The potential energy of | Unit & Dimensions | JEE MAIN_2026

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JEE MAIN_2026

06-04-26_S1

Question

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 constant with appropriate dimensions. The dimensions of $A B$ are $\_\_\_\_$。

Select the correct option:

A

$\left[\mathrm{M}^1 \mathrm{~L}^{5 / 2} \mathrm{~T}^{-2}\right]$

B

$\left[\mathrm{M}^{3 / 2} \mathrm{~L}^{5 / 2} \mathrm{~T}^{-2}\right]$

C

$\left[\mathrm{M}^1 \mathrm{~L}^2 \mathrm{~T}^{-2}\right]$

D

$\left[\mathrm{M}^1 \mathrm{~L}^{7 / 2} \mathrm{~T}^{-2}\right]$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

$\begin{aligned} & {[\mathrm{B}]=[\mathrm{x}]=\mathrm{L}} \\ & {[\mathrm{U}]=\left[\frac{\mathrm{A} \sqrt{\mathrm{x}}}{\mathrm{B}+\mathrm{x}}\right] \Rightarrow \mathrm{ML}^2 \mathrm{~T}^{-2}=\frac{[\mathrm{A}] \mathrm{L}^{1 / 2}}{\mathrm{~L}}} \\ & {[\mathrm{~A}]=\mathrm{ML}^{5 / 2} \mathrm{~T}^{-2}}\end{aligned}$

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