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 dimension of AB are ______.

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

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

JEE Main · 2026 · 06 Apr 2026 · Shift 1 · Units & Measurement

The potential energy of a particle changes with distance

x

from a fixed origin as

V = \frac{A x}{x + B}

, where A and B are constant with appropriate dimensions. The dimension of AB are ______.

Answer: (D)

Since

x

and

B

are added,

[B] = [x] = L

. Potential energy has

[V] = M L^{2} T^{- 2}

. From

V = \frac{A x}{x + B}

, dimensionally

[V] = \frac{[ A ] L ^{1/2}}{L} = [A] L^{- 1/2}

, so

[A] = M L^{2} T^{- 2} \cdot L^{1/2} = M L^{5/2} T^{- 2}

. Therefore

[A B] = M L^{5/2} T^{- 2} \cdot L = M L^{7/2} T^{- 2}

JEE Main · 2026 · 06 Apr 2026 · Shift 1 · Single correct option