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JEE Advanced 2020 Solutions | 🤔 FAIR or UNFAIR ? | Thermodynamics #BonusQuestion #IIT
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A question yet again given BONUS in JEE Advanced 2020. This question is from Thermodynamics and came in paper 2.
JEE Advanced 2020 | Paper 2 | Thermodynamics Question
A thermally isolated cylindrical closed vessel of height 8 m is kept vertically. It is divided into two equal parts by a diathermic (perfect thermal conductor) frictionless partition of mass 8.3 kg. Thus the partition is held initially at a distance of 4 m from the top, as shown in the schematic figure below. Each of the two parts of the vessel contains 0.1 mole of an ideal gas at temperature 300 K. The partition is now released and moves without any gas leaking from one part of the vessel to the other. When equilibrium is reached, the distance of the partition from the top (in m) will be _______ (take the acceleration due to gravity =10 ms−2 and the universal gas constant = 8.3 J mol−1K −1 )
Timestamp :
00:00 Introduction
00:32 Question Statement
02:18 Correct Solution
09:14 Incorrect Solution but Same Result
10:54 Why Bonus ? Unfair to many ?
JEE Advanced 2020 Solution | JEE Advanced Bonus Question | JEE Advanced 2020 Thermodynamics Question | How to Solve JEE Advanced 2020 bonus question ?
Target JEE Advanced 2021 | Target JEE Advanced 2022 | Target JEE Advanced 2023
#JEEAdvancedPYQs #Thermodynamics #Eduniti
About Faculty-
Completed Bachelors and Masters from IIT Kharagpur. Having an experience of +4 Years in teaching I reckoned the urge within me to help students appreciate the subject by delivering the toughest concept in simplest manner through practical demonstrations and visual aids.
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JEE Advanced 2020 | Paper 2 | Thermodynamics Question
A thermally isolated cylindrical closed vessel of height 8 m is kept vertically. It is divided into two equal parts by a diathermic (perfect thermal conductor) frictionless partition of mass 8.3 kg. Thus the partition is held initially at a distance of 4 m from the top, as shown in the schematic figure below. Each of the two parts of the vessel contains 0.1 mole of an ideal gas at temperature 300 K. The partition is now released and moves without any gas leaking from one part of the vessel to the other. When equilibrium is reached, the distance of the partition from the top (in m) will be _______ (take the acceleration due to gravity =10 ms−2 and the universal gas constant = 8.3 J mol−1K −1 )
Timestamp :
00:00 Introduction
00:32 Question Statement
02:18 Correct Solution
09:14 Incorrect Solution but Same Result
10:54 Why Bonus ? Unfair to many ?
JEE Advanced 2020 Solution | JEE Advanced Bonus Question | JEE Advanced 2020 Thermodynamics Question | How to Solve JEE Advanced 2020 bonus question ?
Target JEE Advanced 2021 | Target JEE Advanced 2022 | Target JEE Advanced 2023
#JEEAdvancedPYQs #Thermodynamics #Eduniti
About Faculty-
Completed Bachelors and Masters from IIT Kharagpur. Having an experience of +4 Years in teaching I reckoned the urge within me to help students appreciate the subject by delivering the toughest concept in simplest manner through practical demonstrations and visual aids.
Follow us on
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