If force [F], acceleration [A ] and time [T] are chosen as the fundamental physical quantities. Find the dimensions of energy.

The correct option is [F] [A] [T²].

To find the dimensions of energy in terms of force [F], acceleration [A], and time [T], we can express energy (E) as a proportional relationship to these quantities raised to some powers:
$E=k{F}^a{A}^b{T}^c$
where $k$ is a dimensionless constant, and $a$, $b$, and $c$ are the powers to be determined.

Let us write the dimensional formulas for each of these physical quantities in the standard fundamental system of mass [M], length [L], and time [T]:
1. Dimensions of force: $\left[F\right]=\left[ML{T}^{-2}\right]$
2. Dimensions of acceleration: $\left[A\right]=\left[L{T}^{-2}\right]$
3. Dimensions of time: $\left[T\right]=\left[T\right]$
4. Dimensions of energy: $\left[E\right]=\left[M{L}^2{T}^{-2}\right]$

Now, substitute these dimensions into the proportional relationship:
$\left[M{L}^2{T}^{-2}\right]={\left[ML{T}^{-2}\right]}^a{\left[L{T}^{-2}\right]}^b{\left[T\right]}^c$
Simplifying the right-hand side by combining the powers of [M], [L], and [T]:
$\left[M{L}^2{T}^{-2}\right]=\left[M^a{L}^{a+b}{T}^{-2a-2b+c}\right]$

Equating the exponents of [M], [L], and [T] on both sides, we obtain the following equations:
For [M]: $a=1$
For [L]: $a+b=2$
For [T]: $-2a-2b+c=-2$

Let's solve these equations step-by-step:
From the mass equation, we have:
$a=1$
Substituting $a=1$ into the length equation:
$1+b=2\Rightarrow b=1$
Substituting $a=1$ and $b=1$ into the time equation:
$-2\left(1\right)-2\left(1\right)+c=-2$
$-2-2+c=-2$
$-4+c=-2$
$c=2$

Substituting the values of $a$, $b$, and $c$ back into our expression, the dimensions of energy are:
$\left[E\right]=\left[F\right]\left[A\right]\left[T^2\right]$