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Question #82134

Difficulty: 8/10 | 🟢 Free

The Question

A metallic rod of initial length $L_0$ and cross-sectional area $A$ is initially at temperature $T_0$ . Its Young's modulus varies with temperature as $Y(T) = \frac{Y_0}{1 + \beta (T - T_0)}$ , where $Y_0$ and $\beta$ are positive constants. The coefficient of linear thermal expansion of the rod is $\alpha$ . The rod is subjected to a thermodynamic process where a tensile force $F$ is gradually increased from $0$ to $F_0$ while the temperature is simultaneously increased from $T_0$ to $T_0 + \Delta T$ . The force and temperature are coupled such that $F(T) = F_0 \frac{T - T_0}{\Delta T}$ . Assuming the strains remain small and the rod is always in mechanical equilibrium, what is the total work done by the external force on the rod during this process?

Options

A.
$\frac{1}{2} F_0 L_0 \alpha \Delta T + \frac{F_0^2 L_0}{2 A Y_0} \left( 1 + \frac{4}{3} \beta \Delta T \right)$
B.
$\frac{1}{2} F_0 L_0 \alpha \Delta T + \frac{F_0^2 L_0}{2 A Y_0} \left( 1 + \beta \Delta T \right)$
C.
$F_0 L_0 \alpha \Delta T + \frac{F_0^2 L_0}{2 A Y_0} \left( 1 + \frac{4}{3} \beta \Delta T \right)$
D.
$\frac{1}{2} F_0 L_0 \alpha \Delta T + \frac{F_0^2 L_0}{2 A Y_0} \left( 1 + \frac{2}{3} \beta \Delta T \right)$

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

Work Done in Stretching|Thermal Expansion of Area and Volume|Integration

Why This Question Is Tricky

Examiner Levers

Visualization Stress

Requires mental rotation or 3D mapping (Optics/Geometry/Vectors).

Difficulty Boost : +4

Concept Bridge

Cross-chapter synthesis (e.g., using Logarithms in Electrochemistry).

Difficulty Boost : +5

Dimensional Analysis

Tests if the student verifies units/dimensions before selecting.

Difficulty Boost : +2

Common Mistakes

This question tests the fundamental definition of work $W = \int F \, dL$ in a multi-variable state space $(F, T)$ .

Students commonly misapply the elastic potential energy formula $W = \frac{1}{2} F_{max} \Delta L_{total}$ , incorrectly assuming that the work done is independent of the path taken in the force-temperature plane (leading to Option B).

Furthermore, by making Young's modulus temperature-dependent, the total strain becomes a non-linear superposition.

A secondary trap catches students who integrate $\int \frac{F L_0}{A Y(F)} dF$ but forget the product rule when differentiating the state variable $L(F, T)$ , missing the $F \, d(1/Y)$ term and arriving at Option D.

The correct approach requires a rigorous line integral parameterized by the coupled variable.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Mechanical Properties of Solids - Class 11 Part 2
Concept	Work Done in Stretching
Difficulty	8
Cognitive Load	8
Target Solve Time	180 sec
Access	Free