Start Learning Free

Before You Begin

Question #43767

Difficulty: 8/10 | 🟢 Free

The Question

A thin opaque cylindrical disk of initial density ρ0\rho_0 and initial face area A0A_0 at temperature T0T_0 is suspended in a transparent coolant gas maintained at T0T_0. A laser beam of constant total power PP is completely absorbed by one flat face of the disk. The disk loses heat to the gas exclusively from this illuminated face via convection, with a constant convective heat transfer coefficient hh. The coefficient of linear expansion of the disk material is α\alpha. Assume the face area expands as A=A0(1+2αΔT)A = A_0(1 + 2\alpha\Delta T) and the density varies as ρ=ρ0(1−3αΔT)\rho = \rho_0(1 - 3\alpha\Delta T), where ΔT=T−T0\Delta T = T - T_0. Which of the following expressions gives the steady-state density of the disk?

Options

  • A.

    ρ0[1−34(1+8αPhA0−1)]\rho_0 \left[ 1 - \frac{3}{4} \left( \sqrt{1 + \frac{8\alpha P}{h A_0}} - 1 \right) \right]

  • B.

    ρ0[1−3αPhA0]\rho_0 \left[ 1 - \frac{3\alpha P}{h A_0} \right]

  • C.

    ρ0[1−32(1+4αPhA0−1)]\rho_0 \left[ 1 - \frac{3}{2} \left( \sqrt{1 + \frac{4\alpha P}{h A_0}} - 1 \right) \right]

  • D.

    ρ0[1−12(1+12αPhA0−1)]\rho_0 \left[ 1 - \frac{1}{2} \left( \sqrt{1 + \frac{12\alpha P}{h A_0}} - 1 \right) \right]

Check Your Answer

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

Intensity and Photon Flux|Thermal Expansion of Area and Volume|Variation of Density with Temperature

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to handle non-linear feedback loops in thermal steady-state systems.

The primary trap lies in treating the target's area as a static parameter (A0A_0) during the energy balance phase, which incorrectly yields ΔT=P/(hA0)\Delta T = P / (h A_0) and leads to Option B.

Students must recognize that as the laser heats the target, thermal expansion increases the surface area available for convective heat loss, dynamically altering the cooling rate.

By forcing the solution of a quadratic equation coupling the temperature-dependent area A(ΔT)A(\Delta T) with the convective heat flux, the problem exposes the misconception of decoupling thermal expansion from thermodynamic energy balance.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterDual Nature of Radiation and Matter - Class 12
ConceptIntensity and Photon Flux
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree