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 ($E$) in terms of force ($F$), acceleration ($A$), and time ($T$), we can establish a relationship using dimensional analysis.

Let the dimension of energy be proportional to powers of force, acceleration, and time:
$\left[E\right]={\left[F\right]}^a{\left[A\right]}^b{\left[T\right]}^c$

First, let's write the dimensions of each quantity in terms of the standard fundamental dimensions of mass ($M$), length ($L$), and time ($T$):
1. Energy ($E$) represents capacity to do work (Force × displacement):
$\left[E\right]=\left[M{L}^2{T}^{-2}\right]$
2. Force ($F$) is mass multiplied by acceleration:
$\left[F\right]=\left[ML{T}^{-2}\right]$
3. Acceleration ($A$) is the rate of change of velocity:
$\left[A\right]=\left[L{T}^{-2}\right]$
4. Time ($T$) has the dimension:
$\left[T\right]=\left[T\right]$

Now, substitute these dimensions into the initial relationship equation:
$\left[M{L}^2{T}^{-2}\right]={\left[ML{T}^{-2}\right]}^a{\left[L{T}^{-2}\right]}^b{\left[T\right]}^c$

Simplifying the right-hand side by grouping like terms, we get:
$\left[M{L}^2{T}^{-2}\right]=\left[M^a{L}^{a+b}{T}^{-2a-2b+c}\right]$

By equating the exponents of $M$, $L$, and $T$ on both sides, we get a system of linear equations:
For $M$: $a=1$
For $L$: $a+b=2$
For $T$: $-2a-2b+c=-2$

Solving this system of equations:
Substitute $a=1$ into the equation for $L$:
$1+b=2\Rightarrow b=1$
Now substitute $a=1$ and $b=1$ into the equation for $T$:
$-2\left(1\right)-2\left(1\right)+c=-2$
$-4+c=-2\Rightarrow c=2$

Therefore, substituting the values of $a$, $b$, and $c$ back into the proportional relation gives the dimensions of energy as:
$\left[E\right]=\left[F\right]\left[A\right]{\left[T\right]}^2$