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PhysicsJEE Main 6 Apr 2026, Shift 1·Q2·MCQ·Easy–moderate

Units and Measurements · Dimensional analysis

The potential energy of a particle changes with distance x from a fixed origin as V=Axx+BV = \dfrac{A\sqrt{x}}{x + B}, where A and B are constants with appropriate dimensions. The dimensions of AB 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}}]
We corrected a printing error in the original paper

Source printed “A and B are constant with appropriate dimensions. The dimension of AB are”.

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