(b) $L \propto {v^x}{A^y}{F^z}$ $ \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 $f$
$[L] = [F{v^3}{A^{ – 2}}]$