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 ______
Answer: (D) $[\text{M}^1\text{L}^{7/2}\text{T}^{-2}]$
Concept. In $x+B$, the constant $B$ is added to a distance, so by homogeneity $B$ is a length. With $B$ known, the dimensions of $A$ follow from requiring the whole expression to be an energy.
Step 1. Dimensions of $B$.
$$[B]=[x]=[L]$$
Step 2. Dimensions of $A$. The denominator $x+B$ has dimensions $L$, so
$$[V]=\frac{[A]\,L^{1/2}}{L}=[A]\,L^{-1/2}\ \Rightarrow\ [A]=[V]\,L^{1/2}$$
With $[V]=ML^2T^{-2}$ (energy):
$$[A]=ML^{2}T^{-2}\cdot L^{1/2}=[ML^{5/2}T^{-2}]$$
Step 3. Multiply.
$$[AB]=ML^{5/2}T^{-2}\cdot L=[ML^{7/2}T^{-2}]$$
Answer. $[M^1L^{7/2}T^{-2}]$ (Option D). Option A is the trap: it is $[A]$ alone.
Solution by Sreeraj P, M.Sc Physics
