JEE Main 2026 Apr 6 Shift 1, Physics Q2: Units And Dimensions

JEE Main2026Apr 6, Shift 1Physics

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 $A B$ are _______.

A
$[M^{1} L^{5/2} T^{- 2}]$
B
$[M^{3/2} L^{5/2} T^{- 2}]$
C
$[M^{1} L^{2} T^{- 2}]$
D
$[M^{1} L^{7/2} T^{- 2}]$

Solution

Since $B$ adds to $x$ , $[B] = [x] = L$ .

From $V = \frac{A x}{x + B}$ : $[V] = M L^{2} T^{- 2}$ , $[x] = L^{1/2}$ , $[x + B] = L$ .

$[A] = \frac{[ V ] [ x + B ]}{[ x ]} = \frac{M L ^{2} T ^{- 2} \cdot L}{L ^{1/2}} = M L^{5/2} T^{- 2}$ .

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

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JEE Main 2026 Apr 6 Shift 1, Physics Q2: Units And Dimensions

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