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Question #83696

Difficulty: 8/10 | 🟢 Free

The Question

A mechanical damper exhibiting elastic hysteresis is modeled as an ideal gas (with dimensionless adiabatic index $\gamma$ ) undergoing a thermodynamic cycle. The cycle consists of three reversible processes:

Polytropic expansion from state A with pressure $P_0$ (in $\text{Pa}$ ) and volume $V_0$ (in $\text{m}^3$ ) to state B at volume $\eta V_0$ with dimensionless polytropic index $n_1$ .
Isochoric pressure increase from state B to state C.
Polytropic compression from state C back to state A with dimensionless polytropic index $n_2$ . It is given that $1 < n_2 < n_1 < \gamma$ and $\eta > 1$ (where $\eta$ is a dimensionless constant). During the cycle, net mechanical work is converted into heat. What is the total heat rejected to the surroundings per cycle?

Options

A.
$P_0 V_0 \left[ \frac{1 - \eta^{1-n_2}}{n_2 - 1} - \frac{1 - \eta^{1-n_1}}{n_1 - 1} \right] \text{ J}$
B.
$P_0 V_0 \frac{(1 - \eta^{1-n_2})(\gamma - n_2)}{(\gamma - 1)(n_2 - 1)} \text{ J}$
C.
$P_0 V_0 \frac{(1 - \eta^{1-n_1})(\gamma - n_1)}{(\gamma - 1)(n_1 - 1)} \text{ J}$
D.
$P_0 V_0 \frac{1 - \eta^{1-n_2}}{n_2 - 1} \text{ J}$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Variable_Constraint

Develop pure symbolic logic skills by mastering algebraic constraints without relying on numerical data.

Concepts Behind This Question

Elastic Hysteresis Loop|Polytropic Process|Approximation (First-order expansion)

Why This Question Is Tricky

Examiner Levers

Variable Manipulation

Algebraic solutions only (No numbers); tests pure symbolic logic.

Difficulty Boost : +4

Common Mistakes

This question tests the 'Internal Energy Entropy Coupling' trap in thermodynamic cycles modeling mechanical hysteresis.

Students often incorrectly assume that the total heat rejected to the surroundings equals the net mechanical energy dissipated by the damper (the area of the P-V loop, Option A).

They fail to recognize that during the expansion (loading) phase, the material undergoes thermoelastic cooling and absorbs heat from the surroundings.

According to the First Law of Thermodynamics for a cycle ( $Q_{net} = W_{net}$ ), the total heat rejected during compression must equal the sum of the mechanical energy dissipated AND the heat absorbed during expansion.

By identifying that heat is only rejected during the compression phase, students must calculate the specific heat exchange for that path rather than the net work of the cycle.

Quick Facts About This Question

Question Information

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