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Friday, 9 October

JEE Main 2026 · PhysicsMultiple choiceSingle correctMediumApplication

JEE Main 6 April 2026, Shift 1, Physics Q27

Question 27 of 75 in this shift, Physics question 2 of 25, Section A.

The potential energy of a particle changes with distance xx from a fixed origin as V=Axx+BV=\frac{A\sqrt{x}}{x+B}, where AA and BB are constant with appropriate dimensions. The dimensions of ABAB are ________.
  1. (1)[M1L5/2T−2][\mathrm{M^1L^{5/2}T^{-2}}]
  2. (2)[M3/2L5/2T−2][\mathrm{M^{3/2}L^{5/2}T^{-2}}]
  3. (3)[M1L2T−2][\mathrm{M^1L^2T^{-2}}]
  4. (4)[M1L7/2T−2][\mathrm{M^1L^{7/2}T^{-2}}]Official answer

Official answer

Option 4

NTA final key.

Chapter
Units and Measurements
Topic
No close match in our taxonomy
Idea tested
No close match in our taxonomy

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.