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(B) The principle of homogeneity of dimensions states that only quantities with the same dimensions can be added or subtracted.Since $B$ is added to $

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$M^1 L^{3/2} T^{-2}$

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(B) The principle of homogeneity of dimensions states that only quantities with the same dimensions can be added or subtracted.Since $B$ is added to $x^2$,the dimensions of $B$ must be equal to the dimensions of $x^2$.$[B] = [x^2] = L^2$.The potential energy $U$ has the dimensions of work,which is $[U] = M L^2 T^{-2}$.The given equation is $U = \frac{A \sqrt{x}}{x^2 + B}$.Considering the dimensions of the denominator,$[x^2 + B] = L^2$.Thus,$[U] = \frac{[A] [x]^{1/2}}{L^2}$.Substituting the dimensions: $M L^2 T^{-2} = \frac{[A] L^{1/2}}{L^2}$.$[A] = (M L^2 T^{-2}) 	imes (L^{3/2}) = M L^{7/2} T^{-2}$.Now,calculate the dimensions of $A/B$:$[A/B] = \frac{M L^{7/2} T^{-2}}{L^2} = M L^{7/2 - 2} T^{-2} = M L^{3/2} T^{-2}$.

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. Find the dimensional formula for $A/B$.

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