Find dimensions of A/B if

A/B ( P + At²/B ) + 1/2 ρv² = constant

where P → pressure, ρ → density, V → speed.

The correct option is ML^-1T^-4.

To find the dimensions of $\frac{A}{B}$, we can apply the principle of homogeneity of dimensions. According to this principle, physical quantities can only be added or subtracted if they have the same dimensions.

Let's analyze the given equation:
$\frac{A}{B}(P+\frac{A{t}^2}{B})+\frac{1}{2}\rho v^2=\text{constant}$

Inside the parentheses, the physical quantity pressure $P$ is added to the term $\frac{A{t}^2}{B}$. Therefore, the dimensions of $P$ must be equal to the dimensions of $\frac{A{t}^2}{B}$:
$\left[P\right]=\left[\frac{A{t}^2}{B}\right]$

We can rewrite the right side as:
$\left[P\right]=\left[\frac{A}{B}\right]·\left[t^2\right]$

Let's write down the dimensional formulas of the known variables:
1. Pressure ($P$) is force per unit area:
$\left[P\right]=\frac{\left[\text{Force}\right]}{\left[\text{Area}\right]}=\frac{\left[ML{T}^{-2}\right]}{\left[L^2\right]}=\left[M{L}^{-1}{T}^{-2}\right]$
2. Time ($t$):
$\left[t\right]=\left[T\right]\Rightarrow \left[t^2\right]=\left[T^2\right]$

Now, substitute these dimensions back into our equation to solve for the dimensions of $\frac{A}{B}$:
$\left[M{L}^{-1}{T}^{-2}\right]=\left[\frac{A}{B}\right]·\left[T^2\right]$

Rearranging the terms:
$\left[\frac{A}{B}\right]=\frac{\left[M{L}^{-1}{T}^{-2}\right]}{\left[T^2\right]}$
$\left[\frac{A}{B}\right]=\left[M{L}^{-1}{T}^{-4}\right]$

Thus, the dimensional formula of $\frac{A}{B}$ is ML^-1T^-4.