Question

What is the stress developed when a steel rod of radius 10 mm and length 1.0 m is stretched by a 100 kN force?

• A. $2.13 \times 10^8 \ N/m^2$
• B. $2.10 \times 10^8 \ N/m^2$
• C. $2.71 \times 10^8 \ N/m^2$
• D. $2.31 \times 10^8 \ N/m^2$

Hint:

Use Stress =Force/πr^2 for cylindrical rods, and convert all units to SI for consistency.

Answer 1

The Correct Option is A

Stress is determined using the formula:
\[ \text{Stress} = \frac{\text{Force}}{\text{Area}} \]
The cross-sectional area is calculated as:
\[ A = \pi r^2 = \pi (10 \times 10^{-3})^2 = 3.14 \times 10^{-4} \text{ m}^2 \]
Consequently, the stress is:
\[ \text{Stress} = \frac{100 \times 10^3}{3.14 \times 10^{-4}} \approx 2.13 \times 10^8 \text{ N/m}^2 \]