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Heat and Thermodynamics question

2007 · Shift 1 · Q59
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Heat and Thermodynamics question

2007 · Shift 1 · Q59

JEE AdvancedPhysicsHeat and ThermodynamicsMCQ+3 / −1
A fixed thermally conducting cylinder has a radius RRR and height L0L_0L0​. The cylinder is open at its bottom and has a small hole at its top. A piston of mass MMM is held at a distance LLL from the top surface, as shown in the figure. The atmospheric pressure is P0P_0P0​. IIT-JEE 2007 Paper 1 Offline Physics - Heat and Thermodynamics Question 19 English ComprehensionThe piston is now pulled out slowly and held at a distance 2L from the top. The pressure in the cylinder between its top and the piston will then be
  1. A
    P 0_00​
  2. B
    P02{{{P_0}} \over 2}2P0​​
  3. C
    P02+MgπR2{{{P_0}} \over 2} + {{Mg} \over {\pi {R^2}}}2P0​​+πR2Mg​
  4. D
    P02−MgπR2{{{P_0}} \over 2} - {{Mg} \over {\pi {R^2}}}2P0​​−πR2Mg​
View written solutionFree

Correct answer: A

Step-by-step Solution:

  1. Analyze the System: The problem describes a cylinder with a piston inside. We need to find the pressure in the region between the top of the cylinder and the piston after the piston is moved from a distance LLL to 2L2L2L from the top.

  2. Identify the Key Feature: The most crucial piece of information provided is that the cylinder has a "small hole at its top".

  3. Apply the Concept of Pressure Equalization: This small hole connects the space inside the cylinder (above the piston) directly to the outside atmosphere. In any static or quasi-static (slowly changing) situation, when a volume of gas is connected to an external reservoir (like the atmosphere) through an opening, the pressure inside that volume will equalize with the pressure of the external reservoir.

  4. Initial State Analysis: In the initial state, the piston is at a distance LLL from the top. The volume above the piston is connected to the atmosphere (at pressure P0P_0P0​) through the hole. Therefore, the pressure in this region is equal to the atmospheric pressure, P0P_0P0​.

  5. Final State Analysis: The piston is then "pulled out slowly" to a distance of 2L2L2L from the top. The term "slowly" implies that the process is quasi-static, meaning the system is always in or very close to equilibrium. As the piston moves down, the volume of the space above it increases from πR2L\pi R^2 LπR2L to πR2(2L)\pi R^2 (2L)πR2(2L). Because of the hole at the top, air from the atmosphere will flow into this expanding volume to maintain the pressure inside equal to the atmospheric pressure.

  6. Conclusion: Since the space above the piston is always connected to the atmosphere through the hole, its pressure will remain constant and equal to the atmospheric pressure, P0P_0P0​, regardless of the piston's position (as long as it doesn't block the hole). Therefore, the final pressure in the cylinder between its top and the piston is P0P_0P0​.

  7. Evaluating Other Information (Distractors):

    • The mass of the piston (MMM) and the fact that the cylinder is open at the bottom would be relevant for calculating the pressure below the piston. They do not affect the pressure above the piston because of the hole at the top.
    • The fact that the cylinder is "thermally conducting" and the process is "slow" would suggest an isothermal process. If the gas were trapped (no hole), the pressure would change according to Boyle's Law (P1V1=P2V2P_1V_1 = P_2V_2P1​V1​=P2​V2​), leading to a final pressure of P0/2P_0/2P0​/2. However, the presence of the hole makes this line of reasoning incorrect as the gas is not trapped.

Based on the direct connection to the atmosphere via the hole, the pressure must be P0P_0P0​.

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