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 at both ends, a change in temperature attempts to alter its length. Since its length is constrained, a thermal stress is developed within the rod. Cooling the rod creates tension because it wants to contract but is prevented from doing so. 
Step 2: Key Formula or Approach: 
The thermal strain is given by: $\frac{\Delta L}{L} = \alpha \Delta T$ 
The thermal stress is: $\text{Stress} = Y \times \text{Strain} = Y \alpha \Delta T$ 
The tension (force) developed is: $F = \text{Stress} \times A = Y A \alpha \Delta T$ 
Step 3: Detailed Explanation: 
Given values: Diameter $d = 1.0 \text{ cm} = 10^{-2} \text{ m}$ 
Young's modulus $Y = 2 \times 10^{11} \text{ N/m}^2$ 
Coefficient of linear expansion $\alpha = 1 \times 10^{-5} \text{ /}^\circ\text{C}$ 
Change in temperature $\Delta T = 25^\circ\text{C} - 0^\circ\text{C} = 25^\circ\text{C}$ Calculate the cross-sectional area $A$: \[ A = \frac{\pi d^2}{4} = \frac{3.14 \times (10^{-2})^2}{4} = \frac{3.14 \times 10^{-4}}{4} = 0.785 \times 10^{-4} \text{ m}^2 \] Now compute the tension $F$: \[ F = Y A \alpha \Delta T \] \[ F = (2 \times 10^{11}) \times (0.785 \times 10^{-4}) \times (1 \times 10^{-5}) \times 25 \] \[ F = (2 \times 25) \times 0.785 \times 10^{11} \times 10^{-9} \] \[ F = 50 \times 0.785 \times 10^2 \] \[ F = 5000 \times 0.785 = 3925 \text{ N} \] Step 4: Final Answer: 
The tension in the rod is 3925 N.