The potential energy of a particle varies with distance x from...

Given, $v=\frac{A \sqrt{x}}{x+B}$....(i) Dimensions of $v=$ dimensions of potential energy =\left[$\mathrm{ML}^2 \mathrm{~T}^{-2}$\right]$From Eq. (i) Dimensions of$B=$dimensions of$x=\left[\mathrm{M}^0 \mathrm{LT}^0\right] \therefore$Dimensions of$A $$\begin{aligned} & =\frac{\text{ dimension of } v \times \text{ dimensions of }(x+B)}{\text{ dimensions of } \sqrt{x}} \\ & =\frac{\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]\left[\mathrm{M}^0 \mathrm{LT}^0\right]}{\left[\mathrm{M}^0 \mathrm{~L}^{1 / 2} \mathrm{~T}^0\right]}=\left[\mathrm{ML}^{5 / 2} \mathrm{~T}^{-2}\right] \end{aligned}$$$Hence, dimensions of$A B =\left[\mathrm{ML}^{5 / 2} \mathrm{~T}^{-2}\right]\left[\mathrm{M}^0 \mathrm{LT}^0\right]=\left[\mathrm{ML}^{7 / 2} \mathrm{~T}^{-2}\right]