0.001

2. **Calculate the Cross-Sectional Area (A) of the Rod:**
The area (A) can be calculated using the formula for the area of a circle:
\[ A = \frac{\pi}{4} d^2 \]
Substituting the value of d:
\[ A = \frac{\pi}{4} (4 \times 10^{-3})^2 = \frac{\pi}{4} (16 \times 10^{-6}) = 4\pi \times 10^{-6} \, \text{m}^2 \]

3. **Use Young's Modulus Formula:**
Young's modulus (Y) is defined as:
\[ Y = \frac{F/A}{\Delta L/L} \]
Rearranging this formula to find the force (F):
\[ F = Y \cdot A \cdot \frac{\Delta L}{L} \]

4. **Substituting Values:**
We need to express ΔL/L in terms of the original length (L). Since we are looking for the force required to stretch by 0.1%, we can substitute:
\[ F = Y \cdot A \cdot 0.001 \]

Now substituting the values of Y and A:
\[ F = (9 \times 10^{10}) \cdot (4\pi \times 10^{-6}) \cdot 0.001 \]

5. **Calculate the Force:**
First, calculate the area:
\[ A = 4\pi \times 10^{-6} \approx 12.5664 \times 10^{-6} \, \text{m}^2 \]
Then substitute this back into the force equation:
\[ F = 9 \times 10^{10} \cdot 12.5664 \times 10^{-6} \cdot 0.001 \]
\[ F = 9 \times 12.5664 \times 10^{4} \approx 113.097 \times 10^{4} \, \text{N} \]
\[ F \approx 1.13097 \times 10^{6} \, \text{N} \approx 1130.97 \, \text{N} \]

### Final Answer:
The force required to stretch the brass rod by 0.1% of its length is approximately **1130.97 N**.