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

Correct Answer: The correct option is [F] [A] [T²].

To find the dimensions of energy in terms of force ($\left[F\right]$), acceleration ($\left[A\right]$), and time ($\left[T\right]$), we can represent energy ($\left[E\right]$) as a product of these quantities raised to some powers:

$\left[E\right]={\left[F\right]}^a{\left[A\right]}^b{\left[T\right]}^c$

Here, $a$, $b$, and $c$ are the exponents we need to find. Let's write the dimensions of each quantity in the standard Mass ($M$), Length ($L$), and Time ($T$) system:

1. Energy ($E$), which is work done (Force × Displacement):
$\left[E\right]=\left[M{L}^2{T}^{-2}\right]$

2. Force ($F$):
$\left[F\right]=\left[ML{T}^{-2}\right]$

3. Acceleration ($A$):
$\left[A\right]=\left[L{T}^{-2}\right]$

4. Time ($T$):
$\left[T\right]=\left[T\right]$

Now, substitute these dimensional formulas back into the relation:

$\left[M{L}^2{T}^{-2}\right]={\left[ML{T}^{-2}\right]}^a{\left[L{T}^{-2}\right]}^b{\left[T\right]}^c$

Group the terms on the right-hand side by the fundamental bases $M$, $L$, and $T$:

$\left[M{L}^2{T}^{-2}\right]=\left[M^a{L}^{a+b}{T}^{-2a-2b+c}\right]$

By comparing the powers of $M$, $L$, and $T$ on both sides, we obtain three 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 first equation, we have $a=1$.
Substitute $a=1$ into the equation for $L$:
$1+b=2$
$b=1$

Now substitute both $a=1$ and $b=1$ into the equation for $T$:
$-2\left(1\right)-2\left(1\right)+c=-2$
$-2-2+c=-2$
$-4+c=-2$
$c=2$

Substituting these values back into our original assumption:

$\left[E\right]={\left[F\right]}^1{\left[A\right]}^1{\left[T\right]}^2$

Hence, the dimensions of energy are [F] [A] [T²].