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Question #84001

Difficulty: 5/10 | 🟢 Free

The Question

A thin spherical shell of radius $R$ and thickness $t$ ( $t \ll R$ ) is made of a material with Young's modulus $Y$ and coefficient of linear thermal expansion $\alpha$ . Assume the Poisson's ratio of the material is zero. The shell contains an ideal gas having $f$ degrees of freedom. Initially, the system is in thermal equilibrium at temperature $T_0$ , the shell is completely unstressed (internal pressure equals the external atmospheric pressure $P_0$ ). A total heat $Q$ is slowly supplied to the gas. Assume the heat capacity of the shell is negligible, and the work done by the gas during expansion is negligible compared to the heat supplied. Assuming the change in radius $\Delta R \ll R$ , which of the following expressions correctly represents the tensile hoop stress $\sigma$ developed in the shell?

Options

A.
$\frac{3 Q (1 - 3 \alpha T_0)}{4 \pi R^2 t f \left( 1 + \frac{3 P_0 R}{2 t Y} \right)}$
B.
$\frac{3 Q (1 + 3 \alpha T_0)}{4 \pi R^2 t f \left( 1 - \frac{3 P_0 R}{2 t Y} \right)}$
C.
$\frac{3 Q}{4 \pi R^2 t f \left( 1 + \frac{3 P_0 R}{2 t Y} \right)}$
D.
$\frac{3 Q (1 - \alpha T_0)}{4 \pi R^2 t f}$

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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

Molecular Degrees of Freedom|Thermal Stress Calculation|Molecular Forces of Cohesion

Why This Question Is Tricky

Examiner Levers

Dimensional Analysis

Tests if the student verifies units/dimensions before selecting.

Difficulty Boost : +2

Common Mistakes

This question tests the coupling of thermodynamic state equations with solid mechanics, specifically targeting the misconception of treating gas containers as perfectly rigid bodies.

Students often calculate the pressure rise using $\Delta P = \frac{nR\Delta T}{V}$ assuming constant volume, which leads directly to incorrect options (ignoring thermal expansion or elastic expansion).

By applying the First Law of Thermodynamics and the Equipartition Theorem, the temperature rise is found.

However, the true pressure depends dynamically on the shell's volume, which changes due to both thermal expansion ( $\alpha\Delta T$ ) and elastic stretching ( $\sigma/Y$ ).

The correct approach requires setting up a simultaneous equilibrium where the internal gas pressure is a function of the shell's strain, and the shell's stress is a function of the gas pressure.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Kinetic Theory - Class 11 Part 2
Concept	Molecular Degrees of Freedom
Difficulty	5
Cognitive Load	6
Target Solve Time	140 sec
Access	Free