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Saturday, 31 October

JEE Main 2021 · PhysicsMultiple choiceSingle correctMediumMulti-step

JEE Main 24 February 2021, Shift 1, Physics Q1

Question 1 of 90 in this shift, Physics question 1 of 30, Section A.

The workdone by a gas molecule in an isolated system is given by, W=αβ2e−x2αkTW = \alpha\beta^{2} e^{-\frac{x^{2}}{\alpha kT}}, where xx is the displacement, kk is the Boltzmann constant and T is the temperature. α\alpha and β\beta are constants. Then the dimensions of β\beta will be :
  1. (1)[M2 L T2][\mathrm{M^{2}\,L\,T^{2}}]
  2. (2)[M0 L T0][\mathrm{M^{0}\,L\,T^{0}}]
  3. (3)[M L T−2][\mathrm{M\,L\,T^{-2}}]Official answer
  4. (4)[M L2 T−2][\mathrm{M\,L^{2}\,T^{-2}}]

Official answer

Option 3

NTA final key.

Topic
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Idea tested
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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 108 of the 174 JEE Main shifts held from 2017 to 2026.