Q52 / 75 · JEE Main 2026 (06 April Shift 1) · Physics
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 _______.
- A$[M^1 L^{5/2} T^{-2}]$
- B$[M^{3/2} L^{5/2} T^{-2}]$
- C$[M^1 L^2 T^{-2}]$
- Correct answerD$[M^1 L^{7/2} T^{-2}]$
Solution
By the principle of dimensional homogeneity, quantities added or subtracted must have the same dimensions. In the denominator, $B$ is added to $x$ (distance).
$[B] = [x] = [L]$
The dimensions of potential energy $V$ are $[M^1 L^2 T^{-2}]$.
From the given equation $V = \dfrac{A\sqrt{x}}{x + B}$, we can write the dimensional formula as:
$[V] = \dfrac{[A][x]^{1/2}}{[x + B]}$
$[M^1 L^2 T^{-2}] = \dfrac{[A][L]^{1/2}}{[L]}$
$[M^1 L^2 T^{-2}] = [A][L]^{-1/2}$
$[A] = [M^1 L^2 T^{-2}][L]^{1/2} = [M^1 L^{5/2} T^{-2}]$
Now, the dimensions of $AB$ are:
$[AB] = [A][B] = [M^1 L^{5/2} T^{-2}][L] = [M^1 L^{7/2} T^{-2}]$
Answer: $[M^1 L^{7/2} T^{-2}]$
$[B] = [x] = [L]$
The dimensions of potential energy $V$ are $[M^1 L^2 T^{-2}]$.
From the given equation $V = \dfrac{A\sqrt{x}}{x + B}$, we can write the dimensional formula as:
$[V] = \dfrac{[A][x]^{1/2}}{[x + B]}$
$[M^1 L^2 T^{-2}] = \dfrac{[A][L]^{1/2}}{[L]}$
$[M^1 L^2 T^{-2}] = [A][L]^{-1/2}$
$[A] = [M^1 L^2 T^{-2}][L]^{1/2} = [M^1 L^{5/2} T^{-2}]$
Now, the dimensions of $AB$ are:
$[AB] = [A][B] = [M^1 L^{5/2} T^{-2}][L] = [M^1 L^{7/2} T^{-2}]$
Answer: $[M^1 L^{7/2} T^{-2}]$
Hence, the correct answer is option: 'D'.