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Question #81259

Difficulty: 5/10 | 🟢 Free

The Question

A sealed spherical elastomeric shell contains air (assumed to be an ideal gas) initially at atmospheric pressure P0=1.0×105 PaP_0 = 1.0 \times 10^5 \text{ Pa} and temperature T0=300 KT_0 = 300 \text{ K}. The shell is submerged in a lake to a depth h=2.0 mh = 2.0 \text{ m}, where the water temperature is Tf=294 KT_f = 294 \text{ K}. The fractional change in the cavity's volume δ=ΔVV0\delta = \frac{\Delta V}{V_0} is governed by the relation δ=Pint−PextKeff−γΔT\delta = \frac{P_{int} - P_{ext}}{K_{eff}} - \gamma \Delta T, where PintP_{int} and PextP_{ext} are the final internal and external pressures, Keff=9.0×105 PaK_{eff} = 9.0 \times 10^5 \text{ Pa} is the effective elastic modulus of the shell, γ=5.0×10−4 K−1\gamma = 5.0 \times 10^{-4} \text{ K}^{-1} is the coefficient of volume expansion, and ΔT=T0−Tf\Delta T = T_0 - T_f. Assuming ∣δ∣≪1|\delta| \ll 1, calculate the magnitude of the fractional change in volume ∣δ∣|\delta| of the cavity. (Take the density of water ρ=103 kg/m3\rho = 10^3 \text{ kg/m}^3, g=10 m/s2g = 10 \text{ m/s}^2, and use the first-order approximation for small changes, ignoring cross terms).

Options

  • A.

    2.47 \times 10^{-2}

  • B.

    2.52 \times 10^{-2}

  • C.

    2.27 \times 10^{-2}

  • D.

    1.93 \times 10^{-2}

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Atmospheric Pressure Reference|Expansion of Cavities in Solids|Stress vs. Strain Definition

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 synthesis of mechanical elasticity, thermal expansion, and the ideal gas law.

The primary trap lies in treating the internal pressure of the cavity as a constant (atmospheric) or failing to couple the ideal gas law with the elastic deformation of the shell.

Students often calculate the mechanical compression using only the external hydrostatic pressure, ignoring that the cavity's volume reduction increases the internal gas pressure, which in turn resists further compression (the P0δP_0 \delta term).

Furthermore, they frequently miss the dual effect of temperature: it causes thermal contraction of the shell material AND reduces the internal gas pressure, both of which must be simultaneously resolved in the equilibrium equation.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterMechanical Properties of Fluids - Class 11 Part 2
ConceptAtmospheric Pressure Reference
Difficulty5
Cognitive Load6
Target Solve Time140 sec
AccessFree