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Question #42500

Difficulty: 8/10 | 🟢 Free

The Question

A vertical cylinder of cross-sectional area A=0.1 m2A = 0.1 \, \text{m}^2 is completely submerged in a fluid bath. The top of the cylinder is fixed at a depth H0=2.0 mH_0 = 2.0 \, \text{m} below the fluid surface. The bottom of the cylinder is sealed by a movable piston of mass MM and volume Vp=0.05 m3V_p = 0.05 \, \text{m}^3. An external mechanism moves the piston to perform a cycle.

The cycle consists of four steps:

  1. The piston is slowly pushed down, increasing the cylinder's internal volume from V1=0.1 m3V_1 = 0.1 \, \text{m}^3 to V2=0.3 m3V_2 = 0.3 \, \text{m}^3, while the fluid is maintained at TH=350 KT_H = 350 \, \text{K}.
  2. The piston is held in place while the fluid is cooled to TC=300 KT_C = 300 \, \text{K}.
  3. The piston is slowly pulled up, decreasing the volume from V2V_2 to V1V_1, while the fluid is maintained at TCT_C.
  4. The piston is held in place while the fluid is heated back to THT_H.

The fluid density is ρ(300 K)=1000 kg/m3\rho(300 \, \text{K}) = 1000 \, \text{kg/m}^3 and ρ(350 K)=960 kg/m3\rho(350 \, \text{K}) = 960 \, \text{kg/m}^3. Calculate the magnitude of the net work done BY THE FLUID on the piston during one complete cycle. (Assume g=10 m/s2g = 10 \, \text{m/s}^2 and Patm=105 PaP_{\text{atm}} = 10^5 \, \text{Pa})

Options

  • A.

    360 J360 \, \text{J}

  • B.

    320 J320 \, \text{J}

  • C.

    280 J280 \, \text{J}

  • D.

    440 J440 \, \text{J}

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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

Refrigerator Coefficient of Performance|Mechanical Energy Conservation|Archimedes' Buoyancy Dynamics

Why This Question Is Tricky

Common Mistakes

This question tests the precise application of Archimedes' principle and hydrostatic pressure on a partially exposed submerged boundary.

A common trap is assuming the external pressure is simply Patm+ρghtopP_{\text{atm}} + \rho g h_{\text{top}} and neglecting the buoyant contribution of the piston's own volume (its thickness), which leads to an incorrect pressure head (yielding 320 J).

Another trap is incorrectly subtracting the buoyant force (yielding 280 J).

Additionally, students often fail to integrate the pressure over the varying depth, incorrectly treating the hydrostatic pressure as constant during the expansion phase.

By asking for the work done specifically by the fluid, it isolates the mechanical-hydrostatic coupling from the thermodynamic gas laws.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterThermodynamics - Class 11 Part 2
ConceptRefrigerator Coefficient of Performance
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree