The elastic limit of brass is `379MPa`. What should be the minimum diameter of a brass rod if it is to support a 400 N load without exceeding its elastic limit?

Options

A. 1.00 mm
B. 1.16 mm
C. 0.90 mm
D. 1.36 mm

Answer

To solve the problem, we need to determine the minimum diameter of a brass rod that can support a load of 400 N without exceeding its elastic limit of 379 MPa.

### Step-by-Step Solution:

1. **Understand the Concept of Stress**:
Stress is defined as the force applied per unit area. The formula for stress (σ) is given by:
\[ \sigma = \frac{F}{A} \]
where \( F \) is the force applied, and \( A \) is the cross-sectional area.

2. **Relate Area to Diameter**:
For a circular cross-section (like a rod), the area \( A \) can be expressed in terms of the diameter \( d \):
\[ A = \frac{\pi d^2}{4} \]

3. **Substitute Area into the Stress Formula**:
Substituting the expression for area into the stress formula gives:
\[ \sigma = \frac{F}{\frac{\pi d^2}{4}} = \frac{4F}{\pi d^2} \]

4. **Set Stress Equal to the Elastic Limit**:
We know the elastic limit of brass is 379 MPa, which can be converted to Pascals:
\[ 379 \text{ MPa} = 379 \times 10^6 \text{ Pa} \]
Setting the stress equal to the elastic limit:
\[ \frac{4F}{\pi d^2} = 379 \times 10^6 \]

5. **Substitute the Load into the Equation**:
The load \( F \) is given as 400 N. Substituting this value into the equation:
\[ \frac{4 \times 400}{\pi d^2} = 379 \times 10^6 \]

6. **Rearranging the Equation**:
Rearranging the equation to solve for \( d^2 \):
\[ d^2 = \frac{4 \times 400}{\pi \times (379 \times 10^6)} \]

7. **Calculate the Right Side**:
Calculate the right side of the equation:
\[ d^2 = \frac{1600}{\pi \times 379 \times 10^6} \]
Using \( \pi \approx 3.14 \):
\[ d^2 = \frac{1600}{3.14 \times 379 \times 10^6} \approx \frac{1600}{1.187 \times 10^9} \approx 1.347 \times 10^{-6} \]

8. **Taking the Square Root**:
Now, take the square root to find \( d \):
\[ d = \sqrt{1.347 \times 10^{-6}} \approx 0.00115 \text{ m} = 1.15 \text{ mm} \]

### Final Answer:
The minimum diameter of the brass rod should be approximately **1.15 mm**.