If velocity v, acceleration A and force F are chosen as fundamental quantities, then the dimensional formula of angular momentum in terms of v, A and F would be

The correct option is Fv��A⁻².

To find the dimensional formula of angular momentum in terms of velocity ($v$), acceleration ($A$), and force ($F$), we can express angular momentum ($L$) as a power function of these fundamental quantities:
$L=k{F}^x{v}^y{A}^z$
where $k$ is a dimensionless constant, and $x$, $y$, and $z$ are the exponents we need to determine.

First, let's write the dimensional formulas of all the quantities involved in the basic MLT (Mass, Length, Time) system:
1. Angular Momentum ($L$): Angular momentum is given by $L=r\times p=r\times mv$. Therefore, its dimensions are:
$\left[L\right]=\left[M\right]\left[L{T}^{-1}\right]\left[L\right]=\left[M^1{L}^2{T}^{-1}\right]$
2. Force ($F$):
$\left[F\right]=\left[M^1{L}^1{T}^{-2}\right]$
3. Velocity ($v$):
$\left[v\right]=\left[L^1{T}^{-1}\right]$
4. Acceleration ($A$):
$\left[A\right]=\left[L^1{T}^{-2}\right]$

Now, substitute these dimensions into the relationship:
$\left[M^1{L}^2{T}^{-1}\right]={\left[ML{T}^{-2}\right]}^x{\left[L{T}^{-1}\right]}^y{\left[L{T}^{-2}\right]}^z$

Combine the powers of $M$, $L$, and $T$ on the right-hand side:
$\left[M^1{L}^2{T}^{-1}\right]=\left[M^x{L}^{x+y+z}{T}^{-2x-y-2z}\right]$

By comparing the exponents of $M$, $L$, and $T$ on both sides, we get the following system of linear equations:
For $M$: $x=1$
For $L$: $x+y+z=2$
For $T$: $-2x-y-2z=-1$

Substitute $x=1$ into the equations for $L$ and $T$:
1. From the $L$ equation:
$1+y+z=2\Rightarrow y+z=1$ (Equation 1)
2. From the $T$ equation:
$-2\left(1\right)-y-2z=-1\Rightarrow -y-2z=1\Rightarrow y+2z=-1$ (Equation 2)

Subtract Equation 1 from Equation 2 to find $z$:
$(y+2z)-(y+z)=-1-1$
$z=-2$

Substitute $z=-2$ back into Equation 1 to find $y$:
$y+(-2)=1\Rightarrow y=3$

Substituting the values of $x=1$, $y=3$, and $z=-2$ back into our original formula yields:
$\left[L\right]=\left[F^1{v}^3{A}^{-2}\right]$

Therefore, the dimensional formula of angular momentum in terms of $v$, $A$, and $F$ is Fv³A⁻².