Dimensions of Constant Product AB in Potential Energy Function

To find the dimensions of the product $AB$, we use the principle of homogeneity of dimensions for the given potential energy equation:

$V = \frac{A\sqrt{x}}{x + B}$

1. Dimensions of Potential Energy ($V$): Potential energy has the same dimensions as work: $[V] = [\text{M}^1 \text{L}^2 \text{T}^{-2}]$

2. Dimensions of Distance ($x$): $[x] = [\text{L}]$

3. Dimensions of Constant $B$: Since $B$ is added to $x$ in the denominator $(x + B)$, they must have the same physical dimensions by the principle of homogeneity: $[B] = [x] = [\text{L}]$

4. Dimensions of Constant $A$: Rewriting the dimension equation for potential energy: $[V] = \frac{[A] [x]^{1/2}}{[x + B]}$

Substitute the known dimensions into the equation: $[\text{M}^1 \text{L}^2 \text{T}^{-2}] = \frac{[A] [\text{L}]^{1/2}}{[\text{L}]}$

Simplifying the right side: $[\text{M}^1 \text{L}^2 \text{T}^{-2}] = [A] [\text{L}]^{-1/2}$

Solving for $[A]$: $[A] = [\text{M}^1 \text{L}^2 \text{T}^{-2}] \cdot [\text{L}]^{1/2} = [\text{M}^1 \text{L}^{5/2} \text{T}^{-2}]$

1. Dimensions of the Product $AB$: Now, we multiply the dimensions of $A$ and $B$: $[AB] = [A] \times [B]$ $[AB] = [\text{M}^1 \text{L}^{5/2} \text{T}^{-2}] \times [\text{L}^1]$ $[AB] = [\text{M}^1 \text{L}^{7/2} \text{T}^{-2}]$

Therefore, the dimensions of $AB$ are $[\text{M}^1 \text{L}^{7/2} \text{T}^{-2}]$, which corresponds to option D.