Question

A compressive force, F is applied at the two ends of a long thin steel rod. It is heated, simultaneously, such that its temperature increases by ΔT. The net change in its length is zero. Let L be the length of the rod, A is its area of cross-section. Y is Young’s modulus, and α is its coefficient of linear expansion. Then, F is equal to

• (A) ${\mathrm{L}}^2\mathrm{Y}\mathrm{a}\mathrm{\Delta }\mathrm{T}$
• (B) $\frac{\mathrm{AY}}{\mathrm{a}\mathrm{\Delta }\mathrm{T}}$
• (C) $\mathrm{AYa}\mathrm{\Delta }\mathrm{T}$
• (D) $\mathrm{LAYa}\mathrm{\Delta }\mathrm{T}$

Answer 1

The Correct Option is C

This problem involves both thermal expansion and mechanical stress. The rod experiences a change in length due to both thermal expansion and the applied compressive force. The condition given is that the net change in its length is zero.

To solve the problem, we use two concepts:

1. Thermal Expansion: When a rod is heated, its length increases. The change in length due to heating is given by:

\(\Delta L_{\text{thermal}} = \alpha L \Delta T\)

2. Mechanical Compression: The compressive force reduces the length of the rod. The change in length due to the compressive force is given by Hooke's Law:

\(\Delta L_{\text{compression}} = \frac{F L}{A Y}\)

According to the problem, the net change in length is zero. Thus:

\(\Delta L_{\text{thermal}} + \Delta L_{\text{compression}} = 0\)

Substituting the expressions for each change in length gives:

\(\alpha L \Delta T = \frac{F L}{A Y}\)

Solving for the compressive force \( F \), we get:

\(F = A Y \alpha \Delta T\)

Therefore, the correct answer is:

(C) AY\alpha\Delta T