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Question #43534

Difficulty: 8/10 | 🟢 Free

The Question

A straight rod of length $L$ is rigidly fixed between two unyielding walls. The rod has a uniform Young's modulus $E$ . Its cross-sectional area $A(x)$ and coefficient of linear thermal expansion $\alpha(x)$ vary along its length $x$ (from $x=0$ to $x=L$ ) as: $A(x) = A_0 2^{-x/L}$ $\alpha(x) = \alpha_0 2^{x/L}$ where $A_0$ and $\alpha_0$ are positive constants. The temperature of the rod is uniformly increased by $\Delta T$ . If the material fails when the local compressive stress reaches a critical value $\sigma_b$ , at what minimum temperature rise $\Delta T$ will the rod break?

Options

A.
$\frac{\sigma_b}{2 E \alpha_0}$
B.
$\frac{\sigma_b \ln 2}{E \alpha_0}$
C.
$\frac{\sigma_b}{E \alpha_0}$
D.
$\frac{\sigma_b \ln 2}{2 E \alpha_0}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Multi-Concept_Bridge

Master problems that require linking physics and math concepts to solve complex competitive questions.

Concepts Behind This Question

Breaking Stress Concept|Thermal Stress in Constrained Rods|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to handle non-uniform thermal expansion and varying cross-sections in constrained structures.

A common trap is assuming the standard formula $\sigma = E \alpha \Delta T$ applies globally using average values, or failing to recognize that while the internal equilibrium force $F$ is uniform, the mechanical stress $\sigma(x) = F/A(x)$ varies inversely with the cross-sectional area.

Students must correctly apply the superposition of thermal and mechanical strains inside an integral, and then evaluate the local stress at the weakest point (minimum area) rather than averaging the stress.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Mechanical Properties of Solids - Class 11 Part 2
Concept	Breaking Stress Concept
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free