Physics · 2026
JEE Main · 6 April 2026, Shift 1 · Q27
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 constant with appropriate…
The potential energy of a particle changes with distance $\displaystyle x$ from a fixed origin as $\displaystyle V=\frac{A \sqrt{x}}{x+B}$, where $\displaystyle A$ and $\displaystyle B$ are constant with appropriate dimensions. The dimensions of $\displaystyle A B$ are $\displaystyle \_\_\_\_$.
Official answer
From NTA’s final answer key for this paper.
(4)
$\displaystyle \left[\mathrm{M}^1 \mathrm{~L}^{7 / 2} \mathrm{~T}^{-2}\right]$
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JEE Main 2026 Physics question, with the answer from NTA’s final answer key. Where our answers come from.