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(D) The dimension of distance $x$ is $[L]$.The dimension of potential energy $V$ is $[ML^2T^{-2}]$.From the principle of homogeneity, in the denominat

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$[M^1L^{7/2}T^{-2}]$

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(D) The dimension of distance $x$ is $[L]$.The dimension of potential energy $V$ is $[ML^2T^{-2}]$.From the principle of homogeneity, in the denominator $(x + B)$, the dimension of $B$ must be equal to the dimension of $x$. Therefore, $[B] = [L]$.The given equation is $V = \frac{A\sqrt{x}}{x + B}$.Substituting the dimensions: $[ML^2T^{-2}] = \frac{[A][L^{1/2}]}{[L]}$.$[ML^2T^{-2}] = [A][L^{-1/2}]$.Therefore, $[A] = [ML^2T^{-2}] 	imes [L^{1/2}] = [ML^{5/2}T^{-2}]$.Now, the dimension of $AB$ is $[AB] = [ML^{5/2}T^{-2}] 	imes [L] = [ML^{7/2}T^{-2}]$.

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 constants with appropriate dimensions. The dimensions of $AB$ are . . . . . . .

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