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Friday, 9 October

JEE Main 2021 · PhysicsMultiple choiceSingle correctEasyApplication

JEE Main 25 July 2021, Shift 1, Physics Q20

Question 20 of 90 in this shift, Physics question 20 of 30, Section A.

A monoatomic ideal gas, initially at temperature T1T_1 is enclosed in a cylinder fitted with a frictionless piston. The gas is allowed to expand adiabatically to a temperature T2T_2 by releasing the piston suddenly. If l1l_1 and l2l_2 are the lengths of the gas column, before and after the expansion respectively, then the value of T1T2\frac{T_1}{T_2} will be :
  1. (1)l1l2\frac{l_1}{l_2}
  2. (2)l2l1\frac{l_2}{l_1}
  3. (3)(l1l2)23\left(\frac{l_1}{l_2}\right)^{\frac{2}{3}}
  4. (4)(l2l1)23\left(\frac{l_2}{l_1}\right)^{\frac{2}{3}}Official answer

Official answer

Option 4

NTA final key.

Chapter
Thermodynamics
Topic
Thermodynamic Processes
Idea tested
Adiabatic T-V and P-T Relations

Same idea in other shifts

Asked 3× in all
  1. 9 Jan 2020, Shift 2 · Q21Starting at temperature 300 K, one mole of an ideal diatomic gas (γ=1.4\gamma = 1.4) is first compressed adiabatically from volume V1V_1 to…MediumNumerical value
  2. 2 Sep 2020, Shift 1 · Q22An engine takes in 5 moles of air at 20∘20^\circC and 1 atm, and compresses it adiabaticaly to 1/10th1/10^{\mathrm{th}} of the original volume.…MediumNumerical value

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.