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(A) The potential energy $U$ is given by $U = \frac{A \sqrt{x}}{x^2 + B}$.According to the principle of homogeneity of dimensions,the dimensions of te

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$[ML^{11/2} T^{-2}]$

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(A) The potential energy $U$ is given by $U = \frac{A \sqrt{x}}{x^2 + B}$.According to the principle of homogeneity of dimensions,the dimensions of terms added or subtracted must be the same. Therefore,the dimensions of $B$ must be equal to the dimensions of $x^2$.Since $x$ is distance,$[x] = [L]$,so $[B] = [L^2]$.The dimensions of potential energy $U$ are $[ML^2 T^{-2}]$.From the equation,$A = \frac{U(x^2 + B)}{\sqrt{x}}$.Substituting the dimensions:$[A] = \frac{[ML^2 T^{-2}] [L^2]}{[L^{1/2}]} = [ML^2 T^{-2} L^2 L^{-1/2}] = [ML^{7/2} T^{-2}]$.Now,we need the dimensions of $AB$:$[AB] = [A][B] = [ML^{7/2} T^{-2}] [L^2] = [ML^{11/2} T^{-2}]$.Thus,the correct option is $A$.

The potential energy of a particle varies with distance $x$ from a fixed origin as $U = \frac{A \sqrt{x}}{x^2 + B}$,where $A$ and $B$ are dimensional constants. Then,the dimensional formula for $AB$ is:

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