The potential energy of a particle changes with distance x from a fixed origin as V = (A√(x))/(x+B), where A and B are constants with appropriate dimensions. The dimensions of AB are ______ (a) [M¹L5/2T⁻²] (b) [M3/2L5/2T⁻²] (c) [M¹L²T⁻²] (d) [M¹L7/2T⁻²]

Answer: (d). B is added to x: [B] = [L]. [V] = [A][L1/2]/[L] ⇒ [A] = [ML²T⁻²][L1/2] = [ML5/2T⁻²]. [AB] = [ML7/2T⁻²]. So the answer is (d).

The potential energy of a particle changes with distance x from a… | JEE Main 2026 (6 Apr, Shift 1)

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Applications of Dimensional Analysis

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

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Answer: (d)

$B$ is added to $x$ : $[B] = [L]$ .
$[V] = [A][L^{1/2}]/[L] \Rightarrow [A] = [ML^2T^{-2}][L^{1/2}] = [ML^{5/2}T^{-2}]$ .
$[AB] = [ML^{7/2}T^{-2}]$ .
So the answer is (d).

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