Question

A steel rod of diameter 1.0 cm is clamped firmly at each end when its temperature is 25°C so that it cannot contract on cooling. The tension in the rod at 0°C is (\(\alpha = 1 \times 10^{-5} /^\circ\)C, \(Y = 2 \times 10^{11}\) N/m²)

• A. 3925 N
• B. 7000 N
• C. 7400 N
• D. 4700 N

Hint:

Thermal force is independent of the length of the rod! It only depends on the material properties ($Y, \alpha$), the temperature change, and the cross-sectional area.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
When a rod is clamped and cooled, it tries to contract. Since the clamps prevent this, a thermal stress is developed within the rod, which manifests as tension.

Step 2: Key Formula or Approach:
1. Thermal Stress: \(\sigma = Y \alpha \Delta T\)
2. Tension (Force): \(F = \text{Stress} \times \text{Area} = Y \alpha \Delta T A\)
3. Area (\(A\)): \(\pi r^2 = \pi (d/2)^2\)

Step 3: Detailed Explanation:
Given: \(d = 1.0 \text{ cm} = 10^{-2} \text{ m} \implies r = 0.5 \times 10^{-2} \text{ m}\). \(\Delta T = 25^\circ\text{C} - 0^\circ\text{C} = 25^\circ\text{C}\). \(Y = 2 \times 10^{11} \text{ N/m}^2\), \(\alpha = 1 \times 10^{-5} /^\circ\text{C}\). 1. Calculate Area (\(A\)): \[ A = \pi (0.5 \times 10^{-2})^2 = \pi (0.25 \times 10^{-4}) \approx 3.14 \times 0.25 \times 10^{-4} \text{ m}^2 \] \[ A \approx 0.785 \times 10^{-4} \text{ m}^2 \] 2. Calculate Tension (\(F\)): \[ F = (2 \times 10^{11}) \times (1 \times 10^{-5}) \times 25 \times (0.785 \times 10^{-4}) \] \[ F = 2 \times 25 \times 0.785 \times (10^{11} \times 10^{-5} \times 10^{-4}) \] \[ F = 50 \times 0.785 \times 10^2 = 50 \times 78.5 \] \[ F = 3925 \text{ N} \]

Step 4: Final Answer
The tension in the rod is 3925 N.