JEE Main 2026 · 6 Apr · Shift 1 · Physics Q27

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PhysicsSingle correct+4 / -1Question 27 of 75

The potential energy of a particle changes with distance $x$ from a fixed origin as $V=\dfrac{A\sqrt x}{x+B}$ , where $A$ and $B$ are constant with appropriate dimensions. The dimensions of $AB$ are ________.

[M^1L^{5/2}T^{-2}]

[M^{3/2}L^{5/2}T^{-2}]

[M^1L^2T^{-2}]

[M^1L^{7/2}T^{-2}]

Answer and worked solution

Official answer: D

B. It is added to x, so $[B]=L$ .

A. $V\sim\dfrac{A\sqrt x}x$ , so $[A]=[V]L^{1/2}=ML^{5/2}T^{-2}$ .

Product: $[AB]=ML^{7/2}T^{-2}$ .

(D)

Topics:Dimensional analysis

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JEE Main 2026 · 6 Apr · Shift 1, Physics question 27