Explanation:
Dimension of the given physical quantities are:
Force \([F]=M L T^{-2}\)
Velocity, \([V]=L T^{-1}\)
Time \([T]=T\)
Density \([D]=M L^{-3}\)
We can write
\(D=\left[M L^{-3}\right]=[F]^{x}[V]^{y}[T]^{x}\) or
\(M^{1} L^{-3} T^{0}=\left[M L T^{-2}\right]^{x}\left[L T^{-1}\right]^{y}[T]^{2}\)
\(M^{1} L^{-3} T^{0}=[M]^{x}[L]^{x+y}[T]^{-2 x-2 y+z}\)
Now compare both sides
\(x = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 1 \right)\)
\(x + y = - 3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 2 \right)\)
\( - 2{\rm{ }}x - 2{\rm{ }}y + z = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 3 \right)\)
After solving the equations
\(x=1, y=-4, z=-2\)
\(D=F V^{-4} T^{-2}\)