The potential energy of a particle changes with distance \(x\) from a fixed origin as \(V=\frac{A\sqrt{x}}{x+B}\), where…

(d) \(B\) must have the same dimensions as \(x:[B]=[L]\)
The formula for potential energy is \(V=\frac{A\sqrt{x}}{x+B}\).
Rearranging for \(A\) :

\(A=\frac{V(x+B)}{\sqrt{x}}\\\)
\([A]=\frac{\left[{M}^{1}{L}^{2}{T}^{-2}\right][L]}{\left[{L}^{1/2}\right]}=\frac{\left[{M}^{1}{L}^{3}{T}^{-2}\right]}{\left[{L}^{1/2}\right]}\\\)
\(=\left[{M}^{1}{L}^{5/2}{T}^{-2}\right]\)

Dimensions for AB ;

\([AB]=\left[{M}^{1}{L}^{5/2}{T}^{-2}\right]\times \left[{L}^{1}\right]\\\)
\([AB]=\left[{M}^{1}{L}^{7/2}{T}^{-2}\right]\)