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 _______.

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: \([M^1 L^{7/2} T^{-2}]\)