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
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
$F{A^{ - 1}}v$
$F{v^3}{A^{ - 2}}$
$F{v^2}{A^{ - 1}}$
${F^2}{v^2}{A^{ - 1}}$
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
$L \propto {v^x}{A^y}{F^z}$ $ \Rightarrow $ $L = k{v^x}{A^y}{F^z}$
Putting the dimensions in the above relation
$[M{L^2}{T^{ - 1}}] = k{[L{T^{ - 1}}]^x}{[L{T^{ - 2}}]^y}{[ML{T^{ - 2}}]^z}$
$⇒$ $[M{L^2}{T^{ - 1}}] = k[{M^z}{L^{x + y + z}}{T^{ - x - 2y - 2z}}]$
Comparing the powers of $M,\,L$ and $T$
$z = 1$ …$(i)$
$x + y + z = 2$ …$(ii)$
$ - x - 2y - 2z = - 1$ …$(iii)$
On solving $(i)$, $(ii)$ and $(iii)$ $x = 3,\,y = - 2,\,z = 1$
So dimension of $L$ in terms of $v,\,A$ and $f$
$[L] = [F{v^3}{A^{ - 2}}]$
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