Q 11-01-053JEE MainJEE Main 2026 (6 Apr, Shift 1)Medium
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 ______
Answer: (D) $[\text{M}^1\text{L}^{7/2}\text{T}^{-2}]$
$B$ is added to $x$, so $[B] = [\text{L}]$.
Then $[V] = \dfrac{[A]\text{L}^{1/2}}{\text{L}}$, so $[A] = [V]\,\text{L}^{1/2} = \text{ML}^2\text{T}^{-2} \cdot \text{L}^{1/2} = \text{ML}^{5/2}\text{T}^{-2}$.
$[AB] = \text{ML}^{7/2}\text{T}^{-2}$.
Solution by Sreeraj P, M.Sc Physics